MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS...

Monomial Algebras, Second Edition presents algebraic, combina-torial, and computational methods for studying monomial algebras and their ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free mono-mials and the corresponding graphs, clutters, or hypergraphs.

New to the Second Edition• Four new chapters that focus on the algebraic properties of

blowup algebras in combinatorial optimization problems of clut-ters and hypergraphs

• Two new chapters that explore the algebraic and combinatorial properties of the edge ideal of clutters and hypergraphs

• Full revisions of existing chapters to provide an up-to-date ac-count of the subject

Bringing together several areas of pure and applied mathematics, this book shows how monomial algebras are related to polyhedral geom-etry, combinatorial optimization, and combinatorics of hypergraphs. It directly links the algebraic properties of monomial algebras to com-binatorial structures (such as simplicial complexes, posets, digraphs, graphs, and clutters) and linear optimization problems.

Features• Presents computational and combinatorial methods in commuta-

tive algebra • Shows how to solve a variety of problems of monomial algebras• Covers various affine and graded rings, including Cohen–Macau-

lay, complete intersection, and normal• Examines their basic algebraic invariants, such as multiplicity,

Betti numbers, projective dimension, and Hilbert polynomial• Contains more than 550 exercises and over 50 examples, many of

which illustrate the use of computer algebra systems

Mathematics

Monomial Algebras Second Edition

Rafael H. VillarrealV

illarrealSecond EditionM

onomial A

lgebrasK23008

w w w . c r c p r e s s . c o m

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICSMONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

K23008_cover.indd 1 2/13/15 12:23 PM

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS


Rafael H. VillarrealCentro de Investigación y de Estudios Avanzados del IPNMexico City, Mexico

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Series Editors

John A. BurnsThomas J. TuckerMiklos BonaMichael RuzhanskyChi-Kwong Li

Published Titles

Application of Fuzzy Logic to Social Choice Theory, John N. Mordeson, Davender S. Malik and Terry D. Clark

Blow-up Patterns for Higher-Order: Nonlinear Parabolic, Hyperbolic Dispersion and Schrödinger Equations, Victor A. Galaktionov, Enzo L. Mitidieri, and Stanislav Pohozaev

Difference Equations: Theory, Applications and Advanced Topics, Third Edition, Ronald E. Mickens

Dictionary of Inequalities, Second Edition, Peter Bullen

Iterative Optimization in Inverse Problems, Charles L. Byrne

Modeling and Inverse Problems in the Presence of Uncertainty, H. T. Banks, Shuhua Hu, and W. Clayton Thompson

Monomial Algebras, Second Edition, Rafael H. Villarreal

Set Theoretical Aspects of Real Analysis, Alexander B. Kharazishvili

Signal Processing: A Mathematical Approach, Second Edition, Charles L. Byrne

Sinusoids: Theory and Technological Applications, Prem K. Kythe

Special Integrals of Gradshetyn and Ryzhik: the Proofs – Volume l, Victor H. Moll

Forthcoming Titles

Actions and Invariants of Algebraic Groups, Second Edition, Walter Ferrer Santos and Alvaro Rittatore

Analytical Methods for Kolmogorov Equations, Second Edition, Luca Lorenzi

Complex Analysis: Conformal Inequalities and the Bierbach Conjecture, Prem K. Kythe

Computational Aspects of Polynomial Identities: Volume l, Kemer’s Theorems, 2nd Edition Belov Alexey, Yaakov Karasik, Louis Halle Rowen

Cremona Groups and Icosahedron, Ivan Cheltsov and Constantin Shramov

Geometric Modeling and Mesh Generation from Scanned Images, Yongjie Zhang

Groups, Designs, and Linear Algebra, Donald L. Kreher

Handbook of the Tutte Polynomial, Joanna Anthony Ellis-Monaghan and Iain Moffat

Lineability: The Search for Linearity in Mathematics, Juan B. Seoane Sepulveda, Richard W. Aron, Luis Bernal-Gonzalez, and Daniel M. Pellegrinao

Line Integral Methods and Their Applications, Luigi Brugnano and Felice Iaverno

Microlocal Analysis on Rˆn and on NonCompact Manifolds, Sandro Coriasco

Nonlinear Functional Analysis in Banach Spaces and Banach Algebras: Fixed Point Theory Under Weak Topology for Nonlinear Operators and Block Operators with Applications Aref Jeribi and Bilel Krichen

Partial Differential Equations with Variable Exponents: Variational Methods and Quantitative Analysis, Vicentiu Radulescu and Dusan Repovs

Practical Guide to Geometric Regulation for Distributed Parameter Systems, Eugenio Aulisa and David S. Gilliam

Reconstructions from the Data of Integrals, Victor Palamodov

Special Integrals of Gradshetyn and Ryzhik: the Proofs – Volume ll, Victor H. Moll

Stochastic Cauchy Problems in Infinite Dimensions: Generalized and Regularized Solutions, Irina V. Melnikova and Alexei Filinkov

Symmetry and Quantum Mechanics, Scott Corry

Forthcoming Titles (continued)

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

© 2015 by Taylor & Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa business

No claim to original U.S. Government worksVersion Date: 20150202

International Standard Book Number-13: 978-1-4822-3470-1 (eBook - PDF)

This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint.

Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information stor-age or retrieval system, without written permission from the publishers.

For permission to photocopy or use material electronically from this work, please access www.copy-right.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that pro-vides licenses and registration for a variety of users. For organizations that have been granted a photo-copy license by the CCC, a separate system of payment has been arranged.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.com

and the CRC Press Web site athttp://www.crcpress.com

Contents

Preface ix

Preface to First Edition xv

1 Polyhedral Geometry and Linear Optimization 1

1.1 Polyhedral sets and cones . . . . . . . . . . . . . . . . . . . . 1

1.2 Relative volumes of lattice polytopes . . . . . . . . . . . . . . 20

1.3 Hilbert bases and TDI systems . . . . . . . . . . . . . . . . . 28

1.4 Rees cones and clutters . . . . . . . . . . . . . . . . . . . . . 39

1.5 The integral closure of a semigroup . . . . . . . . . . . . . . . 46

1.6 Unimodularity of matrices and normality . . . . . . . . . . . 48

1.7 Normaliz, a computer program . . . . . . . . . . . . . . . . . 50

1.8 Cut-incidence matrices and integrality . . . . . . . . . . . . . 51

1.9 Elementary vectors and matroids . . . . . . . . . . . . . . . . 55

2 Commutative Algebra 61

2.1 Module theory . . . . . . . . . . . . . . . . . . . . . . . . . . 61

2.2 Graded modules and Hilbert polynomials . . . . . . . . . . . 76

2.3 Cohen–Macaulay modules . . . . . . . . . . . . . . . . . . . . 79

2.4 Normal rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

2.5 Valuation rings . . . . . . . . . . . . . . . . . . . . . . . . . . 92

2.6 Krull rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

2.7 Koszul homology . . . . . . . . . . . . . . . . . . . . . . . . . 104

2.8 A vanishing theorem of Grothendieck . . . . . . . . . . . . . . 107

3 Affine and Graded Algebras 111

3.1 Cohen–Macaulay graded algebras . . . . . . . . . . . . . . . . 111

3.2 Hilbert Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . 123

3.3 Grobner bases . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

3.4 Projective closure . . . . . . . . . . . . . . . . . . . . . . . . . 132

3.5 Minimal resolutions . . . . . . . . . . . . . . . . . . . . . . . 134

vi CONTENTS

4 Rees Algebras and Normality 1414.1 Symmetric algebras . . . . . . . . . . . . . . . . . . . . . . . . 1414.2 Rees algebras and syzygetic ideals . . . . . . . . . . . . . . . 1424.3 Complete and normal ideals . . . . . . . . . . . . . . . . . . . 1454.4 Multiplicities and a criterion of Herzog . . . . . . . . . . . . . 1594.5 Jacobian criterion . . . . . . . . . . . . . . . . . . . . . . . . . 165

5 Hilbert Series 1715.1 Hilbert–Serre Theorem . . . . . . . . . . . . . . . . . . . . . . 1715.2 a-invariants and h-vectors . . . . . . . . . . . . . . . . . . . . 1775.3 Extremal algebras . . . . . . . . . . . . . . . . . . . . . . . . 1825.4 Initial degrees of Gorenstein ideals . . . . . . . . . . . . . . . 1895.5 Koszul homology and Hilbert functions . . . . . . . . . . . . . 1965.6 Hilbert functions of some graded ideals . . . . . . . . . . . . . 199

6 Stanley–Reisner Rings and Edge Ideals of Clutters 2016.1 Primary decomposition . . . . . . . . . . . . . . . . . . . . . . 2016.2 Simplicial complexes and homology . . . . . . . . . . . . . . . 2096.3 Stanley–Reisner rings . . . . . . . . . . . . . . . . . . . . . . 2126.4 Regularity and projective dimension . . . . . . . . . . . . . . 2266.5 Unmixed and shellable clutters . . . . . . . . . . . . . . . . . 2346.6 Admissible clutters . . . . . . . . . . . . . . . . . . . . . . . . 2466.7 Hilbert series of face rings . . . . . . . . . . . . . . . . . . . . 2506.8 Simplicial spheres . . . . . . . . . . . . . . . . . . . . . . . . . 2556.9 The upper bound conjectures . . . . . . . . . . . . . . . . . . 258

7 Edge Ideals of Graphs 2617.1 Graph theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 2617.2 Edge ideals and B-graphs . . . . . . . . . . . . . . . . . . . . 2687.3 Cohen–Macaulay and chordal graphs . . . . . . . . . . . . . . 2747.4 Shellable and sequentially C–M graphs . . . . . . . . . . . . . 2827.5 Regularity, depth, arithmetic degree . . . . . . . . . . . . . . 2937.6 Betti numbers of edge ideals . . . . . . . . . . . . . . . . . . . 2987.7 Associated primes of powers of ideals . . . . . . . . . . . . . . 303

8 Toric Ideals and Affine Varieties 3118.1 Binomial ideals and their radicals . . . . . . . . . . . . . . . . 3118.2 Lattice ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 3168.3 Monomial subrings and toric ideals . . . . . . . . . . . . . . . 3268.4 Toric varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . 3358.5 Affine Hilbert functions . . . . . . . . . . . . . . . . . . . . . 3428.6 Vanishing ideals over finite fields . . . . . . . . . . . . . . . . 3458.7 Semigroup rings of numerical semigroups . . . . . . . . . . . . 3478.8 Toric ideals of monomial curves . . . . . . . . . . . . . . . . . 352

CONTENTS vii

9 Monomial Subrings 3659.1 Integral closure of monomial subrings . . . . . . . . . . . . . 3669.2 Homogeneous monomial subrings . . . . . . . . . . . . . . . . 3729.3 Ehrhart rings . . . . . . . . . . . . . . . . . . . . . . . . . . . 3809.4 The degree of lattice and toric ideals . . . . . . . . . . . . . . 3929.5 Laplacian matrices and ideals . . . . . . . . . . . . . . . . . . 3969.6 Grobner bases and normal subrings . . . . . . . . . . . . . . . 4039.7 Toric ideals generated by circuits . . . . . . . . . . . . . . . . 4109.8 Divisor class groups of semigroup rings . . . . . . . . . . . . . 416

10 Monomial Subrings of Graphs 42310.1 Edge subrings and ring graphs . . . . . . . . . . . . . . . . . 42310.2 Incidence matrices and circuits . . . . . . . . . . . . . . . . . 44010.3 The integral closure of an edge subring . . . . . . . . . . . . . 44810.4 Ehrhart rings of edge polytopes . . . . . . . . . . . . . . . . . 45410.5 Integral closure of Rees algebras . . . . . . . . . . . . . . . . 45710.6 Edge subrings of complete graphs . . . . . . . . . . . . . . . . 46110.7 Edge cones of graphs . . . . . . . . . . . . . . . . . . . . . . . 46710.8 Monomial birational extensions . . . . . . . . . . . . . . . . . 477

11 Edge Subrings and Combinatorial Optimization 48311.1 The canonical module of an edge subring . . . . . . . . . . . 48311.2 Integrality of the shift polyhedron . . . . . . . . . . . . . . . 48411.3 Generators for the canonical module . . . . . . . . . . . . . . 48711.4 Computing the a-invariant . . . . . . . . . . . . . . . . . . . . 48911.5 Algebraic invariants of edge subrings . . . . . . . . . . . . . . 493

12 Normality of Rees Algebras of Monomial Ideals 49912.1 Integral closure of monomial ideals . . . . . . . . . . . . . . . 49912.2 Normality criteria . . . . . . . . . . . . . . . . . . . . . . . . 50512.3 Rees cones and polymatroidal ideals . . . . . . . . . . . . . . 50812.4 Veronese subrings and the a-invariant . . . . . . . . . . . . . 51312.5 Normalizations of Rees algebras . . . . . . . . . . . . . . . . . 51912.6 Rees algebras of Veronese ideals . . . . . . . . . . . . . . . . . 52412.7 Divisor class group of a Rees algebra . . . . . . . . . . . . . . 52912.8 Stochastic matrices and Cremona maps . . . . . . . . . . . . 531

13 Combinatorics of Symbolic Rees Algebrasof Edge Ideals of Clutters 53713.1 Vertex covers of clutters . . . . . . . . . . . . . . . . . . . . . 53713.2 Symbolic Rees algebras of edge ideals . . . . . . . . . . . . . 54013.3 Blowup algebras in perfect graphs . . . . . . . . . . . . . . . 55113.4 Algebras of vertex covers of graphs . . . . . . . . . . . . . . . 55513.5 Edge subrings in perfect matchings . . . . . . . . . . . . . . . 558

viii CONTENTS

13.6 Rees cones and perfect graphs . . . . . . . . . . . . . . . . . . 56113.7 Perfect graphs and algebras of covers . . . . . . . . . . . . . . 564

14 Combinatorial Optimization and Blowup Algebras 56714.1 Blowup algebras of edge ideals . . . . . . . . . . . . . . . . . 56814.2 Rees algebras and polyhedral geometry . . . . . . . . . . . . . 57014.3 Packing problems and blowup algebras . . . . . . . . . . . . . 58314.4 Uniform ideal clutters . . . . . . . . . . . . . . . . . . . . . . 59914.5 Clique clutters of comparability graphs . . . . . . . . . . . . . 61014.6 Duality and integer rounding problems . . . . . . . . . . . . . 61514.7 Canonical modules and integer rounding . . . . . . . . . . . . 62814.8 Clique clutters of Meyniel graphs . . . . . . . . . . . . . . . . 633

Appendix Graph Diagrams 635A.1 Cohen–Macaulay graphs . . . . . . . . . . . . . . . . . . . . . 635A.2 Unmixed graphs . . . . . . . . . . . . . . . . . . . . . . . . . 638

Bibliography 639

Notation Index 669

Index 673

Preface

The main purpose of this book is to introduce algebraic, combinatorial, andcomputational methods to study monomial algebras and their presentationideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings,blowup algebras, emphasizing square-free monomials and its correspondinggraphs, clutters, or hypergraphs.

Monomial algebras are related to various fields, for instance to numericalsemigroups, semigroup rings, algebraic geometry, commutative algebra andcombinatorics, integer programming and polyhedral geometry, graph theoryand combinatorial optimization. We develop links between the areas shownas the vertices of the following graph.

��

��

��

��

��

��

CombinatorialOptimization

��

��

��

��

��

��

Combinatoricsof Hypergraphs

��

��

��

��

��

��

PolyhedralGeometry

��

��

��

��

��

��

��

��

��

��

MonomialAlgebras

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

This allows us to solve a variety of problems of monomial algebras usingthe methods of the other areas and vice versa. An effort has been made togive a unifying presentation of the techniques and notions of these areas. Inthis book we are interested in the algebraic properties of monomial algebrasthat can be directly linked to combinatorial structures—such as simplicial

x Preface

complexes, posets, digraphs, graphs, and clutters—and to linear optimiza-tion problems. We study various types of affine and graded rings (Cohen–Macaulay, sequentially Cohen–Macaulay, Gorenstein, Artinian, completeintersection, unmixed, normal, reduced) and examine their basic algebraicinvariants (type, multiplicity, a-invariant, regularity, Betti numbers, Krulldimension, projective dimension, h-vector, Hilbert polynomial).

In recent years the algebraic properties of blowup algebras have beenlinked to combinatorial optimization problems of clutters and hypergraphs.In this edition, we include four new chapters (Chapters 1, 11, 13, and 14) tointroduce this area of research and to present some of the main advances.

The study of algebraic and combinatorial properties of the edge ideal ofclutters and hypergraphs has attracted a great deal of interest in the lasttwo decades [163, 189, 224, 326, 408]. In the present edition we include twochapters with some of the advances in the area (Chapters 6 and 7).

In order to present an up-to-date account of the subject, we have madea full revision of all chapters in the first edition. The chapters have beenreorganized and arranged in a different order. In particular Chapters 9 and12 were originally a single chapter, while Chapters 8 and 10 were originallydivided into two chapters for each of them.

This book brings together several areas of pure and applied mathemat-ics. It contains over 550 exercises and over 50 examples, many of themillustrating the use of computer algebra systems. It has extensive indices ofterminology and notation.

The contents of this book are as follows. In Chapter 1 we begin byintroducing several notions and results coming from polyhedral geometry,combinatorial optimization, and linear programming. Relative volumes oflattice polytope are studied here. We present relations between Hilbertbases, TDI linear systems of inequalities, max-flow min-cut properties ofclutters, and normality of affine and Rees semigroups. For instance themax-flow min-cut property is classified in terms of Hilbert bases and theintegrality of certain polyhedra. This is used in Chapter 14 to prove thatthe edge ideal of a clutter is normally torsion-free if and only if the clutterhas the max-flow min-cut property. We present a slight generalization of atheorem of Lucchesi and Younger [298] that is useful to detect TDI systemsarising from incidence matrices of digraphs. This is used in Chapter 11 toexpress the a-invariant of the edge subring of a bipartite graph in terms ofdirected cuts. Elementary integral vectors and matroids are introduced atthe end of the chapter. The notion of a matroid will appear in several placesin this book in connection with monomial rings and ideals. The computerprogram we have used to study linear systems of inequalities and to computeHilbert bases of rational cones is Normaliz [68]. The prerequisite for thischapter is a course on linear algebra and familiarity with point set topology.

Chapter 2 discusses certain topics and results on commutative algebra

Preface xi

(module theory, normal and graded rings). It includes some detailed proofs,and points the reader to the appropriate references when proofs are omitted.However we make free use of the standard terminology and notation ofhomological algebra (including Tor and Ext) as described in [363] and [428].This edition has three new sections on valuation rings, Krull rings, and avanishing theorem of Grothendieck (Sections 2.5, 2.6, and 2.8).

A number of topics connected to affine and graded algebras are studied inChapter 3, e.g., Grobner bases, Hilbert Nullstellensatz and affine varieties,projective closure, minimal resolutions, and Betti numbers. We present twoversions of the Noether normalization lemma and some of its applicationsto affine algebras and to Cohen–Macaulay graded algebras.

In Chapter 4 a thorough presentation of complete and normal idealsis given. Here the systematic use of blowup algebras makes clear theirimportance for the area. In this chapter we introduce symbolic powersof ideals and normally torsion-free ideals. Then we present a few caseswhere equality of symbolic and ordinary powers can be described in termsof properties of the associated graded ring. An elegant and useful Cohen–Macaulay criterion due to Herzog is included here.

Chapter 5 deals with the Hilbert series of graded modules and algebras,a topic that is quite useful in Stanley’s proof of the upper bound conjecturefor simplicial spheres. The h-vector and a-invariant of a graded algebra aredefined through the Hilbert–Serre theorem. For Cohen–Macaulay gradedalgebras we present the main properties of their h-vectors and a-invariants.Some optimal upper bounds for the number of generators in the least degreeof Gorenstein and Cohen–Macaulay graded ideals are given, which naturallyleads to the notion of an extremal algebra. As an application the Koszulhomology of Cohen–Macaulay ideals with pure resolutions is studied usingHilbert function techniques. This edition has a new section on Hilbertfunctions of a certain type of graded ideals that occur in algebraic codingtheory (Section 5.6).

Chapter 6 is an introduction to monomial ideals, Stanley–Reisner rings,and edge ideals of clutters. An understated goal here is to highlight some ofthe works of T. Hibi, J. Herzog, M. Hochster, G. Reisner, and R. Stanley. Westudy algebraic and combinatorial properties of edge ideals and simplicialcomplexes. In particular we examine shellable, unmixed, and sequentiallyCohen–Macaulay simplicial complexes and their corresponding edge idealsand invariants (projective dimension, regularity, depth, Hilbert series). Asfor applications, the proofs of the upper bound conjectures for polytopesare discussed to give a flavor to some of the methods and ideas of thearea. Monomial ideals can also be used to solve certain problems of generalpolynomial ideals using the theory of Grobner bases.

In Chapter 7 we give an introduction to graph theory and study algebraicproperties and invariants of edge ideals using the combinatorial structure of

xii Preface

graphs. The study of edge ideals of graphs has been a very active area ofresearch in the last decade because of its connections to graph theory. Wepresent classifications of the following families: unmixed bipartite graphs,Cohen–Macaulay bipartite graphs, Cohen–Macaulay trees, shellable bipar-tite graphs, sequentially Cohen–Macaulay bipartite graphs. The invariantsexamined here include regularity, depth, Krull dimension, and multiplicity.Edge ideals of graphs are shown to have the persistence property.

In Chapter 8 we study algebraic and geometric aspects of three specialtypes of polynomial ideals and their quotient rings:

toric ideals ↪→ lattice ideals ↪→ binomial ideals.

These ideals are interesting from a computational point of view and arerelated to diverse fields, such as combinatorics, algebraic geometry, integerprogramming, semigroup rings, coding theory, and algebraic statistics.

Chapter 9 deals with point configurations and their lattice polytopes.We consider monomial subrings and binomial ideals associated with them,e.g., Ehrhart rings, Rees algebras, homogeneous subrings, lattice ideals, andtoric ideals. Using Hilbert functions, polyhedral geometry, and Grobnerbases, we study normalizations of monomial subrings, initial ideals of toricideals, normal monomial subrings, primary decompositions and multiplici-ties of lattice ideals. Algebraic graph theory is used to study matrix idealsof Laplacian matrices. The reciprocity law of Ehrhart for integral poly-topes and the Danilov–Stanley formula for canonical modules of monomialsubrings will be introduced here. Applications of these results will be given.

In Chapter 10 we study monomial subrings associated with graphs andtheir toric ideals. We relate the even closed walks and circuits of the vectormatroid of a graph with Grobner bases theory. A description of the integralclosure of the edge subring of a multigraph will be presented along with adescription of the circuits of its toric ideal. We study the family of graphswhose number of primitive cycles equals its cycle rank. It is shown that thisfamily is precisely the family of ring graphs. These graphs are characterizedin algebraic and combinatorial terms. We classify edge subrings of bipartitegraphs which are complete intersections. Then we present sharp upperbounds for the multiplicity of edge subrings. Several connections betweenmonomial subrings, graph theory, and polyhedral geometry will occur inthis chapter. We study in detail the irreducible representation of an edgecone of a graph and show some applications to graph theory (for instancewe show the marriage theorem).

Chapter 11 focuses on edge subrings of connected bipartite graphs andtheir algebraic invariants. We show how to compute the canonical moduleand the a-invariant of an edge subring using linear programming techniquesand introduce an integral polyhedron whose vertices correspond to minimalgenerators of the canonical module. The a-invariant of an edge subring will

Preface xiii

be interpreted in combinatorial optimization terms as the maximum numberof edge disjoint directed cuts. We study the Gorenstein property and thetype of an edge subring.

Chapter 12 is about Rees algebras of monomial ideals. We study theintegral closure of a monomial ideal, and the normality and invariants ofits Rees algebra. The normalization of a Rees algebra is examined usingthe Danilov–Stanley formula, Caratheodory’s theorem, and Hilbert bases ofRees cones. Interesting classes of normal ideals, such as ideals of Veronesetype and polymatroidal ideals are introduced in the chapter. The divisorclass group of a normal Rees algebra is computed using polyhedral geometry.

Chapter 13 shows the interaction between graph theory, combinatorialoptimization, commutative algebra, and the theory of blowup algebras.In this chapter, we give a description—using notions from combinatorialoptimization—of the minimal generators of the symbolic Rees algebra ofthe edge ideal of a clutter and show a graph theoretical description of theminimal generators of the symbolic Rees algebra of the ideal of covers of agraph. The minimal sets of generators of symbolic Rees algebras of edgeideals are studied using polyhedral geometry. Indecomposable graphs arerelated to the strong perfect graph theorem. We give a description—interms of cliques—of the symbolic Rees algebra and the Simis cone of theedge ideal of a perfect graph.

In Chapter 14 we relate combinatorial optimization with commutativealgebra, and present applications to both areas. We establish some linksbetween the algebraic properties of blowup algebras of edge ideals and thecombinatorial optimization properties of clutters and polyhedra. A long-standing conjecture of Conforti and Cornuejols about packing problems isexamined from an algebraic point of view. We study max-flow min-cutproblems of clutters, packing problems, and integer rounding propertiesof various systems of linear inequalities to gain insight about the algebraicproperties of blowup algebras. Systems with integer rounding properties andclutters with the max-flow min-cut property come from linear optimizationproblems. The equality between the Rees algebra and the symbolic Reesalgebra of an edge ideal is characterized in combinatorial and algebraicterms. A number of properties of clutters and edge ideals are shown to beclosed (under taking) minors, Alexander duals, and parallelizations. Themax-flow min-cut property of a clutter is characterized in algebraic andcombinatorial terms. The structure of ideal uniform clutters is presentedhere. If a clutter satisfies the max-flow min-cut property, we prove that allinvariant factors of its incidence matrix are equal to 1 and that the columnsof this matrix form a Hilbert basis. It is shown that the clique clutter of acomparability graph satisfies the max-flow min-cut property. The normalityof an ideal is described in terms of the integer rounding property of a linearsystem and we establish a duality theorem for monomial subrings. We show

xiv Preface

that the Rees algebra of the ideal of covers of a perfect graph is normal andthat clique clutters of Meynel graphs are Ehrhart clutters.

This book stresses the use of computational and combinatorial methodsin commutative algebra because they have been a major factor in discoveringnew and interesting results. The main computer algebra programs that weuse in this book are Normaliz [68] and Macaulay2 [199]. These programsprovide an invaluable tool to study monomial algebras and their algebraicinvariants. As a handy reference we include a section summarizing the typeof computations that can be done using Normaliz (see Section 1.7). Wealso occasionally use other computer algebra systems, such as CoCoA [88],

MapleTM

[80], and Mathematica R© [431]. Singular [200] is another systemthat can be used for computations in commutative algebra with Grobnerbases. Porta [84] and polymake [177] are two systems that can be used forcomputations in convex polyhedra and finite simplicial complexes.

Combinatorial commutative algebra is an extensive area of mathematics.The main references for the area are the books of Bruns and Herzog [65], Hibi[240], Miller and Sturmfels [317], and Stanley [395]. This book emphasizesthe use of discrete mathematics and combinatorial optimization methods incombinatorial commutative algebra. There are a number of excellent recentbooks that offer a complementary view of the subject, namely, Beck andRobins [21], Bruns and Gubelazde [61], De Loera, Hemmecke and Koppe[105], Ene and Herzog [142], Herzog and Hibi [224], Huneke and Swanson[259], and Vasconcelos [414].

Outstanding references for computational and combinatorial aspects thatcomplement and—in some cases—extend some of the material included hereare the books of Berge [25, 27], Brøndsted [57], Chvatal [87], Cornuejols [93],De Loera, Rambau and Santos [106], Diestel [111], Cox, Little and O’Shea[99], Eisenbud [128], Ewald [151], Gitler and Villarreal [189], Godsil andRoyle [190], Golumbic [191], Greuel and Pfister [200], Harary [208], Kreuzerand Robbiano [282], Schenck [369], Schrijver [372, 373], Stanley [394, 396],Sturmfels [400], Vasconcelos [413], and Ziegler [438].

We would like to thank Winfried Bruns, Enrique Reyes, and Aron Simisfor their helpful comments and corrections. Thanks to executive editorRobert B. Stern for suggesting we prepare an up-to-date second editionof Monomial Algebras for the new series in mathematics Monographs andResearch Notes in Mathematics, Taylor & Francis (Chapman and Hall/CRCPress Group). The support of Marsha Pronin and Samantha White atTaylor & Francis is much appreciated. Finally, we are grateful to KarenSimon for her careful editorial work on the manuscript.

Rafael H. VillarrealCinvestav-IPN

Mexico City, D.F.

Preface to First Edition

Let R = K[x] = K[x1, . . . , xn] be a polynomial ring in the indeterminatesx1, . . . , xn, over the field K. Let

fi = xvi = xvi11 · · ·xvinn , i = 1, . . . , q,

be a finite set of monomials of R. We are interested in studying severalalgebras and ideals associated with these monomials. Some of these are:

• the monomial subring: K[f1, . . . , fq] ⊂ K[x],

• the Rees algebra: K[x, f1t, . . . , fqt] ⊂ K[x, t], which is also a mono-mial subring,

• the face ring or Stanley–Reisner ring: K[x]/(f1, . . . , fq), if the mono-mials are square-free, and

• the toric ideal: the ideal of relations of a monomial subring.

In the following diagram we stress the most relevant relations betweenthe properties of those algebras that will take place in this text.

Toric ideal

Face ring

Rees algebra

Monomial subring

��

��

��

��

��

��

If such monomials are square-free they are indexed by a hypergraphbuilt on the set of indeterminates, which provides a second combinatorialstructure in addition to the associated Stanley–Reisner simplicial complex.

xvi Preface to First Edition

This book was written with the aim of providing an introduction to themethods that can be used to study monomial algebras and their presentationideals, with emphasis on square-free monomials. We have striven to providemethods that are effective for computations.

A substantial part of this volume is dedicated to the case of monomialalgebras associated to graphs, that is, those defined by square-free quadraticmonomials defining a simple graph. We will systematically use graph theoryto study those algebras. Such a systematic treatment is a gap in the litera-ture that we intend to fill. Two outstanding references for graph theory are[48] and [208].

In the text, special attention is paid to providing means to determinewhether a given monomial algebra or ideal is Cohen–Macaulay or normal.Those means include diverse characterizations and qualities of those twoproperties.

Throughout this work base rings are assumed to be Noetherian andmodules finitely generated.

An effort has been made to make the book self-contained by includinga chapter on commutative algebra (Chapter 2) that includes some detailedproofs and often points the reader to the appropriate references when proofsare omitted. However, we make free use of the standard terminology andnotation of homological algebra (including Tor and Ext) as described in[363] and [428].

The first goal is to present basic properties of monomial algebras. Forthis purpose in Chapter 3 we study affine and graded algebras. The topicsinclude Noether normalizations and their applications, diverse attributes ofCohen–Macaulay graded algebras, Hilbert Nullstellensatz and affine vari-eties, some Grobner bases theory, and minimal resolutions.

In Chapter 4 a thorough presentation of complete and normal ideals isgiven. Here the systematic use of Rees algebras and associated graded ringsmakes clear their importance for the area.

Chapter 5 deals with the Hilbert series of graded modules and algebras,a topic that is quite useful in Stanley’s proof of the upper bound conjecturefor simplicial spheres. Here we introduce the h-vector and a-invariant ofgraded algebras and give several interpretations of the a-invariant when thealgebra is Cohen–Macaulay. Some optimal upper bounds for the number ofgenerators in the least degree of Gorenstein and Cohen–Macaulay ideals arepresented, which naturally leads to the notion of an extremal algebra. Asan application the Koszul homology of Cohen–Macaulay ideals with pureresolutions is studied using Hilbert function techniques.

General monomial ideals and Stanley–Reisner rings are examined inChapter 6. The first version of this chapter was some notes originally pre-pared to teach a short course during the XXVII Congreso Nacional de laSociedad Matematica Mexicana in October of 1994. In this course we pre-

Preface to First Edition xvii

sented some applications of commutative algebra to combinatorics. We haveexpanded these notes to include a more complete treatment of shellable andCohen–Macaulay complexes. The presentation of the last three sections ofthis chapter, discussing the Hilbert series of face rings and the upper boundconjectures, was inspired by [41, 65] and [393].

Since monomial algebras defined by square-free monomials of degree twohave an underlying graph theoretical structure it is natural that some inter-action will occur between monomial algebras, graph theory, and polyhedralgeometry. We have included three chapters that focus on monomial algebrasassociated to graphs. One of them is Chapter 7, where we present connec-tions between graphs and ideals and study the Cohen–Macaulay property ofthe face ring. Another is Chapter 10, where we present a combinatorial des-cription of the integral closure of the corresponding monomial subring andgive some applications to graph theory. In Chapter 10 we consider mono-mial subrings and toric ideals of complete graphs with the aim of computingtheir Hilbert series, Noether normalizations, and Grobner bases.

The central topic of Chapters 9 and 12 is the normality of monomialsubrings and ideals; some features of toric ideals are presented here.

Chapter 8 is devoted to the study of monomial curves and their toricideals, where the focus of our attention will be on monomial space curvesand monomial curves in four variables. Affine toric varieties and their toricideals are studied in Chapter 8.

Most of the material in this textbook has been written keeping in minda typical graduate student with a basic knowledge of abstract algebra anda non-expert who wishes to learn the subject. We hope that this book canbe read by people from diverse subjects and fields, such as combinatorics,graph theory, and computer algebra. Various units are accessible to upperundergraduates.

In the last fifteen years a dramatic increase in the number of researcharticles and books in commutative algebra that stress its connections withcomputational issues in algebraic geometry and combinatorics has takenplace. Excellent references for computational and combinatorial aspectsthat complement some of the material included here are [99, 128, 413],[65, 395, 400] and [57, 438].

A constant concern during the writing of this text was to give appro-priate credits for the proofs and results that were adapted from printedmaterial or communicated to us. We apologize for any involuntary omis-sion and would appreciate any comments and suggestions in this regard.

During the fall of 1999 a course on monomial algebras associated tographs was given at the University of Messina covering Chapter 7 to Chapter10 with the support of the Istituto Nazionale Di Alta Matematica FrancescoSeveri . It is a pleasure to thank Vittoria Bonanzinga, Marilena Crupi,Gaetana Restuccia, Rossana Utano, Maurizio Imbesi, Giancarlo Rinaldo,

xviii Preface to First Edition

Fabio Ciolli, and Giovanni Molica for the opportunity to improve thosechapters and for their hospitality.

We thank Wolmer V. Vasconcelos for his comments and encouragementto write this book. A number of colleagues and students provided helpfulannotations to some early drafts. We are especially grateful to AdrianAlcantar, Joe Brennan, Alberto Corso, Jose Martınez-Bernal, Susan Morey,Carlos Renterıa, Enrique Reyes, and Aron Simis. We are also grateful toLaura Valencia for her competent secretarial assistance.

The Consejo Nacional de Ciencia y Tecnologıa (CONACyT) and theSistema Nacional de Investigadores (SNI) deserve special acknowledgmentfor their generous support. It should be mentioned that the development ofthis book was included in the project Estudios sobre Algebras Monomiales ,which was supported by the CONACyT grant 27931E.

In the homepage “http://www.math.cinvestav.mx/∼vila” we will main-tain an updated list of corrections.

Rafael H. VillarrealCinvestav-IPN

Mexico City, D.F.

Chapter 1

Polyhedral Geometry andLinear Optimization

In this chapter we introduce several notions and results from polyhedralgeometry, combinatorial optimization and linear programming. Excellentgeneral references for these areas are [35, 87, 281, 372, 373, 438]. Thenwe study relative volumes of lattice polytopes. We present various relationsbetween Hilbert bases, TDI linear systems of inequalities, the max-flow min-cut property of clutters, integral linear systems, and the normality of affinesemigroups and Rees semigroups. Elementary integral vectors and matroidsare introduced at the end of the chapter.

1.1 Polyhedral sets and cones

An affine space or linear variety in Rn is by definition a translation of alinear subspace of Rn. Let A be a subset of Rn. The affine space generatedby A, denoted by aff(A), is the set of all affine combinations of points in A:

aff(A) = {a1v1 + · · ·+ arvr| vi ∈ A, ai ∈ R, a1 + · · ·+ ar = 1}.

There is a unique linear subspace V of Rn such that aff(A) = x0 + V , forsome x0 ∈ Rn. The dimension of A is defined as dim A = dimR(V ).

Definition 1.1.1 An affine map is a function between two affine spacesthat preserves affine combinations.

A point x ∈ Rn is called a convex combination of v1, . . . , vr ∈ Rn if thereare a1, . . . , ar in R such that ai ≥ 0 for all i, x =

∑i aivi and

∑i ai = 1.

Let A be a subset of Rn. The convex hull of A, denoted by conv(A), is theset of all convex combinations of points in A. If A = conv(A) we say thatA is a convex set.

2 Chapter 1

Definition 1.1.2 A convex polytope P ⊂ Rn is the convex hull of a finiteset of points v1, . . . , vr in Rn, that is, P = conv(v1, . . . , vr).

The inner product of two vector x = (x1, . . . , xn) and y = (y1, . . . , yn)in Rn is defined by

〈x, y〉 = x · y = x1y1 + · · ·+ xnyn.

Definition 1.1.3 Let x = (x1, . . . , xn) be a point in Rn. The Euclideannorm of x is defined as

‖x‖ =√〈x, x〉.

We shall always assume that a subset of Rn has the topology inducedby the usual topology of Rn.

Definition 1.1.4 A point a of a set A in Rn is said to be a relative interiorpoint of A if there exists r > 0 such that

Br(a) ∩ aff(A) ⊂ A,

where Br(a) = {x| ‖x− a‖ < r}.

Let A ⊂ Rn. The set of relative interior points of A, denoted by ri(A),is called the relative interior of A. We denote the closure of A by A. Theset A \ ri(A) is called the relative boundary of A and is denoted by rb(A),points in rb(A) are called the relative boundary points of A. If dim(A) = n,the relative interior of A is sometimes denoted by Ao.

An affine space of Rn of dimension n− 1 is called an affine hyperplane.Given a ∈ Rn \ {0} and c ∈ R, define the affine hyperplane

H(a, c) = {x ∈ Rn| 〈x, a〉 = c}.

Notice that any affine hyperplane of Rn has this form. The two closedhalfspaces bounded by H(a, c) are

H+(a, c) = {x ∈ Rn| 〈x, a〉 ≥ c} and H−(a, c) = H+(−a,−c).

If a is a rational vector and c is a rational number, the affine hyperplaneH(a, c) (resp. the halfspace H+(a, c)) is called a rational hyperplane (resp.rational halfspace). If c = 0, for simplicity the set Ha will denote thehyperplane of Rn through the origin with normal vector a, that is,

Ha := H(a, 0) = {x ∈ Rn| 〈x, a〉 = 0}.

The two closed halfspaces bounded by Ha are denoted by

H+a = {x ∈ Rn| 〈x, a〉 ≥ 0} and H−

a = {x ∈ Rn| 〈x, a〉 ≤ 0}.

Polyhedral Geometry and Linear Optimization 3

Definition 1.1.5 A polyhedral set or convex polyhedron is a subset of Rn

which is the intersection of a finite number of closed halfspaces of Rn. Theset Rn is considered a polyhedron.

The transpose of a matrix A (resp. vector x) will be denoted by At orA� (resp. xt or x�). Often a vector x will denote a column vector or a rowvector, from the context the meaning should be clear. Thus, a polyhedralset Q can be represented as

Q = {x ∈ Rn|Ax ≤ b}

for some matrix A and for some vector b. As usual, if a = (ai) and c = (ci)are vectors in Rq, then a ≤ c means ai ≤ ci for all i. If A and b have rationalentries, Q is called a rational polyhedron.

Since the intersection of an arbitrary family of convex sets in Rn isconvex, one derives that any polyhedral set is convex and closed.

Definition 1.1.6 Let Q be a closed convex set in Rn. A hyperplane H ofRn is called a supporting hyperplane of Q if Q is contained in one of the twoclosed halfspaces bounded by H and Q∩H �= ∅.

Definition 1.1.7 A proper face of a polyhedral set Q is a set F ⊂ Q suchthat there is a supporting hyperplane H(a, c) satisfying the conditions:

(a) F = Q∩H(a, c) �= ∅,(b) Q �⊂ H(a, c), and Q ⊂ H+(a, c) or Q ⊂ H−(a, c).

The improper faces of a polyhedral set Q are Q itself and ∅.

Definition 1.1.8 A proper face F of a polyhedral set Q ⊂ Rn is called afacet of Q if dim(F ) = dim(Q)− 1.

Proposition 1.1.9 If Q is a polyhedral set in Rn and F1, F2 are faces ofQ, then their intersection F = F1 ∩ F2 is a face of Q.

Proof. Let Hi = H(ai, ci) be a supporting hyperplane of Q, where ci ∈ Rand 0 �= ai ∈ Rn, such that Fi = Q∩Hi and Q ⊂ H+

i for i = 1, 2. Next weprove the equality

F = Q∩H(a1 + a2, c1 + c2).

The left-hand side is clearly contained in the right-hand side. On the otherhand if x ∈ Q ∩H(a1 + a2, c1 + c2), then using 〈x, ai〉 ≥ ci one has:

c1 + c2 = 〈x, a1 + a2〉 = 〈x, a1〉+ 〈x, a2〉 ≥ c1 + c2,

hence 〈x, ai〉 = ci for i = 1, 2 and x ∈ F . As Q ⊂ H+(a1 + a2, c1 + c2), theset F is a face of Q. �

4 Chapter 1

Definition 1.1.10 A partially ordered set or poset is a pair P = (V,�),where V is a finite set of vertices and � is a binary relation on V satisfying:

(a) u � u, ∀u ∈ V (reflexivity).

(b) u � v and v � u, imply u = v (antisymmetry).

(c) u � v and v � w, imply u � w (transitivity).

A poset P = (V,�) with vertex set V = {x1, . . . , xn} can be displayedby an inclusion diagram , where xi is joined to xj by raising a line if xi ≺ xjand there is no other vertex in between.

Example 1.1.11 The inclusion diagram of the divisors of 12 is:

�6

�3�1

�2

�4

�12

��

��

��

A poset in which any two elements x, y have a greatest lower boundinf{x, y} and a lowest upper bound sup{x, y} is called a lattice. A latticeis called complete if inf S and supS exist for any subset S. A mapping ϕfrom one lattice L1 = (V1,�) onto another lattice L2 = (V2,�) is calledan isomorphism when it is one-to-one and we have x � y if and only ifϕ(x) � ϕ(y) for all x, y ∈ V1.

Corollary 1.1.12 Let Q be a polyhedral set in Rn and let F be the set ofall faces of Q. The partially ordered set (F ,⊂) is a complete lattice withthe lattice operations

inf G = ∩{F |F ∈ G} and sup G = ∩{G ∈ F| ∀F ∈ G;F ⊂ G}.

Proof. It follows from Proposition 1.1.9 and from the fact that a convexpolyhedron has only finitely many faces [427, Theorem 3.2.1(v)]. �

Definition 1.1.13 The lattice (F ,⊂) is called the face-lattice of Q.

Definition 1.1.14 A polytopal complex C is a finite collection of polytopesin Rn such that (i) ∅ ∈ C, (ii) if P ∈ C, then all faces of P are in C, and (iii)the intersection P ∩Q of two polytopes P ,Q ∈ C is a face of both P and Q.

Similarly, one can define the notion of a polyhedral complex replacingpolytope by polyhedron. The dimension of a polytopal complex C, denotedby dim(C), is the largest dimension of a polytope in C.

Definition 1.1.15 Let P be a convex polytope. All proper faces of P formthe boundary complex C(∂P), whose facets are the facets of P .


The boundary complex C(∂P) of a polytope P of dimension d + 1 is apure polytopal complex of dimension d [57, Chapter 2].

Definition 1.1.16 Let P be a convex polytope of dimension d + 1. Theboundary complex C(∂P) is shellable if there is a linear ordering F1, . . . , Fsof the facets of P such that for every 1 ≤ i < j ≤ s, there is 1 ≤ � < jsatisfying Fi ∩ Fj ⊂ F� ∩ Fj and F� ∩ Fj is a facet of Fj .

Theorem 1.1.17 [58] The boundary complex of a polytope is shellable.

The set of nonnegative real numbers and the set of nonnegative integersare denoted by:

R+ = {x ∈ R|x ≥ 0} and N = {0, 1, 2, . . .}

respectively, R+ is also denoted by R≥0. If A is a set of points in Rn, thecone generated by A, denoted by R+A or cone(A), is defined as

R+A =

{q∑i=1

aiβi

∣∣∣∣∣ ai ∈ R+, βi ∈ A for all i

}.

The vector space spanned by A is denoted by RA. Given a vector v ∈ Rn,we define:

R+v := {λv|λ ∈ R+}.

Theorem 1.1.18 (Caratheodory’s theorem) Let v1, . . . , vq be a sequenceof vectors in Rn not all of them zero. If x ∈ R+v1 + · · · + R+vq, thenthere is a linearly independent set V ⊂ {v1, . . . , vq} such that x ∈ R+V.

Proof. By induction on q. The case q = 1 is clear. If q ≥ 2, we canwrite x = a1v1 + · · ·+ aqvq, ai ≥ 0 for i = 1, . . . , q. One may assume thatv1, . . . , vq are linearly dependent and ai > 0 for all i, otherwise the resultfollows by induction. There are real numbers b1, . . . , bq such that at leastone bi is positive and

∑qi=1 bivi = 0. Setting

0 < c = min{ai/bi| bi > 0} = aj/bj,

we get x = x− c(b1v1 + · · ·+ bqvq) =∑q

i=1(ai− cbi)vi, with ai− cbi ≥ 0 forall i and aj − cbj = 0. Hence, by induction, the result follows. �

Definition 1.1.19 A set of vectors α1, . . . , αq ∈ Rn is affinely independentif for every sequence λ1, . . . , λq of real numbers satisfying

∑qi=1 λiαi = 0

and∑q

i=1 λi = 0, one has λi = 0 for all i.

Corollary 1.1.20 Let v1, . . . , vq be a set of vectors in Rn and let x be in itsconvex hull. Then there exists an affinely independent set V ⊂ {v1, . . . , vq}such that x ∈ conv(V).

6 Chapter 1

Proof. Consider B = {(v1, 1), . . . , (vq, 1)}. Since (x, 1) belongs to R+B,the result follows from Caratheodory’s theorem and Exercise 1.1.67. �

Definition 1.1.21 If C ⊂ Rn is closed under linear combinations withnonnegative real coefficients, we say that C is a convex cone. A polyhedralcone is a convex cone which is also a polyhedral set.

Proposition 1.1.22 If C is a closed convex cone and H is a supportinghyperplane of C, then H is a hyperplane passing through the origin.

Proof. Let H = H(a, c). Since 0 ∈ C ⊂ H+, one has c ≤ 0. If C∩H = {0},then c = 0 as required. Assume C ∩H �= {0} and pick 0 �= z ∈ C such that〈z, a〉 = c. Using that tz ∈ C ⊂ H+(a, c) for all t ≥ 0, one derives

tc = t〈z, a〉 = 〈tz, a〉 ≥ c ∀ t ≥ 0⇒ c = 0. �

An affine space of dimension 1 is called a line. The following result isquite useful in determining the facets of a polyhedral cone.

Proposition 1.1.23 Let A be a finite set of points in Zn and let F be aface of R+A. The following hold.

(a) If F �= {0}, then F = R+V for some V ⊂ A.(b) If dim(F ) = 1 and R+A contains no lines, then F = R+α with α ∈ A.(c) If dim(R+A) = n and F is a facet defined by the supporting hyperplane

Ha, then Ha is generated by a linearly independent subset of A.

Proof. Let F = R+A∩Ha with R+A ⊂ H−a . Then F is equal to the cone

generated by the set V = {α ∈ A| 〈α, a〉 = 0}. This proves (a). Parts (b)and (c) follow from (a). �

Most of the notions and results considered thus far make sense if wereplace R by an intermediate field Q ⊂ K ⊂ R, i.e., we can work in theaffine space Kn. However with very few exceptions, like for instance Theo-rem 1.1.24, we will always work in the Euclidean space Rn or Qn.

For convenience we state the fundamental theorem of linear inequalities :

Theorem 1.1.24 [372, Theorem 7.1] Let Q ⊂ K ⊂ R be an intermediatefield and let C ⊂ Kn be a cone generated by A = {α1, . . . , αq}. If α ∈ Kn\Cand t = rank{α1, . . . , αq, α}, then there exists a hyperplane Ha containingt−1 linearly independent vectors from A such that 〈a, α〉 > 0 and 〈a, αi〉 ≤ 0for i = 1, . . . , q.

Theorem 1.1.25 (Farkas’s Lemma) Let A be an n× q matrix with entriesin a field K and let α ∈ Kn. Assume Q ⊂ K ⊂ R. Then either there existsx ∈ Kq with Ax = α and x ≥ 0, or there exists v ∈ Kn with vA ≥ 0 and〈v, α〉 < 0, but not both.


Proof. Let A = {α1, . . . , αq} be the set of column vectors of A. Assumethat there is no x ∈ Kq with Ax = α and x ≥ 0, i.e., α is not in R+A.By Theorem 1.1.24 there is a hyperplane Hv such that 〈v, α〉 < 0 and〈v, αi〉 ≥ 0 for all i. Hence vA ≥ 0, as required. If both conditions hold,then 0 > 〈v, α〉 = 〈v,Ax〉 = 〈vA, x〉 ≥ 0, a contradiction. �

The reader is referred to [438] for other versions of Farkas’s lemma. Thenext result tells us how to separate a point from a cone.

Corollary 1.1.26 Let C ⊂ Kn be a cone generated by A = {a1, . . . , am}.If γ ∈ Kn and γ /∈ C, then there is a hyperplane H through the origin suchthat γ ∈ H−\H and C ⊂ H+.

Proof. Let A be the matrix with column vectors a1, . . . , am. By Farkas’slemma (see Theorem 1.1.25) there exists μ ∈ Kn such that μA ≥ 0 and〈γ, μ〉 < 0. Thus 〈μ, ai〉 ≥ 0 for all i. If H is the hyperplane through theorigin with normal vector μ we get C ⊂ H+, as required. �

Corollary 1.1.27 Let A be a finite set in Zn, then

ZA ∩ R+A = ZA ∩Q+A and Zn ∩R+A = Zn ∩Q+A,

where ZA is the subgroup of Zn spanned by A and Q+ = {x ∈ Q|x ≥ 0}.

Proof. It follows at once from Theorem 1.1.25. �

Definition 1.1.28 We say that C is a finitely generated cone if C = R+A,for some finite set A = {v1, . . . , vq}.

The proof of the next result yields the duality theorem for cones .

Theorem 1.1.29 If C ⊂ Rn, then C is a finitely generated cone (resp.finitely generated cone by rational vectors) if and only if C is a polyhedralcone (resp. rational polyhedral cone) in Rn.

Proof. ⇒) Assume that C �= (0) is a cone generated by A = {α1, . . . , αm}.We set r = dim(R+A). Notice that aff(C) is the real vector space generatedby A, because 0 ∈ C. If aff(C) = C, then C is a polyhedral cone, becauseC is the intersection of n − r hyperplanes of Rn through the origin. Nowassume that C � aff(C). Consider the family

F = {F |F = Ha ∩C; dim(F ) = r − 1; C ⊂ H−a }.

By Theorem 1.1.24 the family F is non-empty. Notice that F is a finiteset because each F in F is a cone generated by a subset of A; see Proposi-tion 1.1.23. Assume that F = {F1, . . . , Fs}, where Fi = Hai ∩C. We claimthat the following equality holds

C = H−a1 ∩ · · · ∩H

−as ∩ aff(C). (1.1)

8 Chapter 1

The inclusion “⊂” is clear. To show the inclusion “⊃”, we proceed bycontradiction. Assume that there exists α �∈ C such that α belongs tothe right-hand side of Eq. (1.1). By Theorem 1.1.24 and by reorderingthe elements of A if necessary, there is a hyperplane Ha containing linearlyindependent vectors α1, . . . , αr−1 such that (i) 〈a, α〉 > 0, and (ii) 〈a, αi〉 ≤ 0for i = 1, . . . ,m. Thus F = Ha ∩ C = Hak ∩ C for some 1 ≤ k ≤ s.We may assume that α1, . . . , αr form a basis of aff(C). Since α ∈ aff(C)there are scalars λ1, . . . , λr with α = λ1α1 + · · · + λrαr. Using (ii) we get〈a, α〉 = λr〈αr , a〉 > 0. Hence λr < 0. On the other hand 〈ak, αr〉 < 0because C ⊂ H−

akand dim(Fk) = r − 1. Therefore

〈ak, α〉 = λr︸︷︷︸<0

〈ak, αr〉︸︷︷︸<0

> 0,

a contradiction to the fact that α ∈ H−ak.

⇐) Assume that C = H−b1∩ · · · ∩H−

br, where b1, . . . , br ∈ Rn. Consider

the cone C′ generated by b1, . . . , br. From the first part of the proof we canwrite

C′ = R+{b1, . . . , br} = H−α1∩ · · · ∩H−

αm, (∗)

for some set of vectors A = {α1, . . . , αm} in Rn. Next we show the equalityC = R+A. Notice that 〈bi, αj〉 ≤ 0 for all i, j, because bi ∈ C′ for all i.Thus R+A ⊂ C. Assume that there is α ∈ C \ R+A. By Corollary 1.1.26,there exists a hyperplane Ha such that R+A ⊂ H−

a and 〈a, α〉 > 0. Hence〈αi, a〉 ≤ 0 for all i, and by Eq. (∗) we conclude that a ∈ R+{b1, . . . , br}.Therefore, we can write a = λ1b1 + · · ·+λrbr, λi ≥ 0 for all i. Since α ∈ C,we have 〈α, a〉 = λ1〈α, b1〉 + · · · + λr〈α, br〉 ≤ 0, contradicting 〈a, α〉 > 0.Thus C = R+A. The respective statement about the rationality characterof the representations is left as an exercise. �

Corollary 1.1.30 (Duality theorem for cones) Let B = {β1, . . . , βr} be asubset of Rn, and let {α1, . . . , αm} and {c1, . . . , cs} be subsets of Rn \ {0}.

(a) If R+B = ∩mi=1H−αi, then H−

β1∩ · · · ∩H−

βr= R+α1 + · · ·+ R+αm.

(b) If R+B = ∩si=1H+ci , then H

+β1∩ · · · ∩H+

βr= R+c1 + · · ·+ R+cs.

Proof. Part (a) follows from the second part of the proof of Theorem 1.1.29.Part (b) follows using the equality H+

ci = H−−ci and using part (a). �

By Theorem 1.1.29 a polyhedral cone C � Rn has two representations:

Minkowski representation C = R+B with B = {β1, . . . , βr} a finite set, and

Implicit representation C = H+c1 ∩ · · · ∩H+

cs for some c1, . . . , cs ∈ Rn \ {0}.


From the duality theorem for cones these two representations satisfy:

H+β1∩ · · · ∩H+

βr= R+c1 + · · ·+ R+cs. (1.2)

The dual cone of C is defined as

C∗ :=⋂c∈C

H+c =

⋂a∈B

H+a .

By the duality theorem one has C∗∗ = C. An implicit representationof C is called irredundant or irreducible if none of the closed half-spacesH+c1 , . . . , H

+cs can be omitted from the intersection.

Remark 1.1.31 The left-hand side of Eq. (1.2) is an irredundant repre-sentation of C∗ if and only if no proper subset of B generates C.

Corollary 1.1.32 Let C ⊂ Rn be a finitely generated cone and let

F = {F |F = Ha ∩C; dim(F ) = r − 1; C ⊂ H−a } = {F1, . . . , Fs} �= ∅,

where Fi = C ∩Hai , dim(C) = r. Then C = aff(C) ∩H−a1 ∩ · · · ∩H−

as .

Proof. It follows from the first part of the proof of Theorem 1.1.29. �

One of the fundamental results in polyhedral geometry is the followingremarkable decomposition theorem for polyhedra. See [372, Corollary 7.1b]and [427, Theorem 4.1.3] for historical comments and for more informationabout this result.

Theorem 1.1.33 (Finite basis theorem) If Q is a set in Rn, then Q is apolyhedron (resp. rational polyhedron) if and only if Q can be expressed asQ = P + C, where P is a convex polytope (resp. rational polytope) and Cis a finitely generated cone (resp. finitely generated rational cone).

Proof. ⇒) Let Q = {x|Ax ≤ b} be a polyhedron in Rn, where A is amatrix and b is a vector. Consider the set

C′ =

{(xλ

)∣∣∣∣ x ∈ Rn;λ ∈ R+;Ax − λb ≤ 0

}.

Notice that C′ can be written as

C′ =

{(xλ

)∣∣∣∣x ∈ Rn;λ ∈ R;

(A −b0 −1

)(xλ

)≤ 0

},

where −b is a column vector. Thus C′ is a polyhedral cone in Rn+1. UsingCorollary 1.1.29 we get that C′ can be expressed as

C′ = R+

{(x1λ1

), . . . ,

(xmλm

)}(λi ≥ 0; xi ∈ Rn).

10 Chapter 1

We may assume that λi ∈ {0, 1} for all i. Set

A = {xi|λi = 1} = {x1, . . . , xr}, B = {xi|λi = 0} = {xr+1, . . . , xm},

P = conv(A), and C = R+B. Notice that x ∈ Q if and only if (x, 1) ∈ C′.Thus x ∈ Q if and only if (x, 1) can be written as(

x1

)= μ1

(x11

)+ · · ·+ μr

(xr1

)+ μr+1

(xr+1

0

)+ · · ·+ μm

(xm0

)with μi ≥ 0 for all i. Consequently x ∈ Q if and only if x ∈ P+C. Thereforewe obtain Q = P + C.⇐) Assume that Q is equal to P + C with P = conv(x1, . . . , xr) and

C = R+(xr+1, . . . , xm). Consider the following finitely generated cone

C′ = R+

{(x11

), . . . ,

(xr1

),

(xr+1

0

), . . . ,

(xm0

)}.

By Corollary 1.1.29 the cone C′ is a polyhedron. Thus there exists a matrixA and a vector b such that C′ can be written as

C′ =

{(xλ

)∣∣∣∣x ∈ Rn;λ ∈ R;Ax+ λb ≤ 0

}.

Since x ∈ Q if and only if (x, 1) ∈ C′ we conclude that Q = {x|Ax ≤ −b},that is Q is a polyhedron. This proof is due to Schrijver [372]. �

Thus, by the finite basis theorem, a polyhedron has two representations.The computer program PORTA [84] can be used to switch between these tworepresentations. This program is available for Unix and Windows systems.For polyhedral cones one can use Normaliz [68].

Corollary 1.1.34 A set Q ⊂ Rn is a convex polytope if and only if Q is abounded polyhedral set.

Proof. If Q = conv(α1, . . . , αm) is a polytope, then by the triangle inequal-ity for all x ∈ Q we have ‖x‖ ≤ ‖α1‖+ · · ·+ ‖αm‖. Thus Q is bounded.

Conversely if Q is a bounded polyhedron, then by Theorem 1.1.33 we candecompose Q as Q = P +C, with P a polytope and C a finitely generatedcone. Notice that C = {0}, otherwise fixing p0 ∈ P and c0 ∈ C \ {0} weget p0 +λc0 ∈ Q for all λ > 0, a contradiction because Q is bounded. ThusP = Q, as required. �

Definition 1.1.35 Let Q be a polyhedral set and x0 ∈ Q. The point x0 iscalled a vertex or an extreme point of Q if {x0} is a proper face of Q.

Proposition 1.1.36 [57, Theorem 7.2] If P = conv(α1, . . . , αr) ⊂ Rn andV is the set of vertices of P, then V ⊂ {α1, . . . , αr} and P = conv(V ).


Lemma 1.1.37 [57, Theorem 5.6] If Q is a polyhedral set in Rn and v ∈ Qis not a vertex of Q, then there is a face F of Q such that v ∈ ri(F ).

Lemma 1.1.38 Let Q be a convex polyhedron in Rn and x0 ∈ Q. Then x0is a vertex of Q if and only if Q \ {x0} is a convex set.

Proof. ⇒) Let H(a, c) be a proper supporting hyperplane of Q such that{x0} = Q ∩H(a, c) and Q ⊂ H+(a, c). Take x, y ∈ Q \ {x0} and considerz = ty + (1− t)x with 0 < t < 1. Note that x, y are not in H(a, c), thus

〈z, a〉 = t〈y, a〉+ (1− t)〈x, a〉 > tc+ (1− t)c = c.

Hence 〈z, a〉 > c, that is, z �= x0 and z ∈ Q, as required.⇐) Let Q = ∩ri=1H

+(ai, ci) be a decomposition of Q as an intersectionof closed halfspaces, where 0 �= ai ∈ Rn and ci ∈ R for all i. First observethat x0 is in H(ai, ci) for some i, otherwise if 〈x0, ai〉 > ci for all i, thereis an open ball Bδ(x0) in Rn, of radius δ centered at x0, whose closure liesin Q. Hence taking two antipodal points x1, x2 in the boundary of Bδ(x0)one obtains x0 ∈ Q \ {x0}, a contradiction. One may now assume there isk ≥ 1 such that x0 ∈ H(ai, ci) for i ≤ k and 〈x0, ai〉 > ci for i > k. Set

A = H(a1, c1) ∩ · · · ∩H(ak, ck).

We claim that A = {x0}. If there is x1 ∈ A \ {x0}, pick Bδ(x0) whoseclosure is contained in H+(ai, ci) for all i > k. Since z = tx0 + (1 − t)x1is in A for all t ∈ R, making t = 1 + δ/‖x1 − x0‖ one derives z ∈ A and‖z − x0‖ = δ, thus z ∈ Q. Note that if k = r, then x1 is already in Qand in this case we set z = x1. Making t = −1 in z1 = tz + (1 − t)x0 oneconcludes z1 ∈ A and ‖z1 − x0‖ = δ. Altogether z, z1 are in Q \ {x0} andx0 = (z + z1)/2, a contradiction because Q \ {x0} is a convex set.

From the equality {x0} = A = A∩Q one has that {x0} is an intersectionof faces. Using Proposition 1.1.9 yields that {x0} is a face, as required. �

Proposition 1.1.39 If Q = C + P ⊂ Rn with C a polyhedral cone and Pa polytope, then every vertex of Q is a vertex of P.

Proof. Let x be a vertex of Q and write x = c + p for some c ∈ C andp ∈ P . We claim that p is a vertex of P . If p is not a vertex of P , then byLemma 1.1.37 there is a face F of P such that p ∈ ri(F ). Pick p �= v ∈ F .By [57, Theorem 3.5], there is y ∈ F with p ∈]y, v[. Hence p = λy+(1−λ)vwith 0 < λ < 1. Note that x �= y + c ∈ Q, x �= v + c ∈ Q, and

x = λ(y + c) + (1− λ)(v + c) (0 < λ < 1),

thus Q \ {x} is not a convex set, a contradiction to Lemma 1.1.38. Thisproves that p is a vertex of P . Assume c �= 0, then

p+ λc �= x = p+ c �= p+ 2c (0 < λ < 1).

12 Chapter 1

Notice that x = p+c = λ0(p+λc)+(1−λ0)(p+2c) is a convex combination,where 0 < λ0 = 1/(2− λ) < 1. Since p+ λc and p+ 2c are in Q \ {x}, thisshows that Q \ {x} is not a convex set, a contradiction. Thus c = 0 and xis a vertex of P , as required. �

Theorem 1.1.40 [372, Theorem 8.4] Let Q ⊂ Rn be a polyhedral set. ThenQ has at least one vertex if and only if Q does not contain any lines in Rn.

Proposition 1.1.41 Let Q be a polyhedron containing no lines and letf(x) = 〈x, a〉. If max{f(x)|x ∈ Q} <∞, then the maximum is attained ata vertex of Q.

Proof. Let x0 ∈ Q be an optimal solution and let λ = f(x0) be the optimalvalue. Note that F = Q∩H(a, λ) is a face of Q because Q ⊂ H−(a, λ) andx0 ∈ F . Since F contains no lines, by Theorem 1.1.40, the face F containsat least one vertex z0, which is also a vertex of Q by transitivity. Thus theoptimal value λ is attained at the vertex z0, as required. �

Theorem 1.1.42 If Q ⊂ Rn is a polyhedral set containing no lines andα1, . . . , αr are the vertices of Q, then there are β1, . . . , βs ∈ Rn such that

Q = conv(α1, . . . , αr) + (R+β1 + · · ·+ R+βs).

Proof. It follows from the proof of [427, Theorem 4.1.3]. �

Definition 1.1.43 If a polyhedron Q in Rn is represented as

Q = aff(Q) ∩H+(a1, c1) ∩ · · · ∩H+(ar, cr) (1.3)

with ai ∈ Rn \ {0}, ci ∈ R for all i, and none of the closed halfspacesH+(a1, c1), . . . , H

+(ar, cr) can be omitted from the intersection, we saythat Eq. (1.3) is an irreducible representation of Q.

Theorem 1.1.44 [427, Theorem 3.2.1] Let Q be a polyhedral set in Rn

which is not an affine space and let

Q = aff(Q) ∩H+(a1, c1) ∩ · · · ∩H+(ar, cr)

be an irreducible representation. If Fi = Q∩H(ai, ci), i = 1, . . . , r, then

(a) ri(Q) = {x ∈ Q| 〈x, a1〉 > c1, . . . , 〈x, ar〉 > cr};

(b) rb(Q) = F1 ∪ · · · ∪ Fr ;

(c) the facets of Q are precisely the sets F1, . . . , Fr;

(d) each face F � Q is the intersection of all Fi such that F ⊂ Fi.


Corollary 1.1.45 Let Q be a polyhedron and let β be a vector in ri(Q).If F is a proper face of Q, then β �∈ F .

Proposition 1.1.46 Let Q =⋂ri=1H

+(ai, ci) be a polyhedral set in Rn

which is not an affine space, where 0 �= ai ∈ Rn and ci ∈ R for all i. Ifx0 ∈ Rn, then x0 is a vertex of Q if and only if

(a) {x0} = ∩i∈IH(ai, ci) for some I ⊂ {1, . . . , r}, and

(b) 〈x0, ai〉 ≥ ci for all i = 1, . . . , r.

Proof. ⇒) This direction follows at once from the proof of Lemma 1.1.38.⇐) Note x0 ∈ Q. Since the intersection of faces of Q is a face (see

Proposition 1.1.9), one has that {x0} is a face. �

Corollary 1.1.47 Let A = (aij) be an r×n matrix, let c = (ci) be a columnvector and let a1, . . . , ar be the rows of A. If Q = {x ∈ Rn|Ax ≥ c} andx0 ∈ Q, then x0 is a vertex of Q if and only if there is J ⊂ {1, . . . , r} with|J | = n such that the set {ai| i ∈ J} is linearly independent and

{x0} = {(xi) ∈ Rn| ai1x1 + · · ·+ ainxn = ci for all i ∈ J}.

Proof. Assume x0 is a vertex of Q. Hence, by Proposition 1.1.46, x0 is theunique solution of a system of linear equations:

ai1x1 + · · ·+ ainxn = ci (i ∈ I)

for some I ⊂ {1, . . . , r}. Let [A′|c′] be the augmented matrix of this system,by Gaussian elimination one obtains that this matrix reduces to[

In | c′′

0 | 0

].

Hence the rank of A′ is equal to n. Thus there are n linearly independentrows of A′ and x0 is the unique solution of the system

ai1x1 + · · ·+ ainxn = ci (i ∈ J ⊂ I)

for some J ⊂ I with |J | = n. The converse is clear. �

Let Q be a polyhedron in Rn represented by a system of linear con-straints

〈ai, x〉 ≥ bi (i = 1, . . . , r),

where ai ∈ Rn and bi ∈ R for all i. With a slight abuse of language, we willsay that the constraints are linearly independent if the corresponding ai arelinearly independent.

14 Chapter 1

Definition 1.1.48 A vector x0 in Rn is called a basic feasible solution of asystem of linear constraints 〈ai, x〉 ≥ bi, i = 1, . . . , r, if

(a) x0 satisfies all linear constraints, and

(b) out of the constraints that satisfy 〈ai, x0〉 = bi, there are n of themthat are linearly independent.

The next result is a restatement of the corollary above.

Corollary 1.1.49 Let Q be a polyhedron in Rn. A vector x0 in Rn is avertex of Q if and only if x0 is a basic feasible solution of any system oflinear constraints that represents Q.

This result yields a method to find the vertices of a polyhedron whichis by no means the best in practice. See [10, 11] for a thorough discussionon how to find all the vertices of a polyhedron.

Corollary 1.1.50 If C �= Rn is a polyhedral cone of dimension n, thenthere is a unique irreducible representation

C = H+a1 ∩ · · · ∩H

+ar , where ai ∈ Rn \ {0}.

Proof. According to Proposition 1.1.22 a set F is a proper face of C ifthere is a supporting hyperplane Ha of C such that F = C ∩ Ha �= ∅ andC �⊂ Ha. In particular the facets of C are defined by hyperplanes throughthe origin. Therefore by Theorem 1.1.44 the irreducible representation of Chas the required form and is unique. �

Proposition 1.1.51 If Q �= Rn is a rational polyhedral cone of dimensionn in Rn, then there are unique (up to sign) a1, . . . , ar in Zn with relativelyprime entries such that the irreducible representation of Q is

Q = H+a1 ∩H

+a2 ∩ · · · ∩H

+ar

Proof. By the finite basis theorem, there are α1, . . . , αq in Qn such thatQ = R+α1+ · · ·+R+αq. First note that if Hb is a supporting hyperplane ofQ generated by a set of n− 1 linearly independent vectors in {α1, . . . , αq},then by the Gram–Schmidt process Hb has an orthogonal basis of vectorsin Qn, and consequently there is a normal vector a to Hb such that a ∈ Qn

and Ha = Hb. Hence multiplying a by a suitable integer and then dividingby the greatest common divisor of the entries, one may assume Ha = Hb,where a is in Zn and has relative prime entries. Observe that a, b are linearlydependent because the orthogonal complement of Ha is one dimensional. Itis readily seen that a is uniquely determined up to sign.

To complete the proof use Proposition 1.1.23 (c) to see that any facetof Q is defined by a supporting hyperplane Hb as above. �


Definition 1.1.52 A polyhedron containing no lines is called pointed.

If F is a face of dimension 1 of a pointed polyhedral cone, then F = R+vfor some v (see Proposition 1.1.23). The face F is called an extremal ray.

Lemma 1.1.53 [372, Section 8.8] If C is a pointed polyhedral cone in Rn,then C is generated by non-zero representatives of its extremal rays.

Theorem 1.1.54 Let A be a finite set of non-zero points in Zn and let Vbe the set of all α ∈ A such that R+α is a face of R+A of dimension 1. IfR+A is a pointed cone, then R+A = R+V.

Proof. We set A = {α1, . . . , αq}. Let F be a face of R+A of dimension 1.Then, by Proposition 1.1.23, F = R+αi for some i. Thus, the result followsfrom Lemma 1.1.53. �

Definition 1.1.55 Let A be an n × q real matrix and let b, c be two realvectors of sizes q and n, respectively. The primal problem is defined as

Maximize f(x) =∑n

i=1 cixi (∗)Subject to xA ≤ b and x ≥ 0.

Its dual problem is defined as

Minimize g(y) =∑q

i=1 biyi (∗∗)Subject to Ay ≥ c and y ≥ 0.

A solution x0 ∈ Rn of the primal (resp. dual) problem that maximizes theobjective function f is called and optimal solution, the corresponding valuef(x0) of the objective function is called an optimal value.

The most important theorem in Linear Programming (LP) theory is:

Theorem 1.1.56 (LP duality theorem, [372, Corollary 7.1.g]) If the primalproblem (∗) has an optimal solution (x1, . . . , xn), then the dual problem (∗∗)has an optimal solution (y1, . . . , yq) such that

∑ni=1 cixi =

∑qi=1 biyi.

Theorem 1.1.57 [372, Corollary 7.1g, p. 92] Let A be a matrix and b, cvectors. Then

max{〈c, x〉 |xA ≤ b} = min{〈b, y〉 | y ≥ 0, Ay = c} (1.4)

provided that at least one of the sets in (1.4) is non-empty.

An immediate consequence of the duality theorem is:

16 Chapter 1

Corollary 1.1.58 Consider the LP primal-dual pair

max{〈x, c〉|xA ≤ b} = min{〈y, b〉| y ≥ 0;Ay = c}.

Let x and y be feasible solutions, i.e., xA ≤ b, y ≥ 0 and Ay = c. Then thefollowing conditions are equivalent :

(a) x and y are both optimum solutions.

(b) 〈x, c〉 = 〈y, b〉.(c) (b− xA)y = 0.

Condition (c) is called complementary slackness. See Exercise 1.1.81 foranother form of complementary slackness

Proposition 1.1.59 [281, Proposition 3.3] Let Q = {x ∈ Rn |xA ≤ b} �= ∅be a polyhedral set and let F �= ∅ be a proper subset of Q. Then the followingconditions are equivalent :

(a) F = {x ∈ Q |A′x = b′} for some subsystem xA′ ≤ b′ of xA ≤ b.(b) F is a proper face of Q.(c) F = {x ∈ Q | 〈x, c〉 = δ} for some vector 0 �= c ∈ Rn such that the

maximum value δ = max{〈x, c〉 |x ∈ Q} is finite.

Definition 1.1.60 A minimal face of a polyhedron is a face not containingany other face.

Lemma 1.1.61 [372, p. 104] Let Q = {x ∈ Rn |xA ≤ b} �= ∅ be a convexpolyhedron. A set F is a minimal face of Q if and only if ∅ �= F � Q and Fis equal to F = {x ∈ Rn |A′x = b′} for some subsystem xA′ ≤ b′ of xA ≤ b.

Definition 1.1.62 A rational polyhedron Q ⊂ Rn is called integral if Qis equal to conv(Zn ∩ Q).

If Q is a pointed rational polyhedron, then Q is integral if and only if Qhas only integral vertices. This is shown below.

Theorem 1.1.63 [281, Theorem 5.12] If Q is a rational polyhedron, thenQ is integral if and only if any of the following statements hold:

(a) Each face of Q contains integral vectors.

(b) Each minimal face of Q contains integral vectors.

(c) max{〈c, x〉|x ∈ Q} is attained by an integral vector for each c for whichthe maximum is finite.

(d) max{〈c, x〉|x ∈ Q} is an integer for each integral vector c for whichthe maximum is finite.


Corollary 1.1.64 If Q is a pointed polyhedron, then Q is integral if andonly if all vertices of Q are integral.

Proof. Let x0 be a vertex of Q. There is a supporting hyperplane H =H(a, c) such that {x0} = H ∩ Q and Q ⊂ H+. Since x0 is a convexcombination of points in Q∩Zn, it is seen that x0 is integral. The conversefollows from Theorem 1.1.63 and Proposition 1.1.41. �

Proposition 1.1.65 Let Q ⊂ Rn be a rational polyhedron and let QI =conv(Q ∩ Zn) be its integral hull. Then

(a) QI is a polyhedron.

(b) If Q is a pointed polyhedron, then QI is integral.

Proof. We may assume that QI is non-empty. By the finite basis theoremwe can write Q = P + C, where C is a cone generated by integral vectorsα1, . . . , αs and P is a polytope. Consider the linear map T : Rs → Rn

induced by T (ei) = αi. Notice that B := T ([0, 1]s) is a polytope whoseelements have the form λ1α1+ · · ·+λsαs, 0 ≤ λi ≤ 1 for all i. It is not hardto see thatQI = conv((P+B)∩Zn)+C. Since P+B is bounded, we get thatQI is a polyhedron by the finite basis theorem. This proves (a). To provepart (b) observe that the vertices of QI are contained in (P +B)∩Zn, thisfollows from Proposition 1.1.39. Hence by Corollary 1.1.64, QI is integral.This proof was adapted from [372]. �

Exercises

1.1.66 If f : Rq → Rn is an affine map, then f(x) = Ax+ b, for some n× qmatrix A and some vector b.

1.1.67 Let A = {α0, . . . , αr} be a sequence of distinct vectors in Rn. Prove:

(a) A is affinely independent ⇔ α1−α0, . . . , αr −α0 are linearly indepen-dent ⇔ (α0, 1), . . . , (αr, 1) are linearly independent.

(b) If A is contained in an affine hyperplane not containing the origin, thenA is affinely independent if and only if it is linearly independent.

1.1.68 If V ⊂ Rn, prove that V is an affine space in Rn if and only ifλ1V + λ2V ⊂ V for all λ1, λ2 in R such that λ1 + λ2 = 1.

1.1.69 If C ⊂ Rn, prove that C is a convex cone if and only if λx+μy ∈ Cfor all x, y ∈ C and λ, μ ≥ 0.

1.1.70 Let π1 : R2 → R be the projection π1(x1, x2) = x1. Prove that π1 isnot a closed linear map.

Hint Consider the closed set {(x, 1/x)|x > 0}.

18 Chapter 1

1.1.71 Let Q be a polyhedral set in Rn and let f : Rn → Rm be a linearfunction. Prove that f(Q) is a polyhedral set.

1.1.72 Let Q be a polyhedron in Rn and let f : Q → R be a linear function.If f is bounded from above and Q contains no lines, prove that f attainsits maximum at a vertex of Q. Notice that f(Q) is a closed convex set.

1.1.73 Determine the facets of the convex polytope

P = conv(±e1, . . . ,±en) ⊂ Rn.

1.1.74 Let C be a polyhedral cone in Rn and let C∗ be its dual cone. ThenC is a pointed cone if and only if dim(C∗) = n.

1.1.75 If C = Rn and C∗ is its dual cone, then C∗ = (0).

1.1.76 If C = H+e2 ⊂ R2, then C = R+e1 + R+(−e1) + R+e2. Use the

duality theorem to show that C∗ = R+e2.

1.1.77 If C = R+(1, 1), prove that C = H+(1,−1) ∩ H

+(−1,1) ∩ H+

e1 and that

the dual cone is given by C∗ = H+(1,1).

1.1.78 If C is a rational cone and CI := conv(C ∩ Zn), then CI = C.

1.1.79 Let Q = {x| Ax ≤ b} �= ∅ be a polyhedron. If Q = P +C, where Pis a polytope and C is a polyhedral cone, prove that

{y|Ay ≤ 0} = {y|x+ y ∈ Q, ∀x ∈ Q}.

The cone C = {y|Ay ≤ 0} is called the characteristic cone of Q.

1.1.80 Let Q = {x|Ax ≤ b} be a polyhedron. The lineality space of Q isthe linear space lin.space(Q) = {x|Ax = 0}. Prove that Q is pointed if andonly if lin.space(Q) = (0).

1.1.81 (Complementary slackness) Consider the LP primal-dual pair

min{〈x, c〉|x ≥ 0;xA ≥ b} = max{〈y, b〉| y ≥ 0;Ay ≤ c}.

Let x and y be feasible solutions, i.e., xA ≥ b, Ay ≤ c and x, y ≥ 0. UseTheorem 1.1.56 to show that the following conditions are equivalent:

(a) x and y are both optimum solutions.

(b) 〈x, c〉 = 〈y, b〉.(c) (b− xA)y = 0 and x(c−Ay) = 0.


1.1.82 (P. Gordan [194]) The system Ax < 0 is unsolvable if and only if thesystem yA = 0, y ≥ 0, y �= 0 is solvable. The vector inequality (ai) < (bi)means that ai < bi for all i.

In the following exercises K will denote an intermediate field Q ⊂ K ⊂ R.

1.1.83 Let H ⊂ Kn be a linear subspace and {v1, v2, . . . , vm} a basis of H .Prove that H has an orthogonal basis using the Gram–Schmidt process:

1. Set u1 := v1 and u2 := v2− ((v2 ·u1)/(u1 ·u1))u1. Verify that {u1, u2}is an orthogonal set and Ku1 +Ku2 = Kv1 +Kv2.

2. If u1, u2, . . . , uk (k < m) have been constructed so that they form anorthogonal basis for the linear space generated by v1, . . . , vk, setting

uk+1 = vk+1 −[vk+1 · u1u1 · u1

u1 + · · ·+vk+1 · uiui · ui

ui + · · ·+vk+1 · ukuk · uk

uk

],

verify that {u1, u2, . . . , uk, uk+1} is an orthogonal set and that this setgenerates Kv1 +Kv2 + · · ·+Kvk +Kvk+1.

1.1.84 If A = (aij) is a matrix with entries in K, then dim(RA) is equal todim(im(A)), where RA is the row space of A and im(A) is the image of thelinear map defined by A.

1.1.85 If H is a linear subspace of Kn, then H ⊕H⊥ = Kn, where

H⊥ = {x ∈ Kn|x · h = 0 for all h ∈ H}.

1.1.86 If A ∈Mn×q(K), then

(im(A))⊥ = ker(At), im(A)⊕ ker(At) = Kn, and (im(At))⊥ = ker(A).

1.1.87 Let α, β, γ be three vectors inKn such that 〈α, β〉 ≤ 0 and 〈β, γ〉 > 0.

Prove that the vector α′ = α− 〈α,β〉〈γ,β〉γ is in Kn, 〈α′, β〉 = 0, and α′ belongs

to the cone generated by α and γ.

1.1.88 Using the previous exercises show that the proof of Farkas’s lemmagiven in [304, p. 86] is valid for any intermediate field Q ⊂ K ⊂ R.

1.1.89 Show that the proof of the fundamental theorem of linear inequali-ties given in [372, Theorem 7.1] works for any intermediate field Q ⊂ K ⊂ R.

1.1.90 If A ∈ Mn×q(K) and γ ∈ Kn, then either there exists a vectorx ∈ Kq with Ax ≤ γ, or there exists a vector α ∈ Kn with α ≥ 0, αA = 0and 〈α, γ〉 < 0 , but not both.

20 Chapter 1

1.2 Relative volumes of lattice polytopes

In this section we introduce relative volumes and unimodular coverings.Two main references for relative volumes are [21, 146].

Let A = {v1, . . . , vq} be a set of distinct vectors in Zn and let

P = conv(v1, . . . , vq)

be the convex hull of A. A point in Zn is called a lattice point and P iscalled a lattice polytope. Here a lattice point (resp. lattice polytope) is usedas a synonym of integral point (resp. integral polytope).

In the sequel to simplify the exposition and the proofs we set m = q− 1and α0 = v1, . . . , αm = vq.

If V = {0, α1−α0, . . . , αm−α0} is the image of A under the translation

f : Rn −→ Rn, xf�−→ x− α0,

then aff(A) = α0 + RV . In particular, from this expression we get

d = dim(P) = dimR(RV),

where RV is the linear space spanned by V .

Theorem 1.2.1 [261, Theorem 3.7] Let Zn be the free Z-module of rank n.Then any submodule of Zn is free of rank at most n.

Using this result, and the inclusions ZV ⊂ Zn ∩ RV ⊂ RV , it followsthat Zn ∩RV is a free abelian group of rank d. A constructive proof of thisfact, which is useful for computing relative volumes, is included below inthe proof of Lemma 1.2.3.

Theorem 1.2.2 [332, Theorem II.9, pp. 26-27] Let A be an integral matrixof rank d. Then there are invertible integral matrices U and Q such that

U−1AQ = diag{s1, . . . , sd, 0, . . . , 0},

si > 0 for 1 ≤ i ≤ d and si divides si+1 for 1 ≤ i ≤ d− 1.

The matrix D = diag{s1, . . . , sd, 0, . . . , 0} is called the Smith normalform of A and s1, . . . , sd are called the invariant factors of A.

Lemma 1.2.3 RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd for some γ1, . . . , γd in Zn.

Proof. Consider the matrix A of size n ×m whose column vectors corre-spond to the non-zero vectors in V . By Theorem 1.2.2, there are unimodularintegral matrices U and Q, of orders n and m, respectively, such that

U−1AQ = D = diag{s1, . . . , sd, 0, . . . , 0} (si �= 0; si|si+1 ∀ i).


Let u1, . . . , un be the columns of U . We claim that the leftmost d columnsof U can serve as γ1, . . . , γd. We regard some vectors as column vectors.

Take α ∈ RV ∩ Zn, thus α = Aβ for some β ∈ Rm. Set β′ = Q−1β anddenote the ith entry of β′ by β′

i. Notice the equalities

α = U(Dβ′) and U−1α = Dβ′ = (s1β′1, . . . , sdβ

′d, 0, . . . , 0)

t.

Hence siβ′i ∈ Z for all i, and consequently α is in Zu1 + · · ·+ Zud. For the

reverse inclusion it suffices to prove that ui belongs to RV for i = 1, . . . , d.Using the equality (AQ)ei = (UD)ei (ei=ith unit vector) we derive

Aqi = U(siei) = si(Uei) = siui (1 ≤ i ≤ d),

where qi = Qei is the ith column of Q. Hence ui ∈ RV for i = 1, . . . , d. �

Lemma 1.2.4 There is an affine isomorphism φ : aff(A) → Rd such thatφ(aff(A) ∩ Zn) = Zd, where d = dim(P).

Proof. By Lemma 1.2.3 one has RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd for someγ1, . . . , γd in Zn. Therefore there is a linear map

ψ : RV −→ Rd, γiψ�−→ ei.

Hence φ = ψf : aff(A) −→ Rd satisfies the required condition. �

Observe that φ(P) is an integral polytope of dimension d with a positiveLebesgue measure denoted by m(φ(P)).

Definition 1.2.5 The relative volume of the integral polytope P is:

vol(P) := m(φ(P)).

If P has dimension n, then the relative volume of P agrees with its usualvolume. To see that the relative volume is independent of φ we need thefollowing fact about the standard volume.

Theorem 1.2.6 [427, Theorem 6.2.14] Let T : Rn → Rn be an affine mapgiven by T (x) = Ax+ x0, where A is a real matrix of order n and x0 ∈ Rn.If P is a bounded convex set in Rn, then T (P) has volume |det(A)|vol(P).

Lemma 1.2.7 Let T : Zn → Zn be the affine map given by T (x) = Ax+β,where A is an integral matrix of order n and β ∈ Zn. If T is bijective, thendet(A) = ±1.

Proof. Note that the matrix A determines a bijective linear map, thus Ahas an inverse with integral entries. Indeed, for each i there is a columnvector βi in Zn such that Aβi = ei. Therefore A (β1 · · ·βn) = I, andconsequently det(A) = ±1. �

22 Chapter 1

Proposition 1.2.8 The relative volume of P is independent of φ.

Proof. If φ1 is another affine isomorphism φ1 : aff(A) → Rd such thatφ1(aff(A) ∩ Zn) = Zd, then the affine map φ1φ

−1 : Zd −→ Zd is bijective.Using Theorem 1.2.6 and Lemma 1.2.7 one obtains

vol(φ(P)) = vol((φ1φ−1)(φ(P))) = vol(φ1(P)). �

In practice one can compute volumes of lattice polytopes using Normaliz[68] and Polyprob [104].

Example 1.2.9 We give an illustration in R3 of the procedure outlinedabove to compute relative volumes. We begin by setting

A = {v1, v2, v3, v4} = {(1, 3, 1), (1, 5, 3), (4, 9, 4), (2, 8, 5)}

and P = conv(A). Then V = {(0, 0, 0), (0, 2, 2), (3, 6, 3), (1, 5, 4)} and

f : R3 −→ R3, xf−→ x− α0,

where α0 = (1, 3, 1). Consider the matrix

A =

⎡⎣ 0 3 12 6 52 3 4

⎤⎦.Next using Maple [80] to compute the Smith normal form of A we obtaininvertible integral matrices U and Q such that UAQ = D, where

D =

⎡⎣ 1 0 00 1 00 0 0

⎤⎦; U =

⎡⎣ −2 1 01 0 01 −1 1

⎤⎦; U−1 =

⎡⎣ 0 1 01 2 01 1 1

⎤⎦.By the proof of Lemma 1.2.3, the first two columns of U−1 form a Z-basisfor RV∩Z3. Using the affine map ψ : RV → R3 induced by ψ(0, 1, 1) = (1, 0)and ψ(1, 2, 1) = (0, 1), and the map φ = ψf : aff(A)→ R, we get

φ(P) = conv((0, 0), (2, 0), (0, 3), (3, 1)) and vol(P) = m(φ(P)) = 11/2.

Proposition 1.2.10 ([176], [394, Proposition 4.6.30]) The relative volumeof P is given by:

vol(P) = limi→∞

|Zn ∩ iP|id

.

Definition 1.2.11 For an abelian group (M,+) its torsion subgroup T (M)is the set of all x in M such that px = 0 for some 0 �= p ∈ N. The group Mis torsion-free if T (M) = (0).


Lemma 1.2.12 If B = {β1, . . . , βs} ⊂ Zn, then

(a) (RB ∩ Zn)/ZB = T (Zn/ZB), and (∗)

(b) RB ∩ Zn = ZB if and only if Zn/ZB is torsion-free.

Proof. (a): If z ∈ T (Zn/ZB) ⊂ Zn/ZB, then kz ∈ ZB for some 0 �= k ∈ N,so z ∈ QB ⊂ RB. Since also z ∈ Zn, z is in the left-hand side of Eq. (∗).This shows the inclusion “⊃”. The other inclusion is left as an exercise.

(b): This follows from (a). �

Lemma 1.2.13 [397, pp. 32-33] If H ⊂ G are free abelian groups of thesame rank d with Z-bases δ1, . . . , δd and γ1, . . . , γd related by δi =

∑j gijγj,

where gij ∈ Z for all i, j, then |G/H | = | det(gij)|.

Theorem 1.2.14 [146] If A = {v1, . . . , vq} ⊂ Zn is a set of vectors lyingon an affine hyperplane not containing the origin and P = conv(A) hasdimension d, then

vol(P) = |T (Zn/(v2 − v1, . . . , vq − v1))| limi→∞

|ZA ∩ iP|id

.

Proof. As in the beginning of this section for convenience of notation we setm = q− 1 and α0 = v1, . . . , αm = vq. Let x0 ∈ Qn be such that 〈αi, x0〉 = 1for all i. As 〈αi − α0, x0〉 = 0, there is a decomposition ZA = Zα0 ⊕ ZVwith V = {α1 − α0, . . . , αm − α0}. Pick δ1, . . . , δd ∈ Zn such that

ZV = Zδ1 ⊕ · · · ⊕ Zδd. (1.5)

Therefore one can write

αi = α0 + fi1δ1 + · · ·+ fidδd (fij ∈ Z). (1.6)

Consider the lattice polytope

P1 = conv((1, 0, . . . , 0), (1, f11, . . . , f1d), . . . , (1, fm1, . . . , fmd)) ⊂ Rd+1.

There is a bijective map ZA∩iP ϕ−→ Zd+1∩iP1, namely ϕ is the restrictionof the linear map from ZA to Zd+1 that maps each vector into its coordinatevector with respect to the basis {α0, δ1, . . . , δd}. Therefore

limi→∞

|ZA ∩ iP|id

= limi→∞

|Zd+1 ∩ iP1|id

= vol(P1).

To estimate vol(P1) consider the lattice polytope

P2 = conv((0, . . . , 0), (f11, . . . , f1d), . . . , (fm1, . . . , fmd)) ⊂ Rd

24 Chapter 1

and note that applying the translation α �→ α− e1 to P1 gives:

vol(P1) = vol(conv((0, . . . , 0), (0, f11, . . . , f1d), . . . , (0, fm1, . . . , fmd))).

Thus vol(P1) = vol(P2). The next step is to relate vol(P2) with vol(P) byobtaining a second expression for vol(P2). According to Lemma 1.2.3, thereare γ1, . . . , γd ∈ Zn such that

RV ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd. (1.7)

Writing

δi = gi1γ1 + · · ·+ gidγd (gij ∈ Z; i, j = 1, . . . , d), (1.8)

from Eqs. (1.6) and (1.8) one derives the matrix equality

(fij)(gij) = (cij), (1.9)

where αi − α0 = ci1γ1 + · · ·+ cidγd, 1 ≤ i ≤ m. By definition

vol(P) = vol(conv((0, . . . , 0), (c11, . . . , c1d), . . . , (cm1, . . . , cmd))).

By Eq. (1.9), the linear transformation

σ : Zd −→ Zd, xσ�−→ x(gij),

satisfies σ(fi1, . . . , fid) = (ci1, . . . , cid). Hence by Theorem 1.2.6 we get

vol(P) = vol(σ(P2)) = | det(gij)|vol(P2).

To finish the proof it suffices to show that | det(gij)| is the order of thetorsion subgroup of Zn/ZV . Now we have

T (Zn/ZV) (a)= (RV ∩ Zn)/ZV (b)

= (Zγ1 ⊕ · · · ⊕ Zγd)/(Zδ1 ⊕ · · · ⊕ Zδd),

where in (a) we use Lemma 1.2.12, and in (b) the identities (1.5) and (1.7).Hence, from the identity (1.8) and Lemma 1.2.13, we get

|T (Zn/ZV)| = | det(gij)|. �

Proposition 1.2.15 Let α0, . . . , αn be a set of affinely independent pointsin Rn and Δ = conv(α0, . . . , αn). Then the volume of the simplex Δ is

vol(Δ) =

∣∣∣∣∣∣∣det⎛⎜⎝ α0 1

......

αn 1

⎞⎟⎠∣∣∣∣∣∣∣

n!=

∣∣∣∣∣∣∣det⎛⎜⎝ α1 − α0

...αn − α0

⎞⎟⎠∣∣∣∣∣∣∣

n!.


Proof. The result follows using linear algebra or applying the change ofvariables formula. See Theorem 1.2.6. �

Definition 1.2.16 A set Δ in Rn is called a lattice d-simplex if Δ is theconvex hull of a set of d+ 1 affinely independent points in Zn.

Let Δ be a lattice d-simplex in Rn. The normalized volume of Δ isdefined as d!vol(Δ). If d = n, then from Proposition 1.2.15 one has

vol(Δ) ≥ 1

n!.

Definition 1.2.17 A lattice d-simplex Δ in Rn is called unimodular ifd!vol(Δ) = 1.

Proposition 1.2.18 Let Δ = conv(α0, . . . , αd) be a lattice d-simplex. If

R(α1 − α0, . . . , αd − α0) ∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd

for some γ1, . . . , γd ∈ Zn and αi − α0 =∑

j cijγj, then

vol(Δ) =| det(cij)|

d!.

Proof. Note vol(Δ) = vol(conv(0, c1, . . . , cd)), where ci = (ci1, . . . , cin).Hence the formula follows from Proposition 1.2.15. �

Lemma 1.2.19 If β1, . . . , βm ∈ Zn, then

R(β1, . . . βm) ∩ Zn = Zβ1 + · · ·+ Zβm + Zδ1 + · · ·+ Zδs,

where δ1, . . . , δs generate the torsion subgroup of Zn/(β1, . . . , βm).

Proof. It is straightforward and will be left as an exercise. �

Lemma 1.2.20 If α0, . . . , αm ∈ Zn, then there is an isomorphism of groups

ϕ : T (Zn/(α1 − α0, . . . , αm − α0)) −→ T(Zn+1/((α0, 1), . . . , (αm, 1))

),

given by ϕ(α) = (α, 0). Here T (M) denotes the torsion subgroup of M .

Proof. The map ϕ is clearly well-defined and injective. To prove that ϕis onto consider an equation s(α, b) = λ0(α0, 1) + · · · + λm(αm, 1), whereλi ∈ Z, 0 �= s ∈ N, α ∈ Zn and b ∈ Z. Then

sα = λ0α0 + · · ·+ λmαm,

sb = λ0 + · · ·+ λm.

Hence s(α − bα0) = λ1(α1 − α0) + · · · + λm(αm − α0). It follows readilythat ϕ(α− bα0) = (α, b), as required. �

26 Chapter 1

Proposition 1.2.21 Let α = α0, . . . , αd be a set of affinely independentvectors in Zn defining a simplex Δ. Then the following are equivalent:

(a) d!vol(Δ) = 1.

(b) R(α1 − α0, . . . , αd − α0) ∩ Zn = Z(α1 − α0) + · · ·+ Z(αd − α0).

(c) Zn/Z{α1 − α0, . . . , αd − α0} is torsion-free.

(d) Zn+1/Z{(α0, 1), . . . , (αd, 1)} is torsion-free.

Proof. (a) ⇔ (b) follows from Proposition 1.2.18 and (b) ⇔ (c) ⇔ (d)follows from Lemmas 1.2.19 and 1.2.20. �

Corollary 1.2.22 If Δ = conv(α0, α1, . . . , αd) is a unimodular lattice d-simplex in Rn, then Δ ∩ Zn = {α0, α1, . . . , αd}.

Proof. Let α ∈ Δ ∩ Zn. Then α = λ0α0 + · · ·+ λdαd, where λi ∈ Q+ forall i and λ0 + · · ·+ λd = 1. Using the equality

α− α0 = λ1(α1 − α0) + · · ·+ λd(αd − α0)

and Proposition 1.2.21(c) we get

α− α0 = η1(α1 − α0) + · · ·+ ηd(αd − α0)

= λ1(α1 − α0) + · · ·+ λd(αd − α0),

where ηi ∈ Z for all i. Since α1 − α0, . . . , αd − α0 are linearly independentwe obtain λi = ηi for all i. Hence exactly one of the λi’s is equal to 1 andthe others are equal to zero, that is, α ∈ {α0, α1, . . . , αd}. �

Definition 1.2.23 A lattice polytope P = conv(A) of dimension d is saidto have a unimodular covering with support in A if there are simplicesΔ1, . . . ,Δm of dimension d such that the following two conditions hold

(i) P = ∪mi=1Δi and the vertices of Δi are contained in A for all i.

(ii) d!vol(Δi) = 1.

If condition (ii) is replaced by

(ii)’ [ZA : ZAi] = 1 for all i,

where Ai denotes the vertex set of Δi, then we say that Δ1, . . . ,Δm is aweakly unimodular covering of P with support in A.

For a discussion on the existence of unimodular covers of rational conessee [60] and the references therein.


Proposition 1.2.24 If A = {v1, . . . , vq} ⊂ Zn and the vectors in A lie ina hyperplane of Rn not containing the origin, then any unimodular coveringof P = conv(A) with support in A is weakly unimodular.

Proof. Let Δi = conv(Ai) be any unimodular simplex of dimension d withvertex set Ai ⊂ A and d = dim(P). For convenience of notation assumethat Ai = {v1, . . . , vd+1}. To show the equality ZA = ZAi it suffices toprove ZA ⊂ ZAi. Take an arbitrary vector α in ZA and write:

α = η1v1 + · · ·+ ηqvq (ηi ∈ Z),

(α, η) = η1(v1, 1) + · · ·+ ηq(vq, 1) (η = η1 + · · ·+ ηq).

Since A lies in an affine hyperplane not containing the origin and Ai isaffinely independent, the setAi is seen to be linearly independent. Therefore

d+ 1 = dim(R{(v1, 1), . . . , (vd+1, 1)}) = dim(R{(v1, 1), . . . , (vq, 1)}).

Thus one can write (α, η) = λ1(v1, 1) + · · · + λd+1(vd+1, 1), λi ∈ Q. ByProposition 1.2.21 the group Zn+1/Z{(v1, 1), . . . , (vd+1, 1)} is torsion-free.Hence it follows from Lemma 1.2.12(b) that λi ∈ Z for all i, and conse-quently α is in ZAi. �

Example 1.2.25 Let A = {v1, . . . , v5} be the set of vectors in Z6 given by

v1 = (1, 0, 1, 0, 0, 1), v2 = (1, 0, 0, 1, 1, 0), v3 = (0, 1, 1, 0, 1, 0),v4 = (0, 1, 0, 1, 0, 1), v5 = (1, 0, 0, 1, 0, 1).

Consider A1 = {v2, v3, v4, v5}, A2 = {v1, v3, v4, v5}, A3 = {v1, v2, v3, v5},Δ1 = conv(A1), Δ2 = conv(A2), Δ3 = conv(A3),

and P = conv(A). The groups Z6/ZA and Z6/ZAi are torsion-free forall i. This readily implies that ZA = ZAi, i = 1, 2, 3. Using the relationv1 + v2 + v4 = v3 + 2v5 it is not hard to see that P = Δ1 ∪Δ2 ∪Δ3. ThusΔ1,Δ2,Δ3 is a weakly unimodular covering. The relative volume of P canbe computed using the procedure described before. It is left to the readerto verify that dim(P) = 3 and vol(P) = 1/2.

Exercises

1.2.26 Let Δ be an n-dimensional simplex in Rn with vertices α0, . . . , αnin Zn. Prove that Δ is unimodular if and only if any of the following twoequivalent conditions hold

(a) Zn+1 = Z(α0, 1) + · · ·+ Z(αn, 1).

(b) Zn = Z(α1 − α0) + · · ·+ Z(αn − α0).

28 Chapter 1

1.2.27 Construct a lattice simplex of normalized volume greater than 1.Then prove that the converse of Proposition 1.2.24 fails.

1.2.28 If P is the integral polytope with vertices (0, 0) and (1, 3):

�

�

��

1

2

3

y

0 1 2 x

�

�

Prove that the relative volume of P is equal to 1.

1.2.29 If P = conv((3, 1), (1, 3)), then vol(P) = 2.

1.2.30 Consider the set A = {e1 + e2, e2 + e3, e3 + e4, e1 + e4} and thepolytope P = conv(A). Prove that dim(P) = 2 and vol(P) = 1.

1.2.31 Let P be the tetrahedron in R3 whose vertices are v1 = (1, 1, 0),v2 = (2, 1,−1), v3 = (2,−5, 0), v4 = (7, 1,−8). Prove that vol(P) = 2.Then verify this using Maple [80].

1.2.32 If A = {v1, . . . , vq} ⊂ Zn, prove that the rank of RA∩Zn, as a freeabelian group, is equal to the dimension of RA as a real vector space.

1.3 Hilbert bases and TDI systems

The set of integral points of a rational polyhedral cone form a semigroup thatarises in many branches of mathematics, such as combinatorial commutativealgebra [65], toric varieties [176], and integer programming [372].

A finite set H ⊂ Rn is called a Hilbert basis if

Zn ∩ R+H = NH,

where NH is the semigroup spanned by H consisting of all linear combina-tions of H with coefficients in N. Notice that all vectors in a Hilbert basisare integral. A nice introduction to Hilbert bases can be found in [215].

Let C ⊂ Rn be a rational polyhedral cone. A finite set H is called aHilbert basis of C if C = R+H and H is a Hilbert basis. A Hilbert basis ofC is minimal if it does not strictly contain any other Hilbert basis of C.


Hilbert bases of rational polyhedral cones always exist. For a pointedcone there is only one minimal Hilbert basis. To prove these two facts, weneed Gordan’s lemma. Below we give two versions of this lemma.

Definition 1.3.1 Given a = (ai) ∈ Rn, we define the value of a as:

|a| = a1 + · · ·+ an.

Lemma 1.3.2 (Gordan, version 1) If A = {v1, . . . , vq} ⊂ Zn and ZA isthe subgroup generated by A, then there exist γ1, . . . , γm in Zn such that

ZA ∩ R+A = Nγ1 + · · ·+ Nγm

and γi ∈ [N,M ]n for all i, where N = −q max1≤i≤q

|v−i | and M = q max1≤i≤q

|v+i |.

Proof. Recall that C = ZA∩R+A = ZA∩Q+A; see Corollary 1.1.27. Letβ ∈ C. Then one can write

β =

q∑i=1

(xiyi

)vi,

where xi ∈ N and 0 �= yi ∈ N. By the division algorithm there are ri, ni inN such that xi = niyi + ri and 0 ≤ ri < yi. Therefore one can write

β =

q∑i=1

nivi +

q∑i=1

aivi,

with ai ∈ [0, 1]∩Q. As∑q

i=1 aivi ∈ C ∩ [N,M ]n, the set A∪ (C ∩ [N,M ]n)is a generating set for C with the required property. Our argument wasbased on the proof of Gordan’s lemma given in [65]. �

Definition 1.3.3 A semigroup (S,+, 0) of Zn is said to be finitely generatedif there exists a finite set Γ = {γ1, . . . , γr} ⊂ S such that:

S = NΓ := Nγ1 + · · ·+ Nγr.

A set of generators Γ of S is called minimal if γi �= 0 ∀i and none of itselements is a linear combination with coefficients in N of the others.

Remark 1.3.4 There are examples of subsemigroups of Nn, with n ≥ 2,which are not finitely generated; see Exercise 1.3.34.

Lemma 1.3.5 (Gordan, version 2) If R+A is a cone in Rn generated bya finite set A ⊂ Zn, then the semigroup Zn ∩ R+A is finitely generated.

30 Chapter 1

Proof. If A = {v1, . . . , vq}, consider the finite set of integral points:

{a1v1 + · · ·+ aqvq| 0 ≤ ai ≤ 1} ∩ Zn = {γ1, . . . , γr}.

It is left to the reader to prove that γ1, . . . , γr is the required set of generatorsfor Zn ∩ R+A. See [372, p. 233]. �

Let A be a finite set in Zn and let G = Zn or G = ZA. Then, byGordan’s lemma (versions 1 and 2) there exists γ1, . . . , γr ∈ Zn such that:

G ∩R+A = Nγ1 + · · ·+ Nγr.

Computing the γi’s is in general difficult [69]. Fortunately, the γi’s can becomputed using Normaliz [68].

Therefore Hilbert bases of rational polyhedral cones always exist:

Proposition 1.3.6 Let A be a finite set in Zn. Then there exist γ1, . . . , γrsuch that

R+A ∩ Zn = Nγ1 + · · ·+ Nγr,

and H = {γ1, . . . , γr} is a Hilbert basis of R+A.

Proof. The existence follows from Lemma 1.3.5. That H is a Hilbert basisof R+A follows from the equality R+A = R+γ1 + · · ·+ R+γr. �

In Definition 1.1.52 we considered pointed polyhedra. For cones we havethe following equivalent definition (see Exercise 1.3.36).

Definition 1.3.7 A polyhedral cone C = {x|Ax ≤ 0} is called pointed ifthe lineality space {x|Ax = 0} is equal to {0}.

Lemma 1.3.8 Let C be a rational polyhedral cone. If C is pointed, thenthere exists an integral vector b such that 〈b, x〉 > 0 for all 0 �= x ∈ C.

Proof. There is a rational matrix A such that C = {x|Ax ≤ 0}; seeTheorem 1.1.29. We may assume that A is integral. If u1, . . . , ut are therows of A, then b = −u1 − · · · − ut satisfies the required condition. �

Theorem 1.3.9 [371] Let A be a finite set in Zn and let C = R+A. If Cis pointed, then there exists a unique minimal Hilbert basis of C given by

H = {x ∈ C ∩ Zn| 0 �= x /∈ Ny1 + Ny2; ∀y1, y2 ∈ (C \ {0}) ∩ Zn }.

Proof. Let Γ = {γ1, . . . , γr} be an arbitrary Hilbert basis of C. We claimthat H ⊂ Γ. Let x ∈ H. Since Zn ∩ C = NΓ we can write x =

∑ri=1 aiγi,

ai ∈ N. Thus, by construction of H, we get x = γi for some i, which provesthe claim. To finish the proof it suffices to prove NH = NΓ, because this


equality implies R+H = R+Γ and consequently Zn ∩ R+H = NH. Weproceed by contradiction by assuming that the set:

V = {x|x ∈ NΓ = C ∩ Zn; x /∈ NH}

is not empty. By Lemma 1.3.8 there exists b ∈ Zn such that 〈b, x〉 > 0 forall 0 �= x ∈ C. Let x0 ∈ V such that

〈x0, b〉 = min{〈x, b〉|x ∈ V}.

Since x0 /∈ H we can write x0 = x1+x2 with x1, x2 in (C \{0})∩Zn. Thus,since 〈xi, b〉 < 〈x0, b〉 for i = 1, 2, we have x1, x2 ∈ NH and x1 + x2 ∈ NH,a contradiction. �

Definition 1.3.10 (Edmonds and Giles [125]) A rational system xA ≤ bof linear inequalities is called totally dual integral, abbreviated TDI, if theminimum in the LP-duality equation

max{〈x, c〉|xA ≤ b} = min{〈y, b〉| y ≥ 0;Ay = c} (1.10)

has an integral optimum solution y for each integral vector c for which theminimum is finite.

Note that there are rational systems of linear inequalities which definethe same polyhedron and such that one system is totally dual integral whilethe other is not (see Exercise 1.3.39).

TDI systems occur in the theory of Grobner bases of toric ideals [252],in the theory of perfect graphs [296, 86], and in combinatorial commutativealgebra [147, 185, 188]. See Theorems 9.6.21, 13.6.8, and 14.3.6.

Proposition 1.3.11 [372, Corollary 22.1c] If xA ≤ b is a TDI-system andb is integral, then the polyhedron {x |xA ≤ b} is integral.

Proof. By hypothesis min{〈y, b〉| y ≥ 0;Ay = c} has an integral optimumsolution y for each integral vector c for which the minimum is finite. Thenby the LP duality theorem (see Theorem 1.1.57), we get that

max{〈x, c〉|xA ≤ b}

is an integer for each integral vector c for which the maximum is finite.Thus the polyhedron {x |xA ≤ b} is integral by Theorem 1.1.63. �

Definition 1.3.12 A set P in Rn is called a parallelotope if it is the imageof [0, 1]n under a non-singular linear transformation, i.e., P has the form

P = {λ1v1 + · · ·+ λnvn| 0 ≤ λi ≤ 1}

for some linearly independent vectors v1, . . . , vn in Rn.

32 Chapter 1

Lemma 1.3.13 Let v1, . . . , vn be a basis of Rn and let

P = {λ1v1 + · · ·+ λnvn| 0 ≤ λi ≤ 1}

be a parallelotope. Then vol(P) = | det[v1, . . . , vn]|.

Proof. Since P is the image of [0, 1]n under the linear map T induced byei �→ vi, the formula follows from Theorem 1.2.6. �

Proposition 1.3.14 Let A be an integral matrix of order n × n and letA = {v1, . . . , vn} be the set of columns of A. If det(A) �= 0, then A is aHilbert basis if and only if | det(A)| = 1.

Proof. ⇒) Let Q = [0, 1]n be the unit cube and let P be the parallelotope

P = {λ1v1 + · · ·+ λnvn| 0 ≤ λi ≤ 1}.

By Lemma 1.3.13 one has vol(P) = | det(A)|. As A is a Hilbert basis andA is linearly independent, we have

(k + 1)n = |kQ ∩ Zn| = |kP ∩ Zn|, ∀ k ∈ N.

Therefore by Proposition 1.2.10 and the fact that [0, 1]n has volume 1, weconclude that 1 = vol(P) = | det(A)|, as required.⇐) Let α be a vector in R+A ∩ Zn. As | det(A)| = 1 and since A is

linearly independent, by Cramer’s rule, it follows that α ∈ NA. Thus A isa Hilbert basis. �

Notation Let A �= (0) be an integral matrix. The greatest common divisorof all the non-zero r × r sub determinants of A will be denoted by Δr(A).

Theorem 1.3.15 [261, Theorem 3.9] Let A be an integral matrix of rank rand let d1, . . . , dr be the invariant factors of A. Then

d1 = Δ1(A), d2 = Δ2(A)Δ1(A)−1, . . . , dr = Δr(A)Δr−1(A)

−1.

The next result is called the fundamental structure theorem for finitelygenerated abelian groups .

Theorem 1.3.16 [261, pp. 187-188] LetM be a finitely generated Z-modulewith a presentation M � Zn/(a1, . . . , aq). If A is the matrix of rank r withcolumns a1, . . . , aq and d1, . . . , dr are the invariant factors of A, then

M � Z/(d1)⊕ Z/(d2)⊕ · · · ⊕ Z/(dr)⊕ Zn−r.

Lemma 1.3.17 If A is an integral matrix of size n× q and rank r, then

Δr(A) = |T (Zn/(a1, . . . , aq))|,

where ai is the ith column of A. In particular Δr(A) = 1 if and only if thequotient group Zn/(a1, . . . , aq) is torsion-free.


Proof. Let d1, . . . , dr be the invariant factors of the matrix A. On onehand using Theorem 1.3.15, we get the equality d1d2 · · · dr = Δr(A). Onthe other hand, by Theorem 1.3.16, we obtain an isomorphism

T (Zn/(a1, . . . , aq)) � Z/(d1)⊕ · · · ⊕ Z/(dr).

Therefore the order of the torsion subgroup is Δr(A), as required. �

Lemma 1.3.18 Let A = {v1, . . . , vd} ⊂ Zn be a set of linearly independentvectors and let A be the matrix with column vectors v1, . . . , vd. Then

d!vol(conv(0, v1, . . . , vd)) = Δd(A).

Proof. Let Γ = {γ1, . . . , γd} be a set of vectors such that

RA∩ Zn = Zγ1 ⊕ · · · ⊕ Zγd, (1.11)

see Lemma 1.2.3. Then we can write

vi = ci1γ1 + · · ·+ cidγd (i = 1, . . . , d) (1.12)

where C = (cij) is an integral matrix. By Proposition 1.2.18, we have

d!vol(conv(0, v1, . . . , vd)) = | det(C)|.

Using Eqs. (1.11) and (1.12), we get that the map

T (Zn/ZA) −→ T (Zd/Z{c1, . . . , cd}), α �→ [α]Γ,

is an isomorphism of groups, where ci is the ith row of C and [α]Γ is thecoordinate vector of α in the basis Γ. Hence, by Lemma 1.3.17, we get

Δd(A) = |T (Zn/ZA)| = |T (Zd/Z{c1, . . . , cd})| = | det(C)|. �

Proposition 1.3.19 Let A = {v1, . . . , vd} ⊂ Zn be a linearly independentset and let A be the matrix with column vectors v1, . . . , vd. Then A is aHilbert basis if and only if Δd(A) = 1.

Proof. ⇒) Let C = (cij) be as in the proof of Lemma 1.3.18. As A isa Hilbert basis, it is seen that the rows of C form a Hilbert basis. Then| det(C)| = 1 by Proposition 1.3.14. Therefore Δd(A) = 1 by Lemma 1.3.18.⇐) If Δd(A) = 1, then Zn/ZA is torsion-free. Since the vi’s are linearly

independent, it follows readily that A is a Hilbert basis. �

Theorem 1.3.20 [179] Let A = {v1, . . . , vq} ⊂ Zn be a Hilbert basis andlet r = rank (ZA). If R+A is pointed, then there exist H ⊂ A such that His linearly independent, H is a Hilbert basis and |H| = r.

34 Chapter 1

Corollary 1.3.21 If A = {v1, . . . , vq} ⊂ Zn and R+A is a pointed cone,then A is a Hilbert basis if and only if R+A ∩ ZA = NA and Zn/ZA is atorsion-free group.

Proof. Assume that A is a Hilbert basis. Let A be the matrix with columnvectors v1, . . . , vq and let r = rank (A). Then clearly R+A∩ZA = NA. ByTheorem 1.3.20 and Proposition 1.3.19, we get Δr(A) = 1. Thus Zn/ZAis a torsion-free group. The converse follows readily. Notice that for theconverse the hypothesis that the cone R+A is pointed is not needed. �

Let P = {x |xA ≤ b} be a rational polyhedron and let F be a face of P .A column of A is active in F if the corresponding inequality in xA ≤ b issatisfied with equality for all vectors x in F . A minimal face of P is a facenot containing any other face. A face F of P is minimal if and only if F isan affine subspace [372].

Theorem 1.3.22 [372, Theorem 22.5] A rational system xA ≤ b is TDI ifand only if for each minimal face F of the polyhedron P = {x |xA ≤ b}, thecolumns of A which are active in F form a Hilbert basis.

Proof. ⇒) Let v1, . . . , vq be the column vectors of A and let b = (bi).Assume that v1, . . . , vr are the columns of A which are active in F . Take anintegral vector c in R+{v1, . . . , vr}. Then the maximum in the LP-dualityequation

max{〈x, c〉|xA ≤ b} = min{〈y, b〉| y ≥ 0;Ay = c} (1.13)

is attained by each vector of F . Indeed if x0 ∈ F and c = λ1v1 + · · ·+λrvr,with λi ≥ 0 for all i, then 〈x0, c〉 = λ1b1 + · · ·+ λrbr and

min{〈y, b〉| y ≥ 0;Ay = c} ≤ λ1b1 + · · ·+ λrbr ≤ max{〈x, c〉|xA ≤ b}.

Thus we have equality everywhere and the maximum is attained at anyvector of F . The minimum has an integral optimum solution y. Then bycomplementary slackness (see Corollary 1.1.58), we get (b − xA)y = 0 forany x ∈ F . Therefore (bi − 〈x, vi〉)yi = 0 for i = 1, . . . , q. If vi is inactive inF , then 〈x, vi〉 < bi for some x ∈ F . Then yi = 0. This proves that yi = 0for i > r, i.e., c ∈ N{v1, . . . , vr}. Altogether v1, . . . , vr is a Hilbert basis.⇐) Let c be an integral vector for which the optimum values of Eq. (1.13)

are finite. Let F be a face of P such that each vector in F attains themaximum in Eq. (1.13). We may assume that F is a minimal face of P .By the minimality of F if a vector x0 in F satisfies 〈x0, vi〉 = bi for somei, then any other vector x in F satisfies 〈x0, vi〉 = bi, i.e., vi is active in F .Thus we may assume that there is 1 ≤ r ≤ q such that 〈x, vi〉 = bi for i ≤ rand x ∈ F , and 〈x, vi〉 < bi for i > r and x ∈ F . Thus by hypothesis theset A1 = {v1, . . . , vr} is a Hilbert basis. Let x and y = (yi) be optimumsolutions of Eq. (1.13) with x ∈ F . Then by complementary slackness (see


Corollary 1.1.58), we get (b − xA)y = 0, Ay = c and y ≥ 0. If yi > 0, then〈x, vi〉 = bi, i.e., i ≤ r. Therefore c is in R+A1. Hence c is in NA1 becauseA1 is a Hilbert basis. Then we can write c =

∑ri=1 ηivi, ηi ∈ N. Consider

the vector y0 = (η1, . . . , ηr, 0 . . . , 0) in Nq whose last q − r entries are equalto zero. Thus c = Ay0. Since bi−〈x, vi〉 = 0 for i ≤ r, we get (b−xA)y0 = 0.Thus the minimum in Eq. (1.13) is attained in y0, as required. �

Notice that we have in fact shown that Theorem 1.3.22 is valid if wereplace “each minimal face F” by “each face F”.

Definition 1.3.23 Let A be an integral matrix and let w be an integralvector. The system xA ≤ w is said to have the integer rounding property if

�min{〈y, w〉| y ≥ 0; Ay = a}� = min{〈y, w〉|Ay = a; y ≥ 0 ; y integral}

for each integral vector a for which min{〈y, w〉| y ≥ 0; Ay = a} is finite.

Theorem 1.3.24 [372, Theorem 22.18] Let A be an integral matrix of sizen × q with column vectors v1, . . . , vq and w = (wi) ∈ Zq. Then the systemxA ≤ w has the integer rounding property if and only if the set

H′ = {(v1, w1), . . . , (vq , wq), en+1} ⊂ Zn+1

is a Hilbert basis.

Corollary 1.3.25 Let A be an integral matrix and let w = (wi) be anintegral vector. The system xA ≤ w is TDI if and only if {x|xA ≤ w} isan integral polyhedron and the set H′ ⊂ Zn+1 is a Hilbert basis.

Proof. It follows from Proposition 1.3.11, Theorem 1.3.24, and using thedefinition of a TDI system. �

Theorem 1.3.26 Let A be an integer matrix with column vectors v1, . . . , vqand let w = (wi) be an integral vector. If the polyhedron P = {x|xA ≤ w} isintegral and H = {(v1, w1), . . . , (vq, wq)} is a Hilbert basis, then the systemxA ≤ w is TDI.

Proof. Let F be a minimal face of P . Recall that a column of A is activein F if the corresponding inequality in xA ≤ w is satisfied with equality forall vectors in F . We may assume that v1, . . . , vr are the columns of A whichare active in F . Then 〈x, vi〉 = wi for x ∈ F and 1 ≤ i ≤ r.

If 〈y, vi〉 < wi for some y ∈ F , then 〈x, vi〉 < wi for any other x ∈ F .Indeed if 〈x, vi〉 = wi for some x ∈ F , consider the supporting hyperplaneof P given by H = {x|〈x, vi〉 = wi}, then x ∈ F ∩ H � F because y ∈ Fand y /∈ F ∩H , a contradiction to the minimality of the face F . Thus wemay also assume that 〈x, vi〉 < wi for x ∈ F and i > r.

36 Chapter 1

Since P is integral, by Theorem 1.1.63, each face of P contains integralvectors. Pick an integral vector x0 ∈ F . By Theorem 1.3.22, it suffices toprove that B = {v1, . . . , vr} is a Hilbert basis. Take a ∈ R+B ∩ Zn. Thena = λ1v1 + · · ·+ λrvr with λi ≥ 0 for all i. Thus we have

b = 〈a, x0〉 = λ1〈v1, x0〉+ · · ·+ λr〈vr , x0〉= λ1w1 + · · ·+ λrwr.

Hence b is an integer and (a, b) =∑ri=1 λi(vi, wi). As H is a Hilbert basis,

we can write (a, b) =∑q

i=1 ηi(vi, wi), ηi ∈ N for all i. Therefore on onehand 0 = 〈(a, b), (x0,−1)〉. On the other hand 〈(a, b), (x0,−1)〉 is equal to

r∑i=1

ηi 〈(vi, wi), (x0,−1)〉︸︷︷︸=0

+

q∑i=r+1

ηi 〈(vi, wi), (x0,−1)〉︸︷︷︸<0

.

Hence ηi = 0 for i > r and a = η1v1 + · · ·+ ηrvr. Thus a ∈ NB. �

The converse is not true in general. However there are some interestinglinear systems where the converse holds.

Definition 1.3.27 Let A be an integral matrix and w an integral vector.The system x ≥ 0;xA ≤ w is TDI if the minimum in the LP-duality equation

max{〈a, x〉|x ≥ 0;xA ≤ w} = min{〈y, w〉| y ≥ 0;Ay ≥ a} (1.14)

has an integral optimum solution y for each integral vector a with finiteminimum.

Proposition 1.3.28 Let A be a nonnegative integral matrix of size n × qwith column vectors v1, . . . , vq and let w = (wi) ∈ Nq. Then, the systemx ≥ 0;xA ≤ w is TDI if and only if the polyhedron P = {x|x ≥ 0;xA ≤ w}is integral and H = {(v1, w1), . . . , (vq, wq),−e1, . . . ,−en} is a Hilbert basis.

Proof. Assume the system x ≥ 0;xA ≤ w is TDI. By Proposition 1.3.11,we get that P is integral. Next we prove that H is a Hilbert basis. Take(a, b) ∈ R+H∩Zn+1, where a ∈ Zn and b ∈ Z. By hypothesis, the minimumin Eq. (1.14) has an integral optimum solution y = (yi) such that 〈y, w〉 ≤ b.Since y ≥ 0 and a ≤ Ay, we can write

a = y1v1 + · · ·+ yqvq − δ1e1 − · · · − δnen (δi ∈ N) =⇒(a, b) = y1(v1, w1) + · · ·+ yq−1(vq−1, wq−1)

+(yq + b− 〈y, w〉)(vq , wq)− (b − 〈y, w〉)vq − δ,

where δ = (δi). As the entries of A are in N, the vector −vq can be writtenas a nonnegative integer combination of −e1, . . . ,−en. Thus (a, b) ∈ NH.This proves that H is a Hilbert basis. The converse follows readily fromTheorem 1.3.26. �


Lemma 1.3.29 Let A be an integral matrix of size n×q with column vectorsv1, . . . , vq and w = (wi) ∈ Zq. If P = {x|xA ≤ w} is pointed, then α is avertex of P if and only if H−

(α,−1) is a closed halfspace defining a facet of

the cone C = R+{(v1, w1), . . . , (vq, wq), en+1}.

Proof. Since P is pointed, the rank of A is equal to n and C is a coneof full dimension n + 1. Using that α is a vertex of P if and only if α isa basic feasible solution of the system xA ≤ w (see Corollary 1.1.49), theproof follows from Theorem 1.1.44. �

Procedure 1.3.30 If P = {x|xA ≤ w} is pointed, using Corollary 1.3.25and Lemma 1.3.29 we can check whether a system xA ≤ w is TDI. UsingNormaliz [68] this can be achieved using a single input file of the form:

q+1

n+1

v1,w1

...

vq,wq

0 1

0

The first and third block of the output file contain the Hilbert basis andthe support hyperplanes. To verify if P is integral, by Lemma 1.3.29, itsuffices to verify that all rows of the support hyperplanes having its lastentry positive are integral and its last entry is equal to 1.

Example 1.3.31 To illustrate Procedure 1.3.30 consider the integer vectorw = (3, 0, 0, 1, 0) and the matrix

A =

⎛⎝ 1 1 1 1 10 1 0 1 20 0 1 1 2

⎞⎠We now verify that the system xA ≤ w is not TDI. Using the followinginput file for Normaliz

6

4

1 0 0 3

1 1 0 0

1 0 1 0

1 1 1 1

1 2 2 0

0 0 0 1

0

38 Chapter 1

We obtain the following output file (we only show the first and third block):

6 Hilbert basis elements: 6 support hyperplanes:

1 0 0 3 -3 3 3 1

1 1 0 0 0 0 0 1

1 0 1 0 0 0 1 0

0 0 0 1 0 1 0 0

1 2 2 0 2 -2 1 0

1 1 1 0 2 1 -2 0

Thus the polyhedron {x|xA ≤ w} is integral and the rows in the input filedo not form a Hilbert basis. For instance the vector (1, 1, 1, 0) is in the conegenerated by the rows of the input file but it is not an N-linear combinationof the rows.

Corollary 1.3.32 Let xA ≤ w be a system satisfying the integer roundingproperty and let A = {v1, . . . , vq} be the set of column vectors of A. Then

(a) A is a Hilbert basis.

(b) Zn/ZA is a torsion-free group provided that R+A is a pointed cone.

Proof. (a): Take a ∈ R+A ∩ Zn, then we can write a =∑qi=1 λivi, for

some λ1, . . . , λq in R+. Hence

(a, �∑

i λiwi�) = λ1(v1, w1) + · · ·+ λq(vq, wq) + δ(0, 1),

where δ ≥ 0. Therefore, by Theorem 1.3.24, there are λ′1, . . . λ′q ∈ N and

δ′ ∈ N such that

(a, �∑i wiλi�) = λ′1(v1, w1) + · · ·+ λ′q(vq, wq) + δ′(0, 1),

Thus a ∈ NA. (b): It follows at once from (a) and Corollary 1.3.21. �

Exercises

1.3.33 Prove that any subsemigroup of N is finitely generated.

1.3.34 Prove that the subsemigroup of N2 given by S = (N+×N)∪{(0, 0)}is not finitely generated, where N+ = {1, 2, . . .}.

1.3.35 (Gordan’s lemma) Let L be a lattice in Rn, i.e., L is an additivesubgroup of Zn, and let C be a rational polyhedral cone in Rn. Prove thatL ∩C is a finitely generated semigroup.

Hint Adapt the proof of Lemma 1.3.2.

1.3.36 Let C = {x|Ax ≤ 0} be a rational polyhedral cone. Prove that Ccontains no lines if and only if {x|Ax = 0} = (0).


1.3.37 Let A be an integral matrix. If C = {x|x ≥ 0;Ax = 0}, prove thatC is a rational polyhedral cone.

1.3.38 If w = (3, 0, 0,−1, 0) and A is the matrix

A =

⎛⎝ 1 1 1 1 10 1 0 1 10 0 1 1 2

⎞⎠,use Normaliz to verify that the system xA ≤ w is TDI.

1.3.39 Consider the two systems

(x1, x2, x3)

⎛⎝ 1 0 1 1

1 1 0 10 1 1 1

⎞⎠ ≤

⎛⎜⎜⎝

2223

⎞⎟⎟⎠ ; (x1, x2, x3)

⎛⎝ 1 0 1

1 1 00 1 1

⎞⎠ ≤

⎛⎝ 2

22

⎞⎠

Prove that they define the same polyhedron. Then prove that the firstsystem is TDI but the second is not. Use Normaliz to verify these assertions.

1.3.40 Let A be an n × d integral matrix of rank d whose set of columnvectorsA = {v1, . . . , vd} form a Hilbert basis. Prove that the system xA ≤ bis TDI for each integral vector b.

1.4 Rees cones and clutters

In this section we characterize the max-flow min-cut property of clutters interms of Hilbert bases of Rees cones and the integrality of polyhedra.

Let A be an n× q matrix with entries in N, let A = {v1, . . . , vq} be theset of columns of A, and let

A′ := {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Rn+1,

where ei is the ith unit vector. The cone R+A′, generated by A′, is calledthe Rees cone of A or the Rees cone of A.

The term Rees cone was coined in [147] because this cone encodes someinformation about the Rees algebra of the monomial ideal associated to A;see Chapters 12–14. The first aim of this section is to study the irreduciblerepresentation, as an intersection of closed halfspaces, of a Rees cone.

We shall always assume that all rows and columns ofA are non-zero. Theset covering polyhedron (in Proposition 13.1.2 we clarify this terminology)is by definition the rational polyhedron:

Q(A) := {x ∈ Rn | x ≥ 0, xA ≥ 1},

40 Chapter 1

where 0 and 1 are vectors whose entries are equal to 0 and 1, respectively.Often we denote the vectors 0, 1 simply by 0, 1. As is seen below, one canexpress the irreducible representation of R+A′ in terms of Q(A).

Some of the facets of a Rees cone are easy to identify. Consider theindex set

J = {i | 1 ≤ i ≤ n and 〈ei, vj〉 = 0 for some j} ∪ {n+ 1}.

Notice that R+A′ has dimension n+ 1. It is not hard to see that the set

F = R+A′ ∩Hei (1 ≤ i ≤ n+ 1)

defines a facet of R+A′ if and only i ∈ J . Therefore, by Theorem 1.1.44and Proposition 1.1.51, the Rees cone has the following unique irreduciblerepresentation

R+A′ =

(⋂i∈J

H+ei

)⋂(r⋂i=1

H+αi

)(1.15)

such that 0 �= αi ∈ Qn+1 and 〈αi, en+1〉 = −1 for all i. In many interestingcases, from the viewpoint of commutative algebra, one has the equalityJ = {1, . . . , n}; see Section 14.2 and Exercise 14.2.34.

Lemma 1.4.1 Let A be a matrix with entries in N and let a = (ai1, . . . , aiq)be its ith row. Set k = min{aij | 1 ≤ j ≤ q}. If aij > 0 for all j, then ei/kis a vertex of Q(A).

Proof. Set x0 = ei/k. Clearly x0 ∈ Q(A), 〈x0, vj〉 = 1 for some j, and〈x0, e�〉 = 0 for � �= i. It is seen that x0 is a vertex of Q(A). �

The irreducible representation of a Rees cone can be expressed in termsof the vertices of the set covering polyhedron.

Theorem 1.4.2 Let V be the vertex set of Q(A). Then the irreduciblerepresentation of the Rees cone is given by

R+A′ =

(⋂i∈J

H+ei

)⋂(⋂u∈V

H+(u,−1)

).

Proof. We set B = {ei| i ∈ J } ∪ {(u,−1)|u ∈ V } and V = {u1, . . . , up}.First we dualize Eq. (1.15) and use the duality theorem for cones to obtain

(R+A′)∗ = {y ∈ Rn+1| 〈y, x〉 ≥ 0, ∀x ∈ R+A′}= H+

e1 ∩ · · · ∩H+en ∩H

+(v1,1)

∩ · · · ∩H+(vq,1)

=∑i∈J

R+ei + R+α1 + · · ·+ R+αr. (1.16)


Next we show the equality (R+A′)∗ = R+B. The right-hand side is clearlycontained in the left-hand side because a vector α belongs to Q(A) if andonly if (α,−1) is in (R+A′)∗. To prove the reverse inclusion observe that byEq. (1.16) it suffices to show that αk ∈ R+B for all k. Writing αk = (ck,−1)and using αk ∈ (R+A′)∗ gives ck ∈ Q(A). The set covering polyhedron canbe written as

Q(A) = R+e1 + · · ·+ R+en + conv(V ),

where conv(V ) denotes the convex hull of V, this follows from the structureof polyhedra (see Theorem 1.1.33) by noticing that the characteristic coneof Q(A) is precisely Rn+. Thus we can write

ck = λ1e1 + · · ·+ λnen + μ1u1 + · · ·+ μpup,

where λi ≥ 0, μj ≥ 0 for all i, j and μ1+· · ·+μp = 1. If 1 ≤ i ≤ n and i /∈ J ,then the ith row of A has all its entries positive. Thus by Lemma 1.4.1 weget that ei/ki is a vertex of Q(A) for some ki > 0. To avoid cumbersomenotation we denote ei and (ei, 0) simply by ei, from the context the meaningof ei should be clear. Therefore from the equalities

∑i/∈J

λiei =∑i/∈J

λiki

(eiki

)=∑i/∈J

λiki

(eiki,−1

)+

(∑i/∈J

λiki

)en+1

we conclude that∑

i/∈J λiei is in R+B. From the identities

αk = (ck,−1) = λ1e1 + · · ·+ λnen + μ1(u1,−1) + · · ·+ μp(up,−1)

=∑i/∈J

λiei +∑

i∈J\{n+1}λiei +

p∑i=1

μi(ui,−1)

we obtain αk ∈ R+B, as required. Taking duals in (R+A′)∗ = R+B yields

R+A′ =⋂a∈B

H+a . (1.17)

Thus, by Remark 1.1.31, the proof reduces to showing that β /∈ R+(B\{β})for all β ∈ B. To show this we will assume that β ∈ R+(B \ {β}) for someβ ∈ B and derive a contradiction.

Case (I): β = (uj,−1). For simplicity assume β = (up,−1). Then

(up,−1) =∑i∈J

λiei +

p−1∑j=1

μj(uj ,−1), (λi ≥ 0;μj ≥ 0) ⇒

up =∑

i∈J\{n+1}λiei +

p−1∑j=1

μjuj (1.18)

−1 = λn+1 − (μ1 + · · ·+ μp−1). (1.19)

42 Chapter 1

To derive a contradiction we claim that Q(A) = Rn+ + conv(u1, . . . , up−1),which is impossible because by Proposition 1.1.39 the vertices ofQ(A) wouldbe contained in {u1, . . . , up−1}. To prove the claim note that the right-handside is clearly contained in the left-hand side. For the other inclusion takeγ ∈ Q(A) and write

γ =

n∑i=1

biei +

p∑i=1

ciui (bi, ci ≥ 0;

p∑i=1

ci = 1)

(1.18)= δ +

p−1∑i=1

(ci + cpμi)ui (δ ∈ Rn+).

Therefore using the inequality

p−1∑i=1

(ci + cpμi) =

p−1∑i=1

ci + cp

(p−1∑i=1

μi

)(1.19)= (1− cp) + cp(1 + λn+1) ≥ 1

we get γ ∈ Rn+ + conv(u1, . . . , up−1). This proves the claim.Case (II): β = ek for some k ∈ J . We only consider the subcase k ≤ n.

The subcase k = n+ 1 can be treated similarly. We can write

ek =∑

i∈J\{k}λiei +

p∑i=1

μi(ui,−1), (λi ≥ 0;μi ≥ 0).

From this equality we get ek =∑pi=1 μiui. Hence ekA ≥ (

∑pi=1 μi)1 > 0, a

contradiction because k ∈ J and 〈ek, vj〉 = 0 for some j. �

As a consequence we obtain:

Theorem 1.4.3 [188, Theorem 3.2] The mapping ϕ : Qn → Qn+1 given byϕ(α) = (α,−1) induces a bijective mapping

ϕ : V −→ {α1, . . . , αr}

between the set V of vertices of Q(A) and the set {α1, . . . , αr} of vectorsthat occur in the irreducible representation of R+A′ given in Eq. (1.15).

Definition 1.4.4 Let A be a matrix with entries in {0, 1}. The systemx ≥ 0; xA ≥ 1 is called totally dual integral (TDI) if the maximum in theLP-duality equation

min{〈α, x〉|x ≥ 0;xA ≥ 1} = max{〈y, 1〉| y ≥ 0;Ay ≤ α}

has an integral optimum solution y for each integral vector α with finitemaximum.


Proposition 1.4.5 If the system x ≥ 0; xA ≥ 1 is TDI, then Q(A) hasonly integral vertices.

Proof. It follows from Theorem 1.1.63. �

Definition 1.4.6 A clutter C with vertex set X = {x1, . . . , xn} is a familyof subsets of X , called edges, none of which is included in another. The setof vertices and edges of C are denoted by V (C) and E(C), respectively.

Let C be a clutter with vertex set X = {x1, . . . , xn} and let f1, . . . , fqbe the edges of C. The incidence matrix of C is the n× q matrix A = (aij)given by aij = 1 if xi ∈ fj and aij = 0 otherwise.

Definition 1.4.7 A clutter C, with incidence matrix A, has the max-flowmin-cut (MFMC) property if both sides of the LP-duality equation

min{〈α, x〉|x ≥ 0;xA ≥ 1} = max{〈y, 1〉| y ≥ 0;Ay ≤ α} (1.20)

have integral optimum solutions x, y for each nonnegative integral vector α.

It turns out that the max-flow min-cut property is equivalent to requirethat a certain blowup ring has no nilpotent elements; see Theorem 14.3.6.

Proposition 1.4.8 Let C be a clutter and let A be its incidence matrix.The system x ≥ 0; xA ≥ 1 is TDI if and only if C has the max-flow min-cutproperty.

Proof. If follows from Propositions 1.4.5 and 1.1.41. �

Let C be a clutter with n vertices and let A = {v1, . . . , vq} be the set ofcolumns of its incidence matrix. For use below we denote by A′ the Reesconfiguration of C:

A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Nn+1,

where ei is the ith unit vector of Rn

Theorem 1.4.9 [188, Theorem 3.4] If C is a clutter and A is its incidencematrix, then C has the max-flow min-cut property if and only if Q(A) is anintegral polyhedron and A′ is a Hilbert basis of R+A′.

Proof. ⇒) By Proposition 1.4.5 the polyhedron Q(A) is integral. Next weshow that A′ is an integral Hilbert basis. Take (α, αn+1) ∈ Zn+1 ∩ R+A′.Then Ay ≤ α and 〈y,1〉 = αn+1 for some vector y ≥ 0. Therefore oneconcludes that the optimal value of the linear program

max{〈y,1〉| y ≥ 0; Ay ≤ α}

44 Chapter 1

is greater than or equal to αn+1. Since C has the MFMC property, thislinear program has an optimal integral solution y0. By Exercise 1.4.12,there exists an integral vector y′0 such that

0 ≤ y′0 ≤ y0 and |y′0| = αn+1.

Therefore(α

αn+1

)=

(A1

)y′0 +

(A0

)(y0 − y′0) +

(α0

)−(A0

)y0

and (α, αn+1) ∈ NA′, as required.⇐) Assume that C does not satisfy the MFMC property. Then there is

an α0 ∈ Nn such that if y0 is an optimal solution of the linear program:

max{〈y,1〉| y ≥ 0; Ay ≤ α0}, (∗)

then y0 is not integral. We claim that also the optimal value |y0| = 〈y0,1〉of this linear program is not integral. If |y0| is integral, then (α0, |y0|) isin Zn+1 ∩ R+A′. As A′ is a Hilbert basis we get that (α0, |y0|) is in NA′,but this readily yields that the linear program (∗) has an integral optimalsolution, a contradiction. This completes the proof of the claim.

Now, consider the dual linear program:

min{〈x, α0〉| x ≥ 0, xA ≥ 1}.

By Proposition 1.1.41 the optimal value of this linear program is attainedat a vertex x0 of Q(A). Then by the duality theorem (Theorem 1.1.56) weget 〈x0, α0〉 = |y0| /∈ Z. Hence x0 is not integral, a contradiction to theintegrality of the set covering polyhedron Q(A). �

Remark 1.4.10 Normaliz [68] computes the irreducible representation andthe minimal integral Hilbert basis of a Rees cone. Thus we can effectivelyuse Theorems 1.4.2 and 1.4.9 to determine whether a given clutter has themax-flow min-cut property; see example below.

Example 1.4.11 Let A be the transpose of the matrix:⎡⎢⎢⎣1 0 0 0 10 1 0 1 00 0 1 1 11 1 1 0 0

⎤⎥⎥⎦and let C be the clutter with incidence matrix A. Consider the followinginput file for Normaliz that we call “example.in”:


4

5

1 0 0 0 1

0 1 0 1 0

0 0 1 1 1

1 1 1 0 0

3

Applying Normaliz , i.e., typing “normaliz example” in the directory whereone has the file normaliz.exe we get the output file “example.out”:

9 generators of integral closure of Rees algebra:

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

1 0 0 0 1 1

0 1 0 1 0 1

0 0 1 1 1 1

1 1 1 0 0 1

10 support hyperplanes:

0 0 1 1 1 -1

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0 0 1

0 0 1 0 0 0

1 0 0 1 0 -1

0 0 0 1 0 0

0 0 0 0 1 0

0 1 0 0 1 -1

1 1 1 0 0 -1

semigroup is not homogeneous

The first block shows thatA′ is a Hilbert basis for the Rees cone. The secondblock shows the irreducible representation of the Rees cone of A, thus usingTheorem 1.4.2 we obtain that Q(A) is integral. Altogether Theorem 1.4.9proves that the clutter C has the max-flow min-cut property.

Exercises

1.4.12 Let b, η1, . . . , ηq be a sequence in N. If η1 + · · ·+ ηq ≥ b, then thereare ε1, . . . , εq ∈ N such that 0 ≤ εi ≤ ηi for all i and ε1 + · · ·+ εq = b.

1.4.13 Let A be the incidence matrix of a clutter C and let v1, . . . , vq be itscolumn vectors. The system x ≥ 0; xA ≥ 1 is TDI if and only if the set

Au = {vi| 〈vi, u〉 = 1} ∪ {ei| 〈ei, u〉 = 0}is a Hilbert basis for any vertex u of Q(A).

46 Chapter 1

1.5 The integral closure of a semigroup

Let A = {v1, . . . , vq} be a set of vectors in Nn \ {0}. The integral closure ornormalization of the affine semigroup

NA := Nv1 + · · ·+ Nvq ⊂ Nn,

is defined as NA := ZA∩R+A, where ZA is the subgroup of Zn generatedby A. The semigroup NA is called normal or integrally closed if NA = NA.The Rees semigroup of A is by definition NA′ ⊂ Nn+1, where

A′ = {(v1, 1), . . . , (vq, 1), e1, . . . , en}.

Notice that en+1 = (v1, 1) − 〈(v1, 1), e1〉e1 − · · · − 〈(v1, 1), en〉en. HenceZA′ = Zn+1. As a consequence we get

NA′ = R+A′ ∩ Zn+1.

From this equality we obtain a characterization of the normality of a Reessemigroup in terms of Hilbert bases:

Proposition 1.5.1 NA′ is normal if and only if A′ is a Hilbert basis.

The algebraic invariants of semigroup rings of semigroups generated bya Hilbert basis are easier to understand; see [405, Section 6]. This family ofsemigroups and their cones are studied throughout this book.

Proposition 1.5.2 If A is a Hilbert basis, then NA is normal.

Proof. Since NA ⊂ R+A∩ ZA ⊂ R+A ∩ Zn = NA we obtain the equalityNA = R+A ∩ ZA. �

Example 1.5.3 If A = {(1, 1, 0), (0, 1, 1), (1, 0, 1)}, the affine semigroupNA is normal because A is linearly independent, but A is not a Hilbertbasis because NA � Z3 ∩ R+A (see Exercise 1.5.8).

Notation Given β = (βi) ∈ Rn+1, we define degn+1(β) = βn+1.

Proposition 1.5.4 [147] Let R+A′ be the Rees cone of A and let B be itsminimal integral Hilbert basis. If β ∈ B, then degn+1(β) < n.

Proof. We set β = (α, b), with α ∈ Nn and b = degn+1(β) ∈ N. Notethat R+A′ ∩ Zn+1 = Q+A′ ∩ Zn+1 and dim(R+A′) = n + 1. Thus, since(α, b) ∈ R+A′, by Caratheodory’s theorem (Theorem 1.1.18) we can write

(α, b) = λ1(vi1 , 1)+ · · ·+λr(vir , 1)+μ1ej1 + · · ·+μsejs (λi, μk ∈ Q+), (∗)


where {(vi1 , 1), . . . , (vir , 1), ej1 , . . . , ejs} is a linearly independent subset ofA′. By the minimality of B it is seen that 0 ≤ λi < 1 and 0 ≤ μk < 1 for alli, k. From Eq. (∗) we get b = λ1 + · · ·+ λr < r. Thus b ≤ r − 1. If r ≤ n,then b ≤ n− 1. Thus from now on we may assume r = n + 1 and Eq. (∗)takes the simpler form

(α, b) = λ1(vi1 , 1) + · · ·+ λn+1(vin+1 , 1).

Consider the cone C generated by A′′ = {(vi1 , 1), . . . , (vin+1 , 1)}. Since −e1is not in C, using Exercise 1.5.9, we obtain a point

x0 = (1− λ0)(α, b) + λ0(−e1) (0 ≤ λ0 < 1)

in the relative boundary of C. By Theorem 1.1.44 the relative boundaryof C is the union of its facets. Hence using that any facet of C is ann-dimensional cone generated by a subset of A′′ (see Proposition 1.1.23)together with the minimality of B it follows that we can write (α, b) as:

(α, b) = ρ0e1 + ρ1(vj1 , 1) + · · ·+ ρn(vjn , 1) (0 ≤ ρi < 1 ∀i),

and consequently b ≤ n− 1, as required. �

Definition 1.5.5 A matrix A is called totally unimodular if each i × isubdeterminant of A is 0 or ±1 for all i ≥ 1.

Examples of totally unimodular matrices include incidence matrices ofclutters without odd cycles [27, Chapter 5], incidence matrices of bipartitegraphs and digraphs [372, p. 274], and network matrices [372, Chapter 19].

Theorem 1.5.6 (Hoffman, Kruskal) If A is a totally unimodular matrix,then Q = {x |x ≥ 0; Ax ≤ b} is integral for each integral vector b.

Proof. Assume that A has size n×q. Notice that Q contains no lines, thusit suffices to show that Q has only integral vertices. Let x0 be a vertex ofQ. By Corollary 1.1.47, A′x0 = b′ for some subsystem A′x ≤ b′ of(

A−Iq

)≤(b0

)such that A′ is a square matrix of order q with linearly independent rows.By hypothesis det(A′) = ±1, hence x0 is integral because the inverse of A′

is an integral matrix by Cramer’s rule. �

Corollary 1.5.7 [49] If A = (aij) is a totally unimodular {0, 1}-matrix ofsize n× q with column vectors v1, . . . , vq, then R+A′ is normal.

Proof. By the Hoffman–Kruskal theorem, the system x ≥ 0;xA ≥ 1 isTDI. Thus, by Theorem 1.4.9, the set A′ is a Hilbert basis, i.e., the Reessemigroup R+A′ is normal. �

48 Chapter 1

Exercises

1.5.8 If A = {(1, 1, 0), (0, 1, 1), (1, 0, 1)}, use Normaliz , with option 0, toverify that the minimal integral Hilbert basis of R+A is A ∪ {(1, 1, 1)}.

1.5.9 Let a be a point of a set V in Rn and let x be a point in aff(V) notin V . If λ0 = sup{λ ∈ [0, 1]| (1− λ)a+ λx ∈ V}, then x0 = (1− λ0)a+ λ0xis a relative boundary point of V lying between a and x.

1.5.10 If A is a totally unimodular matrix and A′ is obtained from A byadding columns of unit vectors, then the matrix A′ is totally unimodular.

1.6 Unimodularity of matrices and normality

In this section, we characterize when a vector belongs to a subgroup of Zn

and prove Heger’s theorem for the existence of solutions of an integer linearsystem. Then we show the normality of semigroups arising from unimodularmatrices and unimodular coverings.

A minor of order r (r-minor for short) of a matrix A is defined as thedeterminant of a square submatrix of A of order r.

Definition 1.6.1 An integral matrix A �= (0) is t-unimodular if all the non-zero r-minors of A have absolute value equal to t, where r is the rank of A.If A is 1-unimodular, we say that A is a unimodular matrix .

Theorem 1.6.2 [372, Theorem 19.2] Let A be an integral matrix of full rowrank. Then the polyhedron {x|x ≥ 0;Ax = b} is integral for each integralvector b if and only if A is unimodular.

Theorem 1.6.3 Let A be an n × q integral matrix whose set of columnvectors is A = {v1, . . . , vq} and let b ∈ Zn be a column vector such thatrank(A) = rank([Ab]). The following conditions are equivalent:

(a) b ∈ ZA.(b) Zn/ZA and Zn/Z(A ∪ {b}) have the same invariant factors.

(c) The matrices [A 0] and [Ab] have the same Smith normal form.

(d) Δr(A) = Δr([Ab]), where r = rank(A) and Δr(A) is the gcd of allthe non-zero r-minors of A.

Proof. There is k ∈ N \ {0} such that kb =∑qi=1 λivi, λi ∈ Z. Therefore,

there is a canonical epimorphism of finite groups

ϕ : T (Zn/ZA) −→ T (Zn/ZB) (α+ ZA ϕ�−→ α+ ZB),

where B = A∪ {b}. Notice that ϕ is injective if and only if (a) holds.


The implication (a)⇒ (b) is straightforward. There are invertible inte-gral matrices Pi and Qi such that

D1 = Q1[A 0]P1 = diag(d1, . . . , dr, 0, . . . , 0),

D2 = Q2[A b]P2 = diag(e1, . . . , er, 0, . . . , 0),

are the Smith normal forms of [A 0] and [Ab], respectively; that is, di, ei arepositive integers satisfying that di divides di+1 and ei divides ei+1 for all i.By the fundamental structure theorem for finitely generated abelian groups(see Theorem 1.3.16) there are isomorphisms:

T (Zn/ZA) � (Z/d1Z)× · · · × (Z/drZ),

T (Zn/ZB) � (Z/e1Z)× · · · × (Z/erZ).

Thus (b) ⇔ (c). Note Δi(A) = d1 · · · di and Δi([Ab]) = e1 · · · ei for all i.Hence (c) ⇒ (d). To prove (d) ⇒ (i) observe |T (Zn/ZA)| = |T (Zn/ZB)|and consequently ϕ must be injective. �

As an immediate consequence of Theorem 1.6.3 we get:

Theorem 1.6.4 (I. Heger, [372, p. 51]) Let A be an n× q integral matrixand let b ∈ Zn be a column vector. If rank(A) = rank([Ab]), then the systemAx = b has an integral solution if and only if Δr(A) = Δr([Ab]).

Remark 1.6.5 Let A ⊂ Zn be a finite set. Note that the finite basistheorem together with Theorem 1.6.3 yield a membership test to decidewhen a given vector b in Zn belongs to NA = ZA ∩ R+A.

Proposition 1.6.6 If A is a t-unimodular matrix with columns v1, . . . , vqand vi1 , . . . , vir is a Q-basis for the column space of A, then

ZA = Zvi1 ⊕ · · · ⊕ Zvir

Proof. For each j one has Δr(vi1 · · · vir ) = Δr(vi1 · · · vir vj) = t. Then byTheorem 1.6.3, we get vj ∈ Zvi1 ⊕ · · · ⊕ Zvir , as required. �

Theorem 1.6.7 If A is a t-unimodular matrix whose set of columns is A,then the affine semigroup NA is normal.

Proof. If b ∈ ZA ∩ R+A, by Theorem 1.1.18 and Proposition 1.6.6 thereare vi1 , . . . , vir linearly independent vectors in A, r = rank(A), such that

b ∈ R+vi1 + · · ·+ R+vir and b ∈ Zvi1 + · · ·+ Zvir . (1.21)

By comparing the coefficients of b with respect to the two representationsobtained from Eq. (1.21), one derives b ∈ NA. �

50 Chapter 1

Corollary 1.6.8 If A is a totally unimodular matrix whose set of columnsis A, then NA = Zn ∩ R+A; that is, A is a Hilbert basis.

Proof. Let A = {v1, . . . , vq} be the set of column vectors of A. Takeb ∈ Zn ∩ R+A, then by Caratheodory’s theorem (see Theorem 1.1.18) andafter permutation of the vi’s we can write b =

∑ri=1 ηivi with ηi ≥ 0 for all

i, where r is the rank of A and v1, . . . , vr are linearly independent.The submatrix A′ = (v1 · · · vr) is totally unimodular. Therefore, by

Theorem 1.6.3, the system of equations A′x = b has an integral solution.Thus b is a linear combination of v1, . . . , vr with coefficients in Z. It followsthat ηi ∈ N for all i, that is, b ∈ NA. The other inclusion is clear. �

Proposition 1.6.9 If A = {v1, . . . , vq} ⊂ Zn lie in a hyperplane of Rn notcontaining the origin and P = conv(A) has a weakly unimodular coveringwith support in A, then NA is a normal semigroup.

Proof. Take 0 �= β ∈ R+A ∩ ZA. There exists λ1 . . . , λq ∈ Q+ such thatβ = λ1v1+· · ·+λqvq. We set |λ| = λ1+· · ·+λq. Since β/|λ| is in P and usingthat P has a weakly unimodular covering, there is an affinely independentset A1 ⊂ A defining a lattice simplex Δ = conv(A1) such that β/|λ| ∈ Δand ZA1 = ZA. Altogether β belongs to ZA1 ∩ R+A1. Note that A1 is alinearly independent set because A1 lie in a hyperplane not containing theorigin. Therefore it follows rapidly that β ∈ NA1, as required. �

Exercises

1.6.10 If A = {v1, . . . , vq} ⊂ Zn and P = conv(A) has a unimodularcovering with support in A, then N(v1, 1) + · · ·+ N(vq, 1) is normal.

1.7 Normaliz, a computer program

Throughout this book we will frequently use Normaliz [68], a program thatprovides an invaluable effective tool to study monomial subrings and theiralgebraic invariants. This program computes the following:

• normalization (or integral closure) of an affine semigroup;

• Hilbert basis of a pointed rational cone;

• lattice points of an integral polytope;

• support hyperplanes of a Rees cone;

• generators of the integral closure of the Rees algebra of a monomialideal I = (xv1 , . . . , xvq ) ⊂ R = K[x1, . . . , xn];

• generators of the integral closure of I;


• Hilbert series and Hilbert polynomial of a homogeneous affine semi-group;

• Ehrhart ring (Theorem 9.3.6) and relative volume (Proposition 1.2.10)of a lattice polytope.

It can be used to check the following integer programming properties:

• integer rounding property of any of the linear systems

x ≥ 0;xA ≥ 1, x ≥ 0;xA ≤ 1, xA ≤ 1,

where A is the matrix with column vectors v1, . . . , vq (Corollary 14.6.9,Theorem 14.6.16, and Proposition 14.6.30);

• integer rounding property and total dual integrality (TDI) of a linearsystem xA ≤ w (Theorem 1.3.24, Procedure 1.3.30);

• perfection of a graph (Remark 13.6.4).

If I is square-free, it can also be used to determine the following:

• minimal primes of I (Remark 13.2.23);

• reducedness of the associated graded ring grI(R) (Example 14.2.20);

• the max-flow min-cut property of a clutter (Example 1.4.11);

• integrality of the set covering polyhedron

Q(A) = {x|xA ≥ 1; x ≥ 0},

where A = (v1, . . . , vq);

• total dual integrality of the system x ≥ 0; xA ≥ 1;

• the generators of the symbolic Rees algebra Rs(I) (Example 13.2.22).

1.8 Cut-incidence matrices and integrality

In this section we introduce incidence matrices of digraphs, and present ageneralization of a theorem of Lucchesi and Younger [298] which is veryuseful to detect TDI systems arising from incidence matrices of digraphs.This theorem will be used in Theorem 11.3.2 to express the a-invariant of theedge subring of a bipartite graph in terms of directed cuts. This illustratesthat some deep results in combinatorial optimization can be used to studyalgebraic invariants of rings.

Next we introduce some more TDI systems which are suited for ourpurposes (see [281, Chapter 5] and [372, p. 311]).

52 Chapter 1

Definition 1.8.1 Let B be an integral matrix. A system Bx ≥ b, x ≥ 0 istotally dual integral (TDI) if the maximum in

min{〈c, x〉|x ≥ 0;Bx ≥ b} = max{〈y, b〉| y ≥ 0; yB ≤ c}

has an integral optimum solution y for each integral vector c with finitemaximum.

Proposition 1.8.2 If the matrix B is totally unimodular, then the systemBx ≥ b; x ≥ 0 is TDI.


Proposition 1.8.3 Let B be an integral matrix and let b be an integralvector. If the system Bx ≥ b; x ≥ 0 is TDI, then Q = {x|Bx ≥ b;x ≥ 0}has only integral vertices.


Definition 1.8.4 A digraph G consists of a finite vertex set V (G) and afamily E(G) of ordered pairs of elements of V (G). The pairs (v1, v2) ∈ E(G)are called directed edges or arrows .

Definition 1.8.5 Let G be a digraph with vertex set V (G) and edge setE(G). Given a family F of subsets of V (G), the one-way cut-incidencematrix of F is the matrix B = (bX,e)X∈F ,e∈E(G), where

bX,e =

{1 if e ∈ δ+(X)0 otherwise.

The two-way cut-incidence matrix is B = (bX,e)X∈F ,e∈E(G), where

bX,e =

⎧⎨⎩1 if e ∈ δ+(X)−1 if e ∈ δ−(X)0 otherwise.

Here δ+(X) = {e = (z, w) ∈ E(G)| z ∈ X, w /∈ X} is the set of edges leavingthe vertex set X and δ−(X) is the set of edges entering the vertex set X .If F = {{v} | v ∈ V (G)}, the matrix B is called the incidence matrix of G.

Remark 1.8.6 An interesting case occurs when the one-way and the two-way cut-incidence matrix of the family F coincide. In this case the rowsof the matrix B correspond to directed cuts only. Recall that δ+(X) is adirected cut of a digraph G if ∅ �= X ⊂ V (G) and δ−(X) = ∅.

It was pointed out to us by Jens Vygen that the next result is a slightgeneralization of the Lucchesi, Younger theorem [298]. It follows using thetechnique of proof of [281, Theorem 19.10].


Theorem 1.8.7 [405] Let G be a digraph and F a family of subsets of V (G)such that the one-way cut incidence matrix B of F is equal to the two-waycut incidence matrix of F . If F satisfies the following three conditions:

(a) the rows of B are non-zero,

(b) if X,Y ∈ F and X ∪ Y �= V (G), then X ∪ Y ∈ F , and

(c) if X,Y ∈ F and X ∩ Y �= ∅, then X ∩ Y ∈ F ,

then the system Bx ≥ 1, x ≥ 0 is totally dual integral and the rationalpolyhedron {x|Bx ≥ 1;x ≥ 0} is integral.

Proof. Let c be an integral vector such that the maximum in

min{〈c, x〉|x ≥ 0;Bx ≥ 1} = max{〈y,1〉| y ≥ 0; yB ≤ c} (1.22)

is attained at y0 = (yF1 , . . . , yFr ), where F = {F1, . . . , Fr}. Note c ≥ 0.Since the set of vectors y for which the maximum in Eq. (1.22) is attainedis a face of the polytope P = {y ∈ Rr| y ≥ 0; yB ≤ c} and since any face ofP is a compact set, one may assume

r∑i=1

yFi |Fi|2 = max

{r∑i=1

yi|Fi|2∣∣∣∣∣ 〈y,1〉 = 〈y0,1〉; y ≥ 0; yB ≤ c

}. (1.23)

We claim that the family G = {Fi| yFi > 0} is cross-free, that is, forany Fi, Fj ∈ G, at least one of the four sets Fi \ Fj , Fj \ Fi, Fi ∩ Fj ,V (G)\Fi∪Fj is empty. To prove the claim suppose F1\F2 �= ∅, F2\F1 �= ∅,F1 ∩ F2 �= ∅, F1 ∪ F2 �= V (G) for some F1, F2 ∈ G (for simplicity we areassuming i = 1, j = 2). Set ε = min{yF1 , yF2} and consider the vector y′0 inRr whose entries are given by

y′F1:= yF1 − ε, y′F2

:= yF2 − ε, y′F1∩F2:= yF1∩F2 + ε, y′F1∪F2

:= yF1∪F2 + ε,

and y′S := yS for all other S ∈ F . Let B = (bFi,αj ) be the one-way cut-incidence matrix of F , where α1, . . . , αq are the edges of G. Consider anyedge αj = (z, w). For each 1 ≤ j ≤ q one has the following inequality

bF1,j(yF1 − ε) + bF2,j(yF2 − ε) + bF1∩F2,j(yF1∩F2 + ε) + bF1∪F2,j(yF1∪F2 + ε)

≤ bF1,jyF1 + bF2,jyF2 + bF1∩F2,jyF1∩F2 + bF1∪F2,jyF1∪F2 .

To show the inequality consider two cases. Case (I): bF1,j = bF2,j = 0. Ifi = 1 or i = 2 and z ∈ Fi, then w ∈ Fi and bF1∪F2,j = 0. If i = 1 or i = 2and z /∈ Fi, then w /∈ Fi and bF1∩F2,j = 0. Hence in this case yF1∩F2,j = 0and yF1∪F2,j = 0. Case (II): bF1,j = 0 and bF2,j = 1. By the previousconsiderations one has that either yF1∩F2,j = 0 or yF1∪F2,j = 0.

54 Chapter 1

From the inequality above we obtain y′0B ≤ y0B. Since 〈y′0,1〉 = 〈y0,1〉,we obtain a contradiction with the choice of y0 because

r∑i=1

yFi |Fi|2 <r∑i=1

y′Fi|Fi|2.

Note that for any numbers a > b ≥ c > d > 0 with a + d = b + c one hasthe inequality a2 + d2 > b2 + c2. In our case to show the strict inequalitywe take a = |F1 ∪ F2|, b = |F1| ≥ c = |F2| and d = |F1 ∩ F2|. Thus we haveshown that the family G is cross-free.

Let B′ be the submatrix of B whose rows correspond to the elements ofthe family G. Then

max{〈y′,1〉| y′ ≥ 0; y′B′ ≤ c} = max{〈y,1〉| y ≥ 0; yB ≤ c}. (1.24)

The inequality ≤ is clear because B′ is a submatrix of B. As the maximumin the right-hand side is attained at y0, where y0 has zero entries in positionscorresponding to rows outside B′, the reverse inequality follows. Now thematrix B′, being the two-way cut-incidence matrix of a cross-free family G,is a network matrix, hence it is totally unimodular; see [281, Theorem 5.27].Hence by the Hoffman–Kruskal theorem the polytope

{y′| y′ ≥ 0; y′B′ ≤ c}

is integral, and so the maximum in the left-hand side of Eq. (1.24) is attainedat an integral vector y′. To complete the proof that the system Bx ≥ 1,x ≥ 0 is TDI observe that y′ can be extended (by adding zero entries)to an integral optimum solution y of the right-hand side of Eq. (1.24).Since the system Bx ≥ 1, x ≥ 0 is totally dual integral it follows fromProposition 1.8.3 that the polyhedron {x|Bx ≥ 1;x ≥ 0} is integral. �

Corollary 1.8.8 Let G be a connected digraph and let

F = {X | ∅ �= X � V (G); δ−(X) = ∅}.

If B is the one-way cut-incidence matrix of F , then {x|Bx ≥ 1;x ≥ 0} isa non-empty integral polyhedron.

Proof. It suffices to verify the hypothesis of Theorem 1.8.7. First we provethat δ+(X) �= ∅ for X ∈ F . Assume δ+(X) = ∅. Pick z ∈ V (G) \ Xand x ∈ X . If (z, w) ∈ E(G) or (w, z) ∈ E(G), then w /∈ X . There is anundirected path {z0 = z, z1, . . . , zr = x}, a contradiction since zi /∈ X forall i. Thus (a) is satisfied. Next take X,Y ∈ F . From the inequality

|δ−(X)|+ |δ−(Y )| ≥ |δ−(X ∩ Y )|+ |δ−(X ∪ Y )|,


see [281, Lemma 2.1(b)], we get δ−(X ∩Y ) = ∅ and δ−(X ∪Y ) = ∅. Thus ifX∩Y �= ∅ (resp. X∪Y �= V (G)), then X∩Y ∈ F (resp. X∪Y ∈ F). Thusconditions (b) and (c) are satisfied. On the other hand by construction ofF the matrix B is also the two-way cut-incidence matrix of F . �

Lemma 1.8.9 Let G be a connected bipartite graph with bipartition (V1, V2)and let F be the family

F = {A ∪ A′| ∅ �= A � V1;N(A) ⊂ A′ ⊂ V2} ∪ {A′| ∅ �= A′ ⊂ V2}.

If G is regarded as the digraph with all its arrows leaving the vertex set V2,then the following equality holds F = {X | ∅ �= X � V (G); δ−(X) = ∅}.

Proof. It follows readily from the definitions. �

Exercises

1.8.10 If A is the incidence matrix of a digraph G, prove that A is totallyunimodular.

1.8.11 Let T : Rq → Rn be a linear map and let Q′ ⊂ Rq be a rationalpolyhedron. If T (Zq) ⊂ Zn, then T (Q′) is a rational polyhedron.

1.9 Elementary vectors and matroids

In this section, we give formulae to compute the circuits of the kernel of anintegral matrix and show that the circuits generate the kernel. We introducethe notion of a matroid and explain the relation with vector matroids.

The notion of an elementary integral vector or circuit occurs in convexanalysis [357], in the theory of toric ideals of graphs [34, 400, 417], and inmatroid theory [338].

Given α = (α1, . . . , αq) ∈ Rq, its support is supp(α) = {i |αi �= 0}. Notethat α can be written uniquely as α = α+ − α−, where α+ and α− are twononnegative vectors with disjoint support which are called the positive partand the negative part of α, respectively. Sometimes we denote the positiveand negative part of α by α+ and α−, respectively.

Remark 1.9.1 If α is a linear combination of β1, . . . , βr ∈ Rq, then onehas the inclusion supp(α) ⊂ supp(β1) ∪ · · · ∪ supp(βr).

Definition 1.9.2 Let V be a linear subspace of Qq. An elementary vectorof V is a non-zero vector α in V whose support is minimal with respect toinclusion, i.e., supp(α) does not properly contain the support of any othernon-zero vector in V .

56 Chapter 1

The concept of an elementary vector arises in graph theory when V isthe kernel of the incidence matrix of a graph G [358, Section 22].

Lemma 1.9.3 If V is a linear subspace of Qq and α, β are two elementaryvectors of V with the same support, then α = λβ for some λ ∈ Q.

Proof. If i ∈ supp(α), then one can write αi = λβi for some scalar λ. Sincesupp(α− λβ) � supp(α), one concludes α− λβ = 0, as required. �

Definition 1.9.4 Two vectors α = (αi) and β = (βi) in Qq are in harmonyif αiβi ≥ 0 for every i.

Lemma 1.9.5 [357] Let V be a linear subspace of Qq. If 0 �= α ∈ V ,then there is an elementary vector γ ∈ V in harmony with α such thatsupp(γ) ⊂ supp(α).

Proof. Let β = (β1, . . . , βq) be an elementary vector of V whose supportis contained in supp(α). By replacing β by −β one may assume αiβi > 0for some i. Consider

λj =αjβj

= min1≤i≤q

{αiβi

∣∣∣∣ αiβi > 0

}.

Note that zi = (αi − λjβi)αi ≥ 0 for all i. Indeed if αiβi ≤ 0, then clearlyzi ≥ 0, and if αiβi > 0 by the minimality of λj one has zi ≥ 0. Thus thevector α−λjβ is in harmony with α and its support is strictly contained inthe support of α. If α− λjβ = 0 we are done, otherwise we apply the sameargument with α − λjβ playing the role of α. Since being in harmony isan equivalence relation, the result follows by a recursive application of thisprocedure. �

Theorem 1.9.6 [357] If V is a vector subspace of Qq and α ∈ V \{0}, thenα can be written as α =

∑ri=1 βi for some elementary vectors β1, . . . , βr of

V with r ≤ dimV such that

(i) β1, . . . , βr are in harmony with α,

(ii) supp(βi) ⊂ supp(α) for all i, and

(iii) supp(βi) is not contained in the union of the supports of β1, . . . , βi−1

for all i ≥ 2.

Proof. By induction on the number of elements in the support of α. If α isnot an elementary vector of V , then by Lemma 1.9.5 there is an elementaryvector α1 ∈ V in harmony with α such that supp(α1) ⊂ supp(α). Using theproof of Lemma 1.9.5 it follows that there is a positive scalar λ1 > 0 suchthat α− λ1α1 is in harmony with α and the support of α− λ1α1 is strictlycontained in supp(α). Therefore the result follows applying induction. Thisproof uses the original argument given in [357]. �


Definition 1.9.7 Let V be a linear subspace of Qq. An elementary integralvector or circuit of V is an elementary vector α of V such that α ∈ Zq andthe non-zero entries of α are relatively prime.

Corollary 1.9.8 If V is a linear subspace of Qq, then there are only a finitenumber of circuits of N and they generate V as a Q-vector space.

Proof. It follows from Lemma 1.9.3 and Theorem 1.9.6, respectively. �

Proposition 1.9.9 [134] Let A be an n× q integral matrix and let α be anon-zero vector in ker(A). If A has rank n, then α is an elementary vectorof ker(A) if and only if there is 0 �= λ ∈ Q such that

α = λ

n+1∑k=1

(−1)k det[vi1 , . . . , vik−1, vik+1

, . . . , vin+1 ]eik , (1.25)

for some column vectors vij of A. Here ek is the kth unit vector in Rq.

Proof. ⇐) Let A′ be the submatrix of A consisting of the column vectorsvi1 , . . . , vin+1 . Since A

′ has size n× (n+1) and rank n, ker(A′) is generatedby a non-zero vector. It follows readily that the vector α given by Eq. (1.25)is an elementary vector.⇒) Let supp(α) = {j1, . . . , jr} and let v1, . . . , vq be the column vectors

of A. Since α ∈ ker(A) one has

α = λj1ej1 + · · ·+ λjrejr and λj1vj1 + · · ·+ λjrvjr = 0, (1.26)

where 0 �= λjk ∈ Q and vjk are columns of A for all k. By the minimality ofsupp(α) we may assume that {vj1 , . . . , vjr−1} is linearly independent. Thus,using that rank(A) = n, there are column vectors vjr+1 , . . . , vjn+1 such thatthe set A′ = {vj1 , . . . , vjr−1 , vjr+1 , . . . , vjn+1} is linearly independent. Afterpermuting j1, . . . , jn+1, one can write

A′ ∪ {vjr} = {vi1 , . . . , vin+1}

with 1 ≤ i1 < · · · < in+1 ≤ q. Consider the vector

β =n+1∑k=1

(−1)k det[vi1 , . . . , vik−1, vik+1

, . . . , vin+1 ]eik . (1.27)

Note that in Eq. (1.26) and Eq. (1.27) the coefficient of ejr is non-zero,hence there are scalars 0 �= λ and cik in Q such that

α− λβ =∑ik �=jr

cikvik .

Since the support of α−λβ is strictly contained in the support of β and sinceβ is an elementary vector of ker(A), one obtains α−λβ = 0, as required. �

58 Chapter 1

Theorem 1.9.10 Let A �= (0) be an n× q integral matrix and let ψ be theZ-linear homomorphism

ψ : Zq −→ Zn

given by ψ(α) = A(α). Then ker(ψ) is generated as a Z-module by thecircuits of ker(ψ).

Proof. By Theorem 1.2.2, there are invertible integral matrices P and Qsuch that PAQ = D, where D is an r × q matrix with r = rank(A). SinceA and DQ−1 have the same kernel and DQ−1 has size r× q and rank r, wemay assume that n = rank(A).

We set N = ker(ψ). Let α1, . . . , αq be the column vectors of A and letei be the ith unit vector of Zq. By Proposition 1.9.9, it follows that the setof circuits of N is the set of vectors of the form

±1h

n+1∑j=1

(−1)j · det(αi1 , . . . , αij−1 , αij+1 , . . . , αin+1) · eij =±1h· v,

where 1 ≤ i1 < · · · < in+1 ≤ q, and h is the greatest common divisor of theentries of the vector v, if v is non-zero. For convenience we introduce thenotation v = v(i1, . . . , in+1) and

[i1, . . . , ij , . . . , in+1] = det(αi1 , . . . , αij−1 , αij+1 , . . . , αin+1).

Set d equal to the greatest common divisor of all the non-zero integershaving the special form [i1, . . . , ij , . . . , in+1] and

B =

{v(i1, . . . , in+1)

d�= 0

∣∣∣∣ 1 ≤ i1 < · · · < in+1 ≤ q}.

Note the inclusions ZB ⊂ L ⊂ N , where L is the Z-module generated bythe circuits of N .

We claim that ZB = N . First observe that L and N have both rankq − n; see Corollary 1.9.8. Since ZB and L have equal rank, one concludesthat N/ZB is a finite group. Thus to show ZB = N it suffices to prove thatN/ZB is torsion-free or equivalently that Zq/ZB is torsion-free of rank n.

Consider the matrix M whose rows are the vectors of B. By the funda-mental theorem for finitely generated abelian groups (see Theorem 1.3.16)the proof reduces to showing that the ideal Iq−n(M) of Z generated by the(q − n)-minors of M is equal to Z.

Fix Δ = [i1, . . . , ij, . . . , in+1]/d. For simplicity of notation one mayconsider the case Δ = [1, . . . , n]/d. Note that Δ occurs in v(1, . . . , n, i)/dwith a ± sign for i = n+ 1, . . . , q. Thus the matrix

M ′ =

⎛⎜⎜⎜⎝m11 · · · m1n Δ 0 0 · · · 0m21 · · · m2n 0 Δ 0 · · · 0...

......

......

...m(q−n)1 · · · m(q−n)n 0 0 0 · · · Δ

⎞⎟⎟⎟⎠


whose ith row is the vector v(1, . . . , n, i)/d is a submatrix of M . ThereforeΔq−n belongs to Iq−n(M). If Iq−n(M) �= Z, there is a prime number pdividing Δq−n for all Δ. Thus p divides Δ for all Δ, and this implies thatpd divides d, a contradiction. �

Corollary 1.9.11 If A is a t-unimodular matrix of size n × q and α is acircuit of ker(A), then α =

∑qi=1 εiei for some ε1, . . . , εq in {0,±1}.

Proof. It follows readily from the explicit description of the circuits givenin Proposition 1.9.9 and from Exercise 1.9.17. �

Matroids Let M = (E, I) be a matroid on E, i.e., there is a collection Iof subsets of E satisfying the following three conditions:

(i1) ∅ ∈ I.

(i2) If I ∈ I and I ′ ⊂ I, then I ′ ∈ I.

(i3) If I1 and I2 are in I and |I1| < |I2|, then there is an element e of I2 \I1such that I1 ∪ {e} ∈ I.

The members of I are the independent sets of M . It is convenient towrite I(M) for I and E(M) for E, particularly when several matroids arebeing considered. A subset of E that is not in I is called dependent . Amaximal independent set of M with respect to inclusion is called a basis.A minimal dependent set in M is called a circuit . The reader is referred to[338] for the general theory of matroids.

Proposition 1.9.12 [338, Proposition 1.1.1, p. 8] Let E be the set of col-umn labels of an m × n matrix A over a field K, and let I be the set ofsubsets B of E for which the multiset of columns labeled by B is linearlyindependent in Km. Then M [A] = (E, I) is a matroid.

The matroid M [A] obtained as above from the matrix A is called thevector matroid of A. Any matroid M arising from a matrix A is calledrepresentable over the field K.

We are especially interested in vector matroid arising from a nonnegativeintegral matrix A with column vectors v1, . . . , vq. By Proposition 1.9.12there is a matroid M [A] on A = {v1, . . . , vq} over the field Q of rationalnumbers, whose independent sets are the independent subsets of A.

Definition 1.9.13 A minimal dependent set or circuit ofM [A] is a depen-dent set all of whose proper subsets are independent. A subset B of A iscalled a basis of M [A] if B is a maximal independent set.

60 Chapter 1

There is a correspondence: Circuits of ker(A) −→ Circuits of M [A]given by α = (α1, . . . , αq) −→ C(α) = {vi| i ∈ supp(α)}. Thus the setof circuits of the kernel of A is the algebraic realization of the set of circuitsof the vector matroid M [A]. The circuits of ker(A) can be used to studythe normality of monomial subrings (see Theorem 9.7.1).

Theorem 1.9.14 [338, Corollary 1.2.5, p. 18] Let B be a non-empty familyof subsets of X. Then B is the collection of bases of a matroid on X ifand only if the following exchange property is satisfied: If B1 and B2 aremembers of B and b1 ∈ B1 \B2, then there is an element b2 ∈ B2 \B1 suchthat (B1 \ {b1}) ∪ {b2} is in B.

The family of bases of any matroid satisfies the following symmetricexchange property (see [286]):

Theorem 1.9.15 If B1 and B2 are bases of a matroid M and b1 ∈ B1\B2,then there is an element b2 ∈ B2 \B1 such that both (B1 \ {b1}) ∪ {b2} and(B2 \ {b2}) ∪ {b1} are bases of M .

It is well known [338] that all bases of a matroidM have the same numberof elements. This common number is called the rank of the matroid.

Exercises

1.9.16 Prove that the circuits of the kernel of the matrix M are preciselythe row vectors of the matrix A with a ± sign

M =

(4 3 1 00 1 3 4

)A =

⎛⎜⎜⎝2 −3 1 03 −4 0 10 1 −3 21 0 −4 3

⎞⎟⎟⎠ .

1.9.17 Let A be an n× q matrix with entries in an integral domain R andlet F1, . . . , Fn be the rows of A. Assume that Fi =

∑j �=i λjFj , for some

λj ’s in the field of fractions of R. Consider the R-linear maps

ϕ : Rq −→ Rn αϕ�−→ Aα

ϕ′ : Rq −→ Rn−1 αϕ′�−→ A′α,

where A′ is the matrix obtained from A by removing its ith row, prove thatker(ϕ) = ker(ϕ′).

1.9.18 If X = {x1, . . . , x4} and

B = {{x1, x2}, {x2, x3}, {x3, x4}, {x1, x4}},

prove that B satisfies the bases exchange property of matroids.

Chapter 2

Commutative Algebra

In this chapter some basic notions and results from commutative algebrawill be introduced. All rings considered in this book are commutative andNoetherian and modules are finitely generated. Our main references arethe books of Bruns and Herzog [65], Eisenbud [128], Matsumura [309, 310],and Vasconcelos [413]. Some of the results presented below are just statedwithout giving proofs, if need be, the reader may locate the missing proofsin those references.

2.1 Module theory

Noetherian modules and localizations Let R be a commutative ringwith unit and let M be an R-module. Recall that M is called Noetherian ifevery submodule N ofM is finitely generated, that is, N = Rf1+ · · ·+Rfq,for some f1, . . . , fq in N .

Theorem 2.1.1 The following conditions are equivalent:

(a) M is Noetherian.

(b) M satisfies the ascending chain condition for submodules; that is, forevery ascending chain of submodules of M

N0 ⊂ N1 ⊂ · · · ⊂ Nn ⊂ Nn+1 ⊂ · · · ⊂M

there exists an integer k such that Ni = Nk for every i ≥ k.

(c) Any family F of submodules of M partially ordered by inclusion hasa maximal element, i.e., there is N ∈ F such that if N ⊂ Ni andNi ∈ F , then N = Ni.

62 Chapter 2

Proof. (a)⇒(b): Consider the submodule N = ∪i≥0Ni. By hypothesisthere are m1, . . . ,mr such that N = Rm1 + · · · + Rmr. Then, there is ksuch that mi ∈ Nk for all i. It follows that Ni = Nk for all i ≥ k.

(b)⇒(c): Let N1 ∈ F . If N1 is not maximal, there is N2 ∈ F suchthat N1 � N2. If N2 is not maximal, there is N3 ∈ F such that N2 � N3.Applying this argument repeatedly we get that F has a maximal element.

(c)⇒(a): Let N be a submodule of M and let F be the family of sub-modules of N that are finitely generated. By hypothesis F has a maximalelement N ′. It follows that N = N ′. �

In particular a Noetherian ring R is a commutative ring with unit withthe property that every ideal of I is finitely generated; that is, given anideal I of R there exists a finite number of generators f1, . . . , fq such that

I = {a1f1 + · · ·+ aqfq| ai ∈ R, ∀ i} .

As usual, if I is generated by f1, . . . , fq, we write I = (f1, . . . , fq).

Proposition 2.1.2 IfM is a finitely generated R-module over a Noetherianring R, then M is a Noetherian module.

Corollary 2.1.3 If R is a Noetherian ring and I is an ideal of R, thenR/I and Rn are Noetherian R-modules. In particular any submodule of Rn

is finitely generated.

Theorem 2.1.4 (Hilbert’s basis theorem [9, Theorem 7.5]) A polynomialring R[x] over a Noetherian ring R is Noetherian.

One of the important examples of a Noetherian ring is a polynomial ringover a field k. Often we will denote a polynomial ring in several variablesby k[x] and a polynomial ring in one variable by k[x]. The letters k and Kwill always denote fields.

In this book, unless otherwise stated, by a ring (resp. module) we shallalways mean a Noetherian ring (resp. finitely generated module).

The prime spectrum of a ring R, denoted by Spec(R), is the set of primeideals of R. The minimal primes of R are the minimal elements of Spec(R)with respect to inclusion and the maximal ideals of R are the maximalelements of the set of proper ideals of R with respect to inclusion.

Let R be a ring and let X = Spec(R) be its prime spectrum. Given anideal I of R, the set of all prime ideals of R containing I will be denotedby V (I). The minimal primes of I are the minimal elements of V (I) withrespect to inclusion. It is not hard to verify that the pair (X,Z) is atopological space, where Z is the family of open sets of X , and where U isin Z iff U = X \ V (I), for some ideal I. This topology is called the Zariskitopology of the prime spectrum of R.

Commutative Algebra 63

A local ring (R,m, k) is a Noetherian ring R with exactly one maximalideal m, the field k = R/m is called the residue field of R.

A homomorphism of rings is a map ϕ : R→ S such that:

(i) ϕ(a+ b) = ϕ(a) + ϕ(b), ∀ a, b ∈ A,(ii) ϕ(ab) = ϕ(a)ϕ(b), ∀ a, b ∈ A, and(iii) ϕ(1) = 1.

Let R be a ring and let ϕ : Z→ R be the canonical homomorphism

ϕ(a) = a · 1R,

then ker(ϕ) = nZ, for some n ≥ 0. The integer n is called the characteristicof R and is denoted by char(R).

Proposition 2.1.5 Let (R,m) be a local ring, then either char(R) = 0 orchar(R) = pk, for some prime number p and some integer k ≥ 1.

Proof. Let n = char(R) ≥ 2. If n = pqm, where p, q are prime numberswith q �= p, then p · 1R and q · 1R are both in m because they are noninvertible, but this is impossible because 1 = ap+ bq and from the equality

1R = (a · 1R)(p · 1R) + (b · 1R)(q · 1R),

one derives 1R ∈ m, a contradiction. �

Modules of fractions and localizations

Let R be a ring, M an R-module, and S a multiplicatively closed subset ofR so that 1 ∈ S. Then the module of fractions of M with respect to S, orthe localization of M with respect to S, is defined by

S−1(M) = {m/s|m ∈M, s ∈ S} ,

where m/s = m1/s1 if and only if t(s1m − sm1) = 0 for some t ∈ S. Inparticular S−1R has a ring structure given by the usual rules of additionand multiplication, and S−1M is a module over the ring S−1R with theoperations:

m1/s1 +m2/s2 = (s2m1 + s1m2)/s1s2 (si ∈ S, mi ∈M),

(a/s) · (m1/s1) = am1/ss1 (a ∈ R, s ∈ S).

There is a map ϕ : M → S−1M , given by ϕ(m) = m/1. If f : M → N isa homomorphism of R-modules, then there is an induced homomorphism

S−1f : S−1M −→ S−1N

of S−1R-modules given by f(m/s) = f(m)/s.

64 Chapter 2

Example 2.1.6 If f is in a ring R and S = {f i | i ∈ N}, then S−1R isusually written Rf . For instance if R = C [x] is a polynomial ring in onevariable over the field C of complex numbers, then Rx = C [x, x−1] is thering of Laurent polynomials.

Definition 2.1.7 Let p be a prime ideal of a ring R and S = R \ p. In thiscase S−1R is written Rp and is called the localization of R at p.

Example 2.1.8 Let p be a prime ideal of a ring R. The local ring

(Rp, pRp, k(p))

is the prototype of a local ring, where k(p) = Rp/pRp denotes the residuefield of Rp.

Krull dimension and height

By a chain of prime ideals of a ring R we mean a finite strictly increasingsequence of prime ideals

p0 ⊂ p1 ⊂ · · · ⊂ pn,

the integer n is called the length of the chain. The Krull dimension of R,denoted by dim(R), is the supremum of the lengths of all chains of primeideals in R. Let p be a prime ideal of R, the height of p, denoted by ht (p)is the supremum of the lengths of all chains of prime ideals

p0 ⊂ p1 ⊂ · · · ⊂ pn = p

which end at p. Note dim(Rp) = ht (p). If I is an ideal of R, then ht (I),the height of I, is defined as

ht(I) = min{ht(p)| I ⊂ p and p ∈ Spec(R)}.In general dim(R/I)+ ht (I) ≤ dim(R). The difference dim(R)− dim(R/I)is called the codimension of I and dim(R/I) is called the dimension of I.

Let M be an R-module. The annihilator of M is given by

annR(M) = {x ∈ R|xM = 0},if m ∈ M the annihilator of m is ann (m) = ann (Rm). It is convenient togeneralize the notion of annihilator to ideals and submodules. Let N1 andN2 be submodules of M , their ideal quotient or colon ideal is defined as

(N1 : RN2) = {x ∈ R|xN2 ⊂ N1}.Let us recall that the dimension of an R-module M is

dim(M) = dim(R/ann (M))

and the codimension of M is codim(M) = dim(R)− dim(M).

Theorem 2.1.9 If R[x] is a polynomial ring over a Noetherian ring R,then dimR[x] = dim(R) + 1.


Primary decomposition of modules Let I be an ideal of a ring R.The radical of I is

rad (I) = {x ∈ R|xn ∈ I for some n > 0},

the radical is also denoted by√I. In particular

√(0), denoted by NR or

nil(R), is the set of nilpotent elements of R and is called the nilradical of R.A ring is reduced if its nilradical is zero. The Jacobson radical of R is theintersection of all the maximal ideals of R.

Proposition 2.1.10 If I is a proper ideal of a ring R, then rad (I) is theintersection of all prime ideals containing I.

Definition 2.1.11 LetM be a module over a ring R. The set of associatedprimes ofM , denoted by AssR(M), is the set of all prime ideals p of R suchthat there is a monomorphism φ of R-modules:

R/pφ↪→M.

Note that p = ann (φ(1)).

Lemma 2.1.12 If M �= 0 is an R-module, then AssR(M) �= ∅.

Proof. Consider the family of ideals F = {ann(m)| 0 �= m ∈ M} orderedby inclusion. By Theorem 2.1.1 this family has a maximal element that wedenote by ann(m). It suffices to show that ann(m) is prime. Let x, y betwo elements of R such that xy ∈ ann(m). Assume that x /∈ ann(m). Thenann(m) ⊂ ann(xm) and consequently ann(m) = ann(xm). This shows thaty ∈ ann(m), as required. �

If M = R/I, it is usual to say that an associated prime ideal of R/I isan associated prime ideal of I and to set Ass(I) = Ass(R/I).

Proposition 2.1.13 Let R be a ring and let S be a multiplicatively closedsubset of R. If M is an R-module and p is a prime ideal of R with S∩p = ∅,then p is an associated prime ofM if and only if S−1p is an associated primeof S−1M .

Proof. If p is in Ass(M), then R/p ↪→ M . Hence S−1R/S−1p ↪→ S−1M .Thus, S−1p is an associated prime of S−1M .

For the converse assume S−1p = ann(m/1). Since R is Noetherian, p isgenerated by a finite set a1, . . . , an. Hence for each i there is si ∈ S suchthat siaim = 0. Set s = s1 · · · sn. We claim p = ann(sm). Clearly onehas p ⊂ ann(sm). To show the other containment take x ∈ ann(sm), thenxsm = 0 and x/1 ∈ ann(m/1) = S−1p. Hence x ∈ p. �

66 Chapter 2

Definition 2.1.14 Let M be an R-module, the support of M , denoted bySupp(M), is the set of all prime ideals p of R such that Mp �= 0.

A sequence 0 → M ′ f→ Mg→ M ′′ → 0 of R-modules is called a short

exact sequence if f is a monomorphism, g is an epimorphism, and im(f) =ker(g). A sequence of R modules and homomorphisms:

· · · −→Mi−1fi−1−→Mi

fi−→Mi+1fi+1−→ · · ·

is said to be exact at Mi if im(fi−1) = ker(fi). If the sequence is exact ateach Mi it is called an exact sequence.

Lemma 2.1.15 If 0 → M ′ → M → M ′′ → 0 is a short exact sequence ofmodules over a ring R, then Supp(M) = Supp(M ′) ∪ Supp(M ′′).

Proof. Let p be a prime ideal of R. It suffices to observe that from theexact sequence

0→M ′p →Mp →M ′′

p → 0,

we get Mp �= 0 if and only if M ′p �= 0 or M ′′

p �= 0. �

Theorem 2.1.16 If M is an R-module, then there is a filtration of sub-modules

(0) =M0 ⊂M1 ⊂ · · · ⊂Mn =M

and prime ideals p1, . . . , pn of R such that Mi/Mi−1 � R/pi for all i.

Proof. By Lemma 2.1.12 there is a prime ideal p1 and a submodule M1

of M such that R/p1 � M1. If M1 � M , then there is an associatedprime ideal p2 of M/M1 such that R/p2 is isomorphic to a submodule ofM/M1, i.e., R/p2 � M2/M1, where M2 is a submodule of M containingM1. If M2 � M , we pick an associated prime p3 of M/M2 and repeat theargument. Since M is Noetherian a repeated use of this procedure yieldsthe required filtration. �

In general the primes p1, . . . , pn that occur in a filtration of the typedescribed in the previous result are not associated primes of the moduleM ; see [115] for a careful discussion of filtrations and for some of theirapplications to combinatorics.

Lemma 2.1.17 If 0 → M ′ f→ Mg→ M ′′ → 0 is a short exact sequence of

modules over a ring R, then Ass(M) ⊂ Ass(M ′) ∪ Ass(M ′′).

Proof. We may assume (without loss of generality) that f is the identitymap by replacing M ′ by f(M ′). Let p be an associated prime of M . Thereis 0 �= m ∈M such that p = ann(m).


Case (I): Rm∩M ′ �= (0). There is r ∈ R such that 0 �= rm ∈M ′. Fromthe equality p = ann(m) we get p = ann(rm). Thus p ∈ Ass(M ′).

Case (II): Rm ∩M ′ = (0). Notice that g(m) �= 0. From the equalityp = ann(g(m)) we get p ∈ Ass(M ′′). �

Corollary 2.1.18 If M is an R-module, then AssR(M) is a finite set.

Proof. Let p1, . . . , pn be prime ideals as in Theorem 2.1.16. By a repeateduse of Lemma 2.1.17 one has AssR(M) ⊂ {p1, . . . , pn} ⊂ Supp(M). �

Let M be an R-module. An element x ∈ R is a zero divisor of M ifthere is 0 �= m ∈ M such that xm = 0. The set of zero divisors of M isdenoted by Z(M). If x is not a zero divisor on M , we say that x is a regularelement of M .

Lemma 2.1.19 If M is an R-module, then

Z(M) =⋃

p∈AssR(M)

p.

Proof. The right-hand side is clearly contained in the left-hand side bydefinition of an associated prime. Let r be a zero divisor of M and considerthe family F = {ann(m)| 0 �= m ∈ M ; rm = 0}. Notice that any maximalelement of this family is a prime ideal. �

Proposition 2.1.20 If M is an R-module, then

Ass(M) ⊂ Supp(M) = V (ann (M)),

and any minimal element of Supp(M) is in Ass(M).

Proof. If p is an associated prime of M , then there is a monomorphismR/p↪→M and thus 0 �= (R/p)p↪→Mp. Hence p is in the support of M , thisshows the first containment.

Next, we show Supp(M) = V (ann (M)). Let p ∈ Supp(M) and letx ∈ ann (M). If x �∈ p, then xm = 0 for all m ∈M and Mp = (0), which isabsurd. Therefore p is in V (ann (M)). Conversely let p be in V (ann (M))and let m1, . . . ,mr be a finite set of generators of M . If Mp = (0), then foreach i there is si �∈ p so that simi = 0, therefore s1 · · · sr is in ann (M) ⊂ p,which is impossible. Hence Mp �= 0 and p is in the support of M .

To prove the last part take a minimal prime p in the support of M . AsMp �= (0) there is an associated prime p1Rp ofMp, where p1 is a prime idealof R contained in p. Since Mp1 � (Mp)p1 �= (0), we get that p1 is in thesupport ofM and p = p1. Therefore using Proposition 2.1.13 one concludesp ∈ Ass(M). �

68 Chapter 2

Let M be an R-module, the minimal primes of M are defined to be theminimal elements of Supp(M) with respect to inclusion. A minimal primeof M is called an isolated associated prime of M . An associated prime ofM which is not isolated is called an embedded prime.

If R is a ring and I is an ideal, note that the minimal primes of Iare precisely the minimal primes of AssR(R/I). In particular the minimalprimes of R are precisely the minimal primes of AssR(R).

Definition 2.1.21 Let M be an R-module. A submodule N of M is saidto be a p-primary submodule if AssR(M/N) = {p}. An ideal q of a ring Rcalled a p-primary ideal if AssR(R/q) = {p}.

Proposition 2.1.22 An ideal q �= R of a ring R is a primary ideal if andonly if xy ∈ q and x /∈ q implies yn ∈ q for some n ≥ 1.

Proof. Assume q is a primary ideal. Let x, y ∈ R such that xy ∈ q andx /∈ q. Hence y is a zero divisor of R/q because yx = 0 and x �= 0. SinceZ(R/q) = {p} and rad(q) = p, we get yn ∈ q for some positive integer n.The converse is left as an exercise. �

Definition 2.1.23 LetM be an R-module. A submodule N ofM is said tobe irreducible if N cannot be written as an intersection of two submodulesof M that properly contain N .

Proposition 2.1.24 Let M be an R-module. If Q �= M is an irreduciblesubmodule of M , then Q is a primary submodule.

Proof. Assume there are p1and p2 distinct associated prime ideals ofM/Qand pick r0 ∈ p1 \ p2 (or vice versa). There is xi in M \ Q such thatpi = ann(xi), where xi = xi +Q. We claim that

(Rx1 +Q) ∩ (Rx2 +Q) = Q.

If z is in the intersection, then z = λ1x1 + q1 = λ2x2 + q2, for some λi ∈ Rand qi ∈ Q. Note that r0z ∈ Q, hence r0λ2x2 is in Q and consequentlyr0λ2 ∈ p2. Thus λ2 ∈ p2 and we get λ2x2 ∈ Q. This shows z ∈ Q andcompletes the proof of the claim. As Q is irreducible one has Q = Rx1 +Qor Q = Rx2 +Q, which is a contradiction because xi �∈ Q for i = 1, 2. �

Lemma 2.1.25 If N1, N2 are p-primary submodules of M , then N1 ∩ N2

is a p-primary submodule.

Proof. Set N = N1 ∩N2. There is an inclusion M/N ↪→ M/N1 ⊕M/N2.Hence using Exercise 2.1.53 we get

Ass(M/N) ⊂ Ass(M/N1 ⊕M/N2) = Ass(M/N1) ∪ Ass(M/N2) = {p}.

Therefore Ass(M/N) = {p}. �


Definition 2.1.26 Let M be an R-module and let N � M be a propersubmodule. An irredundant primary decomposition of N is an expression ofN as an intersection of submodules, say N = N1 ∩ · · · ∩Nr, such that:

(a) (Submodules are primary) AssR(M/Ni) = {pi} for all i.(b) (Irredundancy) N �= N1 ∩ · · · ∩Ni−1 ∩Ni+1 ∩ · · · ∩Nr for all i.

(c) (Minimality) pi �= pj if Ni �= Nj .

Theorem 2.1.27 Let M be an R-module. If N �M is a proper submoduleof M , then N has an irredundant primary decomposition

Proof. First we claim that N is an intersection of a finite number ofirreducible submodules of M . Let F be the family of submodules of Msuch that the claim is false. Assume F �= ∅. By Theorem 2.1.1, F hasa maximal element that we denote by N . Since N is not irreducible, wecan write N = N2 ∩ N3, for some submodules N2, N3 strictly containingN . By the maximality of N we get that N1 and N2 can be written as anintersection of irreducible submodules, a contradiction. Thus F = ∅. Hencethe result follows from Proposition 2.1.24 and Lemma 2.1.25. �

Corollary 2.1.28 If R is a Noetherian ring and I a proper ideal of R, thenI has an irredundant primary decomposition I = q1 ∩ · · · ∩ qr such that qiis a pi-primary ideal and AssR(R/I) = {p1, . . . , pr}.

Proof. Let (0) = I/I = (q1/I)∩ · · ·∩ (qr/I) be an irredundant decomposi-tion of the zero ideal of R/I. Then I = q1 ∩ · · · ∩ qr and qi/I is pi-primary;that is, AssR((R/I)/(qi/I)) = Ass(R/qi) = {pi}. Let us show that qi is aprimary ideal. If xy ∈ qi and x /∈ qi, then y is a zero-divisor of R/qi, butZ(R/qi) = pi, hence y ∈ pi = rad (ann (R/qi)) = rad (qi) and y

n is in qi forsome n > 0. �

Corollary 2.1.29 If M is an R-module, then

rad(ann(M)) =⋂

p∈Ass(M)

p.

Proof. Let p1, . . . , pr be the minimal elements of Supp(M). By Proposi-tion 2.1.20 we have Supp(M) = V (ann (M)) and pi is in Ass(M) for all i.By Corollary 2.1.28 we have that rad(ann(M)) = p1 ∩ · · · ∩ pr. �

Corollary 2.1.30 If N � M and N = N1 ∩ · · · ∩ Nr is an irredundantprimary decomposition of N with AssR(M/Ni) = {pi}, then

AssR(M/N) = {p1, . . . , pr},

and ann (M/Ni) is a pi-primary ideal for all i.

70 Chapter 2

Proof. There is a natural monomorphism

M/(N1 ∩ · · · ∩Nr) ↪→ (M/N1)⊕ · · · ⊕ (M/Nr).

Hence Ass(M/N) ⊂ {p1, . . . , pr}. There are natural monomorphisms

(N2 ∩ · · · ∩Nr)/N ↪→M/N1; (N2 ∩ · · · ∩Nr)/N ↪→M/N.

Since Ass(M/N1) = {p1} we get p1 ∈ Ass(N2 ∩ · · · ∩ Nr/N), consequentlyp1 ∈ Ass(M/N). Similarly one can show that any other pi is an associatedprime of M/N . This proves the asserted equality.

By Corollary 2.1.29 we have rad(ann(M/Ni)) = pi. Thus it suffices toshow that I = ann(M/Ni) is a primary ideal. Assume that xy ∈ I for somex, y ∈ R. If x is not in I, then xM is not contained in Ni. Pick m ∈ Msuch that xm /∈ Ni. Since y(xm) ⊂ Ni, we get that y is a zero divisor ofM/Ni, but the zero divisors of this module are precisely the elements of piaccording to Lemma 2.1.19. Hence yr ∈ I for some r. �

Definition 2.1.31 Let R be a ring and let S be the set of nonzero divisorsof R. The ring S−1R is called the total ring of fractions of R. If R is adomain, S−1R is the field of fractions of R.

Proposition 2.1.32 Let R be a ring and let K be the total ring of fractionsof R. If R is reduced, then K is a direct product of fields.

Proof. Let p1, . . . , pr be the minimal primes of R and S = R \ ∪ri=1pi.Since R is reduced one has (0) = p1 ∩ · · · ∩ pr and K = S−1R. Define

φ : K −→ S−1R/S−1p1 × · · · × S−1R/S−1pr

by φ(x) = (x + S−1p1, . . . , x + S−1pr). As S−1p1, . . . , S−1pr are maximal

ideals and its intersection is zero, it follows from the Chinese remaindertheorem that φ is an isomorphism (see Exercise 2.1.46). �

Lemma 2.1.33 Let M be an R-module and L an ideal of R. If LM =M ,then there is x ∈ R such that x ≡ 1 (modL) and xM = (0).

Proof. Let M = Rα1 + · · · + Rαn, αi ∈ M . As LM = M , there are bijin L such that αi =

∑nj=1 bijαi. Set α = (α1, . . . , αn) and H = (bij) − I,

where I is the identity matrix. Since Hαt = 0 and Hadj(H) = det(H)I,one concludes det(H)αi = 0 for all i. Hence xM = (0) and x ≡ 1(modL),where x = det(H). �

Lemma 2.1.34 (Nakayama) Let R be a ring and let N be a submodule ofan R-module M . If I is an ideal of R contained in the intersection of allthe maximal ideals of R such that M = IM +N , then M = N .


Proof. Note I(M/N) = M/N . By Lemma 2.1.33 there exists an elementx ≡ 1(mod I) such that x(M/N) = (0). To finish the proof note that xis a unit; otherwise x belongs to some maximal ideal m and this yields acontradiction because I ⊂ m. �

Let M be an R-module. The minimum number of generators of M willbe denoted by μ(M). A consequence of Nakayama’s lemma is an expressionfor μ(M), when the ring R is local (cf. Corollary 3.5.2).

Corollary 2.1.35 If M is a module over a local ring (R,m), then

μ(M) = dimk(M/mM), where k = R/m.

Proof. Let α1, . . . , αq be a minimal generating set forM and αi = αi+mM .After a permutation of the αi one may assume that α1, . . . , αr is a basis forM/mM as a k-vector space, for some r ≤ q. Set N = Rα1 + · · · + Rαr.Note the equality M = N + mM , then by Nakayama’s Lemma N = M .Therefore r = q, as required. �

Modules of finite length An R-module M has finite length if there is acomposition series

(0) =M0 ⊂M1 ⊂ · · · ⊂Mn =M, (2.1)

where Mi/Mi−1 is a non-zero simple module (that is, Mi/Mi−1 has noproper submodules other than (0)) for all i. Note that Mi/Mi−1 must becyclic and thus isomorphic to R/m, for some maximal ideal m. The numbern is independent of the composition series and is called the length of M ; itis usually denoted by �R(M) or simply �(M).

Proposition 2.1.36 [9, Proposition 6.9] If 0 → M ′ → M → M ′′ → 0 isan exact sequence of R-modules of finite length, then

�R(M) = �R(M′) + �R(M

′′).

An R-module M is called Artinian if M satisfies the descending chaincondition for submodules; that is, for every chain of submodules of M

· · · ⊂ Nn+1 ⊂ Nn ⊂ · · · ⊂ N2 ⊂ N1 ⊂ N0 =M

there exists an integer k such that Ni = Nk for every i ≥ k. It is easy toverify that M is Artinian if and only if any family F of submodules of Mpartially ordered by inclusion has a minimal element, i.e., there is N ∈ Fsuch that if Ni ⊂ N and Ni ∈ F , then N = Ni.

Proposition 2.1.37 Let M be an R-module. Then �R(M) < ∞ if andonly if M is Noetherian and Artinian.

72 Chapter 2

Proof. ⇒) Set n = �R(M). If M is not Noetherian or Artinian, picksubmodules of M such that

(0) = N0 � N1 � · · · � Nn � Nn+1 =M,

a contradiction because this chain can be refined to a composition series oflength n.⇐) We construct a finite composition series as follows. Set M0 = M .

Consider the family F1 of proper submodules of M and pick a maximalelement M1, which exists because M is Noetherian. By induction considerthe family Fi of proper submodules of Mi and pick a maximal elementMi+1. Notice that this process must stop at the zero module because M isArtinian. �

Lemma 2.1.38 Let M be a k-vector space. Then M is Artinian if andonly if dimk(M) = �k(M) <∞.

Proof. Assume that M is Artinian and dimk(M) =∞. Let B ⊂ M be aninfinite linearly independent set, say B = {α1, . . . , αi, . . . , }. Then

· · · � k(B \ {α1, . . . , αi}) � · · · � k(B \ {α1, α2}) � k(B \ {α1}) �M,

a contradiction. The converse is also easy to show. �

Proposition 2.1.39 If 0→ M ′ →M →M ′′ → 0 is an exact sequence ofmodules over a ring R, then dim(M) = max{dim(M ′), dim(M ′′)}.

Proof. Set d = dim(M), d′ = dim(M ′) and d′′ = dim(M ′′). First note thatd = dim(R/p) for some prime p containing ann (M), by Proposition 2.1.20we obtain Mp �= (0). Therefore using Lemma 2.1.15 one has M ′

p �= (0)or M ′′

p �= (0), thus either p contains ann (M ′) or p contains ann (M ′′).This proves d ≤ max{d′, d′′}. On the other hand ann (M) is contained inann (M ′) ∩ ann (M ′′) and consequently max{d′, d′′} ≤ d. �

Proposition 2.1.40 If M is an R-module, then M has finite length if andonly if every prime ideal in Supp(M) is a maximal ideal.

Proof. ⇒) Let {Mi}ni=0 be a composition series as in Eq. (2.1). Noticethat pi are maximal ideals for all i. Indeed if N is a simple R-module andR/p � N for some prime ideal p, then p is a maximal ideal. Hence fromLemma 2.1.15 we obtain that Supp(M) is equal to {p1, . . . , pn}.⇐) There is a filtration (0) =M0 ⊂M1 ⊂ · · · ⊂Mn =M of submodules

and prime ideals p1, . . . , pn of R such that Mi/Mi−1 � R/pi for all i (seeTheorem 2.1.16). By the proof of Corollary 2.1.18 one has

Ass(M) ⊂ {p1, . . . , pn} ⊂ Supp(M).


Hence by Proposition 2.1.20 we get Ass(M) = {p1, . . . , pn} = Supp(M).Thus the filtration above is a composition series because pi is a maximalideal for all i. �

Theorem 2.1.41 Let R be a ring. Then R is Artinian if and only if

(a) R is Noetherian, and

(b) every prime ideal of R is maximal.

Proof. ⇒) First we show that �R(R) <∞. Let

F = {m1m2 · · ·ms|mi ∈Max(R), ∀ i},

we do not require m1, . . . ,ms to be distinct. The family F has a minimalelement that we denote by I = m1 · · ·mr. We claim that I = (0). Assumethat I �= (0). Clearly I2 ∈ F and I2 ⊂ I, thus I = I2. Consider the family

G = {J | J is an ideal; JI �= (0)}.

This family is non-empty because I ∈ G. Since R is Artinian, there exists aminimal element J of G. Then

(0) �= JI = JI2 = (JI)I ⇒ JI ∈ G.

Since IJ ⊂ J , we get J = IJ . Notice that J is a principal ideal. Indeedsince IJ �= (0), there is x ∈ J such that xI �= (0). Hence (x) ∈ G, andconsequently J = (x). If m is a maximal ideal of R, then mI ∈ F andmI ⊂ I. Hence mI = I and I ⊂ m. Thus I is contained in the Jacobsonradical of I. From the equality (x) = J = IJ = Ix, we obtain that x = λxfor some λ ∈ I. As x(1 − λ) = 0, with 1 − λ a unit of R, we get x = 0. Acontradiction to J �= (0). This proves that I = (0). Set Ik = m1 · · ·mk for1 ≤ k ≤ r and consider the filtration

(0) = I = Ir ⊂ Ir−1 ⊂ Ir−2 ⊂ · · · ⊂ I1 ⊂ I0 = R.

Observe that Ii/Ii+1 is an R/mi+1 vector space. The vector space Ii/Ii+1

is Artinian because the subspaces of Ii/Ii+1 correspond bijectively to theintermediate ideals of Ii+1 ⊂ Ii. Therefore by Lemma 2.1.38 we get thatIi/Ii+1 has finite length as an R/mi+1 vector space and consequently as anR-module. Hence the filtration can be refined to a finite composition series,i.e., �R(R) < ∞. Therefore R is Noetherian by Proposition 2.1.37 and (a)holds. To prove (b) take a minimal prime p of R. Since p contains I = (0)it follows that p = mi for some 1 ≤ i ≤ r.⇐) Let (0) = q1 ∩ · · · ∩ qr be an irredundant primary decomposition.

Then all the associated primes are maximal. By Proposition 2.1.20 weobtain AssR(R) = Supp(R). Hence R is Artinian by Proposition 2.1.40. �

74 Chapter 2

Definition 2.1.42 A ring is called semilocal if it has only finitely manymaximal ideals.

Using the Chinese reminder theorem (see Exercise 2.1.46) and the proofof Theorem 2.1.41 we obtain:

Proposition 2.1.43 If R is an Artinian ring, then R has only a finitenumber of maximal ideals m1, . . . ,ms and

R � R/ma11 × · · · ×R/mass .

Definition 2.1.44 Let R be a ring with total ring of fractions Q. An R-module M has rank r if M ⊗R Q is a free Q-module of rank equal to r.

Lemma 2.1.45 If M is an R-module of positive rank r, then

dim(M) = dim(R).

Proof. Let p1, . . . , pn be the associated primes of R and S = R \ ∪ni=1pi.By hypothesis S−1M � (S−1R)r. Since

Mpi � (S−1M)pi � [(S−1R)r]pi � [(S−1R)pi ]r � (Rpi)

r �= (0),

we obtain that all the minimal primes of R are in the support of M andconsequently one has dim(M) = dim(R). �

Exercises

2.1.46 (Chinese remainder theorem) Let I1, . . . , Ir be ideals of a ring R. IfIi + Ij = R for i �= j, prove:

(a) I1 ∩ · · · ∩ Ir = I1 · · · Ir ,(b) The rings R/I1 ∩ · · · ∩ Ir and R/I1 × · · · ×R/Ir are isomorphic.

2.1.47 Let I be an ideal of a ring R. Prove that rad(I) = I if and only iffor any x ∈ R such that x2 ∈ I one has x ∈ I.

2.1.48 Let I1, . . . , In be ideals of a ring R and let p be a prime ideal of R.(a) If I1 · · · In ⊂ p, prove that Ii ⊂ p for some i. (b) If ∩ni=1Ii ⊂ p, thenIi ⊂ p for some i. (c) If ∩ni=1Ii = p, then p = Ii for some i.

2.1.49 Let I, p1, . . . , pr be ideals of a ring R. If I is contained in ∪ri=1pi andpi is a prime ideal for i ≥ 3, prove that I ⊂ pi for some i (cf. Lemma 3.1.31).

2.1.50 Let R be a ring and let I be an ideal. If p is a prime ideal suchthat I ⊂ p, prove that ht(I) ≤ ht(Ip). Give an example where the strictinequality holds.


2.1.51 Let R be a ring and let I be an ideal of R. If all the minimal primesof I have the same height and p is a prime ideal such that I ⊂ p, prove thatht(I) = ht(Ip).

2.1.52 Let I be an ideal of a ring R. Then a prime ideal p of R is anassociated prime of R/I if and only if p = (I : x) for some x ∈ R.

2.1.53 If M , N are R-modules, prove Ass(M ⊕N) = Ass(M) ∪ Ass(N).

2.1.54 Let M be an R-module and I an ideal of R contained in annR(M).Note that M inherits a structure of R/I-module. Prove that p ∈ AssR(M)if and only if p/I ∈ AssR/I(M).

2.1.55 Let M be an R-module and let p be a prime ideal of R. Prove thatp ∈ AssR(M) if and only if pRp ∈ AssRp(Mp).

2.1.56 Let A be a ring and q a proper ideal of A. Prove that q is a primaryideal if and only if Z(A/q) ⊂ NA/q, where NA/q is the nilradical of A/q.

2.1.57 Let A be a ring and q a proper ideal of A. If every element in A/qis either nilpotent or invertible, prove that q is a primary ideal.

2.1.58 Let A be a ring and q an ideal of A such that rad (q) = m is amaximal ideal. Show that q is a primary ideal.

2.1.59 If B = A[x] is a polynomial ring over a ring A and q a primary idealof A, then qB is a primary ideal of B.

2.1.60 Let M be an R-module and S ⊂ T two multiplicatively closed sub-sets of R. Prove T−1M � T−1(S−1M) as T−1R modules.

2.1.61 Let M be an R-module and S a multiplicatively closed subset of R.If p is a prime ideal such that S ∩ p = ∅, prove Mp � (S−1M)p.

2.1.62 Let R be a ring and I an ideal. If x ∈ R \ I, then there is an exactsequence of R-modules:

0 −→ R/(I : x)ψ−→ R/I

φ−→ R/(I, x) −→ 0,

where ψ(r) = xr is multiplication by x and φ(r) = r.

2.1.63 If 0 → M ′ f→ Mg→ M ′′ → 0 is an exact sequence of R-modules,

prove that the following sequence is also exact

0 −→ S−1M ′ S−1f−→ S−1M

S−1g−→ S−1M ′′ −→ 0.

76 Chapter 2

2.1.64 Let N1 and N2 be submodules of an R-module M . If S is a multi-plicatively closed subset of R, then

(a) S−1(N1 +N2) = S−1(N1) + S−1(N2),

(b) S−1(N1 ∩N2) = S−1(N1) ∩ S−1(N2),

(c) S−1(N1/N2) � S−1(N1)/S−1(N2), as S

−1(R)-modules,

(d) S−1(ann (N2)) = ann (S−1(N2)), and

(e) S−1(N1 : N2) = (S−1(N1) : S−1(N2)).

Hint Use that N2 is a finitely generated R-module.

2.1.65 If I and J are ideals of a ring R and S is a multiplicatively closedsubset of R, then

(a)√S−1(I) = S−1(

√I), and

(b) S−1(IJ) = S−1(I)S−1(J).

2.2 Graded modules and Hilbert polynomials

Let (H,+) be an abelian semigroup. An H-graded ring is a ring R togetherwith a decomposition

R =⊕a∈H

Ra (as a Z-module),

such that RaRb ⊂ Ra+b for all a, b ∈ H . A graded ring is by definition aZ-graded ring.

If R is an H-graded ring and M is an R-module with a decomposition

M =⊕a∈H

Ma,

such that RaMb ⊂ Ma+b for all a, b ∈ H , we say that M is an H-gradedmodule. An element 0 �= f ∈ M is said to be homogeneous of degree a iff ∈ Ma, in this case we set deg(f) = a. The non-zero elements in Ra arealso called forms of degree a. Any element f ∈M can be written uniquelyas f =

∑a∈H fa with only finitely many fa �= 0.

A map ϕ : M → N between H-graded modules is graded if ϕ(Ma) ⊂ Nafor all a ∈ H . Let M = ⊕a∈HMa be an H-graded module and N a gradedsubmodule; that is, N is graded with the induced gradingN = ⊕a∈HN∩Ma.Then M/N is an H-graded R-module with (M/N)a = Ma/N ∩ Ma fora ∈ H , R0 ⊂ R is a subring and Ma is an R0-module for a ∈ H .


Proposition 2.2.1 [310, p. 92] Let M = ⊕a∈HMa be an H-graded moduleand N ⊂M a submodule. Then the following conditions are equivalent:

(g1) N is generated over R by homogeneous elements.

(g2) If f =∑

a∈H fa is in N , fa ∈Ma for all a, then each fa is in N .

(g3) N is a graded submodule of M .

Let R = K[x1, . . . , xn] be a polynomial ring over a field K and letd1, . . . , dn be a sequence in N+. For a = (ai) in Nn we set xa = xa11 · · ·xannand |a| =

∑ni=1 aidi. The induced N-grading on R is given by:

R =

∞⊕i=0

Ri, where Ri =⊕|a|=i

Kxa.

Notice that deg(xi) = di for all i. The induced grading extends to a Z-grading by setting Ri = 0 for i < 0. The homogeneous elements of Rare called quasi-homogeneous polynomials . Let I be a homogeneous idealof R generated by a set f1, . . . , fr of homogeneous polynomials. Settingdeg(fi) = δi, I becomes a graded ideal with the grading

Ii = I ∩Ri = f1Ri−δ1 + · · ·+ frRi−δr .

Hence R/I is an N-graded R-module graded by (R/I)i = Ri/Ii.

Definition 2.2.2 The standard grading or usual grading of a polynomialring K[x1, . . . , xn] is the N-grading induced by setting deg(xi) = 1 for all i.

Hilbert polynomial and multiplicity Let R = ⊕∞i=0Ri be an N-graded

ring. We recall that R is a Noetherian ring if and only if R0 is a Noetherianring and R = R0[x1, . . . , xn] for some x1, . . . , xn in R. If M is a finitelygenerated N-graded R-module and R0 is an Artinian local ring, define theHilbert function of M as

H(M, i) = �R0(Mi) (�R0 = length w.r.t R0).

Definition 2.2.3 An N-graded ring R = ⊕∞i=0Ri is called a homogeneous

ring if R = R0[x1, . . . , xn], where deg(xi) = 1 for all i.

For use below, by convention the zero polynomial has degree −1.

Theorem 2.2.4 (Hilbert [65, Theorem 4.1.3]) Let R = ⊕∞i=0Ri be a ho-

mogeneous ring and let M be a finitely generated N-graded R-module withd = dim(M). If R0 is an Artinian local ring, then there is a unique polyno-mial ϕM (t) ∈ Q[t] of degree d− 1 such that ϕM (i) = H(M, i) for i 0.

Definition 2.2.5 The polynomial ϕM (t) is called the Hilbert polynomial ofM . If ϕM (t) = ad−1t

d−1+ · · ·+a0, the multiplicity or degree ofM , denotedby e(M) or deg(M), is (d− 1)!ad−1 if d ≥ 1 and �R0(M) if d = 0.

78 Chapter 2

Graded primary decomposition Let R = K[x1, . . . , xn] be a polyno-mial ring over a field K endowed with a positive grading induced by settingdeg(xi) = di for all i, where di is a positive integer for i = 1, . . . , n.

Lemma 2.2.6 [65, Lemma 1.5.6] If M is an N-graded R-module and p isin Ass(M), then p is a graded ideal and there is m ∈M homogeneous suchthat p = ann(m).

Proposition 2.2.7 [309, p. 63] Let M be an N-graded R-module and let Qbe a p-primary submodule ofM . If p is graded and Q∗ is the submodule ofMgenerated by the homogeneous elements in Q, then Q∗ is again a p-primarysubmodule.

Theorem 2.2.8 Let M be an N-graded R-module and let N be a propergraded submodule of M . Then N has an irredundant primary decompositionN = N1 ∩ · · · ∩Nr such that Ni is a graded submodule for all i.

Proof. Use Lemma 2.2.6, Proposition 2.2.7, and Theorem 2.1.27. �

Finding primary decompositions of graded ideals in polynomial ringsover fields is a difficult task. For a treatment of the principles of primarydecomposition consult the book of Vasconcelos [413, Chapter 3].

Notation If R = ⊕∞i=0Ri is an N-graded ring, we set R+ = ⊕i≥1Ri.

Lemma 2.2.9 (Graded Nakayama lemma) Let R be an N-graded ring andlet M be an N-graded R-module. If N is a graded submodule of M andI ⊂ R+ is a graded ideal of R such that M = N + IM , then N =M .

Proof. Since M/N = I(M/N), one may assume N = (0). If x ∈ M isa homogeneous element of degree r, then a recursive use of the equalityM = IM yields that x ∈ Ir+1M and x must be zero. �

Exercises

2.2.10 Let I be a graded ideal of a polynomial ring R over a field K. Provethat the radical of I is also graded.

2.2.11 Let I be a graded ideal of a polynomial ring R = K[x1, . . . , xn] overa field K. If m = (x1, . . . , xn) is the irrelevant maximal ideal of R, provethat m ∈ Ass(R/I) if and only if depth(R/I) = 0.

Hint Use Lemma 2.1.19.

2.2.12 Let R = ⊕i∈ZRi be a graded ring. Prove that R0 is a subring ofR with 1 ∈ R0. A graded ring R with Ri = 0 for i < 0 is called a gradedR0-algebra.

2.2.13 Let R = K[x1, x2] be a polynomial ring over a field K with thegrading induced by deg(xi) = (−1)i. Determine the subring R0.


2.3 Cohen–Macaulay modules

Here we introduce some special types of rings and modules and present thefollowing fundamental result of dimension theory.

Theorem 2.3.1 (Dimension theorem [310, Theorem 13.4]) Let (R,m) be alocal ring and let M be an R-module. Set

δ(M) = min{r| there are x1, . . . , xr ∈ m with �R(M/(x1, . . . xr)M) <∞},

then dim(M) = δ(M).

Definition 2.3.2 Let (R,m) be a local ring and let M be an R-moduleof dimension d. A system of parameters (s.o.p for short) of M is a set ofelements θ1, . . . , θd in m such that �R(M/(θ1, . . . , θd)M) <∞.

Corollary 2.3.3 Let (R,m) be a local ring and let M be an R-module ofdimension d. If h1, . . . , hd ∈ m is a system of parameters of M , then

dim M/(h1, . . . , hi)M = d− i for 1 ≤ i ≤ d.

Proof. Set M = M/(h1, . . . , hi)M . First note that hi+1, . . . , hd is a s.o.pfor M , hence dim(M) ≤ d − i by Theorem 2.3.1. On the other hand ifθ1, . . . , θr is a s.o.p for M , then θ1, . . . , θr, h1, . . . , hi is a s.o.p for M andagain by the dimension theorem one has d ≤ r + i = dim(M) + i. �

Definition 2.3.4 Let M be an R-module. A sequence θ = θ1, . . . , θn in Ris called a regular sequence of M or an M -regular sequence if (θ)M �= Mand θi /∈ Z(M/(θ1, . . . , θi−1)M) for all i.

Proposition 2.3.5 Let M be an R-module and let I be an ideal of R suchthat IM �= M . If θ = θ1, . . . , θr is an M -regular sequence in I, then θ canbe extended to a maximal M -regular sequence in I.

Proof. By induction assume there is an M -regular sequence θ1, . . . , θi inI for some i ≥ r. Set M = M/(θ1, . . . , θi)M . If I �⊂ Z(M), pick θi+1 in Iwhich is regular on M . Since

(θ1) ⊂ (θ1, θ2) ⊂ · · · ⊂ (θ1, . . . , θi) ⊂ (θ1, . . . , θi+1) ⊂ R

is an increasing sequence of ideals in a Noetherian ring R, this inductiveconstruction must stop at a maximal M -regular sequence in I. �

Lemma 2.3.6 Let M be a module over a local ring (R,m). If θ1, . . . , θr isan M -regular sequence in m, then r ≤ dim(M).

80 Chapter 2

Proof. By induction on dim(M). If dim(M) = 0, then m is an associatedprime of M and every element of m is a zero divisor of M . We claim thatdim(M/θ1M) < dim(M). If this equality does not hold, there is a saturatedchain of prime ideals

ann (M) ⊂ ann (M/θ1M) ⊂ p0 ⊂ · · · ⊂ pd,

where d is the dimension of M and p0 is minimal over ann (M). Accordingto Proposition 2.1.20 the ideal p0 consists of zero divisors, a contradictionsince θ1 ∈ ann (M/θ1M) ⊂ p0. This proves the claim. Since θ2, . . . , θr is aregular sequence on M/θ1M by induction one derives r ≤ dim(M). �

Proposition 2.3.7 Let M be an R-module and let I be an ideal of R.

(a) HomR(R/I,M) = (0) iff there is x ∈ I which is regular on M .

(b) ExtrR(R/I,M) � HomR(R/I,M/θM), where θ = θ1, . . . , θr is anyM -regular sequence in I.

Proof. (a) ⇒) Assume I ⊂ Z(M). Using Lemma 2.1.19 one has I ⊂ p forsome p ∈ AssR(M). Hence there is a monomorphism ψ : R/p → M . Toderive a contradiction note that the composition

R/Iϕ−→ R/p

ψ−→M

is a non-zero map, where ϕ is the canonical map from R/I to R/p. Theconverse is left as an exercise.

(b) Consider the exact sequence

0 −→Mθ1−→M −→M =M/θ1M −→ 0.

According to [363, Theorem 7.3] there is a long exact sequence with naturalconnecting homomorphisms

· · · −→ Extr−1R (R/I,M)

θ1−→ Extr−1R (R/I,M) −→ Extr−1

R (R/I,M)∂−→

ExtrR(R/I,M)θ1−→ ExtrR(R/I,M) −→ · · ·

Since θ1 is in I, using [363, Theorem 7.16] it follows that in the last exactsequence the maps given by multiplication by θ1 are zero. Hence

Extr−1R (R/I,M) � ExtrR(R/I,M),

and the proof follows by induction on r. �

Let M �= (0) be a module over a local ring (R,m). The depth of M ,denoted by depth(M), is the length of any maximal regular sequence on Mwhich is contained in m. From Proposition 2.3.7 one derives

depth(M) = inf{r|ExtrR(R/m,M) �= (0)}.

In general by Lemma 2.3.6 we have depth(M) ≤ dim(M).


Definition 2.3.8 An R-module M is called Cohen–Macaulay (C–M forshort) if depth(M) = dim(M), or if M = (0).

Lemma 2.3.9 (Depth lemma [413, p. 305]) If 0→ N →M → L→ 0 is ashort exact sequence of modules over a local ring R, then

(a) If depth(M) < depth(L), then depth(N) = depth(M).

(b) If depth(M) = depth(L), then depth(N) ≥ depth(M).

(c) If depth(M) > depth(L), then depth(N) = depth(L) + 1.

Lemma 2.3.10 If M is a module over a local ring (R,m) and z ∈ m is aregular element of M , then

(a) depth(M/zM) = depth(M)− 1, and

(b) dim(M/zM) = dim(M)− 1.

Proof. As depthM > depthM/zM applying the depth lemma to the exactsequence

0 −→Mz−→M −→M/zM −→ 0

yields depth(M) = depth(M/zM)+1. To prove the second equality firstobserve the inequality dim(M) > dim(M/zM), which follows from the proofLemma 2.3.6 by making z = θ1. On the other hand the reverse inequalitydim(M) ≤ dim(M/zM) + 1 follows from Theorem 2.3.1. �

Proposition 2.3.11 [65, p. 58] If M is a Cohen–Macaulay R-module, thenS−1M is Cohen–Macaulay for every multiplicatively closed set S of R.

Proposition 2.3.12 [65, p. 58] Let M be an R-module and let x be anM -regular sequence. If M is Cohen–Macaulay, then M/xM is Cohen–Macaulay (over R or R/(x)). The converse holds if R is local.

Proposition 2.3.13 [309, Theorem 30] If M �= (0) is a Cohen–MacaulayR-module over a local ring R and p ∈ Ass(M), then dim(R/p) = depth(M).

Definition 2.3.14 Let (R,m) be a local ring of dimension d. A system ofparameters (s.o.p) of R is a set θ1, . . . , θd generating an m-primary ideal.

Theorem 2.3.15 Let (R,m, k) be a local ring. If

δ(R) := min{μ(I) = dimk(I/mI) | I is an ideal of R with rad (I) = m},

then dim(R) = δ(R).

Proof. This result follows readily from Theorem 2.3.1. �

82 Chapter 2

Theorem 2.3.16 (Krull principal ideal theorem) Let I be an ideal of aring R generated by a sequence h1, . . . , hr. Then

(a) ht (p) ≤ r for any minimal prime p of I.

(b) If h1, . . . , hr is a regular sequence, then ht (p) = r for any minimalprime p of I.

Proof. (a): Since√IRp = pRp from Theorem 2.3.15 one has the inequality

ht (p) = dim(Rp) ≤ r.(b): Set J = (h1, . . . , hi−1) and L = (h1, . . . , hi). Assume ht (P ) = i− 1

for any minimal prime P over J . Since hi is regular on R/J and L/J is aprincipal ideal one has ht (L/J) = 1, thus there is a prime ideal p0 minimalover J such that J ⊂ p0 ⊂ p and consequently ht (p) ≥ i. Using (a) onegets ht (p) = i. �

Corollary 2.3.17 If (R,m, k) is a local ring, then dim(R) ≤ dimk(m/m2)

and R has finite Krull dimension.

Proof. Let x1, . . . , xq be a set of elements in m whose images in m/m2

form a basis of this vector space. Then m = (x1, . . . , xq) + m2. Henceby Nakayama’s lemma we get m = (x1, . . . , xq). Hence dim R ≤ q byTheorem 2.3.15. �

Remark If R is a Noetherian ring, then dim Rp < ∞ for all p ∈ Spec(R).There are examples of Noetherian rings of infinite Krull dimension [9].

Definition 2.3.18 A local ring (R,m, k) is called regular if

dim(R) = dimk(m/m2).

A ring R is regular if Rp is a regular local ring for every p ∈ Spec(R).

Cohen–Macaulay rings A local ring (R,m) is called Cohen–Macaulayif R is Cohen–Macaulay as an R-module. If R is non local and Rp is a C–Mlocal ring for all p ∈ Spec(R), then we say that R is a Cohen–Macaulay ring.An ideal I of R is Cohen–Macaulay if R/I is a Cohen–Macaulay R-module.

If R is a Cohen–Macaulay ring and S is a multiplicatively closed subsetof R, then S−1(R) is a Cohen–Macaulay ring (see [65, Theorem 2.1.3]).

Proposition 2.3.19 Let M be a module of dimension d over a local ring(R,m) and let θ = θ1, . . . , θd be a system of parameters of M . Then M isCohen–Macaulay if and only if θ is an M -regular sequence.

Proof. ⇒) Let p be an associated prime of M . By Proposition 2.3.13one has dim(R/p) = d. We claim that θ1 is not in p. If θ1 ∈ p, then byNakayama’s lemma one has (M/θ1M)p �= (0). Hence p is in the support


of M/θ1M , a contradiction because by Corollary 2.3.3 one has the equalitydim(M/θ1M) = d− 1. Therefore θ1 �∈ p and θ1 is regular on M . Thus theproof follows by induction, because according to Lemma 2.3.10 one has thatM/θ1M is a C–M module of dimension d − 1 and the images of θ2, . . . , θdin R/θ1 form a system of parameters of M/(θ1)M .⇐) By Corollary 2.3.3 one has dim(M/θM) = 0 and M/(θ)M is C–M.

Therefore M is Cohen–Macaulay by Lemma 2.3.10. �

Lemma 2.3.20 Let (R,m) be a local ring and let (f1, . . . , fr) be an ideal ofheight equal to r. Then there are fr+1, . . . , fd in m such that f1, . . . , fd is asystem of parameters of R.

Proof. Set d = dim(R) and I = (f1, . . . , fr). One may assume r < d,otherwise there is nothing to prove. Let p1, . . . , ps be the minimal primesof I. Note that ht (pi) = r for all i by Theorem 2.3.16. Hence if we pickfr+1 in m \ ∪si=1pi, one has the equality ht (I, fr+1) = r + 1, and the resultfollows by induction. �

Definition 2.3.21 Let R be a ring and let I be an ideal of R. If I isgenerated by a regular sequence we say that I is a complete intersection.

Definition 2.3.22 An ideal I of a ring R is called a set-theoretic completeintersection if there are f1, . . . , fr in I such that rad (I) = rad (f1, . . . , fr),where r = ht (I).

Definition 2.3.23 An ideal I of a ring R is height unmixed or unmixed ifht (I) = ht (p) for all p in AssR(R/I).

Proposition 2.3.24 Let (R,m) be a Cohen–Macaulay local ring and letI be an ideal of R. If I is a complete intersection, then R/I is Cohen–Macaulay and I is unmixed.

Proof. Set d = dim(R) and r = ht (I). By Lemma 2.3.10 R/I is Cohen–Macaulay and dim(R/I) = d−r. Let p be an associated prime of R/I, thenusing Proposition 2.3.13 yields

dim(R)− ht (p) ≥ dim(R/p) = depth(R/I) = d− r.

As a consequence ht (p) ≤ r and thus ht (p) = r. Hence I is unmixed. �

Theorem 2.3.25 Let R be a Cohen–Macaulay ring and I a proper ideal ofR of height r. If I is generated by r elements f1, . . . , fr, then I is unmixed.

Proof. It is enough to prove that I has no embedded primes, becauseby Krull’s theorem all minimal primes of I have height r. Let p and q betwo associated primes of R/I and assume p ⊂ q, then IRq ⊂ pRq ⊂ qRq.

84 Chapter 2

Since IRq has height r and is generated by r elements, one obtains (usingLemma 2.3.20) that f1/1, . . . , fr/1 is part of a system of parameters of Rq.Therefore by Proposition 2.3.19 the sequence f1/1, . . . , fr/1 is a regularsequence and thus Proposition 2.3.24 proves that IRq is unmixed. Noticingthat pRq and qRq are both associated primes of IRq, one derives pRq = qRq

and hence p = q. �

Remark 2.3.26 The case r = 0 in the statement of Theorem 2.3.25 meansthat (0) is unmixed.

Theorem 2.3.27 (Unmixedness theorem [310, Theorem 17.6]) A ring Ris Cohen–Macaulay if and only if every proper ideal I of R of height rgenerated by r elements is unmixed.

Proposition 2.3.28 If I is an ideal of height r in a Cohen–Macaulay ringR, then there is a regular sequence f1, . . . , fr in I.

Proof. By induction assume that f1, . . . , fs is a regular sequence in Isuch that s < r. Note that (f1, . . . , fs) is unmixed by Theorem 2.3.25. IfI ⊂ Z(R/(f1, . . . , fs)), then ht (I) = s, which is a contradiction. Thereforethere is an element fs+1 in I which is regular modulo (f1, . . . , fs). �

Let A be a (Noetherian) ring one says that A is a catenary ring if forevery pair p ⊂ q of prime ideals ht (q/p) is equal to the length of anymaximal chain of prime ideals between p and q. If A is a domain, then Ais catenary if and only if ht (q/p) = ht (q) − ht (p) for every pair of primeideals p ⊂ q.

Theorem 2.3.29 [65, Theorem 2.1.12] If R is a Cohen–Macaulay ring,then R is catenary.

Proposition 2.3.30 Let A and B be two rings and let C = A × B. If Cis a Cohen–Macaulay ring, then A is Cohen–Macaulay.

Proof. From the ring homomorphism A × B → A, (a, b) �→ a, we getA×B/{0}×B � A. It follows that P is a prime ideal of C containing theideal {0} ×B if and only if P = p×B for some prime ideal p of A.

Let I � A be an ideal and let I ′ = I × B. We are going to establisha correspondence between the associated primes of I and I ′. If p is anassociated prime ideal of I, there is z ∈ A such that p = (I : z). Notice thatp × B = (I ′ : (z, 1)), thus p × B is an associated prime of I ′. Conversely ifP is an associated prime of I ′, then there is a prime ideal p containing Iand a pair (z1, z2) in C such that P = p × B = (I ′ : (z1, z2)). Thus p is anassociated prime of I because p = (I : z1).

Assume that I = (x1, . . . , xr) is an ideal of A of height r generated by relements. By Theorem 2.3.27 it suffices to prove that I is unmixed.


Case (I): Assume r = 0, that is I = (0). Any associated prime ideal of I ′

consists of zero divisors only. Since C is C–M, the ideal I ′ has no embeddedprimes. Hence I has no embedded primes by the correspondence betweenassociated primes of I and I ′ established above.

Case (II): Assume r ≥ 1 and set s = ht(I ′). The ideal I ′ is generated byr elements. Indeed if (x, y) ∈ I ′, then

(x, y) = (a1, y)(x1, 1) + (a2, 0)(x2, 0) + · · ·+ (ar, 0)(xr, 0) (ai ∈ A).

This means that I ′ = ((x1, 1), (x2, 0), . . . , (xr , 0)) and by Krull theorem weobtain s ≤ r. Pick a minimal prime ideal P of I ′ such that s = ht(P ). Thereis a prime ideal p such that P = p×B. It is seen that p is a minimal primeof I and by Krull theorem ht(p) = r. Thus there is a strictly descendingchain of prime ideals

pr � · · · � p0 = p =⇒ pr ×B � · · · � p0 ×B = p×B.

Therefore s = ht(P ) ≥ r, and we obtain that ht(I ′) = r. Thus since C isCohen–Macaulay, the ideal I ′ has no embedded primes and consequently Ihas no embedded primes either; by the correspondence between associatedprimes of I and I ′. �

Gorenstein rings Let M �= (0) be a module over a local ring (R,m) andlet k = R/m be the residue field of R. The socle of M is defined as

Soc(M) = (0: Mm) = {z ∈M |mz = (0)},

and the type of M is defined as type(M) = dimk Soc(M/xM), where x isa maximal M -sequence in m. Observe that the type of M is well-definedbecause by Proposition 2.3.7 one has:

ExtrR(k,M) � HomR(k,M/xM) � Soc(M/xM),

where r = depth(M). The ring R is said to be Gorenstein if R is a Cohen–Macaulay ring of type 1. An ideal I ⊂ R is called Gorenstein if R/I is aGorenstein ring. For a thorough study of Gorenstein rings see [18, 65].

Exercises

2.3.31 Let φ : A → B be an epimorphism of rings. If dim(A) = dim(B) <∞ and A is a domain, prove that φ is injective.

2.3.32 Let R be a catenary ring. If S ⊂ R is a multiplicatively closed setand I is an ideal of R, prove that R/I and S−1R are both catenary rings.

2.3.33 Let I be an ideal of a ring R and let x ∈ R \ I. If I is unmixed andI = rad (I), then (I : x) is unmixed and ht(I) = ht(I : x).

86 Chapter 2

2.3.34 Let M = Zr × Zpα11× · · · × Zpαm

mbe a finitely generated Z-module,

where p1, . . . , pm are prime numbers. Prove the following:

(a) M has finite length iff r = 0.

(b) �(M) = α1 + · · ·+ αm, if �(M) <∞.

(c) Ass(M) �= Supp(M) if r �= 0.

(d) Z(M) = ∪mi=1(pi).

(e) Ass(M) = {(p1), . . . , (pm), (0)} if r �= 0.

(f) Ass(M) = {(p1), . . . , (pm)} if r = 0.

2.3.35 (Cayley–Hamilton) Let ϕ : M →M be an endomorphism of an R-module M . If I is an ideal of R such that ϕ(M) ⊂ IM , then ϕ satisfies anequation of the form ϕn + an−1ϕ

n−1 + · · ·+ a1ϕ+ a0 = 0 (ai ∈ I).

2.3.36 If R is a principal ideal domain, prove that R is a regular ring.

2.3.37 Let M be a module over a local ring (R,m). If θ = θ1, . . . , θr isan M -regular sequence in m, prove that θ can be extended to a system ofparameters of M .

2.3.38 Let M be an R-module and let I be an ideal of R. If there is x ∈ Iwhich is regular on M , show that HomR(R/I,M) = (0).

2.3.39 Let M be an R-module and let I be an ideal of R. If J ⊂ I, then:

HomR(R/I,M/JM) � (JM : MI)/JM.

2.3.40 If A and B are two Cohen–Macaulay rings, prove that A × B is aCohen–Macaulay ring.

2.4 Normal rings

Let A and B be two rings. One says that B is an A-algebra if there is ahomomorphism of rings

ϕ : A −→ B.

Note that B has an A-module structure (compatible with its ring structure)given by a · b = ϕ(a)b for all a ∈ A and b ∈ B.

Thus if A ⊂ B is a ring extension, that is, A is a subring of B, then Bhas in a natural way an A-algebra structure induced by the inclusion map.If A = K is a field and B is a K-algebra, then ϕ is injective; in this caseone may always assume that K ⊂ B is a ring extension.

Given F = {f1, . . . , fq} a finite subset of B, we denote the subring ofB generated by F and ϕ(A) by A[F ] = A[f1, . . . , fq]. Sometimes A[F ] is


called the A-subalgebra or A-subring of B spanned by F . If B = A[F ], onesays that B is an A-algebra of finite type or a finitely generated A-algebra;on the other hand one says that ϕ is finite or that B is a finite A-algebra ifB is a finitely generated A-module.

A homomorphism φ : B → C of A-algebras is a map φ which is botha homomorphism of rings and a homomorphism of A-modules. Note thatφ(a · 1) = a · 1 for all a ∈ A.

Definition 2.4.1 Let A be a subring of B. An element b ∈ B is integralover A, if there is a monic polynomial 0 �= f(x) ∈ A[x] such that f(b) = 0.The set A of all elements b ∈ B which are integral over A is called theintegral closure of A in B.

Let A be a subring of B, one says that A ⊂ B is an integral extensionof rings or that B is integral over A, if b is integral over A for all b ∈ B.A homomorphism of rings ϕ : A→ B is called integral if B is integral overϕ(A); in this case one also says that B is integral over A.

Proposition 2.4.2 Let A be a subring of B. If B is a finitely generatedA-module, then B is integral over A.

Proof. There are α1, . . . , αn ∈ B such that B = Aα1 + · · · + Aαn. Letβ ∈ B. Then, one can write βαi =

∑nj=1mijαj , where mij ∈ A. Set

M = (mij), N =M −βI and α = (α1, . . . , αn). Here I denotes the identitymatrix. Since Nα� = 0, one can use the formula Nadj(N) = det(N)I toconclude αi det(N) = 0 for all i. Hence det(N) = 0. To complete the proofnote that

f(x) = (−1)n det(M − xI)

is a monic polynomial in A[x] and f(β) = (−1)n det(N) = 0. �

Corollary 2.4.3 If A is a subring of B, then the integral closure A of Ain B is a subring of B.

Proof. Let α, β ∈ A. If α and β satisfy monic polynomials with coefficientsin A of degree m and n, respectively, then A[α, β] is a finitely generated A-module with basis {αiβj | 0 ≤ i ≤ m − 1 and 0 ≤ j ≤ n − 1}. Hence, byProposition 2.4.2, A[α, β] is integral over A. In particular α±β and αβ areintegral over A. �

Corollary 2.4.4 If B is an A-algebra of finite type, then B is integral overA if and only if B is finite over A.

Proof. ⇒) It follows by the arguments given in the proof of Corollary 2.4.3.⇐) It follows from Proposition 2.4.2. �

88 Chapter 2

Lemma 2.4.5 Let A ⊂ B be a ring extension. If B is a domain and b ∈ Bis integral over A, then A[b] is a field if and only if A is a field.

Proof. ⇒) Let a ∈ A \ {0} and c = a−1 its inverse in A[b]. Since A[b] isintegral over A, there is an equation

cn + an−1cn−1 + · · ·+ a1c+ a0 = 0 (ai ∈ A),

multiplying by an it follows rapidly that c ∈ A.⇐) Let A[x] be a polynomial ring and ψ : A[x]→ A[b] the epimorphism

given by ψ(f(x)) = f(b). As ker(ψ) is a non-zero prime ideal and A[x] is aprincipal ideal domain, ker(ψ) is a maximal ideal and A[b] is a field. �

Proposition 2.4.6 Let A ⊂ B be an integral extension of rings. If B is adomain, then A is a field if and only if B is a field.

Proof. It follows from Lemma 2.4.5. �

Corollary 2.4.7 Let A ⊂ B be an integral extension of rings. If P is aprime ideal of B and p = P ∩A, then P is a maximal ideal of B if and onlyif p is a maximal ideal of A.

Proof. The result is a direct consequence of Proposition 2.4.6, becauseA/p ⊂ B/P is an integral extension of rings. �

Corollary 2.4.8 Let A ⊂ B be an integral extension of rings. If P ⊂ Qare two prime ideals of B such that P ∩ A = Q ∩ A, then P = Q.

Proof. Set p = P ∩ A. As PBp ∩ Ap = pAp and Bp is integral over Ap,using Corollary 2.4.7 we get that PBp is maximal and hence P = Q. �

Lemma 2.4.9 If A ⊂ B is an integral extension of rings and p is a maximalideal of A, then pB �= B and p = P ∩ A for any maximal ideal P of Bcontaining pB.

Proof. If pB = B, one can write 1 = a1b1 + · · · + aqbq with ai ∈ pand bi ∈ B. Set S = A[b1, . . . , bq], the subring of B generated by b1, . . . , bq.Since S is a finitely generated A-module and pS = S, then by Lemma 2.1.33there is x ≡ 1(mod p) such that xS = (0). As 1 ∈ B, one derives x = 0 and1 ∈ p, which is impossible. Hence pB �= B. Note that this part of the proofholds if p is a proper ideal of A. If pB ⊂ P is a maximal ideal of B, thenp ⊂ P ∩ A and thus p = A ∩ P by the maximality of p. �

Proposition 2.4.10 (Lying over) If A ⊂ B is an integral extension and pis a prime ideal of A, then there is a prime ideal P of B such that p = P ∩A.


Proof. Since Ap ⊂ Bp is an integral extension, then by Lemma 2.4.9pBp �= Bp. Note that P ∩A = p for any prime ideal P of B such that PBp

is prime and contains pBp (see Exercise 2.4.22). �

Proposition 2.4.11 If A ⊂ B is an integral extension of rings and A islocal, then B is semilocal.

Proof. Let m be the maximal ideal of A. By Corollary 2.4.7 every primeideal of B/mB is maximal. Hence, B/mB is an Artinian semilocal ringby Proposition 2.1.43. Since the maximal ideals of B are in one to onecorrespondence with the maximal ideals of B/mB, the result follows. �

Theorem 2.4.12 (Going-up) Let A ⊂ B be an integral ring extension. Ifp ⊂ q are two prime ideals of A and p = P ∩ A for some P in Spec(B),then there is Q ∈ Spec(B) such that q = Q ∩ A and P ⊂ Q.

Proof. Since A/p ⊂ B/P is an integral extension, by Proposition 2.4.10one has q/p = (Q/P )∩ (A/p) for some prime ideal Q of B containing P . Itfollows readily that q = A ∩Q. �

Proposition 2.4.13 Let A ⊂ B be an integral extension of rings. If B isintegral over A, then dim(A) = dim(B).

Proof. The formula follows using Theorem 2.4.12, Corollary 2.4.8, andProposition 2.4.10. �

Let A be an integral domain and let KA be its field of fractions. Theintegral closure or normalization of A, denoted A, is the set of all f ∈ KA

satisfying an equation of the form

fn + an−1fn−1 + · · ·+ a1f + a0 = 0 (ai ∈ A and n ≥ 1).

By Corollary 2.4.3 the integral closure A of A is a subring of KA. If A = Awe say that A is integrally closed or normal . If A is not a domain we saythat A is normal if Ap is a normal domain for every prime ideal p of A.Any normal ring is a direct product of finitely many normal domains [128].

Proposition 2.4.14 Let A ⊂ B be an integral extension of rings. If B isa domain and A is a normal domain, then

(a) (going-down) if p ⊂ q are two prime ideals of A and q = Q ∩ A forsome Q in Spec(B), then there is P ∈ Spec(B) such that p = P ∩ Aand P ⊂ Q.

(b) ht (I) = ht (I ∩A) for any ideal I of B.

(c) ht (I) = ht (IB) for any ideal I of A.

90 Chapter 2

Proof. For parts (a) and (b) see [309, Theorems 5 and 20]. Notice that,by part (a) and Proposition 2.4.10, A ⊂ B satisfies the going down and thelying over conditions. Hence, by [259, Proposition B.2.4], ht (I) = ht (IB)for any ideal I of A. This proves (c). �

Theorem 2.4.15 (Serre’s criterion [310, Theorem 23.8]) A ring A is nor-mal if and only if

(S2) depth(Ap) ≥ inf{2, ht (p)} for all p ∈ Spec(A), and

(R1) Ap is regular for all p ∈ Spec(A) with ht (p) ≤ 1.

Theorem 2.4.16 (Auslander–Buchsbaum [310, Theorem 20.3]) If (A,m)is a regular local ring, then A is a unique factorization domain.

Definition 2.4.17 Let R be a ring and let G ≤ Aut(R) be a group of ringautomorphisms of R. The invariant ring of G is:

RG = {r ∈ R|φ(r) = r, ∀φ ∈ G}.

Proposition 2.4.18 If R is a normal domain and G is a group of ringautomorphisms of R, then RG is a normal domain.

Proof. Let f be an element in the field of fractions of RG which is integralover RG. Since R is normal f ∈ R and r2f = r1 for some r1, r2 ∈ RG. Thus

r2φ(f) = φ(r2)φ(f) = φ(r1) = r1

for all φ ∈ RG. Hence f ∈ RG, as required. �

Example 2.4.19 If R = K[x1, . . . , xn] is a polynomial ring over a fieldK and G is a subgroup of GLn(K), then for every A in G there is a ringautomorphism φA : R→ R, f �→f(Ax), where x = (x1, . . . , xn)

t. Thus G is agroup of ring automorphisms of R and RG = {f ∈ R| f(Ax) = f, ∀A ∈ G}.

Flat and faithfully flat algebras Let φ : A → B be a homomorphismof rings and consider an exact sequence of A-modules

· · · −→ Ni−1fi−1−→ Ni

fi−→ Ni+1fi+1−→ · · · (N )

If the sequence

· · · −→ B ⊗A Ni−11⊗fi−1−→ B ⊗A Ni

1⊗fi−→ B ⊗A Ni+11⊗fi+1−→ · · · (B ⊗A N )

is exact for any N , B is called a flat A-algebra and φ is said to be flat .The homomorphism φ is said to be faithfully flat if any sequence N ofA-modules is exact if and only if B ⊗A N is exact. In a similar way onecan define flat and faithfully flat A-modules.

A basic property is that any free A-module (not necessarily finitely gen-erated) is faithfully flat. In particular a polynomial ring A[x] in severalvariables with coefficients in A is a faithfully flat A-algebra.


Theorem 2.4.20 [309, Sections 3.H, 5.D and 9.C] Let φ : A→ B be a flathomomorphism of rings. The following hold:

(a) (I1 ∩ I2)B = I1B ∩ I2B, for any two ideals I1, I2 of A.

(b) The going down theorem holds for φ.

(c) If q is a p-primary ideal of A such that pB is prime, then qB is apB-primary ideal.

Theorem 2.4.21 [309, Sections 4.C and 13.B] Let φ : A→ B be a faithfullyflat homomorphism of rings. The following hold:

(a) The map φ∗ : Spec(B)→ Spec(A), φ∗(P ) = φ−1(P ) is surjective.

(b) IB ∩ A = I and ht (I) = ht (IB) for any ideal I of A.

Exercises

2.4.22 Let A ⊂ B be a ring extension and p a prime ideal of A. If S = A\pand Bp = S−1B, prove that the map P �−→ PBp gives a bijection betweenthe set of prime ideals of P of B such that P ∩ A = p and the set of primeideals of Bp containing pBp.

2.4.23 Let K be a field and ϕ : A→ B an isomorphism of K-algebras, noteϕ(r) = r for all r ∈ K. If A = ⊕i≥0Ai is a graded K-algebra, then B is alsoa graded K-algebra graded by B = ⊕i≥0ϕ(Ai).

2.4.24 Let ϕ : A→ B be an isomorphism between two integral domains. IfKA (resp. KB) is the field of fractions of A (resp. B), then ϕ extends toan isomorphism ϕ from KA to KB such that ϕ(A) = B.

2.4.25 If A is a unique factorization domain, show that A is normal.

2.4.26 Show an example of an integral extension of rings A ⊂ B, such thatA is a field and B is not an integral domain (cf. Lemma 2.4.5).

2.4.27 Let B = A[x] be a polynomial ring over a ring A and let I be anideal ofA. Prove (A/I)[x] � B/IB, where the left-hand side is a polynomialring with coefficients in A/I.

2.4.28 If B = A[x] is a polynomial ring over a ring A and q is a p-primaryideal of A, then qB is a pB-primary ideal of B.

2.4.29 Let B = A[x] be a polynomial ring over a ring A and let I be an idealof A. If I = ∩ri=1qi is a primary decomposition of I, then IB = ∩ri=1qiB isa primary decomposition of IB.

2.4.30 If A ⊂ B is an integral extension of rings and A is semilocal, thenB is semilocal.

2.4.31 Let A be a domain and let x be an indeterminate over A. Then Ais normal if and only if A[x] is normal.

92 Chapter 2

2.5 Valuation rings

In this part we introduce valuation rings. Let A be an integral domain andlet K be its field of fractions.

Definition 2.5.1 A is a valuation ring of K if for each 0 �= x ∈ K eitherx ∈ A or x−1 ∈ A.

Proposition 2.5.2 Let A be a valuation ring of K. The following hold.

(a) If I, J are two ideals of A, then I ⊂ J or J ⊂ I.(b) A is a local ring.

(c) If A′ is an intermediate ring A ⊂ A′ ⊂ K, then A′ is a valuation ring.

(d) A is integrally closed in K.

Proof. (a) Assume I �⊂ J and pick x ∈ I \ J . Let y ∈ J . One hasy−1x /∈ A, because if y−1x ∈ A, then x = (xy−1)y ∈ J , a contradiction.Thus yx−1 ∈ A and consequently y = x(x−1y) ∈ I. This proves J ⊂ I, asrequired. Part (b) follows from (a).

(c) Let x ∈ K and assume that x /∈ A′. Then x /∈ A and consequentlyx−1 ∈ A. Thus x−1 ∈ A′. Hence, A′ is a valuation ring of K.

(d) Let x ∈ K be an integral element over A. There are a0, a1, . . . , an−1

in A such that xn + an−1xn−1 + · · ·+ a1x + a0 = 0. Assume x /∈ A. Then

x−1 ∈ A. Multiplying this equation by x1−n we get

x = −(an−1 + an−2x−1 + · · ·+ a1x

2−n + a0x1−n).

Thus x ∈ A, a contradiction. Therefore x ∈ A. �

Definition 2.5.3 An abelian group H with a total order ≺ is called anordered group if x ≺ y and u ≺ v implies x+ u ≺ y + v.

Example 2.5.4 Let G1, . . . , Gn be abelian ordered groups and let G be itscartesian product. To make G and ordered group for a = (ai), b = (bi) inG and a �= b we define a ≺ b if the first non-zero entry of b − a is positive.This defines a total order on G called the lexicographic order.

Let H be an abelian ordered group. We order H ∪ {∞} by adding anelement ∞ bigger than any element in H and declaring ∞ + x = ∞ forx ∈ H and ∞+∞ =∞.

Definition 2.5.5 Let H be an ordered group. A map ν : K → H ∪ {∞} iscalled a valuation of K if ν satisfies:

(1) ν(xy) = ν(x) + ν(y),


(2) ν(x + y) ≥ min{ν(x), ν(y)},(3) ν(x) =∞ if and only if x = 0.

The valuation ring of a valuation ν, denoted by Aν , is the set of all a ∈ Kwith ν(a) ≥ 0. The value group of ν is the subgroup ν(K \ {0}).

If ν is a valuation of a field K, then Aν is a valuation ring of K and theideal mν = {x ∈ K|ν(x) > 0} is its maximal ideal.

Example 2.5.6 Let p ≥ 2 be a prime number and let ν : Q→ Z ∪ {∞} bethe map defined by

ν(ab

)=

⎧⎨⎩ n if 0 �= a

b= pn

a1b1

and (a1b1, p) = 1,

∞ ifa

b= 0.

It is easy to see that ν is a valuation of Q whose valuation ring is Z(p), thelocalization of Z at the prime ideal p = (p).

Definition 2.5.7 Let G be an ordered group. A subgroup H of G is calledisolated if for each α ∈ H such that 0 ≺ β ≺ α, one has β ∈ H . If H is anisolated subgroup of G and H �= G we say that H is proper. The rank of Gis the number of isolated proper subgroups of G.

Definition 2.5.8 Let ν be a valuation of a field K. The rank of the valu-ation ring Aν is the rank of the value group of ν.

Example 2.5.9 The group G = Zn with the lexicographical order has rankn because the proper non-zero isolated subgroups of G have the form:

Hi = {(0, . . . , 0, xi, . . . , xn)|xi, . . . , xn ∈ Z} (i = 2, . . . , n).

Definition 2.5.10 A valuation ring whose value group is isomorphic to Zis called a discrete valuation ring (DVR for short).

Theorem 2.5.11 [310, Chapter 4] A is a discrete valuation ring if andonly if any of the following equivalent conditions hold.

(a) A is a Noetherian valuation ring.

(b) A is a normal Noetherian local ring of dimension 1.

(c) A is a Noetherian local ring, dim(A) ≥ 1 and the maximal ideal of Ais principal.

Theorem 2.5.12 [310, p. 82] Let A be a Noetherian domain. Then, thefollowing three conditions are equivalent:

(a) A is normal.

(b) Ap is a discrete valuation ring for each prime ideal p of A of height 1.

(c) All associated primes of non-zero principal ideals of A have height 1.

94 Chapter 2

Exercises

2.5.13 Let A be a valuation ring ofK. Prove that there is an ordered groupG and a valuation ν : K → G ∪ {∞} such that the valuation ring of ν is Aand G is its value group.

2.5.14 Let K(x) be the field of rational functions in one variable over afield K. For f(x) ∈ K[x], write f(x) = xmg(x), where m ∈ N and g(x) is apolynomial with g(0) �= 0 and define ν(f) = m. Prove that extend ν to adiscrete valuation of K(x) by setting ν(f/g) = ν(f)− ν(g).

2.5.15 Let G be an ordered abelian group and let K[{xi|i ∈ G}] be apolynomial ring over a fieldK with quotient fieldQ. Define ν : Q→ G∪{∞}as follows

ν(f) =

⎧⎪⎨⎪⎩k1g1 + · · ·+ kngn if f = axk1g1 · · ·xkngn is a monomial,mini{ν(fi)} if f = f1 + · · ·+ fn is a sum of monomials,

∞ if f = 0.

Prove that ν is a valuation of Q whose value group is G.

2.5.16 If (A,m) is a discrete valuation ring, prove that m = (x) for somex ∈ A and that every non-zero ideal of A is a power of m.

2.6 Krull rings

In this section we study the notion of a Krull ring and introduce its divisorclass group. The main references for this section are [161, 309, 310].

Let A be an integral domain with quotient field K and let Z = X(1)(A)be the set of prime ideals of A of height 1.

Definition 2.6.1 An integral domain A is called a Krull ring if the follow-ing three conditions are satisfied:

(a) Ap is a discrete valuation ring for p ∈ Z.(b) A = ∩p∈ZAp.

(c) Each 0 �= x ∈ A belongs to at most a finite number of ideals in Z.

Theorem 2.6.2 (Mori–Nagata [161], [309]) Let A be an integral domainwith quotient field K and let L be a finite algebraic field extension of K. IfA is Noetherian, then the integral closure A′ of A in L is a Krull ring (notnecessarily Noetherian). If dim(A) = 2, then A′ is Noetherian.

Corollary 2.6.3 If A is Noetherian and normal, then A is a Krull ring.


Theorem 2.6.4 (Krull–Akizuki [309, p. 297]) Let A be a domain withquotient field K and let L be a finite field extension of K. If A is Noetherianand dim(A) = 1, then every subring between A and L is Noetherian.

Theorem 2.6.5 [309, Theorem 38] If A is a Noetherian normal domain,then A = ∩p∈ZAp.

Theorem 2.6.6 [310, Theorem 12.4(iii)] If A is a Krull ring, then A[x]and A[[x]] are Krull rings.

Proposition 2.6.7 Let A be a valuation ring. If dim(A) ≥ 2, then theformal power series ring A[[x]] is not normal.

Proof. Let (0) � p1 � p2 be a chain of prime ideals of A. Pick 0 �= b ∈ p1and a ∈ p2 \ p1. Then ba−n ∈ A for n > 0. Indeed if ba−n /∈ A, thenan ∈ bA ⊂ p1 and a ∈ p1, a contradiction. It is not hard to see that thereis f =

∑∞i=1 uix

i that satisfies the equation z2 + az + x = 0 and such thatu1 = a−1 and ui ∈ a−2i+1A for i ≥ 2. Thus bf ∈ A[[x]] and consequentlyf is in the field of fractions of A[[x]]. Since a−1 /∈ A we get that f /∈ A[[x]].Thus A[[x]] is not normal. �

Corollary 2.6.8 Let A be a valuation ring. If dim(A) ≥ 2, then A is nota Krull ring.

Proof. If A is a Krull ring, then A[[x]] is a Krull ring by Theorem 2.6.6. Inparticular A[[x]] is normal, a contradiction to Proposition 2.6.7. �

Divisor class groups of Krull rings Let p ∈ Z such thatAp is a discretevaluation ring. The maximal ideal pAp of the local ring Ap is principal and∩∞i=1p

i = (0) because Ap is a Noetherian domain. Let 0 �= x ∈ A. Sinceany non-zero ideal of Ap is a power of pAp, there is a unique nonnegativeinteger np satisfying (x)Ap = pnpAp and (x)Ap �= pnp+1Ap. This defines avaluation vp : K

∗ −→ Z of K given by

vp(x) =

{np if 0 �= x ∈ A,vp(a)− vp(b) if x = a/b; a, b ∈ A \ {0}.

The valuation vp is called the associated valuation of p.

Lemma 2.6.9 Let A be a Krull ring and let 0 �= a ∈ A. Then

aA =⋂p∈Z

p(np),

where np = vp(a) and p(n) := pnAp ∩ A is the symbolic power of order n.

96 Chapter 2

Proof. From the equality A = ∩p∈ZAp we get aA = ∩p∈Z(aAp ∩ A). AsaAp = pnpAp, we conclude that p(np) = aAp ∩ A as desired. �

Definition 2.6.10 Let A be a Krull ring and let D(A) be the free abeliangroup on Z. That is D(A) consists of the formal sums

∑p∈Z np · p (with

np ∈ Z and np = 0 except for a finite number) with the sum defined by(∑np · p

)+(∑

n′p · p

)=∑

(np + n′p) · p.

For each 0 �= x in K, we set

div(x) :=∑p∈Z

vp(x) · p.

The function div gives a group homomorphism from K∗ to D(A), becausediv(xy) = div(x) + div(y). The divisor class group of A is defined as

Cl(A) := D(A)/F (A),

where F (A) is the image of K∗ under the homomorphism div.

Lemma 2.6.11 If A is a unique factorization domain and 0 �= p ∈ A is aprime element, then ht(p) = 1.

Proof. Assume there is a prime ideal q such that (0) � q � (p). Take0 �= x ∈ q. There is a prime divisor q of x that belongs to q. Thus q ∈ (p)and p = uq for some unit u ∈ A, a contradiction. �

The next result is a measure, in terms of Cl(A), of how far is A frombeing a unique factorization domain.

Theorem 2.6.12 An integral domain A is a unique factorization domainif and only if A is a Krull ring and Cl(A) = (0).

Proof. ⇒) As A is a unique factorization domain the notions of primeelement and irreducible element coincide. Hence every prime ideal of A ofheight 1 is principal. To prove this assertion take p ∈ Z. Let 0 �= x ∈ p.Notice that some prime factor x1 of x belongs to p. Thus (x1) = p.

Next we show that A is a Krull ring. Let p ∈ Z and let x ∈ A such thatp = (x). Clearly Ap is normal and dim(Ap) = 1. Let 0 � J � Ap be anideal of Ap. Set n equal to the smallest integer n > 0 such that xn ∈ J , itfollows readily that J = (xn). This proves that Ap is a discrete valuationring. Now we show the equality

A =⋂p∈Z

Ap.


Let 0 �= x = a/b with a, b ∈ A. We may assume that a and b are relativelyprime. Assume that x belongs to the right-hand side of the equality above.To prove that x ∈ A it suffices to show that b is a unit of A. If b is not a unitpick a prime divisor p of b and set p = (p), then b ∈ p and by Lemma 2.6.11one has p ∈ Z. Since a/b ∈ Ap we get a ∈ p, a contradiction because a andb are relatively prime. Thus b is a unit and x ∈ A. Since A is a UFD every0 �= x ∈ A is contained in at most a finite number of principal prime ideals.Altogether we have shown that A is a Krull ring. It remains to show thatCl(A) = (0). This follows immediately by observing that if p0 = (x0) ∈ Z,then as an element of D(A) one has p0 = div(x0).⇐) Let p0 ∈ Z. We claim that p0 is a principal ideal. As Cl(A) = (0),

there exists 0 �= x0 ∈ K such that

div(x0) =∑p∈Z

vp(x0) · p = p0.

Hence vp0(x0) = 1 and vp(x0) = 0 for p �= p0. Writing x0 = a/b, witha, b ∈ A∗ we get vp0(a) = vp0(b) + 1 and vp(a) = vp(b) for p �= p0. ByLemma 2.6.9, we have the equalities

aA =⋂p∈Z

p(vp(a)) and bA =⋂p∈Z

p(vp(b)).

Consequently, using p(vp0(b)+1) ⊂ p(vp0(b)), one has aA ⊂ bA. This proves

x0 ∈ A. Again by Lemma 2.6.9 we get x0A = p(1)0 = p0. This proves that

p0 is principal.Let 0 �= x ∈ A be a non-unit of A. Notice there exists at least one

ideal in Z that contains x. Otherwise consider x−1, the inverse of x in K,and observe that x−1 ∈ ∩p∈ZAp = A, a contradiction because x is not aunit of A. Let p1 = (x1), . . . , ps = (xs) be the ideals in Z that contain x.There are elements λ1, . . . , λs in A such that x = λ1x1 = · · · = λsxs. Noticethat λ1x1 ∈ (x2). Since (x2) is a prime ideal of height 1 we get λ1 ∈ (x2).Using this argument we readily obtain that x = λxa11 · · ·xass with λ ∈ A, inaddition since ∩∞i=1p

ik = (0) for all k we may assume that xak+1

k does notdivide x for all k. It follows that λ is a unit of A, because if λ is not a unitit must belong to one of the ideals p1, . . . , ps which contradicts the choiceof a1, . . . , as. Clearly the elements x1, . . . , xs are irreducible and we haveproved that x is a product of irreducible elements. Finally assume that xhas two representations

x = y1 · · · yr = z1 · · · zm

with yi and zj irreducible elements for all i, j. Notice that every irreducibleelement z ∈ A is prime. Indeed since z is contained in at least one primeideal p = (w) of height 1, we get z = λw. As z is irreducible it follows that

98 Chapter 2

λ is a unit and z is prime. Therefore it follows readily that r = m and thereis a permutation σ such that yi is associated of zσ(i) for all i. �

Corollary 2.6.13 If K is a field and K[X ] is a ring of polynomials in anynumber of variables (finite or infinite), then K[X ] is a Krull ring.

Definition 2.6.14 An A-submodule I of K is called a fractional ideal ofA if I �= (0) and xI ⊂ A for some 0 �= x ∈ A (or equivalently 0 �= x ∈ K).If I is a fractional ideal of A define I−1 = (A : KI) = {x ∈ K|xI ⊂ A}. Wesay that a fractional ideal I is divisorial if (I−1)−1 = I.

Remark 2.6.15 An A-submodule I �= (0) of K is a fractional ideal of Aif and only if there is an ideal J of A and z ∈ K such that I = zJ . Thisfollows readily using that if xI ⊂ A for some x ∈ K∗, then J = xI is anideal of A and I = x−1J .

Lemma 2.6.16 If I is a fractional ideal of A, then I−1 is a fractional idealof A and II−1 ⊂ A.

Definition 2.6.17 A fractional ideal I of A is invertible if II−1 = A.

A fractional ideal of the form xA with x ∈ K∗ is called principal . Everyprincipal fractional ideal is invertible and divisorial.

Lemma 2.6.18 Let I be an invertible fractional ideal of A. If IJ = A forsome fractional ideal J of A, then

J = I−1 = (A :K I).

In particular if I, J are invertible fractional ideals, then the usual productIJ is invertible and (IJ)−1 = I−1J−1.

Definition 2.6.19 The Picard group of A is defined as

Pic(A) =

{invertible fractionalideals

}/{principal fractionalideals

}.

Theorem 2.6.20 [310, Theorem 11.3, p. 80] Let I be a fractional ideal ofA. The following conditions are equivalent :

(a) I is invertible.

(b) I is a projective A-module.

(c) I is finitely generated as an A-module and for every maximal ideal mof A, the fractional ideal Im = IAm of Am is principal.

Theorem 2.6.21 Let p �= (0) be a prime ideal of A. If A is Noetherianand p is invertible, then ht(p) = 1 and Ap is a discrete valuation ring.


Lemma 2.6.22 Let I be a fractional ideal of A. Then I−1 is a fractionalideal and

(I−1)−1 = (A : K(A : KI)) =⋂I⊂yA

yA.

In particular I is divisorial if and only if I is an intersection of principalfractional ideals.

Lemma 2.6.23 Let I = x0J be a fractional ideal of A such that 0 �= x0 ∈ Kand J is a fractional ideal of A. Then J is divisorial if and only if I isdivisorial.

Proof. Assume J is divisorial. By Lemma 2.6.22 we have

J = (J−1)−1 = (A : K(A : KJ)) =⋂

J⊂xAxA.

Multiplying by x0 and making the change of variable z = x0x we obtain

I = x0J =⋂

J⊂zx−10 A

zA =⋂

x0J⊂zAzA = (I−1)−1.

Therefore I is divisorial. The converse follows similarly. �

Proposition 2.6.24 Let I, J be two fractional ideals of A. Then everyhomomorphism of A-modules f : J → I is of the form f(x) = αx for someα ∈ (I : KJ). In particular there is an isomorphism of A-modules given by

HomA(J, I) −→ (I : KJ); f �−→ α.

Proof. Let 0 �= x, y ∈ J . There are 0 �= a, b ∈ A such that ax = by.Therefore the element

f(x)

x=af(x)

ax=bf(y)

by=f(y)

y= α

is independent of x. Hence f(x) = αx for x ∈ J and α ∈ (I : KJ). �

Corollary 2.6.25 Let I, J be two fractional ideals of A. If I � J as A-modules, then I is divisorial if and only if J is divisorial.

Proof. By Proposition 2.6.24 we have I = αJ for some α ∈ K. Thus theresult follows from Lemma 2.6.23. �

Definition 2.6.26 Let M be an A-module and let M∗ = HomA(M,A) beits dual. We say that M is reflexive if the canonical homomorphism

ϕ : M →M∗∗ = HomA(M∗, A); ϕ(m)(f) = f(m); ∀ m ∈M, f ∈M∗

is an isomorphism.

100 Chapter 2

Lemma 2.6.27 A direct summand of a reflexive module is reflexive.

Proof. It is left as an exercise. �

Proposition 2.6.28 If F is a free A-module of finite rank, then F is areflexive module.

Corollary 2.6.29 Every finitely generated projective module is reflexive.

Proposition 2.6.30 Let I be a fractional ideal of A. If I is reflexive (asan A-module), then I is divisorial.

Proof. As I is always contained in (I−1)−1, it is enough to show that(I−1)−1 is contained in I. Let x ∈ (I−1)−1. The idea is to show that xdefines an element ψ of I∗∗ and to use that ϕ is onto to obtain that x belongsto I. Let f ∈ I∗. By Proposition 2.6.24 there exists a unique element αf in(A : KI) such that f(z) = zαf for all z ∈ I. Consider the map

ψ : I∗ −→ A; f �−→ xαf .

First let us show that ψ is well-defined, that is we must show xαf ∈ A. Wemay assume αf �= 0. Since αfI ⊂ A we obtain I ⊂ α−1

f A. Hence, using

(I−1)−1 =⋂I⊂yA

yA,

we conclude that x ∈ α−1f A, this proves that xαf ∈ A. It is not hard to see

that ψ is a homomorphism. Hence since ϕ is onto there exists x0 ∈ I suchthat ϕ(x0) = ψ. Therefore

αfx = ψ(f) = ϕ(x0)(f) = f(x0) = x0αf ∀ f ∈ I∗ ⇒ x = x0 ∈ I. �

Proposition 2.6.31 [161, Proposition 5.2, Corollary 5.5.] Let I be a frac-tional ideal of a Krull ring A. Then I is reflexive if and only if I is divisorial.

Proposition 2.6.32 Let A be a Krull ring and let I be an ideal of A. ThenI is divisorial if and only if I = q1 ∩ · · · ∩ qs, where qi is a primary ideal ofA of height 1 for all i.

Example 2.6.33 Let R = K[x1, x2] be a polynomial ring over a field Kand let A = K[F ], the subring of R generated by F = {x21, x1x2, x22}. Letp = x1R ∩ A = (x21, x1x2)A. We claim that p is a divisorial prime idealsuch that p2 is not divisorial. The ring A is a normal Noetherian domain,thus it is a Krull ring. Since A ⊂ R is an integral extension, by the goingdown theorem (see Proposition 2.4.14) we have that p is a prime ideal of Aof height 1. Hence p is divisorial.


Now we show that p2 is not divisorial. By Proposition 2.6.32 it sufficesto observe that from the equality

(x21, x1x2, x22)A = (p2 : Ax1x2)

it follows that the maximal ideal m = (x21, x1x2, x22)A of A is an embedded

associated prime of p2.

Definition 2.6.34 The set of divisorial ideals of A is denoted by Div(A)and the set of principal fractional ideals by Prin(A).

Proposition 2.6.35 The set Div(A) with the binary operation

Div(A) ×Div(A) −→ Div(A); (I, J) �−→ I ∗ J := ((IJ)−1)−1

is a monoid whose identity is A.

Definition 2.6.36 Let A be an integral domain and K its field of fractions.An element x ∈ K is almost integral over A if there is 0 �= a ∈ A such thataxn ∈ A for all n ≥ 0.

Proposition 2.6.37 Let A be an integral domain and let K be its field offractions. Then an element x ∈ K is integral over A if and only if x isalmost integral over A.

Proof. Let x = c/d, where c, d are in A and d �= 0. If x is integral over A,then there is an equation

xm + b1xm−1 + · · ·+ bm−1x+ bm = 0, bi ∈ A.

Setting a = dm one has axn ∈ A for all n > 0.Conversely assume there is 0 �= a ∈ A such that axn ∈ A for all n > 0.

Since a−1A is a Noetherian A-module and A[x] ⊂ a−1A one has that A[x] isa finitely generated A-module. As A is a subring of A[x] by Proposition 2.4.2one derives that x is integral over A. �

Definition 2.6.38 The set of elements x ∈ K that are almost integral overA is denoted by A and this set is called the almost integral closure of A. IfA = A it is said that A is completely normal .

Lemma 2.6.39 [161] An element x ∈ K is almost integral over A if andonly if there exists a fractional ideal I of A such that xI ⊂ I.

Corollary 2.6.40 The set A is a subring of K and

A ⊂ A ⊂ A ⊂ K.

The equality A = A holds if A is Noetherian. In particular if A is completelynormal, then A is normal.

102 Chapter 2

Proof. Taking into account Proposition 2.6.37 it suffices to show that A isa subring of K. Let x, y ∈ A. By Lemma 2.6.39 there are fractional idealsI, J of A such that x ∈ (I :K I), y ∈ (J :K J). Observe that

xy ∈ (IJ :K IJ); x+ y ∈ (I ∩ J :K I ∩ J).

Since IJ , I ∩ J are fractional ideals of A, by Lemma 2.6.39 we get xy ∈ Aand x+ y ∈ A. �

Corollary 2.6.41 If A is a Krull ring, then A is completely normal.

Proof. A is an intersection of discrete valuation rings and these rings arenormal and Noetherian, hence completely normal. �

Theorem 2.6.42 [289, Theorem 5.19, p. 113] If A is a valuation ring withA �= K, then A is completely normal if and only if the value group of A hasrank 1.

Remark 2.6.43 By Exercise 2.5.15 there exists a field Q and a valuationν of Q such that its value group is Z×Z. Thus Aν is not completely normalbecause the rank of Z×Z is 2. Any valuation ring of Krull dimension greaterthan or equal to two is normal but not completely normal. See [436].

Theorem 2.6.44 The set Div(A) with the binary operation

Div(A) ×Div(A) −→ Div(A); (I, J) �−→ I ∗ J := ((IJ)−1)−1

is a group if and only if A is completely normal.

Proposition 2.6.45 Let A be a Krull ring and let I � A be an ideal of A.Then I is divisorial if and only if I can be written as

I = p1 ∗ · · · ∗ pr = (A : K(A : Kp1 · · · pr))

for some pi ∈ Z and this “factorization” is unique.

Corollary 2.6.46 If A is a Krull ring, then there is an isomorphism fromthe group (D(A),+, 0) to the group (Div(A), ∗, A) given by:

ρ : D(A) −→ Div(A); a1p1 + · · ·+ arpr �−→ pa11 ∗ · · · ∗ parr ,

where pa11 means pa11 = p1 ∗ · · · ∗ p1 (a1 times) if a1 > 0.

Proof. Clearly ρ is a homomorphism. To prove that ρ is onto take adivisorial fractional ideal I of A. There is 0 �= a ∈ A such that aI ⊂ A.Notice that aI is a divisorial ideal of A because

I =⋂I⊂xA

xA =⇒ aI =⋂

aI⊂zAzA.


Thus (aA)I = aI = ((aI)−1)−1 = (aA) ∗ I. By Proposition 2.6.45 we have

(aA) ∗ I = pa11 ∗ · · · ∗ parr and aA = pb11 ∗ · · · ∗ pbrr (ai, bi ≥ 0).

Therefore solving for I we get that I is in the image of ρ. To finish the proofnotice that from the uniqueness of the factorization in Proposition 2.6.45we readily get that ρ is injective. �

Another way of representing the divisor class group of a Krull ring is:

Theorem 2.6.47 If A is a Krull ring, then Cl(A) � Div(A)/Prin(A),where Prin(A) is the group of principal fractional ideals of A.

Proposition 2.6.48 If I is an invertible fractional ideal of A, then I isdivisorial.

Proof. Let x ∈ (I−1)−1. Then xI−1 ⊂ A. Now since I is invertible one hasII−1 = A, then there are ai ∈ I and bi ∈ I−1 such that 1 =

∑i aibi. Hence

x =∑i ai(xbi). Notice that xbi ∈ A because bi ∈ I−1, hence x ∈ I. Thus

we have shown (I−1)−1 ⊂ I. To prove the reverse inclusion take x ∈ I.Notice that xy ∈ A for all y ∈ I−1, thus x ∈ (I−1)−1. �

Lemma 2.6.49 If I, J are two invertible fractional ideals, then

IJ = I ∗ J = ((IJ)−1)−1.

Proof. By Lemma 2.6.18 one has (IJ)−1 = I−1J−1. Therefore taking theinverse on both sides gives the required equality. �

As a consequence of the last two results we obtain:

Corollary 2.6.50 The Picard group of A is a subgroup of the semigroupDiv(A)/Prin(A) of divisor classes.

Example 2.6.51 Let R = K[x1, x2, . . . , xn, . . .] be a polynomial ring in aninfinite number of variables over a field K and let A = K[F ] be the subringof R generated by F = {xixj | 1 ≤ i ≤ j}. Consider the ideal

p = x1R ∩ A = (x21, x1x2, x1x3, . . .)A.

We claim that p is a divisorial prime ideal of A and that p is not invertible.The ring A is a Krull ring [161]. Since A ⊂ R is an integral extension by thegoing down theorem one has that p is a height 1 prime ideal of A. Thereforep is divisorial. Since p is not finitely generated it cannot be invertible.

Definition 2.6.52 A is a Dedekind ring if every fractional ideal of A isinvertible.

104 Chapter 2

Notice that by Remark 2.6.15 A is a Dedekind ring if and only if everyideal (0) �= I ⊂ A of A is invertible. Principal ideal domains are Dedekindrings. If S is a multiplicatively closed subset of A and A is a Dedekind ring,then S−1A is a Dedekind ring. If dim(A) = 1, then A is a Dedekind ring ifand only if A is a Krull ring.

Dedekind rings occur in algebraic number theory because the ring ofalgebraic integers in a finite extension of Q is a Dedekind ring. By theKrull–Akizuki theorem if A = Z, K = Q and L is a finite extension of Q,then the integral closure of A in L is a Dedekind ring. More generally if Ais a Dedekind ring and L is a finite extension of K, then the integral closureA of A in L is a Dedekind ring. If A = Z, then Pic(A) is finite. See [329].

Theorem 2.6.53 [310, Theorem 11.6] An integral domain A is a Dedekindring if and only if any of the following equivalent conditions hold:

(a) A is a field or A is a normal Noetherian domain of dimension 1.

(b) Each non-zero ideal I of A has a unique factorization I = p1p2 · · · psinto prime ideals.

Corollary 2.6.54 If A is a Dedekind ring, then Div(A) is a free abeliangroup generated by the non-zero prime ideals of A and Pic(A) = Cl(A).

Proof. It follows from Corollary 2.6.46 and Lemma 2.6.49. �

Exercises

2.6.55 Let A be a domain with quotient fieldK. Prove that the set given byG = {xA|x ∈ K∗} is an ordered abelian group with the usual multiplicationof fractional ideals and where G is ordered by xA ≺ yA if and only ifxA ⊃ yA.

2.6.56 Let K = Z2 be the field with two elements and let F = ⊕i∈NFi,where Fi = K for all i. Prove that F ∗ = HomK(F,K) is uncountable andthat F is not a reflexive K-module.

2.7 Koszul homology

Let a be an element of the ring R and let K (a) be the complex defined as

Ki =

{R for i = 0, 1,0 otherwise,

with d1 : K1(a)→ K0(a) being multiplication by a.


Let I be an ideal of R generated by the sequence x = {x1, . . . , xn}. Theordinary Koszul complex associated to x is defined as

K (x;R) = K (x1)⊗ · · · ⊗K (xn).

For an R-moduleM we shall write K (x;M) for K (x;R)⊗M . The KoszulcomplexK (x;R) is then the exterior algebra complex associated to E = Rn

and the mapθ : E −→ R,

defined as θ(z1, . . . , zn) = z1x1 + · · ·+ znxn. That is, θ defines a differential∂ = dθ on the exterior algebra

∧(E) of E given in degree r by

∂(e1 ∧ · · · ∧ er) =r∑i=1

(−1)i−1θ(ei)e1 ∧ · · · ∧ ei ∧ · · · ∧ er.

From the definition of the differential of K (x;R), we get that if w and w′

are homogeneous elements of∧(E), of degrees p and q, respectively, then

∂(w ∧ w′) = (−1)pw ∧ ∂(w′) + ∂(w) ∧w′.

This implies that the cycles Z(K ) form a subalgebra of∧(E), and that

the boundaries B(K ) form a two-sided ideal of Z(K ). As a consequencethe homology of the Koszul complex, H (x), inherits a skew commutativeR-algebra structure.

One can also see that H (x) is annihilated by I = (x). Indeed, if e ∈ Eand w ∈ Zr(K ), we have from the last formula ∂(e ∧ w) = θ(e)w.

The ordinary Koszul complex K (x) = K (x;R) is simply the complexof free modules

K (x) : 0→∧n

Rn →∧n−1

Rn → · · · →∧1

Rn →∧0

Rn → 0,

where ∧kRn is the kth exterior power of Rn; thus ∧kRn is a free R-moduleof rank

(nk

)with basis {ei1 ∧ · · · ∧ eik |1 ≤ i1 < · · · < ik ≤ n}.

Proposition 2.7.1 [410, Theorem 2.3] If x is a regular sequence in R, thenthe Koszul complex is acyclic; that is, the complex K (x) is exact.

Sliding depth Let (R,m) be a Cohen–Macaulay local ring and let I bean ideal of R generated by x1, . . . , xn, denote by H (x) the homology of theordinary Koszul complex built on the sequence x = {x1, . . . , xn}.

Definition 2.7.2 (i) (SD) I satisfies sliding depth if

depthHi(x) ≥ dim(R)− n+ i, ∀ i ≥ 0.

(ii) (SCM ) I is strongly Cohen–Macaulay if Hi(x) are C–M, ∀ i ≥ 0.

(Depths are computed with respect to maximal ideals. It is usual to setdepth(0) equal to ∞.)

106 Chapter 2

Remark 2.7.3 (a) The (SD) condition localizes [234], (b) If I satisfies (SD)with respect to some generating set, then it will satisfy (SD) with respectto any other generating set of I. This follows from the isomorphisms:

Hi({x1, . . . , xn, 0}) � Hi(x)⊕Hi−1(x), and

Hi({x1, . . . , xn, y}) � Hi({x1, . . . , xn, 0}), where y ∈ (x).

Linkage Let I and J be two ideals in a Cohen–Macaulay local ring R. Theideals I and J are said to be (algebraically) linked if there is an R-sequencex = {x1, . . . , xn} in I ∩ J such that

I = ((x) : J) and J = ((x) : I),

if in addition I and J are unmixed ideals of the same height n withoutcommon components and such that I ∩ J = (x), then I and J are said tobe geometrically linked .

When I and J are linked we shall write I ∼ J . We say that J is in thelinkage class of I if there are ideals I1, . . . , Im such that

I ∼ I1 ∼ · · · ∼ Im ∼ J.

The ideal J is said to be in the even linkage class of I if m is odd.Let R be a Gorenstein ring and let I be a Cohen–Macaulay ideal of

R. If J is linked to I, then Peskine and Szpiro [339] showed that J isCohen–Macaulay. An interesting result of Huneke [253] proves that (SCM)is preserved under even linkage. His method can be adapted to prove thefollowing result.

Proposition 2.7.4 Let I and J be two ideals in a Gorenstein local ring Rof dimension d, and let x = {x1, . . . , xn} be a generating set for I. Assumethat J is evenly linked to I. If I satisfies the condition

(SDk) depth Hi(x;R) ≥ d− n+ i, 0 ≤ i ≤ k,

then J satisfies the (SDk) condition as well.

Exercises

2.7.5 Let R = K[x1, . . . , x6] be a polynomial ring over a field K and letI = I2(X) be the ideal of 2× 2-minors of the symmetric matrix:

X =

⎡⎣ x1 x2 x3x2 x4 x5x3 x5 x6

⎤⎦ .


Prove that I is linked to I2(X′), where X ′ is the symmetric matrix

X ′ =

⎡⎣ x1 x2 −x3x2 x4 x5−x3 x5 x6

⎤⎦ .In particular if char(K) = 2 the ideal I is self-linked. For other character-istics this is an open question.

2.8 A vanishing theorem of Grothendieck

Let R be a ring and let I be an R-module. We say that I is injective if thefunctor HomR( · , I) is exact. Note that this functor is always left exact.

Definition 2.8.1 Let R be a ring and let M be an R-module. A complex

I : 0 −→ I0∂0−→ I1

∂1−→ I2∂2−→ · · ·

of injective R-modules is an injective resolution ofM if Hi(I ) = 0 for i > 0and H0(I ) = ker(∂0) ∼=M .

The injective dimension ofM , denoted inj dim M , is the smallest integern for which there exist an injective resolution I of M with Im = 0 form > n. If there is no such n, the injective dimension of M is infinite.

For the proofs of the next three results and for additional informationon Gorenstein rings and injective resolutions see [18, 65, 255, 363].

Theorem 2.8.2 Let (R,m) be a local ring and let M be an R-module offinite injective dimension. Then

dim M ≤ inj dim M = depth R.

Definition 2.8.3 A local ring R is Gorenstein if inj dim R <∞. A ring Ris Gorenstein if Rm is a Gorenstein ring for all m maximal ideal of R.

Definition 2.8.4 Let R be a ring and let N ⊂M be R-modules. M is anessential extension of N if for any non-zero R-submodule U of M one hasU ∩N �= 0. An essential extension M of N is called proper if N �=M .

Proposition 2.8.5 Let R be a ring. An R-module N is injective if andonly if it has no proper essential extensions.

Definition 2.8.6 Let R be a ring and M an R-module. An injective mod-ule E such that M ⊂ E is an essential extension is called an injective hullof M and is denoted by E = E(M) or E = ER(M).

108 Chapter 2

Let M be an R-module. Then M admits an injective hull and M has aminimal injective resolution E (M):

0→M∂−1→ E0(M)

∂0→ E1(M)→ · · ·

→ Ei−1(M)∂i−1→ Ei(M)

∂i→ Ei+1(M)→ · · ·

where E0(M) = E(M) and Ei+1 = E(coker∂i−1). Here ∂−1 denotes theembedding M → E(M), and ∂i is defined in a natural way. Any twominimal injective resolutions of M are isomorphic. If I is an injectiveresolution of M , then E (M) is isomorphic to a direct summand of I .

Definition 2.8.7 Let (R,m,K) be a local ring R and let M �= 0 be anR-module with depth r. The type of M is the number

type(M) = dimK ExtrR(K,M).

Theorem 2.8.8 Let R be a local ring. Then R is a Gorenstein ring if andonly if R is a Cohen–Macaulay ring of type 1.

Local cohomology Here we present a vanishing theorem of Grothendieck(Theorem 2.8.12) which is useful to prove Reisner criterion (see Theo-rem 6.3.12) for Cohen–Macaulay complexes. Our main references for homo-logical algebra and local cohomology are [65, 202, 245, 363].

Let (R,m) be a local ring and letM be an R-module. Denote by Γm(M)the submodule of M of all the elements with support in {m}, that is,

Γm(M) = {x ∈M |mkx = 0 for some k > 0}.

Let x = {x1, . . . , xn} be a sequence of elements in R generating an m-primary ideal. Set xk = {xk1 , . . . , xkn}. The family xk gives the m-adictopology on R, hence

Γm(M) = {z ∈M | (x)kz = 0 for some k ≥ 0}.

Since HomR(R/I,M) = {x ∈M | Ix = 0} for any ideal I of R, we obtain anatural isomorphism

Γm(M) ∼= lim−→

HomR(R/mk,M) ∼= lim

−→HomR(R/(x

k),M). (2.2)

Proposition 2.8.9 Γm( · ) is a left exact additive functor.

Definition 2.8.10 The local cohomology functors , denoted by Him( · ) are

the right derived functors of Γm( · ).


Remark Let M and I be R-modules. (a) If I is an injective resolutionof M , then Hi

m(M) = Hi(Γm(I )) for i ≥ 0, (b) H0m(M) = Γm(M) and

Him(M) = 0 for i < 0. If I is injective, then Hi

m(I) = 0 for i > 0.

Proposition 2.8.11 If (R,m) is a local ring and M is an R-module, then

Him(M) ∼= lim

−→ExtiR(R/m

k,M) ∼= lim−→

ExtiR(R/(xk),M),

for i ≥ 0, where x is a sequence in R generating an m-primary ideal.

Proof. Recall that if P is a projective resolution of L and I is an injectiveresolution of M , then ExtiR(L,M) can be computed as follows:

ExtiR(L,M) ∼= Hi(HomR(P ,M)) ∼= Hi(HomR(L, I )),

see [245, Proposition 8.1]. Assume I is an injective resolution of M , then

Him(M) ∼= Hi(Γm(I )) and Γm(I ) ∼= lim

−→HomR(R/m

k, I ).

Therefore

Him(M) ∼= Hi(lim

−→HomR(R/m

k, I )) ∼= lim−→

Hi(HomR(R/mk, I ))

∼= lim−→

ExtiR(R/mk,M).

Since

Γm(M) ∼= lim−→

HomR(R/mk,M) ∼= lim

−→HomR(R/(x

k),M),

the second isomorphism follows using the same arguments. �

Next we recall the following vanishing theorem.

Theorem 2.8.12 (Grothendieck [202]) If (R,m) is a local ring and M isan R-module of depth t and dimension d, then

(a) Him(M) = 0 for i < t and i > d.

(b) Htm(M) �= 0 and Hd

m(M) �= 0.

Local cohomology of face rings Let Δ be a simplicial complex and let

R = K[Δ] = K[X1, . . . , Xn]/IΔ

be the Stanley–Reisner ring of Δ, where IΔ is the ideal of K[X1, . . . , Xn]generated by all Xi1 · · ·Xir such that {Xi1 , . . . , Xir} /∈ Δ.

Let m be the maximal ideal generated by the residue classes xi of theindeterminates Xi and let Hi

m(R) be the local cohomology modules of R.

110 Chapter 2

Consider the complex C

C : 0 −→ C0 −→ C1 −→ · · · −→ Cn −→ 0,

Ct =⊕

1≤i1<···<it≤nRxi1 ...xit

,

where Ry denotes R localized at S = {yi}i≥0 and the differentiation mapdt : C

t → Ct+1 is given on the component

Rxi1 ···xit

dt−→ Rxj1 ···xjt+1

to be the natural homomorphism

(−1)s−1 · η : Rxi1 ···xit−→ R(xi1 ···xit )xjs

if {i1, . . . , it} = {j1, . . . , js, . . . , jt+1} and 0 otherwise. If x = xi1 · · ·xir isin K[Δ], then Rx �= 0 iff supp x ∈ Δ. Hence Hi(C ) = 0 for i > dim K[Δ](cf. Theorem 2.8.12). Recall that there is an isomorphism

Him(R)

∼= Hi(C ).

The reader should consult [65] for further details and results about thelocal cohomology of face rings.

Chapter 3

Affine and GradedAlgebras

A few topics connected with affine and graded algebras are studied in thischapter, e.g., Grobner bases, Hilbert Nullstellensatz and affine varieties,projective closure, minimal resolutions, and Betti numbers. We present theaffine and graded versions of the Noether normalization lemma and some ofits applications to affine algebras and Cohen–Macaulay graded algebras.

As before all base rings considered here are Noetherian and modules arefinitely generated.

3.1 Cohen–Macaulay graded algebras

In this section we will emphasize the relationship between graded Cohen–Macaulay rings and their homogeneous Noether normalizations. Then someuseful characterizations of those rings will be given. First we introduce affinealgebras and affine Noether normalizations.

Definition 3.1.1 Let k be a field and let S be a k-algebra. We say that Sis an affine k-algebra if S = k[y1, . . . , yr] for some y1, . . . , yr ∈ S.

Definition 3.1.2 Let α and β be in Nn. The lexicographical order in Nn isobtained by declaring β " α if the last non-zero entry of β − α is positive.

Notation The set of positive integers will be denoted by N+. If α, β ∈ Rn,here α · β will denote the usual inner product of α and β.

Lemma 3.1.3 Let α1 " · · · " αm be a sequence of m distinct points in Nn

ordered lexicographically. Then, there is w = (w1, . . . , wn) ∈ Nn+ such thatw1 = 1 and w · α1 > w · αi for i ≥ 2.

112 Chapter 3

Proof. Let αi = (αi1, . . . , αin) and βi = (αi1, . . . , αi(n−1)). We proceed byinduction on n ≥ 2. There is k so that α1n = · · · = αkn and αkn > αin fori > k. One may assume k < m; for otherwise β1 " · · · " βm and one canuse induction.

Since β1 " · · · " βk by induction there is w′ = (1, w2, . . . , wn−1) suchthat w′ · β1 > w′ · βi for 2 ≤ i ≤ k. On the other hand for every i > kone can choose δi ∈ N+ so that w′ · β1 + α1nδi > w′ · βi + αinδi. Settingwn = max{δi| k < i ≤ m}, w = (w′, wn), we get w · α1 > w · αi for i ≥ 2. �

Proposition 3.1.4 Let R = k[x1, . . . , xn] be a polynomial ring over a fieldk and let f be a polynomial in R \ k. Then, there is a change of variablesxi = xwi

1 + yi for i ≥ 2 such that

f(x1, xw2

1 + y2, . . . , xwn

1 + yn) = c0xr1 + c1x

r−11 + · · ·+ cr−1x1 + cr,

where 0 �= c0 ∈ k, r > 0, and ci ∈ k[y2, . . . , yn] for i ≥ 1.

Proof. The polynomial f can be written as f =∑m

i=1 bixαi , where

0 �= bi ∈ k for all i. One may assume that α1 " · · · " αm are orderedlexicographically. By Lemma 3.1.3 there is w ∈ Nn+ such that xi = xwi + yisatisfies the required properties. Note r = w · α1. �

Lemma 3.1.5 If R = k[x1, . . . , xn] is an affine k-algebra of dimension n,then R is a polynomial ring.

Proof. One may assume R � B/I, where B is a polynomial ring in nvariables with coefficients in the field k and I is an ideal of B. Let

I ⊂ p0 ⊂ · · · ⊂ pn

be a chain of prime ideals of B of length n. If I �= (0), then adding (0)to the chain yields a chain of length n + 1, which is impossible becausedim(B) = n. Hence I = (0). �

Theorem 3.1.6 Let R = k[x1, . . . , xn] be a polynomial ring over a field kand let I �= R be an ideal of R. Then there are z1, . . . , zn in R such that

(a) k[x1, . . . , xn] is integral over k[z1, . . . , zn], and

(b) I ∩ k[z1, . . . , zn] = z1k[z1, . . . , zn] + · · ·+ zgk[z1, . . . , zn].

Proof. The proof is by induction on n. If n = 1 and I = (f(x1)) �= 0, thenone sets z1 = f(x1). Assume n ≥ 2 and I �= 0. One may assume that Icontains a monic polynomial in x1. Otherwise take a non-zero polynomialg in I and apply Proposition 3.1.4 to get an isomorphism of k-algebras

k[x1, . . . , xn]ϕ−→ k[x1, y2 . . . , yn]

Affine and Graded Algebras 113

induced by ϕ(x1) = x1 and ϕ(xi) = xwi1 + yi for i ≥ 2, such that ϕ(g) is

monic in x1. Let

f = xr1 + c1xr−11 + · · ·+ cr−1x1 + cr

be a polynomial in I with ci ∈ k[x2, . . . , xn] and r > 0. Set

I1 = I ∩ k[x2, . . . , xn].

By induction there are z2, . . . , zn such that:

(i) k[x2, . . . , xn] is integral over k[z2, . . . , xn], and

(ii) I1 ∩ k[z2, . . . , zn] = (z2, . . . , zn).

Set z1 = f . It is not hard to see that R is integral over k[z1, . . . , zn] and

k[z1, . . . , zn] ∩ I = (z1, . . . , zg). �

Corollary 3.1.7 If R = k[x1, . . . , xn] is a polynomial ring over a field kand I �= R is an ideal of R, then

dim(R/I) = dim(R)− ht (I).

Proof. One may assume that there are z1, . . . , zn in R and an integer gsuch that the conditions (a) and (b) of Theorem 3.1.6 are satisfied. Wewill show that g is equal to the height of I. By Lemma 3.1.5 the zi’s arealgebraically independent. Hence Proposition 2.4.14 yields

ht (I) = ht (I ∩ k[z1, . . . , zn]) = g.

Note that there is an integral extension

k[zg+1, . . . , zn] � k[z1, . . . , zn]/(z1, . . . , zg)ϕ↪→ k[x1, . . . , xn]/I.

Therefore n− g = dim(R/I). �

Corollary 3.1.8 (Noether normalization lemma) If R = k[x1, . . . , xn] is apolynomial ring over a field k and I �= R is an ideal, then there is an integralextension

k[h1, . . . , hd]↪→R/I,where h1, . . . , hd are in R and d = dim(R/I).

Proof. By Theorem 3.1.6 there is an integral extension

k[zg+1, . . . , zn] � k[z1, . . . , zn]/(z1, . . . , zg)ϕ↪→ R/I.

To conclude the argument set hi = zg+i for i = 1, . . . , d. �

114 Chapter 3

Corollary 3.1.9 If R = k[x1, . . . , xn] is a polynomial ring over a field k,then R is a catenary ring.

Proof. Note ht (q/p) = dim(R/p) − dim(R/q) for any two prime idealsp ⊂ q in R. Hence by Corollary 3.1.7 we get ht (q/p) = ht (q)− ht (p). �

Definition 3.1.10 Let k ⊂ L be a field extension. A subset of L whichis algebraically independent and is maximal with respect to inclusions iscalled a transcendence basis of L over k.

Theorem 3.1.11 [262, Theorem 8.35] If k ⊂ L is a field extension, thenany two transcendence bases have the same cardinality.

Definition 3.1.12 Let k ⊂ L be a field extension. The transcendencedegree of L over k, denoted trdegk(L), is the cardinality of any transcendencebasis of L over k.

Corollary 3.1.13 Let k be a field and let A be a finitely generated k-algebra. If A is a domain with field of fractions L, then

dim(A) = trdegk(L).

Definition 3.1.14 Let k be a field. A standard algebra or homogeneousalgebra is a finitely generated N-graded k-algebra

S =

∞⊕i=0

Si = k[y1, . . . , yr]

such that yi ∈ S1 for all i and S0 = k. If we only require yi homogeneousand deg(yi) > 0 for all i, we say that S is a positively graded k-algebra.

The irrelevant maximal ideal m of S is defined by

m = S+ =

∞⊕i=1

Si.

Definition 3.1.15 Let k be a field and S = k[y1, . . . , yr] a positively gradedk-algebra with yi homogeneous of degree di. There is a graded epimorphism

ϕ : R = k[x1, . . . , xr] −→ S

given by f(x1, . . . , xr)ϕ�−→ f(y1, . . . , yr), where R is a polynomial ring

graded by deg(xi) = di, the presentation of S is the k-algebra R/ ker(ϕ).The graded ideal ker(ϕ) is called the ideal of relations or the presentationideal of S.


Proposition 3.1.16 If S is a Cohen–Macaulay standard algebra over afield k and L is an ideal of S, then

dim(S) = dim(S/L) + ht (L).

Proof. One may assume S = R/I and L = J/I, where R is a polynomialring over k and J is an ideal of R containing I. Set g = ht (J/I) andr = ht (I). Let p be a prime ideal such that J ⊂ p and g = ht (p/I). Thereis a saturated chain of prime ideals

I ⊂ p0 ⊂ p1 ⊂ · · · ⊂ pg = p.

Since p0 is a minimal prime ideal of I using Proposition 2.3.13 one obtainsdim(R/I) = dim(R/p0), that is, r = ht (p0). Hence there is a saturatedchain of prime ideals

q0 = (0) ⊂ q1 ⊂ · · · ⊂ qr = p0 ⊂ p1 ⊂ · · · ⊂ pg = p.

As R is a catenary domain one obtains ht (p) = r + g and consequentlyht (J) ≤ r + g. Therefore, we get

dim(S) ≤ ht (L) + dim(S/L).

To conclude the proof note that the reverse inequality holds in general. �

Corollary 3.1.17 Let R be a positively graded polynomial ring over a fieldk and I a graded ideal of R. If R/I is Cohen–Macaulay, then I is unmixed.

Proof. It follows from Propositions 2.3.13 and 3.1.16. �

Definition 3.1.18 Let k be a field and let S be a positively graded k-algebra. A set of homogeneous elements θ = {θ1, . . . , θd} is called a ho-mogeneous system of parameters (h.s.o.p for short) if d = dim(S) andrad (θ) = S+.

Corollary 3.1.19 Let S be a positively graded algebra over a field k andlet h1, . . . , hd be a h.s.o.p for S. If 1 ≤ i ≤ dim(S), then

dimS/(h1, . . . , hi) = dim(S)− i.

Proof. Set S = S/(h1, . . . , hi) and d = dim(S). Note that the set ofimages of hi+1, . . . , hd in S generate an m-primary ideal, where m = S+

is the irrelevant maximal ideal of S and m = mS. Hence, by the gradedversion of the dimension theorem (Theorem 2.3.15), we get dim(S) ≤ d− i.On the other hand if θ1, . . . , θr is a h.s.o.p for S, then the sequence

θ1, . . . , θr, h1, . . . , hi

116 Chapter 3

generates an m-primary ideal of S. Applying the dimension theorem onceagain one has

d ≤ r + i = dim(S) + i.

Thus dim(S) = d− i. �

Proposition 3.1.20 Let S be a positively graded algebra over a field k andθ = θ1, . . . , θd a h.s.o.p for S. Then S is Cohen–Macaulay if and only if θis a regular sequence.

Proof. By Corollary 3.1.19 dimS/(θ1, . . . , θi) = d − i. If S is C–M, thenby Proposition 3.1.16 one has ht (θ1, . . . , θi) = i. Assume θ1, . . . , θi−1 is aregular sequence. Next we show that θi is regular on A = S/(θ1, . . . , θi−1).Otherwise if θi is a zero divisor of A, then θi belongs to some minimalprime p of (θ1, . . . , θi−1). Since ht (p) ≤ i − 1 (see Theorem 2.3.16) weobtain ht (θ1, . . . , θi) ≤ i − 1, which is impossible. Conversely if θ is aregular sequence, then depth(S) = d and S is C–M. �

Proposition 3.1.21 Let S be a positively graded algebra over a field k. If Sis Cohen–Macaulay and θ = θ1, . . . , θq is a regular sequence of homogeneouselements in S+, then S/(θ) is Cohen–Macaulay.

Proof. It suffices to prove the case q = 1, which is a direct consequence ofLemma 2.3.10. �

Lemma 3.1.22 Let V �= {0} be a vector space over an infinite field K.Then V is not a finite union of proper subspaces of V .

Proof. We proceed by contradiction. Assume that there are proper sub-spaces V1, . . . , Vm of V such that V =

⋃mi=1 Vi, where m is the least positive

integer with this property. Let

v1 ∈ V1 \ (V2 ∪ · · · ∪ Vm) and v2 ∈ V2 \ (V1 ∪ V3 ∪ · · · ∪ Vm).

Pick m+ 1 distinct non-zero scalars k0, . . . , km in K. Consider the vectorsβi = v1 − kiv2 for i = 0, . . . ,m. By the pigeon-hole principle there aredistinct vectors βr, βs in Vj for some j. Since βr − βs ∈ Vj we get v2 ∈ Vj .Thus j = 2 by the choice of v2. To finish the proof observe that βr ∈ V2imply v1 ∈ V2, which contradicts the choice of v1. �

Proposition 3.1.23 Let R = k[x1, . . . , xn] be a polynomial ring over a fieldk with a positive grading and let I be a graded ideal of R. Then there arehomogeneous polynomials h1, . . . , hd in R+ such that

dimR/(I, h1, . . . , hi) = d− i, for i = 1, . . . , d,

where dim(R/I) = d. If k is an infinite field and deg(xi) = 1 for all i, thenh1, . . . , hd can be chosen in R1.


Proof. Assume d > 0. Let p1, . . . , pr be the set of minimal primes of Iof height g = ht (I). We claim that there is a homogeneous polynomial h1not in ∪ri=1pi. To show it we use induction on r. Since p1 · · · pr−1 �⊂ pr andbecause pi is graded we can pick f ∈ p1 · · · pr−1 ∩ Rd1 and f �∈ pr, d1 > 0.By induction there is g ∈ Rd2 and g �∈ ∪r−1

i=1 pi, d2 > 0. Assume Ri ⊂ ∪ri=1pifor all i > 0, hence g ∈ pr. To complete the proof of the claim considerh = fd2 − gd1 to derive a contradiction.

Note dim(R/I) > dimR/(I, h1), by the choice of h1. Hence a repeatedapplication of the claim rapidly yields a sequence h1, . . . , hs of homogeneouspolynomials in R+ with s ≤ d and such that ht (I, h1, . . . , hs) = dim(R).Therefore by Theorem 2.3.15 one concludes d = s, as required.

If k is infinite and deg(xi) = 1 for all i, then there is h1 in R1 and notin ∪ri=1pi; for otherwise one has R1 = ∪ri=1(pi)1 and since R1 cannot bea finite union of proper subspaces (see Lemma 3.1.22) we derive R+ = pifor some i, which is impossible. Thus one may proceed as above to get therequired sequence. �

Theorem 3.1.24 (Homogeneous Noether normalization lemma) Let k bea field and let R be a polynomial ring over k. If R is positively graded andI is a graded ideal of R, then there are homogeneous polynomials h1, . . . , hdin R+, with d = dim(R/I), and a natural embedding

A = k[h1, . . . , hd]ϕ↪→ R/I

such that R/I is a finitely generated A-module. Moreover if k is infiniteand deg(xi) = 1 for all i, then h1, . . . , hd can be chosen in R1.

Proof. We set R = k[x1, . . . , xn] and Bi = {xa| deg(xa) = i}. Accordingto Proposition 3.1.23, there are homogeneous polynomials h1, . . . , hd in mwith rad (I, h1, . . . , hd) = m, where m = R+. Hence, there is r such that

mr ⊂ (I, h1, . . . , hd).

Note that Bi ⊂ (I, h1, . . . , hd) for i ≥ s = rδ, where δ is the maximum ofthe degrees of x1, . . . , xn. We set B = ∪si=1Bi and J = I + Bk[h1, . . . , hd].Using induction on i, we now show that Bi ⊂ J for i ≥ s. The inclusionis clear for i ≤ s. Assume i > s and take xa in Bi. Let f1, . . . , fq bea homogeneous generating set for I. Since (I, h1, . . . , hd) is graded, we

can write xa =∑qj=1 bjfj +

∑dj=1 cjhj, where b1, . . . , bq and c1, . . . , cd are

homogeneous and deg(cjhj) = i. Hence all the monomials in the supportof cj have degree less than i. Thus, by induction, one gets that cj is in Jfor all j. Consequently xa ∈ J .

Observe that the image of B in R/I generates R/I as an A-module.Next we verify that the canonical map ϕ from A to R/I is injective. SinceR/I is integral over ϕ(A), by Proposition 2.4.13, one obtains that d is equal

118 Chapter 3

to dimϕ(A). Using A/ ker(ϕ) � ϕ(A) and the fact that A has dimension atmost d we get ker(ϕ) = 0. If k is infinite and deg(xi) = 1, then accordingto Proposition 3.1.23 one can choose h1, . . . , hd of degree one. �

Definition 3.1.25 Let R be a positively graded polynomial ring over afield k and I �= R a graded ideal. A homogeneous Noether normalization ofS = R/I is an integral extension k[h1, . . . , hd] ↪→ S, where h1, . . . , hd arehomogeneous polynomials in R+ and d = dim(S).

Proposition 3.1.26 (Stanley) Let R = k[x1, . . . , xn] be a polynomial ringover a field k and let I be a monomial ideal with d = dim(R/I). Then

A = k[σ1, . . . , σd]↪→S = R/I

is a Noether normalization, where σi is the ith symmetric polynomial.

Proof. It suffices to prove that S is integral over A. Let xi = xi+ I. Sinced = dim(R/rad(I)), one has σi ∈ rad(I) for i > d. Hence, from the equality

(x − x1) · · · (x− xn) = xn − σ1xn−1 + · · ·+ (−1)nσn,

we get xni − σ1xn−1i + · · · + (−1)dσdxn−di = h, where h is nilpotent in S.

Hence if hm ∈ I, raising the last equality to the mth power yields that xiis integral over A. �

Proposition 3.1.27 Let R = k[x1, . . . , xn] be a positively graded polyno-mial ring over a field k and let I �= R be a graded ideal of R. If S = R/Iis a Cohen–Macaulay ring and A = k[h1, . . . , hd] ↪→ S is a homogeneousNoether normalization of S, then

S = Axβ1 ⊕ · · · ⊕Axβm

for any set of monomials B = {xβ1 , . . . , xβm} whose image in the Artinianring S = R/(I, h1, . . . , hd) is a k-vector space basis of S.

Proof. Set M i = {xα ∈ R| deg(xα) = i}. First we show that the image ofB generate S as an A-module. It suffices to prove M i ⊂ J = I + AB forall i ≥ 0. Let xα ∈ M i. There are homogeneous polynomials μ1, . . . , μd inR and λ1, . . . , λm in k such that xα = f +

∑mj=1 λjx

βj +∑dj=1 μjhj , where

f ∈ Ri ∩ I and deg(μjhj) = i if μj �= 0. Since deg μj < i, by induction oneobtains that μj ∈ J for all j. Consequently xα ∈ J .

We claim that if c1xβ1+· · ·+cmxβm belongs to (I, h1, . . . , hi−1) for some

2 ≤ i ≤ d and c1, . . . , cm are in k[hi, . . . , hd], then cj = 0 for all j. To provethe claim note that h1, . . . , hd is a regular sequence (see Proposition 3.1.20)and use descending induction on i starting with i = d.


Next, we show the equality S = Axβ1 ⊕ · · · ⊕ Axβm . Assume that∑mi=1 aix

βi is in I for some a1, . . . , am in A. If a� �= 0 for some �, write

ai =∑si

j=0 aijhj1, where aij are in k[h2, . . . , hd]. One may assume ar0 �= 0

for some r; otherwise we may factor out h1 to some power and apply thath1 is regular on S. Hence a10x

β1 + · · ·+ am0xβm is in (I, h1), and applying

the claim with i = 2 we derive ar0 = 0, which is a contradiction. Thereforeaj = 0 for all j, as required. �

Lemma 3.1.28 Let k be an infinite field and let S be a standard k-algebra.If S is Cohen–Macaulay, then there exists a h.s.o.p θ = {θ1, . . . , θd} suchthat θ is a regular sequence and θi is homogeneous of degree 1 for all i.

Proof. Let S = ⊕∞i=0Si and Ass(S) = {p1, . . . , ps}. Notice that Ass(S) =

Min(S), because S is Cohen–Macaulay. We may assume d > 0, otherwisethere is nothing to prove. If S1 ⊂ Z(S) = ∪si=1pi, then S1 = ∪si=1(pi)1.Since k is infinite, by Lemma 3.1.22, we obtain that s = 1 and p1 = S+,that is, dim(S) = 0, which is a contradiction. Hence there is θ1 ∈ S1 whichis regular on S. By Proposition 3.1.16 dimS/(h1) = d − 1. Applying thedepth lemma (see Lemma 2.3.9) to the exact sequence

0 −→ S(−1) θ1−→ S −→ S/(θ1) −→ 0

yields depthS/(θ1) = d−1. Altogether S/(θ1) is C–M and the result followsby induction. �

Lemma 3.1.29 Let R be an N-graded polynomial ring over a field k and Ia graded ideal of R of height g. If I is a complete intersection, then thereare homogeneous polynomials f1, . . . , fg such that I = (f1, . . . , fg).

Proof. Let h1, . . . , hg be a regular sequence generating the ideal I. Usingthe graded version of Corollary 2.1.35 one has

r = μ(I) = dimk(I/mI) ≤ g,

wherem = R+. On the other hand, since I is graded, there are homogeneouspolynomials f1, . . . , fr generating I and such that {fi+mI}ri=1 is a k-basisfor I/mI. By Theorem 2.3.16 one concludes r = g. �

Proposition 3.1.30 Let R be a positively graded polynomial ring over afield k and let I be a graded ideal of R. If I is a complete intersection, thenR/I is Cohen–Macaulay.

Proof. By Lemma 3.1.29 the ideal I is generated by a sequence f1, . . . , fgconsisting of homogeneous polynomials, where g denotes the height of I.

120 Chapter 3

According to Proposition 3.1.23 there is a sequence h1, . . . , hn−g consistingof homogeneous polynomials such that

rad (f1, . . . , fg, h1, . . . , hn−g) = m = R+.

Therefore {f1, . . . , fg, h1, . . . , hn−g} is a homogeneous system of parame-ters for R and by Proposition 3.1.20 we derive that f1, . . . , fg is a regularsequence. Finally Lemma 2.3.10 yields that R/I is Cohen–Macaulay. �

Lemma 3.1.31 Let I, p1, . . . , pm be graded ideals of a polynomial ring Rover a field k such that p1, . . . , pm are prime ideals. If f ∈ ∪mi=1pi for anyf ∈ I homogeneous, then I ⊂ pi for some i.

Proof. By induction on m. If m = 2 and I �⊂ pi for i = 1, 2, then for eachi pick a homogeneous polynomial fi ∈ I \ pi of degree ui. Since fu2

1 + fu12

is homogeneous we readily derive a contradiction. Hence I ⊂ pi for some i.Let G be the set of homogeneous elements in I and suppose G ⊂ ∪mi=1pi.

One may assume pi �⊂ pj for i �= j. If I �⊂ pi for all i, then by inductionthere is f1 in G \ ∪m−1

i=1 pi. On the other hand since Ip1 · · · pm−1 �⊂ pm (seeExercise 2.1.48) there is a homogeneous polynomial f2 in Ip1 · · · pm−1 andf2 not in pm. Since fu2

1 + fu12 , deg(fi) = ui, is homogeneous we rapidly

derive a contradiction. Thus I ⊂ pi for some i. �

Proposition 3.1.32 Let R be a polynomial ring over a field k and let I bea graded ideal of height r, then there is a regular sequence f1, . . . , fr in I ofhomogeneous polynomials.

Proof. We will proceed by induction. Assume that f1, . . . , fs is a regularsequence of homogeneous polynomials in I, where s < r. Note that theideal (f1, . . . , fs) is Cohen–Macaulay by Proposition 3.1.30, and hence it isunmixed by Corollary 3.1.17. If all the homogeneous polynomials in I belongto Z(R/(f1, . . . , fs)), then by Lemma 3.1.31 the ideal I must be containedin an associated prime of (f1, . . . , fs) and consequently ht (I) ≤ s, which isimpossible. Thus there is a homogeneous polynomial fs+1 in I such thatfs+1 is regular modulo (f1, . . . , fs). �

Tensor product of affine algebras Let A, B be two affine algebrasover a field k and consider presentations A � R1/I1, B � R2/I2, whereR1 = k[x], R2 = k[y] are polynomial rings in disjoint sets of variables andIi is an ideal of Ri. Set R = k[x,y] and I = I1 + I2. The map

R �−→ R1/I1 ⊗k R2/I2, xi �→ xi ⊗ 1, yi �→ 1⊗ yi,

induces a k-algebra homomorphism:

ϕ : R/I → R1/I1 ⊗k R2/I2, ϕ(f(x)g(y)) = f(x)⊗ g(y).


On the other hand there is a k-bilinear map

ψ : R1/I1 ×R2/I2 → R/I,

given by multiplication ψ(f, g) = fg. By the universal property of thetensor product there is a map ψ that makes the following diagram

R1/I1 ×R2/I2 R1/I1 ⊗k R2/I2�φ

�ψ

R/Iψ

commutative, where φ is the canonical map and ψ = ψφ. As a consequenceψ is the inverse of ϕ. Thus we have proved:

Proposition 3.1.33 If A and B are two affine algebras over a field k withpresentations A � k[x]/I1 and B � k[y]/I2, then

A⊗k B � k[x,y]/(I1 + I2).

Theorem 3.1.34 If A and B are two standard algebras over a field k, then

depth(A⊗k B) = depth(A) + depth(B).

Proof. Pick a regular sequence g = g1, . . . , gr (resp. h = h1, . . . , hs) onA = k[x]/I1 (resp. B = k[y]/I2) with g ⊂ (x) (resp. h ⊂ (y)) and such thatg (resp. h) consist of forms, where r is the depth of A and s is the depth ofB. As B is a faithfully flat k-algebra, applying the functor (·) ⊗k B to theinjective map

0 −→ k[x]/(I1, g1, . . . , gi−1)gi−→ k[x]/(I1, g1, . . . , gi−1)

gives a natural commutative diagram

k[x]/(I1, g1, . . . , gi−1)⊗k Bgi⊗1−→ k[x]/(I1, g1, . . . , gi−1)⊗k B⏐⏐2� ⏐⏐2�

k[x,y]/(I1, I2, g1, . . . , gi−1)gi−→ k[x,y]/(I1, I2, g1, . . . , gi−1)

such that the map in the first row is injective. Since the vertical arrows arenatural isomorphisms by Proposition 3.1.33, one concludes that the map inthe second row is also injective. As a consequence g is a regular sequenceon k[x,y]/(I1, I2). Similarly if we tensor the injective map

0 −→ k[y]/(I2, h1, . . . , hi−1)hi−→ k[y]/(I2, h1, . . . , hi−1)

122 Chapter 3

with the faithfully flat k-algebra k[x]/(I1, g) it follows rapidly that h is aregular sequence on k[x,y]/(I1, I2, g).

Altogether g, h is a regular sequence on k[x,y]/(I1, I2). To finish theproof note that (x,y) is an associated prime ideal of k[x,y]/(I1, I2, g, h) byExercise 3.1.41, thus r + s is the length of a maximal regular sequence in(x,y) and by Proposition 2.3.7 one obtains depth(A⊗k B) = r + s. For analternative proof see [171]. �

Corollary 3.1.35 If A and B are two standard algebras over a field k, thenA⊗k B is Cohen–Macaulay if and only if A and B are Cohen–Macaulay.

Proof. As the dimension is always greater than or equal to the depth byLemma 2.3.6, the result follows at once using Theorem 3.1.34 together withExercise 3.1.36. �

A similar statement holds if one replaces Cohen–Macaulay by Goren-stein; indeed the type of the tensor products of two Cohen–Macaulay stan-dard algebras is the product of their types [171].

Exercises

3.1.36 If A and B are standard algebras over a field k, then

dim(A⊗k B) = dim(A) + dim(B).

3.1.37 Let R = k[x1, . . . , xn] be a polynomial ring over a field k and letA = (aij) be an invertible matrix with entries in k. If yi =

∑nj=1 aijxj

for i = 1, . . . , n, prove that R = k[y1, . . . , yn]. In particular y1, . . . , yn arealgebraically independent.

3.1.38 Let V be a vector space over an infinite field k and W,V1, . . . , Vmvector subspaces of V . If W �⊂ Vi for all i, prove that W �⊂ ∪mi=1Vi.

3.1.39 Let I, I1, . . . , Im be ideals of a ring R and let k be an infinite field.If I ⊂ ∪mi=1Ii and k is a subring of R, then I ⊂ Ii for some i.

3.1.40 Let R be a polynomial ring over an infinite field k and I a gradedideal of height r. If I is minimally generated by forms of degree p ≥ 1,prove that there are forms f1, . . . , fm of degree p in I such that f1, . . . , fris a regular sequence and I is minimally generated by f1, . . . , fm.

3.1.41 Let A = k[x]/I1 and B = k[y]/I2 be affine algebras over a field k. If(x) (resp. (y)) is an associated prime of k[x]/I1 (resp. k[y]/I2), then (x,y)is an associated prime of k[x,y]/(I1, I2).


3.1.42 Let k[x] be a polynomial ring and I an ideal (resp. graded). If k ⊂ Kis a field extension, then there is a natural (resp. graded) isomorphism ofK-algebras:

K[x]/IK[x] � k[x]/I ⊗k K.

3.1.43 Let A be an affine algebra over a field k and let k ⊂ K be a fieldextension. Prove:

(a) dim(A) = dim(A⊗k K).

(b) depth(A) = depth(A⊗k K) if A is a standard algebra.

3.1.44 Let K be a field and let A be an affine K-algebra. Prove that A isArtinian if and only if dimK(A) <∞.

3.2 Hilbert Nullstellensatz

Let K be a field and let S = K[t1, . . . , tq] be a polynomial ring. In whatfollows t stands for the set of variables of S, that is, S = K[t]. We definethe affine space of dimension q over K, denoted by AqK , to be the cartesianproduct Kq = K × · · · ×K.

Given an ideal I ⊂ S, define the zero set or affine variety of I as

V (I) = {α ∈ AqK | f(α) = 0, ∀f ∈ I}.

By the Hilbert’s basis theorem V (I) is the zero locus of a finite collectionof polynomials (see Theorem 2.1.4). Conversely, for any X ⊂ AqK defineI(X), the vanishing ideal of X , as the set of polynomials of S that vanish atall points of X . An affine variety is the zero set of an ideal. The dimensionof a variety X is the Krull dimension of its coordinate ring S/I(X).

Proposition 3.2.1 (Zariski topology [200, Lemma A.2.4])

(a) V (1) = ∅ and V (0) = AqK .

(b) V (I ∩ J) = V (I) ∪ V (J) = V (IJ), for all ideals I and J of S.

(c) ∩V (Iα) = V (∪Iα), where {Iα} is any family of ideals of S.

By the previous result the sets in the family

τ = {AqK \ V (I)| I is an ideal of S}

are the open sets of a topology on AqK , called the Zariski topology.

Definition 3.2.2 Given X ⊂ AqK , the Zariski closure of X , denoted by X,is the closure of X in the Zariski topology of AqK , i.e., X is the smallestaffine variety of AqK containing X .

124 Chapter 3

Proposition 3.2.3 If X ⊂ AqK , then X = V (I(X)).

Proof. As V (I(X)) is closed in the Zariski topology and X ⊂ V (I(X)),taking closures one has X ⊂ V (I(X)). For the other inclusion note thatX is closed; that is, X = V (J), where J is an ideal of S. Applying I toX ⊂ V (J), one gets I(V (J)) ⊂ I(X). Since one also has the inclusionJ ⊂ I(V (J)), by transitivity J ⊂ I(X). Therefore V (I(X)) ⊂ V (J) andconsequently V (I(X)) ⊂ X. �

Lemma 3.2.4 Let X and Y be affine varieties in AqK . If I(X) = I(Y ),then X = Y

Proof. By symmetry it suffices to show the inclusion X ⊂ Y . Take α ∈ X .There are g1, . . . , gr in S such that Y = V (g1, . . . , gr). Clearly gi ∈ I(Y ) forall i. Then any gi vanishes at all points of X because I(Y ) = I(X). Thusgi(α) = 0 for all i, i.e., α ∈ Y . �

Definition 3.2.5 An affine variety X ⊂ AqK is reducible if there are affinevarieties X1 �= X and X2 �= X such that X = X1 ∪ X2; otherwise, X isirreducible.

Theorem 3.2.6 Let K be a field and let X be an affine variety of AqK , thenX is irreducible if and only if I(X) is a prime ideal.

Proof. ⇒) Let I be an ideal of S with X = V (I) and let f, g be polynomialsof S such that fg ∈ I(X). Then V (I, f) ∪ V (I, g) = V (I). As X isirreducible, we get that V (I, f) = V (I) or V (I, g) = V (I). Hence f ∈ I(X)or g ∈ I(X). Therefore I(X) is a prime ideal.⇐) Let X1 and X2 be affine varieties such that X = X1 ∪ X2. Then

I(X) = I(X1) ∩ I(X2). As I(X) is prime, by Exercise 2.1.48, we get thatI(X) = I(X1) or I(X) = I(X2). Hence, by Lemma 3.2.4, we have X = X1

or X = X2. Therefore X is irreducible. �

Proposition 3.2.7 [210, Proposition 1.5] If X ⊂ AqK is an affine varietyover a field K, then there are unique irreducible affine varieties X1, . . . , Xr

in AqK such that Xi �⊂ Xj for all i �= j and X = ∪ri=1Xi.

Proof. The existence follows using that the Zariski topology is Artinian(Exercise 3.2.16). To show uniqueness assume that Y1, . . . , Ys is anotherdecomposition of X into affine varieties such that Yi �⊂ Yj for all i �= j. Wehave the equality ∩ri=1I(Xi) = ∩si=1I(Yi). Fix 1 ≤ i ≤ r. By Theorem 3.2.6,I(Xi) is a prime ideal of S. Then, by Exercise 2.1.48, we obtain thatI(Xi) ⊃ I(Yj) for some j. Hence, by Exercise 3.2.15, Xi ⊂ Yj . Similarly,there is k such that Yj ⊂ Xk. Altogether, we have Xi ⊂ Yj ⊂ Xk. Thus,i = k and Xi = Yj for some j. This means that {X1, . . . , Xr} is containedin {Y1, . . . , Ys}. A symmetric argument shows the reverse inclusion. �


The irreducible affine varieties X1, . . . , Xr of Proposition 3.2.7 are calledthe irreducible components of X .

Theorem 3.2.8 Let S = K[t1, . . . , tq] be a polynomial ring over a field Kand let m be a maximal ideal of S. If K is algebraically closed, then thereare a1, . . . , aq ∈ K such that m = (t1 − a1, . . . , tq − aq).

Proof. There is an integral extension K[h1, . . . , hd]↪→S/m, where d =dim(S/m) = 0; see Corollary 3.1.8. Hence, the canonical map ϕ : K → S/mis an isomorphism because K is algebraically closed. To complete the proofchoose ai ∈ K so that ϕ(ai) = ai = ti = ϕ(ti). �

Proposition 3.2.9 [200, Lemma 1.8.8] If I is an ideal of a ring S andf ∈ S, then f ∈

√I if and only if (I, 1 − tf) = S[t], where t is a new

variable.

Theorem 3.2.10 (Hilbert Nullstellensatz) Let S be a polynomial ring overan algebraically closed field K and let I be an ideal of S, then

I(V (I)) =√I.

Proof. One clearly has I ⊂ I(V (I)), hence√I ⊂ I(V (I)) because I(V (I))

is a radical ideal. For the other inclusion take f ∈ I(V (I)) and consider theideal J = (I, 1− ft) of the ring S[t], where S = K[t1, . . . , tq] and t is a newvariable. Next we show the equality J = S[t]. If J �= S[t] by Theorem 3.2.8one has

(I, 1− ft) ⊂ (t1 − a1, . . . , tq − aq, t− aq+1), ai ∈ K.

Hence β ∈ V (I), where β = (a1, . . . , aq). Using that f ∈ I(V (I)) givesf(β) = 0, but this is impossible because 1− f(β)aq+1 = 0. Hence J = S[t].

Thus f ∈√I by Proposition 3.2.9. �

Corollary 3.2.11 Let X = V (f1, . . . , fr) be an affine variety, defined bypolynomials f1, . . . , fr in q variables over an algebraically closed field K.Then r ≥ q − dim(X).

Proof. By the Nullstellensatz I(X) = rad (f1, . . . , fr). Let p be a minimalprime of (f1, . . . , fr) of height q−dim(X). Applying Theorem 2.3.16 we getq − dim(X) ≤ r. �

Exercises

3.2.12 Let I and J be two ideals in a polynomial ring S over a field K. Ifrad (I) = rad (J), prove that V (I) = V (J).

126 Chapter 3

3.2.13 If K is a field and X ⊂ AqK , then X ⊂ V (I(X)) with equality if Xis an affine variety.

3.2.14 Let J be an ideal of a polynomial ring S over a field K. Prove thatJ ⊂ I(V (J)) with equality if J is the ideal of an affine variety.

3.2.15 Let X and Y be affine varieties in AqK . Then X ⊂ Y if and only ifI(X) ⊃ I(Y ).

3.2.16 (The Zariski topology is Artinian) For any descending chain

X1 ⊃ X2 ⊃ · · · ⊃ Xi ⊃ · · ·

of affine varieties in AqK there is k ∈ N \ {0} such that Xi = Xk for i ≥ k.

3.2.17 Let I be an ideal of a polynomial ring S = K[t] over an algebraicallyclosed field K and let p1, . . . , ps be the minimal primes of I. Prove that Yis an irreducible component of V (I) if and only if Y = V (pi) for some i.

3.2.18 Let K be an algebraically closed field. Then there is a one to onecorrespondence between affine varieties (resp. irreducible varieties) in AqKand radical ideals (resp. prime ideals) in the polynomial ring K[t1, . . . , tq].

3.2.19 Let I be a monomial ideal in K[t1, . . . , tq] and X = V (I) ⊂ AqK theassociated monomial variety. If K is infinite, prove that X is irreducible ifand only if X = V (ti1 , . . . , tir ).

3.2.20 Let K be an infinite field. Prove: (a) AqK is an irreducible variety,(b) any two non-empty open sets of AqK intersect, (c) any non-empty openset of AqK is dense, (d) (K∗)q is an open set of AqK , (e) (K∗)q is not an affinevariety.

3.2.21 Let S = K[t1, . . . , tq] be a polynomial ring over a field K and let mbe a maximal ideal of S. If K is uncountable, then S/m is a finite extensionof K.

3.3 Grobner bases

In this section we review some basic facts and definitions on Grobner bases.The reader may consult [1, 99, 142] for a detailed discussion of Grobnerbases and for the missing proofs of this section.

Let K be a field and let R = K[x1, . . . , xn] be a polynomial ring. Themonomials or terms of R are the power products

xa = xa11 · · ·xann , a = (a1, . . . , an) ∈ Nn.

The set of monomials of R is denoted by Mn = {xa | a ∈ Nn}.


Definition 3.3.1 A total order " of Mn is called a monomial order orterm order if

(a) xa # 1 for all xa ∈Mn, and

(b) for all xa, xb, xc ∈Mn, xa " xb implies xaxc " xbxc.

Two examples of monomial orders of Mn are the lexicographical orderor lex order defined as xb " xa iff the last non-zero entry of b−a is positive,and the reverse lexicographical order or revlex order given by xb " xa iffthe last non-zero entry of b− a is negative.

In the sequel we assume that a monomial order ≺ for Mn has been fixed.Let f be a non-zero polynomial in R. One can write

f =

r∑i=1

aiMi,

with ai ∈ K∗ = K \ {0}, Mi ∈ Mn and M1 " · · · " Mr. The leading termM1 of f is denoted by in≺(f) or lt≺(f), or simply by in(f). The leadingcoefficient a1 of f and a1M1 are denoted by lc(f) and lm(f), respectively.

Definition 3.3.2 Let I be an ideal of R. The initial ideal of I, denoted byin≺(I) or simply by in(I), is given by

in≺(I) = ({in≺(f)| f ∈ I}).

Lemma 3.3.3 (Dickson) If {Mi}∞i=1 is a sequence in Mn, then there is aninteger k so that Mi is a multiple of some term in the set {M1, . . . ,Mk} forevery i > k.

Proof. Let I ⊂ K[x1, . . . , xn] be the ideal generated by {Mi}∞i=1. By theHilbert’s basis theorem I is finitely generated (see Theorem 2.1.4). It isseen that I can be generated by a finite set of terms M1, . . . ,Mk. Hence foreach i > k, there is 1 ≤ j ≤ k such that Mi is a multiple of Mj . �

Definition 3.3.4 Let F = {f1, . . . , fq} ⊂ R \ {0} be a set of polynomialsin R. One says that f reduces to g modulo F , denoted f →F g , if

g = f − (au/ lc(fi))fi

for some fi ∈ F , u ∈ Mn, a ∈ K∗ such that a · u · in≺(fi) occurs in f withcoefficient a.

Proposition 3.3.5 The reduction relation “−→F ” is Noetherian, that is,any sequence of reductions g1 −→F · · · −→F gi −→F · · · is stationary.

128 Chapter 3

Proof. Notice that at the ith step of the reduction some term of gi isreplaced by terms of lower degree. Therefore if the sequence above is notstationary, then there is a never ending decreasing sequence of terms in Mn,but this is impossible according to Dickson’s lemma. �

Theorem 3.3.6 (Division algorithm [142, Theorem 2.11]) If f, f1, . . . , fqare polynomials in R, then f can be written as

f = a1f1 + · · ·+ aqfq + r,

where ai, r ∈ R and either r = 0 or r �= 0 and no term of r is divisible byone of in(f1), . . . , in(fq). Furthermore if aifi �= 0, then in(f) ≥ in(aifi).

Definition 3.3.7 The polynomial r in the division algorithm is called aremainder of f with respect to F = {f1, . . . , fq}.

Definition 3.3.8 Let I �= (0) be an ideal of R and let F = {f1, . . . , fr} bea subset of I. The set F is called a Grobner basis of I if

in≺(I) = (in≺(f1), . . . , in≺(fr)).

Definition 3.3.9 A Grobner basis F = {f1, . . . , fr} of an ideal I is calleda reduced Grobner basis for I if:

(i) lc(fi) = 1 ∀i, and(ii) none of the terms occurring in fi belongs to in≺(F \ {fi}) ∀i.

Theorem 3.3.10 [142, Theorem 2.17] Each ideal I has a unique reducedGrobner basis.

Definition 3.3.11 Let f, g ∈ R and let [f, g] = lcm(f, g) be its least com-mon multiple. The S-polynomial of f and g is given by

S(f, g) =[in(f), in(g)]

lm(f)f − [in(f), in(g)]

lm(g)g,

Given a set of generators of a polynomial ideal one can determine aGrobner basis using the next fundamental procedure:

Theorem 3.3.12 (Buchberger [74]) If F = {f1, . . . , fq} is a set of genera-tors of an ideal I of R, then one can construct a Grobner basis for I usingthe following algorithm:

Input: FOutput: a Grobner basis G for IInitialization: G := F , B := {{fi, fj}| fi �= fj ∈ G}while B �= ∅ do


pick any {f, g} ∈ BB := B \ {{f, g}}r := remainder of S(f, g) with respect to Gif r �= 0 thenB := B ∪ {{r, h}|h ∈ G}G := G ∪ {r}

Proposition 3.3.13 Let I be an ideal of R and let F = {f1, . . . , fq} be aGrobner basis of I. If

B = {u |u ∈Mn and u �∈ (in(f1), . . . , in(fq))},

then B is a basis for the K-vector space R/I.

Proof. First we show that B is a generating set for R/I. Take f ∈ R/I.Since “−→F ” is Noetherian, we can write f =

∑qi=1 aifi+

∑ri=1 λiui, where

λi ∈ K∗ and such that every ui is a term which is not a multiple of any of theterms in(fj). Accordingly ui is in B for all i and f is a linear combinationof the ui’s.

To prove that B is linearly independent assume h =∑s

i=1 λiui ∈ I,where ui ∈ B and λi ∈ K. We must show h = 0. If h �= 0, then we can labelthe ui’s so that u1 " · · · " us and λ1 �= 0. Hence in(h) = u1 ∈ in(I), butthis is a clear contradiction because in(I) = (in(f1), . . . , in(fq)). Thereforeh = 0, as required. �

Definition 3.3.14 A monomial in B is called a standard monomial withrespect to f1, . . . , fq.

Corollary 3.3.15 (Macaulay) If I is a graded ideal of R, then R/I andR/ in≺(I) have the same Hilbert function.

Lemma 3.3.16 [142, Proposition 2.15] Let f , g be polynomials in R andlet F = {f, g}. If in(f) and in(g) are relatively prime, then S(f, g)→F 0.

Theorem 3.3.17 [74] Let I be an ideal of R and let F = {f1, . . . , fq} be aset of generators of I, then F is a Grobner basis for I if and only if

S(fi, fj) −→F 0 for all i �= j.

Definition 3.3.18 Let I be an ideal of R generated by F = {f1, . . . , fr}and consider the homomorphism of R-modules ϕ : Rr → I, ei �−→ fi, whereei is the ith unit vector. The kernel of ϕ, denoted by Z(F ), is called thesyzygy module of I with respect to F .

130 Chapter 3

Theorem 3.3.19 [1, Theorem 3.4.1] Let G = {g1, . . . , gr} be a Grobnerbasis and write

S(gi, gj) =[in(gi), in(gj)]

lm(gi)· gi −

[in(gi), in(gj)]

lm(gj)· gj =

r∑k=1

aijkgk,

where aijk ∈ R and in(S(gi, gj)) # in(aijkgk) for all i, j, k. Then the set{[in(gi), in(gj)]

lm(gi)· ei −

[in(gi), in(gj)]

lm(gj)· ej −

r∑k=1

aijkek

}1≤i, j≤r

generates the syzygy module of I = (g1, . . . , gr) with respect to G.

Elimination of variables Let K[x1, . . . , xn, t1, . . . , tq] be a polynomialring over a field K. A useful monomial order is the elimination order withrespect to the variables x1, . . . , xn. This order is given by

xatc " xbtd

if and only if deg(xa) > deg(xb), or both degrees are equal and the last non-zero entry of (a, c)−(b, d) is negative. The elimination order with respect toall variables x1, . . . , xn, t1, . . . , tq is defined accordingly. This order is calledthe GRevLex order.

Theorem 3.3.20 Let B = K[x1, . . . , xn, t1, . . . , tq] be a polynomial ringover a field K with a term order ≺ such that terms in the xi’s are greaterthan terms in the ti’s. If I is an ideal of B with a Grobner basis G, thenG ∩K[t1, . . . , tq] is a Grobner basis of I ∩K[t1, . . . , tq].

Proof. Set S = K[t1, . . . , tq] and Ic = I ∩S. If M ∈ in(Ic), there is f ∈ Ic

with lt(f) =M . Hence M = λ lt(g) for some g ∈ G, because G is a Grobnerbasis. Since M ∈ S and xα " tβ for all α and β we obtain g ∈ G ∩ S, thatis, M ∈ (in(G ∩ S)). Thus in(Ic) = (in(G ∩ S)), as required. �

Example 3.3.21 Let≺ be the elimination order with respect to x1, . . . , x4.Using Macaulay2 [199], we can compute the reduced Grobner basis of

I = (t1 − x1x2, t2 − x1x3, t3 − x1x4, t4 − x2x3, t5 − x2x4, t6 − x3x4).

By Theorem 3.3.20, it follows that I∩K[t1, . . . , t6] = (t3t4−t1t6, t2t5−t1t6).

Definition 3.3.22 Let I and J be two ideals of a ring R. The ideal

(I : J∞) =⋃i≥1

(I : RJi)

is the saturation of I w.r.t J . If f ∈ R, we set (I : (f)∞) = (I : f∞).


The saturation can be computed by elimination of variables using thefollowing result.

Proposition 3.3.23 Let R[t] be a polynomial ring in one variable over aring R and let I be an ideal of R. If f ∈ R, then

(I : f∞) =⋃i≥1

(I : Rfi) = (I, 1 − tf) ∩R.

Proof. Let g ∈ (I, 1− tf) ∩R. Then g =∑q

i=1 aifi + aq+1(1− tf), wherefi ∈ I and ai ∈ R[t]. Making t = 1/f in the last equation and multiplyingby fm, with m large enough, one derives an equality

gfm = b1f1 + · · ·+ bqfq,

where bi ∈ R. Hence gfm ∈ I and g ∈ (I : f∞).Conversely let g ∈ (I : f∞), hence there is m ≥ 1 such that gfm ∈ I.

Since one can write

g = (1− tmfm)g + tmfmg and 1− tmfm = (1− tf)b,

for some b ∈ R[t], one derives g ∈ (I, 1− tf) ∩R. �

Exercises

3.3.24 If I and J are two ideals of a polynomial ring R over a field K andlet t be a new variable, then

I ∩ J = (t · I, (1− t) · J) ∩R.

3.3.25 Let I be an ideal of a ring R and f a non-zero element of R, then

(f)(I : f) = I ∩ (f).

3.3.26 Let R = k[x] be a polynomial ring, where k is a field. Recall thatf is a binomial in R if f = xα − xβ for some α, β in Nn. Use Buchberger’salgorithm to prove that an ideal of R generated by a finite set of binomialshas a Grobner basis consisting of binomials.

3.3.27 Let I be an ideal of a ring R and f ∈ R. Prove that the followingstatements are equivalent:

(a) f is regular on R/I.

(b) (I : Rf) = I.

(c) I = (I, 1− tf) ∩R, where t is a new variable.

3.3.28 Let Nn be the set of n-tuples of nonnegative integers endowed withthe partial order given by (a1, . . . , an) ≥ (b1, . . . , bn) if and only if ai ≥ bifor all i. If A ⊂ Nn, use Dickson’s lemma to prove that A has only a finitenumber of minimal elements.

132 Chapter 3

3.4 Projective closure

Let S = K[t1, . . . , tq] be a polynomial ring over a field K and let u be a newvariable. For f ∈ S of degree e define

fh = uef

(t1u, . . . ,

tqu

),

that is, fh is the homogenization of the polynomial f with respect to u.The homogenization of an ideal I ⊂ S is the ideal Ih of S[u] given by

Ih = ({fh| f ∈ I}).

Proposition 3.4.1 If I is an ideal of S, then√Ih =

√Ih.

Proof. To show the inclusion√Ih ⊂

√Ih, we need only observe the

equality (fh)m = (fm)h for f ∈ S and m ≥ 0. To show the reverse inclusion

let g ∈√Ih. As the ideal

√Ih is a graded ideal one may assume that g is

homogeneous. Write g = usgh1 , where g1 ∈ S. Since gm is in Ih, it followsthat g1 ∈

√I, as required. �

Let " be the elimination order on the monomials of S[u] with respectto t1, . . . , tq, u, this order extends the elimination order with respect tot1, . . . , tq on the monomials of S.

Proposition 3.4.2 Let I be an ideal of S spanned by a finite set G. SettingGh = {gh| g ∈ G} and in{Gh} = {in(gh)| g ∈ G}, the following hold.

(a) If in(Gh) = (in{Gh}), then in(I) = ({in(g)| g ∈ G}) and Ih = (Gh).(b) G is a Grobner basis of I if and only if Gh is a Grobner basis of Ih.

Proof. (a): We set G = {g1, . . . , gr} and in{G} = {in(g)| g ∈ G}. Toshow the first equality we need only show that in(I) ⊂ (in{G}). Let m be amonomial in the ideal in(I). There is g ∈ I, of degree e, such that in(g) = m.Writing g =

∑ri=1 figi for some f1, . . . , fr in S, from the equality

gh

ue=

r∑i=1

fi

(t1u, . . . ,

tqu

)gi

(t1u, . . . ,

tqu

)(3.1)

we get usgh ∈ (Gh) for s 0. As in(gh) = in(g), one has in(usgh) = usm.Hence usm ∈ (in{Gh}). Using in(gi) = in(ghi ) yields m ∈ (in{G}).

To show the second equality it is enough to show that {gh| g ∈ I} ⊂ (Gh).By the first equality G is a Grobner basis of I. Hence any g ∈ I can bewritten as g =

∑ri=1 figi, where in(g) # in(figi) for all i. Notice that

e = deg(g) ≥ deg(figi) = deg(fi)+deg(gi). Since Eq. (3.1) holds, it followsthat gh ∈ (gh1 , . . . , g

hr ).


(b) ⇒) By part (a) we need only show that in(Gh) ⊂ (in{Gh}). Letm ∈ in(Gh) be a term, then m = in(g) for some g ∈ (Gh). We may assumethat g is homogeneous. We can write g = up(m1 +m2u

e2 + · · · +msues),

wherem1, . . . ,ms are monomials in S such thatm1 " m2ue2 " · · · " msu

es .As all miu

ei have the same degree we obtain 0 ≤ e2 ≤ · · · ≤ es. It followsthat g′ = g(t1, . . . , tq, 1) belongs to I and in(g′) = m1. Thereforem1 belongsto in(I) = (in{G}). Since in(gi) = in(ghi ) we obtain that m ∈ (in{Gh}).⇐) This implication follows from part (a). �

Projective closure Let K be a field. We define the projective space ofdimension q over K, denoted by PqK , to be the quotient space

(Kq+1 \ {0})/ ∼

where two points α, β in Kq+1 \ {0} are equivalent under ∼ if α = cβ forsome c ∈ K. It is usual to denote the equivalence class of α by [α].

For any set X ⊂ PqK define I(X), the vanishing ideal of X , as the idealgenerated by the homogeneous polynomials in S[u] that vanish at all pointsof X . Conversely, given a homogeneous ideal I ⊂ S[u] define its zero set as

V (I) = {[α] ∈ PqK | f(α) = 0, ∀f ∈ I homogeneous} .

A projective variety is the zero set of a homogeneous ideal. It is notdifficult to see that the members of the family

τ = {PqK \ V (I)| I is an ideal of S[u]}

are the open sets of a topology on PqK , called the Zariski topology.

Definition 3.4.3 Let Y ⊂ AqK . The projective closure of Y is defined as

Y := ϕ(Y ), where ϕ is the map ϕ : AqK → PqK , α �→ [(α, 1)], and ϕ(Y ) is theclosure of ϕ(Y ) in the Zariski topology of PqK .

Proposition 3.4.4 If Y ⊂ AqK , then Y = V (I(ϕ(Y ))).

Proof. It is left as an exercise; see Proposition 3.2.3 for a similar formulafor the Zariski closure in the affine case. �

Proposition 3.4.5 Let Y ⊂ AqK be a set and let Y ⊂ PqK be its projectiveclosure. If f1, . . . , fr is a Grobner basis of I(Y ), then

I(Y ) = (fh1 , . . . , fhr ).

Proof. Note that I(Y ) = I(Y )h and use Proposition 3.4.2. �

Corollary 3.4.6 Let Y ⊂ AqK be a set and let Y ⊂ PqK be its projectiveclosure. Then the height of I(Y ) in S is equal to the height of I(Y ) in S[u].

134 Chapter 3

Exercises

3.4.7 If I is a prime ideal of a polynomial ring S, prove that Ih is also aprime ideal.

3.4.8 Let Y = {(x31, x21, x1)|x1 ∈ K} ⊂ A3K be a monomial curve and let Y

be its projective closure. If K is an infinite field prove that

I(Y ) = (t23 − t2u, t2t3 − t1u, t22 − t1t3).

3.4.9 Let S = K[t1, . . . , tq] be a polynomial ring over a field K. If I is anideal of S generated by f1, . . . , fm, then ((fh1 , . . . , f

hm) : u∞) = Ih.

3.4.10 Let Y = {(xd11 , . . . , xdq1 )|x1 ∈ K} be a monomial curve in the affine

space AqK . If d1 > d2 > · · · > dq and K is the field of complex numbers,then the projective closure Y of Y is equal to

Y ={[(xd11 , x

d21 u

d1−d21 , . . . , x

dq1 u

d1−dq1 , ud11 )] ∈ PqK

∣∣∣ u1, x1 ∈ K} .3.5 Minimal resolutions

The aim of this section is to study homogeneous resolutions of positivelygraded modules over polynomial rings. We shall be interested in the nu-merical data of these resolutions and in particular in the Betti numbers ofthese modules.

We begin with a result which is a consequence of the graded Nakayama’slemma (see Lemma 2.2.9).

Proposition 3.5.1 Let R = K[x] be a polynomial ring over a field K andletM be an N-graded R-module. If F = {f1, . . . , fq} is a set of homogeneouselements of M and m = (x), then F is a minimal set of generators for Mif and only if the image of F in M/mM is a K-basis of M/mM .

Corollary 3.5.2 Let R = K[x] be a polynomial ring over a field K and letm = (x). If M is an N-graded R-module, then

μ(M) = dimR/m(M/mM) = dimK(M/mM),

where μ(M) is the minimum number of generators of M .

Proof. Let f1, . . . , fr be a minimum set of generators ofM . The dimensionof M/mM , as a K-vector space, is less than or equal to r = μ(M) becausethe images of f1, . . . , fr generate M/mM . Hence, by Proposition 3.5.1, weget the required equality. �


Lemma 3.5.3 Let R = K[x] be a polynomial ring over a field K and letM be an N-graded module over R with a presentation

0 −→ ker(ϕ) −→ Rqϕ−→M −→ 0, ei

ϕ�−→ fi.

If m = (x) and F = {f1, . . . , fq} is a set of homogeneous elements, then Fis a minimal set of generators for M if and only if L = ker(ϕ) ⊂ mRq.

Proof. ⇐) Assume L ⊂ mRq. By Proposition 3.5.1 we need only showthat the image of F in M/mM is linearly independent over K. If

λ1f1 + · · ·+ λqfq ∈ mM,

for some λi’s in K, then (λi) − (ai) ∈ L ⊂ mRq, for some ai’s in m. Weare using (λi) as a short hand for (λ1, . . . , λq). Hence (λi) is in mRq. Thus,λj = 0 for all j.

⇒) If L �⊂ mRq, pick z = (z1, . . . , zq) ∈ L such that z /∈ mRq. Then ziis not in m for some i. Since the fi’s are homogeneous this readily impliesthat fi is a linear combination of elements in F \ {fi}, a contradiction. �

Definition 3.5.4 Let R = ⊕∞i=0Ri be a positively graded ring and a ∈ N.

The graded R-module obtained by a shift in the graduation of R is given by

R(−a) =∞⊕i=0

R(−a)i,

where the ith graded component of R(−a) is R(−a)i = R−a+i.

Proposition 3.5.5 Let R = ⊕∞i=0Ri be a positively graded polynomial ring

over a field K with maximal irrelevant ideal m = R+ and M an N-gradedR-module. Then there is an exact sequence of graded free modules

F. · · · −→bk⊕i=1

R(−dki)ϕk−→ · · · −→

b0⊕i=1

R(−d0i)ϕ0−→M −→ 0

where ϕk is a degree preserving map and im(ϕk) ⊂ mRbk−1 for k ≥ 1.

Proof. Let f1, . . . , fq be a set of homogeneous elements that minimallygenerate M . Set d0i = deg(fi) and b0 = q. There is an exact sequence

0 −→ Z1 = ker(ϕ0)i−→

b0⊕i=1

R(−d0i)ϕ0−→M −→ 0,

where ϕ0 is a degree preserving homomorphism such that ϕ0(ei) = fi forall i. Note that Z1 ⊂ mRb0 by Lemma 3.5.3. Since Z1 is once again afinitely generated graded R-module one may iterate the process to obtainthe required exact sequence F. �

136 Chapter 3

Theorem 3.5.6 (Hilbert syzygy theorem [128]) Let R = K[x1, . . . , xn] bea polynomial ring over a field K and let M be an N-graded R-module. ThenM has a graded free resolution of length at most n.

Proof. Set m = R+. First notice that TorRi (M,R/m) = 0 for i ≥ n + 1,because the ordinary Koszul complex K (x) is a graded free resolution ofR/m = K of length n. On the other hand assume that

· · · −→ Fkϕk−→ · · · −→ F1

ϕ1−→ F0 −→M −→ 0

is a graded free resolution ofM as in Proposition 3.5.5. Applying the functor(·)⊗R R/m yields the complex

· · · −→ Fk ⊗R/mϕk⊗1−→ · · · −→ F1 ⊗R/m

ϕ1⊗1−→ F0 ⊗R/m −→M ⊗R/m −→ 0.

Using im(ϕk) ⊂ mFk−1 for all k ≥ 1 one obtains that all the maps ϕk ⊗ 1are zero. Hence

TorRi (M,R/m) � Fi ⊗R/mfor i ≥ 1. In particular Fi ⊗ R/m � Fi/mFi = 0 for i ≥ n + 1 and byNakayama’s lemma one obtains Fi = 0 for all i ≥ n+ 1. �

Corollary 3.5.7 Let R = ⊕∞i=0Ri be a polynomial ring of dimension n over

a field K and let M be an N-graded R-module. Then there is a unique (upto complex isomorphism) exact sequence of graded modules

0 −→bg⊕i=1

R(−dgi)ϕg−→ · · · −→

bk⊕i=1

R(−dki)ϕk−→ · · ·

−→b1⊕i=1

R(−d1i)ϕ1−→

b0⊕i=1

R(−d0i)ϕ0−→M −→ 0,

such that

(a) g = sup{i |TorRi (M,K) �= 0} ≤ n, and(b) im(ϕk) ⊂ mRbk−1 for all k ≥ 1, where m = R+.

Proof. Notice the following isomorphisms:

TorRj (M,K) �bj⊕i=1

K(−dji) � Fj/mFj ,

where Fj is the jth free module in the resolution of M . Hence the bj ’s andthe dji’s are uniquely determined and so is the length of the resolution. �


Remark 3.5.8 The entries of the matrices ϕk are in m. This condition isequivalent to require that at each stage we use a minimal generating set.

Remark 3.5.9 If dk1 ≤ · · · ≤ dkbk for all k, then

d11 < d21 < · · · < dg1.

Indeed if ϕk(ei) =∑j rijej , where rij ∈ m and rij homogeneous, then

deg(ϕk(ei)) > d(k−1)1 because rij have positive degree. Hence one obtainsdk1 > d(k−1)1.

Definition 3.5.10 The integers b0, . . . , bg are the Betti numbers ofM . Thedji’s are the twists , they indicate a shift in the graduation.

In general the Betti numbers and twists of M may depend on the basefield K; see for instance [149, Example 2.10] and [346, Remark 3].

Definition 3.5.11 The R-module Zk = ker(ϕk−1) is called the k-syzygymodule of M .

Definition 3.5.12 The homogeneous resolution or minimal resolution ofM is the unique graded resolution of M by free R-modules described inCorollary 3.5.7.

IfM has a minimal free resolution as above, then note that pdR(M), theprojective dimension ofM , is equal to g. In our situation the notions of “freeresolution” and “projective resolution” coincide because of the theorem ofQuillen-Suslin: if K is a principal ideal domain, then all finitely generatedprojective K[x1, . . . , xn]-modules are free [363].

Theorem 3.5.13 (Auslander–Buchsbaum [150, Theorem 3.1]) Let M bean R-module. If R is a regular local ring, then

pdR(M) + depth(M) = dim(R).

Corollary 3.5.14 If R is a polynomial ring over a field K and I is a gradedideal, then

pdR(R/I) ≥ ht (I),

with equality if and only if R/I is a Cohen–Macaulay R-module.

Proof. The result follows from an appropriate graded version of theAuslander–Buchsbaum formula. �

Proposition 3.5.15 Let S be a positively graded algebra of dimension dover a field K. If A = K[h1, . . . , hd] ↪→ S is a homogeneous Noether nor-malization of S, then S is a Cohen–Macaulay ring if and only if S is a freeA-module.

138 Chapter 3

Proof. By Theorem 3.5.13 pdA(S) + depth(S) = dim(A), where the depthof S is taken with respect to A+. Assume S is a free A-module, thendepth(S) = d and thus the depth of S with respect to S+ is also equal d.Therefore S is Cohen–Macaulay.

Conversely assume S Cohen–Macaulay. Note that A is a polynomialring. Let p ∈ SpecS be a minimal (graded) prime over A+S. There is anintegral extension

K = A/A+ ↪→ S/p,

hence S/p is a zero dimensional domain and consequently p is a maximalideal. Therefore rad (A+S) = S+, that is, h1, . . . , hd is a h.s.o.p for S. Nowuse Proposition 3.1.20 to conclude that h1, . . . , hd is a regular sequence inS, that is, depthS = d. Another application of the Auslander–Buchsbaumformula yields that S is a free A-module. �

Theorem 3.5.16 (Hilbert–Burch [128, Theorem 20.15]) Let R be a poly-nomial ring over a field K and let I be a graded Cohen–Macaulay ideal ofheight two. If

0 −→ Rq−1 ϕ−→ Rq −→ R −→ R/I −→ 0

is the minimal resolution of R/I, then I is generated by all the minors ofsize q − 1 of the matrix ϕ.

Example Let R = Q[x, y, z, w] and let I = (f1, f2, f3), where

f1 = y2 − xz, f2 = x3 − yzw, f3 = x2y − z2w.Let us construct the minimal resolution ofR/I. Consider the exact sequence

0 −→ ker(ϕ1) = Z1id−→ R(−2)⊕R2(−3) ϕ1−→ I −→ 0, ei

ϕ1�−→ fi.

Here Z1 is the module of syzygies of I. Using Macaulay we find that Z1 isgenerated as an R-modulo by the column vectors, ψ1, ψ2, of the matrix:

A =

⎡⎣ zw x2

y z−x −y

⎤⎦Notice that R(−2)⊕R2(−3) is graded by

(R(−2)⊕R2(−3))i = R(−2)i ⊕R(−3)i ⊕R(−3)i.Hence ψ1, ψ2 ∈

[R(−2)⊕R2(−3)

]4. Next consider the exact sequence

0 −→ ker(ϕ2) = Z2id−→ R2(−4) ϕ2−→ Z1 −→ 0, ei

ϕ2�−→ ψi.

Since ψ1R + ψ2R is a free R-module, we have Z2 = (0). Altogether theminimal homogeneous resolution of R/I is:

0 −→ R2(−4) ϕ2−→ R(−2)⊕R2(−3) ϕ1−→ R −→ R/I −→ 0.

In this example R/I is Cohen–Macaulay because pdR(R/I) = 2 = ht (I).


Pure and linear resolutions Let R = ⊕∞i=0Ri be a polynomial ring over

a field K with its usual graduation and I a graded ideal of R. Let

0→bg⊕j=1

R(−dgj)→ · · · →b1⊕j=1

R(−d1j)→ R→ S = R/I → 0 (3.2)

be the minimal graded resolution of S by free R-modules. The ideal I (orthe algebra S) has a pure resolution if there are constants

d1 < d2 < · · · < dg

such that d1j = d1, . . . , dgj = dg for all j. If in addition dj = d1 + j − 1 for2 ≤ j ≤ g the resolution is said to be d1-linear .

Theorem 3.5.17 (Herzog–Kuhl [230]) If S is a Cohen–Macaulay algebrawith a pure resolution, then

bi =∏j �=i

dj|dj − di|

(1 ≤ i ≤ g).

Theorem 3.5.18 (Huneke-Miller [256]) If S is a Cohen–Macaulay algebrawith a pure resolution, then the multiplicity e(S) of S is given by

e(S) = d1 · · · dg/g!.

Eisenbud and Schreyer proved the following long-standing multiplicityconjecture stated in [237, Conjecture 1, p. 2880].

Theorem 3.5.19 (Huneke–Srinivasan Multiplicity Conjecture [133]) If Sis a Cohen–Macaulay algebra of codimension g, then

1

g!

g∏i=1

mi ≤ e(S) ≤1

g!

g∏i=1

Mi,

where Mi = max{dij | j = 1, . . . , bi} and mi = min{dij | j = 1, . . . , bi}.

Herzog and Srinivasan [237, Conjecture 2, p. 2881] conjectured thatthe second inequality holds for non-Cohen–Macaulay algebras. This conjec-ture was shown by Boij and Soderberg [47]. Since the Betti numbers andthe twists are positive integers, these formulas impose restrictions on thenumbers that can occur in the resolution of S.

140 Chapter 3

Exercises

3.5.20 Let R be a polynomial ring and let I be a graded Cohen–Macaulayideal of height 3 with a pure resolution:

0→ Rb3(−(d+ a+ b))→ Rb2(−(d+ a))→ Rb1(−d)→ I → 0,

where a, b ≥ 1. Assume b1 = 6 and b = 1. Show that x = a + 1 andy = d+ a+ 1 is a solution of the diophantine equation

6x(x− 1) = y(y − 1).

Then prove that the integral solutions of this equation are given by

x = (X + 1)/2 and y = (Y + 1)/2,

where√6X + Y = ±(1±

√6)(5 + 2

√6)n, n ∈ Z.

3.5.21 Let R = K[x1, . . . , xn] be a polynomial ring over a field K and let Ibe a graded ideal, then (x1, . . . , xn) is an associated prime of I if and onlyif pdR(R/I) = n.

3.5.22 Let R = K[x, y, z] be a polynomial ring and let I be the ideal

(xn + yn − zn|n ≥ 2).

If char(K) �= 2, then I = (x2 + y2 − z2, x3 + y3 − z3, x2y2). If char(K) �= 2,prove that q1 ∩ q2 ∩ q3 is a primary decomposition of I, where

q1 = (y2, x− z), q2 = (x2, y − z), q3 = (y3, z3, x2y2, x2 + y2 − z2).

3.5.23 Let R = K[x] be a polynomial ring over a field K. If I is a homo-geneous ideal of R of height r which is generated by r polynomials, then Iis generated by r homogeneous polynomials.

Chapter 4

Rees Algebras andNormality

In this part a detailed presentation of complete and normal ideals is given.The systematic use of Rees algebras and associated graded rings will makeclear their importance for the area. Some outstanding references for blowupalgebras and the normality of Rees algebras are [59, 412, 414]. A generalreference for integral closure of ideals, rings, and modules is [259].

4.1 Symmetric algebras

Let M be an R-module. Given n ≥ 0, we define

T n(M) =M ⊗ · · · ⊗M︸︷︷︸n−times

and T 0(M) = R.

The tensor algebra T (M) of M is the noncommutative graded algebra

T (M) =

∞⊕n=0

T n(M),

where the product in T (M) is induced by juxtaposition, that is, the productof x1 ⊗ · · · ⊗ xm and y1 ⊗ · · · ⊗ yn is x1 ⊗ · · · ⊗ xm ⊗ y1 ⊗ · · · ⊗ yn.

The symmetric algebra of M , denoted by SymR(M) or simply Sym(M),is defined as the quotient algebra

Sym(M) = T (M)/J

where J is the two-sided ideal generated by all xy − yx = x ⊗ y − y ⊗ xwith x and y running through M . Observe that Sym(M) is commutative.

142 Chapter 4

Since J is a graded ideal generated by homogeneous elements of degreetwo, the symmetric algebra is graded by

Symn(M) = T n(M)/J ∩ T n(M) and Sym0(M) = R.

Note that Sym2(M) =M⊗M/(x⊗y−y⊗x), with x and y running throughall the elements of M .

Exercises

4.1.1 If M is a free R-module of rank n prove that the symmetric algebraof M is a polynomial ring in n variables with coefficients in R.

4.2 Rees algebras and syzygetic ideals

Let I be an ideal of a ring R generated by f1, . . . , fq. The Rees algebra ofI, denoted by R[It] or R(I), is the subring of R[t] given by

R[It] = R[f1t, . . . , fqt] ⊂ R[t],

where t is a new variable. Note

R[It] = R⊕ It⊕ · · · ⊕ Intn ⊕ · · · ⊂ R[t].

There is an epimorphism of R-algebras

ϕ : B = R[t1, . . . , tq] −→ R[It] −→ 0, tiϕ�−→ fit,

where B = R[t] is a polynomial ring over the ring R. The kernel of ϕ,denoted by J , is the presentation ideal of R[It] with respect to f1, . . . , fq.Notice that J is a graded ideal in the ti-variables: J =

⊕∞i=1 Ji, where B

has the standard grading induced by setting deg(ti) = 1.The mapping ψ : Rq −→ I given by ψ(z1, . . . , zq) =

∑qi=1 zifi induces

an R-algebra epimorphism

β : R[t1, . . . , tq] −→ SymR(I).

Thus, the symmetric algebra of I is:

SymR(I) � R[t1, . . . , tq]/ker(β),

where ker(β) is an ideal of R[t] generated by linear forms:

ker(β) =

({q∑i=1

biti

∣∣∣∣∣q∑i=1

bifi = 0 and bi ∈ R})

.

Rees Algebras and Normality 143

On the other hand the kernel of ϕ is generated by all forms F (t1, . . . , tq)such that F (f1, . . . , fq) = 0. In particular, one may factor ϕ throughSymR(I) and obtain the commutative diagram:

R[t1, . . . , tq]ϕ−→ R[It]

↓β α ↗

SymR(I)

We say that I is an ideal of linear type if α is an isomorphism.

An important module-theoretic obstruction to “SymR(I) � R[It]” isgiven by the following result.

Proposition 4.2.1 (Herzog–Simis–Vasconcelos [233]) Let I be an ideal ofa ring R. If SymR(I) � R[It], then for each prime p containing I, Ip canbe generated by ht (p) elements.

Syzygetic ideals Let I be an ideal of a ring R and let H1(I) be thefirst homology module of the Koszul complex H (x,R) associated to a setx = x1, . . . xn of generators of I. In [382] it is pointed out that H1(I) isrelated to I/I2, the conormal module of I, by the following exact sequence:

H1(I)f−→ Rn ⊗ (R/I)

h−→ I ⊗ (R/I) = I/I2 −→ 0. (∗)

Here f([z]) = z ⊗ 1 and h(ei ⊗ 1) = xi ⊗ 1. Set δ(I) = ker(f).

Definition 4.2.2 The ideal I is called syzygetic if δ(I) = 0.

Although H1(I) may depend on the set of generators for I, δ(I) dependsonly on I. Indeed Simis and Vasconcelos [382] proved that

δ(I) � ker(Sym2(I)→ I2),

where Sym2(I)→ I2 is the surjection induced by the multiplication map.

Definition 4.2.3 An ideal I of a ring R is said to be generically a completeintersection if IRp is a complete intersection for all p ∈ AssR(R/I).

Remark 4.2.4 If I is unmixed and generically a complete intersection,then the R/I-torsion of H1(I) equals δ(I). To prove this, notice that Ip issyzygetic for all p ∈ AssR(R/I); consequently, δ(I) is a torsion module.

144 Chapter 4

Computing the presentation ideal of a Rees algebra Let I be agraded ideal of a polynomial ring R over a field k and let f1, . . . , fq be agenerating set for I.

Consider the presentation of the Rees algebra:

ϕ : B = R[t1, . . . , tq] −→ R[It] −→ 0, (ti �−→ tfi).

The kernel J of ϕ can be obtained as follows:

J = (t1 − tf1, . . . , tq − tfq) ∩B.

Since J is a graded ideal in the ti-variables, J = ⊕i≥1Ji. The relationshipbetween J and the Koszul homology of I is very tight. The exact sequenceof Eq. (∗) can be made precise:

0 −→ J2/B1J1 = δ(I) −→ H1(I) −→ (R/I)q −→ I/I2 −→ 0.

In particular one can decide whether I is syzygetic – that is, J2 = B1J1–orof linear type–that is, J = J1B.

Exercises

4.2.5 Let A[x] be a polynomial ring over a ring A and let f1, . . . , fq be formsof degree d ≥ 1 in A[x]. Then there is a graded isomorphism of A-algebras

ϕ : A[f1, . . . , fq] −→ A[tf1, . . . , tfq], with ϕ(fi) = tfi,

where t is a new variable and both rings have an appropriate grading suchthat deg(fi) = 1 and deg(tfi) = 1.

4.2.6 Let I = (f1, . . . , fq) be an ideal of a polynomial ring R = K[x] overa field K. Prove that

R[It] � R[t1, . . . , tq]/((f2t1 − f1t2, . . . , fqt1 − f1tq) : f∞1 ).

4.2.7 Let I = I2(X) be the ideal of 2× 2-minors of the symmetric matrix:

X =

⎡⎣ x1 x2 x3x2 x4 x5x3 x5 x6

⎤⎦ ,where the entries of X are indeterminates over a field K. Prove that I is asyzygetic ideal that satisfies sliding depth. See [238].


4.3 Complete and normal ideals

Let R be a ring and let I be an ideal of R, an element z ∈ R is integral overI if z satisfies an equation

z� + a1z�−1 + · · ·+ a�−1z + a� = 0, ai ∈ Ii,

the integral closure of I is the set of all elements z ∈ R which are integralover I. This set will be denoted by I or Ia.

Definition 4.3.1 If I = I, I is said to be integrally closed or complete. Ifall the powers Ik are complete, the ideal I is said to be normal.

The simplest kinds of integrally closed monomial ideals are the Stanley–Reisner ideals and the powers of face ideals.

Lemma 4.3.2 Let R be a ring and I an ideal of R. An element α ∈ R isin the integral closure of I if and only if there is an ideal L of R such that

(i) αL ⊂ IL, and(ii) αrann(L) = (0) for some integer r ≥ 0.

Proof. ⇒) As α is in integral closure of I there is an equation:

αn = a1αn−1 + · · ·+ an−1α+ an, where ai ∈ Ii,

and n ≥ 1 is some integer. The required L is obtained by setting

L = Rαn−1 + Iαn−2 + · · ·+ In−2α+ In−1.

Indeed, from the equation above

Rαn ⊂ Iαn−1 + I2αn−2 + · · ·+ In−1α+ In = IL,

and α(Iiαn−i−1) = Iiαn−i = I(Ii−1αn−i) ⊂ IL for i ≥ 1. Hence αL ⊂ IL.To finish this part of the proof note αn−1ann(L) = (0).

⇐) As R is Noetherian, L = (f1, . . . , fn). Then αfi =∑n

j=1 bijfj, wherebij ∈ I. Set B = (bij), C = B − αIn and f = (f1, . . . , fn). Here In denotesthe identity matrix. As Cf t = 0, one can use the formula

C · adj(C) = det(C)In

to conclude fi det(C) = 0 for all i. Therefore det(C) is in ann(L) and byhypothesis αr det(C) = 0 for some r. Expanding in powers of α gives thatα is integral over I. �

Proposition 4.3.3 If I is an ideal of a ring R, then I is an ideal of R.

146 Chapter 4

Proof. Let αi ∈ I, i = 1, 2. By the proof of Lemma 4.3.2, there is an idealLi such that αiLi ⊂ ILi, α

rii ann(Li) = (0) and Iri ⊂ Li for some ri ≥ 0.

Since αi is integral over I, one has αni

i ∈ I for some ni ≥ 1. Therefore forn large enough βn is in L1L2, where β = α1 + α2. Thus

βnann(L1L2) = (0).

On the other hand one clearly has the inclusion βL1L2 ⊂ IL1L2. Hence,by Lemma 4.3.2, one concludes β ∈ I.

To finish proving that I is an ideal note xα1 ∈ I for x ∈ R, this followsat once from the definition of I. �

Proposition 4.3.4 If I is an ideal of a ring R and S is a multiplicativelyclosed subset of R, then

S−1(I) = S−1(I).

Proof. To prove S−1(I) ⊂ S−1(I) take f ∈ S−1(I). One may assumef = x/1 with x ∈ R because the integral closure of an ideal is again anideal. There is an equation

fn +a1s1fn−1 + · · ·+ an−1

sn−1f +

ansn

=0

1,

where ai/si ∈ S−1(I)i = S−1(Ii). Thus one may assume ai ∈ Ii and si ∈ Sfor all i. Clearing denominators and multiplying by an appropriate elementof S one has an equality (in the ring R) of the form

sxn + t1a1xn−1 + · · ·+ tn−1an−1x+ tnan = 0 (s, ti ∈ S),

multiplying both sides of this equation by sn−1 yields sx is in I. Thereforef = (xs)/s is in S−1(I). The other inclusion follows readily using thatlocalizations commute with powers of ideals. �

Lemma 4.3.5 Let A be a domain and x ∈ A \ {0} such that Ax is normal.Then A is normal if and only if (x) is a complete ideal.

Proof. Assume that A is a normal domain. Take z ∈ (x). Since z satisfiesan equation of the form

zm + (a1x)zm−1 + · · ·+ (am−1x

m−1)z + (amxm) = 0, ai ∈ A,

dividing by xm yields that z/x is integral over A and z ∈ (x).Conversely assume (x) is complete. Let z be an element of the quotient

field of A which is integral over A. As A and Ax have the same field offractions and Ax is normal one may assume z = a/xr, for some a ∈ A andr ≥ 1. To show z ∈ A it suffices to verify a ∈ (x). There is an equation

zm + b1zm−1 + · · ·+ bm−1z + bm = 0, bi ∈ A,


multiplying by xrm yields the identity

am + (b1xr)am−1 + · · ·+ (bm−1x

r(m−1))a+ (bmxrm) = 0,

hence a ∈ (x) = (x), as required. �

Proposition 4.3.6 If A is a domain and x ∈ A \ {0}, then

A = Ax⋂⎛⎝ ⋂

p∈AssA/(x)

Ap

⎞⎠ .

Proof. Note that the left-hand side of the equality is contained in theright-hand side because A is a domain.

Conversely take an element z ∈ Ax and z ∈ Ap, for all p ∈ AssA/(x).Write z = a/xn, a ∈ A and n ≥ 1. It is enough to prove that a ∈ (x). If ais not in (x) note that

((x) : a) ⊂ Z(A/(x)).Hence ((x) : a) ⊂ p, for some p ∈ AssA/(x). Since z ∈ Ap one may writez = a/xn = b/s, b ∈ A and s �∈ p. Therefore s ∈ ((x) : a) ⊂ p, which yieldsa contradiction. �

Proposition 4.3.7 If A is an integral domain and S is a multiplicativelyclosed subset of A, then

S−1(A) = S−1(A).

Proof. Note that A and S−1(A) have the same field of fractions K. Firstwe prove S−1(A) ⊂ S−1(A). Take any x in K integral over S−1(A). Thereis an equation

xn +a1s1xn−1 + · · ·+ an−1

sn−1x+

ansn

=0

1,

where ai ∈ A and si ∈ S for all i. Set s = s1 · · · sn. If we multiply by sn, itfollows that sx ∈ A and x ∈ S−1(A).

Conversely take x ∈ S−1(A). There is s ∈ S such that sx is integralover A. Hence sx satisfies

(sx)n + a1(sx)n−1 + · · ·+ an−1(sx) + an = 0,

for some a1, . . . , an in A, dividing by sn immediately yields that x is integralover S−1(A), as required. �

Corollary 4.3.8 If A is a normal domain and S is a multiplicatively closedsubset of A, then S−1(A) is a normal domain.

148 Chapter 4

Corollary 4.3.9 Let R = k[x1, . . . , xn] be a polynomial ring over a field kand let KR be its field of fractions. If R′ = k[x±1

1 , . . . , x±1n ] ⊂ KR is the

ring of Laurent polynomials, then R′ is a normal domain.

Proof. Notice that R′ is the localization of R at the multiplicative set ofmonomials of R. Hence, by Corollary 4.3.8, we get that R′ is normal. �

Corollary 4.3.10 Let A be a domain and x ∈ A \ {0}. Then A is normalif and only if Ax and Ap are normal for every p ∈ AssA/(x).

Proof. If A is normal, by Corollary 4.3.8, Ax and Ap are normal for everyx ∈ A \ {0} and p ∈ Spec(A). For the converse use Proposition 4.3.6 andobserve that Ax and Ap have the same quotient field as A. �

To state a useful criterion of normality of Rees algebras, it is convenientto introduce the extended Rees algebra of an ideal I in a ring R:

A = R[It, u], u = t−1.

Part of its usefulness is derived from the equality Au = R[t, t−1].

Proposition 4.3.11 Let I be an ideal of a ring R and A = R[It, t−1] itsextended Rees algebra. If R is a normal domain, then A is normal if andonly if Ap is normal for each associated prime p of t−1A.

Proof. Note At−1 = R[t, t−1]. Hence At−1 is a normal domain and one canuse Corollary 4.3.10. �

Lemma 4.3.12 Let A be a ring and x a regular element of A. If A/(x) isreduced and p ∈ AssAA/(x), then Ap is a normal domain.

Proof. Let AssAA/(x) = {p1, . . . , pr}. Since A/(x) is reduced

(x) = p1 ∩ · · · ∩ pr,

hence (x)Api = piApi for all i. Note that ht (pi) = 1, by Krull’s principalideal theorem. Therefore the maximal ideal of Api is generated by a systemof parameters, that is, Api is a regular local ring and consequently Api is anormal domain by Theorem 2.4.16. �

Proposition 4.3.13 Let A be a ring and x a regular element of A. If Axis a normal domain and A/(x) is reduced, then A is a normal domain.

Proof. As Ax is a domain and x is a regular element on A one obtainsthat A is a domain. By Lemma 4.3.12 Ap is normal for all p ∈ AssAA/(x),hence A must be normal according to Corollary 4.3.10. �


There is another graded algebra associated to an ideal I of a ring Rwhose properties are related to the normality of the Rees algebra of I, it iscalled the associated graded ring of I and is defined as:

grI(R) = R/I ⊕ I/I2 ⊕ · · · ⊕ Ii/Ii+1 ⊕ · · · ,

with multiplication

(a+ Ii)(b+ Ij) = ab+ Ii+j−1 (a ∈ Ii−1, b ∈ Ij−1).

Given a generating set f1, . . . , fq of I, it is not difficult to verify thatgrI(R) is a graded algebra over R/I generated by the following elements ofdegree one:

f1 = f1 + I2, . . . , fq = fq + I2.

Thus one has grI(R) = (R/I)[ f1, . . . , fq].

Lemma 4.3.14 If I is an ideal of a ring R, then

R[It]/IR[It] � grI(R) and A/t−1A � grI(R),

where A = R[It, t−1] is the extended Rees algebra of I.


Theorem 4.3.15 Let I be an ideal of a ring R. If I is generated by aregular sequence f1, . . . , fq, then the epimorphism of graded algebras :

ϕ : (R/I)[t1, . . . , tq] −→ grI(R), ϕ(ti) = f i = fi + I2,

is an isomorphism, where t1, . . . , tq are indeterminates over R/I.

Proof. The proof is by induction on q, the case q = 1 is easy to show. LetJ be the ideal (f1, . . . , fq−1) and consider the epimorphism:

ϕ′ : (R/J)[t1, . . . , tq−1] −→ grJ (R) = (R/J)[ f1, . . . , fq−1],

ϕ′(ti) = f i = fi + J2. Note (J : fq) = J because fq is regular on R/J ;moreover, since ϕ′ is an isomorphism it follows (by induction on m) thatone has the following equalities

(Jm : fq) = Jm for all m ≥ 1.

Let F ∈ R[t] = R[t1, . . . , tq] be a homogeneous polynomial of degree dsuch that its image in (R/I)[t] is in the kernel of ϕ, that is,

F (f) = F (f1, . . . , fq) ∈ Id+1.

150 Chapter 4

As ϕ is graded it suffices to prove F ∈ IR[t]. To show F ∈ IR[t] we proceedby induction on d, the case d = 0 is clear. There is W ∈ R[t] of degreed + 1 with F (f) = W (f), write W =

∑qi=1 tiWi, where Wi = 0 or Wi is a

homogeneous polynomial in R[t] of degree d. There are polynomials G inR[t1, . . . , tq−1] and H in R[t] of degrees d and d− 1, respectively, such that

F ′ = F −q∑i=1

fiWi = G+ tqH.

Observe that F ′(f) = 0, hence H(f) ∈ (Jd : fq) = Jd ⊂ Id and by inductionon d one concludes that H has coefficients in I. It only remains to provethat G has coefficients in I. There is a polynomial H ′ ∈ R[t1, . . . , tq−1] ofdegree d with H(f) = H ′(f). Set

F ′′ = G+ fqH′ ∈ R[t1, . . . , tq−1],

noting F ′′(f) = F ′(f) = 0 and using that ϕ′ is an isomorphism one derivesthat F ′′ has coefficients in J , which implies that G has coefficients in I asrequired. �

Theorem 4.3.16 Let I be an ideal of a ring R and A = R[It, t−1] itsextended Rees algebra. If R is a normal domain and grI(R) is reduced, thenA is normal.

Proof. As A/t−1A � grI(R) is reduced and At−1 = R[t, t−1] is a normaldomain, by Proposition 4.3.13, we get that A is normal. �

Theorem 4.3.17 [236] Let I be an ideal of a normal domain R. Then thefollowing are equivalent:

(a) I is a normal ideal of R.

(b) The Rees algebra R[It] is normal.

(c) The ideal IR[It] ⊂ R[It] is complete.

(d) The ideal (t−1) ⊂ R[It, t−1] is complete.

(e) The extended Rees algebra R[It, t−1] is normal.

Proof. (a)⇒ (b) Set A = R[It]. Let z ∈ A ⊂ R[t] and write z =∑s

i=0 biti.

It suffices to prove bsts ∈ A. First we prove that bst

s ∈ A. As z is almostintegral over A there is 0 �= f ∈ A such that fzn ∈ A for all n > 0. Hencethere is 0 �= fm ∈ Im such that (fmt

m)(bsts)n is in A for all n > 0; that is,

bsts is almost integral over A. As R is a Noetherian integral domain using

Proposition 2.6.37 one derives that bsts is integral over A. Thus there is an

equation of the form

(bsts)m + a1(bst

s)m−1 + · · ·+ am−1(bsts) + am = 0,


where ai =∑rij=0 aijt

j and aij ∈ Ij . Grouping all the terms of t-degreeequal to sm one has the equation

bms +

m∑i=1

ai,sibm−is = 0.

Thus bs is integral over Is and consequently bs ∈ Is, as required.(b) ⇒ (c) Let z = b0 + b1t + · · · + bst

s be an element of R[It] which isintegral over IR[It]. Then z satisfies an equation of the form

zm + a1zm−1 + · · ·+ am−1z + am = 0, ai ∈ IiR[It],

multiplying by tm one obtains that tz is integral over R[It]. Thereforetz ∈ R[It], which proves that z ∈ IR[It].

(c) ⇒ (d) Set B = R[It, t−1]. Let z be an element of B integral overt−1B. As the negative part of the Laurent expansion of z is in t−1B, onemay assume z =

∑si=0 bit

i, s ≥ 0 and bi ∈ Ii for all i ≥ 0. By descendinginduction on s it suffices to prove that bs ∈ Is+1. There is an equation

zm + a1zm−1 + · · ·+ am−1z + am = 0, ai ∈ (t−1)iB,

hence zt is almost integral over B. It follows rapidly that bsts+1 is almost

integral over B and thus bsts+1 is integral over B. A direct calculation

yields that bsts+1 is integral over IR[It], which shows that bs is in Is+1.

(d) ⇒ (e) Set u = t−1 and note R[It, u]u = R[t, u] is a normal domain.Therefore, by Lemma 4.3.5, one concludes that R[It, u] is normal.

(e)⇒ (a) If z ∈ Ir, then z satisfies a polynomial equation

zm + a1zm−1 + · · ·+ am−1z + am = 0, ai ∈ Iri,

thus multiplying by trm we get that the element ztr is integral over the ringR[It, t−1]. Hence z ∈ Ir. �

Corollary 4.3.18 Let I be an ideal of a normal domain R and R[It] itsRees algebra. If grI(R) is reduced, then R[It] is normal.

Proof. It follows from Theorems 4.3.16 and 4.3.17. �

Theorem 4.3.19 (Huneke [254]) Let R be a Cohen–Macaulay ring and letI be an ideal of R containing regular elements. If R[It] is Cohen–Macaulay,then grI(R) is Cohen–Macaulay.

An ideal I of a ring R is a radical ideal if I = rad (I). Note: (i) a properideal I is radical if and only if I is an intersection of finitely many primes,and (ii) I ⊂ rad (I) and equality occurs if I is a radical ideal.

152 Chapter 4

Corollary 4.3.20 Let I be a radical ideal of a normal domain R. If Iis generated by a regular sequence, then grI(R) is reduced and R[It] is anormal domain.

Proof. It follows from Theorem 4.3.15 and Corollary 4.3.18. �

Example 4.3.21 Let R = Q[x, y] be a polynomial ring over the field Qand let I be the ideal (x2, y2). Observe that I is equal to (x2, y2, xy). ThusR[It] is not normal because I is not even complete.

Definition 4.3.22 Let I be an ideal of a ring R and p1, . . . , pr the minimalprimes of I. Given an integer n ≥ 1, the nth symbolic power of I is definedto be the ideal

I(n) = q1 ∩ · · · ∩ qr,

where qi is the primary component of In corresponding to pi.

Let (R,m) be a regular local ring and I an unmixed ideal. An interestingproblem is whether I(2) is contained in mI, although the answer is negativein general, the problem remains open in characteristic zero. For some insightinto this problem see [132, 257].

Proposition 4.3.23 Let I be a proper ideal of a ring R and S = R\∪ri=1pi,where p1, . . . , pr are the minimal primes of I. Then

I(n) = S−1In ∩R for n ≥ 1.

Proof. Let In = q1 ∩ · · · ∩ qr ∩ qr+1 ∩ · · · ∩ qs be a primary decomposition,where qi is the primary component of pi for i ≤ r and qi is a primarycomponent of the embedded prime pi for i > r. Since pi ∩ S �= ∅ for i > rone has S−1qi = S−1R. Hence

S−1In = S−1

(s⋂i=1

qi

)=

r⋂i=1

S−1qi.

To finish the argument note that S−1qi ∩ R = qi and intersect with R theequality above. �

Proposition 4.3.24 Let I be a radical ideal of a ring R and p1, . . . , pr theminimal primes of I. Then

I(n) = p(n)1 ∩ · · · ∩ p(n)r for n ≥ 1.

Proof. Let In = q1 ∩ · · · ∩ qr ∩ qr+1 ∩ · · · ∩ qs be a primary decompositionof In, where qi is pi-primary for i ≤ r and qi is an embedded primary


component of In for i > r. Localizing at pi yields InRpi = qiRpi and from

I = p1 ∩ · · · ∩ pr one obtains:

InRpi = (IRpi)n = (piRpi)

n = pni Rpi .

Thus pni Rpi = qiRpi and contracting one has p(n)i = qi, as required. �

This result and also Exercise 6.1.25 have interesting generalizations tomonomial ideals [92, Theorem 3.7].

Proposition 4.3.25 Let R be a polynomial ring over a field K and let I bean ideal of R generated by square-free monomials. If n ≥ 1 and p1, . . . , prare the minimal primes of I, then

I(n) = pn1 ∩ · · · ∩ pnr .

Proof. From Proposition 6.1.7 one has pni = p(n)i . Since I is a radical ideal

the result follows from Proposition 4.3.24. �

Corollary 4.3.26 Let R be a polynomial ring over a field K. If I is anideal of R generated by square-free monomials, then I(n) is integrally closedfor n ≥ 1.

Proof. Set J = I(n). Let p1, . . . , pr be the associated primes of I. If f ∈ J ,then there is m ≥ 1 so that fm ∈ pnmi for all i. By Corollary 4.3.20 pni iscomplete, thus f ∈ pni for all i. Hence Proposition 4.3.25 shows f ∈ J . �

Corollary 4.3.27 Let R be a polynomial ring over a field K and let I bea monomial ideal. If I is a radical ideal, then In ⊂ I(n) for n ≥ 1.

Proof. Let x ∈ In, there is k ≥ 1 so that xk ∈ (In)k ⊂ (I(n))k, thus x is in

∈ I(n). Note that I(n) is complete by Corollary 4.3.26, hence x ∈ I(n). �

Definition 4.3.28 An ideal I of a ring R is called normally torsion-free ifAss(R/Ii) is contained in Ass(R/I) for all i ≥ 1 and I �= R.

Brodmann [56] showed that when R is a Noetherian ring and I is anideal of R the sets Ass(R/In) stabilize for large n. If I is a radical idealwhich is normally torsion-free, then Ass(R/In) = Ass(R/I) for all n ≥ 1;that is, the notion of normally torsion-free is a strong form of stability.

An aspect of symbolic powers that has attracted a lot of attention is todescribe when the symbolic and ordinary powers of a given ideal I coincide;see [249, 333] and [185, 188]. Later in this chapter, we present a few caseswhere equality of symbolic and ordinary powers can be described in termsof properties of the associated graded ring.

In Chapter 14 we characterize normally torsion-free monomial ideals inalgebraic and combinatorial optimization terms (see Theorem 14.3.6).

For a certain type of ideals, the next result characterizes normally torsion-free ideals as those ideals whose ordinary and symbolic powers coincide.

154 Chapter 4

Proposition 4.3.29 Let I be an ideal of a ring R. If I has no embeddedprimes, then I is normally torsion-free if and only if In = I(n) for all n ≥ 1.

Proof. ⇒) Note Ass(R/In) = Ass(R/I) for n ≥ 1, because any associatedprime p of I is a minimal prime of In and thus p ∈ Ass(R/In). As In has noembedded primes one concludes that In has a unique irredundant minimalprimary decomposition and In = I(n) for n ≥ 1.⇐) Let p1, . . . , pr be the associated primes of I. Since pi is a minimal

prime of I for all i, one derives

Ass(R/In) = Ass(R/I(n)) = {p1, . . . , pr}. �

Proposition 4.3.30 Let R be a polynomial ring with coefficients in a field.If I is a radical monomial ideal of R which is normally torsion-free, thenits Rees algebra R[It] is a normal domain.

Proof. By Proposition 4.3.29 In = I(n) for n ≥ 1. Hence, thanks toCorollary 4.3.27, In is integrally closed for n ≥ 1, i.e., R[It] is normal. �

Proposition 4.3.31 Let I be an ideal of a Cohen–Macaulay ring R. If Iis generated by a regular sequence, then In = I(n) for n ≥ 1.

Proof. Let f1, . . . , fr be an R-regular sequence that generates I. By Krull’sprincipal ideal theorem ht (I) = r. Hence, by Theorem 2.3.25, I is unmixed.From Theorem 4.3.15 there is an isomorphism of graded rings

ϕ : B = (R/I)[t1, . . . , tr] −→ grI(R), ti �−→ fi + I2,

where B is a polynomial ring over R/I. Hence Ii/Ii+1 is a free R/I-modulobecause it is isomorphic to Bi, the ith graded component of B. Thus

AssR(Ii/Ii+1) = AssR(R/I).

Using the exact sequence 0 −→ Ii/Ii+1 −→ R/Ii+1 −→ R/Ii −→ 0, itfollows by induction that AssR(R/I

n) ⊂ AssR(R/I) for n ≥ 1. To finishthe proof, we apply Proposition 4.3.29 to conclude the equality between theordinary and symbolic powers of I. �

Definition 4.3.32 A proper ideal I of a ring R is said to be locally acomplete intersection if IRp is a complete intersection for all p ∈ V (I).

Proposition 4.3.33 Let R be a Cohen–Macaulay ring and I a prime ideal.If I is locally a complete intersection, then In = I(n) for all n ≥ 1.

Proof. By Proposition 4.3.29 it suffices to prove Ass(R/In) ⊂ Ass(R/I) forn ≥ 1. If p is an associated prime of R/In, then pRp is an associated primeof Rp/I

np . Using Proposition 4.3.31 yields that Ip is normally torsion-free,

that is, the only associated prime of Rp/Inp is Ip. Therefore pRp = Ip and

p = I. �


Proposition 4.3.34 Let p be a prime ideal of a ring R such that pRp is acomplete intersection, then p(n) = pn for all n ≥ 1 if and only if grp(R) isa domain.

Proof. ⇒) Since grp(R) = R[pt]/pR[pt], it is enough to show that pR[pt] isa prime ideal of R[pt]. Let x = a0 + a1t+ · · ·+ art

r be an element in R[pt]and x′ = (a0/1)+ (a1/1)t+ · · ·+(ar/1)t

r the image of x in Rp[pRpt]. Notethat ai is in pi+1 if and only if (ai/1) is in pi+1Rp, because the ordinaryand symbolic powers of p coincide, thus x is in pR[pt] if and only if x′ isin pRp[pRpt]. As a consequence pR[pt] is prime if and only if pRp[pRpt] isprime. To finish the argument use Theorem 4.3.15 and the hypothesis thatpRp is generated by a regular sequence to get that grpRp

(Rp) is a domain.

⇐) First we prove AssR(pi/pi+1) = {p} for i ≥ 1. Let p1 be an associated

prime of pi/pi+1, that is, p1 = ann (x+ pi+1), for some x in pi \ pi+1. Notea ∈ p1 iff ax ∈ pi+1, and since grp(R) is a domain, one readily concludesthat a ∈ p1 iff a ∈ p. Hence p1 = p. There is an exact sequence:

0 −→ pi/pi+1 −→ R/pi+1 −→ R/pi −→ 0.

Hence, using induction on i ≥ 1, Lemma 2.1.17 and the exact sequenceabove, we get Ass(R/pi) ⊂ Ass(R/p) = {p}. Thus p is normally torsion-free and by Proposition 4.3.29 one has pn = p(n) for all n ≥ 1. �

Theorem 4.3.35 [383] Let R be a normal domain and I a radical idealwhich is generically a complete intersection. If I is normally torsion-free,then its Rees algebra R[It] is a normal domain.

Proof. By Corollary 4.3.18 we need only show that the associated gradedring grI(R) = R[It]/IR[It] is reduced. Let

f = a0 + a1t+ · · ·+ asts + IR[It]

be a nilpotent element of grI(R). Hence (asts)m is in IR[It] for somem ≥ 1,

by induction it suffices to verify that asts is in IR[It]. Let p1, . . . , pr be the

minimal primes of I. Since IRpi = piRpi is a complete intersection onederives that grpiRpi

(Rpi) is reduced (see Corollary 4.3.20). Therefore the

image of asts in grpiRpi

(Rpi) is zero for all i, and one readily concludes thatas belongs to the following intersection:

(R ∩ ps+11 Rp1) ∩ · · · ∩ (R ∩ ps+1

r Rpr ) = p(s+1)1 ∩ · · · ∩ p(s+1)

r .

As I is normally torsion-free one can write In = q1 ∩ · · · ∩ qr, where qiis pi-primary, localizing yields InRpi = pni Rpi = qiRpi and consequently

qi = p(n)i . Making n equal to s+ 1 proves that as is in Is+1. �

156 Chapter 4

Gorenstein Rees algebras For use below R = K[x1, . . . , xn] will denotea polynomial ring over a field K and S = R[It] will denote the Rees algebraof an ideal I of R, which is assumed to be Cohen–Macaulay.

Definition 4.3.36 If (1, t)m = R⊕Rt⊕· · ·⊕Rtm⊕ Itm+1 · · · , we say thatthe canonical module ωS of S has the expected form if ωS = ωR(1, t)

m forsome m ≥ −1.

Theorem 4.3.37 [235] If ωS has the expected form, then ωS � ωR(1, t)g−2

if and only if I is generically a complete intersection.

Lemma 4.3.38 [412, Page 142] If S is a regular local ring and J is an idealof S generated by a regular sequence h1, . . . , hg, then the Rees algebra S[Jt]is determinantal:

S[Jt] � S[z1, . . . , zg]/I2(z1 · · · zgh1 · · · hg

)and its canonical module is ωS(1, t)

g−2.

Proposition 4.3.39 If I is an ideal of height g generated by square-freemonomials and S = R[It] is Gorenstein, then g = 1 or g = 2.

Proof. Assume g > 1. Then ωS � S = (1, t)0 = R(1, t)0 = ωR(1, t)0 and

therefore ωS has the expected form with m = 0. Clearly I is generically acomplete intersection because I is the intersection of prime ideals of R; seeProposition 6.1.4. Thus, by Theorem 4.3.37, we get

S � ωS � ωR(1, t)g−2 = R⊕Rt⊕ · · · ⊕Rtg−2 ⊕ Itg−1 ⊕ · · · (4.1)

Take a minimal prime p of I of height g. Then Sp = Rp[Ipt] is the Reesalgebra of the ideal Ip, which is generated by a regular sequence. Thuslocalizing the extremes of Eq. (4.1) at p and using Lemma 4.3.38 we obtain

Sp = Rp[Ipt] � ωRp(1, t)g−2 � ωSp .

Note that it is important to know a priori that the canonical module of Sp

is ωRp(1, t)g−2. Hence Sp is Gorenstein. To finish the proof note that the

only Gorenstein determinantal rings that occur in Lemma 4.3.38 are thosewith g = 2; see [65, 71, 73]. �

Descent of normality Let A ⊂ B be an extension of rings such that Bis normal; in general the normality of A is not inherited from B. We discusssome sufficient conditions for A to be normal.

Lemma 4.3.40 Let A ⊂ B be an extension of rings. If B = A ⊕ C (asA-modules), then IB ∩ A = I for every ideal I of A.


Proof. Let z ∈ IB ∩ A and write z =∑q

i=1 bifi, where bi ∈ B and fi ∈ I.By hypothesis bi = ai + ci, ai ∈ A and ci ∈ C. Since z ∈ A it follows thatz =

∑qi=1 aifi ∈ I. This proves the containment IB ∩ A ⊂ I, the reverse

containment is clear. �

Proposition 4.3.41 Let A ⊂ B be integral domains with field of fractionsKA and KB, respectively. If B = A⊕C (as A-modules), then KA∩B ⊂ A.In particular if B is normal, then A is normal.

Proof. By Lemma 4.3.40 one has IB ∩ A = I for every ideal I of A. Letb = a/c ∈ B, a, c ∈ A, then a ∈ (c)B ∩ A = (c), hence a = λc = bc, withλ ∈ A. Therefore b = λ ∈ A. �

Proposition 4.3.42 Let R be a polynomial ring over a field K and Ian ideal of R generated by homogeneous polynomials f1, . . . , fq. Assumedeg(fi) = d for all i. If R[It] is normal, then K[f1, . . . , fq] is normal.

Proof. Let m = R+ be the irrelevant maximal ideal of the polynomial ringR and A = K[tf1, . . . , tfq]. Observe that there is a decomposition of A-modules: R[It] = K[tf1, . . . , tfq]⊕mR[It]. As A � K[f1, . . . , fq] (as rings),by Proposition 4.3.41, the result follows. �

Exercises

4.3.43 Let R = K[x1, . . . , xn] be a polynomial ring over a field K and letm be the maximal ideal (x1, . . . , xn). If I = md and J = (xd1 , . . . , x

dn) for

some positive integer d, then R[Jt] = R[It].

4.3.44 Let I, J be ideals of a normal domain R. Define the multi Reesalgebra of I, J as R(I ⊕ J) = R[uI, vJ ], here u, v are new variables. Then

R(I ⊕ J) = ⊕In Jmunvm.

4.3.45 Let R be a ring and F = {Ii}i∈N a family of ideals in R. It is saidthat F is a filtration of R, if Ii+1 ⊂ Ii, I0 = R, and IiIj ⊂ Ii+j for alli, j ∈ N. If I is an ideal of R, prove that the following are filtrations of R:

(a) Ii = Ii, the ordinary powers of I,

(b) Ii = Ii, the integral closure of Ii,

(c) Ii = I(i), the symbolic powers of I, and

(d) In = ⊕i≥nRi, where R = ⊕i≥0Ri is a graded ring.

Hint For (b) use the method of proof of Proposition 4.3.3.

158 Chapter 4

4.3.46 Let R be a normal domain and F = {Ii}i∈N a filtration of R. If Iiis integrally closed for all i, prove that the Rees algebra

R(F) =∞⊕i=0

Iiti ⊂ R[t]

of the filtration F is integrally closed. See [198] for a study of the Cohen–Macaulay and Gorenstein property of R(F).

4.3.47 Let R be an integral domain. If {Rα}α∈Λ is a collection of normalsubrings of R, then ∩α∈ΛRα is normal.

4.3.48 Let R be a ring and I, J ideals of R. If I and J are integrally closed,prove that I ∩ J is integrally closed.

4.3.49 If R is a domain and I ⊂ R is an ideal, then⊕∞

i=0 Iiti ⊂ R[It] with

equality if R is a normal domain.

4.3.50 Let R be a domain and let K be its field of fractions. If K[x] is apolynomial ring in one variable, then R[x] ⊂ R[x] ⊂ K[x] and R[x] = R[x].

4.3.51 Let F and G be two finite sets of monomials in a polynomial ringR over a field K. If (F ) = (G) prove that R[Ft] = R[Gt].

4.3.52 If F = {x1x2, x3x4x5, x1x3, x2x4, x2x5, x1x5} ⊂ K[x1, . . . , x5] andI = (F ), prove that K[F ] is not normal and R(I) is normal.

4.3.53 If R = Q[x1, . . . , x7] and I is the principal ideal generated by thebinomial f = x1x3x5x7−x2x24x6, prove that R/I is a normal domain. Givea description of the irreducible binomials f such that R/(f) is normal.

4.3.54 Let R be a Noetherian ring and a an element in the Jacobson radicalof R. If a is regular and R/(a) is an integral domain, prove that R is anintegral domain.

4.3.55 Let R = K[x] be a polynomial ring over a field K. If I is an idealgenerated by square-free monomials of degree 2, then I and I2 are complete.

4.3.56 LetR = K[x1, . . . , xn] be a polynomial ring and letR(I) be the Reesalgebra of an ideal I = (f1, . . . , fq), where K is a field. If fi is homogeneousof degree d ≥ 1 for all i. Prove

(a) R(I) � R(I)/mR(I)⊕mR(I) as R(I) modules, where m = R+,

(b) K[f1, . . . , fq] � R(I)/mR(I) as K-algebras,

(c) if F = {f1, . . . , fq} is a set of monomials, then R(I) � K[F, tx] asK-algebras, where x = {x1, . . . , xn}.


4.4 Multiplicities and a criterion of Herzog

In this section we present an elegant and useful Cohen–Macaulay criteriondue to Herzog (Theorem 4.4.13). Once more we recall that all rings consid-ered in this book are Noetherian and modules are finitely generated.

Quasi regular sequences It is useful to generalize Theorem 4.3.15 tomodules by introducing a polynomial with coefficients in a module. Let Mbe an R-module and R[t1, . . . , tq] a polynomial ring over the ring R. Set

M [t] =M [t1, . . . , tq] =M ⊗R R[t1, . . . , tq],

and note that an element of M [t] can be regarded as a polynomial in the tivariables with coefficients in M , thus M [t] is naturally graded. If f1, . . . , fqis a sequence in R and I = (f1, . . . , fq), there is a degree preserving map ofadditive groups

ϕ : (M/IM)[t1, . . . , tq] −→ grI(M) =∞⊕i=0

IiM/Ii+1M,

such that ϕ(F (t)) = F (f1, . . . , fq) + Ii+1M , for all F (t) in M [t] homoge-

neous of degree i, where the notation F (t) means reducing the coefficientsof F (t) modulo IM . It is not hard to verify the equivalence between thefollowing two conditions:

(a) ϕ is injective.

(b) For every homogeneous polynomial F inM [t] of positive degree n suchthat F (f) ∈ In+1M , one has F ∈ IM [t].

Definition 4.4.1 If the map ϕ is an isomorphism and IM �= M , the se-quence f1, . . . , fq is called an M -quasi-regular sequence.

Theorem 4.4.2 Let M be an R-module and f = f1, . . . , fq a sequence inR. If f is an M -regular sequence, then f is an M -quasi-regular sequence.

Proof. It follows adapting the proof of Theorem 4.3.15. �

Theorem 4.4.3 (Krull’s intersection theorem [310, Theorem 8.9]) Let Mbe an R-module and let I be an ideal of R. Then there is a ∈ R such thata ≡ 1mod (I) and a ·

(⋂∞i=1 I

iM)= 0.

Lemma 4.4.4 Let (R,m) be a local ring (resp. N-graded) and M an R-module (resp. N-graded). If N is a submodule ofM (resp. graded submodule)and I ⊂ m is an ideal (resp. graded ideal), then N =

⋂∞i=1(N + IiM).

160 Chapter 4

Proof. Using Krull’s intersection theorem one has⋂∞i=1 I

iM ′ = (0) (thisequality holds also in the graded case), where M ′ = M/N . Hence therequired equality follows at once. �

Proposition 4.4.5 Let (R,m) be a local ring (resp. N-graded ring) andM an R-module (resp. N-graded module). If f = f1, . . . , fq is an M -quasiregular sequence of elements in m (resp. homogeneous elements in m = R+),then f is an M -regular sequence.

Proof. Fix an integer 1 ≤ r ≤ q and set I = (f1, . . . , fq). To begin with wesplit the ideal I as I = J+L, where J = (f1, . . . , fr−1) and L = (fr, . . . , fq),it is convenient to set J = (0) if r = 1. It suffices to show the equality(JM : Mfr) = JM , because this is equivalent to prove that fr is not a zerodivisor of M/JM . Let m ∈ (JM : Mfr), according to Lemma 4.4.4 one has⋂∞n=1(JM + LnM) = JM, thus the proof reduces to proving by induction

on n that m ∈ JM + LnM for n ≥ 1. Since m ∈ (JM : Mfr), there are mi

in M such that

mfr = m1f1 + · · ·+mr−1fr−1 (m,mi ∈M). (4.2)

Consider the polynomial F = mtr −m1t1 − · · · −mr−1tr−1 ∈ M [t]. Notedegt(F ) = 1 and F (f) = 0. As f is an M -quasi regular sequence we getm ∈ IM = JM + LM . By induction assume m ∈ JM + LnM , that is,there is G ∈M [tr, . . . , tq] homogeneous of degree n such that

m = b1f1 + · · ·+ br−1fr−1 +G(fr, . . . , fq) (bi ∈M). (4.3)

From Eqs. (4.2) and (4.3) we get frG(fr, . . . , fq) = a1f1 + · · · + ar−1fr−1,ai ∈ M . Using that f is an M -quasi regular sequence one may assume(after an induction argument) that ai ∈ InM , and thus trG(tr, . . . , tq) is inIM [t], that is, G(tr, . . . , tq) is in IM [t]. One can write G = G1+G2, whereG1, G2 are homogeneous polynomials in M [t] of degree n with G1 ∈ JM [t]and G2 ∈ LM [t]. Since G1(fr, . . . , fq) is in JM and G2(fr, . . . , fq) is inLn+1M , from Eq. (4.3) we get that m is in JM + Ln+1M , as required. �

Lemma 4.4.6 Let f(t) ∈ Q[t] be a polynomial of degree d − 1 such thatf(n) ∈ Z for n ∈ Z, then there are unique integers a0, . . . , ad−1 such that

f(t) =

d−1∑i=0

aifi(t), where fi(t) =

(t+ i

i

).

Proof. The polynomials fi(t), i ∈ N, are a basis for Q[t] as a Q-vector

space. Hence f(t) =∑d−1i=0 aifi(t), for some ai ∈ Q. Using the Pascal

triangle we get

f(t)− f(t− 1) =d−1∑i=0

ai

[(t+ i

i

)−(t+ i− 1

i

)]=d−1∑i=0

ai

(t+ i− 1

i− 1

),


thus by induction on the degree it follows that ai ∈ Z for all i. �

Multiplicities Let (R,m) be a local ring, M a finitely generated R-module and q an ideal with rad (q) = m. As grq(M) is a finitely generatedgrq(R)-module and A0 = R/q is Artinian, by Theorem 2.2.4 there exists apolynomial P (t) ∈ Q[t] of degree d− 1, where d = dim(grq(M)), such that

P (i) = �A0(qiM/qi+1M), for i ≥ n0.

By Lemma 4.4.6 there are integers a0, . . . , ad−1 such that

P (i) =d−1∑j=0

aj

(i+ j

j

), for all i ≥ 0.

From the short exact sequences

0 −→ qiM/qi+1M −→M/qi+1M −→M/qiM −→ 0,

one derives �(M/qi+1M)− �(M/qiM) = P (i) for i ≥ n0. Hence, using theidentity of Exercise 4.4.14, we get

χqM (i) := �(M/qi+1M) = �(M/qn0M) +

i∑j=n0

P (j)

= c0 +

d−1∑j=0

aj

(j + i+ 1

j + 1

),

for i ≥ n0. Thus the function χqM (i) = �(M/qi+1M) is polynomial of degree

d, and is called the Samuel function of q with respect to M . The integerad−1, is the multiplicity of q on M and is denoted by e(q,M) or simply bye(q) if M = R. Note that e(q,M)/d! is the leading coefficient of χq

M (i).By [259, Proposition 11.2.1], we have e(q,M) = e(q,M). If R is a

polynomial ring with coefficients in a field K, there are efficient methods tocompute the multiplicity of a zero dimensional monomial ideal; see [104, 424]and Section 12.5.

We recall the following result in dimension theory that relates the degreeof the Samuel function with the dimension of the module.

Theorem 4.4.7 [310, Theorem 13.4] χqM is a polynomial function of degree

equal to the dimension of M .

Definition 4.4.8 Let (R,m) be a local ring of dimension d,M an R-moduleand q an ideal of R with rad (q) = m. We define

ed(q,M) =

{e(q,M) if dim(M) = d,0 if dim(M) < d.

162 Chapter 4

In order to show how the “multiplicity” behaves under short exact se-quences we need to recall a result of E. Artin and D. Rees.

Theorem 4.4.9 (Artin–Rees lemma [310, Theorem 8.5]) Let M be a mod-ule over a ring R. If N is a submodule of M and I an ideal of R, then thereis a positive integer c such that

InM ∩N = In−c(IcM ∩N), ∀n > c.

Proposition 4.4.10 Let (R,m) be a local ring of dimension d and let q bean m-primary ideal of R. If 0 −→ N −→ M −→ N ′ −→ 0 is an exactsequence of R-modules, then ed(q,M) = ed(q, N) + ed(q, N

′).

Proof. By Proposition 2.1.39, dim(M) = max{dim(N), dim(N ′)}. Hence,one may assume d = dim(M). Tensoring with R/qn+1 the exact sequenceabove yields an exact sequence

0→ (N ∩ qn+1M)/qn+1N → N/qn+1N →M/qn+1M → N ′/qn+1N ′ → 0.

Taking lengths with respect to R/q gives

χqM (n) = χq

N (n) + χqN ′(n)− �(N ∩ qn+1M/qn+1N). (∗)

By Theorem 4.4.9, N ∩ qn+1M ⊂ qn+1−cN , for some integer c > 0. Hence

�(N ∩ qn+1M/qn+1N) ≤ �(qn+1−cN/qn+1N) = χqN (n)− χq

N (n− c),

as χqN (n)− χq

N(n− c) is a polynomial function of degree at most d− 1, theresult follows by dividing Eq. (∗) by nd and taking limits when n goes toinfinity. �

Next we show that the multiplicity is additive (cf. Proposition 8.5.9).

Proposition 4.4.11 Let (R,m) be a local ring of dimension d and q anm-primary ideal. If M is a finitely generated R-module and A is the set ofall prime ideals p of R with dim(R/p) = d, then

ed(q,M) =∑p∈A

�(Mp)ed(q, R/p).

Proof. Set B = A∩ Supp(M). If B = ∅, then dim(M) < d and ed(q,M) isequal to 0, thus in this case the identity above holds. Hence we may assumeB �= ∅, this yields the equality dim(M) = d. By Theorem 2.1.16 there areprime ideals p1, . . . , pn of R and a filtration of submodules:

(0) =M0 ⊂M1 ⊂ · · · ⊂Mn =M,


such that Mi/Mi−1 � R/pi for all i. Note B = {pi| dim(R/pi) = d}. Toshow this equality observe that

Ass(M) ⊂ {p1, . . . , pn} ⊂ Supp(M),

see Corollary 2.1.18 and its proof, and recall that the minimal elements ofSupp(M) are in Ass(M). Using Lemma 4.4.10 and the exact sequences

0 −→Mi−1 −→Mi −→ R/pi −→ 0 (i = 1, . . . , n),

we get e(q,M) =∑e(q, R/pi), where the sum is taken over all pi, in the

multiset {p1, . . . , pn}, such that dim(R/pi) = d. Let p ∈ B, then

(Mi/Mi−1)p �{

(0) if p �= piRp/pRp if p = pi.

(4.4)

Since the second module is a field we get that the length of Mp is equal tothe number of times that p occurs in the multiset {p1, . . . , pn}. Therefore

e(q,M) =∑pi∈B

�(Mpi)e(q, R/pi),

as required. This proof was adapted from [65]. �

Proposition 4.4.12 Let (S,m) be a local ring and M an S-module with apositive rank r = rank(M). If q is an m-primary ideal, then

e(q,M) = e(q, S)rank(M).

Proof. First note d = dim(M) = dim(S) by Lemma 2.1.45. Let A be theset of all prime ideals p of S with dim(S/p) = d. Since M has rank r onehas Mp � (Sp)

r for any associated prime p of S, thus Mp � (Sp)r for any

p ∈ A. Therefore using Proposition 4.4.11 one obtains

ed(q,M) =∑p∈A

�(Mp)ed(q, S/p) =∑p∈A

�(Srp)ed(q, S/p)

=∑p∈A

r�(Sp)ed(q, S/p) = rank(M)ed(q, S).

As e(q,M) = ed(q,M) and e(q, S) = ed(q, S), the proof is complete. �

In Chapter 5 we use the following important Cohen–Macaulay criterionto study the Koszul homology of graded ideals.

Theorem 4.4.13 (Herzog [217]) Let (S,m) be a Cohen–Macaulay local ringand let M be a finitely generated S-module with a well-defined and positiverank. If y = y1, . . . , yd is a system of parameters of S, then

�(S/(y)) · rank(M) ≤ �(M/(y)M).

Furthermore equality holds if and only if M is Cohen–Macaulay.

164 Chapter 4

Proof. By Lemma 2.1.45 one has d = dim(M) = dim(S). Let q be them-primary ideal generated by y. There is a (graded) epimorphism

ϕ : (M/qM)[t] = (M/qM)[t1, . . . , td]→ grq(M) =

∞⊕i=0

qiM/qi+1M (4.5)

such that ϕ(F ) = F (y1, . . . , yd) + qi+1M , for all F in M [t] homogeneousof degree i, where F means reducing the coefficients of F modulo qM . Byrestriction of ϕ to the ith graded component of (M/qM)[t] gives

�(M/qM)

(i+ d− 1

d− 1

)≥ �(qiM/qi+1M) for i ≥ 0, (4.6)

observe that both sides of the inequality are polynomial functions of degreed − 1 and consequently �(M/qM) ≥ e(q,M). On the other hand y is aregular sequence, because S is Cohen–Macaulay (see Proposition 2.3.19),this forces ϕ to be an isomorphism when M = S (see Theorem 4.4.2).It follows rapidly that �(S/q) = e(q, S). To finish the first part of theproof note that making use of Proposition 4.4.12 one obtains the requiredinequality.

At this point one should observe that, by the arguments above, theproof reduces to show that �(M/qM) = e(q,M) if and only if M is Cohen–Macaulay. Next we show both implications.⇒) Thanks to Propositions 4.4.5 and 2.3.19 it suffices to prove that ϕ is

injective; because this implies that y is an M -regular sequence and henceM is Cohen–Macaulay. Using Eq. (4.5) we obtain an exact sequence

0 −→ ker(ϕ)i −→ ((M/qM)[t])i −→ qiM/qi+1M −→ 0,

hence we get

�(ker(ϕ)i) = �(M/qM)

(i+ d− 1

d− 1

)− �(qiM/qi+1M) for i ≥ 0.

As the polynomial functions on the right-hand side have both degree d− 1and their leading terms cancel out, one concludes that �(ker(ϕ)i) grows asa polynomial function of degree at most d− 2.

Let F =∑mαt

α be a homogeneous polynomial inM [t] of degree p, withmα in M , and denote mα = mα + qM . Assume F is in ker(ϕ) and F �= 0.Since mk ⊂ q, there is an s ≥ 1 such that ms−1F �= 0 and msF = 0, thusone may assume mF = 0 and F �= 0. Hence there is a graded epimorphismψ induced by multiplication by F

ψ : (S/m)[t](−p) −→ (S/q)[t]F .

We now show that ψ is injective. Let g =∑bβt

β be an element in the kernel

of ψ, where bβ ∈ S and bβ = bβ + m. Thus (∑bβt

β)(∑mαt

α) = 0, where


bβ = bβ + q. One may assume that bβtβ and mαt

α are the leading terms ofg and F , respectively, w.r.t the lexicographical ordering of the ti variables.Thus bβmαt

α+β is the leading term of gF and bβmα ∈ qM . Hence bβ + q isa zero divisor of M/qM and since AssR/q(M/qM) = {m/q}, one has that

bβ + q is in m/q, which proves bα ∈ m and bβ = 0. Altogether one derivesg = 0 and ψ is injective. Using that ψ is a graded isomorphism yields(

i− p+ d− 1

d− 1

)= �(((S/q)[t]F )i) ≤ �(ker(ϕ)i),

which is a contradiction because the left-hand side is a polynomial functionof degree d− 1. Therefore F = 0 and ϕ is injective, as required.⇐) As M is Cohen–Macaulay, the sequence y is an M -regular sequence,

an application of Theorem 4.4.2 yields that ϕ is an isomorphism. ThereforeEq. (4.6) becomes an equality and we get �(M/qM) = e(q,M). �

Exercises

4.4.14 Let d,m ∈ N. Prove the equality(d+m

d

)=

m∑j=0

(j + d− 1

d− 1

).

4.4.15 Let f : Z→ Z be a numerical function, f is said to be a polynomialfunction of degree d if there is a polynomial P (t) ∈ Q[t] such that P (i) =f(i) for i 0. Prove that f is a polynomial function of degree d if andonly if the numerical function g : Z→ Z given by g(i) = f(i)− f(i− 1) is apolynomial function of degree d− 1.

4.5 Jacobian criterion

Here we introduce the Jacobian criterion and present some applications andexamples to illustrate its use. Along the way some facts about regular localrings are discussed. This type of rings are used in algebraic geometry toshow for instance that the coordinate ring of a smooth irreducible affinevariety is normal [100, Proposition 1.0.9].

Let (R,m) be a local ring. By Nakayama’s Lemma it follows readilythat R is a regular local ring if and only if m is generated by a system ofparameters of R. If R is a polynomial ring or a formal power series ringover a field k, then R is a regular ring [310, Theorem 19.5], that is, Rp is aregular local ring for all p in Spec(R) or equivalently Rm is a regular localring for any maximal ideal m of R (see [310, Theorem 19.3]).

Proposition 4.5.1 If R is a regular local ring, then R is Cohen–Macaulay.

166 Chapter 4

Proof. Let m be the maximal ideal of R and let k = R/m be its residuefield. Assume x1, . . . , xd is a set of generators of m, where d = dim(R). Ifk[t1, . . . , td] is a ring of polynomial over the field k, there is an epimorphismof graded k-algebras

ϕ : k[t] = k[t1, . . . , td] −→ grm(R) (4.7)

induced by ϕ(ti) = xi +m2 for all i. It suffices to prove that the map ϕ isinjective. Indeed if ϕ is injective, then x1, . . . , xd is a regular sequence byProposition 4.4.5, and hence R is Cohen–Macaulay by Proposition 2.3.19.

If I = ker(ϕ) �= 0, pick a homogeneous polynomial f in I of degrees ≥ 1. Using the isomorphism k[t](−s) � fk[t], together with Eq. (4.7),one concludes:

dimk(mi/mi+1) = dimk k[t]i − dimk Ii ≤

(i+ d− 1

d− 1

)−(i− s+ d− 1

d− 1

).

By Theorem 4.4.7 the left-hand side is a polynomial function of degree d−1while the right-hand side is a polynomial function of degree d−2. ThereforeI must be zero and ϕ is injective. �

Corollary 4.5.2 If (R,m) is a regular local ring, then R is a domain.

Proof. By Proposition 4.5.1 m is generated by a regular system of param-eters. Thus grm(R) is a domain. Let x, y ∈ R such that xy = 0. If x �= 0and y �= 0, then by Theorem 4.4.3 there are r, s ∈ N with x ∈ mr \mr+1 andy ∈ ms \ ms+1. Multiplying the images x, y of x, y in grm(R), one obtainsxy = 0. Hence x = 0 or y = 0, which is impossible. This proves R is adomain. �

Proposition 4.5.3 Let (R,m) be a regular local ring and I �= R an idealof R. If R/I is a regular local ring, then I is a complete intersection.

Proof. Let x1, . . . , xd be a generating set of m = m/I, where xi = xi + Iand d is the dimension of R/I. Note that the set of images of x1, . . . , xd inm/m2 is a basis for m/m2 as a vector space over k = R/m. Hence

x1 +m2, . . . , xd +m2

are linearly independent in m/m2. Set n = dim(R) and m′ = (x1, . . . , xd).As R is a regular local ring n = dimk(m/m

2), hence using the proof ofCorollary 2.1.35 and the equality m = m′ + I one derives m = m′ + J , forsome ideal J = (xd+1, . . . , xn) contained in I.

By Proposition 4.5.1 x1, . . . , xn is a regular sequence, thus it sufficesto show J = I, to prove this equality observe that R/J is a regular localring of dimension d, and therefore R/J is a domain by Corollary 4.5.2.Since R/I has also dimension d, we see that the canonical homomorphismϕ : R/J → R/I is an isomorphism, thus I = J , as asserted. �


Definition 4.5.4 Let R = k[x1, . . . xn] be a polynomial ring over a field kand I = (f1, . . . , fq) an ideal. The Jacobian matrix of I is the matrix

J = (∂fi/∂xj).

We denote the Jacobian matrix of I taken modulo an ideal P by

(∂fi/∂xj)(P ).

Theorem 4.5.5 (Jacobian criterion [309, p. 213]) Let B = R/I be a quo-tient ring, where R = k[x1, . . . , xn] is a polynomial ring in n variables overa field k, and let I = (f1, . . . , fq) ⊂ R be an ideal. If P ⊂ R is a prime idealcontaining I and p = P/I, one has:

(a) rank(∂fi/∂xj)(P ) ≤ ht(IP ).

(b) If rank(∂fi/∂xj)(P ) = ht(IP ), then Bp is a regular ring.

(c) If k is a perfect field and Bp is regular, then

rank(∂fi/∂xj)(P ) = ht(IP ).

Lemma 4.5.6 Let R be a ring and I an ideal. If I is height unmixed andP is a prime ideal such that I ⊂ P , then IP is also height unmixed andht(I) = ht(IP ).

Proof. Let I = q1∩· · ·∩qr be a primary decomposition of I, we may assume√qi ⊂ P for i = 1, . . . s. Localizing at P gives a primary decomposition

IRP =⋂si=1 qiRP . Hence the associated primes of IP are

√qiRP , where

i ≤ s. To finish the proof note ht(√qiRP ) = ht(

√qi) = ht(I) for i ≤ s. �

Definition 4.5.7 Let k be a field and I = (f1, . . . , fq) an ideal of heightg of k[x1, . . . , xn]. The Jacobian ideal J of I is the ideal generated by theg × g minors of the Jacobian matrix J = (∂fi/∂xj).

Corollary 4.5.8 Let R = k[x1, . . . , xn] be a polynomial ring over a perfectfield k and I = (f1, . . . , fq) ⊂ R an unmixed ideal. If P is a prime ideal ofR containing I and p = P/I, then (R/I)p is regular if and only if J/I �⊂ p,where J is the Jacobian ideal of I.

Proof. ⇒) By Theorem 4.5.5 and Lemma 4.5.6 one has

rank(∂fi/∂xj)(P ) = ht(I),

hence there is f ∈ J \ P and J/I �⊂ P/I.⇐) If (R/I)p is not regular, then by Theorem 4.5.5 one has the inequality

rank(∂fi/∂xj)(P ) < ht(I), hence J/P = (0) and J/I ⊂ P/I, which isimpossible. Note that this part of the proof is valid for any field k. �

168 Chapter 4

Proposition 4.5.9 Let R be a polynomial ring over a field k and I anunmixed ideal of height g. If J is the Jacobian ideal of I and ht (J, I) ≥ g+2,then (R/I)p is regular for any prime p of R/I such that ht (p) ≤ 1.

Proof. Let B = R/I and p = P/I, where P is a prime ideal of R. IfBp is not regular, then according to Corollary 4.5.8 one has J/I ⊂ p andhence (J, I) ⊂ P . Since R is a catenary ring and I is unmixed one obtainsht (P ) = g if ht (p) = 0 and ht (P ) = g + 1 if ht (p) = 1. Thereforeht (J, I) ≤ ht (P ) ≤ g + 1, which is impossible. Thus Bp must be regular,as required. �

Another consequence of the Jacobian criterion is the following radicaltest. Let k be a field and I = (f1, . . . , fq) ⊂ R = k[x1, . . . , xn] an ideal ofheight g. Let J be the Jacobian ideal of I generated by the g× g minors ofthe Jacobian matrix J = (∂fi/∂xj).

Theorem 4.5.10 (Vasconcelos [411]) If k is a perfect field and I is un-mixed, then I is a radical ideal if and only if there is an element f in J \ Isuch that (I : f) = I.

Proof. ⇒) Let P1, . . . , Pr be the associated primes of I. If J ⊂ Z(R/I),then J ⊂ Pi for some i because Z(R/I) = ∪ri=1Pi. Hence by Corollary 4.5.8one derives that (R/I)Pi is not a regular ring, which is a contradictionbecause (R/I)Pi = RPi/PiRPi is a field.⇐) Let I = q1 ∩ · · · ∩ qr be an irredundant primary decomposition of I

and Pi =√qi an associated prime ideal of I. Note (R/I)Pi � (R/I)pi and

IPi = (qi)Pi , where pi = Pi/I. First we observe that (R/I)pi is a regularring; otherwise by Corollary 4.5.8 one has J/I ⊂ Pi/I, hence f ∈ Pi, whichis impossible because f is regular modulo I. Therefore (R/I)Pi is a regularlocal ring and thus an integral domain by Theorem 2.4.16. As IPi = (qi)Pi

must be prime, one concludes that qi is also prime and consequently qi = Pi.Note that here the hypothesis k perfect is not being used. �

Example 4.5.11 Let R = k[x1, . . . , x7] and I = (x1x7−x2x6, x3x5−x4x7),where k is a field. We now show that B = R/I is a normal ring usingTheorem 2.4.15. The Jacobian matrix of I is:

J =

(x7 −x6 0 0 0 −x2 x10 0 x5 −x7 x3 0 −x4

).

Note that x27, x5x6, x3x6, x4x6, x2x5, x1x5, x2x4 are in the Jacobian idealJ of I. Hence ht (J, I) ≥ g + 2, where g = ht (I) = 2. By Proposition 4.5.9,B satisfies (R1) and, since I is a complete intersection, it follows by Propo-sition 3.1.30 that B satisfies (S2). Thus, by Serre’s normality criterion, thering B is normal.


Lemma 4.5.12 Let R be a ring and I a prime ideal. If f ∈ R \ I and

I(n)f = Inf for some n ≥ 1, then I(n) = (In : f∞).

Proof. If z ∈ (In : f∞), then zfk ∈ In for some k ≥ 1, hence z ∈ I(n).Conversely if z ∈ I(n), then sz ∈ In for some s /∈ I. Since s/1 /∈ If and(s/1)(z/1) ∈ Inf , then z/1 is in the nth symbolic power of If . Thus byhypothesis z/1 ∈ Inf and this means that z ∈ (In : f∞). �

There are a few methods to compute symbolic powers of prime ideals inpolynomial rings, see [378] and [413, Chapter 3]. The following is a subtleapplication of the Jacobian criterion due to Vasconcelos.

Proposition 4.5.13 Let R be a polynomial ring over a field k and I aprime ideal. If f is an element in the Jacobian ideal J of I which is not inthe ideal I, then

I(n) = (In : f∞) for n ≥ 1,

and such element f exist if k is a perfect field.

Proof. Since the localization Rf is Cohen–Macaulay, by Proposition 4.3.33and Lemma 4.5.12 it is enough to prove that If is locally a complete inter-section. Let P be a prime ideal such that I ⊂ P and f �∈ P . Set p = P/I.One has

(Rf/If )Pf� (Rf )Pf

/(If )Pf� RP /IP � (R/I)P � (R/I)p.

As J/I �⊂ p, using Corollary 4.5.8 we conclude that RP /IP is a regular ringand consequently (If )Pf

is a complete intersection (see Proposition 4.5.3).The last assertion follows from Theorem 4.5.10. �

Example 4.5.14 Let I = (x3 − yz, y2 − xz, z2 − x2y) ⊂ Q[x, y, z]. Theideal I is prime because it is the kernel of the homomorphism:

ϕ : Q[x, y, z] −→ Q[t], where ϕ(h(x, y, z)) = h(t3, t4, t5).

Using CoCoA [88] and Proposition 4.5.13 we get

I(2) = (I2 : f∞) = (I2,Δ),

where f = 2y2 + xz and Δ = −x5 + 3x2yz − xy3 − z3. See [280].

Exercises

4.5.15 Let R = Q[x, y, z] and I the prime ideal generated by

f1 = x3 − yz, f2 = y2 − xz, f3 = z2 − x2y.

Prove that Ix = (f1, f2) and I(n) = (In : x∞) for n ≥ 1.

170 Chapter 4

4.5.16 Use the following procedure in CoCoA [88] to verify the formula forthe second symbolic power of Example 4.5.14

F1 := x3− y ∗ z; - - variables must begin with capital lettersF2 := y2− x ∗ z;F3 := z2− x2 ∗ y;Jac := Jacobian([F1, F2, F3]);J := Minors(2, Mat(Jac)); - - 2× 2 minors of the Jacobian matrixI := Ideal(F1, F2, F3);H := Elim(t, I 2+ Ideal(1− t(2 ∗ y2+ x ∗ z)));Print(H);

4.5.17 Let p be a prime ideal of a ring R such that pRp is a completeintersection, then p(n) = pn for all n ≥ 1 if and only if grp(R) is a domain.

Hint Use Theorem 4.3.15, Lemma 2.1.17, and Proposition 4.3.29.

4.5.18 Let R be a Cohen–Macaulay ring and I an ideal of R which isheight unmixed. If I is a complete intersection, then I is locally a completeintersection.

Hint Use Lemma 2.3.20.

4.5.19 Let R be a Cohen–Macaulay ring and I a prime ideal. If f ∈ R \ Iand If is a complete intersection, then I(n) = (In : f∞) for n ≥ 1.

4.5.20 Let I = I2(X) be the ideal of 2×2-minors of the symmetric matrix:

X =

⎡⎣ x1 x2 x3x2 x4 x5x3 x5 x6

⎤⎦ ,where the entries of X are indeterminates over a field k. Prove that Ix1 isa complete intersection. Find a set of generators for I(2) and prove thatdepth(H1) = 1, where H1 is the first Koszul homology module of I.

Chapter 5

Hilbert Series

Hilbert series of graded modules and graded algebras are introduced andstudied in this chapter. The h-vector and the a-invariant of a graded algebraare defined using the Hilbert–Serre theorem. For Cohen–Macaulay algebras,we present their main properties. Some features of Cohen–Macaulay andGorenstein extremal algebras will be presented.

5.1 Hilbert–Serre Theorem

Unless otherwise stated we shall always assume that modules are finitelygenerated and N-graded. We refer to Section 2.2 for an introduction tograded modules, Hilbert polynomials, and multiplicities.

Let R = R0[x1, . . . , xn] = ⊕∞i=0Ri be an R0-algebra of finite type over

an Artinian local ring R0 with the grading induced by deg(xi) = di, wheredi is a positive integer for i = 1, . . . , n. If

M =M0 ⊕M1 ⊕ · · · ⊕Mi ⊕ · · ·

is a finitely generated N-graded module over R, its Hilbert function andHilbert series are defined by

H(M, i) = �(Mi) and F (M, t) =

∞∑i=0

H(M, i)ti

respectively, where �(Mi) denotes the length of Mi as an R0-module, if R0

is a field �(Mi) = dimR0(Mi).

Lemma 5.1.1 If 0 → M ′ → M → M ′′ → 0 is a degree preserving shortexact sequence of N-graded R-modules, then

(a) H(M, i) = H(M ′, i) +H(M ′′, i) for all i, and

(b) F (M, t) = F (M ′, t) + F (M ′′, t).

172 Chapter 5

Proof. It follows Proposition 2.1.36. �

If j ∈ N, then M(−j) is the regrading of M obtained by a shift of thegraduation of M ; more precisely

M(−j) =∞⊕i=0

M(−j)i,

where M(−j)i = M−j+i. Note that we are assuming Mi = 0 for i < 0. Inthis way M(−j) becomes an N-graded R-module.

Lemma 5.1.2 F (M(−j), t) = tjF (M, t).

Proof. Since M(−j)i =Mi−j one has:

F (M(−j), t) = tj∞∑i=j

�(Mi−j)ti−j = tjF (M, t),

where the first equality follows using that Mi−j = 0 for i = 0, . . . , j − 1. �

Lemma 5.1.3 If z ∈ Rj, there is a degree preserving exact sequence

0 −→ (M/(0 : z))(−j) z−→Mφ−→M/zM −→ 0 (φ(m) = m+ zM),

where (0 : z) = {m ∈M | zm = 0} and the first map is multiplication by z.

Proof. As the map ψ : M(−j)→M , given by ψ(m) = zm, is a degree zerohomomorphism one has that (0 : z)(−j) is a graded submodule of M(−j).The exactness of the sequence above follows because ψ induces an exactsequence

0 −→ (0 : z)(−j) ı−→M(−j) ψ−→Mφ−→M/zM −→ 0,

where ı is an inclusion. �

Theorem 5.1.4 (Hilbert–Serre Theorem) The Hilbert series F (M, t) of Mis a rational function that can be written as

F (M, t) =h(t)

n∏i=1

(1− tdi)for some h(t) ∈ Z[t].

If di = 1 for all i, there is a unique polynomial h(t) ∈ Z[t] such that

F (M, t) =h(t)

(1 − t)d and h(1) �= 0.

Hilbert Series 173

Proof. If n = 0, the Hilbert function of M is zero for i 0 and F (M, t) isa polynomial. If n > 0, consider the exact sequence of graded modules

0 −→ (0 : xn)(−dn) −→M(−dn)xn−→M −→M/xnM −→ 0.

Since the ends of this exact sequence are finitely generated modules overR0[x1, . . . , xn−1] the proof follows readily by induction on n. �

The integer d in the Hilbert–Serre theorem is denoted by d(M). Thisinteger is the Krull dimension of M ; see Proposition 5.1.6 and its proof.

Definition 5.1.5 The degree of F (M, t) as a rational function is denotedby a(M); it is called the a-invariant of M .

If R has the standard grading, below we see how the a-invariant measuresthe difference between the Hilbert polynomial and the Hilbert function.

For the rest of this section we shall always assume that the ring R hasthe standard grading induced by setting deg(xi) = 1 for i = 1, . . . , n. Wedenote the Hilbert polynomial of M by ϕM (t).

Proposition 5.1.6 IfM is a graded R-module of dimension d. Then, thereare integers a−d, . . . , a−1 so that

H(M, i) =d−1∑j=0

a−(d−j)

(i+ d− j − 1

d− j − 1

), ∀ i ≥ a(M) + 1,

where a(M) is the degree of F (M, t) as a rational function.

Proof. By the Hilbert–Serre Theorem there is a polynomial h(t) ∈ Z[t]such that

F (M, t) =h(t)

(1− t)d

and h(1) �= 0. If d = 0, then a(M) is equal to deg(h) and H(M, i) = 0for i ≥ a(M) + 1. Thus one may assume d > 0. Observe that by thedivision algorithm we can find e(t) ∈ Z[t] so that the Laurent expansion ofF (M, t)− e(t), in negative powers of (1− t), is equal to

F (M, t)− e(t) =d−1∑j=0

a−(d−j)(1− t)d−j , where a−(d−j) =

(−1)jh(j)(1)j!

,

where h(j) is the jth derivative of h. Next we expand (1 − t)d−j in powersof t to obtain

F (M, t) = e(t) +

∞∑i=0

⎡⎣d−1∑j=0

a−(d−j)

(i+ d− j − 1

d− j − 1

)⎤⎦ ti = e(t) +

∞∑i=0

ϕ(i)ti,

174 Chapter 5

observe that ϕ(i) = H(M, i) for i ≥ deg e(t)+1, where the degree of the zeropolynomial is set equal to −1. To complete the proof note that a−d, . . . , a−1

are integers by Lemma 4.4.6, and that d = d(M) is the dimension of M byTheorem 2.2.4. �

Remark 5.1.7 The leading coefficient of the Hilbert polynomial ϕM (t) ofM is equal to h(1)/(d − 1)!, where h(t) is the polynomial F (M, t)(1 − t)d.If d = 0, then h(1) = �(M) and �(M) = dimK(M) if R is a polynomial ringover a field K.

Definition 5.1.8 The index of regularity of M is the least integer � ≥ 0such that H(M, i) = ϕM (i) for i ≥ �.

Corollary 5.1.9 Let M be a graded R-module. If

r0 = min{r ∈ N |H(M, i) = ϕM (i), ∀ i ≥ r},

then r0 = 0 if a(M) < 0 and r0 = a(M) + 1 otherwise.

Corollary 5.1.10 Let S be a standard graded algebra of dimension d overa field K. If S is Cohen–Macaulay and h = {h1, . . . , hd} is a system ofparameters of S consisting of linear forms, then the multiplicity of S is:

e(S) = �(S/hS).

Proof. By Proposition 3.1.20 {h1, . . . , hd} is a regular sequence. Thusthere are exact sequences of K-vector spaces

0 −→ (S[−1])i = Si−1h1−→ Si −→ (S/h1S)i −→ 0.

Therefore one has the relation ϕS(i) = ϕS(i) − ϕS(i − 1) between Hilbertpolynomials for i 0, where S = S/h1S. Let

ϕS(t) =e(S)

(d− 1)!td−1 + ad−2t

d−2 + · · ·+ a0,

ϕS(t) =e(S)

(d− 2)!td−2 + bd−3t

d−3 + · · ·+ b0.

Now, observe that thanks to ϕS(i) = ϕS(i)− ϕS(i − 1) one derives:

ϕS(t) =e(S)(d− 1)

(d− 1)!td−2 + lower order terms,

consequently e(S) = e(S). Since S is Cohen–Macaulay of dimension d − 1(see Lemma 2.3.10), the result follows by induction. �

Hilbert Series 175

Proposition 5.1.11 If A, B are standard algebras over a field K, then

dim(A⊗K B) = dim(A) + dim(B).

Proof. One has the next equality between Hilbert series:

F (A, t)F (B, t) =

[ ∞∑i=0

(dimK Ai)ti

][ ∞∑i=0

(dimK Bi)ti

]

=

∞∑i=0

(dimK Ci)ti = F (A⊗K B, t),

whereA⊗KB =⊕∞

r=0 Cr and Cr =⊕

i+j=r Ai⊗KBj . Thus, Theorem 5.1.4yields the asserted equality. �

Using the Noether normalization lemma it is not difficult to prove thatthe result above holds for arbitrary affine algebras over a field K.

Computation of Hilbert series Let R be a polynomial ring over a fieldK with a monomial order ≺ and let I ⊂ R be a graded ideal. Since R/Iand R/ in≺(I) have the same Hilbert function (see Corollary 3.3.15), theactual computation of the Hilbert series of R/I is a two-step process:

• first one finds a Grobner basis of I using Buchberger’s algorithm, and

• second one computes the Hilbert series of R/ in(I) using eliminationof variables (see [20, 37]).

Other approaches to compute Hilbert series use minimal resolutions andStanley decompositions [398]. An ad hoc method to compute Hilbert seriesof Cohen–Macaulay N-graded algebras is given by Proposition 3.1.27. Someexamples are given below.

Example 5.1.12 Let R = K[x1, x2, x3] be a polynomial ring and let Ibe the ideal (x21, x

22x3, x

32). Let us compute the Hilbert series of R/I using

elimination. Pick any monomial involving more than one variable, say x22x3.The idea is to eliminate x22 from the monomials containing more than onevariable. From the exact sequence of graded modules:

0 −→ R/(x21, x3, x2)(−2)x22−→ R/I−→R/(x21, x22) −→ 0,

and applying Exercise 5.1.20 to the ends of this sequence, we get

F (R/I, t) = t2[(1− t)2(1− t2)

(1 − t)3

]+

(1− t2)2(1 − t)3 =

−t4 + 2t2 + 2t+ 1

(1− t) .

176 Chapter 5

Example 5.1.13 Let R = Q[x, y, z, w] and I = (f1, f2, f3), where

f1 = y2 − xz, f2 = x3 − yzw, f3 = x2y − z2w.

Using Macaulay2 [199] one finds that the minimal resolution of R/I is:

0 −→ R2(−4) ϕ2−→ R(−2)⊕R2(−3) ϕ1−→ R −→ R/I −→ 0.

Let us compute the Hilbert series of R/I using its minimal resolution:

F (R/I, t) = F (R, t)− F (R(−2)⊕R2(−3), t) + F (R2(−4), t)

=1

(1 − t)4 −t2

(1− t)4 −2t3

(1− t)4 +2t4

(1− t)4

=1 + 2t+ 2t2

(1 − t)2 .

Example 5.1.14 Let R = K[x1, . . . , x5] and I = (x1x2, x3x4x5). Notethat I is a complete intersection. Hence, by Exercise 5.1.20, one has

F (R/I, t) =(1− t2)(1 − t3)

(1− t)5 =1 + 2t+ 2t2 + t3

(1 − t)3 = −1 + 5t2 − t+ 2

(1− t)3 .

The Laurent expansion of F (R/I, t) + 1, in negative powers of (1− t), is

5t2 − t+ 2

(1− t)3 =a−3

(1− t)3 +a−2

(1− t)2 +a−1

(1− t) ,

where a−1 = 5, a−2 = −9, a−3 = 6. Therefore the Hilbert polynomial ofS = R/I is ϕS(t) = 3t2 + 2 because by Proposition 5.1.6 one has:

H(S, i) = 6

(i+ 2

i

)− 9

(i+ 1

i

)+ 5

(i

i

)= 3i2 + 2 for i ≥ 1.

Exercises

5.1.15 If n ≥ 1 is an integer, then

1

(1 − t)n =

∞∑m=0

(m+ n− 1

m

)tm.

5.1.16 Let R be a polynomial ring in n variables over a field K. If R hasthe standard grading, then

F (R, t) =1

(1− t)n and H(R,m) =

(m+ n− 1

n− 1

).

Hilbert Series 177

5.1.17 Let K be a field and R = K[x1, . . . , xn] a polynomial ring gradedby deg(xi) = di ∈ N+ for i = 1, . . . , n. Then the Hilbert series of R is

F (R, t) = 1/∏ni=1(1 − tdi).

5.1.18 Let R = K[x, y, z] be a polynomial ring over a field K and I is theideal (xa, yb, zc, xa1yb1zc1). If a ≥ a1, b ≥ b1 and c ≥ c1, then

e(R/I) = dimK(R/I) = abc1 + (c− c1)(a1b+ (a− a1)b1).

5.1.19 Let R = K[x, y] be a polynomial ring over a field K. If I is the ideal(xn, xayb, yn) and a < n, b < n, then e(R/I) = n(a+ b)− ab.

5.1.20 Let R = K[x1, . . . , xn] be a polynomial ring over a field K graded bydeg(xi) = di ∈ N+ for i = 1, . . . , n. If f1, . . . , fr is a homogeneous regularsequence in R and ai = deg(fi), then

F (R/(f1, . . . , fr), t) =∏ri=1 (1− tai)/

∏ni=1

(1− tdi

).

5.1.21 Let R = K[x1, x2, x3] and let I be the ideal (x21, x22x3, x

32). Prove

that the a-invariant of R/I is 3 and H(R/I, i) = 4 for i ≥ 4.

5.1.22 Let A, B be two standard algebras over a field K and define theirSegre product as the graded algebra

S = A⊗S B = (A0 ⊗K B0)⊕ (A1 ⊗K B1)⊕ · · · ⊂ A⊗K B,

where (A⊗K B)p =∑

i+j=p Ai ⊗K Bj . Use Hilbert functions to prove

dim(S) = dim(A) + dim(B) − 1.

Note dim(Ai ⊗K Bi) = dim(Ai) dim(Bi).

5.2 a-invariants and h-vectors

Our goal here is to relate the a-invariant of a Cohen–Macaulay N-gradedalgebra with its minimal resolution and to prove that h-vectors of suchalgebras are nonnegative. We include Stanley’s characterization of Cohen–Macaulay algebras in terms of Hilbert series.

a-invariants of positively graded algebras Let R = K[x1, . . . , xn] bea positively graded polynomial ring over a field K, let I be a graded idealof R, and let

F : 0→bg⊕i=1

R(−dgi)ϕg→ · · · →

bk⊕i=1

R(−dki)ϕk→ · · · →

b1⊕i=1

R(−d1i)ϕ1→ R

178 Chapter 5

be the minimal resolution of S = R/I (see Section 3.5). From the resolutionof S, using Lemmas 5.1.1 and 5.1.2, one can write the Hilbert series as:

F (S, t) =

(1 +

g∑k=1

(−1)k(

bk∑i=1

tdki

))n∏i=1

(1− tdeg(xi))

. (5.1)

We begin with a general inequality, that follows from Eq. (5.1), relatinga(S), the a-invariant S, with the shifts of F , the minimal resolution of S.

Proposition 5.2.1 a(S) ≤ max{dki| 1 ≤ k ≤ g, 1 ≤ i ≤ bk}+ a(R).

Definition 5.2.2 If I ⊂ R is a Cohen–Macaulay graded ideal of height g,the canonical module of S = R/I is defined as

ωS = ExtgR(R/I, ωR),

where ωR = R(−δ) and δ =∑n

i=1 deg(xi).

Proposition 5.2.3 If I ⊂ R is a Cohen–Macaulay graded ideal and a(S)is the a-invariant of S = R/I, then

(a) maxi{d1i} < · · · < maxi{dgi}, and

(b) a(S) = max1≤i≤bg{dgi}+ a(R) = −min{i | (ωS)i �= 0}.

Proof. (a): We may assume that dk1 ≤ · · · ≤ dkbk for k = 1, . . . , g.Since S is Cohen–Macaulay, the height of I is equal g, this follows fromCorollary 3.5.14. Recall that by Proposition 3.1.32 there exists a regularsequence f1, . . . , fg inside I. Hence, using Proposition 2.3.7, one has:

ExtiR(S, ωR) = 0 for i < g.

We set δ =∑n

i=1 deg(xi) and ωR = R(−δ). Therefore dualizing F , theminimal resolution of S, with respect to the canonical module of R, oneobtains the (exact) complex:

Hom(F , ωR) : 0→ Hom(R,ωR)→ · · · → Hom

⎛⎝ bg⊕i=1

R(−dgi), ωR

⎞⎠→ 0.

Hilbert Series 179

Observe the following natural isomorphisms:

Hom

⎛⎝ bg⊕i=1

R(−dgi), R(−δ)

⎞⎠ � Hom

⎛⎝ bg⊕i=1

R(−dgi), R

⎞⎠ (−δ)

�bg⊕i=1

R(dgi)(−δ)

�bg⊕i=1

R(−(δ − dgi)).

Since Hom(F , ωR) is a minimal resolution of ExtgR(S, ωR), by Remark 3.5.9,one has d1b1 < d2b2 < · · · < dgbg .

(b): By part (a), we have d1b1 < d2b2 < · · · < dgbg . As a consequence,from the expression for F (S, t) given in Eq. (5.1), one concludes the equalitya(S) = dgbg + a(R). In particular one has a surjection

bg⊕i=1

R(−(δ − dgi))→ ExtgR(S,R(−δ)) = ωS → 0.

To complete the proof note that since δ − dgbg ≤ · · · ≤ δ − dg1, one has theequalities min{i| (ωS)i �= 0} = δ − dgbg = −a(S). �

h-vectors of standard algebras Next we introduce the notion of h-vector of standard algebras through the Hilbert–Serre theorem for Hilbertseries (see Theorem 5.1.4).

Definition 5.2.4 Let S be a standard algebra and let h(t) be the (unique)polynomial with integral coefficients such that h(1) �= 0 and

F (S, t) =h(t)

(1− t)d ,

where d = dim(S). If h(t) = h0 + h1t + · · · + hrtr, the h-vector of S is

defined as h(S) = (h0, . . . , hr).

Theorem 5.2.5 [392] Let S be a standard algebra and let θ1, . . . , θd be ahomogeneous system of parameters for S with ai = deg(θi). Then S isCohen–Macaulay if and only if

F (S, t) =F (S/(θ1, . . . , θd), t)

d∏i=1

(1 − tai). (5.2)

180 Chapter 5

Proof. ⇒) Assume S is Cohen–Macaulay. Hence, by Proposition 3.1.20,θ1, . . . , θd is a regular sequence. It is enough to prove the equality

F (S/(θ1, . . . , θs), t) =s∏i=1

(1 − tai)F (S, t), (5.3)

for 1 ≤ s ≤ d. We proceed by induction on s. Assume s = 1. From theexact sequence of graded S-modules

0 −→ S[−a1]θ1−→ S −→ S/(θ1) −→ 0,

we obtain F (S/(θ1), t) = F (S, t)− F (S[−a1], t). Since

F (S[−a1], t) =

∞∑i=0

S[−a1]iti =∞∑i=a1

S[−a1]iti

= ta1∞∑i=a1

Si−a1ti−a1 = ta1F (S, t),

we conclude F (S/(θ1), t) = (1 − ta1)F (S, t). Now assume s > 1 and thatthe equality (5.3) is true for s− 1. From the exact sequence

0→ (S/(θ1, . . . , θs−1)) [−as]θs−→ S/(θ1, . . . , θs−1) −→ S/(θ1, . . . , θs)→ 0

we obtain F (S/(θ1, . . . , θs), t) = (1 − tas)F (S/(θ1, . . . , θs−1), t). Then, theinduction hypothesis yields the required equality.⇐) Let f(t) =

∑i≥0 ait

i and g(t) =∑i≥0 bit

i be two formal powerseries, we say that f(t) ≥ g(t) if ai ≥ bi for all i. From the exact sequence

0 −→ (S/ann (θ1)) [−a1]θ1−→ S −→ S/(θ1) −→ 0,

we obtain F (S/(θ1), t) − (1 − ta1)F (S, t) = ta1F (ann (θ1), t). Hence, thepower series F (S/(θ1), t) is greater than or equal to (1− ta1)F (S, t), and byinduction it follows that

F (S/(θ1, . . . θr), t) ≥r∏i=1

(1 − tai)F (S, t) for 1 ≤ r ≤ d.

Set S′ = S/(θ1, . . . , θd−1). Using the exact sequence

0 −→ (S′/ann (θd))[−ad]θd−→ S′ −→ S′/(θd) = S/(θ1, . . . , θd) −→ 0,

we get F (S′/(θd), t) = (1 − t)adF (S′, t) + tadF (ann (θd), t). By hypothesis,

we have F (S′/(θd), t) =∏di=1(1 − tai)F (S, t). Therefore

(1− t)ad︸︷︷︸≥0

(F (S′, t)− (1 − ta1) · · · (1− tad−1)F (S, t)︸︷︷︸≥0

)

= −tadF (ann (θd), t).

Hilbert Series 181

Hence F (ann (θd), t) = 0. This shows that θd is a regular element on S′. Byinduction it follows that θ1, . . . , θd is a regular sequence, Then S is Cohen–Macaulay by Proposition 3.1.20. This proof was adapted from [392]. �

Theorem 5.2.6 Let S be a Cohen–Macaulay standard algebra over a fieldk and h(S) = (hi) the h-vector of S, then hi ≥ 0 for all i.

Proof. One may assume that k is infinite, otherwise one may change thecoefficient field k using the functor (·) ⊗k K, where K is an infinite fieldextension of k. There is a h.s.o.p y = {y1, . . . , yd} for S, where each yi is aform in S of degree one. Set S = S/(y)S. From Theorem 5.2.5 we derivehi = H(S, i) for all i and consequently hi ≥ 0. �

Theorem 5.2.7 (Stanley) Let K be a field and let S be a positively gradedK-algebra of dimension d. If S is a Cohen–Macaulay domain and

F (S, t) = (−1)dtaF (S, t−1)

for some a ∈ Z, then S is Gorenstein and a is the a-invariant of S.

Proof. By Exercise 5.2.10, F (S, t) = taF (ωS , t) = F (ωS [−a], t). Pick anelement 0 �= x ∈ ωS[−a] of degree zero and define

ϕ : S −→ ωS[−a] (rϕ�−→ rx),

the map ϕ is injective by Exercise 5.2.12. Therefore the equality above givesthat ϕ is an isomorphism. Thus ωS � S[a], as required. �

Proposition 5.2.8 Let S be a standard graded algebra and let M ⊂ N befinitely generated graded S-modules of the same dimension d and multiplicitye. If M and N are Cohen–Macaulay and

HM (t) =f(t)

(1− t)d , HN (t) =g(t)

(1− t)d

are their Hilbert series, then deg f(t) ≥ deg g(t).

Proof. There is an exact sequence 0 → M → N → N/M → 0 of gradedmodules. If M �= N , N/M is a module of dimension < d since M and Nhave the same multiplicity. Since M and N are Cohen–Macaulay, standarddepth chasing (see Lemma 2.3.9) implies that N/M is Cohen–Macaulay ofdimension d− 1.

We have the equality of Hilbert series,

g(t)

(1− t)d =f(t)

(1− t)d +h(t)

(1− t)d−1,

so g(t) − f(t) = (1 − t)h(t). The assertion follows since the h-vectors ofthese two modules are positive (see Exercise 5.2.9). �

182 Chapter 5

Exercises

5.2.9 Let R be a polynomial ring over a field K with the usual gradingandM a finitely generated N-graded module. IfM is Cohen–Macaulay andh(t) = h0 + h1t+ · · ·+ hrt

r the polynomial in Z[t] so that h(1) �= 0 and

F (M, t) = h(t)/(1− t)d, where d = dim(M).

Prove that hi ≥ 0 for all i.

5.2.10 (Stanley) Let K be a field and let S be a positively graded K-algebra. If S is Cohen–Macaulay of dimension d prove the equality:

F (ωS , t) = (−1)dF (S, t−1).

Hint Note S � R/I, with R a polynomial ring. Compute the Hilbert seriesof S and ωS using the minimal resolution of R/I as an R-module.

5.2.11 (Stanley) Let K be a field and let S be a positively graded K-algebra. If S is Gorenstein of dimension d and a(S) is its a-invariant, prove:

F (S, t) = (−1)dta(S)F (S, t−1).

Hint ωS � S[a(S)].

5.2.12 Let S be a Cohen–Macaulay local domain and let M be a Cohen–Macaulay S-module. If dim(S) = dim(M), prove that Z(M) = (0).

Hint Note dim(M) = dim(S/p) for p ∈ Ass(M) and ann(M) = 0.

5.2.13 Let K be a field and let S be a positively graded K-algebra. If S isa Cohen–Macaulay domain and the h-vector of S is symmetric, prove thatS is a Gorenstein ring.

Hint Use Theorem 5.2.7.

5.2.14 Let R be a polynomial ring over a fieldK with the standard grading,and let M1,M2 be two finitely generated graded R-modules. Then

depth(M1 ⊕M2) = max{depth(M1), depth(M2)}.

5.3 Extremal algebras

The concept of an extremal Cohen–Macaulay—or Gorenstein—standard al-gebra (resp. ideal) has its origins in the works of Sally [366] and Schenzel[370]. Those algebras (resp. ideals) have the smallest possible reductionnumber (resp. smallest possible number of generators of the least degree).We will study with certain detail those two classes of algebras.

Hilbert Series 183

The Cohen–Macaulay case Let R be a polynomial ring over a field kwith its usual grading and let I =

⊕∞i=p Ii be a graded ideal with Ip �= (0).

The integer p is the initial degree of I. As usual μ(I) denotes the minimalnumber of generators of I, and μ(Ip) stands for the minimal number ofgenerators of I in degree p.

Proposition 5.3.1 Let R be a polynomial ring over a field k and I a gradedideal in R of height g and initial degree p. If I is Cohen–Macaulay, then

μ(Ip) = H(I, p) ≤(p+ g − 1

p

).

Proof. We may assume that k is infinite, since a change of the coefficientfield can be easily carried out using the functor (·) ⊗k K, where K is aninfinite field extension of k. This change of the coefficient field preservesthe hypothesis on I and leaves both the height of I and the dimensions ofthe vector spaces of forms of a given degree in I unchanged.

We set S = R/I. By Lemma 3.1.28, there is a homogeneous system ofparameters y = {y1, . . . , yd} for S, where each yi is a linear form in R. Ifwe tensor the minimal resolution of S with R = R/(y) it follows that

0 ≤ H(S, p) = H(R, p)−H(I, p),

where S = S/(y)S. To get the desired inequality it suffices to observe thatR is a polynomial ring in g variables over the field k. �

Definition 5.3.2 If S is an Artinian positively graded algebra the socle ofS is given by

Soc(S) = (0: S S+).

Lemma 5.3.3 [174] Let S = R/I be a quotient ring of a polynomial ringR over a field k modulo a graded ideal I. Let

F : 0→bg⊕i=1

R(−dgi)→ · · · →b1⊕i=1

R(−d1i)→ R→ S = R/I → 0,

be the minimal resolution of S. If S is Artinian, then there is a degree zero

isomorphism of graded k-vector spaces Soc(S) �⊕bg

i=1 k[g − dgi].

Proof. Let R = k[x1, . . . , xg] and x = {x1, . . . , xg}. The ordinary Koszulcomplex K associated to the regular sequence x is an R-free resolution ofk = R/(x) and thus

TorRi (S, k) � Hi(K ⊗R S).

184 Chapter 5

On the other hand, we have TorRg (S, k) �⊕bg

i=1 k[−dgi] (see the proof ofCorollary 3.5.7). To complete the argument note

Hg(K ⊗R S)[g] � Zg(K ⊗R S)[g]

=

{ze1 ∧ · · · ∧ eg

∣∣∣∣∣g∑i=1

(−1)i−1zxie1 ∧ · · · ∧ ei ∧ · · · ∧ eg = 0

}[g]

� Soc(S).

Altogether one has Soc(S) �⊕bg

i=1 k[g − dgi], as required. �

Proposition 5.3.4 If S is a Cohen–Macaulay positively graded k-algebraover a field k with presentation R/I, then the type of S is equal to the lastBetti number in the minimal free resolution of R/I as an R-module.

Proof. By making a change of coefficients using the functor (·)⊗kK, whereK is an infinite field extension of k, one may assume that k is infinite.

By Theorem 3.1.24 there exists a system of parameters y = {y1, . . . , yd}for S = R/I with each yi a form of degree one ofR. Since Tori(S,R/(y)) = 0for i ≥ 1, the minimal resolution of S/(yS) as an R/(y)-module has thesame twists and Betti numbers as the minimal resolution of S over R. SinceS/(yS) is Artinian the result follows from Lemma 5.3.3. �

Corollary 5.3.5 If R is a polynomial ring over a field k, then a gradedideal I of R is Gorenstein iff R/I is Cohen–Macaulay and the last Bettinumber in the minimal graded resolution of R/I is equal to 1.

Corollary 5.3.6 Let S be a Cohen–Macaulay positively graded k-algebraover a field k and let ωS be its canonical module. Then

type(S) = μ(ωS),

where μ(ωS) is the minimum number of generators of ωS.

Proof. It follows from the proofs of Propositions 5.2.3 and 5.3.4. �

Theorem 5.3.7 ([79], [350]) Let R be a polynomial ring over a field k andlet I be a graded ideal in R of height g and initial degree p. If S = R/I isa Cohen–Macaulay ring, then S has a p-linear resolution if and only if thefollowing equality holds

μ(Ip) = H(I, p) =

(p+ g − 1

p

). (5.4)

Proof. Let F be the minimal resolution of S as in Lemma 5.3.3. Wecan order the shifts so that dk1 ≤ · · · ≤ dkbk for k = 1, . . . , g; by the

Hilbert Series 185

minimality of the resolution d11 < d21 < · · · < dg1 (see Remark 3.5.9).Observe that if the resolution of S is linear, then the Herzog–Kuhl formulasof Theorem 3.5.17 give the required equality.

Conversely assume the equality above. Since we may assume that k isinfinite, by Lemma 3.1.28 there is a h.s.o.p y = {y1, . . . , yd} for S, whereeach yi is a form of degree one in R. Set S = S/(y)S and R = R/(y). FromTheorem 5.2.5 we get that the h-vector of S satisfies hp = 0, and we alsoget hi = H(S, i) for all i. Therefore hi = 0 for all i ≥ p. Using Lemma 5.3.3one obtains a degree zero isomorphism

Soc(S) �bg⊕i=1

k[g − dgi]

of graded k-vector spaces. In particular, the socle of S can only live indegrees dgi − g, hence we conclude the inequality dgi − g ≤ p − 1. On theother hand the minimality of the resolution of S gives p + (g − 1) ≤ dgifor all i. Altogether we have dgi = p + g − 1. Because I is C–M, then byProposition 5.2.3 we must have p ≤ d1b1 < d2b2 < · · · < dgbg = p + g − 1,which implies that the resolution is p-linear, as required. �

Example 5.3.8 [346] Let R = k[a, . . . , f ] be a polynomial ring over a fieldk and let I be the ideal

I = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf).

Then R/I is C–M if char(k) �= 2 and non C–M otherwise. Thus in the firstcase R/I has a 3-linear resolution.

Other nice examples of Cohen–Macaulay algebras with linear resolutionsinclude rings of minimal multiplicity [366], the coordinate ring of a varietydefined by the submaximal minors of a generic symmetric matrix [264], thecoordinate ring of a variety defined by the maximal minors of a genericmatrix [123], and some face rings [170].

Cohen–Macaulay rings with linear resolutions have been studied by Sally[366] for the case p = 2, and by Schenzel [370] for the general case; moregeneral rings with linear resolutions have been examined in [130, 233].

The Gorenstein case To give a sharp bound for the number of gener-ators in the least degree of a graded Gorenstein ideal is harder than theCohen–Macaulay case. We deal with this problem in the next section.

The aim here is to introduce and characterize extremal Gorenstein ringsin terms of their minimal resolutions and to determine their Betti numbers.Those Betti numbers are our “natural candidates” to bound the initial Bettinumbers of graded Gorenstein ideals.

186 Chapter 5

Proposition 5.3.9 If S = S0 + · · · + Ss is a positively graded ArtinianGorenstein algebra over a field k, then the multiplication maps

Si × Ss−i → Ss = Soc(S) � k

are a perfect pairing, that is, the homomorphisms:

ϕi : Si → Homk(Ss−i, Ss) � Ss−i, ϕi(r)(x) = rx

are isomorphisms of k-vector spaces.

Proof. To show that ϕi is one to one take r ∈ ker(ϕ) and assume r �= 0and i < s. Since r is not in the socle of S, one may pick z ∈ S homogeneousof maximal positive degree such that zr �= 0. Note that 0 < deg(z) < s− i,because rx = 0 for x ∈ Ss−i. Hence zr has degree less than s, using thatthe socle of S lives only in degree s one obtains a homogeneous elementw with deg(w) > 0 such that wzr = 0, which contradicts the choice ofz. Therefore r = 0 and ϕ is injective. Using the injectivity of ϕi one hasdimk Si ≤ dimk Ss−i ≤ dimk Si, and thus ϕi is an isomorphism. �

Corollary 5.3.10 Let (h0, . . . , hs) be the h-vector of a standard Gorensteink-algebra, where k is a field. Then h is symmetric, that is,

hi = hs−i for all i.

Proposition 5.3.11 Let R be a polynomial ring over a field k with thestandard grading and I a graded Gorenstein ideal of initial degree p andheight g. If the minimal resolution of S = R/I by free R-modules is:

F : 0→bg⊕i=1

R(−dgi)→ · · · →bk⊕i=1

R(−dki)→ · · · →b1⊕i=1

R(−d1i)→ R,

then⊕bg−k

i=1 R(−(dg − d(g−k)i)) �⊕bk

i=1 R(−dki) for 1 ≤ k ≤ g − 1.

Proof. One may always order the shifts so that dk1 ≤ · · · ≤ dkbk fork = 1, . . . , g. As bg = 1 and dgi = dg, by Proposition 5.2.3, one has thatthe a-invariant of S is a(S) = dg − n, where n is the number of variables ofR. Since Hom(F , ωR) is a minimal resolution of

ωS � ExtgR(S, ωR)

and using ωS � S(dg − n), one obtains that F is self dual. �

With the same assumptions and notation of Proposition 5.3.11 one has:

Hilbert Series 187

Proposition 5.3.12 If (h0, . . . , hs) is the h-vector of S, then s ≥ 2(p− 1)with equality if and only if the minimal resolution of S is of the form

0→ R(−(2p+ g − 2))→ Rbg−1(−(p+ g − 2))→· · · → Rb2(−(p+ 1))→ Rb1(−p)→ R.

Proof. As in the proof of Theorem 5.2.6, one may assume that S is Artinianand that hi is equal to H(S, i). Note that the socle of S lives in degreedg − g by Lemma 5.3.3. Hence s ≥ dg − g. By Proposition 5.3.11 and theminimality of the resolution one concludes

s ≥ dg − g = d1b1 + d(g−1)1 − g ≥ p− g + d(g−1)1 ≥ 2(p− 1),

where the last inequality uses dk1 ≥ p + k − 1 for 1 ≤ k ≤ g − 1. Forthe second assertion use the chain of inequalities above and compute theh-vector of S using the resolution of S. �

Definition 5.3.13 If s = 2(p− 1), S is called an extremal Gorenstein ringand its resolution is called pure and almost linear .

Proposition 5.3.14 [370] Let R be a polynomial ring over a field k and Ia graded ideal of initial degree p and height g. If S = R/I is an extremalGorenstein algebra, then for 1 ≤ m < g the mth Betti number of S is:

bm =

(p+ g − 1

p+m− 1

)(p+m− 2

m− 1

)+

(p+ g − 1

m

)(p+ g −m− 2

p− 1

)−(g

m

)(p+ g − 2

p− 1

).

Proof. As in the proof of Theorem 5.3.7, one may assume that S is Artinianand that R a polynomial ring in g variables. Set s = 2(p − 1). From theminimal resolution of S of Proposition 5.3.12, and using the additivity ofHilbert series (see Lemma 5.1.1), we get(

s∑i=0

hiti

)(1− t)g = 1 +

g−1∑i=1

(−1)ibitp+i−1 + (−1)gbgts+g,

where hi = H(S, i) = hs−i. Therefore

bm =

p+m−1∑i=0

(−1)p−1−i(

g

p+m− 1− i

)hi, 1 ≤ m ≤ g − 1

Using the identity

p−1∑i=0

(−1)p−1−i(

g

p+m− 1− i

)(g − 1 + i

i

)=

(p+ g − 1

p+m− 1

)(p+m− 2

m− 1

)(5.5)

188 Chapter 5

and the symmetry of the h-vector of S (see Corollary 5.3.10) one has

bm =

(p+ g − 1

p+m− 1

)(p+m− 2

m− 1

)+

p+m−1∑i=p

(−1)p−1−i(

g

p+m− 1− i

)hs−i.

Note

p+m−1∑i=p

(−1)p−1−i(

g

p+m− 1− i

)hs−i

=m−1∑i=0

(−1)i+1

(g

m− 1− i

)(p+ g − 3− i

g − 1

),

since the terms in the right-hand side of the equality are zero for i > p− 1the last summation reduces to

p−2∑i=0

(−1)i+1

(g

m− 1− i

)(p+ g − 3− i

g − 1

)

=

p−1∑i=0

(−1)p−1−i(

g

g −m+ p− 1− i

)(g − 1 + i

i

)−(g

m

)(g + p− 2

p− 1

)=

(p+ g − 1

m

)(p+ g −m− 2

p− 1

)−(g

m

)(g + p− 2

p− 1

),

the last equality follows from Eq. (5.5) making m equal to g −m. �

Exercises

5.3.15 Let R be a polynomial ring over a field k and let I be a Cohen–Macaulay graded ideal in R of height g and initial degree p. Let (h0, . . . , hs)be the h-vector of R/I. Prove that s ≥ p − 1, with equality if and only ifthe minimal resolution of R/I is p-linear. If s = p− 1, the algebra R/I iscalled an extremal Cohen–Macaulay algebra.

5.3.16 Let R be a polynomial ring over a field k and let I be a C–M gradedideal in R of height g and initial degree p. Prove that the Hilbert series ofS = R/I can be written as

F (S, t) =

p−1∑i=0

(i+ g − 1

g − 1

)ti

(1− t)d , where d = dim S,

if and only if S has a p-linear resolution.

Hilbert Series 189

5.3.17 Let R be a polynomial ring over a field k and I a Cohen–Macaulaygraded ideal in R of height g and initial degree p. If S has a linear resolution,show that the Betti numbers of S = R/I and its multiplicity are given by

bk =

(p+ k − 2

k − 1

)·(p+ g − 1

g − k

)and e(S) =

(p+ g − 1

g

).

5.3.18 Let S = R/I be a standard Gorenstein algebra so that I has initialdegree p and codimension g. Show that the multiplicity of S satisfies:

e(S) ≥ e(g, p) =(g + p− 1

g

)+

(g + p− 2

g

),

with equality if and only if S is an extremal Gorenstein algebra.

Hint The h-vector h(S) = (h0, . . . , hs) is symmetric and s ≥ 2(p− 1). Then

compute explicitly hp−1 + 2∑p−2i=0 hi.

5.3.19 Let R be a polynomial ring over a field k and I a graded Gorensteinideal of height 4 generated by forms of degree p. Let h(S) = (h0, . . . , hs) bethe h-vector of S = R/I. Then the minimal resolution of S has the form:

0→ R(−(s+4))→ Rb(−(s−p+4))→s−2p+3⊕i=1

Rai(−(p+i))→ Rb(−p)→ R,

where r = s− 2p+ 3 and ai = ar−i+1 for i ≥ 1.

5.3.20 If x > 1 is an integer, prove that x(x+ 8) is never a perfect square.

5.3.21 Let I be a graded Cohen–Macaulay ideal of a polynomial ring Rwith a minimal resolution:

0→ Rb3(−(p+ a+ b))→ Rb2(−(p+ a))→ Rb1(−p)→ R→ R/I.

If b1 = 9, use the Herzog–Kuhl formulas to prove that I is Gorenstein.

Hint Note b3 − b2 + b1 − 1 = 0 and prove that b1b2b3 = 9b3(b3 + 8) is aperfect square, then use Exercise 5.3.20.

5.4 Initial degrees of Gorenstein ideals

Assume that R is a polynomial ring over a field and that I is a homogeneousGorenstein ideal of codimension g ≥ 3 and initial degree p ≥ 2. We set

e(g, p) =

(g + p− 1

g

)+

(g + p− 2

g

)and μ0 =

(p+ g − 1

g − 1

)−(p+ g − 3

g − 1

).

190 Chapter 5

By the symmetry of the h-vector of R/I given in Corollary 5.3.10, themultiplicity of R/I satisfies

e(R/I) ≥ e(g, p),

see Exercise 5.3.18. Given codimension g and initial degree p, a gradedGorenstein algebra R/J with multiplicity e(R/J) = e(g, p) is extremal.This has strong structural implications: the minimal free resolution of R/Jmust be pure and almost linear; hence all the Betti numbers bi(R/J) aredetermined, and J is generated by μ0 forms of degree p (see Section 5.3).

Results in the literature dealing with Cohen–Macaulay ideals (such as[140, 361]) give upper bounds for the minimal number of generators μ(I) ofI in terms of codimension, initial degree, and multiplicity e(R/I). One ofthe aims below is to elucidate the multiplicity information that is alreadydetermined by the codimension and initial degree (see Theorem 5.4.3). Ingeneral one expects no such conclusion, but for graded Gorenstein algebrasthe symmetry of the h-vector h(R/I) can be effectively exploited.

Macaulay’s Theorem

First we introduce the binomial expansion, and then state a famous resultof Macaulay on the growth of Hilbert functions that will be needed later.

Lemma 5.4.1 Let h, j be positive integers. Then h can be written uniquely

h =

(ajj

)+

(aj−1

j − 1

)+ · · ·+

(aii

), (5.6)

where aj > aj−1 > · · · > ai ≥ i ≥ 1.

Proof. By induction on j. If j = 1, then h =(h1

). Assume j > 1, then

there is a unique aj such that(ajj

)≤ h <

(aj+1j

). Let r = h −

(ajj

). By

induction

r = h−(ajj

)=

(aj−1

j − 1

)+ · · ·+

(aii

)where aj−1 > · · · > ai ≥ i ≥ 1.

From the inequality

h =

(ajj

)+

(aj−1

j − 1

)+ · · ·+

(aii

)<

(aj + 1

j

)=

(ajj

)+

(ajj − 1

)we obtain

(aj−1

j−1

)<(ajj−1

)and this implies aj−1 < aj . Let us show the

uniqueness. Assume h can be written as

h =

(bjj

)+

(bj−1

j − 1

)+ · · ·+

(bkk

),

Hilbert Series 191

where bj > bj−1 > · · · > bk ≥ k ≥ 1. By induction on j one has(bjj

)≤ h <

(bj + 1

j

).

Since aj is unique we derive aj = bj . Hence by induction hypothesis k = iand aj−1 = bj−1, . . . , ai = bi. �

The unique expression for h in Eq. (5.6) is called the binomial expansionof h in base j. We set

h(j) =

(ajj + 1

)+

(aj−1

j

)+ · · ·+

(aii+ 1

), (5.7)

h〈j〉 =

(aj + 1

j + 1

)+

(aj−1 + 1

j

)+ · · ·+

(ai + 1

i+ 1

), (5.8)

sometimes h〈j〉 is called the Macaulay symbol.

Theorem 5.4.2 (Macaulay [413, Appendix B]) Let h : N→ N be a numer-ical function and k a field. The following conditions are equivalent :

(a) There exists a homogeneous k-algebra S with H(S, i) = h(i) for i ≥ 0.

(b) h(0) = 1 and h(i+ 1) ≤ h(i)〈i〉 for all i ≥ 1.

Theorem 5.4.3 [318] If I is a graded Gorenstein ideal of height g ≥ 3 andinitial degree p ≥ 2, then

μ(Ip) ≤ μ0 =

(p+ g − 1

g − 1

)−(p+ g − 3

g − 1

),

and I is itself extremal if equality holds.

Proof. If either μ(Ip) > μ0, or μ(Ip) = μ0 and I is not extremal, then bythe symmetry of the h-vector h(R/I) there is some j ≥ p so that

h(R/I) = (h0, h1, . . . , hp−1, hp, . . . , hj , hp−1, . . . , h1, h0),

where hj = hp ≤(p+g−3g−1

)= hp−2. The idea of the argument is to use the

Macaulay bound h〈j〉j for hj+1 to see that such a small value of hj cannot

grow to such a large value of

hj+1 = hp−1 =

(p+ g − 2

g − 1

).

Recall that this estimate is calculated from the binomial expansion for hj :

hj =

(ajj

)+

(aj−1

j − 1

)+ · · ·+

(aii

), (5.9)

192 Chapter 5

where aj > aj−1 > · · · > ai ≥ i ≥ 1. Then

hj+1 ≤ h〈j〉j =

(aj + 1

j + 1

)+

(aj−1 + 1

j

)+ · · ·+

(ai + 1

i+ 1

). (5.10)

We may assume that hj > j, for if not, then hj ≤ j would imply that a� = �for all �, and hence hj ≥ hj+1, which contradicts our assumption. Noticethat by grouping the terms of (5.9) according to the value of a� − � thebinomial expansion for hj can be written as

hj =

r∑n=0

[(jn + knjn

)+

(jn − 1 + knjn − 1

)+ · · ·+

(jn − in + knjn − in

)]

=

r∑n=0

[(jn + kn + 1

jn

)−(jn − in + knjn − in − 1

)], (5.11)

where aj − j = k0 > k1 > · · · > kr ≥ 0, j = j0 > j1 > · · · > jr, jr − ir = i,and jn = jn−1 − in−1 − 1 for 1 ≤ n ≤ r. Set k = k0. Since p ≤ j and(

j + k

j

)≤ hj ≤

(p+ g − 3

g − 1

)it follows that k ≤ g − 2. From (5.9), (5.10), and (5.11), together withPascal’s identity and(

a+ b+ 1

b + 1

)=a+ 1

b+ 1

(a+ b+ 1

b

),

we have

hj+1 − hj ≤r∑

n=0

[(jn + kn + 1

jn + 1

)−(jn − in + knjn − in

)]

=

r∑n=0

[kn + 1

jn + 1

(jn + kn + 1

jn

)− kn + 1

jn − in

(jn − in + knjn − in − 1

)].

On the other hand from the upper bound on hj and hj+1 = hp−1 we get

g − 1

p− 1hj ≤

(p+ g − 3

g − 2

)≤ hj+1 − hj .

Since (g − 1)/(p − 1) > (k + 1)/(j + 1) it follows from (5.11) and the lasttwo inequalities that F0 < 0, where for 0 ≤ s ≤ r we set

Fs =

r∑n=s

[(ks + 1

js + 1− kn + 1

jn + 1

)(jn + kn + 1

jn

)]

+

r∑n=s

[(kn + 1

jn − in− ks + 1

js + 1

)(jn − in + knjn − in − 1

)].

Hilbert Series 193

To derive a contradiction we are going to show the following inequalities

F0 > F1 > · · · > Fr =

(kr + 1

jr − ir− kr + 1

jr + 1

)(aii− 1

)> 0.

Assume 1 ≤ s+ 1 ≤ r. Notice that

0 ≤r−1∑n=s

[(jn − in + knjn − in − 1

)−(jn+1 + kn+1 + 1

jn+1

)]+

(jr − ir + krjr − ir − 1

)

=

(js − is + ksjs − is − 1

)−

r∑n=s+1

[(jn + kn + 1

jn

)−(jn − in + knjn − in − 1

)].

Therefore(js − is + ksjs − is − 1

)≥

r∑n=s+1

(jn + kn + 1

jn

)−

r∑n=s+1

(jn − in + knjn − in − 1

).

Let As denote the first summation and Bs the second in this last inequality;clearly As −Bs > 0. Also note that

ks + 1

js − is>ks + 1

js + 1and

ks + 1

js − is>ks+1 + 1

js+1 + 1.

Putting these together, we compute

Fs − Fs+1 =

(ks + 1

js − is− ks + 1

js + 1

)(js − is + ksjs − is − 1

)+

(ks + 1

js + 1− ks+1 + 1

js+1 + 1

)As +

(ks+1 + 1

js+1 + 1− ks + 1

js + 1

)Bs

≥(ks + 1

js − is− ks + 1

js + 1

)(As −Bs) +

(ks + 1

js + 1− ks+1 + 1

js+1 + 1

)As

+

(ks+1 + 1

js+1 + 1− ks + 1

js + 1

)Bs

=

(ks + 1

js − is− ks+1 + 1

js+1 + 1

)(As −Bs) > 0.

Hence F0 > Fr > 0, which contradicts the observation that F0 < 0. �

One might hope for even stronger estimates than suggested by the resultabove, for instance that μ(I) ≤ μ0. Next we present a counterexampleshowing that Theorem 5.4.3 cannot be extended to bound the number ofgenerators in all degrees.

194 Chapter 5

Example 5.4.4 If I = ((x21, x42, x

33, x

44) : (x1x2 − x3x4)), using Macaulay2

[199] one readily obtains that I is given by:

(x21, x1x2x3 + x23x4, x33, x1x

23, x

42, x

44, x1x

34, x1x2x

24 + x3x

34, x

32x

23, x

32x

34).

Then R/I is a Gorenstein Artin algebra with h-vector (1, 4, 9, 13, 13, 9, 4, 1)and Betti sequence (1, 10, 18, 10, 1), whereas the h-vector for an extremalGorenstein algebra of codimension four and initial degree two is (1, 4, 1) andthe Betti sequence is (1, 9, 16, 9, 1); notice that the multiplicity e(R/I) = 54is far greater than the minimal value of six exhibited by an extremal algebra.

In height three one can bound the Betti numbers, thanks to the structuretheorem of Buchsbaum and Eisenbud. For initial degree p, the extremalGorenstein algebra of codimension three has b1 = b2 = 2p+ 1.

Theorem 5.4.5 Let R be a polynomial ring over a field k and let I be ahomogeneous Gorenstein ideal of height 3 and initial degree p. Then

μ(I) ≤ 2p+ 1 and b2(R/I) ≤ 2p+ 1.

Proof. We may assume, without loss of generality, that R is equal tok[x1, x2, x3] and A = R/I is Artinian and local with socle(A) = Aσ. Thenby [75] the minimal free resolution of A has the form

0→ R(−σ − 3)→μ⊕j=1

R(−nj) Y−→μ⊕i=1

R(−mi)→ R,

where Y is an alternating matrix, and the generators f1, . . . , fμ of I are themaximal pfaffians of Y . If the theorem fails every generator has degree atleast (2p + 1)/2 > p, which is a contradiction since at least one minimalgenerator has degree p. �

Conjecture 5.4.6 (Miller–Villarreal) Let I ⊂ R be a graded Gorensteinideal of initial degree p and height g. If 1 ≤ i < g and γi is the ith initialBetti number of R/I, then

γi ≤(p+ g − 1

p+ i− 1

)(p+ i − 2

i− 1

)+

(p+ g − 1

i

)(p+ g − i− 2

p− 1

)−(g

i

)(p+ g − 2

p− 1

).

Proposition 5.4.7 ([39], [65, Lemma 4.2.13]) Let n be a given positiveinteger and f, g : N→ N the numerical functions given by

g(x) = n〈x〉 and f(x) = x〈n〉,

where h〈j〉 is the Macaulay symbol. Then g(x) is non increasing and f(x)is strictly increasing.

Hilbert Series 195

There is a short and elegant argument, based on the behavior of thecombinatorial functions g(x) = n〈x〉 and f(x) = x〈n〉, to show the inequalityof Theorem 5.4.3. We include the details of this argument in the proof below.

Corollary 5.4.8 If R is a polynomial ring over a field k and I is a gradedGorenstein ideal of R of height g ≥ 3 and initial degree p ≥ 2, then

μ(Ip) ≤ μ0 =

(p+ g − 1

g − 1

)−(p+ g − 3

g − 1

).

Proof. First we make a change of coefficients using the functor (·) ⊗k K,where K is an infinite field extension of k. Hence we may assume that k isinfinite. Let A be an Artinian reduction of S = R/I, that is,

A = S/(h1, . . . , hd),

where h = h1, . . . , hd is a system of parameters of S with hi a form ofdegree 1 for all i. Note that one can write A = R/I, where R = R/(h)is a polynomial ring in g variables and I = IR is a graded Gorensteinideal of R of codimension g and initial degree p. By Proposition 5.3.9 theHilbert function of A is symmetric. Since the h-vector h(S) = (hi) satisfyhi = dimk(Ai) there is some j ≥ p so that

h(S) = (h0, h1, . . . , hp−1, hp, . . . , hj , hp−1, . . . , h1, h0),

thus

hj+1 = hp−1 =

(p+ g − 2

g − 1

)and hj = hp (1)

The idea of the argument is to use the Macaulay bound

hj+1 ≤ h〈j〉j , (2)

see Theorem 5.4.2. From the inequalities:

hj+1(1)= hp−1 =

(p+ g − 2

p− 1

)=

(p+ g − 3

p− 2

)〈p−2〉

(2)

≤ h〈j〉j

(1)= h〈j〉p =

[(p+ g − 1

p

)− μ(Ip)

]〈j〉use g(x)

≤[(p+ g − 1

p

)− μ(Ip)

]〈p−2〉

we conclude (p+ g − 3

p− 2

)〈p−2〉≤[(p+ g − 1

p

)− μ(Ip)

]〈p−2〉.

Using that f(x) is non decreasing yields the required inequality. �

196 Chapter 5

Exercises

5.4.9 Let I be a graded Gorenstein ideal of height 4 and initial degree p,prove μ(Ip) ≤ (p+ 1)2.

5.4.10 Let I be a graded Gorenstein ideal of height 4 generated by formsof degree p and let R/J be an extremal algebra of the same codimensionand initial degree. Prove that their Betti numbers satisfy

bi(R/I) ≤ bi(R/J) for all i.

5.5 Koszul homology and Hilbert functions

The first Koszul homology module of a graded Cohen–Macaulay ideal isstudied here using the Cohen–Macaulay criterion of Herzog introduced inSection 4.4. Our approach makes use of Hilbert functions.

Lemma 5.5.1 Let p and g be positive integers and let

ψ(p) =

(p+ g − 1

g − 1

)2

− 2

(2p+ g

g

).

If p ≥ 2 and g ≥ 6 then ψ(p) > 0.

Proof. Notice the equality 6ψ(2) = (g + 1)(g3 − 3g2 − 13g − 12), whichis certainly positive for g ≥ 6. We proceed by induction on p. Assumeψ(p) > 0. It is easy to check that ψ(p+ 1) is greater than

2

(2p+ g

g

)[(p+ 1)(4(g − 2)p2 + (3g2 − 3g − 8)p+ g2 − 3g − 2)

(p+ 1)2(2p+ 1)(2p+ 2)

].

Since the right-hand side of this inequality is positive for p ≥ 2 and g ≥ 3,the induction step is complete. �

Theorem 5.5.2 [350] Let R = ⊕∞i=0Ri be a polynomial ring with its usual

graduation over an infinite field k and let I be a Cohen–Macaulay gradedideal of R of height g with a p-linear resolution

0→ Rbg(−(p+g−1))→ · · · → Rbk(−(p+k−1))→ · · · → Rb1(−p)→ I → 0.

If I is generically a complete intersection with p ≥ 2 and g ≥ 3, then H1(I),the first Koszul homology module of I, is not Cohen–Macaulay.

Proof. The module H1 = H1(I) has a well-defined rank equal to b1 − g.Thus, by Theorem 4.4.13, it suffices to prove that for some homogeneoussystem of parameters y of S = R/I, one has

�(H1(I)/yH1(I)) > rank(H1(I)) · �(S/yS).

Hilbert Series 197

We start by making a specialization to the case of a polynomial ring ofdimension g. Since k is infinite according to Theorem 3.1.24, there exists asystem of parameters y = {y1, . . . , yd} for S with each yi a form of degreeone of R. We make two observations: (i) Because Tori(S,R/(y)) = 0 fori ≥ 1, it is clear that the minimal resolution of S = S/(yS) as an R(= R/(y))-module has the same twists and Betti numbers as the minimalresolution of S over R. (ii) With H1 Cohen–Macaulay it is easy to see thatthe first Koszul homology module of I ⊗ R over R is precisely H1(I) ⊗ R.We may then in the sequel assume that S is zero dimensional.

We will compare the integer r=rank(original H1(I))·�(S), with partialHilbert sums contributing to �(H1). First notice (see Exercise 5.3.17) thatthe length of S and its Betti numbers can be calculated using:

bk =

(p+ k − 2

k − 1

)·(p+ g − 1

g − k

)and �(S) =

(p+ g − 1

g

). (5.12)

From the Koszul complex we obtain the exact sequences

0→ B1 → Z1 → H1 → 0,

0→ Z2 → R(b12 )(−2p)→ B1 → 0,

where Zi and Bi are the modules of cycles and boundaries defining Hi(I).To simplify notation we set �(M)i = H(M, i), the dimension of the ithcomponent of the graded module M . One may write

�(H1)i = �(Z1)i − �(B1)i = �(Z1)i −(b12

)�(R(−2p))i + �(Z2)i,

and therefore

�(H1) ≥2p∑i=0

�(H1)i ≥2p∑i=0

�(Z1)i −(b12

) 2p∑i=0

�(R(−2p))i. (5.13)

From the minimal resolution of S we obtain

�(S)i = �(R)i − b1�(R(−p))i + �(Z1)i ≥ 0.

Hence �(Z1)i ≥ b1�(R(−p))i − �(R)i and �(Z1)i = 0 for 0 ≤ i ≤ p, whichleads to

2p∑i=0

�(Z1)i ≥ b12p∑

i=p+1

�(R(−p))i −2p∑

i=p+1

�(R)i. (5.14)

Finally by (5.13) and (5.14) we have

�(H1) ≥ b12p∑

i=p+1

�(R(−p))i −2p∑

i=p+1

�(R)i −(b12

). (5.15)

198 Chapter 5

The proof now reduces to showing that the right-hand side of (5.15) isgreater than rank(H1(I)) · �(S) = (b1− g) · �(S); that is, we must show thatthe following inequality holds for p ≥ 2 and g ≥ 3

f(p, g) = b1

2p∑i=p+1

�(R(−p))i −2p∑

i=p+1

�(R)i −(b12

)− (b1 − g) · �(S) > 0.

It is not hard to see that f(p, g) simplifies to

f(p, g) =

(b1 + 1

2

)+ (1 + g)

(p+ g − 1

p− 1

)−(2p+ g

g

).

Observe that from this equality we obtain the inequality:

2f(p, g) >

(p+ g − 1

p

)2

− 2

(2p+ g

g

).

It is easy to check that f(p, g) > 0 for g ∈ {3, 4, 5} and p ≥ 2. The requiredinequality is now a direct consequence of Lemma 5.5.1. �

The result above complements one of Ulrich [404] in which the twistedconormal module ωS ⊗R I = ωS ⊗R I/I2 was shown not to be Cohen–Macaulay; either result implies that I is not in the linkage class of a completeintersection.

Corollary 5.5.3 Let I be a graded Cohen–Macaulay ideal of height g ≥ 3in a polynomial ring R. If I is generically a complete intersection and hasa p-linear resolution with p ≥ 2, then I is not in the linkage class of acomplete intersection.

Proof. If I is in the linkage class of a complete intersection, then I isa strongly Cohen–Macaulay ideal, a contradiction because H1(I) is notCohen–Macaulay. See [253] and [412, Corollary 4.2.4]. �

Exercises

5.5.4 Let I be a graded ideal of a polynomial ring R over a field K with apure resolution:

0→ R3(−15)→ R27(−10)→ R25(−9)→ I → 0.

Prove that H1(I) is not Cohen–Macaulay.

Hilbert Series 199

5.6 Hilbert functions of some graded ideals

This section is about the behavior of the Hilbert function in some specialcases of interest in algebraic coding theory [349, 348] (see Section 8.6).

Throughout this section R = K[x1, . . . , xn] = ⊕∞i=0Ri is a polynomial

ring over a fieldK with the standard grading induced by setting deg(xi) = 1for i = 1, . . . , n and m = (x1, . . . , xn) is the irrelevant maximal ideal of R.

Proposition 5.6.1 Let I be a graded ideal of R. If K is infinite and m isnot in Ass(R/I), then there is h1 ∈ R1 such that h1 /∈ Z(R/I).

Proof. Let p1, . . . , pm be the associated primes of R/I. As R/I is graded,p1, . . . , pm are graded ideals by Lemma 2.2.6. We proceed by contradiction.Assume that R1, the degree 1 part of R, is contained in Z(R/I). Thanksto Lemma 2.1.19, one has that Z(R/I) = ∪mi=1pi. Hence

R1 ⊂ (p1)1 ∪ (p2)1 ∪ · · · ∪ (pm)1 ⊂ R1,

where (pi)1 is the homogeneous part of degree 1 of the graded ideal pi.Since K is infinite, from Lemma 3.1.22, we get R1 = (pi)1 for some i.Hence, pi = m, a contradiction. �

The hypothesis that m /∈ Ass (R/I) is equivalent to require that R/Ihas positive depth. This follows from Lemma 2.1.19 (see Exercise 2.2.11).

Theorem 5.6.2 Let I be a graded ideal of R. If depth(R/I) > 0, and HI

is the Hilbert function of R/I, then HI(i) ≤ HI(i+ 1) for i ≥ 0.

Proof. Case (I): If K is infinite, by Proposition 5.6.1, there exists h ∈ R1

a non-zero divisor of R/I. The homomorphism of K-vector spaces

(R/I)i −→ (R/I)i+1, z �→ hz

is injective, therefore HI(i) = dimK(R/I)i ≤ dimK(R/I)i+1 = HI(i+ 1).

Case (II): If K is finite, consider the algebraic closure K of K. We set

R = R⊗K K, I = IR.

Hence, from [392, Lemma 1.1], one has that HI(i) = HI(i). This meansthat the Hilbert function does not change when the base field is extendedfrom K to K. Applying Case (I) to HI(i) we obtain the result. �

Lemma 5.6.3 Let I be a graded ideal of R. The following hold.

(a) If Ri = Ii for some i, then R� = I� for all � ≥ i.(b) If dim R/I ≥ 2, then dimK(R/I)i > 0 for i ≥ 0.

200 Chapter 5

Proof. (a): It suffices to prove the case � = i + 1. As Ii+1 ⊂ Ri+1, weneed only show Ri+1 ⊂ Ii+1. Take a non-zero monomial xa in Ri+1. Then,xa = xa11 · · ·xann with aj > 0 for some j. Thus, xa ∈ R1Ri. As R1Ii ⊂ Ii+1,we get xa ∈ Ii+1.

(b): If dimK(R/I)i = 0 for some i, then Ri = Ii. Thus, by (a), HI(j)vanishes for j ≥ i, a contradiction because the Hilbert polynomial of R/Ihas degree dim(R/I)− 1 ≥ 1; see Theorem 2.2.4. �

Theorem 5.6.4 [178] Let I be a graded ideal with depth(R/I) > 0.

(i) If dim(R/I) ≥ 2, then HI(i) < HI(i+ 1) for i ≥ 0.

(ii) If dim(R/I) = 1, then there is an integer r and a constant c such that

1 = HI(0) < HI(1) < · · · < HI(r − 1) < HI(i) = c for i ≥ r.

Proof. Consider the algebraic closure K of K. Notice that |K| = ∞. Asin the proof of Theorem 5.6.2, we make a change of coefficients using thefunctor (·)⊗K K. Hence we may assume that K is infinite. By Proposition5.6.1, there is h ∈ R1 a non-zero divisor of R/I. From the exact sequence

0 −→ (R/I)[−1] h−→ R/I −→ R/(h, I) −→ 0,

we get HI(i+ 1)−HI(i) = HS(i + 1), where S = R/(h, I).(i): If HS(i+ 1) = 0, then, by Lemma 5.6.3, dimK(S) <∞. Hence S is

Artinian (see Exercise 3.1.44). Thus dim(S) = 0, a contradiction.(ii): Let r ≥ 0 be the first integer such that HI(r) = HI(r + 1), thus

Sr+1 = (0) and Rr+1 = (h, I)r+1. Then, by Lemma 5.6.3, Sk = (0) fork ≥ r + 1. Hence, the Hilbert function of R/I is constant for k ≥ r andstrictly increasing on [0, r − 1]. �

Exercises

5.6.5 Let Pn−1 be a projective space over a field K and let [P1], . . . , [Pm]be a set of distinct points in Pn−1, then there exists F ∈ K[x1, . . . , xn] aform of degree m− 1 such that F (P1) �= 0 and F (Pi) = 0 for 2 ≤ i ≤ m.

5.6.6 Let X be a finite set in a projective space Pn−1 over a field K. IfI = I(X) ⊂ R is the vanishing ideal of X and HI is the Hilbert function ofR/I(X), then HI(i) = |X | for all i ≥ |X | − 1.

Chapter 6

Stanley–Reisner Ringsand Edge Ideals ofClutters

This chapter is intended as an introduction to the study of combinatorialcommutative algebra, some of the topics were chosen in order to motivate thesubject. Part of our exposition was inspired by [41, 393]. An understatedgoal here is to highlight some of the works of T. Hibi, J. Herzog, M. Hochster,G. Reisner, and R. Stanley.

This chapter deals with monomial ideals, Stanley–Reisner rings andedge ideals of clutters. We study algebraic and combinatorial properties ofedge ideals of clutters and simplicial complexes. In particular we examineshellable, unmixed, and sequentially Cohen–Macaulay simplicial complexesand their corresponding edge ideals and invariants (projective dimension,regularity, depth, Hilbert series). As applications, the proofs of the upperbound conjectures for convex polytopes and simplicial spheres are discussedin order to give a flavor of some of the methods and ideas of the area.

6.1 Primary decomposition

General monomial ideals will be examined first, subsequently we specializeto the case of square-free monomial ideals and edge ideals of clutters, andpresent some of their relevant properties.

Let R = K[x] = K[x1, . . . , xn] be a polynomial ring over a field K. Tomake notation simpler, we will use the following multi-index notation:

xa := xa11 · · ·xann for a = (a1, . . . , an) ∈ Nn.

202 Chapter 6

Definition 6.1.1 An ideal I of R is called a monomial ideal if there isA ⊂ Nn+ such that I is generated by {xa|a ∈ A}. If I is a monomial idealthe quotient ring R/I is called a monomial ring.

Note that, by Dickson’s lemma (see Lemma 3.3.3), a monomial ideal Iis always minimally generated by a unique finite set of monomials. Thisunique set of generators of I is denoted by G(I).

Definition 6.1.2 A face ideal is an ideal p of R generated by a subset ofthe set of variables, that is, p = (xi1 , . . . , xik ) for some variables xij .

Definition 6.1.3 A monomial f in R is called square-free if f = xi1 . . . xirfor some 1 ≤ i1 < · · · < ir ≤ n. A square-free monomial ideal is an idealgenerated by square-free monomials.

Any square-free monomial ideal is a finite intersection of face ideals:

Theorem 6.1.4 Let I ⊂ R be a monomial ideal. The following hold.

(i) Every associated prime of I is a face ideal.

(ii) If I is square-free, then I = ∩si=1pi, where p1, . . . , ps are the associatedprimes of I. In particular pi is a minimal prime of I for all i.

Proof. (i): By induction on the number of variables that occur in G(I).Set m = (x1, . . . , xn). Let p be an associated prime of I. If rad(I) = m,then p = m. Hence we may assume rad(I) �= m. Pick a variable x1 not inrad(I) and consider the ascending chain of ideals

I0 = I and Ii+1 = (Ii : x1) (i ≥ 0).

Since R is Noetherian, one has Ik = (Ik : x1) for some k. There are twocases to consider. If p is an associated prime of (Ii, x1) for some i, then byinduction p is a face ideal because one can write (Ii, x1) = (I ′i , x1), where I

′i

is an ideal minimally generated by a finite set of monomials in the variablesx2, . . . , xn (cf. Exercise 6.1.26). Assume we are in the opposite case. ByLemma 5.1.3 for each i there is an exact sequence

0 −→ R/(Ii : x1)x1−→ R/Ii −→ R/(Ii, x1) −→ 0,

hence making a recursive application of Lemma 2.1.17 one obtains thatp is an associated prime of Ii for all i. Since x1 is regular on R/Ik oneconcludes that Ik is an ideal minimally generated by monomials in thevariables x2, . . . , xn, thus by induction p is a face ideal.

(ii): We only have to check ∩si=1pi ⊂ I because I is contained in any of itsassociated primes. Take a monomial f in ∩si=1pi and write f = xa1i1 · · ·x

arir,

where i1 < · · · < ir and ai > 0 for all i. By Corollary 2.1.29, fk ∈ I for

Stanley–Reisner Rings and Edge Ideals of Clutters 203

some k ≥ 1. Then, using that I is generated by square-free monomials, weobtain xi1 · · ·xir ∈ I. Hence f ∈ I. To finish the proof observe that, bypart (i), ∩si=1pi is a monomial ideal because the intersection of monomialideals is again a monomial ideal (see Exercise 6.1.22). �

Definition 6.1.5 Let xa = xa11 · · ·xann be a monomial in R. The supportof xa is given by supp(xa) := {xi | ai > 0}.

Proposition 6.1.6 [169] Let I ⊂ R be a monomial ideal. If S = R/I,then there is a polynomial ring R′ and a square-free monomial ideal I ′ ofR′ such that S = S′/(h), where S′ = R′/I ′ and h is a regular sequence onS′ of forms of degree one.

Proof. Let G(I) = {f1, . . . , fr} be the set of monomials that minimallygenerate I. Assume that one of the variables, say x1, occurs in at least oneof the monomials in G(I) with multiplicity greater than 1. We may assumethat the elements of G(I) can be written as

f1 = xa11 g1, . . . , fs = xas1 gs,

where ai ≥ 1 for all i, a1 ≥ 2, x1 /∈ supp(gi) for all i ≤ s and x1 /∈ supp(fi)for i > s. We set

I ′ = (x0xa1−11 g1, . . . , x0x

as−11 gs, fs+1, . . . , fr) ⊂ R′ = R[x0],

where x0 is a new variable. We claim that x0 − x1 is a non-zero divisor ofS′ = R′/I ′. On the contrary assume that x0 − x1 belongs to an associatedprime p of I ′ and write p = (I ′ : h), for some monomial h. Since p isa face ideal, one has xih ∈ I ′ for i = 0, 1. Hence using h /∈ I ′ givesx1h = x0x

ai−11 gih1 for some i and consequently h = x0x

ai−21 gih1. Hence

x0h = x20xai−21 gih1 =

{x0x

aj−11 gjM or,

fjM for some j > s.

In both cases one obtains x0 ∈ supp(M) and h ∈ I ′, a contradiction. Since

S′/(x0 − x1) = S

one can repeat the construction to obtain the asserted monomial ideal I ′.This proof is due to R. Froberg. �

The ideal I ′ constructed above is called the polarization of I. Thus anymonomial ring is a deformation by linear forms of a monomial ring withsquare-free relations. Note that I is Cohen–Macaulay (resp. Gorenstein) iffI ′ is Cohen–Macaulay (resp. Gorenstein).

204 Chapter 6

Proposition 6.1.7 A monomial ideal q ⊂ R is primary if and only if, afterpermutation of the variables, q has the form:

q = (xa11 , . . . , xarr , x

b1 , . . . , xbs),

where ai ≥ 1 and ∪si=1supp(xbi ) ⊂ {x1, . . . , xr}.

Proof. If Ass(R/q) = {p}, then by permuting the variables x1, . . . , xn andusing Theorem 6.1.4 one may assume that p is equal to (x1, . . . , xr). Sincerad (q) = p, the ideal q is minimally generated by a set of the form:

{xa11 , . . . , xarr , xb1 , . . . , xbs}.

Let xj ∈ supp(xbi), then xbi = xjxc, where xc is a monomial not in q. Since

q is primary, a power of xj is in q. Thus xj ∈ (x1, . . . , xr) and consequently1 ≤ j ≤ r, as required.

For the converse note that any associated prime p of R/q can be writtenas p = (q : f), for some monomial f . It follows readily that (x1, . . . , xr) isthe only associated prime of q. �

Corollary 6.1.8 If p is a face ideal, then pn is a primary ideal for all n.

Proposition 6.1.9 If I ⊂ R is a monomial ideal, then I has an irredundantprimary decomposition I = q1 ∩ · · · ∩ qr, where qi is a primary monomialideal for all i and rad (qi) �= rad (qj) if i �= j.

Proof. Let G(I) = {f1, . . . , fq} be the set of monomials that minimallygenerate I. We proceed by induction on the number of variables that occurin the union of the supports of f1, . . . , fq.

One may assume that one of the variables in ∪qi=1supp(fi), say xn, satisfyxin /∈ I for all i, otherwise I is a primary ideal and there is nothing to prove.Next we permute the fi in order to find integers 0 ≤ a1 ≤ · · · ≤ aq, withaq ≥ 1, such that fi is divisible by x

ain but by not higher power of xn. If we

apply this procedure to (I, xaqn ), instead of I, note that one must choose a

variable different from xn.Since (I : x

aqn ) is generated by monomials in less than n variables and

because of the equality

I = (I, xaqn ) ∩ (I : xaqn )

one may apply the argument above recursively to the two monomial idealsoccurring in the intersection— and use induction—to obtain a decomposi-tion of I into primary monomial ideals q′1, . . . , q

′s of R. Finally we remove

redundant primary ideals from q′1, . . . , q′s and group those primary ideals

with the same radical. �

Even for monomial ideals a minimal irredundant primary decompositionis not unique. What is unique is the number of terms in such a decomposi-tion and the primary components that correspond to minimal primes [9].


Example 6.1.10 If I = (x2, xy) ⊂ K[x, y], then

I = (x) ∩ (x2, xy, y2) = (x) ∩ (x2, y),

are two minimal irredundant primary decompositions of I.

The computation of a primary decomposition of a monomial ideal canbe carried out by successive elimination of powers of variables, as describedin the proof of Proposition 6.1.9. Now we illustrate this procedure with aspecific ideal.

Example 6.1.11 If R = K[x, y, z] and I = (yz2, x2z, x3y2).

I

(I : x3) = (z, y2)

J = (I, x3) = (x3, zx2, z2y)

(J : z2) = (x2, y)

(J, z2) = (z2, x3, x2z).

��

��

��

Thus I = (z, y2) ∩ (x2, y) ∩ (z2, x3, x2z).

Corollary 6.1.12 If I ⊂ R is a monomial ideal, then there is a primarydecomposition I = q1 ∩ · · · ∩ qm, such that qi is generated by powers ofvariables for all i.

Proof. Let q be a primary ideal. Then, by Proposition 6.1.7, q is minimallygenerated by a set of monomials xa11 , . . . , x

arr , f1, . . . , fs such that

s⋃i=1

supp(fi) ⊂ {x1, . . . , xr},

where ai > 0 for all i. Note that if f1 = xb11 · · ·xbrr and b1 > 0, then a1 > b1and one has a decomposition:

q = (xb11 , xa22 , . . . , x

arr , f2, . . . , fs) ∩ (xa11 , x

a22 , . . . , x

arr , x

b22 · · ·xbrr , f2, . . . , fs),

where in the first ideal of the intersection we have lower the degree of xa11 andhave eliminated f1, while in the second ideal we have eliminated the variablex1 from f1. Applying the same argument repeatedly it follows that one canwrite q as an intersection of primary monomial ideals such that for each ofthose ideals the only minimal generators that contain x1 are pure powersof x1. Therefore, by induction, q is the intersection of ideals generated bypowers of variables. Hence the result follows from Proposition 6.1.9. �

Proposition 6.1.13 Let I be an ideal of R = K[x1, . . . , xn] generated bymonomials in the variables x1, . . . , xr with r < n. If

I = I1 ∩ · · · ∩ Isis an irredundant decomposition of I into monomial ideals, then none of theideals Ii can contain a monomial in K[xr+1, . . . , xn].

206 Chapter 6

Proof. Set X = {x1, . . . , xn} and X ′ = X \ {x1, . . . , xr}. Assume someof the Ii’s contain monomials in K[X ′]. One may split the Ii’s into twosets so that I1, . . . , Im do not contain monomials in K[X ′] (note m couldbe zero), while Im+1, . . . , Is contain monomials in the set of variables X ′.For i ≥ m + 1 pick a monomial gi in Ii whose support is contained in X ′.Since the decomposition of I is irredundant there is a monomial f ∈ ∩mi=1Iiand f �∈ ∩si=m+1Ii, where we set f = 1 if m = 0. To derive a contradictionconsider f1 = fgm+1 · · · gs. As f1 ∈ I, we get f ∈ I which is absurd. �

Corollary 6.1.14 Let I ⊂ R = K[x1, . . . , xn] be an ideal generated bymonomials in the variables x1, . . . , xr with r < n. If

I = q1 ∩ · · · ∩ qs

is an irredundant primary decomposition into monomial ideals, then qi isgenerated by monomials in K[x1, . . . , xr].

Proof. It follows from Propositions 6.1.7 and 6.1.13. �

Definition 6.1.15 An ideal I of a ring R is called irreducible if I cannotbe written as an intersection of two ideals of R that properly contain I.

Proposition 6.1.16 If I ⊂ R = K[x1, . . . , xn] is a monomial ideal, then Iis irreducible if and only if up to permutation of the variables

I = (xa11 , . . . , xarr ),

where ai > 0 for all i.

Proof. ⇒) Since I must be primary (see Proposition 2.1.24) from the proofof Corollary 6.1.12 one derives that I is generated by powers of variables.⇐) If I is reducible, then I = I1 ∩ I2 for some ideals I1 and I2 that

properly contain I. Pick f ∈ I1 \ I (resp. g ∈ I2 \ I) with the smallestpossible number of terms. We can write

f = λ1xγ1 + · · ·+ λsx

γs ; 0 �= λi ∈ K for all i.

Let 1 ≤ k ≤ r be an integer and let xbik be the maximum power of xk that

divides xγi , i.e., we can write xγi = xbik xδi , where xk does not divide xδi .

After permuting terms we may assume that b1 ≥ · · · ≥ bs. Note that xakkdoes not divide xγi for all i, k, otherwise we can find a polynomial in I1 \ I,namely f − λixγi , with less than s terms. Thus bi < ak for all i. We claimthat b1 = · · · = bs. We proceed by contradiction assuming that bp > bp+1

for some p. From the equality

xak−bpk f = x

ak−bpk (λ1x

γ1 + · · ·+ λpxγp) + x

ak−bpk (λp+1x

γp+1 + · · ·+ λsxγs)


we obtain that the polynomial xak−bpk (λp+1x

γp+1 + · · ·+ λsxγs) is in I1 \ I

and has fewer terms than f , a contradiction. This completes the proof ofthe claim. Therefore we can write f = xc11 · · ·xcrr f1 and g = xd11 · · ·xdrr g1,where f1 and f2 are in R′ = K[xr+1, . . . , xn]. Setting ei = max{ci, di} weget that h = xe11 · · ·xerr f1g1 is in I1 ∩ I2, i.e., h ∈ I, a contradiction becauseei < ai for all i and f1g1 ∈ R′. Thus I is irreducible. �

Theorem 6.1.17 If I ⊂ R is a monomial ideal, then there is a uniqueirredundant decomposition I = q1 ∩ · · · ∩ qr such that qi is an irreduciblemonomial ideal.

Proof. The existence follows from Corollary 6.1.12. For the uniquenessassume one has two irredundant decompositions:

q1 ∩ · · · ∩ qr = q′1 ∩ · · · ∩ q′s,

where qi and q′j are irreducible for all i, j. Using the arguments given inthe proof of Proposition 6.1.16 one concludes that for each i, there is σisuch that qσi ⊂ q′i and vice versa for each j there is πj such that q′πj

⊂ qj.Therefore r = s and qi = q′ρi for some permutation ρ. �

For further information on the primary decomposition of more generalmonomial ideals, e.g., monomial ideals obtained from a regular sequence orwith coefficients in a ring other than a field, consult [122, 213, 214].

Example 6.1.18 Let I = (xy2, x2y) ⊂ K[x, y] and V (I) the affine varietydefined by I. From the primary decomposition

I = (x) ∩ (y) ∩ (x2, y2)

one has V (I) = {(0, 0)}∪ V (x)∪ V (y) = V (x) ∪ V (y) ⊂ A2K , where (x2, y2)

corresponds to {(0, 0)} which is embedded in V (x) ∪ V (y). For this reason(x2, y2) is said to be an embedded primary component. If K is infinite, thenthe coordinate axes V (x) and V (y) are the irreducible components of V (I).

Monomial ideals form a lattice Let L be a lattice, that is, L is a posetin which any two elements x, y have a greatest lower bound or a meet x∧ yand a lowest upper bound or a join x ∨ y. The lattice L is distributive if

(a) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), and(b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z), ∀x, y, z in L.

Proposition 6.1.19 If L is the family of monomial ideals of R, order byinclusion, then L is a distributive lattice under the operations I ∧ J = I ∩Jand I ∨ J = I + J .

Proof. It follows from Exercises 6.1.22 and 6.1.23. �

208 Chapter 6

Exercises

6.1.20 If I is a monomial ideal of R, then any associated prime p of R/Ican be written as p = (I : f), for some monomial f .

6.1.21 If L is a lattice, then the following conditions are equivalent:

(a) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z), ∀x, y, z in L.

(b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z), ∀x, y, z in L.

6.1.22 Let I and J be two ideals generated by finite sets of monomials Fand G, respectively, prove that the intersection I∩J is generated by the set

{lcm(f, g)| f ∈ F and g ∈ G}.

6.1.23 If I, J , L are monomial ideals, prove the equalities

I ∩ (J + L) = (I ∩ J) + (I ∩ L),I + (J ∩ L) = (I + J) ∩ (I + L).

6.1.24 If I and J are two monomial ideals, then (I : J) is a monomial ideal.

6.1.25 If q1, . . . , qr are primary monomial ideals of R with non-comparableradicals and I is an ideal such that I = q1 ∩ · · · ∩ qr, then

I(n) = qn1 ∩ · · · ∩ qnr .

Hint Proceed as in the proof of Proposition 4.3.24 (cf. [92, Theorem 3.7]).

6.1.26 Let R′ = k[x2, . . . , xn] and R = R′[x1] be polynomial rings over afield k. If I ′ is an ideal of R′ and p ∈ AssR(R/(I

′, x1)), then

(a) p = x1R+ p′R, where p′ is a prime ideal of R′, and

(b) p′ is an associated prime of R′/I ′.

6.1.27 If I = (x31, x22x

23, x1x

22x3) ⊂ K[x1, x2, x3], show that I = q1 ∩ q2 is

a primary decomposition of I, where q1 = (x31, x22) and q2 = (x31, x1x3, x

23).

Prove that I2 = I(2) and I3 �= I(3).

Hint If z = x31x42x

33 and a = x1 + x22 + x3, then az ∈ I3 and z �∈ I3.

6.1.28 If I = (x1x22, x

21x2, x1x

23, x

21x3, x2x

23, x

22x3, x1x2x3), find a primary

decomposition of I. Notice that (x1, x2, x3) is an associated prime of I.

6.1.29 Show that I = (x21, x1x2) is a non-primary ideal with a prime radical.

6.1.30 If n ≥ 5 prove that the number of monomials of degree n in nvariables whose support is at least three is given by

μ(n) =

(2n− 1

n− 1

)− n− (n− 1)

(n

2

).


6.2 Simplicial complexes and homology

A finite simplicial complex consists of a finite set V of vertices and a collec-tion Δ of subsets of V called faces or simplices such that

(i) If v ∈ V , then {v} ∈ Δ.

(ii) If F ∈ Δ and G ⊂ F , then G ∈ Δ.

Let Δ be a simplicial complex and F a face of Δ. Define the dimensionsof F and Δ by

dim(F ) = |F | − 1 and dim(Δ) = sup{dim(F )|F ∈ Δ}

respectively. A face of dimension q is sometimes referred to as a q-face oras a q-simplex.

Example 6.2.1 A triangulation of a disk with a whisker attached.

�x2��x3

�x4��

��x1

0-simplices: x1, x2, x3, x41-simplices: x1x2, x2x3, x3x4, x2x42-simplices: x2x3x4

Let F be a q-simplex in Δ with vertices v0, v1, . . . , vq. We say that twototal orderings of the vertices

vi0 < · · · < viq and vj0 < · · · < vjq

are equivalent if (i0, . . . , iq) is an even permutation of (j0, . . . , jq). Thisis an equivalence relation because the set of even permutations forms agroup, and for q > 1, it partitions the total orderings of v0, . . . , vq into twoequivalence classes. An oriented q-simplex of Δ is a q-simplex F with achoice of one of these equivalence classes. If v0, . . . , vq are the vertices ofF , the oriented simplex determined by the ordering v0 < · · · < vq will bedenoted by [v0, . . . , vq].

Suppose A is a commutative ring with unit. Let Cq(Δ) be the free A-module with basis consisting of the oriented q-simplices in Δ, modulo therelations

[v0, v1, v2, . . . , vq] + [v1, v0, v2, . . . , vq].

In particular Cq(Δ) is defined for any field K and dimK Cq(Δ) equals thenumber of q-simplices of Δ. For q ≥ 1 we define the homomorphism

∂q : Cq(Δ) −→ Cq−1(Δ)

210 Chapter 6

induced by

∂q([v0, v1, . . . , vq]) =

q∑i=0

(−1)i[v0, v1, . . . , vi−1, vi, vi+1, . . . , vq],

where vi means that the symbol vi is to be deleted. Since ∂q∂q+1 = 0 weobtain the chain complex C(Δ) = {Cq(Δ), ∂q}, which is called the orientedchain complex of Δ. The augmented oriented chain complex of Δ is thecomplex

0 −→ Cd(Δ)∂d−→ Cd−1(Δ) −→ · · · −→ C0(Δ)

ε−→ A −→ 0,

where d = dim(Δ) and ε(v) = 1 for every vertex v of Δ. This chain complexis denoted by C (Δ) and we set ∂0 = ε and C−1(Δ) = A.

Setting Zq(Δ, A) = ker(∂q), Bq = im(∂q+1) and

Hq(Δ;A) = Zq(Δ, A)/Bq(Δ, A), for q ≥ 0,

the elements of Zq(Δ, A) and Bq(Δ, A) are called cycles and boundaries,

respectively, and Hq(Δ;A) is the qth reduced simplicial homology group ofΔ with coefficients in A.

Remark 6.2.2 Note that if Δ �= ∅, then Hi(Δ;A) = 0 for i < 0.

Let C (Δ) be the augmented chain complex of Δ over the ring A. Theq-reduced singular cohomology group with coefficients in A is defined as

Hq(Δ;A) = Hq(HomA(C (Δ), A)).

If K is a field, there are canonical isomorphisms

Hq(Δ;K) ∼= HomK(Hq(Δ,K),K) and

Hq(Δ;K) ∼= HomK(Hq(Δ,K),K).

Thus, in particular, we have Hq(Δ;K) ∼= Hq(Δ;K).

Proposition 6.2.3 If Δ is a non-empty simplicial complex, then H0(Δ;A)is a free A-module of rank r − 1, where r is the number of connected com-ponents of Δ.

Proof. Let V = {x1, . . . , xn} be the vertex set of Δ and let Δ1, . . . ,Δr

be its connected components. Denote the vertex set of Δi by Vi and pick avertex in each one of the Vi; one may assume xi ∈ Vi for i = 1, . . . , r.

Set βi = xi−xr and βi = βi+im(∂1) for i = 1, . . . , r−1. We claim that

the set B = {β1, . . . , βr−1} is an A-basis for H0(Δ;A) = ker(∂0)/im(∂1).


First we prove that B is linearly independent. Assume∑r−1

i=1 aiβi is inim(∂1) for some ai in A, by making an appropriate grouping of terms onecan write

r−1∑i=1

aiβi =∑

xi,xj∈V1

bij(xi − xj) + · · ·+∑

xi,xj∈Vr

bij(xi − xj),

for some bij in A. Since C0(Δ) is a free A-module with basis V and the Viare mutually disjoint one has

akxk =∑

xi,xj∈Vk

bij(xi − xj) for 1 ≤ k ≤ r − 1,

applying ∂0 to both sides of this equality we obtain ak = 0 for all k.Next we show that B is a set of generators. Let xi1 and xj1 be two

vertices of Δ, it suffices to prove that xi1 −xj1 +im(∂1) is in the submodulegenerated by B because ker(∂0) is generated by the elements of the formxi − xj . Assume xi1 ∈ Vm and xj1 ∈ Vs. There is a path xi1 , . . . , xit , xmin Δm and a path xj1 , . . . , xjp , xs in Δs such that every two consecutivevertices of those paths form a 1-face of Δ. From the equalities

(xi1 − xi2 ) + (xi2 − xi3) + · · ·+ (xit − xm) = xi1 − xm(xj1 − xj2 ) + (xj2 − xj3) + · · ·+ (xjp − xs) = xj1 − xs,

one concludes xi1 − xj1 + im(∂1) = (xm − xr)− (xs − xr) + im(∂1). �

Definition 6.2.4 The reduced Euler characteristic χ(Δ) of a simplicialcomplex Δ of dimension d is given by

χ(Δ) = −1 +d∑i=0

(−1)ifi = −1 + χ(Δ),

where fi is the number of i-faces in Δ and χ(Δ) is its Euler characteristic.

Exercises

6.2.5 Let Δ be a simplicial complex of dimension d and fi the number ofi-faces in Δ. If K is a field, then

d∑i=−1

(−1)irank Hi(Δ;K) = −1 +d∑i=0

(−1)ifi.

6.2.6 Let Δ be a simplicial complex, prove that there is an exact sequence

0 −→ H0(Δ;A) −→ H0(Δ;A) −→ A −→ 0.

Then, show H0(Δ;A) � H0(Δ, A)⊕A.

212 Chapter 6

6.2.7 Let Δ be a simplicial complex with vertex set V = {x1, . . . , xn} and

· · · ∂1−→ C0(Δ)∂0−→ C−1(Δ) = A −→ 0,

the right end of the augmented chain complex over a ring A. Prove thatker(∂0) is generated by the cycles of the form xi − xj .

6.2.8 Let Δ be a simplicial complex and fi the number of i-faces of Δ. If

· · · −→ C1(Δ)∂1−→ C0(Δ)

∂0−→ C−1(Δ) = K −→ 0,

is the augmented chain complex of Δ over a field K, then dimK(Z1) is equalto f1 − f0 + r; where r is the number of connected components of Δ andZ1 = ker(∂1).

6.3 Stanley–Reisner rings

In this section we study algebraic and combinatorial properties of edge idealsof clutters and simplicial complexes. In particular we examine shellable,unmixed, and sequentially Cohen–Macaulay simplicial complexes and theircorresponding edge ideals.

Let R = K[x1, . . . , xn] be a polynomial ring over a fieldK. If I is an idealof R generated by square-free monomials, the Stanley–Reisner simplicialcomplex ΔI associated to I has vertex set V = {xi|xi /∈ I} and its facesare defined by

ΔI = {{xi1 , . . . , xik}| i1 < · · · < ik , xi1 · · ·xik �∈ I}.

Conversely if Δ is a simplicial complex with vertex set V contained in{x1, . . . , xn}, the Stanley–Reisner ideal IΔ is defined as

IΔ = ({xi1 · · ·xir | i1 < · · · < ir, {xi1 , . . . , xir} /∈ Δ}) ,

and its Stanley–Reisner ring K[Δ] is defined as the quotient ring R/IΔ.

Definition 6.3.1 Let I ⊂ R be a square-free monomial ideal. The quotientring R/I is called a face ring.

Example 6.3.2 A simplicial complex Δ of dimension 1 and its Stanley–Reisner ideal: �x2

��x4

�x1

��

��

��

x3��

IΔ = (x1x3, x1x2x4, x2x3x4)


Definition 6.3.3 A face F of a simplicial complex Δ is said to be a facetif F is not properly contained in any other face of Δ.

Proposition 6.3.4 If Δ is a simplicial complex with vertices x1, . . . , xn,then the primary decomposition of the Stanley–Reisner ideal of Δ is :

IΔ =⋂F

pF ,

where the intersection is taken over all facets F of Δ, and pF denotes theface ideal generated by all xi such that xi /∈ F .

Proof. Let F be a face of Δ and pF the face ideal generated by the xi suchthat xi /∈ F . Note that pF is a minimal prime of IΔ if and only if F is afacet. Therefore the result follows from Theorem 6.1.4. �

Corollary 6.3.5 Let Δ be a simplicial complex of dimension d on the vertexset V = {x1, . . . , xn} and let K be a field. Then

dimK[Δ] = d+ 1 = max{s |xi1 · · ·xis /∈ IΔ and i1 < · · · < is}.

Proof. Let F be a facet of Δ with d+1 vertices. By Proposition 6.3.4 theface ideal pF generated by the variables not in F has height equal to theheight of IΔ. Hence ht(IΔ) = n − d − 1, and thanks to Proposition 3.1.16one has dimK[Δ] = d+ 1. �

Definition 6.3.6 Let K be a field. A simplicial complex Δ is said tobe Cohen–Macaulay over K if the Stanley–Reisner ring K[Δ] is a Cohen–Macaulay ring.

Corollary 6.3.7 A Cohen–Macaulay simplicial complex Δ is pure; that is,all its maximal faces have the same dimension.

Proof. It follows from Proposition 6.3.4 and Corollary 3.1.17. �

Definition 6.3.8 A pure d-dimensional complex Δ is called strongly con-nected or connected in codimension 1 if each pair of facets F,G can beconnected by a sequence of facets F = F0, F1, . . . , Fs = G, such thatdim(Fi−1 ∩ Fi) = d− 1 for 1 ≤ i ≤ s.

Theorem 6.3.9 [43, Proposition 11.7] Every Cohen–Macaulay complex isstrongly connected.

Definition 6.3.10 Let Δ be a simplicial complex. For F ∈ Δ, define lk(F ),the link of F , as

lk(F ) := {H ∈ Δ| H ∩ F = ∅ and H ∪ F ∈ Δ}.

214 Chapter 6

Theorem 6.3.11 (Hochster [65, Theorem 5.3.8]) Let Δ be a simplicialcomplex. If K is a field, then the Hilbert series of the local cohomologymodules of K[Δ] with respect to the fine grading is given by

F (Hpm(K[Δ]), t) =

∑F∈Δ

dimK Hp−|F |−1(lk F ;K)∏xj∈F

t−1j

1− t−1j

.

Theorem 6.3.12 (Reisner [346]) Let Δ be a simplicial complex. If K is afield, then the following conditions are equivalent:

(a) Δ is Cohen–Macaulay over K.

(b) Hi(lk F ;K) = 0 for F ∈ Δ and i < dim lk F .

Proof. Let Him(K[Δ]) be the local cohomology modules of K[Δ]. Recall

that Him(K[Δ]) ∼= Hi(C ); see Section 2.8 for the definition of C . By

Theorem 2.8.12, Δ is C–M if and only if Hi(C ) = 0 for i < d + 1, whered = dim(Δ).

⇒) If Δ is C–M, then by Hochster’s theorem Hi−|F |−1(lk(F );K) = 0for F ∈ Δ and i < d + 1. By Corollary 6.3.7 we get that Δ is pure. IfF ∈ Δ, there is a face F1 of dimension d containing F , since F1 \F ∈ lk(F )

it follows that dim lk(F ) = |F1 \ F | − 1 = d− |F |. Hence Hi(lk(F );K) = 0for all F ∈ Δ and all i < dim lk(F ).

⇐) Using dim lk(F ) ≤ d − |F |, the hypothesis Hi(lk(F );K) = 0 for

F ∈ Δ and i < dim lk(F ) implies Hi−|F |−1(lk(F );K) = 0 for F ∈ Δ andi < d+ 1. Hence by Hochster’s theorem Hi(C ) = 0 for i < d+ 1. Thus Δis Cohen–Macaulay. This proof was adapted from [65]. The original proofcan be found in [346]. �

In particular a Cohen–Macaulay complex Δ, being the link of its emptyface, must satisfy Hi(Δ;K) = 0 for i < dimΔ.

In general the Cohen–Macaulay property of a face ring may depend onthe characteristic of the base field; classical examples of this dependence aretriangulations of the projective plane [346]. See the exercises at the end ofthis section.

Example 6.3.13 If Δ is a discrete set with n vertices

�x1 �x2�xn

�x4�x6�x5

�x7 �x3

then Δ is Cohen–Macaulay and IΔ = (xixj |1 ≤ i < j ≤ n).


Corollary 6.3.14 If Δ is a 1-dimensional simplicial complex, then Δ isconnected if and only if Δ is Cohen–Macaulay.

Proof. By Reisner’s theorem Δ is Cohen–Macaulay iff H0(Δ;K) = 0,because the link of a vertex of Δ consists of a discrete set of vertices. Tocomplete the argument apply Proposition 6.2.3. �

Proposition 6.3.15 [250] If K is a field, then a simplicial complex Δ isCohen–Macaulay over K if and only if lk(F ) is Cohen–Macaulay over Kfor all F ∈ Δ.

Proof. ⇒) Let F be a fixed face in Δ and Δ′ = lk(F ). If G ∈ Δ′, usingthe equality

lkΔ′(G) = lkΔ(G ∪ F )

and Reisner’s theorem we get Hi(lkΔ′(G),K) = 0 for i < dim lkΔ′(G), thusΔ′ is Cohen–Macaulay.⇐) Since all the links are Cohen–Macaulay, it suffices to take F = ∅ and

observe Δ = lk(∅) to conclude that Δ is Cohen–Macaulay. �

Definition 6.3.16 The q-skeleton of a simplicial complex Δ is the simpli-cial complex Δq consisting of all p-simplices of Δ with p ≤ q.

Proposition 6.3.17 Let Δ be a simplicial complex of dimension d and Δq

its q-skeleton. If Δ is Cohen–Macaulay over a field K, then Δq is Cohen–Macaulay for q ≤ d.

Proof. Set Δ′ = Δq and take F ∈ Δ′ with dim(F ) < q ≤ d. Note

lkΔ′(F ) = (lkΔ(F ))q−|F |.

Since Δ is pure there is a facet F ′ of Δ of dimension d containing F . HenceF ′ \F is a face of lkΔ(F ) of dimension d−|F |. It follows that the dimensionof lkΔ′(F ) is q − |F |. Using Proposition 6.3.15 we get

Hi(lkΔ(F );K) = 0 for i < dim(lkΔ(F )) = d− |F |,

and consequently

Hi((lkΔ(F ))q−|F |;K) = Hi(lkΔ(F );K) = 0 for i < q − |F | ≤ d− |F |.

Observe that the equality between the homology modules holds because theaugmented oriented chain complexes of (lkΔ(F ))

q−|F | and lkΔ(F ) are equalup to degree q − |F | and thus their homologies have to agree up to degreeone less. Hence by Reisner’s theorem Δq is Cohen–Macaulay. �

216 Chapter 6

Example 6.3.18 [170] Let R = K[x1, . . . , xn] be a polynomial ring over afield K and Δ the (p − 2)-skeleton of a simplex of dimension n− 1, wherep ≥ 2. Then IΔ is a Cohen–Macaulay ideal and the face ring

K[Δ] = R/({xi1 · · ·xip | 1 ≤ i1 < · · · < ip ≤ n})

has a p-linear resolution by Theorem 5.3.7.

Face rings with pure and linear resolutions have been studied in [66]and [170, 172]; special results have been obtained when the correspondingStanley–Reisner ideal is generated in degree two.

Definition 6.3.19 Let Δ1, Δ2 be two simplicial complexes with vertex setsV1, V2. The join Δ1 ∗Δ2 is the complex on the vertex set V1 ∪V2 with facesF1 ∪ F2, where Fi ∈ Δi for i = 1, 2.

Proposition 6.3.20 If Δ1 and Δ2 are simplicial complexes, then their joinΔ1∗Δ2 is Cohen–Macaulay if and only if Δi is Cohen–Macaulay for i = 1, 2.

Proof. Since one has the equality IΔ1∗Δ2 = IΔ1 + IΔ2 , there is a gradedisomorphism of K-algebras

K[Δ1 ∗Δ2] � K[Δ1]⊗K K[Δ2],

see Proposition 3.1.33. To complete the proof use Corollary 3.1.35. �

Given a simplicial complex Δ and Δ1, . . . ,Δs sub complexes of Δ, wedefine their union

s⋃i=1

Δi

as the simplicial complex with vertex set ∪si=1Vi and such that F is a faceof ∪si=1Δi if and only if F is a face of Δi for some i. The intersection canbe defined similarly.

Definition 6.3.21 A simplicial complex Δ of dimension d is called pureshellable if Δ is pure and the facets (maximal faces) of Δ can be orderF1, . . . , Fs such that

F i⋂⎛⎝i−1⋃

j=1

F j

⎞⎠is pure of dimension d− 1 for all i ≥ 2. Here F i = {σ ∈ Δ|σ ⊂ Fi}. If Δ ispure shellable, F1, . . . , Fs is called a shelling.

The next definition of shellable is due to Bjorner and Wachs [44] and isusually referred to as non-pure shellable, although in this book we will dropthe adjective “non-pure.”


Definition 6.3.22 A simplicial complex Δ is shellable if the facets (max-imal faces) of Δ can be ordered F1, . . . , Fs such that for all 1 ≤ i < j ≤ s,there exists some v ∈ Fj \Fi and some � ∈ {1, . . . , j−1} with Fj \F� = {v}.We call F1, . . . , Fs a shelling of Δ. For a fixed shelling of Δ, if F, F ′ ∈ Δthen we write F < F ′ to mean that F appears before F ′ in the ordering.

If Δ is a pure simplicial complex, then Δ is shellable if and only if Δ ispure shellable (Exercise 6.3.60).

Theorem 6.3.23 Let Δ be a simplicial complex. If Δ is pure shellable,then Δ is Cohen–Macaulay over any field K.

Proof. Let V = {x1, . . . , xn} be the vertex set of Δ and let R = K[V ] bea polynomial ring over a field K. Assume that F1, . . . , Fs is a shelling of Δsuch that F i ∩ (∪i−1

j=1F j) is pure of dimension d− 1 for all i ≥ 2, where d isthe dimension of Δ. Set

σ =

⎧⎨⎩xi ∈ Fs∣∣∣∣∣∣Fs \ {xi} ∈ F s

⋂⎛⎝s−1⋃j=1

F j

⎞⎠⎫⎬⎭ ,

one may assume σ = {x1, . . . , xr}. There is a short exact sequence

0 −→ R/(I : f)f−→ R/I −→ R/(I, f) −→ 0, (∗)

where f = x1 · · ·xr and I = IΔ. First we show the equality

R/(I : f) = K[F s],

note that K[F s] is a polynomial ring in d+1 variables. By Proposition 6.3.4one can write I = ∩si=1pi, where pi is generated by V \ Fi for all i. Hence

(I : f) =

s⋂i=1

(pi : f) =⋂σ⊂Fi

(pi : f).

Observe that if σ ⊂ Fi for some i, then i = s; otherwise if i < s, thenσ belongs to F s ∩ (∪s−1

j=1F j), but since this simplicial complex is pure ofdimension d − 1, there is v ∈ σ such that σ ⊂ Fs \ {v}, a contradiction.Hence i = s and (I : f) = (ps : f) = ps, as asserted. Next, we show

R/(I, f) = K[F 1 ∪ · · · ∪ F s−1].

Note that f ∈ pi for i = 1, . . . , s−1; otherwise if f /∈ pi for some i < s, thenσ ⊂ Fi and the previous argument yields a contradiction. If p is a minimalprime of (I, f) different from p1, . . . , ps−1, then ps ⊂ p and xi ∈ p for somexi ∈ σ; thus by construction of σ one has Fs \ {xi} ⊂ Fk for some k < s.

218 Chapter 6

Therefore V \Fk ⊂ {xi}∪(V \Fs) ⊂ p, and by the minimality of p one derivespk = ps, which is impossible. Altogether we conclude that p1, . . . , ps−1 arethe minimal primes of (I, f) and (I, f) =

⋂s−1i=1 pi, as required.

Note that the two ends of Eq. (∗) are Cohen–Macaulay R-modules ofdimension d + 1. Hence, using induction on s and Eq. (∗), by the depthlemma it follows that R/I is Cohen–Macaulay. Our proof follows ideas ofHibi [239]. �

Definition 6.3.24 A simplicial complex Δ is constructible if Δ can beobtained by the following recursive procedure:

(i) any simplex is constructible,

(ii) if Δ1, Δ2 are constructible complexes of dimension d, and if Δ1 ∩Δ2

is constructible of dimension d− 1, then Δ1 ∪Δ2 is constructible.

Proposition 6.3.25 If Δ is a pure shellable complex of dimension d, thenΔ is constructible.

Proof. We proceed by induction on d. If d = 0, then Δ is a discrete set ofvertices, which is constructible. Assume d > 0 and that Δ has at least twofacets. Let F1, . . . , Fr be a shelling of Δ. Thus, the complex

Δm = (F 1 ∪ · · · ∪ Fm) ∩ Fm+1

is pure of dimension d − 1 for 1 ≤ m ≤ r − 1. Next we prove that Δm ispure shellable. Consider the set

F = {Fi ∩ Fm+1| dim(Fi ∩ Fm+1) = d− 1; i ≤ m}= {F�1 ∩ Fm+1, . . . , F�s ∩ Fm+1}

where �1 < · · · < �s < m+ 1 and |F| = s. We claim that a face of the form

F = F�i ∩ F�j ∩ Fm+1

has dimension d− 2 for �j < �i < m+1. Clearly dim(F ) ≤ d− 2, otherwiseone has F�i ∩ Fm+1 = F�j ∩ Fm+1, a contradiction because |F| = s. ByExercise 6.3.59, there are vi ∈ Fm+1 \ F�i and vj ∈ Fm+1 \ F�j such that

Fm+1 ∩ F�i ⊂ Fm+1 ∩ Fk1 = Fm+1 \ {vi}; (k1 < m+ 1),

Fm+1 ∩ F�j ⊂ Fm+1 ∩ Fk2 = Fm+1 \ {vj}; (k2 < m+ 1).

By dimension considerations we get

Fm+1 ∩ F�i = Fm+1 \ {vi},Fm+1 ∩ F�j = Fm+1 \ {vj}.


Hence F = Fm+1\{vi, vj}, consequently dim(F ) is equal to d−2, as claimed.It follows readily that F gives a shelling of Δm because the complex(

(F �1 ∩ Fm+1) ∪ · · · ∪ (F �k ∩ Fm+1))∩ (F �k+1

∩ Fm+1).

is pure of dimension d − 2 for 1 ≤ k ≤ s − 1. Therefore by inductionhypothesis Δm is constructible for 1 ≤ m ≤ r − 1. Hence, using inductionon m, it follows that the complex

F 1 ∪ · · · ∪ Fm ∪ Fm+1

is constructible for 1 ≤ m ≤ r − 1, as required. �

Definition 6.3.26 Let R = K[x1, . . . , xn]. A graded R-moduleM is calledsequentially Cohen–Macaulay (over K) if there exists a finite filtration ofgraded R-modules

0 =M0 ⊂M1 ⊂ · · · ⊂Mr =M

such that each Mi/Mi−1 is Cohen–Macaulay, and the Krull dimensions ofthe quotients are increasing:

dim(M1/M0) < dim(M2/M1) < · · · < dim(Mr/Mr−1).

A filtration with these properties is called a C–M filtration of M .

As first observed by Stanley [395, p. 87], shellable implies sequentiallyCohen–Macaulay. In [207] Haghighi, Terai, Yassemi, and Zaare-Nahandiintroduce and study the notion of a sequentially Sr module which is veryreminiscent of the definition of a sequentially Cohen–Macaulay module.

Theorem 6.3.27 If Δ is a shellable simplicial complex, then its associatedStanley–Reisner ring R/IΔ is sequentially Cohen–Macaulay.

If Δ is a simplicial complex and v is a vertex of Δ, the deletion of v,denoted by delΔ(v), is the subcomplex consisting of the faces of Δ that donot contain v. The deletion of a face is defined similarly.

Definition 6.3.28 [43] A simplicial complex Δ is vertex decomposable ifeither Δ is a simplex, or Δ = ∅, or Δ contains a vertex v, called a sheddingvertex, such that both the link lkΔ(v) and the deletion delΔ(v) are vertex-decomposable, and such that every facet of delΔ(v) is a facet of Δ.

Definition 6.3.29 Let Δ be a simplicial complex. The pure i-skeleton ofΔ is defined as:

Δ[i] = 〈{F ∈ Δ| dim(F ) = i}〉; −1 ≤ i ≤ dim(Δ),

where 〈F〉 denotes the subcomplex generated by F . Notice that Δ[i] isalways pure of dimension i.

220 Chapter 6

We say that a simplicial complex Δ is sequentially Cohen–Macaulay ifits Stanley–Reisner ring has this property.

Theorem 6.3.30 [121, Theorem 3.3] A simplicial complex Δ is sequentiallyCohen–Macaulay if and only if the pure i-skeleton Δ[i] is Cohen–Macaulayfor −1 ≤ i ≤ dim(Δ).

Corollary 6.3.31 A simplicial complex Δ is Cohen–Macaulay if and onlyif Δ is sequentially Cohen–Macaulay and Δ is pure.

Proposition 6.3.32 The following hold for simplicial complexes:

(I) pure shellable ⇒ constructible⇒ Cohen–Macaulay ⇒ pure.

(II) vertex decomposable ⇒ shellable⇒ sequentially Cohen–Macaulay.

Proof. (I): The first implication is Proposition 6.3.25. The second impli-cation follows using Exercise 6.3.62 together with the depth lemma. Thethird implication follows at once from Corollary 3.1.17.

(II): The first implication is shown in [45, Theorem 11.3]. The secondimplication is Theorem 6.3.27. �

Definition 6.3.33 A clutter C is a finite ground set X together with afamily E of subsets of X such that if f1, f2 ∈ E, then f1 �⊂ f2. The groundset X is called the vertex set of C and E is called the edge set of C; they aredenoted by V (C) and E(C), respectively.

Clutters are simple hypergraphs (see below) and they are called Spernerfamilies in the literature. One example of a clutter is a graph with thevertices and edges defined in the usual way.

Definition 6.3.34 A hypergraph H is a pair (V,E) such that V is a finiteset and E is a subset of the set of all subsets of V . The elements of Eare called edges and the elements of V are called vertices. A hypergraph issimple if f1 �⊂ f2 for any two edges f1, f2.

An excellent reference for hypergraph theory is the 3-volume book ofSchrijver on combinatorial optimization [373].

Definition 6.3.35 Let C be a clutter with vertex set X = {x1, . . . , xn}.The edge ideal of C, denoted by I(C), is the ideal of R generated by allmonomials xe =

∏xi∈e xi such that e ∈ E(C).

Edge ideals of graphs, clutters, and hypergraphs were introduced in[416], [188], and [206], respectively. The assignment

C �−→ I(C)


establishes a one-to-one correspondence between the family of clutters andthe family of square-free monomial ideals. Edge ideals of clutters are in one-to-one correspondence with simplicial complexes via the Stanley–Reisnercorrespondence

I(C) �−→ ΔI(C).

We shall be interested in studying the relationships between the algebraicand combinatorial properties of C, ΔI(C) and R/I(C).

Let C be a clutter with vertex set X = {x1, . . . , xn}. A subset F of X iscalled independent or stable if e �⊂ F for any e ∈ E(C). The dual concept ofa stable vertex set is a vertex cover , i.e., a subset C of X is a vertex cover ifand only if X \C is a stable vertex set. A minimal vertex cover is a vertexcover which is minimal with respect to inclusion. The number of verticesin any smallest vertex cover, denoted by α0(C), is called the vertex coveringnumber. The vertex independence number, denoted by β0(C), is the numberof vertices in any largest independent set of vertices.

Notice that the Stanley–Reisner complex of I(C) is given by

ΔI(C) = ΔC ,

where ΔC is the simplicial complex whose faces are the independent vertexsets of C. Thus

K[ΔC ] = R/I(C),

where K[ΔC ] is the Stanley–Reisner ring of ΔC .

Definition 6.3.36 The simplicial complex ΔC whose faces are the inde-pendent vertex sets of C is called the independence complex of C.

Lemma 6.3.37 Let C be a set of vertices of a clutter C and let p be theface ideal of R generated by C. The following are equivalent:

(a) C is a minimal vertex cover of C.(b) p is a minimal prime of I(C).

(c) V (C) \ C is a maximal face of ΔC.

Proof. Note that I(C) ⊂ p if and only if C is a vertex cover of C. Takingthis into account, the proof follows readily from the fact that any minimalprime of I(C) is a face ideal; see Theorem 6.1.4. �

Definition 6.3.38 The clutter of minimal vertex covers of C, denoted byC∨ or b(C) is called the Alexander dual clutter or blocker of C. The edgeideal of C∨, denoted by Ic(C), is called the ideal of covers of C. The idealIc(C) is also called the Alexander dual of I(C) and is also denoted by I(C)∨.

222 Chapter 6

Theorem 6.3.39 Let e1, . . . , eq and c1, . . . , cs be the edges and minimalvertex covers of a clutter C. Then, there is a duality given by:

I(C) = (xe1 , xe2 , . . . , xeq ) = (c1) ∩ (c2) ∩ · · · ∩ (cs)' '

Ic(C) = (e1) ∩ (e2) ∩ · · · ∩ (eq) = (xc1 , xc2 , . . . , xcs),

where xek =∏xi∈ek xi and xck =

∏xi∈ck xi for 1 ≤ k ≤ s. In particular

the heights of I(C) and Ic(C) are mini{|ci|} and mini{|ei|}, respectively.

Proof. If pi is the prime ideal generated by ci, then p1, . . . , ps are theminimal primes of I(C) (see Lemma 6.3.37). Thus, by Theorem 6.1.4, weget I(C) = ∩si=1pi. As C is a clutter, we have (C∨)∨ = C and Ic(C∨) = I(C).Hence, the minimal vertex covers of C∨ are e1, . . . , eq. Thus, by the previousargument, we have I(C∨) = ∩qi=1(ei). �

The survey article [218] explains the role of Alexander duality to provecombinatorial and algebraic theorems.

Definition 6.3.40 Let Δ be a simplicial complex on the vertex set V , theAlexander dual Δ∨ of Δ is the simplicial complex given by

Δ∨ = {G ⊂ V |V \G /∈ Δ}.

Theorem 6.3.41 (Eagon–Reiner [124]) If Δ is a simplicial complex, thenthe Alexander dual Δ∨ is Cohen–Macaulay if and only if the Stanley–Reisnerideal IΔ has a linear resolution.

This result can be rephrased as:

Theorem 6.3.42 Let C be a clutter. Then, R/I(C) is Cohen–Macaulay ifand only if Ic(C) has a linear resolution.

This result has been generalized [220] replacing linear resolution by thenotion of componentwise linear ideal , and Cohen–Macaulay by sequentiallyCohen–Macaulay. Next, we introduce the result of Herzog and Hibi thatlink these two notions.

Definition 6.3.43 Let (Id) denote the ideal generated by all degree d ele-ments of a homogeneous ideal I. Then I is called componentwise linear if(Id) has a linear resolution for all d.

If I is a square-free monomial ideal we write I[d] for the ideal generatedby all the square-free monomial ideals of degree d in I.

Theorem 6.3.44 [220] Let I be a square-free monomial ideal of R. Then(a) R/I is sequentially Cohen–Macaulay if and only if I∨ is componentwiselinear. (b) I is componentwise linear if and only if I[d] has a linear resolutionfor all d ≥ 0.


Definition 6.3.45 A monomial ideal I has linear quotients if the mono-mials that minimally generate I can be ordered g1, . . . , gq such that for all1 ≤ i ≤ q − 1, ((g1, . . . , gi) : gi+1) is generated by linear forms xi1 , . . . , xit .

Recall that a clutter is called uniform if all its edges have the samecardinality.

Proposition 6.3.46 [155, Lemma 5.2] If I is an edge ideal of a uniformclutter and I has linear quotients, then I has a linear resolution.

Theorem 6.3.47 [228, Theorem 1.4(c)] If C is a uniform clutter, then ΔCis shellable if and only if Ic(C) has linear quotients.

Exercises

6.3.48 Let R be a polynomial ring over a field K. Prove that the family ofideals of R generated by square-free monomials is a sublattice of the latticeof monomial ideals (cf. Proposition 6.1.19).

6.3.49 A monomial ideal I is square-free monomials iff any of the followingconditions hold:

(a) I is an intersection of prime ideals.

(b) rad (I) = I.

(c) A monomial f is in I iff x1 · · ·xr ∈ I, where supp(f) = {xi}ri=1.

6.3.50 If Δ is the simplicial complex:�x2�

�x4�x1

��

��

��

x3��

then IΔ = (x1, x2) ∩ (x1, x3) ∩ (x1, x4) ∩ (x2, x3) ∩ (x3, x4).

6.3.51 Let Δ be a Cohen–Macaulay simplicial complex and F ∈ Δ. Iflk(F ) is not a discrete set of vertices, prove that lk(F ) is connected.

6.3.52 Let Δ be a simplicial complex on the vertex set V and I its Stanley–Reisner ideal. Take x ∈ V and set J = (I : x).

(a) Prove that J = ∩x/∈pipi, where Ass(I) = {p1, . . . , pr}.

(b) Find a relation between Δ and ΔJ .

6.3.53 Let R be a polynomial ring over a field K and I an ideal minimallygenerated by square-free monomials f1, . . . , fq. If R/I is Cohen–Macaulayand x is a variable not in I such that x ∈ ∪qi=1supp(fi), then R/(I : x) isCohen–Macaulay.

224 Chapter 6

6.3.54 Let Δ be a pure simplicial complex on the vertex set V and I = IΔ.Assume that x is a variable in ∪qi=1supp(fi), where the fi’s are monomialsthat minimally generate I. If (I : x) and (I, x) are Cohen–Macaulay, thenI is Cohen–Macaulay.

Hint Show ht(I) = ht(I : x) = ht(I, x).

6.3.55 Let Δ be a simplicial complex and let σ ∈ Δ, define the star of theface σ as

star(σ) = {G ∈ Δ|σ ∪G ∈ Δ}.Prove that the facets of star(σ) are the facets of Δ that contain σ.

6.3.56 Let R = K[x] be a polynomial ring over a field K and Δ a simplicialcomplex with vertex set x. If σ = {x1, . . . , xr} ∈ Δ, prove the equality

K[star(σ)] = R/(IΔ : f),

where f =∏ri=1 xi and star(σ) is the star of σ.

6.3.57 If Δ is a Cohen–Macaulay complex and F a face of Δ, then star(F )is a Cohen–Macaulay complex.

Hint star(F ) = F ∗ lk(F ).

6.3.58 A simplicial complex Δ of dimension d is pure shellable if and onlyif Δ is pure and the facets of Δ can be listed F1, . . . , Fs such that forevery 1 ≤ i < j ≤ s, there is 1 ≤ � < j with Fi ∩ Fj ⊂ F� ∩ Fj anddim(F� ∩ Fj) = d− 1.

6.3.59 A simplicial complex Δ is pure shellable if and only if Δ is pure andthe facets of Δ can be ordered F1, . . . , Fs such that for all 1 ≤ j < i ≤ s,there is v ∈ Fi \ Fj and k < i with Fi ∩ Fj ⊂ Fi ∩ Fk = Fi \ {v}.

6.3.60 A simplicial complex Δ is pure shellable if and only if Δ is pure andthe facets of Δ can be listed F1, . . . , Fs such that for all 1 ≤ j < i ≤ s, thereexist some v ∈ Fi \ Fj and some k ∈ {1, . . . , i− 1} with Fi \ Fk = {v}.

6.3.61 Let R be a ring and I1, I2 ideals of R. Prove that there is an exactsequence of R-modules:

0 −→ R/(I1 ∩ I2)ϕ−→ R/I1 ⊕R/I2

φ−→ R/(I1 + I2) −→ 0,

where ϕ(r) = (r,−r) and φ(r1, r2) = r1 + r2.

6.3.62 Let R = K[x] be a polynomial ring over a field K and Δ1, Δ2

simplicial complexes whose vertex sets are contained in x. Prove that thereis an exact sequence of R-modules:

0 −→ K[Δ1 ∪Δ2] −→ K[Δ1]⊕K[Δ2] −→ K[Δ1 ∩Δ2] −→ 0.


6.3.63 Prove that any constructible simplicial complex is Cohen–Macaulay.

6.3.64 (Reisner) Let R = K[a, . . . , f ] be a polynomial ring over a field Kand let Δ be the following triangulation of the real projective plane P2:

� b

� c

� a�b

�c

�a

�

f

�

d�e

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Show that IΔ = (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf). The ringK[Δ] is Cohen–Macaulay and has a 3-linear resolution if K has characteris-tic other than 2, and is not Cohen–Macaulay and has a non-linear resolutionotherwise. Prove that Δ is not pure shellable and dimΔ = 2.

The figure below gives a subdivision of the minimal triangulation of theprojective plane given above (cf. Reisner [346, Remark 3]).

6.3.65 (N. Terai) If K is a field and Δ is the following triangulation of thereal projective plane P2

�x3

�x2

�x4

x1x5

x4

x3

x2

x1 x5

x9

x6

x11 x7x10

� �

�

�

��

�

�

�

�

� �x8

�

�

�

��

��

��

��

��

��

��

��

��

�

��

��

��

��

��

��

��

��

��

��

��

��

then the link of any vertex is a cycle and K[Δ] is Cohen–Macaulay if andonly if char(K) �= 2.

Hint H1(Δ;K) � H1(P2;K) = K/2K and use Theorem 6.3.12.

6.3.66 [228] Let Δ be a simplicial complex with vertex set X and facetsF1, . . . , Fs. Then, Δ is shellable if and only if the ideal (xF c

1, . . . , xF c

s) has

linear quotients, where F ci = X \ Fi and xF c =∏xi∈F c xi.

226 Chapter 6

6.3.67 Let Δ be a simplicial complex and let v be a vertex. The followingare equivalent: (i) v is a shedding vertex, i.e., every facet of delΔ(v) is afacet of Δ, (ii) No face of lkΔ(v) is a facet of delΔ(v).

6.3.68 Let Δ be a simplicial complex on the vertex set V = {x1, . . . , xn}and Δ∨ its Alexander dual. Prove that the Stanley–Reisner ideal of Δ∨ isequal to the ideal of minimal covers of I = IΔ, that is:

IΔ∨ = ({xi1 · · ·xir | (xi1 , . . . , xir ) is a minimal prime of I}) = I∨.

6.4 Regularity and projective dimension

Let R = K[x1, . . . , xn] be a polynomial ring over a fieldK with the standardgrading and let C be a clutter with vertex set X = {x1, . . . , xn}. In thissection we study the regularity, depth, and projective dimension of R/I(C),where I = I(C) is the edge ideal of C. There are several well-known resultsrelating these invariants. We collect some of them here for ease of reference.

The first result is a basic relation between the dimension and the depth(see Lemma 2.3.6):

depthR/I(C) ≤ dimR/I(C). (6.1)

The deviation from equality in the above relationship can be quantified usingthe projective dimension, as is seen in a formula discovered by Auslanderand Buchsbaum (see Theorem 3.5.13):

pdR(R/I(C)) + depthR/I(C) = dim(R). (6.2)

Another invariant of interest also follows from a closer inspection of aminimal projective resolution of R/I. Consider the minimal graded freeresolution of M = R/I as an R-module:

F : 0→⊕j

R(−j)bgj → · · · →⊕j

R(−j)b1j → R→ R/I → 0.

The Castelnuovo–Mumford regularity of M (regularity of M for short) isdefined as

reg(M) = max{j − i| bij �= 0}.

The a-invariant, the regularity, and the depth of M are closely related.

Theorem 6.4.1 [413, Corollary B.4.1] a(M) ≤ reg(M) − depth(M), withequality if M is Cohen–Macaulay.


An excellent reference for the regularity of graded ideals is the bookof Eisenbud [129]. There are methods to compute the regularity of R/Iavoiding the construction of a minimal graded free resolution; see [31, 32]and [200, p. 614]. These methods work for any homogeneous ideal over anarbitrary field.

Theorem 6.4.2 [401] Let C be a clutter. If ht(I(C)) ≥ 2, then

reg I(C) = 1 + regR/I(C) = pdR/Ic(C).

This formula also holds for edge ideals of height one:

Corollary 6.4.3 If ht(I(C)) = 1, then regR/I(C) = pdR/Ic(C)− 1.

Proof. We set I = I(C). The formula clearly holds if I is a principal ideal.Assume that I is not principal. By permuting variables, we may assumethat the primary decomposition of I has the form

I = (x1) ∩ · · · ∩ (xr) ∩ p1 ∩ · · · ∩ pm,

where L = p1 ∩ · · · ∩ pm is an edge ideal of height at least 2. Notice thatI = fL, where f = x1 · · ·xr. Then, the Alexander dual of I is

I∨ = (x1, . . . , xr) + L∨.

The multiplication map L[−r] f→ fL induces an isomorphism of gradedR-modules. Thus reg(L[−r]) = r + reg(L) = reg(I). By the Auslander–Buchsbaum formula, one has the equality pd(R/I∨) = r + pd(R/L∨).Therefore, using Theorem 6.4.2, we get

reg(R/I) = reg(R/L) + r = (pd(R/L∨)− 1) + r = pd(R/I∨)− 1. �

Proposition 6.4.4 [434, Lemma 7] Let R1 = K[x] and R2 = K[y] be twopolynomial rings over a field K and let R = K[x,y]. If I1 and I2 are edgeideals of R1 and R2, respectively, then

regR/(I1R + I2R) = reg(R1/I1) + reg(R2/I2).

Proof. By abuse of notation, we will write Ii in place of IiR for i = 1, 2when it is clear from context that we are using the generators of Ii butextending to an ideal of the larger ring. Let x = {x1, . . . , xn} and y ={y1, . . . , ym} be two disjoint sets of variables. Notice that

(I1 + I2)∨ = I∨1 I

∨2 = I∨1 ∩ I∨2

where I∨i is the Alexander dual of Ii. Hence, by Theorem 6.4.2 and usingthe Auslander–Buchsbaum formula, we need only show the equality

depth(R/(I∨1 ∩ I∨2 )) = depth(R1/I∨1 ) + depth(R2/I

∨2 ) + 1.

228 Chapter 6

The equality depth(R/(I∨1 + I∨2 )) = depth(R1/I∨1 ) + depth(R2/I

∨2 ) holds

thanks to Theorem 3.1.34. Thus the proof reduces to showing the equality

depth(R/(I∨1 ∩ I∨2 )) = depth(R/(I∨1 + I∨2 )) + 1. (6.3)

We may assume that depth(R/I∨1 ) ≥ depth(R/I∨2 ). There is an exact se-quence of graded R-modules:

0 −→ R/(I∨1 ∩ I∨2 )ϕ−→ R/I∨1 ⊕R/I∨2

φ−→ R/(I∨1 + I∨2 ) −→ 0, (6.4)

where ϕ(r) = (r,−r) and φ(r1, r2) = r1 + r2. From the inequality

depth(R/I∨1 ⊕R/I∨2 ) = max{depth(R/I∨i )}2i=1 = depth(R/I∨1 )

= depth(R1/I∨1 ) +m

> depth(R1/I∨1 ) + depth(R2/I

∨2 )

= depth(R/(I∨1 + I∨2 ))

and applying the depth lemma to Eq. (6.4), we obtain Eq. (6.3). �

Definition 6.4.5 Let S be a set of vertices of a clutter C. The inducedsubclutter on S, denoted by C[S], is the maximal subclutter of C with vertexset S. A clutter of the form C[S] for some S ⊂ V (C) is called an inducedsubclutter of C.

Thus, the vertex set of C[S] is S and the edges of C[S] are exactly theedges of C contained in S. Notice that C[S] may have isolated vertices, i.e.,vertices that do not belong to any edge of C[S]. If C is a discrete clutter,i.e., all the vertices of C are isolated, we set I(C) = 0 and α0(C) = 0.

Proposition 6.4.6 If D is an induced subclutter of C, then

reg(R/I(D)) ≤ reg(R/I(C)).

Proof. There is S ⊂ V (C) such that D = C[S]. Let p be the prime ideal ofR generated by S. By duality (see Theorem 6.3.39), we have

Ic(C) =⋂

e∈E(C)(e) =⇒ Ic(C)p =

⋂e∈E(C)

(e)p =⋂

e∈E(D)

(e)p = Ic(D)p.

Therefore, using Theorem 6.4.2 and Exercise 6.4.30, we get

reg(R/I(C)) = pd(R/Ic(C))− 1

≥ pd(Rp/Ic(C)p)− 1 = pd(Rp/Ic(D)p)− 1

= pd(R′/Ic(D))− 1 = pd(R/Ic(D)) − 1 = reg(R/I(D)),

where R′ = K[S]. Thus, reg(R/I(C)) ≥ reg(R/I(D)). �


Definition 6.4.7 An induced matching in a clutter C is a set of pairwisedisjoint edges f1, . . . , fr such that the only edges of C contained in ∪ri=1fiare f1, . . . , fr. The induced matching number , denoted by im(C), is thenumber of edges in the largest induced matching.

Corollary 6.4.8 Let C be a clutter and let f1, . . . , fr be an induced match-ing of C with di = |fi| for i = 1, . . . , r. Then(

r∑i=1

di

)− r ≤ reg(R/I(C)).

Proof. Let D = C[∪ri=1fi]. Notice that I(D) = (xf1 , . . . , xfr ). Thus I(D)is a complete intersection and the regularity of R/I(D) is the degree of itsh-polynomial. The Hilbert series of R/I(D) is given by

F (t) =

∏ri=1(1 + t+ · · ·+ tdi−1)

(1− t)n−r .

Thus, the degree of the h-polynomial equals (∑r

i=1 di) − r. Therefore, theinequality follows from Proposition 6.4.6. �

Corollary 6.4.9 If C is a clutter and R/Ic(C) is Cohen–Macaulay, thenim(C) = 1.

Proof. Let r be the induced matching number of C and let d be thecardinality of any edge of C. Using Theorem 6.4.2 and Corollary 6.4.8, weobtain d− 1 ≥ r(d − 1). Thus r = 1, as required. �

Lemma 6.4.10 [128, Corollary 20.19] If 0→ N →M → L→ 0 is a shortexact sequence of graded finitely generated R-modules, then

(a) reg(N) ≤ max(reg(M), reg(L) + 1).

(b) reg(M) ≤ max(reg(N), reg(L)).

(c) reg(L) ≤ max(reg(N)− 1, reg(M)).

The following result was shown by Kalai and Meshulam for square-freemonomial ideals and by Herzog for arbitrary monomial ideals.

Proposition 6.4.11 [269, 219] If I1, I2 are monomial ideals of R. Then

(a) regR/(I1 + I2) ≤ reg(R/I1) + reg(R/I2),

(b) regR/(I1 ∩ I2) ≤ reg(R/I1) + reg(R/I2) + 1.

230 Chapter 6

Corollary 6.4.12 If C1, . . . , Cs are clutters on the vertex set X, then

reg(R/I(∪si=1Ci)) ≤ reg(R/I(C1)) + · · ·+ reg(R/I(Cs)).

Proof. The set of edges of C = ∪si=1Ci is ∪si=1E(Ci). By Proposition 6.4.11,it suffices to notice the equality I(∪si=1Ci) =

∑si=1 I(Ci). �

A clutter C is called co-CM if Ic(C) is Cohen–Macaulay. A co-CM clutteris uniform because Cohen–Macaulay clutters are unmixed.

Corollary 6.4.13 If C1, . . . , Cs are co-CM clutters with vertex set X, then

reg(R/I(∪si=1Ci)) ≤ (d1 − 1) + · · ·+ (ds − 1),

where di is the number of elements in any edge of Ci.

Proof. By Theorem 6.4.2, we get that regR/I(Ci) = di − 1 for all i. Thusthe result follows from Corollary 6.4.12. �

Definition 6.4.14 Let C be a clutter. We will denote the cardinality of asmallest maximal independent set of C by β′

0(C). This number is called thesmall independence number of C.

Lemma 6.4.15 Let C be a clutter and let Δ = ΔC be its independencecomplex. Then Δ[i] = Δi for i ≤ β′

0(C)− 1.

Proof. First we prove the inclusion Δ[i] ⊂ Δi. Let F be a face of Δ[i]. ThenF is contained in a face of Δ of dimension i, and so F is in Δi. Conversely,let F be a face of Δi. Then

dim(F ) ≤ i ≤ β′0(C)− 1 =⇒ |F | ≤ i+ 1 ≤ β′

0(C).

Since β′0(C) is the cardinality of any smallest maximal independent set of

C, we can extend F to an independent set of C with i+ 1 vertices. Thus Fis in Δ[i]. �

Let us recall the following expression for the depth of K[ΔC ]; a simpleproof can be found in [172].

Theorem 6.4.16 [388] Let C be a clutter and let Δ = ΔC be its indepen-dence complex. Then

depthR/I(C) = 1 +max{i |K[Δi] is Cohen–Macaulay},

where Δi = {F ∈ Δ | dim(F ) ≤ i} is the i-skeleton and −1 ≤ i ≤ dim(Δ).

Definition 6.4.17 Let C be a clutter. The cardinality of a largest minimalvertex cover of C, denoted by bight(I(C)), is called the big height of I(C).


Theorem 6.4.18 [326] Let C be a clutter with n vertices. Then

(a) regR/Ic(C) ≥ bight(I(C))− 1,

(b) pdR(R/I(C)) ≥ bight(I(C)),

(c) depthR/I(C) ≤ n− bight(I(C)),with equality everywhere if R/I(C) is sequentially Cohen–Macaulay.

Proof. (a): We set α′0(C) = bight(I(C)). Note that β′

0(C) = n− α′0(C). By

Theorem 6.4.2 and the Auslander–Buchsbaum formula, the proof reducesto showing: depthR/I(C) ≤ β′

0(C), with equality if R/I(C) is sequentiallyCohen–Macaulay.

First we show depthR/I(C) ≤ β′0(C). Assume Δi is Cohen–Macaulay

for some −1 ≤ i ≤ dim(Δ), where Δ is the independence complex of C.By Theorem 6.4.16, it suffices to prove that 1 + i ≤ β′

0(C). We can pick amaximal independent set F of C with β′

0(C) vertices. Since Δi is Cohen–Macaulay all maximal faces of Δ have dimension i. If 1 + i > β′

0(C), thenF is a maximal face of Δi of dimension β′

0(C)− 1, a contradiction.Assume that R/I(C) is sequentially Cohen–Macaulay. By Lemma 6.4.15

Δ[i] = Δi for i ≤ β′0(C) − 1. Then by Theorem 6.3.30, the ring K[Δi] is

Cohen–Macaulay for i ≤ β′0(C)−1. Therefore, applying Theorem 6.4.16, we

get that the depth of R/I(C) is at least β′0(C). Consequently, in this case

one has the equality depthR/I(C) = β′0(C).

(b): It follows from the proof of (a).(c): It follows from (b) and the Auslander–Buchsbaum formula. �

The inequality in part (a) of Theorem 6.4.18 also follows directly fromthe definition of regularity because reg(Ic(C)) is an upper bound for thelargest degree of a minimal generator of Ic(C).

Definition 6.4.19 Let I be a monomial ideal of R. The big height of I,denoted by bight(I), is max{ht(p)| p ∈ Ass(R/I)}.

The following result underlines the advantage of viewing a square-freemonomial ideal as the edge ideal of a clutter. Herzog pointed out that thisresult holds for any sequentially Cohen–Macaulay module, as is seen below.

Corollary 6.4.20 [326] If I is a monomial ideal and R/I is sequentiallyCohen–Macaulay, then pdR(R/I) = bight(I).

Proof. Let I ′ ⊂ R′ be the polarization of I ⊂ R. Since R/I and R′/I ′ havethe same projective dimension and bight(I) = bight(I ′), we may assumethat I is a square-free monomial ideal. As any square-free monomial idealis the edge ideal of a clutter. The formula follows from Theorem 6.4.18. �

This formula can be applied to a wide variety of square-free monomialideals as is seen below.

232 Chapter 6

Definition 6.4.21 A clutter C is called sequentially Cohen–Macaulay ifR/I(C) is sequentially Cohen–Macaulay.

Theorem 6.4.22 The following clutters are sequentially Cohen–Macaulay:

(a) [432] graphs with no chordless cycles of length other than 3 or 5,

(b) [228] clutters whose ideal of covers has linear quotients,

(c) [211] clutters of paths of length t of directed rooted trees,

(d) [155] totally balanced clutters,

(e) [205] uniform admissible clutters whose covering number is 3.

A clutter C is called shellable if ΔC is shellable. The clutters of parts(a)–(e) are in fact shellable, and the clutters of part (a) are in fact vertexdecomposable; see [113, 228, 407, 409, 433, 432]. The families of clutters in(a)–(e) will be studied later in this book (see Chapter 7 and the index).

Theorem 6.4.23 If M is a sequentially Cohen–Macaulay R-module, then

pdR(M) = max{ht(p)| p ∈ AssR(M)}.

Proof. By Theorem 3.5.13 and [413, Proposition A.7.3] one has

pdR(M) = dim(R)− depth(M) = sup{r |ExtrR(M,R) �= 0}, (6.5)

for the second equality see also [65, Exercise 3.1.24, p. 96]. Since M issequentially Cohen–Macaulay, there is a filtration of graded R-modules

0 =M0 ⊂M1 ⊂ · · · ⊂Mr =M

such that each Mi/Mi−1 is C–M and d1 < · · · < dr, where di is equal todim(Mi/Mi−1) for i = 1, . . . , r. By [231, Proposition 2.5] one has

AssR(Mi/Mi−1) = {p ∈ AssR(M) | dim(R/p) = di}.

In particular AssR(M) = ∪iAssR(Mi/Mi−1). Hence

n− d1 = max{ht(p)| p ∈ AssR(M)},

where n = dim(R). By [231, Proposition 2.2], Extn−diR (M,R) is C–M of

dimension di for al i, ExtjR(M,R) = (0) for j /∈ {n− d1, . . . , n− dr} and

Extn−diR (Extn−diR (M,R), R) �Mi/Mi−1

for all i. Hence, by Eq. (6.5), we get pdR(M) = n− d1. �

Definition 6.4.24 The number of maximal independent sets of C, denotedby arith-deg(I(C)), is called the arithmetic degree of I(C).


The arithmetic rank of an ideal I, denoted by ara(I), is the least num-ber of elements of R which generate the ideal I up to radical. By Krull’sprincipal ideal theorem ara(I) ≥ ht(I). If equality occurs, I is called aset-theoretic complete intersection.

Theorem 6.4.25 [401] If ht(I(C)) ≥ 2, then reg(I(C)) ≤ arith-deg(I(C)).

Theorem 6.4.26 [301, Proposition 3] pd(R/I(C)) ≤ ara(I(C)).

There are some instances where the equality pd(R/I) = ara(I) holds; see[15, 16, 143, 279] and the references therein. Barile [15] conjectured that theequality holds for edge ideals of forests. This conjecture was recently shownby Kimura and Terai [278]. Since edge ideals of forests are sequentiallyCohen–Macaulay (see Theorem 6.5.25), by Theorem 6.4.18, it follows thatbight(I) = ara(I) if I is the edge ideal of a forest [278].

Conjecture 6.4.27 (Eisenbud–Goto [130]) If p ⊂ (x1, . . . , xn)2 is a prime

graded ideal, then reg(R/p) ≤ deg(R/p)− codim(R/p).

The following gives a partial answer to the monomial version of theEisenbud–Goto regularity conjecture.

Theorem 6.4.28 [401] Let I be the edge ideal of a clutter C. If ΔC isconnected in codimension 1, then reg(R/I) ≤ deg(R/I)− codim(R/I).

The multiplicity or degree of the edge-ring R/I(C) is equal to the numberof independent sets of C with β0(C) vertices (see Exercise 6.7.10).

Theorem 6.4.29 If C is a k-uniform clutter, then deg(R/I(C)) ≤ kα0(C).

Proof. Let I be the edge ideal of C. We set r = α0(C). One may assumethat K is an infinite field by considering a field extension K ⊂ L, withL an infinite field, and using the functor (·) ⊗K L. It is not hard to seethat I is minimally generated by forms f1, . . . , fq of degree k such thatf1, . . . , fr is a regular sequence (see Exercise 3.1.40). Consider the subidealI ′ = (f1, . . . , fr). There is a graded epimorphism R/I ′ → R/I → 0, wheredim(R/I ′) = dim(R/I). Therefore deg(R/I) ≤ deg(R/I ′). On the otherhand by Exercise 5.1.20 the Hilbert series of R/I ′ is given by

F (R/I ′, t) =(1 + t+ · · ·+ tk−1)r

(1 − t)n−r .

Making t = 1 in the numerator and using Remark 5.1.7 yields the requiredbound. �

234 Chapter 6

Exercises

6.4.30 Let R = K[x1, . . . , xn] and I an ideal of R. If I ⊂ (x1, . . . , xn−1),and R′ = R/(xn) ∼= K[x1, . . . , xn−1], then reg(R/I) = reg(R′/I) andpdR(R/I) = pdR′(R′/I). Similarly, if xn ∈ I and I ′ = I/(xn), thenreg(R/I) = reg(R′/I ′) and pdR(R/I) = pdR′(R′/I ′) + 1.

6.4.31 Let C be a clutter. If α′0(C) := bight(I(C)), then

α′0(C) = max{|e| : e ∈ E(C∨)} and α′

0(C∨) = max{|e| : e ∈ E(C)}.

6.4.32 Let C be a clutter. If I(C) has linear quotients, then

regR/I(C) = max{|e| : e ∈ E(C)} − 1.

6.4.33 Let I ⊂ R be a monomial ideal and let I ′ ⊂ R′ be its polarization.Prove that the following equalities hold:

pd(R/I) = pd(R′/I ′), reg(R/I) = reg(R′/I ′) and bight(I) = bight(I ′).

6.5 Unmixed and shellable clutters

In this section we study unmixed, Cohen–Macaulay and shellable cluttersusing some notions that come from combinatorial optimization. We areinterested in determining what families of clutters have these properties.

Let C be a clutter with vertex set X = {x1, . . . , xn}, let I = I(C) beits edge ideal, and let ΔC be its independence complex. We shall alwaysassume that C has no isolated vertices , i.e., each vertex occurs in at leastone edge.

Definition 6.5.1 If ΔC is pure (resp. Cohen–Macaulay, shellable), we saythat C is unmixed (resp. Cohen–Macaulay, shellable).

The following notions of contraction, deletion, and minor come fromcombinatorial optimization [373].

Definition 6.5.2 For xi ∈ X , the contraction C/xi and deletion C \ xiare the clutters constructed as follows: both have X \ {xi} as vertex set,E(C/xi) is the set of minimal elements of {e\ {xi}| e ∈ E(C}, minimal w.r.tto inclusion, and E(C \ xi) is the set {e|xi /∈ e ∈ E(C)}.

The edge ideals of a deletion and a contraction have a nice algebraicinterpretation. For xi ∈ X , define the contraction and deletion of I as theideals:

(I : xi) and Ic = I ∩K[x1, . . . , xi, . . . , xn],

respectively. Notice that the clutter associated to the square-free monomialideal (I : xi) (resp. Ic) is the contraction C/xi (resp. deletion C \ xi), i.e.,the edge ideals of C/xi and C \ xi are (I : xi) and I

c, respectively.


Definition 6.5.3 A minor of a clutter C is a clutter obtained from C by asequence of deletions and contractions in any order.

A c-minor of I is any ideal generated by the set of monomials obtainedfrom G(I), the minimal set of generators of I, by making any sequence ofvariables equal to 1 in the monomials of G(I). If a c-minor I ′ contains avariable xi and we remove this variable from I ′, we still consider the newideal a c-minor of I. A c-minor of C is any clutter that corresponds to ac-minor of I.

Proposition 6.5.4 Let C1, . . . , Cs be the minimal vertex covers of a clutterC. Then the primary decomposition of I(C) is (C1) ∩ (C2) ∩ · · · ∩ (Cs), andthe facets of ΔC are X \ C1, . . . , X \ Cs.

Proof. This follows from Theorem 6.3.39. �

Corollary 6.5.5 ΔC is pure if and only if all minimal vertex covers of Chave the same cardinality.

Definition 6.5.6 A perfect matching of Konig type of C is a collectione1, . . . , eg of pairwise disjoint edges whose union is X and such that g is theheight of I(C).

A set of pairwise disjoint edges is called independent or a matchingand a set of independent edges of C whose union is X is called a perfectmatching. A clutter C satisfies the Konig property if the maximum numberof independent edges of C equals the height of I(C). For uniform clutters,it is easy to check that if C has the Konig property and a perfect matching,then the perfect matching is of Konig type.

A vertex x of C is called isolated if x does not occur in any edge of C.A clutter with a perfect matching of Konig type has the Konig property.Next we show the converse to be true for unmixed clutters.

Lemma 6.5.7 If C is an unmixed clutter with the Konig property and with-out isolated vertices, then C has a perfect matching of Konig type.

Proof. Let X be the vertex set of C. By hypothesis there are e1, . . . , egindependent edges of C, where g is the height of I(C). If e1 ∪ · · · ∪ eg � X ,pick xr ∈ X \ (e1 ∪ · · · ∪ eg). Since the vertex xr occurs in some edge of C,there is a minimal vertex cover C containing xr. Thus using that e1, . . . , egare mutually disjoint we conclude that C contains at least g + 1 vertices, acontradiction. �

Proposition 6.5.8 Let C be an unmixed clutter with a perfect matchinge1, . . . , eg of Konig type and let C1, . . . , Cr be any collection of minimalvertex covers of C. If C′ is the clutter associated to I ′ = ∩ri=1(Ci), then C′has a perfect matching e′1, . . . , e

′g of Konig type such that: (a) e′i ⊂ ei for all

i, and (b) every vertex of ei \ e′i is isolated in C′.

236 Chapter 6

Proof. We denote the minimal set of generators of the ideal I = I(C) byG(I). There are monomials xv1 , . . . , xvg in G(I) so that supp(xvi) = ei fori = 1, . . . , g. Since xvi is in I and I ⊂ I ′, there is e′i ⊂ ei such that e′iis an edge of C′. Let x be any vertex in ei \ e′i. If x is not isolated in C′,there would a minimal vertex cover Ck of C′ containing x. As Ck containsa vertex of e′j for each 1 ≤ j ≤ g and since e′1, . . . , e

′g are pairwise disjoint,

we get that Ck contains at least g + 1 vertices, a contradiction. Thus (a)and (b) are satisfied. Clearly g is the height of I ′ by construction of I ′.Let X ′ be the vertex set of C′. To finish the proof we need only show thatX ′ = e′1 ∪ · · · ∪ e′g. Let x ∈ X ′, then x ∈ ei for some i and x belongs to atleast one edge of C′. By part (b) we get that x ∈ e′i, as required. �

Corollary 6.5.9 Let C be an unmixed clutter. If C has a perfect matching{ei}gi=1 of Konig type, then C/xj has a perfect matching {e′i}

gi=1 of Konig

type such that: (a) e′i ⊂ ei for all i, and (b) every vertex of ei \ e′i is isolatedin C/xj.

Proof. Let C1, . . . , Cs be the minimal vertex covers of C. Since I(C) isequal to ∩si=1(Ci), one has (I(C) : xj) = ∩xj /∈Ci

(Ci) for any vertex xj /∈I(C). The clutter associated to (I(C) : xj) is the contraction C/xj . Hencefrom Proposition 6.5.8, we get that C/xj has a perfect matching e′1, . . . , e

′g

satisfying (a) and (b). �

Lemma 6.5.10 Let C be an unmixed clutter with a perfect matching {ei}gi=1

of Konig type and let I = I(C). If e1 = {x1, . . . , xr} and C1, . . . , Cs are theminimal vertex covers of C, then⋂

x1∈Ci

(Ci) = (((· · · (((I : x2) : x3) : x4) · · · ) : xr−1) : xr) ,

Proof. Let I ′ be the ideal on the right-hand side of the equality. Ifxv1 , . . . , xvq generate I and we make xi = 1 for i = 2, . . . , r in xv1 , . . . , xvq ,we obtain a generating set for I ′. Notice that I ′ = (I : x2 · · ·xr) by thedefinition of the colon operation.

“⊃”: Take a xa = xa11 xar+1

r+1 · · ·xann in I ′. We may assume a1 = 0,otherwise xa is already in the left-hand side. Then x2 · · ·xrxar+1

r+1 · · ·xann isin I. Let Ci be any minimal vertex cover of C containing x1. Observe thatCi cannot contain xj for 2 ≤ j ≤ r. Indeed if xj ∈ Ci for some 2 ≤ j ≤ r,then Ci would contain {x1, xj} plus at least one vertex of each edge inthe collection e2, . . . , eg, a contradiction because Ci has exactly g vertices.Hence, using that x2 · · ·xrxar+1

r+1 · · ·xann is in I, we get that xar+1

r+1 · · ·xann isin (Ci). Consequently x

a is in the left-hand side of the equality.“⊂”: Let xa be a minimal generator in the left-hand side of the equality.

Then xa ∈ (Ci) if x1 ∈ Ci. If x1 �∈ Ci, then x2 · · ·xr ∈ (Ci) since Ci coverse1. Thus xax2 · · ·xr ∈ (Ci) for all i, and so xax2 · · ·xr ∈ ∩si=1(Ci) = I.Thus xa is in the right-hand side of the equality. �


Definition 6.5.11 We say xi is a free variable (resp. free vertex) of I (resp.C) if xi only appears in one of the monomials of G(I) (resp. in one of theedges of C), where G(I) denotes the minimal set of generators of I = I(C)consisting of monomials.

Theorem 6.5.12 [325] Let C be a clutter with a perfect matching e1, . . . , egof Konig type. If all c-minors of C have a free vertex and C is unmixed, thenΔC is pure shellable.

Proof. The proof is by induction on the number of vertices. We mayassume that C is a non-discrete clutter, i.e., it contains an edge with atleast two vertices. Let z be a free vertex of C and let C1, . . . , Cs be theminimal vertex covers of C. We may also assume that z ∈ em for someem = {z1, . . . , zr}, with r ≥ 2. For simplicity of notation assume thatz = z1 and m = g. Consider the clutters C1 and C2 associated with

I1 =⋂

z1 /∈Ci

(Ci) and I2 =⋂

z1∈Ci

(Ci) (6.6)

respectively. By Proposition 6.5.8, the clutter C2 has a perfect matchinge′1, . . . , e

′g of Konig type such that: (a) e′i ⊂ ei for all i, and (b) every

vertex x of ei \ e′i is isolated in C2, i.e., x does not occur in any edge ofC2. In particular all vertices of eg \ {z1} are isolated vertices of C2. Similarstatements hold for C1 because of Proposition 6.5.8. By Lemma 6.5.10 andCorollary 6.5.9 we get

I1 = (I : z1) and I2 = (((· · · (((I : z2) : z3) : z4) · · · ) : zr−1) : zr) ,

that is, C1 = C/z1 and C2 = C/{z2, . . . , zr}. Hence the ideals I1 and I2 arec-minors of I. The number of vertices of Ci is less than that of C for i = 1, 2.Thus ΔC1 and ΔC2 are shellable by the induction hypothesis. Consider theclutter C′i whose edges are the edges of Ci and whose vertex set is X . Theminimal vertex covers of C′i are exactly the minimal vertex covers of Ci.Thus it follows that ΔC′

iis shellable for i = 1, 2. Let F1, . . . , Fp be the

facets of ΔC that contain z1 and let G1, . . . , Gt be the facets of ΔC thatdo not contain z1. Notice that the edge ideals of Ci and C′i coincide, thevertex set of C′i is equal to the vertex set of C, and I = I1 ∩ I2. Hence fromEq. (6.6) we get that F1, . . . , Fp are the facets of ΔC′

1and G1, . . . , Gt are

the facets of ΔC′2. By the induction hypothesis we may assume F1, . . . , Fp

is a shelling of ΔC′1and G1, . . . , Gt is a shelling of ΔC′

2. We now prove that

F1, . . . , Fp, G1, . . . , Gt

is a shelling of ΔC . We need only show that given Gj and Fi there isv ∈ Gj \ Fi and F� such that Gj \ F� = {v}. We can write

Gj = X \ Cj and Fi = X \ Ci,

238 Chapter 6

where Cj (resp. Ci) is a minimal vertex cover of C containing z1 (resp. notcontaining z1). Notice that z2, . . . , zr are not in Cj because e1, . . . , eg is aperfect matching and |Cj | = g. Thus z2, . . . , zr are in Gj . Since z1 ∈ Fiand Fi cannot contain the edge eg, there is a zk so that zk /∈ Fi and k �= 1.Set v = zk and F� = (Gj \ {zk})∪{z1}. Clearly F� is an independent vertexset because z1 is a free vertex in eg and Gj is an independent vertex set.Thus F� is a facet because C is unmixed. To complete the proof observethat Gj \ F� = {zk}. �

In what follows, we set xe =∏xi∈e xi for any e ⊂ X . Next we give a

characterization of the unmixed property of C. This characterization canbe formulated combinatorially or algebraically.

Theorem 6.5.13 [325] Let C be a clutter with a perfect matching e1, . . . , egof Konig type. Then the following conditions are equivalent:

(a) C is unmixed.

(b) For any two edges e �= e′ and for any two distinct vertices x ∈ e,y ∈ e′ contained in some ei, (e \ {x}) ∪ (e′ \ {y}) contains an edge.

(c) For any two edges e �= e′ and for any T ⊂ ei such that xT dividesxexe′ , supp(xexe′/xT ) contains an edge.

(d) For any two edges e �= e′ and for any ei, (xexe′ : xei ) ⊂ I(C).(e) I(C) = (I(C)2 : xe1) + · · ·+ (I(C)2 : xeg ).

Proof. (a) ⇒ (c): We may assume i = 1. Let T be a subset of e1such that xT divides xexe′ . If T ⊂ e, then e′ is an edge contained inS = supp(xexe′/xT ) and there is nothing to show. The proof is similarif T ⊂ e′. So we can define T1 = e ∩ T and T2 = T \ T1 and we mayassume neither T1 nor T2 is empty. Note that T1 ⊂ e and T2 ⊂ e′. Infact, T2 ⊂ T ∩ e′, but equality does not necessarily hold. Notice thatS = (e \ T1) ∪ (e′ \ T2). If S does not contain an edge, its complementcontains a minimal vertex cover C. We use c to denote complement. Then

C ⊂ X \ S = Sc = (e \ T1)c ∩ (e′ \ T2)c = (ec ∪ T1) ∩ (e′c ∪ T2).

Now C ∩ e �= ∅, so there is an x ∈ C ∩ e. Then x ∈ ec ∪ T1. This forcesx ∈ T1. Similarly there is a y ∈ C ∩ e′, and so y ∈ e′c ∩ T2. Thus y ∈ T2.By the definition of T2, x �= y. To derive a contradiction pick zk ∈ ek ∩ Cfor k ≥ 2 and notice that x, y, z2, . . . , zg is a set of g + 1 distinct vertices inC, which is impossible because C is unmixed.

(c) ⇒ (b): Let x ∈ e and y ∈ e′ be two distinct vertices contained insome ei. Let T = {x, y}. Then xT divides xexe′ and

S = supp(xexe′/xT ) ⊂ (e \ {x}) ∪ (e′ \ {y}).


By (c), S contains an edge. Thus (e \ {x}) ∪ (e′ \ {y}) contains an edge.(b) ⇒ (a): Let C be a minimal vertex cover of C. Since the matching is

perfect, there is a partition C = (C ∩e1)∪· · ·∪ (C ∩eg). Hence it suffices toprove that |C∩ei| = 1 for all i. We proceed by contradiction. For simplicityof notation assume i = 1 and |C ∩ e1| ≥ 2. Pick x �= y in C ∩ e1. Since C isminimal, there are edges e, e′ such that

e ∩ (C \ {x}) = ∅ and e′ ∩ (C \ {y}) = ∅. (6.7)

Clearly x ∈ e, y ∈ e′, and e �= e′ because y /∈ e. Then by hypothesis theset S = (e \ {x}) ∪ (e′ \ {y}) contains an edge e′′. Take z ∈ e′′ ∩ C, thenz ∈ e \ {x} or z ∈ e′ \ {y}, which is impossible by Eq. (6.7).

(c) ⇒ (d): Let xa ∈ (xexe′ : xei) be a monomial generator of the colonideal. Then xaxei = mxexe′ for some monomial m. Let T ⊂ ei bemaximal such that xT divides xexe′ . Then xei\T divides m, and xa =(m/xei\T )(xexe′/xT ). Since supp(xexe′/xT ) contains an edge, we havexexe′/xT ∈ I. Thus xa ∈ I(C) as desired.

(d) ⇒ (c): Suppose T ⊂ ei is such that xT divides xexe′ . Then

(xexe′/xT )xei = xexe′xei\T ,

and so (xexe′/xT ) ∈ (xexe′ : xei ) ⊂ I(C). Thus (xexe′/xT ) is a multiple ofa monomial generator of I(C). Hence supp(xexe′/xT ) contains an edge.

(e) ⇒ (d): If equality in (e) holds, then (I(C)2 : xei) = I(C) for all i.Hence from (I(C)2 : xei) ⊂ I(C) we get condition (d).

(d) ⇒ (e): We set I = I(C). It suffices to show (I2 : xei) = I for all i.Since I is contained in (I2 : xei ), we need only show the inclusion (I2 : xei) ⊂I. Take xa ∈ (I2 : xei), then x

axei = mxexe′ for some edges e, e′ of C andsome monomial m. If e �= e′, then by hypothesis xa ∈ (xexe′ : xei ) ⊂ I, i.e.,xa ∈ I. If e = e′, then xaxei = mx2e. Thus xe divides xa because xei is asquare-free monomial, but this means that xa ∈ I, as required. �

Definition 6.5.14 A matrix A with entries in {0, 1} is balanced if A has nosquare submatrix of odd order with exactly two 1’s in each row and column.

Definition 6.5.15 Let A be the incidence matrix of a clutter C. A clutterC has a special cycle of length r if there is a square submatrix of A of orderr ≥ 3 with exactly two 1’s in each row and column. A clutter with nospecial odd cycles is called balanced and a clutter with no special cycles iscalled totally balanced .

This definition of special cycle is equivalent to the usual definition ofspecial cycle in hypergraph theory [8, 226] (see Exercise 6.5.34).

The next result classifies all unmixed balanced clutters because balancedclutters have the Konig property (Corollary 14.3.11). In particular it givesa classification of all unmixed bipartite graphs (cf. Theorem 7.4.19).

240 Chapter 6

Corollary 6.5.16 [325] A clutter C with the Konig property is unmixed ifand only if there is a perfect matching e1, . . . , eg of Konig type such thatfor any two edges e �= e′ and for any two distinct vertices x ∈ e, y ∈ e′

contained in some ei, one has that (e \ {x}) ∪ (e′ \ {y}) contains an edge.

Proof. ⇒) Assume that C is unmixed. By Theorem 6.5.13 it suffices toobserve that any unmixed clutter with the Konig property and withoutisolated vertices has a perfect matching of Konig type; see Lemma 6.5.7.⇐) This implication follows at once from Theorem 6.5.13. �

Faridi [155] introduced the notion of a leaf for a simplicial complex Δ.Precisely, a facet F of Δ is a leaf if F is the only facet of Δ, or there existsa facet G �= F such that F ∩F ′ ⊂ F ∩G for all facets F ′ �= F . Δ is called asimplicial forest if every non-empty subcollection, i.e., a subcomplex whosefacets are also facets of Δ, of Δ contains a leaf.

A graph G is called chordal if every cycle Cr of G of length r ≥ 4 hasa chord in G. A chord of Cr is an edge joining two non-adjacent verticesof Cr. A chordal graph is called strongly chordal if every cycle Cr of evenlength at least six has a chord that divides Cr into two odd length paths.

A clique of a graph G is a set of vertices inducing a complete subgraph.The clique clutter of G, denoted by cl(G), is the clutter on V (G) whoseedges are the maximal cliques of G (maximal with respect to inclusion).

If all the minors of a clutter C have free vertices, we say that C has thefree vertex property. Note that if C has the free vertex property, then so doall of its minors.

Theorem 6.5.17 A clutter C is totally balanced if and only if any of thefollowing equivalent conditions hold:

(a) [226, Theorem 3.2] C is the clutter of the facets of a simplicial forest.

(b) [389, Corollary 3.1] C has the free vertex property.

(c) [152] C is the clique clutter of a strongly chordal graph.

Theorem 6.5.18 [325] Let C be a clutter with a perfect matching e1, . . . , egof Konig type. If C has no special cycles of length 3 or 4 and C is unmixed,then for any two edges f1, f2 of C and for any ei, one has that f1∩ei ⊂ f2∩eior f2 ∩ ei ⊂ f1 ∩ ei.

Proof. For simplicity assume i = 1. We proceed by contradiction. Assumethere are x1 ∈ f1 ∩ e1 \ f2 ∩ e1 and x2 ∈ f2 ∩ e1 \ f1 ∩ e1. As C is unmixed,by Theorem 6.5.13(b) there is an edge e of C such that

e ⊂ (f1 \ {x1}) ∪ (f2 \ {x2}) = (f1 ∪ f2) \ {x1, x2}.

Since e �⊂ e1, there is x3 ∈ e \ e1. Then either x3 ∈ f1 or x3 ∈ f2. Withoutloss of generality we may assume x3 ∈ f1 \ e1. For use below we denote theincidence matrix of C by A.


Case(I): x3 ∈ f2. Then the matrix

f1 f2 e1x1 1 0 1x2 0 1 1x3 1 1 0

is a submatrix of A, a contradiction.Case(II): x3 /∈ f2. Notice that e �⊂ f1, otherwise e = f1 which is impos-

sible because x1 ∈ f1 \ e. Thus there is x4 ∈ e \ f1 and x4 ∈ (e ∩ f2) \ f1.Subcase(II.a): x4 ∈ e1. Then the matrix

f1 e e1x1 1 0 1x3 1 1 0x4 0 1 1

is a submatrix of A, a contradiction.Subcase(II.b): x4 /∈ e1. Then the matrix

f1 e f2 e1x1 1 0 0 1x2 0 0 1 1x3 1 1 0 0x4 0 1 1 0

is a submatrix of A, a contradiction. �

Proposition 6.5.19 Let C be an unmixed clutter with no special cycles oflength 3 or 4. If e1, . . . , eg is a perfect matching of C of Konig type, then eihas a free vertex for all i.

Proof. Fix an integer i in [1, g]. We may assume that ei has at least onenon-free vertex. Consider the set: F = {f ∈ E(C)| ei ∩ f �= ∅; f �= ei}.By Theorem 6.5.18, the edges of F can be listed as f1, . . . , fr so that theysatisfy the inclusions f1 ∩ ei ⊂ f2 ∩ ei ⊂ · · · ⊂ fr ∩ ei � ei. Thus any vertexof ei \ (fr ∩ ei) is a free vertex of ei. �

Theorem 6.5.20 Let C be an unmixed clutter with a perfect matchinge1, . . . , eg of Konig type. If C has no special cycles of length 3 or 4, then ΔCis pure shellable.

Proof. All hypotheses are preserved under contractions, i.e., under c-minors. This follows from Corollary 6.5.9 and the fact that the incidencematrix of a contraction of C is a submatrix of the incidence matrix of C.Thus by Proposition 6.5.19 any c-minor has a free vertex and the resultfollows from Theorem 6.5.12. �

242 Chapter 6

Theorem 6.5.21 [325] Let C be a clutter with a perfect matching e1, . . . , egof Konig type. If for any two edges f1, f2 of C and for any edge ei of theperfect matching, one has that f1 ∩ ei ⊂ f2 ∩ ei or f2 ∩ ei ⊂ f1 ∩ ei, thenΔC is pure shellable.

Proof. First we show that ΔC is pure or equivalently that C is unmixed.It suffices to verify condition (b) of Theorem 6.5.13. Let f1 �= f2 be twoedges and let x ∈ f1, y ∈ f2 be two distinct vertices contained in some ei.For simplicity we assume i = 1. Set B = (f1 \ {x}) ∪ (f2 \ {y}). Thenf2 ∩ e1 ⊂ f1 ∩ e1 or f1 ∩ e1 ⊂ f2 ∩ e1. In the first case we have that f2 ⊂ B.Indeed let z ∈ f2. If z �= y, then z ∈ f2 \ {y} ⊂ B, and if z = y, thenz ∈ f2 ∩ e1 ⊂ f1 ∩ e1 and z �= x, i.e., z ∈ f1 \ {x} ⊂ B. In the second casef1 ⊂ B. This proves that C is unmixed.

Next we show that ΔC is shellable. Notice that (i) ei has a free vertexfor all i, which follows from the proof of Proposition 6.5.19. Thus, as C isunmixed, by Theorem 6.5.12 we need only show that any c-minor has afree vertex. By (i) it suffices to show that our hypotheses are closed undercontractions. Let x be a vertex of C and let C′ = C/x. By Corollary 6.5.9,we get that C/x has a perfect matching e′1, . . . , e

′g satisfying: (a) e′i ⊂ ei

for all i, and (b) every vertex of ei \ e′i is isolated in C′. Let e, e′ be twoedges of C′ and let e′i be an edge of the perfect matching of C′. Thereare edges f, f ′ of C such that one of the following is satisfied: e = f ande′ = f ′ \ {x}, e = f \ {x} and e′ = f ′, e = f \ {x} and e′ = f ′ \ {x}, e = fand e′ = f ′. We may assume f ∩ ei ⊂ f ′ ∩ ei. To finish the proof we nowshow that e∩e′i ⊂ e′∩e′i. Take z ∈ e∩e′i. Then z ∈ f ∩ei and consequentlyz ∈ f ′ ∩ e′i. Since x /∈ e′i, one has z �= x. It follows that z ∈ e′ ∩ e′i. �

Let G be a graph and let V be its vertex set. For use below consider thegraph G∪W (V ) obtained from G by adding new vertices {yi |xi ∈ V } andnew edges {{xi, yi} |xi ∈ V }.

Corollary 6.5.22 If G is a graph and H = G ∪W (V ), then ΔH is pureshellable.

Proof. It follows at once from Theorem 6.5.21. Indeed if V = {x1, . . . , xn},then {x1, y1}, . . . , {xn, yn} is a perfect matching ofH satisfying the orderingcondition in Theorem 6.5.21. �

Lemma 6.5.23 Let C be a clutter with minimal vertex covers C1, . . . , Cs.If ΔC is shellable, A ⊂ VC is a set of vertices, and

I ′ =⋂

Ci∩A=∅(Ci),

then ΔI′ is shellable with respect to the linear ordering of the facets of ΔI′

induced by the shelling of the simplicial complex ΔC.


Proof. Let H1, . . . , Hs be a shelling of ΔC . We may assume that Hi isequal to VC \Ci for all i. Let Hi and Hj be two facets of ΔI′ with i < j, i.e.,A∩Ci = ∅ and A∩Cj = ∅. By the shellability of ΔC , there is an x ∈ Hj \Hi

and an � < j such that Hj \H� = {x}. It suffices to prove that C� ∩A = ∅.If C� ∩ A �= ∅, pick z ∈ C� ∩ A. Then z /∈ Ci ∪ Cj and z ∈ Hi ∩Hj . Sincez /∈ H� (otherwise z /∈ C�, a contradiction), we get z ∈ Hj \H�, i.e., z = x,a contradiction because x /∈ Hi. �

Lemma 6.5.24 Let xn be a free variable of I(C) = (xv1 , . . . , xvq−1 , xvq ),and let xvq = xnx

u. Then the following hold.

(a) If C1 is the clutter associated to (xv1 , . . . , xvq−1 ), then C is a minimalvertex cover of C containing xn if and only if C ∩ supp(xu) = ∅ andC = {xn} ∪C′ for some minimal vertex cover C′ of C1.

(b) If C2 is the clutter associated to (xv1 , . . . , xvq−1 , xu), then C is a min-imal vertex cover of C not containing xn if and only if C is a minimalvertex cover of C2.

Proof. (a) Assume that C is a minimal vertex cover of C containing xn.If C ∩ supp(xu) �= ∅, then C \ {xn} is a vertex cover of C, a contradiction.Thus C ∩ supp(xu) = ∅. Hence it suffices to notice that C′ = C \ {xn} is aminimal vertex cover of C1. The converse also follows readily.

(b) Assume that C is a minimal vertex cover of C not containing xn. Letxa be a minimal generator of I(C2), then either xu divides xa or xa = xvi

for some i < q. Then clearly C ∩ supp(xa) �= ∅ because C ∩ A �= ∅, whereA = supp(xu). Thus C is a vertex cover of C2. To prove that C is minimaltake C′ � C. We must show that there is an edge of C2 not covered by C′.As C is a minimal vertex cover of C, there is xvi such that supp(xvi)∩C′ = ∅.If xvi is a minimal generator of C2 there is nothing to prove, otherwise xu

divides xvi and the edge A of C2 is not covered by C′. The converse alsofollows readily. �

If all c-minors have a free vertex and C is unmixed, then ΔC is pureshellable (see Theorem 6.5.12). The next result complements this fact.

Theorem 6.5.25 [409] If the clutter C has the free vertex property, thenthe independence complex ΔC is shellable.

Proof. We proceed by induction on the number of vertices of C. Let xnbe a free variable of I = I(C) = (xv1 , . . . , xvq−1 , xvq ). We may assume thatxn occurs in xvq . Hence we can write xvq = xnx

u for some xu such thatxn /∈ supp(xu). For use below we set A = supp(xu). Consider the idealsJ = (xv1 , . . . , xvq−1 ) and L = (J, xu). Then J = I(C1) and L = I(C2),where C1 and C2 are the clutters defined by the ideals J and L, respectively.Notice that J and L are minors of the ideal I obtained by setting xn = 0

244 Chapter 6

and xn = 1, respectively. The vertex set of Ci is VCi = X \ {xn} fori = 1, 2. Thus ΔC1 and ΔC2 are shellable by the induction hypothesis. LetF1, . . . , Fr be the facets of ΔC that contain xn and let G1, . . . , Gs be thefacets of ΔC that do not contain xn. Set Ci = X \Gi and C′

i = Ci \{xn} fori = 1, . . . , s. Then C1, . . . , Cs is the set of minimal vertex covers of C thatcontain xn, and by Lemma 6.5.24(a) C′

1, . . . , C′s is the set of minimal vertex

covers of C1 that do not intersect A. One has the equality Gi = VC1 \ C′i

for all i. Hence, by the shellability of ΔC1 and using Lemma 6.5.23, we mayassume that G1, . . . , Gs is a shelling for the simplicial complex generatedby G1, . . . , Gs. By Lemma 6.5.24(b) one has that C is a minimal vertexcover of C not containing xn if and only if C is a minimal vertex coverof C2. Thus, F is a facet of ΔC that contains xn, i.e., F = F ′ ∪ {xn} ifand only if F ′ is a facet of ΔC2 . By induction we may also assume thatF ′1 = F1 \ {xn}, . . . , F ′

r = Fr \ {xn} is a shelling of ΔC2 . We now prove that

F1, . . . , Fr, G1, . . . , Gs with Fi = F ′i ∪ {xn}

is a shelling of ΔC . We need only show that given Gj and Fi there isa ∈ Gj \ Fi and F� such that Gj \ F� = {a}. We can write

Gj = X \ Cj and Fi = X \ Ci,

where Cj (resp. Ci) is a minimal vertex cover of C containing xn (resp.not containing xn). Recall that A = supp(xu) is an edge of C2. Notice thefollowing: (i) Cj = C′

j ∪ {xn} for some minimal vertex cover C′j of C1 such

that A∩C′j = ∅, and (ii) Ci is a minimal vertex cover of C2. From (i) we get

that A ⊂ Gj . Observe that A �⊂ Fi, otherwise A ∩ Ci = ∅, a contradictionbecause Ci must cover the edge A = supp(u). Hence there is a ∈ A\Fi anda ∈ Gj \Fi. Since C′

j ∪ {a} is a vertex cover of C, there is a minimal vertexcover C� of C contained in C′

j ∪{a}. Clearly a ∈ C� because C� has to coverxu and C′

j ∩ A = ∅. Thus F� = X \ C� is a facet of ΔC containing xn. Tofinish the proof we now prove that Gj \ F� = {a}. We know that a ∈ Gj .If a ∈ F�, then a /∈ C�, a contradiction. Thus a ∈ Gj \ F�. Conversely takez ∈ Gj \ F�. Then z /∈ C′

j ∪ {xn} and z ∈ C� ⊂ C′j ∪ {a}. Hence z = a, as

required. �

According to Theorem 6.5.17, a totally balanced clutter satisfies the freevertex property. Thus, we obtain:

Corollary 6.5.26 If C is a totally balanced clutter, then ΔC is shellable.

By Theorem 6.5.17, a clutter C is a simplicial forest if and only if C is atotally balanced clutter. Thus, we also obtain:

Corollary 6.5.27 If C is the clutter of facets of a simplicial forest, thenΔC is shellable.


Definition 6.5.28 [155] Let Δ be a simplicial complex. The facet ideal ofΔ, denoted by I(Δ), is the edge ideal of the clutter of facets of Δ.

Facet ideals were introduced and studied by Faridi in a series of papers[153, 154, 155, 156, 157].

Corollary 6.5.29 [155] Let I = I(Δ) be the facet ideal of a simplicialforest. Then R/I(Δ) is sequentially Cohen–Macaulay.

Proof. If Δ = 〈F1, . . . , Fs〉, then I(Δ) is also the edge ideal of the clutterC whose edge set is EC = {F1, . . . , Fs}. Now apply Corollary 6.5.27 andTheorem 6.3.27. �

Exercises

6.5.30 Prove that a clutter with a perfect matching of Konig type has theKonig property.

6.5.31 Let C be a uniform clutter. If C has the Konig property and a perfectmatching, then the perfect matching is of Konig type.

6.5.32 Consider the clutter C whose edges are

e1 = {x1, x2}, e2 = {x3, x4, x5, x6}, e3 = {x7, x8, x9}, f4 = {x1, x3},f5 = {x2, x4}, f6 = {x5, x7}, f7 = {x6, x8}.

Prove that C has the Konig property and that C has no perfect matching ofKonig type.

6.5.33 [185, Proposition 5.8] If C is a uniform totally balanced clutter, thereis a partition X1, . . . , Xd of X such that any edge of C intersects any X i inexactly one vertex.

6.5.34 Let H be a hypergraph and let A be its incidence matrix. Recallthat a cycle (resp. special cycle) of H is a sequence x1e1x2e2 · · ·xrerx1 of rdistinct vertices xi and r distinct edges ej (r ≥ 3) such that {xi, xi+1} ⊂ ei(resp. {x1, . . . , xr} ∩ ei = {xi, xi+1}) for i = 1, . . . , r (xr+1 = x1). Thevalue r is the length of the cycle. Prove that H has a special cycle of lengthr if and only if there is a square submatrix of A of order r ≥ 3 with exactlytwo 1’s in each row and column.

6.5.35 If e1 = {x1, x2, x4}, e2 = {x2, x3, x5}, e3 = {x1, x3, x6}, the clutterwith edges e1, e2, e3 has a special cycle x1e1x2e2x3e3x1 of order 3.

6.5.36 [8] If C is a totally balanced clutter with n vertices, then C has atmost

(n2

)+ n edges.

246 Chapter 6

6.6 Admissible clutters

Let X1, . . . , Xd and e1, . . . , eg be two partitions of a finite set X such that|ei ∩ Xj | ≤ 1 for all i, j. The variables of K[X ] are linearly ordered by:x ≺ y iff (x ∈ X i, y ∈ Xj, i < j) or (x, y ∈ X i, x ∈ ek, y ∈ e�, k < �).

Let e ⊂ X such that |e| = k and |e∩X i| ≤ 1 for all i. There are uniqueintegers 1 ≤ i1 < · · · < ik ≤ d and integers j1, . . . , jk ∈ [1, g] such that

∅ �= e ∩X i1 = {x1}, ∅ �= e ∩X i2 = {x2}, . . . , ∅ �= e ∩X ik = {xk}

and x1 ∈ ej1 , . . . , xk ∈ ejk . We say that e is an admissible set if one hasi1 = 1, i2 = 2, . . . , ik = k and j1 ≤ · · · ≤ jk.

We can represent an admissible set e = {x1, . . . , xk} as e = x1j1 · · ·xkjk,

i.e., xi = xiji and xiji ∈ X i ∩ eji for all i. A monomial xa is admissibleif supp(xa) is admissible. A clutter C is called admissible if e1, . . . , eg areedges of C, ei is admissible for all i, and all other edges are admissible setsnot contained in any of the ei’s. We can think of X1, . . . , Xd as color classesthat color the edges.

Lemma 6.6.1 If C is an admissible clutter, then e1, . . . , eg is a perfectmatching of Konig type.

Proof. It suffices to prove that g = ht I(C). Clearly ht I(C) ≥ g becauseany minimal vertex cover of C must contain at least one vertex of each eiand the ei’s form a partition of X . For each 1 ≤ i ≤ g there is yi = x1i sothat ei ∩X1 = {yi}. Since the ei’s form a partition we have the equality

(e1 ∩X1) ∪ · · · ∪ (eg ∩X1) = X1.

Thus |X1| = g. To complete the proof notice that X1 is a vertex cover of Cbecause all edges of C are admissible. This shows ht I(C) ≤ g. �

Admissible clutters with two color classes X1, X2 are special types ofbipartite graphs (see Section 7.4).

Example 6.6.2 Let C be the Cohen–Macaulay admissible balanced clutterwith color classes X1, X2, X3 and edges e1, e2, e3, f1, f2, f3.

X1 X2 X3

e1 = x1 y1e2 = x2 y2 z2e3 = x3 y3

X1 X2 X3

f1 = x1 y2 z2f2 = x1 y3f3 = x2 y3

The edges e1, e2, e3 form a perfect matching of Konig type of C.


Example 6.6.3 Consider the clutter C whose edge ideal is generated by:

a1b1c1d1g1h1k1, a2b2c2d2g2h2k2, a3b3c3d3g3h3k3,a4b4c4d4g4h4k4, a1b1c1d1g2h3k4, a1b2c3d4g2h3k4,

where ai, bi, . . . are variables. This clutter is balanced because its incidencematrix is totally unimodular and I(C) is Cohen–Macaulay. However, wecannot order its vertices so that it becomes an admissible uniform clutter.

Definition 6.6.4 If e1, . . . , eg are admissible subsets of X , the clutter C onX whose set of edges is:

E(C) ={e ⊂ X

∣∣∣∣ ei �⊂ e for i = 1, . . . , g, e is admissible,e �⊂ e′ for any admissible set e′ �= e

}∪ {e1, . . . , eg}

is called a complete admissible clutter (cf. Definition 14.5.11).

The edges of this clutter are the maximal admissible sets with respect toinclusion. By Lemma 6.6.1, e1, . . . , eg is a perfect matching of Konig type.In Section 14.5 we show an optimization property of complete admissibleuniform clutters (see Theorems 14.3.6 and 14.5.12).

Proposition 6.6.5 If C is a complete admissible clutter, then C is unmixed.

Proof. To show that C is unmixed it suffices to verify condition (b) ofTheorem 6.5.13. Let e �= e′ be two edges of C and let x �= y be two verticessuch that {x, y} ⊂ ei for some ei, x ∈ e, and y ∈ e′. Since e, e′, ei areadmissible we can write

e = {x1, . . . , xk}, e′ = {y1, . . . , yk′}, ei = {z1, . . . , zk′′},

where xi ∈ X i, yi ∈ X i, zi ∈ X i. There are i1, i2 such that x = xi1 , y = yi2 ,x = zi1 , and y = zi2 . We may assume i1 < i2. One has i1 < k, because ifk = i1, then e � e ∪ {zi1+1, . . . , zi2} and the right-hand side is admissible,a contradiction. Set f = {y1, . . . , yi1 , xi1+1, . . . , xk}. Then

f ⊂ e \ {x} ∪ e′ \ {y}.

Thus to finish the proof we need only show that f is an edge of C. Sinceyi2 ∈ ei and xi1 ∈ ei, then yi1 ∈ e� for some � ≤ i and xi1+1 ∈ et for somei ≤ t. Hence f is admissible. Next we show that f is maximal. Assume thatf is not maximal. Then there exists an admissible subset f ′ that properlycontains f . Then there is z ∈ f ′ ∩Xk+1 and since f ∪{z} ⊂ f ′, we get thate ∪ {z} = {x1, . . . , xk, z} is admissible, but e � e ∪ {z}, a contradiction.Hence f is maximal. �

Theorem 6.6.6 [325] If C is a complete admissible uniform clutter, thenthe simplicial complex generated by the edges of C is pure shellable.

248 Chapter 6

Proof. Order the variables of K[X ] as in the beginning of Section 6.6. Wecan represent the edges of C as Fi = x1i1x

2i2· · ·xdid , where x

iij∈ X i ∩ eij .

Then, we order the edges of C lexicographically, that is

Fi = x1i1x2i2 · · ·x

did< Fj = x1j1x

2j2 · · ·x

djd

if the first non-zero entry of (j1, j2, . . . , jd)−(i1, i2, . . . , id) = j−i is positive.Under this order, we show that C is shellable.

Suppose Fi and Fj are edges of C with Fi < Fj . Suppose the first non-zero entry of j− i is jt−it. Then 1 ≤ it < jt. Let Fk = Fj \{xtjt}∪{x

tit} and

let v = xtjt . Since j1 = i1 ≤ · · · ≤ jt−1 = it−1 ≤ it < jt ≤ jt+1 ≤ · · · ≤ jd,Fk is maximal admissible, v ∈ Fj \ Fi, Fk < Fj and Fj\Fk = {v}. �

The next example illustrates the construction of the lexicographicalshelling used in the proof of Theorem 6.6.6.

Example 6.6.7 Let C be the complete admissible uniform clutter withcolor classes X1 = {x1, x2, x3}, X2 = {y1, y2, y3}, X3 = {z1, z2, z3}. Thenthe shelling of the simplicial complex generated by the edges of C is:

F1 = {x1, y1, z1} < F2 = {x1, y1, z2} < F3 = {x1, y1, z3} <F4 = {x1, y2, z2} < F5 = {x1, y2, z3} < F6 = {x1, y3, z3} <F7 = {x2, y2, z2} < F8 = {x2, y2, z3} < F9 = {x2, y3, z3} <F10 = {x3, y3, z3}.

Lemma 6.6.8 If C is a complete admissible uniform clutter, then the sim-plicial complex ΔC∨ generated by {X \ F |F ∈ E(C)} is pure shellable.

Proof. Let F1, . . . Fr be the shelling of the edges of C defined in Theorem6.6.6. Let G1 = X \ F1, . . . , Gr = X \ Fr be the facets of ΔC∨ . We claimthat G1, . . . , Gr is the desired shelling. Suppose Gi < Gj . Then Fi < Fj .Using the notation defined in Theorem 6.6.6, let v = xtjt and define u = xtit .Then u ∈ Gj \Gi and Gj \Gk = {u} as required. �

Theorem 6.6.9 [325] If C is a complete admissible d-uniform clutter, thenthe face ring R/I(C) is a Cohen–Macaulay ring with a d-linear resolutionand |E(C)| =

(d+g−1g−1

).

Proof. Consider the clutter C∨ of minimal vertex covers of C. By Lemma6.6.8 and Exercise 6.6.14 we have that ΔC∨ is pure shellable. Now recall thatthe Stanley–Reisner ideal of ΔC∨ is I(C∨) and that I(C∨) is the Alexanderdual of I(C). Thus R/I(C∨) is Cohen–Macaulay, and by Theorem 6.3.41 theideal I(C) has a linear resolution. Since the Alexander dual of a completeadmissible uniform clutter is also a complete admissible uniform clutter andsince (C∨)∨ = C it follows that R/I(C) is Cohen–Macaulay. The formulafor the number of edges of C follows from the explicit formula given in


Theorem 3.5.17 for the Betti numbers of a Cohen–Macaulay ideal with apure resolution; see also Exercise 5.3.17. �

Let C be a complete admissible uniform clutter. For each edge of C,e = x1j1x

2j2 · · ·xdjd , consider all pairs (xiji , x

kjk) with i < k and consider the

union of all these pairs with e = x1j1x2j2· · ·xdjd running through all edges of

C. This defines a poset P = (X,≺) on X whose comparability graph G isdefined by all the unordered pairs {xiji , xkjk}.

Corollary 6.6.10 If G′ is the complement of the comparability graph Gdefined above, then R/I(G′) is Cohen–Macaulay.

Proof. Notice that ΔG′ = {Kr| Kr is a clique of G} = O(P ), where O(P )is the order complex of P . Since the maximal faces of O(P ) are preciselythe edges of C, by Theorem 6.6.6, we obtain that O(P ) is a pure shellablecomplex whose Stanley–Reisner ring is equal to R/I(G′). Hence R/I(G′) isCohen–Macaulay by Theorem 6.3.23. �

Let C be a clutter and let xv1 , . . . , xvq be the minimal set of generatorsof I(C). Consider the ideal I∗ = (xw1 , . . . , xwq ), where vi +wi = (1, . . . , 1).Following the terminology of matroid theory we call I∗ the dual of I. If I∗

has linear quotients and all xwi have the same degree, then I∗ has a linearresolution (see Proposition 6.3.46).

Corollary 6.6.11 If C is a complete admissible uniform clutter, then I(C)∗has linear quotients.

Proof. Let xv1 , . . . , xvq be the minimal set of generators of I = I(C). Weset Fi = supp(xvi) for i = 1, . . . , q. By Theorem 6.6.6, we may assume thatF1, . . . , Fq is a shelling for the simplicial complex 〈F1, . . . , Fq〉 generated bythe Fi’s. Thus, according to Exercise 6.3.66, the ideal I∗ = (xF c

1, . . . , xF c

q)

has linear quotients, where F ck = X \ Fk and xF ck=∏xi∈F c

kxi. �

Exercises

6.6.12 Consider the following clutter with edges e1, e2, f1, f2 and colorclasses X1, X2, X3

X1 X2 X3

e1 = x1 y1e2 = y2 z2

X1 X2 X3

f1 = y1 z2f2 = x1 y2

Prove that this clutter is unmixed, is not Cohen–Macaulay, has a perfectmatching e1, e2 of Konig type, and the height of I(C) is two.

250 Chapter 6

6.6.13 Prove that the uniform admissible clutters with three color classes

X1 = {x1, . . . , xg}, X2 = {y1, . . . , yg}, X3 = {z1, . . . , zg}

are, up to permutation of variables, exactly the clutters with a perfectmatching ei = {xi, yi, zi} for i = 1, . . . , g such that all edges of C havethe form {xi, yj, zk}, with 1 ≤ i ≤ j ≤ k ≤ g.

6.6.14 [325] If C is a complete admissible uniform clutter, then the blockerC∨ of C is also a complete admissible uniform clutter.

6.7 Hilbert series of face rings

For a Stanley–Reisner ring K[Δ] there are formulas for its Hilbert functionand Hilbert series in terms of the combinatorial data of the complex Δ. IfI(C) is the edge ideal of a clutter C, its Hilbert series can be obtained fromthe so-called edge induced polynomial of C [193, 347].

Hilbert series with the fine and standard grading Let K be a field.Note that the polynomial ring R = K[x1, . . . , xn] can be endowed with afine Zn-grading as follows. For a = (a1, . . . , an) ∈ Zn, set

Ra =

{Kxa, if ai ≥ 0 for i = 1, . . . , n,0, if ai < 0 for some i.

Let I ⊂ R be an ideal generated by monomials. Since I is Zn-graded,the quotient ring R/I inherits the Zn-grading given by (R/I)a = Ra/Ia forall a ∈ Zn. In particular Stanley–Reisner rings have a fine grading.

Let M be a Zn-graded R-module. Each homogeneous componentMa ofM is an R0-module. Define the Hilbert function H(M,a) = �(Ma), providedthat the length �(Ma) of Ma is finite for all a, and call

F (M, t) =∑a∈Zn

H(M,a)ta

the Hilbert–Poincare series of M . Here t = (t1, . . . , tn), where the ti areindeterminates and ta = ta11 · · · tann for a = (a1, . . . , an) ∈ Zn.

By induction on n it follows that the polynomial ring R = K[x1, . . . , xn]with the fine grading has Hilbert–Poincare series:

F (R, t) =∑a∈Nn

ta =

n∏i=1

1

1− ti.

Let Δ be a simplicial complex with vertices X1, . . . , Xn, permitting anabuse of notation we also denote by Xi the residue class of xi in K[Δ].


Thus K[Δ] = K[X1, . . . , Xn]. The support of a ∈ Zn, denoted by supp(a),is defined as the set {Xi| ai > 0}.

Let Xa ∈ K[Δ] and let supp(a) = {Xi1 , . . . , Xim}. Since IΔ is generatedby square free monomials we have

Xa �= 0 ⇔ Xi1 · · ·Xim �= 0 ⇔ xi1 · · ·xim /∈ IΔ ⇔ supp(a) ∈ Δ.

Hence the non-zero monomials Xa form a K-basis of K[Δ]. Therefore

F (K[Δ], t) =∑a∈Nn

supp(a)∈Δ

ta =∑F∈Δ

∑a∈Nn

supp(a)=F

ta

Let F ∈ Δ. If F = ∅, then∑

supp a=F ta = 1, and if F �= ∅, then

∑a ∈ Nnsupp a⊂F

ta =∏xi∈F

1

1− ti=⇒

∑a ∈ Nnsupp a=F

ta =∏xi∈F

ti1− ti

.

Altogether we obtain that the expression for F (K[Δ], t) simplifies to

F (K[Δ], t) =∑F∈Δ

∏xi∈F

ti1− ti

, (6.8)

where the product over an empty index set is equal to 1.

Definition 6.7.1 Given a simplicial complex Δ of dimension d its f -vectoris the (d+ 1)-tuple:

f(Δ) = (f0, . . . , fd),

where fi is the number of i-faces of Δ. Note f−1 = 1.

To compute the Hilbert series of K[Δ], as a standard N-graded algebra,note that for i ∈ Z we have

K[Δ]i =⊕

a∈Zn |a|=iK[Δ]a

where |a| = a1 + · · · + an for a = (a1, . . . , an). Observe that the Hilbertseries of K[Δ] with the fine grading specializes to the Hilbert series of K[Δ]with the Z-grading, that is, if ti = t for all i, then

F (K[Δ], t) = F (K[Δ], t).

Thus, by Eq. (6.8), we obtain a formula for the Hilbert series of K[Δ]:

252 Chapter 6

Theorem 6.7.2 [395] If Δ is a simplicial complex and f(Δ) = (f0, . . . , fd)is its f -vector, then the Hilbert series of K[Δ] is given by

F (K[Δ], z) =

d∑i=−1

fizi+1

(1− z)i+1, d = dim(Δ).

The Hilbert function of K[Δ] can be read off from its Hilbert series:

Proposition 6.7.3 If Δ is a simplicial complex of dimension d and (fi) isits f -vector, then the Hilbert function of K[Δ] is:

H(K[Δ], j) =d∑i=0

(j − 1

i

)fi, for j ≥ 1 and H(K[Δ], 0) = 1.

A result of Kruskal–Katona

Kruskal and Katona [283] showed that (f0, f1, . . . , fd) ∈ Zd+1 is the f -vectorof some d-dimensional simplicial complex if and only if

0 < fi+1 ≤ f (i+1)i , 0 ≤ i ≤ d− 1,

where f(i+1)i is defined according to Eq. (5.7), cf. [395, Theorem 2.1, p. 55].

Simplicial complexes and their h-vectors Next we derive formulasfor the h-vector of K[Δ]. If Δ is Cohen–Macaulay, we present some nu-merical constraints for the h-vector which are central for applications ofcommutative algebra to combinatorics [391] (see Theorem 6.7.7).

Definition 6.7.4 The h-vector of a simplicial complex Δ, denoted by h(Δ),,is defined as the h-vector of the standard graded algebra K[Δ].

Theorem 6.7.5 [395, p. 58] Let Δ be a simplicial complex of dimension dwith h-vector (hi) and f -vector (fi). Then hk = 0 for k > d+ 1 and

hk =

k∑i=0

(−1)k−i(d+ 1− ik − i

)fi−1 for 0 ≤ k ≤ d+ 1 . (6.9)

Proof. The idea is to write the Hilbert series of K[Δ] in two ways. ByTheorems 6.7.2 and 5.1.4 there is a polynomial h(t) = h0 + h1t+ · · ·+ hrt

r

with integral coefficients so that h(1) �= 0 and satisfying

F (K[Δ], t) =d∑

i=−1

fiti+1

(1− t)i+1=

h(t)

(1 − t)d+1. (6.10)

Comparing the series yields the asserted equalities. �


Theorem 6.7.6 Let Δ be a simplicial complex of dimension d with h-vector(hi) and f -vector (fi). Then

fk−1 =

k∑i=0

(d+ 1− ik − i

)hi, for 1 ≤ k ≤ d+ 1, (6.11)

fd =∑d+1

i=0 hi, h1 = f0 − (d+ 1) and hd+1 = (−1)d∑d

i=−1(−1)ifi.

Proof. By the substitution t = z/(1 + z) in the equality

h0 + h1t+ · · ·+ hd+1td+1

(1− t)d+1=

d+1∑i=0

fi−1ti

(1− t)i ,

we obtain∑d+1

i=0 hizi(1 + z)d+1−i =

∑d+1i=0 fi−1z

i. Hence

fk−1 =

d+1∑i=0

(d+ 1− ik − i

)hi =

k∑i=0

(d+ 1− ik − i

)hi, for 1 ≤ k ≤ d+ 1.

From Eqs. (6.9) and (6.11) we derive the formulas for fd, h1 and hd+1. �

Computation of the h-vector of face rings

To compute the h-vector (of face rings) the following procedure of Stanley[393] can be used. Consider the 3-dimensional convex polytope P consistingof two solid tetrahedrons joined by a 2-face:

�x4�x3 �x2�x1

�x5��

��

��

��

� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

��

one has f(P) = (5, 9, 6). Write down the f -vector on a diagonal, and put a1 to the left of f0.

6

9

51

Complete this array constructing a table, by placing below a pair of consec-utive entries their difference, and placing a 1 on the left edge:

254 Chapter 6

6

9

51

41

531

2(1 2 1)h(P) =After completing the table the next row of differences will be the h-vector.Hence the boundary simplicial complex Δ = Δ(P) is 2-dimensional, K[Δ]has multiplicity 6, and IΔ = (x1x5, x2x3x4).

Now we present a result that plays a role in the proof of the upper boundconjecture for spheres [391].

Theorem 6.7.7 Let Δ be a simplicial complex of dimension d with n ver-tices. If K[Δ] is Cohen–Macaulay and K is an infinite field, then the h-vector of Δ satisfies

0 ≤ hi(Δ) ≤(i+ n− d− 2

i

)for 0 ≤ i ≤ d+ 1. (6.12)

Proof. Set S = K[Δ] = R/I, where R = K[x1, . . . , xn] and I = IΔ.By Lemma 3.1.28 there exists a regular homogeneous system of parametersθ = θ1, . . . , θd+1 for S such that each θi has degree one. Since A = S/(θ)Sis Artinian, in this case Theorem 5.2.5 says that hi(Δ) = H(A, i).

Note that S/(θ)S � R/I, where R = R/(θ) is a polynomial ring inn− d− 1 variables and I is the image of I in R. Therefore we have

hi(Δ) = H(A, i) = H(R/I, i) ≤ H(R, i) =

(i+ n− d− 2

i

). �

Example 6.7.8 Let Δ be a 2-dimensional complex with f(Δ) = (5, 6, 2).ThusK[Δ] is not Cohen–Macaulay because its h vector has a negative entry:

2

6

51

41

231

2(1 −1 0).h(Δ) =

Exercises

6.7.9 Let S = K[Δ] be the Stanley–Reisner ring of a simplicial complex Δover a field K and ϕ(t) the Hilbert polynomial of S. If χ(Δ) is the reducedEuler characteristic, prove:

(a) ϕ(i) = dimK(Si) for all i ≥ 1, and

(b) ϕS(0) = 1 if and only if χ(Δ) = 0.


6.7.10 Let Δ be a simplicial complex of dimension d and (f0, . . . , fd) itsf -vector. Prove that fd is equal to deg(K[Δ]), the degree of K[Δ].

6.7.11 Let S = K[Δ] be the Stanley–Reisner ring of a simplicial complexΔ with vertices x1, . . . , xn and A = S/(x21, . . . , x

2n). Prove that Δ is pure if

and only if all the non-zero elements of the socle of A have the same degree.

6.7.12 If Δ is a connected simplicial complex having all its vertices of degreetwo, then Δ is a cycle.

6.7.13 [23] Let R = K[x1, . . . , xn] be a polynomial ring over a field K. If Iis a monomial Gorenstein ideal and rad (I) = (x1, . . . , xn), prove that I isgenerated by xa11 , . . . , x

ann for some a1, . . . , an in N+.

6.7.14 Give an example of a monomial Gorenstein ideal which is not acomplete intersection.

6.7.15 Let R = K[x1, . . . , xn] be a polynomial ring over a field K andlet Δ be a simplicial complex with vertices x1, . . . , xn. Associated to theStanley–Reisner ideal IΔ define another ideal

I ′(Δ) = (IΔ, x1y1, . . . , xnyn) ⊂ K[x1, . . . , xn, y1, . . . , yn],

and denote by S(Δ) the Stanley–Reisner complex corresponding to thisideal I ′(Δ). Prove the following:

(a) The simplicial complex S(Δ) is pure shellable.

(b) The f -vector of Δ is related to the h-vector of S(Δ) by

fi(Δ) = hi+1(S(Δ)) for all i.

6.7.16 Let k and n be two positive integers. A k-partition of n is an elementα = (α1, . . . , αk) in Nk+ so that n = α1 + · · · + αk. Prove that the number

of k-partitions of n is equal to(n−1k−1

).

6.8 Simplicial spheres

Let Δ be a finite simplicial complex with vertices x1, . . . , xn and let ei bethe ith unit vector in Rn. Given a face F ∈ Δ, we set

|F | = conv{ei|xi ∈ F},

where “conv” is the convex hull. Define the geometric realization of Δ as

|Δ| =⋃F∈Δ

|F |.

Then |Δ| is a topological space with the induced usual topology of Rn. Notethat there is a canonical isomorphism Hi(Δ, A) ∼= Hi(|Δ|;A) for all i, whereHi are the corresponding reduced homology modules.

256 Chapter 6

Definition 6.8.1 A q-simplex is a polytope generated by a set of q + 1affinely independent points. A polytope is simplicial if everyone of its properfaces is a simplex.

Let P be a polytope of dimension d+1 and denote the number of faces ofdimension i of P by fi, thus f0 is the number of vertices of P . The f -vectorof P is the vector f(P) = (f0, . . . , fd) and we set f−1 = 1. The boundarycomplex Δ(P) of P is by definition the “abstract simplicial complex” whosevertices are the vertices of P and whose faces are those sets of vertices thatspan a proper face of P , thus Δ(P) has dimension d. Notice that P andΔ(P) have the same f -vector and accordingly one defines the h-vector of Pas the h-vector of Δ(P).

The geometric realization |Δ(P)| is therefore homeomorphic to a sphereSd of dimension d. This suggests the following more general concept.

Definition 6.8.2 A simplicial complex Δ of dimension d is a simplicialsphere if |Δ| � Sd, in this case Δ is said to be a triangulation of Sd.

Remark 6.8.3 There are simplicial d-spheres with d = 3 and n = 8 whichare not the boundary complex of a simplicial (d+ 1)-polytope [393]. More-over it follows from a result of Steinitz (a graph is the 1-skeleton of a 3-polytope in R3 if and only if it is a 3-connected planar graph) that such anexample does not exist for d = 2; see [438, Theorem 4.1].

Theorem 6.8.4 [65, Chapter 5] Let Δ be a simplicial complex of dimensiond. If |Δ| ∼= Sd, then

Hi(lkF ;K) ∼={K for i = dim(lk(F )),0 otherwise.

Proposition 6.8.5 If Δ is a simplicial sphere, then Δ is Cohen–Macaulayand the following equality holds

χ(lk F ) = (−1)dim lk F for F ∈ Δ.

Proof. If follows from Reisner’s criterion and Theorem 6.8.4. �

Remark 6.8.6 If P is a simplicial d-polytope, then Δ(P) is a simplicialsphere of dimension d − 1. Hence, from Proposition 6.8.5, we obtain theEuler’s formula

χ(P) = χ(Δ(P)) =d−1∑i=−1

(−1)ifi = (−1)d−1 = −1 + χ(P),

where f = (f0, . . . , fd−1) is the f -vector of P . The Euler’s formula is validfor any d-polytope, not just the simplicial ones (see [57, Theorem 16.1]).


Corollary 6.8.7 If Δ is a simplicial sphere, then K[Δ] is Gorenstein.

Proof. It follows from [395, Theorem 5.1] and Theorem 6.8.4. �

Theorem 6.8.8 If Δ is a simplicial sphere of dimension d, then its h-vector satisfies hi = hd+1−i for i = 0, . . . , d+ 1.

Proof. Since K[Δ] is a Gorenstein ring, its h-vector is symmetric accordingto Proposition 5.3.9 and Corollary 5.3.10. �

Remark 6.8.9 If Δ is a simplicial sphere, then hi = hd+1−i and one canrecover Euler’s formula (make i = 0 and use Theorem 6.7.6). Those relationsare the Dehn–Sommerville equations.

Corollary 6.8.10 [314] If h = (h0, . . . , hd+1) is the h-vector of a simplicialpolytope of dimension d+ 1, then hi = hd+1−i for 0 ≤ i ≤ d+ 1.


The next two conjectures were raised around 1993 in connection with theproblem of bounding the Betti numbers of graded Gorenstein ideals [318].

Conjecture 6.8.11 Let Δ be a simplicial sphere and I the Stanley–Reisnerideal of Δ. If I has initial degree p and height g, then for 1 ≤ i < g the ithBetti number βi of K[Δ] satisfies:

βi ≤(p+ g − 1

p+ i− 1

)(p+ i− 2

i− 1

)+

(p+ g − 1

i

)(p+ g − i− 2

p− 1

)−(g

i

)(p+ g − 2

p− 1

).

Conjecture 6.8.12 Let Δ be a simplicial sphere and I its Stanley–Reisnerideal. If I has initial degree p and height g, then

μ(I) ≤(p+ g − 1

g − 1

)−(p+ g − 3

g − 1

).

Exercises

6.8.13 If Δ is a simplicial sphere on n vertices whose geometric realization|Δ| is isomorphic to the unit circle S1, then

(a) Δ is a cycle of length.

(b) μ(IΔ) =(n2

)− n for n ≥ 4.

258 Chapter 6

6.9 The upper bound conjectures

In this section we present the main steps in the proof of the upper boundconjecture for simplicial spheres. An excellent reference for a detailed proofof this conjecture is [65].

Let P be a polytope of dimension d + 1 and let fi = fi(P) be thenumber of faces of dimension i of P . In 1893 Poincare proved the Eulercharacteristic formula

d∑i=0

(−1)ifi = 1 + (−1)d, (6.13)

see [57, Theorem 16.1]. In [393] there are some historical comments aboutthis formula. It follows from Proposition 6.8.5 that this formula also holdsfor simplicial spheres. Are there any optimal bounds for fi(P)? As it willbe seen there is a positive answer to this question which is valid in a moregeneral setting.

In order to formulate the upper bound theorem for simplicial spheresand the upper bound theorem for convex polytopes we need to introducesome results on cyclic polytopes , the reader is referred to [57] for a detaileddiscussion on this topic.

Consider the monomial curve Γ ⊂ Rd+1 given parametrically by

Γ = {(τ, τ2, . . . , τd+1)| τ ∈ R}.

A cyclic polytope, denoted by C(n, d+1), is the convex hull of any n distinctpoints in Γ such that n > d+ 1. The f -vector of C(n, d + 1) depends onlyon n and d and not on the points chosen, and dim C(n, d+ 1) = d+ 1.

The cyclic polytope C(n, d + 1) is simplicial and has the remarkableproperty that its f -vector satisfies:

fi(C(n, d+ 1)) =

(n

i+ 1

)for 0 ≤ i <

⌊d+ 1

2

⌋, (6.14)

this means that C(n, d+1) has the highest possible number of i-faces wheni is within the specified rank. Hence for any polytope P of dimension d+1with n vertices we have

fi(P) ≤(

n

i+ 1

)= fi(C(n, d + 1)),

for all 0 ≤ i < ((d+1)/2). The upper bound conjecture for polytope (UBCPfor short) states that

fi(P) ≤ fi(C(n, d+ 1)) for 0 ≤ i ≤ d.


This conjecture was posed by Motzkin [327] in 1957 and proved by Mc-Mullen [314] in 1970. It was motivated by the performance of the simplexalgorithm in linear programming.

Proposition 6.9.1 If h = (h0, . . . , hd+1) is the h-vector of a cyclic polytopeC(n, d+ 1), then

hk =

(n− d+ k − 2

k

), 0 ≤ k ≤ ((d+ 1)/2).

Proof. Let 0 ≤ k ≤ ((d+ 1)/2). One has

hk =

k∑i=0

(−1)k−i(d+ 1− ik − i

)fi−1 =

k∑i=0

(−1)2(k−i)(k − d− 2

k − i

)(n

i

)=

(n− d+ k − 2

k

).

The second equality follows from Exercise (6.9.5). �

Since a cyclic polytope C(n, d + 1) is simplicial of dimension d + 1 itsboundary complex Δ(C(n, d + 1)) is a d-sphere. In 1964 V. Klee pointedout that the upper bound conjecture for polytopes may as well be madefor simplicial spheres. This conjecture was proved by Stanley [391] in 1975using techniques from commutative and homological algebra. Next we provethe upper bound theorem for simplicial spheres .

Theorem 6.9.2 [391] Let Δ be simplicial complex of dimension d with nvertices. If |Δ| ∼= Sd, then fi(Δ) ≤ fi(Δ(C(n, d + 1))) for i = 0, . . . , d.

Proof. Consider the following four conditions:

(a) hi(Δ) ≤ hi(Δ(C(n, d+ 1))) for 0 ≤ i ≤ ((d+ 1)/2).(b) hi(Δ(C(n, d + 1))) = hd+1−i(Δ(C(n, d + 1))) for 0 ≤ i ≤ d+ 1.

(c) hi(Δ) ≤(i+n−d−2

i

)all i.

(d) hi(Δ) = hd+1−i(Δ) for 0 ≤ i ≤ d+ 1.

We claim that (a), (b) and (d) imply the theorem. To show it noticethat if ((d+ 1)/2) < i ≤ d+ 1, then 0 ≤ d+ 1− i ≤ ((d+ 1)/2) and

hi(Δ) = hd+1−i(Δ) ≤ hd+1−i(Δ(C(n, d+ 1))) = hi(Δ(C(n, d + 1))),

hence hi(Δ) ≤ hi(Δ(C(n, d + 1))) for 0 ≤ i ≤ d+ 1. Therefore

fk−1(Δ) =

d+1∑i=0

(d+ 1− ik − i

)hi(Δ)

≤d+1∑i=0

(d+ 1− ik − i

)hi(Δ(C(n, d + 1)))

= fk−1(Δ(C(n, d + 1))) for 1 ≤ k ≤ d+ 1,

260 Chapter 6

and the proof of the claim is completed.By Proposition 6.9.1 (c)⇒ (a), and by Lemma 6.8.10 (b) is satisfied due

to the fact that C(n, d + 1) is a simplicial polytope. Hence the theorem isreduced to prove (c) and (d). To complete the proof notice that (c) followsfrom Lemma 6.7.7, and (d) follows from Theorem 6.8.8. �

As a particular case of the upper bound theorem for spheres we obtainthe upper bound theorem for convex polytopes :

Theorem 6.9.3 [314] Let P be a convex polytope of dimension d with nvertices, then fi(P) ≤ fi(C(n, d)) for 0 ≤ i ≤ d− 1.

Proof. By pulling the vertices P can be transformed into a simplicialpolytope with the same number of vertices as P and at least as many facesof higher dimension; see [315] and [438, Lemma 8.24]. Hence, we mayassume that P is simplicial. Since P is simplicial, the boundary complexΔ(P) is a simplicial sphere, that is, |Δ(P)| ∼= Sd−1. Using Theorem 6.9.2we obtain

fi(P) = fi(Δ(P)) ≤ fi(Δ(C(n, d))) = fi(C(n, d)), for i = 0, . . . , d. �

Theorem 6.9.4 [57, Corollary 18.3] If P is a simplicial d-polytope with nvertices, then

fj(P) ≤ ϕd−j−1(d, n) (j = 1, . . . , d− 1),

where

ϕd−j−1(d, n) =

�d/2�∑i=0

(i

d− j − 1

)(n− d+ i− 1

i

)

+

�(d−1)/2�∑i=0

(d− i

d− j − 1

)(n− d+ i− 1

i

).

Exercises

6.9.5 If r ∈ R and k ∈ N, then(r

k

)= (−1)k

(−r + k − 1

k

).

6.9.6 [57, p. 79] Prove that a d-polytope P is simplicial if and only if eachfacet of P is a simplex.

Chapter 7

Edge Ideals of Graphs

In this chapter we give an introduction to graph theory and study algebraicproperties and invariants of edge ideals using the combinatorial structureof graphs. We present classifications of the following families: unmixedbipartite graphs, Cohen–Macaulay bipartite graphs, Cohen–Macaulay trees,shellable bipartite graphs, sequentially Cohen–Macaulay bipartite graphs.The invariants examined here include regularity, depth, Krull dimension,and multiplicity. Edge ideals are shown to have the persistence property.

For those readers interested in a comprehensive view of graph theoryitself, we recommend the books of Bollobas [48], Diestel [111], and Harary[208], which we take as our main references for the subject.

7.1 Graph theory

A graph G is an ordered pair of disjoint finite sets (V,E) such that E is asubset of the set of unordered pairs of V . The set V is the set of verticesand the set E is called the set of edges . The number of vertices of a graphis its order. An edge e = {u, v}, with u, v ∈ V , is said to join the verticesu and v. When working with several graphs it is convenient to write V (G)and E(G) for the vertex set and edge set of G, respectively. Sometimes wewill also write VG and EG for the vertex set and edge set of G, respectively.

Let G be a graph on the vertex set V . If e = {u, v} is an edge of G onesays that the vertices u and v are adjacent or neighboring vertices of G. Inthis case it is also usual to say that the edge e is incident with u and v. Thedegree of a vertex v in V denoted by deg(v), is the number of edges incidentwith v. A vertex with degree zero is called an isolated vertex .

Note that by definition a graph does not contain loops, a pair {v, v} (“anedge joining a vertex to itself”); neither does it contain a pair {u, v} thatoccurs several times (“that is, several edges joining the same two vertices”).

262 Chapter 7

If we allow these type of relations as edges we will call G a multigraph.Most results on graphs carry over to multigraphs in a natural way. Thereare areas in graph theory (such as plane duality [111]) where multigraphsarise more naturally than graphs. Terminology introduced earlier for graphscan be used correspondingly for multigraphs.

Given two graphs G and H , a mapping ϕ, from V (G) to V (H) is calleda homomorphism if {ϕ(u), ϕ(v)} ∈ E(H) whenever {u, v} ∈ E(G). Twographs G and H are isomorphic if there is a bijective map ψ from V (G) toV (H) such that {u, v} ∈ E(G) if and only if {ψ(u), ψ(v)} ∈ E(H). A maptaking graphs as arguments is called a graph invariant if it assigns equalvalues to isomorphic graphs. The number of vertices and the number ofedges are two simple examples of graph invariants.

Let H and G be two graphs, then H is called a subgraph of G if V (H) ⊂V (G) and E(H) ⊂ E(G). A subgraph H is called an induced subgraph if Hcontains all the edges {u, v} ∈ E(G) with u, v ∈ V (H). In this case H issaid to be the subgraph induced by V (H). An induced subgraph is denotedby H = G[V (H)] or H = 〈V (H)〉. It is also denoted by H = GV (H). Aspanning subgraph is a subgraph H of G containing all the vertices of G.

A walk of length n in G is an alternating sequence of vertices and edges,written as w = {v0, z1, v1, . . . , vn−1, zn, vn}, where zi = {vi−1, vi} is the edgejoining the vertices vi−1 and vi. A walk may also be written {v0, . . . , vn}with the edges understood, or {z1, z2, . . . , zn} with the vertices understood.If v0 = vn, the walk w is called a closed walk . A path is a walk with all itsvertices distinct.

We say that G is connected if for every pair of vertices u and v thereis a path from u to v. Notice that G has a vertex disjoint decompositionG = ∪ri=1Gi where G1, . . . , Gr are the maximal (with respect to inclusion)connected subgraphs of G, the Gi’s are called the connected components ofG. A component is called even (resp. odd) if its order is even (resp. odd).

A cycle of length n is a closed path {v0, . . . , vn} in which n ≥ 3. A cycleis even (resp. odd) if its length is even (resp. odd). We denote by Cn thegraph consisting of a cycle with n vertices, C3 will be called a triangle, C4 asquare and so on. If all the vertices of G are isolated, G is called a discretegraph. A forest is an acyclic graph and a tree is a connected forest. Thecomplete graph Kn has every pair of its n vertices adjacent. For instance:

� x1�x2

K4

�x3�x4

��

��

�x1��x2��

x3��

x5�

��x4

C5

�x1�x2

�x3� �x5x4

��

��

Tree

A coloring of the vertices of a graph G is an assignment of colors to the

Edge Ideals of Graphs 263

vertices of G in such a way that adjacent vertices have distinct colors. Thechromatic number of a graph G, denoted by χ(G), is the minimal numberof colors in a coloring of G. There are algebraic methods to compute thechromatic number [165] (see Theorem 7.7.19).

A clique of a graph G is a set of vertices that induces a complete sub-graph. We will also call a complete subgraph of G a clique. A maximalclique of G is a clique which is maximal with respect to inclusion.

The clique number of G, denoted by ω(G), is the size of the largestcomplete subgraph of G. The clique number and the chromatic number arerelated by the inequality

ω(G) ≤ χ(G).A graph is called perfect if for every induced subgraph H , the chromatic

number of H is equal to the size of the largest complete subgraph of H , i.e.,ω(H) = χ(H) for every induced subgraph of G. This notion was introducedby Berge [25, Chapter 16]. An excellent reference for the theory of perfectgraphs is the book of Golumbic [191]; see also [93, 373].

A graph G is bipartite if its vertex set V (G) can be partitioned into twodisjoint subsets V1 and V2 such that every edge of G has one vertex in V1and one vertex in V2. The pair (V1, V2) is called a bipartition of G. If G isconnected such a bipartition is uniquely determined.

A bipartite graphG with bipartition (V1, V2) is called a complete bipartitegraph if {u, v} ∈ E(G) for all u ∈ V1 and v ∈ V2. If V1 and V2 have m andn vertices, respectively, we denote a complete bipartite graph by Km,n. Astar is a complete bipartite graph of the form K1,n. For instance:

�V1

V2 ��

��

��

��

��

�

��

� � �

K3,3

Definition 7.1.1 The distance d(u, v) between two vertices u and v of agraph G is defined to be the minimum of the lengths of all possible pathsfrom u to v. If there is no path joining u and v, then d(u, v) =∞.

Proposition 7.1.2 A graph G is bipartite if and only if all the cycles of Gare even. In particular every forest is a bipartite graph.

Proof. ⇒) Let (V1, V2) be a bipartition of G. If {v0, . . . , vn} is a cycle ofG, one may assume v0 ∈ V1. Since v1 ∈ V2, it follows at once that vi is inV1 if and only if i is even, thus n must be even.⇐) It suffices to prove that each connected component of G is bipartite.

Thus one may assume that G is connected. Pick a vertex v0 ∈ V . Set

V1 = {v ∈ V (G)|d(v, v0) is even} and V2 = V (G) \ V1.

264 Chapter 7

It follows that no two vertices of Vi are adjacent for i = 1, 2, otherwise Gwould contain an odd cycle. Therefore (V1, V2) is a bipartition of G and thegraph G is bipartite. �

If e is an edge, we denote by G\{e} the spanning subgraph of G obtainedby deleting e and keeping all the vertices of G. The removal of a vertex vfrom a graph G results in a subgraph G \ {v} of G consisting of all thevertices in G except v and all the edges not incident with v. For instance:

�x1�x2 G

e

�x3�x4

��

�� x2 G \ {x1}

�x3�x4�

��

� x1�x2 G \ {e}

�x3�x4

��

��

Given a set of vertices A, by G \A, we mean the graph formed from Gby deleting all the vertices in A, and all edges incident to a vertex in A.

Let G be a graph. A vertex v (resp. an edge e) of G is called a cutvertexor cutpoint (resp. a bridge) if the number of connected components ofG\{v}(resp. G \ {e}) is larger than that of G. A maximal connected subgraphof G without cut vertices is called a block . A graph G is 2-connected if|V (G)| > 2 and G has no cut vertices. Thus a block of G is either amaximal 2-connected subgraph, a bridge, or an isolated vertex. By theirmaximality, different blocks of G intersect in at most one vertex, which isthen a cutvertex of G. Therefore every edge of G lies in a unique block, andG is the union of its blocks.

Definition 7.1.3 A set of edges in a graph G is called independent or amatching if no two of them have a vertex in common.

Definition 7.1.4 Let A be a set of vertices of a graph G. The neighbor setof A, denoted by NG(A) or simply by N(A) if G is understood, is the setof vertices of G that are adjacent with at least one vertex of A.

In particular, if G is a graph and v ∈ V (G), the neighbor set of v,denoted by NG(v) or N(v), is the set of vertices of G adjacent to v. Noticethat if Cr is a cycle of a graph G, then Cr ⊂ NG(Cr).

Proposition 7.1.5 Let G be a bipartite graph with bipartition (V1, V2) andlet m, n be the number of vertices in V1, V2, respectively. If |A| ≤ |NG(A)|for all A ⊂ V1, then there are m independent edges in G.

Proof. By induction on m. If m = 1, then there is at least one vertex inV2 connected to V1, thus there is one independent edge. Assume m ≥ 2.

Case (I): Assume that |A| < |NG(A)| for all A ⊂ V1 with |A| < m.Let x1 ∈ V1 and y1 ∈ V2 be two adjacent vertices. Consider the subgraph


H = G \ {x1, y1} obtained from G by removing x1 and y1. Note thatfor every A ⊂ V1 \ {x1} one has NH(A) = NG(A) if y1 /∈ NG(A) andNH(A) = NG(A)\{y1} otherwise. Hence by induction there are independentedges z2, . . . , zm in H , which together with z1 = {x1, y1} yield the requirednumber of independent edges.

Case (II): Assume that |A| = |NG(A)| for some A ⊂ V1 with |A| < m.Set r = |A|. Consider the subgraph H = G \ (V1 \A). As NH(S) = NG(S)for all S ⊂ A, by induction there are independent edges z1, . . . , zr in H . Onthe other hand consider F = G \ (A ∪NG(A)). If S ⊂ V1 \A, then

|NG(A ∪ S)| ≥ |A ∪ S| = |S|+ |A|,

since NG(S ∪A) = NG(A)∪NF (S) and r = |NG(A)| we get |NF (S)| ≥ |S|.Applying induction there are independent edges w1, . . . , wm−r in F . Puttingaltogether it is seen that there are m independent edges in G, as required.This proof is due to Halmos and Vaughn. �

Corollary 7.1.6 Let G be a bipartite graph with bipartition (V1, V2). Ifthere is an integer p ≥ 0 such that |A| − p ≤ |NG(A)| for all A ⊂ V1, thenthere are m− p independent edges in G, where m = |V1|.

Proof. By Proposition 7.1.5 the assertion holds for p = 0. Consider thegraph F obtained from G by adding a new vertex y to V2 and joining everyvertex of V1 to y. Let A ⊂ V1, since NF (A) = NG(A) ∪ {y} one concludes|NF (A)| ≥ |A|−(p−1). Hence by induction there arem−p+1 independentedges in F , it follows that there are m− p independent edges in G. �

Let G be a graph with vertex set V . A subset C ⊂ V is a minimal vertexcover of G if: (i) every edge of G is incident with at least one vertex in C,and (ii) there is no proper subset of C with the first property. If C satisfiescondition (i) only, then C is called a vertex cover of G.

The following example illustrates the notion of a minimal vertex cover:

��x1��

C = {x1, x2, x4, x6}is a minimal vertex cover of G.

�x7��

x6��

��

��x2

��

��

x4�� x3�x5

G

It is convenient to regard the empty set as a minimal vertex cover for a graphwith all its vertices isolated. A set of vertices of G is called independent orstable if no two of them are adjacent. Notice the following duality: a setof vertices in G is a maximal independent set (with respect to inclusion) ifand only if its complement is a minimal vertex cover for G.

266 Chapter 7

Definition 7.1.7 Let G be a graph. The vertex covering number , denotedby α0(G), is the number of vertices in any smallest vertex cover. Thematching number , denoted by β1(G), is the number of edges in any largestindependent set of edges. The vertex independence number, denoted byβ0(G), is the number of vertices in any largest independent set of vertices.

In [26] the integer α0(G) (resp. β0(G)) is denoted by τ(G) (resp. α(G))and is called the transversal number (resp. stability number) of G.

Theorem 7.1.8 (Konig) If G is a bipartite graph, then β1(G) = α0(G).

Proof. Let (V1, V2) be a bipartition of G. Set r = β1(G) and m = |V1|.Since there are r independent edges in G one obtains α0(G) ≥ r.

There exists a subset A ⊂ V1 such that |NG(A)| ≤ |A|−(m−r), otherwiseif |NG(A)| > |A|− (m− r) for all A ⊂ V1, then by Corollary 7.1.6 one wouldhave r+ 1 independent edges which is impossible. Hence NG(A) ∪ (V1 \A)is a vertex cover of G with at most r elements, thus we get α0(G) ≤ r. �

A pairing by an independent set of edges of all the vertices of a graph Gis called a perfect matching or a 1-factor. Thus G has a perfect matching ifand only ifG has an even number of vertices and there is a set of independentedges containing all the vertices.

Next we present a criterion for the existence of a perfect matching.

Theorem 7.1.9 (Marriage theorem) Let G be a bipartite graph. Then thefollowing are equivalent

(a) G has a perfect matching.

(b) |A| ≤ |NG(A)| for all A ⊂ VG independent set of vertices.

Proof. Let (V1, V2) be a bipartition of G. Set r = β1(G). It is easyto see that (a) implies (b). To prove that (b) implies (a), we begin bynoticing that |V1| = |V2| because Vi is an independent set for i = 1, 2. ByKonig theorem there are z1, . . . , zr independent edges and a minimal vertexcover A of G with r elements. Hence zi ∩ A has exactly one vertex forany i and A is an independent set. Using that VG \ A is an independentset of vertices and from the equality NG(VG \ A) = A we conclude that|VG \A| ≤ |NG(VG \ A)| = |A|. It follows that |V1| = r, and thus z1, . . . , zryields a perfect matching. �

What happens if the graphG is not bipartite? The answer is given by thefollowing theorem of Tutte. A very readable and comprehensive referenceabout matchings in finite graphs is the book of Lovasz and Plummer [297].

Theorem 7.1.10 (Tutte; see [111, Theorem 2.2.1]) A graph G has a perfectmatching if and only if c0(G \S) ≤ |S| for all S ⊂ VG, where c0(G) denotesthe number of odd components (connected components with an odd numberof vertices) of G.


A directed graph or digraph D consists of a finite set V (D) of verticestogether with a prescribed collection E(D) of ordered pairs of distinct pointscalled edges or arrows. An oriented graph is a digraph having no-cycles oflength two. In other words an oriented graph is a graph together with anorientation of its edges. A tournament is a complete oriented graph. Anytournament has a spanning directed path [14, Theorem 1.4.5].

Exercises

7.1.11 Let G be a graph with vertex set VG = {x1, . . . , xn} and edge setEG, prove Euler’s identity: 2|EG| =

∑ni=1 deg(xi).

7.1.12 If G is a connected bipartite graph, prove that G has a uniquebipartition.

7.1.13 If u, v are two vertices at maximum distance in a connected graphG, then u, v are not cutpoints.

7.1.14 If G is a regular bipartite graph (all vertices have the same degree),then G has a perfect matching.

7.1.15 An edge cover in a graph G is a set of edges collectively incidentwith each vertex of G. The edge covering number of G denoted by α1(G),is the number of edges in any smallest edge cover in G. Prove:

(a) If G has no isolated vertices, then β1(G) + α1(G) = |VG|.(b) For any graph G, β1(G) ≤ α0(G). If G is bipartite, α1(G) = β0(G).

7.1.16 Let G be a bipartite graph with minimum degree r. Then G is theunion of r edge disjoint edge covers.

7.1.17 Define a 2-coloration of a set A as a surjective mapping from Aonto a set of two elements. A 2-coloration of a graph is a 2-coloration ofits vertex set. Prove that a graph is connected if for any 2-coloration of thegraph there exists a heterochromatic edge.

7.1.18 If G is a bipartite graph with bipartition (V1, V2), prove that themapping from V (G) to V (K2) that sends all the vertices from Vi to vertexi is a homomorphism from G to K2.

7.1.19 Prove that the clique number and the chromatic number of a graphG satisfy ω(G) ≤ χ(G).

7.1.20 Let G be a connected perfect graph with vertex set V (G). Thenthere are cliques K1, . . . ,Ks of G such that V (G) is the disjoint union ofK1, . . . ,Ks and α0(G) = α0(G[K1]) + · · ·+ α0(G[Ks]).

7.1.21 Prove that any tournament has a spanning directed path.

268 Chapter 7

7.2 Edge ideals and B-graphs

In this section we introduce edge ideals and edge rings of graphs. We willcompare several families of graphs and examine some of their numericalinvariants. Some bounds for the Krull dimension are given.

Let G be a graph with vertices x1, . . . , xn and let R = K[x1, . . . , xn]be a polynomial ring over a field K. Here we are permitting an abuse ofnotation by using xi to denote a vertex of G and also a variable of R.

Definition 7.2.1 The edge ideal I(G) associated to the graph G is theideal of R generated by the set of all square-free monomials xixj such thatxi is adjacent to xj . If all the vertices of G are isolated we set I(G) = (0).The ring R/I(G) is called the edge ring of G.

Edge ideals are algebraic representations of graphs. These ideals oc-cur in computational chemistry [291], physics [51], and combinatorics [395].The non-zero edge ideals are precisely those ideals generated by square-freemonomials of degree two.

A prime example of an edge ideal comes from the theory of posets:

Definition 7.2.2 Given a poset P with vertex set X = {x1. . . . , xn}, itsorder complex , denoted by Δ(P), is the simplicial complex on X whosefaces are the chains (linearly ordered sets) in P , thus

K[Δ(P)] = K[X ]/(xixj |xi �∼ xj)

is the Stanley–Reisner ring of Δ(P). Here xi �∼ xj means that xi is notcomparable to xj .

The next result establishes a one-to-one correspondence between theminimal vertex covers of G and the minimal primes of I(G).

Proposition 7.2.3 A prime ideal p ⊂ R is a minimal prime of I(G) if andonly if p is generated by a minimal vertex cover of G.

Proof. It follows at once from Lemma 6.3.37 and Theorem 6.1.4. �

Corollary 7.2.4 If G is a graph, then the vertex covering number α0(G)is equal to the height of the edge ideal I(G).

Proof. It follows at once from Proposition 7.2.3. �

Corollary 7.2.5 Let G be a graph with n vertices and let I(G) be its edgeideal. Then n = α0(G) + β0(G) = ht(I(G)) + dim R/I(G).

Proof. It follows from Corollary 7.2.4. �


Definition 7.2.6 Let I ⊂ R be a graded ideal. The quotient ring R/I iscalled Cohen–Macaulay (C–M for short) if depth(R/I) = dim(R/I). Theideal I is called Cohen–Macaulay if R/I is Cohen–Macaulay.

Definition 7.2.7 A graph G is called Cohen–Macaulay over the field K ifR/I(G) is a Cohen–Macaulay ring.

Definition 7.2.8 A graph G is called unmixed graph or well-covered if anytwo minimal vertex covers of G have the same cardinality.

Proposition 7.2.9 If G is a Cohen–Macaulay graph, then G is unmixed.

Proof. By Theorem 6.1.4 I(G) is the intersection of its minimal primes.Therefore I(G) is an unmixed ideal by Corollary 3.1.17, thus G is unmixedby the correspondence between minimal covers and minimal primes. �

Definition 7.2.10 A vertex v is critical if α0(G \ {v}) < α0(G). A graphG is called vertex-critical if all its vertices are critical.

Proposition 7.2.11 Let G be a graph. If α0(G \ {v}) < α0(G) for somevertex v, then α0(G \ {v}) = α0(G)− 1.

Proof. Set r = α0(G \ {v}) and note that r + 1 ≤ α0(G). Pick a minimalvertex cover C of G \ {v} with r vertices. Since α0(G \ {v}) < α0(G), v isnot an isolated vertex and NG(v) �= C. Thus C ∪ {v} is a minimal vertexcover of G. Hence α0(G) ≤ r + 1 and one has the asserted equality. �

Proposition 7.2.12 Let G be a graph without isolated vertices. Then G isvertex critical if and only if any of the following equivalent conditions hold.

(a) Any vertex is in some minimal vertex cover with α0(G) vertices.

(b) α0(G) = α0(G \ {v}) + 1 for any vertex v of G.

(c) β0(G) = β0(G \ {v}) for any vertex v of G.

Proof. It follows readily from Proposition 7.2.11. �

Proposition 7.2.13 Let I be an ideal of a catenary local domain (R,m)and let x ∈ m. If ht (I) < ht (I, x), then ht (I, x) = ht (I) + 1.

Proof. Let p be a minimal prime of I such that ht (I) = ht (p). Note thatthe image of (x) + p in R/p is a principal ideal, hence by Krull’s principalideal theorem (see Theorem 2.3.16) there is a minimal prime ideal q of (x)+psuch that ht (q/p) ≤ 1. As R is a catenary domain one has

ht (I, x)− ht (I) ≤ ht (q)− ht (p) = ht (q/p) ≤ 1.

Therefore ht (I, x) ≤ ht (I) + 1, and the required equality follows. �

270 Chapter 7

Definition 7.2.14 A family F consisting of independent sets of a graphG is said to be a maximal independent cover of G if V (G) =

⋃C∈F C and

|C| = β0(G) for all C ∈ F . A graph G without isolated vertices having amaximal independent cover is called a B-graph [26].

Proposition 7.2.15 [26] If G is a B-graph, then G is vertex critical.

Proof. Let v be a vertex of G and let e = {v, w} be an edge of G. Since wis contained in an independent set A of G with β0(G) vertices, then v is inV (G) \A. Hence α0(G \ {v}) < α0(G). �

Proposition 7.2.16 [26] If G is an unmixed graph, then G has a maximalindependent cover.

Proof. The result follows readily because any two maximal independentsets have the same cardinality. �

Let us summarize some of the results obtained thus far. The followingimplications hold for any graph without isolated vertices:

C–M =⇒ unmixed =⇒ B-graph =⇒ vertex-critical

Proposition 7.2.17 If G is a vertex critical graph and G\{v} is not vertexcritical for all v ∈ V (G), then G has a maximal independent cover.

Proof. We use induction on the number of vertices of G. If G has twovertices or α0(G) = 1, then for both cases the assumptions on G imply thatG has a single edge and the result is clear. Thus one may assume G has atleast three vertices and α0(G) ≥ 2.

Assume G has no maximal independent cover. Let C1, . . . , Cr be theset of all minimal vertex covers of G with α0(G) vertices. Pick a vertexv ∈ V (G) such that v is in Ci for all i. By hypothesis V (G) = ∪ri=1Cibecause G is vertex-critical (Proposition 7.2.12). Hence degG(x) ≥ 2 for allx ∈ V (G) adjacent to v, indeed if degG(x) = 1, then x ∈ Cj for some j andCj \ {x} is a vertex cover of G which is impossible. Thus G \ {v} has noisolated vertices. Since α0(G \ {v}) is equal to α0(G)− 1, the set Ci \ {v} isa minimal vertex cover of G \ {v} with α0(G \ {v}) vertices for all i. Thus,from the equality V (G) \ {v} = (C1 \ {v}) ∪ · · · ∪ (Cr \ {v}) we concludethat G \ {v} is vertex critical, a contradiction. �

Remark 7.2.18 Let G be a graph and let A be the intersection of allminimal vertex covers of G with α0(G) vertices. Note that G has a maximalindependent cover if and only if A = ∅.

Proposition 7.2.19 Let G be a graph and let A be the intersection of allthe minimal vertex covers of G with α0(G) vertices. If G is vertex critical,then the graph G \ A is vertex-critical, has a maximal independent cover,and β0(G) = β0(G \A).


Proof. Let A = {v1, . . . , vs}. If A = ∅, then G has a maximal independentcover and there is nothing to prove. Assume A �= ∅. By the argumentsin the proof of Proposition 7.2.17 the graph G \ {v1} is vertex-critical andβ0(G) = β0(G\{v1}). Since the intersection of all the minimal vertex coversof G \ {v1} with α0(G \ {v1}) vertices is equal to A \ {v1}, the result followsby induction. �

Theorem 7.2.20 [186] If G is a B-graph, then β0(G) ≤ α0(G).

Let G be a graph. In what follows ΔG is the independence complex ofG and K[ΔG] is the Stanley–Reisner ring of ΔG over a field K.

Theorem 7.2.21 (Erdos–Gallai [144]) If G is a vertex-critical graph withn vertices, then β0(G) = dimR/I(G) = dimK[ΔG] ≤

⌊n2

⌋.

Proof. By induction on n. The case n = 2 is clear. Assume n ≥ 3.One may assume that G has no maximal independent cover, otherwise theresult follows directly from Theorem 7.2.20 and α0(G) + β0(G) = n. ByProposition 7.2.17 there is v ∈ V (G) such that G \ {v} is vertex-critical.Thus by the induction hypothesis β0(G \ {v}) ≤ (n− 1)/2. Since β0(G) isβ0(G \ {v}) (see Proposition 7.2.12), the assertion follows. �

Theorem 7.2.22 Let G be a graph and let A be the intersection of all theminimal vertex covers of G with α0(G) vertices. If G is vertex critical, thendimK[ΔG] ≤ n− |A|/2.

Proof. It follows from Proposition 7.2.19 and Theorem 7.2.21. �

Example 7.2.23 Given n, k positive integers such that n ≥ 2(k− 1), let Gbe the following graph G on n vertices.

��

�

�

�

�

�

�

�

�

�

� �

� ��

��

��

��

��

��

��

��

��

��

��

��

x1 x2 x3 xk−2 xk−1

xk xk+1 xk+2 x2k−3 x2k−2

.

xn

xn−1

xn−2 x2k+1

x2k

x2k−1

Kn−2k+2

G

Using Proposition 7.3.2 it follows that G is Cohen–Macaulay. Notice thatβ0(G) = k. In particular if n = 2k or n = 2k + 1 one has β0(G) = (n2 ).

Lemma 7.2.24 If G is a bipartite graph with n vertices, then α0(G) ≤ (n2 ).

272 Chapter 7

Proof. Let (V1, V2) be a bipartition of G. Since n = |V1| + |V2|, then|Vi| ≤ (n2 ) for some i. Thus α0(G)) ≤ n

2 . �

Example 7.2.25 Let G be the graph below and let I be its edge ideal.

�

��

��

��

��

��

��

��

��

��

x1

x2

x3 x6x4 x5

Using the following simple procedure for Macaulay2

R=ZZ/101[x1,x2,x3,x4,x5,x6]

I=ideal(x1*x2, x1*x3, x1*x4, x1*x5, x1*x6, x2*x3,

x2*x4, x2*x5, x2*x6, x3*x4, x4*x5, x5*x6)

primaryDecomposition I

we obtain that the minimal vertex covers of G are:

{x6, x5, x4, x3, x1}, {x6, x5, x4, x3, x2},

{x5, x3, x2, x1}, {x5, x4, x2, x1}, {x6, x4, x2, x1}.

The graph G is vertex critical, β0(G) = 2, and α0(G) = 4. Thus the setA = {x1, x2} is the intersection of the minimal vertex cover of G with fourvertices. Hence the bound in Theorem 7.2.22 gives β0(G) ≤ 2 and theErdos–Gallai bound gives β0(G) ≤ 3.

Exercises

7.2.26 If I is the ideal (x1x2x3 − 1) ∩ (x1, x2, x3) of the polynomial ringK[x1, x2, x3], then I = (x1x2x

23 − x3, x1x22x3 − x2, x21x2x3 − x1), ht(I, xi) is

strictly greater than ht(I) + 1 and V (I, xi) = {0} for all i.

7.2.27 Let C be a clutter with vertex set X = {x1, . . . , xn} and let I bethe ideal I = I(C) + (x21, . . . , x

2n) ⊂ R, where R = K[X ] is a polynomial

ring over a field K. Prove the following: (a) the standard monomials ofR/I are in one-to-one correspondence with the independent sets of C, (b)the Hilbert series of R/I is a polynomial a0 + a1t + · · · + adt

d of degreed = β0(C), where ai is the number of independent sets of C of size i. Thispolynomial is called the independence polynomial of C.

7.2.28 Let G be a graph. Construct a graph H by appending an isolatedvertex x to G. Prove that R/I(G)[x] = R[x]/I(H), α0(G) = α0(H) andβ0(H) = β0(G) + 1.


7.2.29 Let G be a graph. If e is an edge of G such that α0(G\{e}) < α0(G),then α0(G) = α0(G \ {e}) + 1.

7.2.30 A graph G is edge-critical if α0(G \ {e}) < α0(G) for all e ∈ E(G).If G is an edge-critical graph, then G has a maximal independent cover.

7.2.31 [145] If G is an edge-critical graph, then q ≤ (α0(G) + α0(G)2)/2

where q is the number of edges of G.

7.2.32 If G is a graph, then there is a spanning subgraph H of G such thatH is edge-critical and β0(H) = β0(G).

7.2.33 Prove that every edge-critical bipartite graph is unmixed.

7.2.34 Prove that a cycle with nine vertices is an edge-critical graph whichis not unmixed.

7.2.35 Let G be a graph with q edges. If G has a maximal independentcover, then q ≤ α0(G)

2.

7.2.36 Prove that the complete bipartite graph G = Kg,g is an unmixedgraph with α0(G)

2 edges and has a maximal independent cover.

7.2.37 Let G be a graph with n vertices. If G is an unmixed graph withoutisolated vertices such that 2α0(G) = n, then G has a perfect matching.

7.2.38 Prove that the following graph is neither edge-critical nor vertex-critical, is not unmixed, and {x1, x4, x6, x7}, {x2, x3, x5, x6} is a maximalindependent cover.

��

��

��

��

��

��

��

x4x3x2 x5x6

x1

x7

G

isolated vertex

7.2.39 (N. Terai) Let R = K[a, . . . , j] be a polynomial ring on 10 variablesover a field K and let

I = (abc, abd, acd, bcd, abe, acf, adg, bch, bdi, cdj)

be the edge ideal of the clutter which consists of: the boundary of a tetrahe-dron (abc, abd, acd, bcd) and a new triangle on each edge of the tetrahedronusing a new vertex (e.g., ab → abe). Prove that this ideal is unmixed ofheight 3 (in fact Cohen–Macaulay).

274 Chapter 7

7.2.40 Let G be a bipartite graph with n vertices. If G has no isolatedvertices and G has a maximal independent cover, then n is an even integer.

7.2.41 Prove that the graph G shown below is vertex-critical, α0(G) = 3,β0(G) = 2, and G does not have a maximal independent cover.

� �

��

��

��

��

��

x1 x2 x3

x4 x5

7.2.42 (Dilworth’s theorem) Using Konig’s theorem, prove that in any finiteposet the cardinality of any largest antichain equals the cardinality of anysmallest chain partition.

7.2.43 Prove that in a finite poset the cardinality of any longest chainequals the cardinality of any smallest antichain partition.

7.3 Cohen–Macaulay and chordal graphs

In this section we study Cohen–Macaulay graphs, chordal graphs, and theCohen–Macaulay property of the ideal of covers of a graph. A classificationof Cohen–Macaulay trees is presented.

Constructions of Cohen–Macaulay graphs We begin by showing howlarge classes of Cohen–Macaulay graphs can be produced and give someobstructions for a graph to be Cohen–Macaulay.

� w� z��

x2�

x1�xk

First construction Let H be a graph with vertexset V (H) = {x1, . . . , xn, z, w} and J its edge ideal.Assume that z is adjacent to w with deg(z) ≥ 2 anddeg(w) = 1. We label the vertices of H such thatx1, . . . , xk, w are the vertices of H adjacent to z, as

shown in the figure. The next two results describe how the Cohen–Macaulayproperty of H relates to that of the two subgraphs G = H \ {z, w} andF = G \ {x1, . . . , xk}. One has the equalities

J = (I, x1z, . . . , xkz, zw) and (I, x1, . . . , xk) = (L, x1 . . . , xk),

where I and L are the edge ideals of G and F , respectively.Assume H is unmixed with height of J equal to g + 1. Since z is not

isolated, there is a minimal prime p over I containing {x1, . . . , xk} andsuch that ht(I) = ht(p) = g. It is not difficult to prove that k < n anddeg(xi) ≥ 2 for i = 1, . . . , k.


Proposition 7.3.1 [416] If H is a Cohen–Macaulay graph, then F and Gare Cohen–Macaulay graphs.

Proof. Set A = K[x1, . . . , xn] andR = A[z, w]. By Proposition 3.1.23 thereexists a homogeneous system of parameters {f1, . . . , fd} for A/I, wherefi ∈ A+ for all i. Because of the hypothesis and the equalities

z(z − w) + zw = z2 and w(w − z) + zw = w2,

the set {f1, . . . , fd, z−w} is a regular system of parameters for R/J . Hencef1, . . . fd is a regular sequence on R/I, that is, G is Cohen–Macaulay. Toshow the corresponding property for F we consider the exact sequence

0 −→ R/(I, x1, . . . , xk, w)[−1] z−→ R/Jψ−→ R/(I, z) −→ 0

where the first map is multiplication by z and ψ is induced by a projection.The exactness of this sequence follows from Theorem 6.1.4. Taking depthswith respect to the maximal ideal of R, by the depth lemma one has

n− g + 1 ≤ depth R/(I, x1, . . . , xk, w), where g = ht(I).

Since (I, x1, . . . , xk, w) = (L, x1, . . . , xk, w), F is Cohen–Macaulay. �

Proposition 7.3.2 If F and G are Cohen–Macaulay graphs and x1, . . . , xkare in some minimal vertex cover of G, then H is a Cohen–Macaulay graph.

Proof. Consider the exact sequence of Proposition 7.3.1. Since the ends ofthis sequence have R-depth equal to dim(R/J), by the depth lemma H isa Cohen–Macaulay graph. �

Corollary 7.3.3 If G is Cohen–Macaulay and {x1, . . . , xk} is a minimalvertex cover of G, then H is Cohen–Macaulay.

Proof. Note that in this case F is C–M because I(F ) = (0). �

� z��

x2�

x1�xk

Second construction For the discussion of the sec-ond construction we change our notation. Let H bea graph on the vertex set V (H) = {x1, . . . , xn, z}.Let {x1, . . . , xk} be the vertices of H adjacent to z,as shown in the figure. Taking into account the first

construction one may assume deg(xi) ≥ 2 for i = 1, . . . , k and deg(z) ≥ 2.Setting G = H \ {z} and F = G \ {x1, . . . , xk}, notice that the ideals J , I,and L associated to H , G, and F , respectively are related by the equalitiesJ = (I, x1z, . . . xkz) and (I, x1, . . . , xk) = (L, x1, . . . , xk).

276 Chapter 7

Proposition 7.3.4 [416] If H is a Cohen–Macaulay graph, then F is aCohen–Macaulay graph.

Proof. We set R = K[x1, . . . , xn, z], A = K[x1, . . . , xn] and ht(J) = g + 1.The polynomial f = z − x1 − · · · − xk is regular on R/J because it isclearly not contained in any associated prime of J . Therefore there is asequence {f, f1, . . . , fm} regular on R/J so that {f1, . . . , fm} ⊂ A+, wherem = n− g − 1. Observe that {f1, . . . , fm} is in fact a regular sequence onA/I, which gives depth(A/I) ≥ n− g − 1. Next, we use the exact sequence

0 −→ R/(I, x1, . . . , xk)[−1] z−→ R/J −→ R/(I, z) −→ 0

and ht(I, x1, . . . , xk) = g + 1 to conclude that F is Cohen–Macaulay. �

Proposition 7.3.5 Assume that x1, . . . , xk are not contained in any min-imal vertex cover for G and ht(I, x1, . . . , xk) = ht(I) + 1. If F and G areCohen–Macaulay, then H is Cohen–Macaulay.

Proof. The assumption on {x1, . . . , xk} forces ht(J) = ht(I)+1. From theexact sequence

0 −→ R/(I, x1, . . . , xk)[−1] z−→ R/J −→ R/(I, z) −→ 0

we obtain that H is Cohen–Macaulay. �

Corollary 7.3.6 If G is Cohen–Macaulay and {x1, . . . , xk−1} is a minimalvertex cover for G, then H is Cohen–Macaulay.

Connected components of Cohen–Macaulay graphs An importantproperty of the family of Cohen–Macaulay graphs is their additivity withrespect to connected components.

Lemma 7.3.7 Let K[x] and K[y] be two polynomial rings over a field K.If I1 and I2 are graded ideals of K[x] and K[y], respectively, then

depth(K[x]/I1) + depth(K[y]/I2) = depth(K[x,y]/(I1 + I2)).

Proof. The formula is a direct consequence of Proposition 3.1.33 andTheorem 3.1.34. �

Proposition 7.3.8 A graph G is Cohen–Macaulay if and only if all itsconnected components are Cohen–Macaulay.

Proof. The result follows from Lemma 7.3.7. �


Whisker graphs in Cohen–Macaulay trees In this part we introducewhisker graphs, present a characterization of all Cohen–Macaulay trees, andconstruct a family of graphs containing all Cohen–Macaulay trees.

Proposition 7.3.9 Let R = K[x1, . . . , xn, y1, . . . , yn] be a polynomial ringover a field K and M ⊂ {(i, j)| 1 ≤ i < j ≤ n}. Then, the ideal

I = ({xiyi, yjy�| i = 1, . . . , n and (j, �) ∈M}),

is Cohen–Macaulay.

Proof. The polarization of (x2i , xjx�| 1 ≤ i ≤ n, (j, �) ∈ M) is equal to I.Hence, the result follows applying Proposition 6.1.6. �

Definition 7.3.10 Let G0 be a graph on the vertex set Y = {y1, . . . , yn}and take a new set of variables X = {x1, . . . , xn}. The whisker graph orsuspension of G0, denoted by G0 ∪W (Y ), is the graph obtained from G0

by attaching to each vertex yi a new vertex xi and the edge {xi, yi}. Theedge {xi, yi} is called a whisker.

Proposition 7.3.11 Let G0 be a graph and let G = G0 ∪ W (Y ) be its

whisker graph. Then the multiplicity of R/I(G) is∑di=−1 fi, where fi is the

number of i-faces of the Stanley–Reisner complex of I(G0).

Proof. Let Y = {y1, . . . , yn} be the vertex set of G0. By Proposition 7.3.9,the set of linear forms {yi − xi}ni=1 is a regular system of parameters forR/I(G). The reduction of R/I(G) modulo these forms is

S = K[Y ]/(y21 , . . . , y2n, I(G0)),

whose K-vector space basis is formed by the standard monomials, that is,by the faces of ΔI(G0). To complete the proof notice that the degree ofR/I(G) is the degree of S by Corollary 5.1.10. �

Proposition 7.3.12 Let G0 and G1 be two copies of the same graph ondisjoint sets of vertices {y1, . . . , yn} and {x1, . . . , xn}, respectively. If G∨

1 isthe clutter of minimal vertex covers of G1 and

L = ({xi1 · · ·xis | {xi1 , . . . , xis} ∈ E(G∨1 )}),

then ((z) : I(G)) = (z, L), where z = {xiyi}ni=1 and G = G0 ∪W (Y ).

Proof. Take {xi1 , . . . , xir} ∈ E(G∨1 ) and yky� ∈ I(G0). Since is ∈ {k, �}

for some s we obtain xi1 · · ·xiryky� ∈ (z). This shows (z, L) ⊂ ((z) : I(G)).Conversely, assume M is a monomial in ((z) : I(G)) \ (z). Let {yk, y�} beany edge in G0. Since Myky� = xtytM1, either xk divides M or x� dividesM ; in either case, M ∈ L. �

278 Chapter 7

Corollary 7.3.13 The type of K[X,Y ]/I(G0 ∪W (Y )) is |E(G∨0 )|.

Example 7.3.14 Let I(G0) = (y1y3, y1y4, y2y4, y2y5, y3y5, y3y6, y4y6) andlet G be the whisker graph of G0. The ideal ((z) : I(G)) is generated by

{xiyi}6i=1 ∪ {x2x3x4, x3x4x5, x1x2x5x6, x1x2x3x6, x1x4x5x6}.

The last five monomials correspond to the minimal primes of I(G0), thetype of R/I(G) is 5 and its multiplicity is 17.

Lemma 7.3.15 Let G be a tree. If v and w are two adjacent vertices ofdegree at least two, then there is a minimal vertex cover of G containingboth v and w.

Proof. Let x be the edge joining v and w. Since G \ {x} has exactly twocomponents, say G1 and G2, there are minimal vertex covers A and B forG1 and G2, respectively, so that v ∈ A and w ∈ B. Therefore A ∪B is therequired cover for G. �

Theorem 7.3.16 [416] Let G be a tree. Then, G is Cohen–Macaulay ifand only if |V (G)| ≤ 2 or 2 < |V (G)| = 2r and G has a perfect matching{x1, y1}, . . . , {xr, yr} so that deg(xi) = 1, deg(yi) ≥ 2 for i = 1, . . . , r.

Proof. ⇒) Let (V1, V2) be a bipartition of G. Since V1, V2 are minimalvertex covers of G, and G is unmixed, we have r = |V1| = |V2| = α0(G). ByTheorem 7.1.8 there are r independent edges. Therefore, we may assumeV1 = {zi}ri=1, V2 = {wi}ri=1, and {zi, wi} ∈ E(G) for i = 1, . . . , r. Tocomplete the proof it suffices to show that for each i either zi orwi has degreeone. Assume deg(zi) ≥ 2 and deg(wi) ≥ 2 for some i. By Lemma 7.3.15there is a minimal vertex cover for T , say A, containing zi and wi. Since{zj, wj}

⋂A �= ∅ for j �= i we conclude |A| ≥ r+1, which is a contradiction.

⇐) The sufficiency follows from Proposition 7.3.9. �

The significance of the notion of a whisker graph lies partly in the nextrestatement of Theorem 7.3.16.

Theorem 7.3.17 If G is a graph, then G is a Cohen–Macaulay tree if andonly if G = G0 ∪W (Y ) for some tree G0 with vertex set Y .

Corollary 7.3.18 If G is a tree, then G is Cohen–Macaulay if and only ifG is unmixed.

Proof. If G is Cohen–Macaulay, then G is unmixed by Proposition 7.2.9.If G is unmixed, using the proof above it follows that I(G) is a Cohen–Macaulay ideal of the form described in Proposition 7.3.9. �


Corollary 7.3.19 The only Cohen–Macaulay cycles are the triangle andthe pentagon.

Proof. Let Cn = {x0, x1, . . . , xn = x0} be a Cohen–Macaulay cycle oflength n. The cases n = 4, 6, and 7 can be treated separately, so we assumen ≥ 8. By Proposition 7.3.4 Cn \ {x0, x1, xn−1} is a Cohen–Macaulay pathof length n− 3 which is impossible by Theorem 7.3.16 �

The next result generalizes the Cohen–Macaulay criterion for trees givenin Theorem 7.3.16 and is a generalization of Faridi’s characterization ofunmixed simplicial trees [156].

Theorem 7.3.20 [325] Let C be a clutter with the Konig property and withno special cycles of length 3 or 4. The following conditions are equivalent :

(a) C is unmixed.

(b) There is a perfect matching e1, . . . , eg, g = ht I(C), such that ei hasa free vertex for all i, and for any two edges f1, f2 of C and for anyedge ei, one has that f1 ∩ ei ⊂ f2 ∩ ei or f2 ∩ ei ⊂ f1 ∩ ei.

(c) R/I(C) is Cohen–Macaulay.

(d) ΔC is a pure shellable simplicial complex.

Proof. Using Lemma 6.5.7, Theorems 6.5.18, and 6.5.21, and Proposi-tion 6.5.19 it follows readily that conditions (a) and (b) are equivalent.Since (a) is equivalent to (b), from Theorem 6.5.20 we get that (b) implies(d). That (d) implies (c) and (c) implies (a), were shown in Theorem 6.3.23and Corollary 6.3.7. �

Simplicial complexes of edge ideals Given a graph G on the vertexset V one defines a simplicial complex Simp(G) on the same set of vertices:

Simp(G) = {F | ∃H a subgraph of G such that F = V (H) and H � Kr},

Conversely a simplicial complex Δ defines a graph Skel(Δ), the 1-skeletonof Δ, on the same set of vertices:

Skel(Δ) = {F ∈ Δ| dim(F ) ≤ 1}.

In general one has Δ ⊂ Simp(Skel(Δ)).

Proposition 7.3.21 Let Δ be a simplicial complex and IΔ its Stanley–Reisner ideal. Then Δ = Simp(Skel(Δ)) if and only if IΔ is generated bysquare-free monomials of degree two.

280 Chapter 7

Proof. Let V = {x1, . . . , xn} be the vertex set of Δ. Assume IΔ is gener-ated by square-free monomials of degree two. Take F ∈ Simp(Skel(Δ)), forsimplicity set F = {x1, . . . , xr}, that is, {xi, xj} ∈ Δ for all 1 ≤ i < j ≤ r.If F �∈ Δ, then x1 · · ·xr is in IΔ, and by assumption xixj is in IΔ for some1 ≤ i < j ≤ r. Hence {xi, xj} �∈ Δ, which is a contradiction. This showsΔ = Simp(Skel(Δ)). The converse also follows readily. �

Simplicial complexes associated to edge ideals have been studied in theliterature. They are called flag complexes. (See [395, Chapter III] and thereferences there for a series of amusing related problems).

Definition 7.3.22 The complement G of a graph G has vertex set V (G)and two vertices are adjacent in G if and only if they are not adjacent inG. The complement of G is also denoted by G′.

Proposition 7.3.23 Let G be a graph and let ΔG be its independence com-plex. Then G ⊂ ΔG, with equality if and only if G has no triangles.

Proof. If x = {v1, v2} is an edge of G, then {v1, v2} is a stable set of G andx ∈ ΔG. Hence G ⊂ ΔG. Assume G has no triangles. If A = {v1, . . . , vn}is a stable set of G, then n = 1 or n = 2; otherwise if n > 2, then G wouldcontain the triangle {v1, v2, v3}. Therefore A is either a point or an edge ofG, thus G = ΔG. The converse also follows readily. �

Chordal graphs The use of graph theoretical methods in commutativealgebra is implicit in Lyubeznik thesis [299] and explicit in the work ofFroberg [172]. In both cases the central notion is that of a chordal graph.

Let G be a graph and let G be its complement. Recall that G is said tobe chordal if every cycle Cn in G of length n ≥ 4 has a chord in G. A chordof Cn is an edge joining two non-adjacent vertices of Cn. Chordal graphshave been extensively studied [112, 228, 403].

Lemma 7.3.24 Let G be a connected graph and let d be the minimum de-gree among the vertices of G. If G is not a complete graph, then there isS ⊂ V (G) such that G \ S is disconnected and d = |S|.

Proof. Let v be a vertex of G of degree d and S = NG(v) the neighbor setof v. Since G is not a complete graph and v has minimum degree there is avertex w �∈ S ∪ {v}. Thus G \ S has at least two components. �

Proposition 7.3.25 (Dirac [112]) If H is a chordal non-complete graph,then it can be constructed out of two smaller disjoint chordal graphs H1 andH2 by identifying two (possibly empty) complete subgraphs of the same sizein H1 and H2.


Proof. One may assume that H is connected and has n vertices. Since Hhas at least one vertex of degree less than n− 1, by Lemma 7.3.24 there isa minimal set of vertices S such that H \S is disconnected and |S| ≤ n− 2.Hence H \ S = F1 ∪ F2, where F1 and F2 are disjoint subgraphs of H .Consider the induced subgraphs

H1 = H [V (F1) ∪ S] and H2 = H [V (F2) ∪ S].

Notice that H1 ∪H2 = H and H1 ∩H2 = H [S].Next we show that H [S] is a complete graph. Assume there are vertices

x1 and x2 in S not connected by an edge of H . Since H \ (S \ {xi}) isconnected for i = 1, 2, there are paths p1 and p2 in H1 and H2, respectively,joining x1 with x2, whose only vertices in S are x1 and x2. If p1 and p2are the shortest paths with this property, then p1 ∪ p2 is a cycle of lengthgreater than or equal to four without a chord, a contradiction. �

Theorem 7.3.26 [300] Let G be a graph and let G be its complement. ThenIc(G) is Cohen–Macaulay if and only if G is a chordal graph.

Proposition 7.3.27 If Ic(G) is Cohen–Macaulay, then μ(Ic(G)) ≤ g + 1.where μ(Ic(G)) is the minimum number of generators of Ic(G) and g is theheight of I(G).

Proof. By the Hilbert–Burch Theorem (see Theorem 3.5.16) Ic(G) is gen-erated by the m− 1 minors of an m× (m− 1) matrix A with homogeneousentries, where m = μ(Ic(G)). Take f in Ic(G) of degree g, since any m− 1minor of A has degree at least m− 1, one obtains deg(f) = g ≥ m− 1. �

Exercises

7.3.28 Show that a cycle Cn of length n ≥ 3 is unmixed if and only ifn = 3, 4, 5, 7.

7.3.29 Let G1, G2 be two graphs with vertex set V . If their complementsare triangle free and they have the same number of edges, then the Hilbertseries of K[V ]/I(G1) and K[V ]/I(G2) are identical.

7.3.30 If G is a graph and ht(I(G)) = |V (G)| − 2, then

(a) G is Cohen–Macaulay if and only if ΔG is connected, and

(b) if I(G) is not Cohen–Macaulay, then I(G) is unmixed if and only ifΔG has no isolated vertices.

7.3.31 If G is a Cohen–Macaulay graph with q edges, then q ≤ (g2 + g)/2,where g is the height of I(G).

282 Chapter 7

7.3.32 If G is a Cohen–Macaulay graph and u, v are two vertices of degree1, prove that they cannot have an adjacent vertex in common.

7.3.33 Let G be a connected graph and let {v1, . . . , vn} be a path withn ≥ 4. If deg(v1) = 1, deg(vi) = 2 for i = 2, . . . n − 1 and deg(vn) ≥ 3,prove that G cannot be unmixed.

7.3.34 Prove that H = Skel(Simp(H)), for any graph H .

7.3.35 Let T be a tree with 2r ≥ 4 vertices. Then T is Cohen–Macaulay ifand only if ht I(T ) = r and T has exactly r vertices of degree 1.

7.3.36 Prove that the ideal I = IΔ of Example 6.3.65 is generated by

x1x3, x1x4, x1x7, x1x10, x1x11, x2x4, x2x5, x2x8,x2x10, x2x11, x3x5, x3x6, x3x8, x3x11, x4x6, x4x9,x4x11, x5x7, x5x9, x5x11, x6x8, x6x9, x7x9, x7x10,x8x10.

Let Ri = K[x1, . . . , xi, . . . , x11] and Gi = G\ {xi}. Then I = I(G) for somegraph G. Prove that the f -vector of ΔGi is (10, 25, 15) or (10, 24, 14), theh-vector of Ri/I(Gi) is (1, 7, 8,−1) or (1, 7, 7,−1), and Gi is not a C–Mgraph for all i.

7.3.37 Let I = (xiyj | 1 ≤ i ≤ j ≤ n) ⊂ R = K[X,Y ]. Prove that R/I hastype n and its a-invariant is −(n− 1).

7.4 Shellable and sequentially C–M graphs

In this section we classify all sequentially Cohen–Macaulay bipartite graphs(they are precisely the shellable bipartite graphs) and give a recursive pro-cedure to verify if a bipartite graph is shellable. Then we present somestructure theorems for unmixed and Cohen–Macaulay bipartite graphs.

Definition 7.4.1 Let G be a graph whose independence complex is ΔG.We say G is a shellable graph if ΔG is a shellable simplicial complex.

To prove that a graph G is shellable, it suffices to prove each connectedcomponent of G is shellable.

Lemma 7.4.2 [409] Let G1 and G2 be two graphs with disjoint sets of ver-tices and let G = G1 ∪ G2. Then G1 and G2 are shellable if and only if Gis shellable.


Proof. ⇒) Let F1, . . . , Fr and H1, . . . , Hs be the shellings of ΔG1 and ΔG2 ,respectively. Then if we order the facets of ΔG as

F1 ∪H1, . . . , F1 ∪Hs; F2 ∪H1, . . . , F2 ∪Hs; . . . ; Fr ∪H1, . . . , Fr ∪Hs

we get a shelling of ΔG. Indeed if F ′ < F are two facets of ΔG we have twocases to consider. Case (i): F ′ = Fi ∪ Hk and F = Fj ∪ Ht, where i < j.Because ΔG1 is shellable there is v ∈ Fj \ Fi and � < j with Fj \ F� = {v}.Hence v ∈ F \ F ′, F� ∪ Ht < F , and F \ (F� ∪ Ht) = {v}. Case (ii):F ′ = Fk ∪ Hi and F = Fk ∪ Hj , where i < j. This case follows from theshellability of ΔG2 .⇐) Note that if F is a facet of ΔG, then F ′ = F ∩ VG1 , respectively,

F ′′ = F ∩ VG2 , is a facet of ΔG1 , respectively, ΔG2 . We now show thatG1 is shellable and omit the similar proof for the shellability of G2. LetF1, . . . , Ft be a shelling of ΔG, and consider the subsequence

Fi1 , . . . , Fis with 1 = i1 < i2 < · · · < is

where F1 ∩ VG2 = Fij ∩ VG2 for ij ∈ {i1, . . . , is}, but F1 ∩ VG2 �= Fk ∩ VG2

for any k ∈ {1, . . . , t} \ {i1, . . . , is}. We then claim that

F ′1 = Fi1 \ VG2 , F

′2 = Fi2 \ VG2 , . . . , F

′s = Fis \ VG2

is a shelling of ΔG1 . We first show that this is a complete list of facets;indeed, each F ′

j = Fij ∩VG1 is a facet of ΔG1 , and furthermore, for any facetF ∈ ΔG1 , F ∪ (F1 ∩ VG2) is a facet of ΔG, and hence F ∪ (F1 ∩ VG2) = Fijfor some ij ∈ {i1, . . . , is}.

Because the Fi’s form a shelling, if 1 ≤ k < j ≤ s, there exists v inFij \ Fik = (Fij \ VG2) \ (Fik \ VG2) = F ′

j \ F ′k such that {v} = Fij \ F� for

some 1 ≤ � < ij . It suffices to show that F� is among Fi1 , . . . , Fis . Nowbecause Fij ∩ VG2 ⊂ Fij and v �∈ Fij ∩ VG2 , we must have Fij ∩ VG2 ⊂ F�.So, F� ∩ VG2 ⊃ Fij ∩ VG2 . But F� ∩ VG2 is a facet of ΔG2 , so we must haveF�∩VG2 = Fij ∩VG2 . So F� = Fir for some r < j, and hence, {v} = F ′

j \F ′r,

as desired. �

If F is a face of a simplicial complex Δ, the link of F is defined to belkΔ(F ) = {G | G ∪ F ∈ Δ, G ∩ F = ∅}. When F = {x}, then we shallabuse notation and write lkΔ(x) instead of lkΔ({x}).

Lemma 7.4.3 Let x be a vertex of G and let G′ = G\({x}∪NG(x)). Then

ΔG′ = lkΔG(x),

and F is a facet of ΔG′ if and only if x /∈ F and F ∪ {x} is a facet of ΔG.

Proof. If F ∈ lkΔG(x), then x �∈ F , and F ∪{x} ∈ ΔG implies that F ∪{x}is an independent set of G. So (F ∪{x})∩NG(x) = ∅. But this means that

284 Chapter 7

F ⊂ VG′ because VG′ = VG \ ({x} ∪NG(x)). Thus F ∈ ΔG′ since F is alsoan independent set of the smaller graph G′.

Conversely, if F ∈ ΔG′ , then F is an independent set of G′ that doesnot contain any of the vertices of {x} ∪ NG(x). But then F ∪ {x} is anindependent set of G, i.e., F ∪ {x} ∈ ΔG. So F ∈ lkΔG(x).

The last statement follows readily from the fact that F is a facet oflkΔG(x) if and only if x /∈ F and F ∪ {x} is a facet of ΔG. �

Theorem 7.4.4 [409] Let x be a vertex of a graph G and let G′ be the graphG′ = G \ ({x} ∪NG(x)). If G is shellable, then G′ is shellable.

Proof. Let F1, . . . , Fs be a shelling of ΔG. Suppose the subsequence

Fi1 , Fi2 , . . . , Fit with i1 < i2 < · · · < it

is the list of all the facets with x ∈ Fij . Setting Hj = Fij \ {x} for eachj = 1, . . . , t, Lemma 7.4.3 implies that the Hj ’s are the facets of ΔG′ .

We claim thatH1, . . . , Ht is a shelling of ΔG′ . As the Fi’s form a shelling,if 1 ≤ k < j ≤ t, there is v ∈ Fij \Fik = (Fij \ {x})\ (Fik \ {x}) = (Hj \Hk)such that {v} = Fij \ F� for some 1 ≤ � < ij . It suffices to show that F� isamong the list Fi1 , . . . , Fit . But because x ∈ Fij and x �= v, we must havex ∈ F�. Thus F� = Fik for some k ≤ j. But then {v} = Fij \ F� = Hj \Hk.So, the Hi’s form a shelling of ΔG′ . �

Let G be a graph and let S ⊂ VG. The graph G∪WG(S) obtained fromG by adding new vertices {yi | xi ∈ S} and new edges {{xi, yi} | xi ∈ S} iscalled the whisker graph of G over S. The edges {xi, yi} are called whiskers .

Corollary 7.4.5 [409] Let G be a graph and let S ⊂ VG. If G ∪WG(S) isshellable, then G \ S is shellable.

Proof. We may assume that S = {x1, . . . , xs}. Set G0 = G ∪WG(S) andGi = Gi−1 \({yi}∪NG(yi)) for i = 1, . . . , s. Notice that Gs = G\S. Hence,by repeatedly applying Theorem 7.4.4, the graph G \ S is shellable. �

Lemma 7.4.6 Let G be a bipartite graph with bipartition {x1, . . . , xm},{y1, . . . , yn}. If G is shellable and G has no isolated vertices, then there isv ∈ VG with deg(v) = 1.

Proof. Let F1, . . . , Fs be a shelling of ΔG. We may assume that Fi ={y1, . . . , yn}, Fj = {x1, . . . , xm} and i < j. Then there is xk ∈ Fj \ Fi andF� with � ≤ j − 1 such that Fj \ F� = {xk}. For simplicity assume thatxk = x1. Then {x2, . . . , xm} ⊂ F� and there is yt in F� for some 1 ≤ t ≤ n.Since {yt, x2, . . . , xm} is an independent set of G, we get that yt can onlybe adjacent to x1. Thus deg(yt) = 1 because G has no isolated vertices. �


Theorem 7.4.7 [409] Let G be a graph and let {x1, x2} be an edge of Gwith deg(x1) = 1. If Gi = G \ ({xi} ∪ NG(xi)) for i = 1, 2, then G isshellable if and only if G1 and G2 are shellable.

Proof. If G is shellable, then G1 and G2 are shellable by Theorem 7.4.4.So it suffices to prove the reverse direction. Let F ′

1, . . . , F′r be a shelling of

ΔG1 and let H ′1, . . . , H

′s be a shelling of ΔG2 . It suffices to prove that

F ′1 ∪ {x1}, . . . , F ′

r ∪ {x1}, H ′1 ∪ {x2}, . . . , H ′

s ∪ {x2}

is a shelling of ΔG. One first shows that this is the complete list of facetsof ΔG using Lemma 7.4.3. Indeed, take any facet F of ΔG. If x2 ∈ F , thenx1 �∈ F because {x1, x2} is an edge ofG, and by Lemma 7.4.3, F \{x2} = H ′

i

for some i. On the other hand, if x2 �∈ F , we must have x1 ∈ F , because ifnot, then {x1}∪F is larger independent set of G because x1 is only adjacentto x2. Again, by Lemma 7.4.3, we have F \ {x1} = F ′

i for some i. LetF ′ < F be two facets of ΔG. Consider the case in which F ′ = F ′

i ∪ {x1}and F = H ′

j ∪ {x2}. Since H ′j ∪ {x1} is an independent set of G, it is

contained in a facet of ΔG, i.e., H′j ∪ {x1} ⊂ F ′

� ∪ {x1} for some �. Hence(H ′

j ∪ {x2}) \ (F ′� ∪ {x1}) = {x2}, x2 ∈ F \ F ′, and F ′

� ∪ {x1} < F . Theremaining two cases follow readily from the shellability of ΔG1 and ΔG2 . �

The last two results yield a recursive procedure to verify if a bipartitegraph is shellable. The next result is the first combinatorial characterizationof all shellable bipartite graphs.

Corollary 7.4.8 [409] Let G be a bipartite graph. Then G is shellable ifand only if there are adjacent vertices x and y with deg(x) = 1 such thatthe bipartite graphs G \ ({x} ∪NG(x)) and G \ ({y} ∪NG(y)) are shellable.

Proof. By Lemma 7.4.2 it suffices to verify the statement when G isconnected. By Lemma 7.4.6 there exists a vertex of x1 with deg(x1) = 1.Now apply the previous theorem. �

A clutter is called vertex decomposable if its independence complex isvertex decomposable. The following major result of Woodroofe gives asufficient condition for vertex decomposability of graphs and is an extensionof the fact that chordal graphs are shellable [409].

Theorem 7.4.9 [432, Theorem 1.1] If G is a graph with no chordless cyclesof length other than 3 or 5, then G is vertex decomposable (hence shellableand sequentially Cohen–Macaulay).

Definition 7.4.10 A clutter C is called sequentially Cohen–Macaulay ifR/I(C) is sequentially Cohen–Macaulay.

We now give a sequentially Cohen–Macaulay analog of Theorem 7.4.4.

286 Chapter 7

Theorem 7.4.11 [409] Let x be a vertex of a graph G and let G′ be thegraph G′ = G \ ({x} ∪NG(x)). If G is sequentially Cohen–Macaulay, thenG′ is sequentially Cohen–Macaulay.

Proof. Let F1, . . . , Fs be the facets of Δ = ΔG, and let F1, . . . , Fr be thefacets of Δ that contain x. Set Γ = ΔG′ ; by Lemma 7.4.3, the facets of Γare F ′

1 = F1 \ {x}, . . . , F ′r = Fr \ {x}.

Consider the pure simplicial complexes

Δ[k] = 〈{F ∈ Δ| dim(F ) = k}〉; −1 ≤ k ≤ dim(Δ),

Γ[k] = 〈{F ∈ Γ| dim(F ) = k}〉; −1 ≤ k ≤ dim(Γ),

where 〈F〉 denotes the subcomplex generated by the set of faces F . Recallthat H is a face of 〈F〉 if and only if H is contained in some F in F .Take a facet F ′

i of Γ of dimension d = dim(Γ). Then F ′i ∪ {x} ∈ Δ[d+1]

and consequently {x} ∈ Δ[k+1] for k ≤ d. Because the facets of Γ areF ′1 = F1 \ {x}, . . . , F ′

r = Fr \ {x}, we have the equality

Γ[k] = lkΔ[k+1](x)

for k ≤ d. By Theorem 6.3.30, the complex Δ is sequentially Cohen–Macaulay if and only if Δ[k] is Cohen–Macaulay for −1 ≤ k ≤ dim(Δ).Because Δ[k] is Cohen–Macaulay, by Proposition 6.3.15 lkΔ[k](F ) is Cohen–Macaulay for any F ∈ Δ[k]. Thus, Γ[k] = lkΔ[k+1](x) is Cohen–Macaulay forany −1 ≤ k ≤ dim(Γ) ≤ dim(Δ) − 1. Therefore Γ is sequentially Cohen–Macaulay by Theorem 6.3.30, as required. �

The following result is due to Francisco and Ha.

Corollary 7.4.12 [159, Theorem 4.1] Let G be a graph and let S ⊂ VG. Ifthe whisker graph G∪WG(S) of G over S is sequentially Cohen–Macaulay,then G \ S is sequentially Cohen–Macaulay.

Proof. We may assume that S = {x1, . . . , xs}. Set G0 = G ∪WG(S) andGi = Gi−1 \ ({yi} ∪NG(yi)) for i = 1, . . . , s where yi is the degree 1 vertexadjacent to xi. Notice that Gs = G\S. Hence by Theorem 7.4.11 the graphG \ S is sequentially Cohen–Macaulay. �

Theorem 7.4.13 [229] If I is the edge ideal of a graph, then I has linearquotients if and only if I has a linear resolution if and only if each powerof I has a linear resolution

Lemma 7.4.14 Let G be a bipartite graph with bipartition {x1, . . . , xm},{y1, . . . , yn}. If G is sequentially Cohen–Macaulay and G is not a discretegraph, then there is v ∈ VG with deg(v) = 1.


Proof. We may assume that m ≤ n and that G has no isolated vertices.Let J be the Alexander dual of I = I(G) and let L = J[n] be the monomialideal generated by the square-free monomials of J of degree n. We mayassume that L is generated by g1, . . . , gq, where g1 = y1y2 · · · yn and g2 =x1 · · ·xmy1 · · · yn−m. Consider the linear map

Rqϕ−→ R (ei �→ gi).

The kernel of this map is generated by syzygies of the form

(gj/ gcd(gi, gj))ei − (gi/ gcd(gi, gj))ej .

Since the vector α = x1 · · ·xme1 − yn−m+1 · · · yne2 is in ker(ϕ) and sinceker(ϕ) is generated by linear syzygies (see Theorem 6.3.44), there is a linearsyzygy of L of the form xje1 − zek, where z is a variable, k �= 1. Hencexj(y1 · · · yn) = z(gk) and gk = xjy1 · · · yi−1yi+1 · · · yn for some i. Becausethe support of gk is a vertex cover of G, we get that the complement of thesupport of gk, i.e., {yi, x1, . . . , xj−1, xj+1, . . . , xm}, is an independent set ofG. Thus yi can only be adjacent to xj , i.e., deg(yi) = 1. �

Proposition 7.4.15 If G is a C–M bipartite graph, then G \ {u} is C–Mfor some vertex u of G.

Proof. By Lemma 7.4.14, there is an edge {u, v} of the graph G suchthat deg(v) = 1. Then by Proposition 7.3.1 we obtain that G \ {u} isCohen–Macaulay. �

Theorem 7.4.16 [409] Let G be a bipartite graph. Then G is shellable ifand only if G is sequentially Cohen–Macaulay.

Proof. Assume that G is sequentially Cohen–Macaulay. The proof is byinduction on the number of vertices of G. By Lemma 7.4.14 there is a vertexx1 of G of degree 1. Let x2 be the vertex of G adjacent to x1. For i = 1, 2consider the subgraph Gi = G\({xi}∪NG(xi)). By Theorem 7.4.11 G1 andG2 are sequentially Cohen–Macaulay. Hence ΔG1 and ΔG2 are shellable bythe induction hypothesis. Therefore ΔG is shellable by Theorem 7.4.7. Theconverse follows at once from Theorem 6.3.27. �

Van Tuyl [407] has shown that the independence complex ΔG must bevertex decomposable for any bipartite graph G whose edge ring R/I(G)is sequentially Cohen–Macaulay. Thus Theorem 7.4.16 remains valid if wereplace shellable by vertex decomposable.

As we saw in Corollary 7.4.8, one can verify recursively that a bipartitegraph is shellable. The above theorem, therefore, implies the same forsequentially Cohen–Macaulay bipartite graphs.

288 Chapter 7

Lemma 7.4.17 The complete bipartite graph Km,n is Cohen–Macaulay ifand only if m = n = 1.

Proof. By Exercise 6.3.51 it is enough to observe that the independencecomplex of Km,n is disconnected for m+ n ≥ 3. �

Lemma 7.4.18 Let G be an unmixed bipartite graph and let I(G) be itsedge ideal. If I(G) has height g, then there are disjoint sets of verticesV1 = {x1, . . . , xg} and V2 = {y1, . . . , yg} such that: (i) {xi, yi} is an edge ofG for all i, and (ii) every edge of G joins V1 with V2.

Proof. The statement follows by using that g is equal to the maximumnumber of independent edges of G; see Theorem 7.1.8. �

The following is a combinatorial characterization of all unmixed bipartitegraphs. Unmixed graphs are also called well-covered [340].

Theorem 7.4.19 [421] Let G be a bipartite graph. Then G is unmixed ifand only if there is a perfect matching e1, . . . , eg such that for any two edgese �= e′ and for any two distinct vertices x ∈ e, y ∈ e′ contained in some ei,one has that (e \ {x}) ∪ (e′ \ {y}) is an edge.

Proof. It follows at once from Corollary 6.5.16 because, by Theorem 7.1.8,bipartite graphs satisfy the Konig property. �

This result can be reformulated as:

Theorem 7.4.20 [421] Let G be a bipartite graph without isolated vertices.Then G is unmixed if and only if G has a bipartition V1 = {x1, . . . , xg},V2 = {y1, . . . , yg} such that: (a) {xi, yi} ∈ E(G) for all i, and (b) if {xi, yj}and {xj , yk} are in E(G) and i, j, k are distinct, then {xi, yk} ∈ E(G).

The following nice result of Herzog and Hibi classifies the family ofCohen–Macaulay bipartite graphs. This family is contained in the classof uniform admissible clutters studied in [160, 205, 325].

Theorem 7.4.21 [222] Let G be a bipartite graph without isolated vertices.Then G is a Cohen–Macaulay graph if and only if there is a bipartitionV1 = {x1, . . . , xg}, V2 = {y1, . . . , yg} of G such that: (i) {xi, yi} ∈ E(G) forall i, (ii) if {xi, yj} ∈ E(G), then i ≤ j, and (iii) if {xi, yj} and {xj , yk} arein E(G) and i < j < k, then {xi, yk} ∈ E(G).

Proof. By Proposition 7.3.8 we may assume that G is connected.⇐) The proof is by induction on g. The case g = 1 is clear. Assume

g ≥ 2. Let S = {xr1 , . . . , xrs} be the set of all vertices of G adjacent toyg. Consider the subgraph G′ = G \ S obtained from G be removing the


vertices in S. We claim that yr1 , . . . , yrs are isolated vertices of G′. Indeedif yrj is not isolated, there is an edge {xi, yrj} in G′ with i < rj . Hence bycondition (iii) we get that {xi, yg} is an edge of G and xi must be a vertexin S, a contradiction. Thus by induction the graphs G′ \ {yr1, . . . , yrs} andG \ {xg, yg} are Cohen–Macaulay. Applying Proposition 7.3.2 we get thatG is Cohen–Macaulay.⇒) By Lemma 7.4.18 there is a bipartition (V1, V2) satisfying (i). By

Lemma 7.4.14 we may assume that deg(xg) = 1 and deg(yg) ≥ 2. Since Gis Cohen–Macaulay we get that G \ {yg} is Cohen–Macaulay; see Proposi-tion 7.3.1. Using induction on G it is seen that there is a bipartition of Gsatisfying (i) and (ii). To complete the proof notice that G must also satisfy(iii) according to Theorem 7.4.20. �

The following result characterizes all bipartite graphs with a perfectmatching that satisfies condition (ii) of Theorem 7.4.21.

Theorem 7.4.22 [325, Theorem 4.3] Let G be a bipartite graph with a

perfect matching {ei}gi=1 such that ei = {xi, yi} for all i and let Γ = Δ[g]G be

the pure g-skeleton of ΔG. Then, Γ is pure shellable if and only if we canorder e1, . . . , eg such that {xi, yj} ∈ E(G) implies i ≤ j.

Directed graphs in sequentially C–M graphs Let G be an unmixedbipartite graph without isolated vertices. Then, by Theorem 7.4.20, thereis a bipartition V1, V2 of G such that

(1) if V1 = {x1, . . . , xg} and V2 = {y1, . . . , yg}, where g = α0(G),

(2) for i = 1, . . . , g, (after relabeling) {xi, yi} is an edge of G.

Following Carra Ferro and Ferrarello [78], we define a directed graphD with vertex set V1 as follows: (xi, xj) is a directed edge of D if i �= jand {xi, yj} is an edge of G. We say that a cycle C of D is oriented if allthe arrows of C are oriented in the same direction. Recall that D is calledtransitive if for any two (xi, xj), (xj , xk) in ED with i, j, k distinct, we havethat (xi, xk) ∈ ED.

Example 7.4.23 If G = C4 is a four cycle with edge set

{{x1, y1}, {x2, y2}, {x1, y2}, {x2, y1}},

then D has two vertices x1, x2 and two arrows (x1, x2), (x2, x1) forming anoriented cycle of length two.

Lemma 7.4.24 [208, Theorem 16.3(4), p. 200] Let D be the directed graphdescribed above. D is acyclic, i.e., D has no oriented cycles, if and only ifthere is a linear ordering of the vertex set V1 such that all the edges of Dare of the form (xi, xj) with i < j.

290 Chapter 7

Theorem 7.4.25 Let G be a bipartite graph satisfying (1) and (2). If G issequentially Cohen–Macaulay, then the directed graph D is acyclic.

Proof. We proceed by induction on the number of vertices of G. Assumethat D has an oriented cycle Cr with vertices {xi1 , . . . , xir}. This meansthat the graph G has a cycle

C2r = {yi1 , xi1 , yi2 , xi2 , yi3 , . . . , yir−1 , xir−1 , yir , xir}

of length 2r. By Lemma 7.4.14, the graph G has a vertex v of degree1. Notice that v /∈ {xi1 , . . . , xir , yi1 , . . . , yir}. Furthermore, if w is thevertex adjacent to v, we also have w /∈ {xi1 , . . . , xir , yi1 , . . . , yir}. Hence byTheorem 7.4.11 the graph G′ = G \ ({v} ∪ NG(v)) is sequentially Cohen–Macaulay and DG′ has an oriented cycle, a contradiction to the inductionhypotheses. Thus D has no oriented cycles, as required. �

Cohen–Macaulay trees can also be described in terms of D:

Theorem 7.4.26 Let G be a tree satisfying (1) and (2). Then G is aCohen–Macaulay tree if and only if D is a tree such that every vertex xi ofD is either a source (i.e., has only arrows leaving xi) or a sink (i.e., hasonly arrows entering xi).

Proof. ⇒) Since a tree is bipartite, D is both acyclic and transitive.Suppose there is a vertex xi that is not a sink or source. i.e., there isan arrow entering xi and one leaving xi. Suppose the arrow entering xioriginates at xj , and the arrow leaving xi goes to xk. Note that xj �= xkbecause otherwise we would have a cycle in the acyclic graph D. BecauseD is transitive, the directed edge (xj , xk) also belongs to D. But thenthe induced graph on the vertices {xj , yi, xi, yk} in G forms the cycle C4,contradicting the fact that G is a tree.⇐) The hypotheses on D imply D is acyclic and transitive, so apply

Exercise 7.4.36(d). �

Computing the type Assume that G is a Cohen–Macaulay bipartite graph.Note that {xi− yi}gi=1 is a regular system of parameters for R/I(G), whereR = K[x,y] and R/I(G) modulo ({xi − yi}gi=1) reduces to:

A = K[x]/(x21, . . . , x2g, I(H)),

whereH is a graph on the vertex set V1. As aK-vector space, A is generatedby the image of 1 and the images of all the monomials xi1 · · ·xir such that{xi1 , . . . , xir} is an independent set of H . If H∨ denotes the clutter ofminimal vertex covers of H , then the type of R/I(G) is equal to |E(H∨)|,because Soc(A) is generated by the images of all the monomials xi1 · · ·xirsuch that {xi1 , . . . , xir} is a maximal independent set of H .


Example 7.4.27 Let G be the Cohen–Macaulay bipartite graph whoseedge ideal is

I(G) = (x1y1, x1y2, x1y3, x1y4, x2y2, x2y3, x2y4, x3y3, x4y4).

The graph H has edges x1x2, x1x3, x1x4, x2x3, x2x4. Since this graph hasonly three minimal vertex covers, the type of R/I(G) is equal to three.

Exercises

7.4.28 Let Km,n be the complete bipartite graph. Prove that if m,n ≥ 2,then Km,n is not shellable. Ifm = 1 and n ≥ 1, prove that Km,n is shellable.

7.4.29 If G is a chordal graph, then the ideal I(G)∨ has linear quotients.

7.4.30 [149] Prove that a bipartite graph G is Cohen–Macaulay if and onlyif ΔG is pure shellable.

7.4.31 Prove that no even cycle can be sequentially Cohen–Macaulay.

7.4.32 [167, Proposition 4.1] Prove that C3 and C5 are the only sequentiallyCohen–Macaulay cycles.

7.4.33 LetG be a Cohen–Macaulay bipartite graph. Show that the h-vectorof R/I(G) has length at most g = ht I(G). Can we replace Cohen–Macaulayby unmixed?

7.4.34 (Ravindra [344]) A connected bipartite graph G is unmixed if andonly if there is a perfect matching such that for every edge {x, y} in theperfect matching, the induced subgraph G[NG(x) ∪ NG(y)] is a completebipartite graph.

7.4.35 A bipartite graph G without isolated vertices is unmixed if and onlyif G has a perfect matching {x1, y1}, . . . , {xg, yg} such that (I(G)2 : xiyi) isequal to I(G) for i = 1, . . . , g.

7.4.36 Let G be a connected bipartite graph with bipartition V1 = {xi}gi=1

and V2 = {yi}gi=1 such that {xi, yi} ∈ E(G) for all i and g ≥ 2. Define adirected graph D with vertex set V1 as follows: (xi, xj) is an edge of D ifi �= j and {xi, yj} is an edge of G. Prove that the following hold.

(a) If G is a square, then D has two vertices x1, x2 and two arrows (x1, x2),(x2, x1) forming a cycle of length two.

(b) D is acyclic, i.e., D has no directed cycles, if and only if there is alinear ordering of the vertex set V1 such that all the edges of D are ofthe form (xi, xj) with i < j.

292 Chapter 7

(c) D is called transitive if for any two (xi, xj), (xj , xk) in E(D) with i, j, kdistinct, we have that (xi, xk) ∈ E(D). The digraph D is transitive ifand only if G is unmixed.

(d) (Carra Ferro–Ferrarello [78]) G is Cohen–Macaulay if and only if D isacyclic and transitive.

(e) If G is a Cohen–Macaulay graph, then D has at least one source xi andat least one sink xj .

(f) If every vertex of D is either a source or a sink, then G is Cohen–Macaulay.

(g) If D is transitive and has a cycle, then D has a cycle of length 2.

(h) D is transitive and acyclic if and only if D is transitive and has nocycles of length 2.

(i) D has no cycles of length 2 if and only if for any edge {xi, yj} ∈ E(G)with i �= j, one has that {xj , yi} is not an edge of G.

(j) (Zaare-Nahandi [435]) G is Cohen–Macaulay if and only if the followingconditions hold: (1) The induced subgraph G[N(xi) ∪ N(yi)] is acomplete bipartite graph for all i, and (2) if xi is adjacent to yj fori �= j, then xj is not adjacent to yi.

(k) [435] G is Cohen–Macaulay if and only if G is unmixed and G has aunique perfect matching.

7.4.37 Let G be a connected bipartite graph satisfying the conditions ofTheorem 7.4.21. Prove that deg(y1) = 1 and deg(xg) = 1.

7.4.38 Let G be a Cohen–Macaulay bipartite graph. If G is not a discretegraph, prove that G has at least two vertices of degree one.

7.4.39 Prove that the following is the full list of Cohen–Macaulay connectedbipartite graphs with eight vertices.� �

��

� �

��

� � ��

��

��

� �

��

� �

��

� � ��

��

��

��

� �

��

� �

��

� � ��

��

��

� �

��

� �

��

� � ��

��

��

��

��

� �

��

� � ��

��

��

��

��

� �

��

� �

��

� � ��

��

��

��

��

� �

��

� �

��

� � ��

��

��

��

��

��

7.4.40 Let I = (xiyj | 1 ≤ i ≤ j ≤ n), draw a picture of the bipartite graphG such that I = I(G) and prove that I is a Cohen–Macaulay ideal.


7.5 Regularity, depth, arithmetic degree

In this section we study the regularity and depth of edge ideals of graphs.We introduce a device to estimate the regularity of edge ideals and showsome upper bounds for the arithmetic degree.

Let G be a simple graph with vertex set X = {x1, . . . , xn} and letR = K[X ] be a polynomial ring over a field K. As usual we denote thevertex set and edge set of G by V (G) and E(G), respectively.

Lemma 7.5.1 [274, Lemma 2.2] Let G be a graph and let im(G) be itsinduced matching number. Then im(G) ≤ reg(R/I(G)).

Proof. It follows at once from Corollary 6.4.8. �

Theorem 7.5.2 [326] Let F be a family of graphs containing any discretegraph and let β : F → N be a function satisfying that β(G) = 0 for anydiscrete graph G, and such that given G ∈ F , with E(G) �= ∅, there isx ∈ V (G) such that the following two conditions hold:

(i) G \ {x} and G \ ({x} ∪NG(x)) are in F .(ii) β(G \ ({x} ∪NG(x))) < β(G) and β(G \ {x}) ≤ β(G).

Then reg(R/I(G)) ≤ β(G) for any G ∈ F .

Proof. The proof is by induction on the number of vertices. Let G bea graph in F . If G is a discrete graph, then I(G) = (0), reg(R) = 0 andβ(G) = 0. Assume that G has at least one edge. There is a vertex x ∈ V (G)such that the induced subgraphs G1 = G \ {x} and G2 = G \ ({x}∪NG(x))satisfy (i) and (ii). There is an exact sequence of graded R-modules

0 −→ R/(I(G) : x)[−1] x−→ R/I(G) −→ R/(x, I(G)) −→ 0.

Notice that (I(G) : x) = (NG(x), I(G2)) and (x, I(G)) = (x, I(G1)). Thegraphs G1 and G2 have fewer vertices than G. It follows directly from thedefinition of regularity that reg(M [−1]) = 1 + reg(M) for any graded R-module M . Therefore applying the induction hypothesis to G1 and G2, andusing conditions (i) and (ii) and Exercise 6.4.30, we get

reg(R/(I(G) : x)[−1]) = reg(R′/I(G2)) + 1 ≤ β(G2) + 1 ≤ β(G),reg(R/(x, I(G))) ≤ β(G1) ≤ β(G)

where R′ = K[V (G2)]. Hence, by Lemma 6.4.10, we get that the regularityof R/I(G) is bounded by the maximum of the regularities of R/(x, I(G))and R/(I(G) : x)[−1]. Thus reg(R/I(G)) ≤ β(G), as required. �

Definition 7.5.3 A vertex x of a graph G is called simplicial if the sub-graph G[NG(x)] induced by the neighbor set NG(x) is a complete subgraph.

294 Chapter 7

Lemma 7.5.4 [403] Let H be a chordal graph and K a complete subgraphof H. If K �= H, then there is x �∈ V (K) such that the subgraph H [NH(x)]induced by the neighbor set NH(x) of x is a complete subgraph.

Proof. By induction on V (H). If H is complete, any x �∈ V (K) wouldsatisfy the requirement. Otherwise, using Proposition 7.3.25, H = H1∪H2,where H1, H2 are chordal graphs smaller than H such that H1 ∩ H2 is acomplete subgraph. It follows that K is a subgraph of Hi for some i, sayi = 1. As H1 ∩H2 is a proper complete subgraph of H2, by induction x canbe chosen in H2 \ (H1 ∩H2). �

Let G be a graph. We let β′(G) be the cardinality of any smallestmaximal matching of G. Ha and Van Tuyl proved that the regularity ofR/I(G) is bounded from above by the matching number ofG andWoodroofeimproved this result showing that β′(G) is an upper bound for the regularity.

For edge ideals whose resolutions are k-steps linear and for ideals ofcovers of graphs some new upper bounds for the regularity are given byDao, Huneke, and Schweig [103]. There are also upper bounds given byBanerjee [13] for the regularity of powers of edge ideals of graphs whosecomplement does not have any induced four cycle.

Theorem 7.5.5 Let G be a graph and let R = K[V (G)]. Then

(a) [206, Corollary 6.9] reg(R/I(G)) = im(G) for any chordal graph G.

(b) ([206, Theorem 6.7], [434]) reg(R/I(G)) ≤ β′(G).

(c) [407, Theorem 3.3] reg(R/I(G)) = im(G) if G is bipartite and R/I(G)is sequentially Cohen–Macaulay.

Proof. (a) Let F be the family of chordal graphs and let G be a chordalgraph with E(G) �= ∅. By Lemma 7.5.1 and Theorem 7.5.2 it suffices toprove that there is x ∈ V (G) such that im(G1) ≤ im(G) and im(G2) <im(G), where G1 and G2 are the subgraphs G \ {x} and G \ ({x}∪NG(x)),respectively. The inequality im(G1) ≤ im(G) is clear because any inducedmatching of G1 is an induced matching of G. We now show the otherinequality. By Lemma 7.5.4, there is y ∈ V (G) such that G[NG(y) ∪ {y}]is a complete subgraph. Pick x ∈ NG(y) and set f0 = {x, y}. Consideran induced matching f1, . . . , fr of G2 with r = im(G2). We claim thatf0, f1, . . . , fr is an induced matching of G. Let e be an edge of G containedin ∪ri=0fi. We may assume that e ∩ f0 �= ∅ and e ∩ fi �= ∅ for some i ≥ 1,otherwise e = f0 or e = fi for some i ≥ 1. Then e = {y, z} or e = {x, z}for some z ∈ fi. If e = {y, z}, then z ∈ NG(y) and x ∈ NG(y). Hence{z, x} ∈ E(G) and z ∈ NG(x), a contradiction because the vertex set of G2

is disjoint from NG(x)∪{x}. If e = {x, z}, then z ∈ NG(x), a contradiction.This completes the proof of the claim. Hence im(G2) < im(G).


(b) Let F be the family of all graphs and letG be a graph with E(G) �= ∅.By Theorem 7.5.2 it suffices to prove that there is x ∈ V (G) such thatβ′(G1) ≤ β′(G) and β′(G2) < β′(G), where G1 and G2 are the subgraphsG \ {x} and G \ ({x} ∪NG(x)), respectively. We leave the proof of this asan exercise.

(c) Let F be the family of all bipartite graphs G such that R/I(G) issequentially Cohen–Macaulay, and let β : F → N be the function β(G) =im(G). Let G be a graph in F with E(G) �= ∅. By Lemma 7.5.1 andTheorem 7.5.2 it suffices to observe that, according to Corollary 7.4.8 andTheorem 7.4.16, there are adjacent vertices x1 and x2 with deg(x2) = 1 suchthat the bipartite graph G\({xi}∪NG(xi)) is sequentially Cohen–Macaulayfor i = 1, 2. Thus conditions (i) and (ii) of Theorem 7.5.2 are satisfied. �

Corollary 7.5.6 [437] If G is a forest, then reg(R/I(G)) = im(G).

Proof. Any forest is a bipartite graph. Thus, by Theorem 6.5.25, ΔG

is shellable and R/I(G) is sequentially Cohen–Macaulay. Hence the resultfollows from Theorem 7.5.5. �

A graph G is weakly chordal if every induced cycle in both G and itscomplement G has length at most 4.

Theorem 7.5.7 [434] If G is weakly chordal, then reg(R/I(G)) = im(G).

Proof. It is shown in [77] that a weakly chordal graph G can be coveredby im(G) co-CM graphs. Thus, the result follows from Corollary 6.4.13. �

If G is an unmixed graph, Kummini [285] showed that reg(R/I(G)) isequal to the induced matching number of G. This result was later extendedto all very well-covered graphs [303] (a graph G is very well-covered if it isunmixed without isolated vertices and 2 ht(I(G)) = |V (G)|).

If G is claw-free and G, the complement of G, has no induced 4-cycles,then reg(R/I(G)) ≤ 2 with equality if G is not chordal [331], in this casereg(R/I(G)) = im(G) + 1. Formulas for the regularity of ideals of mixedproducts are given in [260]. The regularity and depth of lex segment edgeideals are computed in [143]. The regularity and other algebraic propertiesof edge ideal of Ferrers graphs are studied in detail in [96].

Corollary 7.5.8 Let G be a bipartite graph without isolated vertices. If Ghas n vertices, then depthK[ΔG] ≤

⌊n2

⌋.

Proof. Let (V1, V2) be a bipartition of G with |V1| ≤ |V2|. Note 2|V1| ≤ nbecause |V1| + |V2| = n. Since V1 is a maximal independent set of verticesone has β′

0(G) ≤ |V1|. Therefore using Theorem 6.4.18 we conclude

depthK[ΔG] ≤ β′0(G) ≤ |V1| ≤ n/2. �

The bound above does not extend to non-bipartite graphs. The depthof powers of edge ideals of trees is studied in [324].

296 Chapter 7

Proposition 7.5.9 Let Kr be the complete graph with r vertices and letGr,i be the graph obtained by attaching i lines at each vertex of Kr. Then

depth(R/I(Gr,i)) = (r − 1)i+ 1.

Proof. Let R = K[x1, . . . , xn] be a polynomial ring and let x1, . . . , xn bethe vertices of Gr,i. Note that Gr,i has n = r(i+1) vertices. We proceed byinduction on r. If r = 1, then G1,i is a star and clearly K[ΔG1,i ] has depth1. Assume r ≥ 2. Set G = Gr,i and I = I(G). Let x1 be any vertex of Krand consider the exact sequence

0 −→ R/(I : x1)x1−→ R/I −→ R/(x1, I) −→ 0. (∗)

It is not hard to see that (I : x1) is a face ideal generated by i + (r − 1)variables. Hence depth(R/(I : x1)) = (r − 1)i+ 1. On the other hand notethe equality (I, x1) = (x1) + I(Gr−1,i). Since x1 is adjacent in G to exactlyi-vertices of degree 1 and those vertices do not occur in (I, x1), by inductionhypothesis we derive depth(R/(x1, I)) = i + [(r − 2)i + 1] = (r − 1)i + 1.Altogether the ends of the exact sequence (∗) have depth equal to (r−1)i+1.Thus, by the depth Lemma 2.3.9, we get depth(R/I) = (r − 1)i+ 1. �

Example 7.5.10 Let K = Q and let G = G4,3 be the following graph

��

��

��

��

��

��

��

��x1

x9x5

x13x4

x3x2

x12

x11x10

x14

x15x16

x6

x7x8

Then, by Proposition 7.5.9, depthR/I(G) = 10 and pd(R/I(G)) = 6.

Theorem 7.5.11 [189, Theorem 2.7.14] Let G be a graph without isolatedvertices and let β′

0(C) be the cardinality of a smallest maximal independentset of G. If G has n vertices, then

depthK[ΔG] ≤ β′0(G) ≤

n+ β0(G)n− 2β0(G)− β0(G)2(n− β0(G))

,

β0(G) ≤ α0(G)[1 + (β0(G)− β′0(G))].

Lemma 7.5.12 Let G be a graph and let G∨ be its blocker. Then thearithmetic degree of G is given by arith-deg(I(G)) = |E(G∨)|.


Proof. It suffices to recall that a set of vertices of G is a maximal stableset if and only if its complement is a minimal vertex cover. �

The basic problem of enumeration and counting of the stable sets in agraph—using an algebraic perspective—is addressed in [110]. The followingestimate is due to E. Sperner (cf. [168, Theorem 3.1]).

Theorem 7.5.13 [390] If G is a graph with n vertices, then

arith-deg(I(G)) = |E(G∨)| ≤(

n

(n/2)

).

Theorem 7.5.14 [212] If G is a graph, then |E(G∨)| ≤ 2α0(G).

Proof. We may assume G has no isolated vertices. We set C = G∨. Let Cbe a fixed vertex cover with α0(G) vertices. For 0 ≤ i ≤ α0(G) consider:

Ci = {C′|C′ ∈ C and |C ∩ C′| = i}.

We claim that the following inequality holds: |Ci| ≤(α0(G)i

)for all i ≥ 0.

We are going to show that if C′ and C′′ are two vertex covers in Ci suchthat C ∩ C′ = C ∩ C′′, then C′ = C′′. Using C ∩ C′ = C ∩ C′′ we obtainthat the neighbors sets satisfy: (∗) NG(C \ C′) = NG(C \ C′′). Since C′

and C′′ are vertex covers of G one has:

NG(C \ C′) ∪ (C ∩ C′) ⊂ C′ and NG(C \ C′′) ∪ (C ∩ C′′) ⊂ C′′.

Observe that NG(C \C′)∪ (C ∩C′) and NG(C \C′′)∪ (C ∩C′′) are vertexcovers of G, because C is a vertex cover of G. From the minimality of C′

and C′′ one concludes the equalities:

NG(C \ C′) ∪ (C ∩ C′) = C′ and NG(C \ C′′) ∪ (C ∩ C′′) = C′′,

which together with (∗) allow us to derive the equality C′ = C′′, as required.

Therefore since there are only(α0(G)i

)subsets in C with i vertices, one

obtains the inequality |Ci| ≤(α0(G)i

), as claimed. Hence

|E(G∨)| = |C| =α0(G)∑i=0

|Ci| ≤α0(G)∑i=0

(α0(G)

i

)= 2α0(G). �

Exercises

7.5.15 A graph whose complement is chordal is called co-chordal. A com-plex Δ is called a quasi-forest if Δ is the clique complex of a chordal graph.Prove that an edge ideal I(C) of a clutter C has regularity 2 if and only ifΔC is the independence complex of a co-chordal graph.

298 Chapter 7

7.5.16 Let G be the complement of a cycle of length six. Use Macaulay2,to show that reg(R/I(G)) = 2 and im(G) = 1.

7.5.17 Let Kr be a complete graph and let G be the graph obtained byattaching ij lines at each vertex vj of Kr. If ij ≤ ij+1 for all j, then

depth(R/I(G)) = i1 + · · ·+ ir−1 + 1.

7.5.18 Let γn = sup {|E(G∨)| | G is a graph with n vertices}. If {Fn} isthe Fibonacci sequence, F0 = F1 = 1 and Fn = Fn−1+Fn−2, then γn ≤ Fnfor n ≥ 1. If Pn is the path {x1, . . . , xn}, n ≥ 3, then Fn = |E(P∨

n )|.

7.5.19 Find e(G), the multiplicity of R/I(G), and arith-deg(I(G)), thearithmetic degree of I(G) if G is a disjoint union of stars.

7.5.20 Let G be the graph which is the disjoint union of k complete graphswith two vertices. Then, |E(G∨)| = e(G) = 2α0(G) and k = α0(G).

7.6 Betti numbers of edge ideals

Let G be a graph and I(G) ⊂ R its edge ideal. An interesting problem is toexpress some of the first initial Betti numbers and invariants of I(G) in termsof the graph theoretical data of G (see [138, 141, 276, 321, 362, 437] andthe references therein). Kimura [277] has studied the projective dimensionof edge ideals of unmixed bipartite graphs.

Let R be a polynomial ring over a field K with the usual grading and I agraded ideal of R. The minimal graded resolution of R/I by free R-modulescan be expressed as:

0 −→cg⊕i=1

Rbgi(−dgi)ϕg−→ · · · −→

c1⊕i=1

Rb1i(−d1i)ϕ1−→ R −→ R/I −→ 0,

where all the maps are degree preserving and the dij are positive integers.One may assume dj1 < · · · < djcj for all j = 1, . . . , g. The integer g is equalto pdR(R/I), the projective dimension of R/I. To simplify notation set

Fj =

cj⊕i=1

Rbji(−dji).

The rank of Fj is the jth Betti number of I. From the minimality of theresolution d11 < · · · < dg1 ; see Remark 3.5.9. We define the jth initial Bettinumber of I as γj = bj1 and the jth initial virtual Betti number of I as

vj = dimK (Fj)d11+j−1.


Note that some of the initial virtual Betti numbers may be zero.For a general monomial ideal I there is an explicit and elegant descrip-

tion of a minimal set of generators for the first module of syzygies of I dueto Eliahou [137]; see also [64].

Definition 7.6.1 The line graph of G, denoted by L(G), has vertex setE = E(G) with two vertices of L(G) adjacent whenever the correspondingedges of G have exactly one common vertex.

Proposition 7.6.2 If G is a graph with vertices x1, . . . , xn and edge setE(G), then the number of edges of the line graph L(G) is given by

|E(L(G))| =n∑i=1

(deg(xi)

2

)= −|E(G)|+

n∑i=1

deg2 xi2

. (7.1)

Proof. To prove the first equality note that each vertex xi of G of degreedi contributes with

(di2

)edges to the number of edges of L(G). The second

equality follows from the Euler’s identity 2|E(G)| =∑n

i=1 deg(xi). �

Let us now present a formula, in graph theoretical terms, for the secondinitial Betti number of I(G).

Proposition 7.6.3 [138] Let I ⊂ R be the edge ideal of a graph G, let Vbe the vertex set of G, and let L(G) be the line graph of G. If

· · · −→ Rc(−4)⊕Rb(−3) −→ Rq(−2) ψ−→ R −→ R/I −→ 0

is the minimal graded resolution of R/I. Then b = |E(L(G))| −Nt, whereNt is the number of triangles of G and c is the number of unordered pairsof edges {f, g} such that f ∩g = ∅ and f and g cannot be joined by an edge.

Proof. Let f1, . . . , fq be the set of all xixj such that {xi, xj} is an edge ofG.We may assume ψ(ei) = fi. Let Z

′1 be the set of elements in ker(ψ) of degree

3. We regard the fi’s as the vertices of L(G). Every edge e = {fi, fj} inL(G) determines a syzygy syz(e) = xjei−xiej , where fi = xiz and fj = xjzfor some z, xi, xj in V . By Theorem 3.3.19 the set of those syzygies generateZ ′1. Given a triangle C3 = {z1, z2, z3} in G we set

φ(C3) = {z1ej − z3ei, z1ej − z2ek, z2ek − z3ei},

where fi = z1z2, fj = z2z3, and fk = z1z3. Notice that φ(C3) ∩ φ(C′3) = ∅

if C3 �= C′3. For every triangle C3 in G choose an element ρ(C3) ∈ φ(C3).

It is not hard to show that the set

B = {syz(e)| e ∈ L(G)} \ {ρ(C3)|C3 is a triangle in G}

300 Chapter 7

is a minimal generating set for Z ′1. Thus b = |E(L(G))| −Nt. The proof of

the expression for c is left as an exercise. �

The 2nd Betti number of a Stanley–Reisner ideal (resp. monomial ideal)is independent of the ground field [244] (resp. [64, 137]); moreover for edgeideals the 3rd and 4th Betti numbers and the 5th and 6th Betti numbersare also independent of the field [243] and [274], respectively.

An application to Hilbert series Let Δ be a simplicial complex ofdimension d and f = (f0, . . . , fd) its f -vector, where fi is the number offaces of dimension i in Δ and f−1 = 1. By Theorem 6.7.2 the Hilbert seriesof the Stanley–Reisner ring S = R/IΔ can be expressed as

F (t) = f−1 +f0t

(1− t) + · · ·+fdt

d+1

(1− t)d+1. (7.2)

We now use this formula to compute a particular instance.

Corollary 7.6.4 Let G be a graph with q edges, V = {x1, . . . , xn} its vertexset and F (t) the Hilbert series of S = R/I(G). If S has dimension 3, thenF (t)(1 − t)3 is equal to

1 + gt+

((g + 1

2

)− q)t2 +

(2g + 3g2 + g3 − 6q − 6gq − 6N + 3v

6

)t3,

where g = n− 3, N is the number of triangles of G, and v =∑n

i=1 deg2xi.

Proof. Let Δ = ΔG be the Stanley–Reisner complex of I = I(G). The

first entries of the f -vector of Δ are f−1 = 1, f0 = n, f1 = n(n−1)2 −q. From

the resolution above and Proposition 6.7.3, dim(R/I)3 =(n+2n−1

)−qn+b and

f2 =(n+2n−1

)− qn+ b− 2f1 − n. The desired formula follows by substitution

of the fi’s in Eq. (7.2) and using Eq. (7.1). �

The number of triangles of a graph Let G be a graph with q edgesand vertex set V = {x1, . . . , xn}. As the number of edges of the line graphL(G) of G is given by the formula

|E(L(G))| =n∑i=1

(deg(xi)

2

)= −q +

n∑i=1

deg2 xi2

,

observe that Proposition 7.6.3 provides a method to compute the number oftriangles of a graph by computing syzygies with Macaulay2. An alternativemethod using algebraic graph theory is recalled next.


The adjacency matrix of G is the n× n matrix A = (aij) with entries

aij =

{1 if xi is adjacent xj ,0 otherwise.

It follows directly that A is symmetric and that its trace is equal to zero. If

f(x) = xn + c1xn−1 + c2x

n−2 + c3xn−3 + · · ·+ cn,

is the characteristic polynomial of A, then c1 = 0, −c2 is the number of edgesof G, and −c3 is twice the number of triangles of G. For an interpretationof all the ci’s see [40].

Example 7.6.5 Let G be the graph whose complement is the union of twodisjoint copies of a path of length two. The edge ideal of G is equal to:

I(G) = (x1x4, x1x5, x1x6, x2x3, x2x4, x2x5, x2x6, x3x4, x3x5, x3x6, x5x6),

and part of its minimal resolution is:

· · · → R24(−3)→ R11(−2)→ R→ R/I(G)→ 0.

Hence the number of triangles of G is equal to 6. The same number isobtained noticing that f(x) = x6 − 11x4 − 12x3 + 3x2 + 4x − 1 is thecharacteristic polynomial of the adjacency matrix of G.

Linear resolutions of edge ideals First we introduce d-trees and statean interesting result of Froberg.

Definition 7.6.6 A d-tree is defined inductively: (i) a complete graphKd+1

is a d-tree, (ii) let G be a d-tree and H a subgraph of G with H � Kd, if vis a new vertex connected to all vertices in H , then G ∪ {v} is a d-tree.Theorem 7.6.7 [172] Let G be a simple graph, let ΔG be its independencecomplex, and let G be the complement of G. The following hold.

(a) K[ΔG] has a 2-linear resolution if and only if G is a chordal graph.

(b) If G is Cohen–Macaulay over a field K, then K[ΔG] has a 2-linearresolution if and only if G is a d-tree, where d = dim ΔG.

The problem of determining the simplicial complexes Δ whose Stanley–Reisner ideal IΔ has a pure resolution was solved by Bruns and Hibi [66, 67].

Corollary 7.6.8 Let G be a connected Cohen–Macaulay bipartite graph. Ifthe number of edges of G is g(g + 1)/2, where g = ht I(G), then

I(G) = (xiyj| 1 ≤ i ≤ j ≤ g).Proof. The ring R/I(G) has a linear resolution by Theorem 5.3.7. Hence,according to Theorem 7.6.7, the 1-skeleton of ΔG is a (g − 1)-tree, whichimplies that there is a vertex w of G of degree g. We now use Exercise 7.4.37to conclude that deg(xi) = 1 and deg(yj) = 1, for some i, j. To finish theproof consider the graph G \ {w} and use induction. �

302 Chapter 7

Exercises

7.6.9 Let G be a graph and let k = max{deg(v)| v ∈ V (G)}. If G isconnected, then G is k-regular, i.e., its vertices have degree k, if and only ifk is a characteristic value of the adjacency matrix of G.

7.6.10 Let G be a graph with vertex set x and FG(t) the Hilbert series ofK[x]/I(G), where K is a field. For any vertex x ∈ V (G) one has:

FG(t) = FG\{x}(t) +t

1− tFG\(N(x)∪{x})(t), with F∅(t) = 1.

7.6.11 If Kr,s is a complete bipartite graph, then

FKr,s(t) =1

(1− t)r +1

(1− t)s − 1.

7.6.12 [426] Let Pn be a path graph with vertices x = x1, . . . , xn. Provethat the Hilbert series and the Hilbert function of K[x]/I(Pn) are given by

FPn(t) =

�n+12 �∑

j=0

(n− j + 1

j

) ∞∑i=0

(j + i− 1

i

)tj+i

HLn(m) =

min(m,�n+12 �)∑

j=0

(n− j + 1

j

)(m− 1

m− j

).

7.6.13 Let G = C7 be a heptagon. Prove the formula:

FG(t) =−t3 + 3t2 + 4t+ 1

(1 − t)3 .

What is the number of minimal primes of I(G)?

7.6.14 Let G = C7 be a heptagon and let ΔG be its independence complex.Prove that ΔG is a triangulation of a Mobius band whose reduced Eulercharacteristic is χ(ΔG) = −1.7.6.15 If G is a graph and I(G) ⊂ R is its edge ideal, then a(R/I(G)) ≤ 0.

7.6.16 If G is a graph and H is its whisker graph, then β0(G) equals theinduced matching number of H .

7.6.17 Let Δ be a simplicial complex of dimension d with n vertices andf = (f0, . . . , fd) its f -vector. If K is a field and K[Δ] is a Cohen–Macaulayring with a 2-linear resolution, then

fk =

(d+ 1

k + 1

)+ (n− d− 1)

(d

k

), 0 ≤ k ≤ d.

7.6.18 Let I be the ideal (abc, abd, ace, adf, aef, bcf, bde, bef, cde, cdf) ofthe polynomial ring K[a, . . . , f ]. Prove that the third Betti number of Idepends on the base field, and that I is equal to its Alexander dual.


7.7 Associated primes of powers of ideals

In this section we show that the sets of associated primes of the powers of theedge ideal of a graph form an ascending chain. Two excellent references forthe general theory of asymptotic prime divisors in commutative Noetherianrings are [259] and [313].

Let G be a simple graph with finite vertex set X = {x1, . . . , xn} andwithout isolated vertices, let R = K[x1, . . . , xn] be a polynomial ring overa field K and let I = I(G) be the edge ideal of G.

In [56], Brodmann showed that there exists a positive integer N1 suchthat Ass(R/Ik) = Ass(R/IN1) for all k ≥ N1. A minimal such N1 is calledthe index of stability of I.

Proposition 7.7.1 [81, Corollary 4.3] If G is a connected non-bipartitegraph with n vertices, s leaves, and the smallest odd cycle of G has length2k + 1, then N1 ≤ n− k − s.

Definition 7.7.2 [165] An ideal I ⊂ R has the persistence property if

Ass(R/Ik) ⊂ Ass(R/Ik+1)

for all k ≥ 1, i.e., the sets Ass(R/Ik) form an ascending chain.

There are some interesting cases where associated primes are known toform ascending chains [36, 163, 165, 232, 326]. The first listed is quitegeneral, but has applications to square-free monomial ideals.

Theorem 7.7.3 ([313, Proposition 3.9], [342, 343]) If R is a Noetherian

ring and I is an ideal, then the sets Ass(R/Ik) form an ascending chain.In particular if I is normal, then I has the persistence property.

Definition 7.7.4 Following Schrijver [373], the duplication of a vertex xiof a graph G means extending its vertex set X by a new vertex x′i andreplacing E(G) by

E(G) ∪ {(e \ {xi}) ∪ {x′i}|xi ∈ e ∈ E(G)}.

The duplication of the vertex xi is denoted by Gxi . The deletion of xi isthe subgraph G\ {xi}. A graph obtained from G by a sequence of deletionsand duplications of vertices is called a parallelization of G.

These two operations commute. If a = (ai) is a vector in Nn, we denoteby Ga the graph obtained from G by successively deleting any vertex xiwith ai = 0 and duplicating ai − 1 times any vertex xi if ai ≥ 1.

Lemma 7.7.5 Any parallelization of G has the form Ga for some a.

304 Chapter 7

Proof. It suffices to notice that, by Exercise 7.7.23, any parallelization ofG can be obtained by duplications and deletions of vertices of G. �

The subring K[G] = K[xixj | {xi, xj} is an edge of G] ⊂ R is called theedge subring of G. This subring will be studied in detail in Chapter 10.

Lemma 7.7.6 Let G be a graph and let K[G] be its edge subring. Then,Ga has a perfect matching if and only if xa ∈ K[G].

Proof. We may assume that a = (a1, . . . , an) and ai ≥ 1 for all i, becauseif a has zero entries we can use the induced subgraph on the vertex set{xi| ai > 0}. The vertex set of Ga is

Xa = {x11, . . . , xa11 , . . . , x1i , . . . , xaii , . . . , x1n, . . . , xann }

and the edges of Ga are exactly those pairs of the form {xkii , xkjj } with

i �= j, ki ≤ ai, kj ≤ aj , for some edge {xi, xj} of G. We can regard xa asan ordered multiset

xa = xa11 · · ·xann = (x1 · · ·x1︸︷︷︸a1

) · · · (xn · · ·xn︸︷︷︸an

)

on the set X ; that is, we can identify the monomial xa with the multiset

Xa = {x1, . . . , x1︸︷︷︸a1

, . . . , xn, . . . , xn︸︷︷︸an

}

on X in which each variable is uniquely identified with an integer between1 and |a|. This integer is the position, from left to right, of xi in Xa. Thereis a bijective map

1 2 · · · a1 · · · a1 + · · ·+ an−1 + 1 · · · a1 + · · ·+ an↓ ↓ · · · ↓ · · · ↓ · · · ↓x1 x1 · · · x1 · · · xn · · · xn↓ ↓ · · · ↓ · · · ↓ · · · ↓x11 x21 · · · xa11 · · · x1n · · · xann .

Hence if Ga has a perfect matching, then the perfect matching induces afactorization of xa in which each factor corresponds to an edge of G, i.e.,xa ∈ K[G]. Conversely, if xa ∈ K[G] we can factor xa as a product ofmonomials corresponding to edges of G and this factorization induces aperfect matching of Ga. �

Given an edge f = {xi, xj} of G, we denote by Gf or G{xi,xj} the graphobtained from G by successively duplicating the vertices xi and xj , i.e.,Gf = G1+ei+ej , where ei is the ith unit vector and 1 = (1, . . . , 1). Recallthat def(G), the deficiency of G, is given by def(G) = |V (G)| − 2β1(G),


where β1(G) is the matching number of G. Hence def(G) is the number ofvertices left uncovered by any maximum matching.

Notation In what follows F = {f1, . . . , fq} denotes the set of all monomialsxixj such that {xi, xj} ∈ E(G). We use f c as an abbreviation for f c11 · · · f

cqq ,

where c = (ci) ∈ Nq.

Lemma 7.7.7 Let G be a graph and let a ∈ Nn and c ∈ Nq. Then

(a) xa = xδf c, where |δ| = def(Ga) and |c| = β1(Ga).

(b) xa belongs to I(G)k \ I(G)k+1 if and only if k = β1(Ga).

(c) (Ga)f = (Ga){xi,xj} for any edge f = {xkii , xkjj } of Ga.

Proof. Parts (a) and (b) follow using the bijective map used in the proofof Lemma 7.7.6. To show (c) we use the notation used in the proof ofLemma 7.7.6. We now prove the inclusion E((Ga)f ) ⊂ E((Ga){xi,xj}). Let

yi and yj be the duplications of xkii and xkjj , respectively. We also denote

the duplications of xi and xj by yi and yj , respectively. The common vertexset of (Ga)f and (Ga){xi,xj} is V (Ga)∪ {yi, yj}. Let e be an edge of (Ga)f .If e = {yi, yj} or e ∩ {yi, yj} = ∅, then clearly e is an edge of (Ga){xi,xj}.

Thus, we may assume that e = {yi, xk�� }. Then {xkii , xk�� } ∈ E(Ga), so

{xi, x�} ∈ E(G). Hence {xi, xk�� } is in E(Ga), so e = {yi, xk�� } is an edgeof (Ga){xi,xj}. This proves the inclusion “⊂”. The other inclusion followsusing similar arguments (arguing backwards). �

Theorem 7.7.8 (Berge [297, Theorem 3.1.14]) Let G be a graph. Then

def(G) = max{c0(G \ S)− |S| |S ⊂ V (G)},

where c0(G) denotes the number of odd components (components with anodd number of vertices) of a graph G.

The theorem of Berge is equivalent to a classical result of Tutte thatdescribes perfect matchings [297] (see Theorem 7.1.10).

Theorem 7.7.9 Let G be a graph. Then def(Gf ) = δ for all f ∈ E(G) ifand only if def(G) = δ and β1(G

f ) = β1(G) + 1 for all f ∈ E(G).

Proof. Assume that def(Gf ) = δ for all f ∈ E(G). In general, def(G) ≥def(Gf ) for any f ∈ E(G). We proceed by contradiction. Assume thatdef(G) > δ. Then, by Berge’s theorem, there is an S ⊂ V (G) such thatc0(G \ S) − |S| > δ. We set r = c0(G \ S) and s = |S|. Let H1, . . . , Hr bethe odd components of G \ S.

Case (I): |V (Hk)| ≥ 2 for some 1 ≤ k ≤ r. Pick an edge f = {xi, xj} ofHk. Consider the parallelization H ′

k obtained from Hk by duplicating the

306 Chapter 7

vertices xi and xj , i.e., H′k = Hf

k . The odd connected components of Gf \Sare H1, H2, . . . , Hk−1, H

′k, Hk+1 . . . , Hr. Thus

c0(Gf \ S)− |S| > δ = def(Gf ).

This contradicts Berge’s theorem when applied to Gf .Case (II): |V (Hk)| = 1 for 1 ≤ k ≤ r. Notice that S �= ∅ because G has

no isolated vertices. Pick f = {xi, xj} an edge of G with {xi} = V (H1) andxj ∈ S. Let yi and yj be the duplications of xi and xj , respectively. Theodd components of Gf \ (S ∪ {yj}) are H1, . . . , Hr, {yi}. Thus

c0(Gf \ (S ∪ {yj}))− |S ∪ {yj}| = c0(G \ S)− |S| > δ = def(Gf ).

This contradicts Berge’s theorem when applied to Gf . Therefore def(G) =def(Gf ) for all f ∈ E(G). Hence β1(G

f ) = β1(G)+1 for all f ∈ E(G). Theconverse follows using the definition of def(G) and def(Gf ). �

Corollary 7.7.10 Let G be a graph. Then G has a perfect matching if andonly if Gf has a perfect matching for every edge f of G.

Proof. Assume that G has a perfect matching. Let f1, . . . , fn/2 be a set ofedges of G that form a perfect matching of V (G), where n is the number ofvertices of G. If f = {xi, xj} is any edge of G and yi, yj are the duplicationsof the vertices xi and xj , respectively, then clearly f1, . . . , fn/2, {yi, yj} forma perfect matching of V (Gf ). Conversely, if Gf has a perfect matching forall f ∈ E(G), then def(Gf ) = 0 for all f ∈ E(G). Hence, by Theorem 7.7.9,we get that def(G) = 0, so G has a perfect matching. �

If I �= (0) is an ideal of a commutative Noetherian domain, Ratliffshowed that (Ik+1 : I) = Ik for all large k [342, Corollary 4.2] and thatequality holds for all k when I is normal [342, Proposition 4.7]. The nextlemma shows that equality holds for all k when I is an edge ideal.

Lemma 7.7.11 Let I be the edge ideal of a graph G. Then (Ik+1 : I) = Ik

for k ≥ 1.

Proof. Let f1, . . . , fq be the set of all xixj such that {xi, xj} ∈ E(G).Given c = (ci) ∈ Nq, we set f c = f c11 · · · f

cqq . The colon ideal (Ik+1 : I) is a

monomial ideal. Clearly Ik ⊂ (Ik+1 : I). To show the other inclusion take amonomial xa in (Ik+1 : I). Then, fix

a ∈ Ik+1 for all i. We may assume thatfix

a /∈ Ik+2, otherwise xa ∈ Ik as required. Thus xa+ei+ej ∈ Ik+1 \ Ik+2

for any ei + ej such that {xi, xj} ∈ E(G). Hence, by Lemma 7.7.7(b),β1(G

a+ei+ej ) = k + 1 for any {xi, xj} ∈ E(G), that is, (Ga){xi,xj} has amaximum matching of size k + 1 for any edge {xi, xj} of G. With the

notation of the proof of Lemma 7.7.6, for any edge {xkii , xkjj } of Ga we have

(Ga){xkii ,x

kjj } = (Ga){xi,xj},


see Lemma 7.7.7(c). Then, (Ga)f has a maximum matching of size k + 1for any edge f of Ga. As a consequence

def((Ga)f ) = (|a|+ 2)− 2(k + 1) = |a| − 2k

for any edge f of Ga. Hence, by Theorem 7.7.9, def(Ga) = |a| − 2k. UsingLemma 7.7.7(a), we can write xa = xδf c, where |δ| = def(Ga) and |c| =β1(G

a). Taking degrees in xa = xδf c gives |a| = |δ|+2|c| = (|a|−2k)+2|c|,that is, |c| = k. Then xa ∈ Ik and the proof is complete. �

Proposition 7.7.12 Let I = I(G) be the edge ideal of a graph G and letm = (x1, . . . , xn). If m ∈ Ass(R/Ik), then m ∈ Ass(R/Ik+1).

Proof. As m is an associated prime of R/Ik, there is xa /∈ Ik such thatmxa ⊂ Ik. By Lemma 7.7.11 there is an edge {xi, xj} of G such thatxixjx

a /∈ Ik+1. Then, x�(xixjxa) ∈ Ik+1 for � = 1, . . . , n; that is, m is an

associated prime of R/Ik+1. �

Lemma 7.7.13 [204, Lemma 3.4] Let I be a square-free monomial ideal inS = K[x1, . . . , xm, xm+1, . . . , xr] such that I = I1S + I2S, where I1 ⊂ S1 =K[x1, . . . , xm] and I2 ⊂ S2 = K[xm+1, . . . , xr]. Then p ∈ Ass(S/Ik) if andonly if p = p1S + p2S, where p1 ∈ Ass(S1/I

k11 ) and p2 ∈ Ass(S2/I

k22 ) with

(k1 − 1) + (k2 − 1) = k − 1.

If pi is an ideal of R, the generators of pi will generate a prime ideal inany ring that contains those variables. We will abuse notation by denotingthe ideal generated by the generators of pi in any other ring by pi as well.

Theorem 7.7.14 [305] Let G be a graph and let I = I(G) be its edge ideal.Then

Ass(R/Ik) ⊂ Ass(R/Ik+1)

for all k. That is, I has the persistence property.

Proof. Take p in Ass(R/Ik). We set m = (x1, . . . , xn). We may assumethat p = (x1, . . . , xr). By Proposition 7.7.12, we may assume that p � m.Write Ip = (I2, I1)p, where I2 is the ideal of R generated by all xixj , i �= j,whose image, under the canonical map R → Rp, is a minimal generator ofIp, and I1 is the ideal of R generated by all xi whose image is a minimalgenerator of Ip, which correspond to the isolated vertices of the graph asso-ciated to Ip. The minimal generators of I2 and I1 lie in S = K[x1, . . . , xr],and the two sets of variables occurring in the minimal generating sets of I1and I2 (respectively) are disjoint and their union is {x1, . . . , xr}.

If I2 = (0), then p is a minimal prime of I so it is an associated primeof R/Ik+1. Thus, we may assume I2 �= (0). An important fact is thatlocalization preserves associated primes; that is p ∈ Ass(R/Ik) if and only

308 Chapter 7

if pRp ∈ Ass(Rp/(IpRp)k). Hence, p is in Ass(R/Ik) if and only if p is in

Ass(R/(I1, I2)k) if and only if p is in Ass(S/(I1, I2)

k). By Proposition 7.7.12and Lemma 7.7.13, p is an associated prime of S/(I1, I2)

k+1. Hence, we canargue backwards to conclude that p is an associated prime of R/Ik+1. �

If G is a graph with loops, then its edge ideal I(G) has the persistenceproperty [353].

Corollary 7.7.15 If I is a square-free monomial ideal and (Ik+1 : I) = Ik

for k ≥ 1. Then, I has the persistence property.

Proof. As in Proposition 7.7.12, we first show that m ∈ Ass(R/Ik) impliesm ∈ Ass(R/Ik+1). Assume m ∈ Ass(R/Ik). Then there is a monomial xa �∈Ik with xix

a ∈ Ik+1 for all i. By the hypothesis, xa �∈ (Ik+1 : I), so there isa square-free monomial generator e of I (which can be viewed as the edgeof a clutter associated to I) with exa �∈ Ik+1. But xiex

a = e(xixa) ∈ Ik+1

for all i, so m ∈ Ass(R/Ik+1).Recall that since I is finitely generated, (Ik+1 : I)p = (Ik+1

p : Ip). Thus

(Ik+1p : Ip) = Ikp . The remainder of the argument now follows from localiza-

tion, as in the proof of Theorem 7.7.14. �

Lemma 7.7.16 If I is a monomial ideal, then Ass(Ik−1/Ik) = Ass(R/Ik).

Proof. Suppose that p ∈ Ass(R/Ik). Then p = (Ik : c) for some monomialc ∈ R. But since p is necessarily a monomial prime, generated by a subsetof the variables, then if xc ∈ Ik for a variable x ∈ p, then c ∈ Ik−1 and sop ∈ Ass(Ik−1/Ik). The other inclusion is automatic. �

Conjecture 7.7.17 (Persistence problem) If I is a square-free monomialideal, then I has the persistence property.

The next counterexample for this conjecture is due to Kaiser, Stehlık,and Skrekovski [268] (cf. Exercise 7.7.25).

Example 7.7.18 [268] Let I∨ be the Alexander dual of the edge ideal

I = (x1x2, x2x3, x3x4, x4x5, x5x6, x6x7, x7x8, x8x9, x9x10, x1x10,

x2x11, x8x11, x3x12, x7x12, x1x9, x2x8, x3x7, x4x6, x1x6, x4x9,

x5x10, x10x11, x11x12, x5x12).

Using Macaulay2, one can readily verify that Ass(R/(I∨)3) is not a subsetof Ass(R/(I∨)4) (see Exercise 7.7.27).

Theorem 7.7.19 [165] If G is a simple graph, then the chromatic numberof G is the minimal k such that (x1 · · ·xn)k−1 ∈ Ic(G)k.


Note that one can use Normaliz [68] inside Macaulay2 to compute theintegral closure of a monomial ideal and also the normalization of the Reesalgebra of a monomial ideal.

Procedure 7.7.20 The following simple procedure for Macaulay2 decideswhether Ass(R/I3) is contained in Ass(R/I4) and whether we have theequality Ass(R/I3) = Ass(R/I4). It also computes I4 and decides whetherAss(R/I4) is equal to Ass(R/I4).

R=QQ[x1,x2,x3,x4,x5,x6,x7,x8,x9];

load "normaliz.m2";

I=monomialIdeal(x1*x2,x2*x3,x1*x3,x3*x4,x4*x5,x5*x6,

x6*x7,x7*x8,x8*x9,x5*x9);

isSubset(ass(I^3),ass(I^4))

ass(I^3)==ass(I^4)

(intCl4,normRees4)=intclMonIdeal I^4;

intCl4’=substitute(intCl4,R);

ass(monomialIdeal(intCl4’))==ass(I^4)

Exercises

7.7.21 Let I be the square-free monomial ideal generated by

x1x2x5, x1x3x4, x1x2x6, x1x3x6, x1x4x5, x2x3x4, x2x3x5,

x2x4x6, x3x5x6, x4x5x6.

Use Normaliz [68] and Macaulay2 [199], to show that I is a non-normalideal such that (I2 : I) = I and (I3 : I) �= I2. Then, prove that I has thepersistence property and that the index of stability of I is equal to 3.

7.7.22 Let I = I(G) be the edge ideal of a graph G. Prove that the setsAss(Ik−1/Ik) form an ascending chain for k ≥ 1.

7.7.23 Let G be a graph and let xi be a vertex. If y ∈ V (Gxi), then(Gxi)y = (Gxi)xj for some xj ∈ V (G).

7.7.24 If I = (I1, . . . , Is), where the Ii are square-free monomial ideals indisjoint sets of variables, then p ∈ Ass(R/Ik) if and only if p = (p1, . . . , ps),where pi ∈ Ass(R/Ikii ) with (k1 − 1) + · · ·+ (ks − 1) = k − 1.

7.7.25 [326] Let I = (x1x22x3, x2x

23x4, x3x

24x5, x4x

25x1, x5x

21x2) be the ideal

obtained by taking the product of consecutive edges of a pentagon. Thenm ∈ Ass(R/Ik) for k = 1, 4, m �∈ Ass(R/Ik) for k = 2, 3, m = (x1, . . . , x5).

7.7.26 [323, Lemma 2.3] If I is a monomial ideal of R and a ∈ I/I2 isa regular element of the associated graded ring grI(R). Then the setsAss(Ik−1/Ik) form an ascending chain.

310 Chapter 7

7.7.27 Use the following procedure for Macaulay2 to verify that the idealI∨ of Example 7.7.18 does not have the persistence property.

R=QQ[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10,x11,x12]

I=monomialIdeal(x1*x2,x2*x3,x3*x4,x4*x5,x5*x6,x6*x7,x7*x8,

x8*x9,x9*x10,x1*x10,x2*x11,x8*x11,x3*x12,x7*x12,x1*x9,x2*x8,

x3*x7,x4*x6,x1*x6,x4*x9,x5*x10,x10*x11,x11*x12,x5*x12)

J=dual I

isSubset(ass(J^3),ass(J^4))

7.7.28 Consider the graph G of the figure below, where vertices are labeledwith i instead of xi. The duplication of the vertices x3 and x4 of G is shownbelow. If f = {x3, x4}, prove that deficiencies of G and Gf are not equal.

� ��

��

�

��

��

��

1

2

3 4

5

6

G

� ��

��

� ��

��

��

��

��

��

��

1

2

3 4

5

6

3′ 4′

Gf

7.7.29 Load the package EdgeIdeals [166] in Macaulay2 [199] to verify thatthe chromatic number χ(G) of the graph G of Example 7.7.18 is equal to 4.

7.7.30 Pick any planar graph and use EdgeIdeals [166] to verify the fourcolor theorem (any planar graph has chromatic number at most 4).

Chapter 8

Toric Ideals and AffineVarieties

In this chapter we study algebraic and geometric aspects of three importanttypes of polynomial ideals and their quotient rings:

toric ideals ↪→ lattice ideals ↪→ binomial ideals.

These ideals are interesting from a computational point of view and arerelated to diverse fields, such as, numerical semigroups [173, 341], semigrouprings [180], commutative algebra and combinatorics [61, 317, 395], algebraicgeometry [176, 317], linear algebra and polyhedral geometry [354, 420], inte-ger programming [400], graph theory and combinatorial optimization [405],algebraic coding theory [348, 367], and algebraic statistics [142].

8.1 Binomial ideals and their radicals

Ideals generated by binomials are an interesting family of polynomial ideals.In this section we study the radical of those ideals.

Definition 8.1.1 Let S = K[t1, . . . , tq] be a polynomial ring over a fieldK. A binomial of S is an element of the form f = tα − tβ, for some α, β inNq. An ideal generated by binomials is called a binomial ideal .

A polynomial with at most two terms, say λta−μtb, where λ, μ ∈ K anda, b ∈ Nq, is called a non-pure binomial. Accordingly an ideal generated bynon-pure binomials is called a non-pure binomial ideal .

The essence of the Eisenbud–Sturmfels paper [134] is that the associatedprimes, the primary components and the radical of a non-pure binomialideal are non-pure binomial ideals if K is algebraically closed. Binomial

312 Chapter 8

primary decompositions can be recovered from the theory of mesoprimarydecomposition of congruences [267].

Computing a primary decomposition of a binomial ideal over a field ofcharacteristic zero can be done using “Binomials”[266], a package for thecomputer algebra system Macaulay2 [199].

To study binomial ideals over arbitrary fields we introduce the notion ofcongruence that comes from semigroup theory, this allows us to link generalsemigroup rings with binomial ideals (see Proposition 8.1.6).

An equivalence relation on a set S is a subset R of S × S which isreflexive: (a, a) ∈ R, symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R, and transitive:(a, b) ∈ R and (b, c) ∈ R⇒ (a, c) ∈ R. As usual we write a ∼ b if (a, b) ∈ R.

Definition 8.1.2 A congruence in a commutative semigroup with identity(S,+) is an equivalence relation ∼ on S compatible with +.

Example 8.1.3 Let L ⊂ Zq be a subgroup. The relation, a ∼L b if a − bis in L, defines a congruence in the additive semigroup Nq.

Given a congruence∼ on a semigroup S, it is usual to denote the quotientsemigroup of S by S/ ∼:= {a| a ∈ S}, where a is the equivalence class of a.

Lemma 8.1.4 If ∼ is a congruence on a semigroup (S,+) and a ∼ b, thenra ∼ rb for r ∈ N.

Proof. By induction on r. If (r − 1)a ∼ (r − 1)b, then (r − 1)a + b ∼ rb.Since a ∼ b, we get ra ∼ b+ (r − 1)a. Thus ra ∼ rb. �

Lemma 8.1.5 Let S be a semigroup with a congruence ∼ and let a, b ∈ S.If m and n are relatively prime positive integers such that ma ∼ mb andna ∼ nb, then ra ∼ rb for all r > mn−m− n.

Proof. By Lemma 8.7.8 we can write r = η1m+ η2n for some η1, η2 in N.As na = nb and ma = mb, it follows that ra = rb as required. �

The radical of a binomial ideal Let S = K[t1, . . . , tq] be a polynomialring over a field K and let I = (ta1− tb1 , . . . , tas− tbs) be a binomial ideal ofS. We set �0 := {(a1, b1), . . . , (as, bs)}. The intersection of all congruencesin (Nq,+) that contain �0 is called the congruence generated by �0 and itwill be denoted by � in what follows.

According to [180] we can construct � as follows:

�1 = �0 ∪ {(a, a)| a ∈ Nq} ∪ {(b, a)| (a, b) ∈ �0},�2 = {(a+ c, b+ c)| (a, b) ∈ �1, c ∈ Nq},� = {(a, b)| ∃a0, . . . , ar ∈ Nq with a0 = a, ar = b, (ai, ai+1) ∈ �2 ∀i}.

Since � is clearly a congruence contained in any congruence containing �0,we have that � is the congruence generated by �0.

Toric Ideals and Affine Varieties 313

Proposition 8.1.6 [180, Corollary 7.3] I = ({ta − tb| (a, b) ∈ �}).

Proof. It follows rapidly from the construction of �. �

Let ∼ be a congruence in Nq. We associate to ∼ an equivalence relationon the monomials in S, by ta ∼ tb if a ∼ b. Note that this relation iscompatible with the product, i.e., ta ∼ tb implies tcta ∼ tctb for all c ∈ Nq.

Definition 8.1.7 A non-zero polynomial f =∑

α λαtα in S, with λα ∈ K,

is called simple (with respect to ∼) if all its monomials, i.e., those tα withnon-zero coefficient λα, are equivalent under ∼.

Given any polynomial f ∈ S \ {0}, we can group together its monomialsby equivalence classes under ∼, thereby obtaining a decomposition

f = h1 + · · ·+ hm

with the property that each summand hi is simple, and that no monomialin hi is equivalent with a monomial in hj if j �= i. Such a decomposition off as a sum of maximal simple subpolynomials is unique up to order. Wewill refer to the hi’s as the simple components of f (with respect to ∼).These concepts were introduced by Eliahou [135].

Lemma 8.1.8 Let ∼ be a congruence in Nq. If J is an ideal of S generatedby a set G of simple polynomials and f ∈ J , then every simple componentof f also belongs to J .

Proof. Each generator g ∈ G is simple. Moreover, tcg remains simplefor any c ∈ Nq, since the relation ∼ on monomials is compatible with theproduct. Let f be any element in the ideal J . Then, f is a finite linearcombination of polynomials of the form tcigi, with gi ∈ G simple and ci ∈ Nq,which are simple. Hence, every simple component of f is also a linearcombination of some tcigi and therefore belongs to J . �

Definition 8.1.9 Let a, b ∈ Nq and let p be a positive prime number. If(pra, prb) ∈ � for some integer r ≥ 0, we say that a and b are p-equivalentmodulo � and write a ∼p b.

The radical of a binomial ideal is generated by binomials as is seen below.If K is algebraically closed, then the radical of a non-pure binomial ideal isa non-pure binomial ideal [134, Theorem 3.1] (see Exercise 8.1.15).

Theorem 8.1.10 [180] If p = char(K) �= 0, then

rad(I) = ({ta − tb| a ∼p b}).

314 Chapter 8

Proof. “⊃”: If a ∼p b, then tpra − tprb ∈ I by Proposition 8.1.6. Thus

(ta − tb)pr belongs to I and ta − tb ∈ rad(I).“⊂”: Take f ∈ rad(I), we can write f =

∑mi=1 λit

αi , with λi ∈ K forall i. For r 0 we have fp

r ∈ I. From the equality fpr

=∑m

i=1 μitprαi ,

μi = λpr

i , and using Lemma 8.1.8 we may assume that fpr

is a simplecomponent with respect to the congruence �. Note that

∑mi=1 μi = 0, hence

fpr

= μ2(tprα2 − tprα1) + · · ·+ μm(tp

rαm − tprα1),

with (prαi, prα1) ∈ � for all i ≥ 2, as required. �

Lemma 8.1.11 Let a, b ∈ Nq. If there is an integer s > 0 such that (ra, rb)is in � for all r ≥ s, then f = ta − tb ∈ rad(I).

Proof. Set r = 2s+ 1. We claim f r ∈ I. From the binomial expansion:

f r =

s∑i=0

[(−1)i

(r

i

)(ta)r−i(tb)i + (−1)r−i

(r

r − i

)(ta)i(tb)r−i

]. (8.1)

Note s ≤ r − i for 0 ≤ i ≤ s. Hence, from ((r − i)a, (r − i)b) ∈ �, we get

(ra, (r − i)b+ ia) ∈ � and ((r − i)a+ ib, rb) ∈ �.

Thus, as (ra, rb) ∈ �, we get ((r− i)b+ ia, (r− i)a+ ib)) ∈ �. Therefore anysummand in Eq. (8.1) is in I, as required. �

Following [180], two elements a, b in Nq that satisfy the conditions ofLemma 8.1.11 are called asymptotically equivalent modulo �.

Theorem 8.1.12 [180] If char(K) = 0, then

rad(I) = ({ta − tb| ∃ 0 �= s ∈ N such that (ra, rb) ∈ � ; ∀ r ≥ s}).

Proof. By Lemma 8.1.11 we need only show the inclusion “⊂”. Take0 �= f ∈ rad (I). We can write f = λ1t

α1 + · · · + λmtαm with 0 �= λi ∈ K

for all i and λ1 + · · · + λm = 0. If α1 is asymptotically equivalent modulo� to α2, from the equality

f1 = f − λ2(tα2 − tα1) = λ3(tα3 − tα1) + · · ·+ λm(tαm − tα1)

and Lemma 8.1.11 we get f1 ∈ rad(I). Thus we may either assume that αiis not asymptotically equivalent modulo � to αj for i �= j or that f = 0. Tocomplete the proof we assume f �= 0 and derive a contradiction.

Let E be the set of prime numbers p > 0 such that αi is p-equivalentmodulo � to αj . By Lemma 8.1.5 the set E is finite. Consider the ringA0 = Z[λ1, . . . , λm] ⊂ K and the set {mi}i∈I of maximal ideals of A0. Since


A0 is a Hilbert ring, using [65, Corollary A.18], it follows that (0) = ∩i∈Imiand A0/mi is a finite field for i ∈ I. If I0 is the subset of I consisting ofall i such that char(A0/mi) is not in E, then (0) = ∩i∈I0mi. Indeed takea ∈ ∩i∈I0mi, then (

∏p∈E p)a = 0 because (

∏p∈E p)a ∈ mi for all i ∈ I,

thus a = 0 because char(A0) = 0.Let i ∈ I0, we set m = mi and p = char(A0/m). Note p /∈ E. Pick

r sufficiently large such that fpr

is in I. We can write fpr

=∑n

i=1 μitβi ,

with μi ∈ A0 \ {0} and tβ1 , . . . , tβn distinct. Since the simple componentsof fp

r

, with respect to �, are again in I using that∑ni=1 μi = 0 (on every

component of fpr

) we readily see that

fpr

= η1(tγ1 − tδ1) + · · ·+ ηs(t

γs − tδs) ((γi, δi) ∈ �)

with ηi ∈ A0 for all i. If h ∈ A0[t], we denote by h the image of h underthe canonical map A0[t]→ (A0/m)[t]. If f �= 0, then from the equation

fpr = η1(tγ1 − tδ1) + · · ·+ ηs(t

γs − tδs) ((γi, δi) ∈ �)

we get αi ∼p αj modulo � for some i �= j, a contradiction because p is notin E. Thus f ∈ m[t]. Therefore λ1, . . . , λm are in the maximal ideal mi forevery i ∈ I0 and consequently λi = 0 for all i, a contradiction. This proofwas adapted from [180]. �

Exercises

8.1.13 Let I = (ta1 − tb1 , . . . , tas − tbs) ⊂ S and let L be the subgroup ofZq generated by a1 − b1, . . . , as − bs. Prove the following:

(a) The congruence � generated by �0 = {(a1, b1), . . . , (as, bs)} is con-tained in the congruence �′ = {(a, b)| a, b ∈ Nq; a− b ∈ L}.

(b) � = �′ if ti is not a zero divisor of S/I for all i.

(c) If char(K) = p > 0, then I is a radical ideal if and only if Nq/� is asemigroup that is p-torsion-free.

8.1.14 Let I = (y2z−x2t, z4−xt3) ⊂ Q[x, y, z, t]. Prove that I is a completeintersection whose primary decomposition is given by I = q1∩q2∩q3, where

q1 = (y2z − x2t, z4 − xt3, xz3 − y2t2, x3z2 − y4t, y6 − x5z),

q2 = (t3, zt2, z2t, z3, y2z − x2t) and q3 = (x, z). Prove that q1 is prime.

8.1.15 (I. Swanson) If K = Z2(t) is a field of rational functions in onevariable and R = K[x, y] is a polynomial ring, prove that the radical of theideal (x2 + t, y2t+ 1) is equal to (x2 + t, x+ y + 1).

316 Chapter 8

8.2 Lattice ideals

In this section we study lattice ideals and their radicals. We present someclassifications of lattice and toric ideals. In Chapter 9 we will study primarydecompositions and the degree of lattice ideals.

The class of lattice ideals has been studied in several places; see forinstance [22, 134, 251, 337] and the references therein.

Let S = K[t1, . . . , tq] be a polynomial ring over a field K and let Lbe a lattice in Zq; that is, L is an additive subgroup of Zq. The rank ofL, denoted by rank(L), is the nonnegative integer r such that L � Zr. Apartial character is a group homomorphism from L to the multiplicativegroup K∗ = K \ {0}.

Definition 8.2.1 Let L be a lattice in Zq and let ρ : L → K∗ be a partialcharacter. The lattice ideal of L relative to ρ is the ideal:

Iρ(L) = ({ta+ − ρ(a)ta− | a ∈ L}) ⊂ S.

If ρ is the trivial character ρ(a) = 1 for a in L, we denote Iρ(L) simply byI(L). If no partial character is specified we shall always assume that ρ isthe trivial character.

Theorem 8.2.2 [134] Let L be a lattice in Zq of rank r and let ρ : L → K∗

be a partial character. The following hold.

(i) Iρ(L) contains no monomials.

(ii) dim(S/Iρ(L)) = q − r, that is, ht(I(L)) = rank(L).

(iii) ti is a non-zero divisor of S/Iρ(L) for all i.

(iv) If Zq/L is torsion-free, then Iρ(L) is prime.

Proof. Let α1, . . . , αr be a Z-basis of L and let A be the r × q matrixwith rows α1, . . . , αr. Consider unimodular integral matrices P = (pij) andQ = (qij) of orders q and r, respectively, and make a change of basis

α′i =

r∑j=1

qijαj and e′i =

q∑j=1

pijej .

By Theorem 1.2.2, QAP−1 = diag{d1, . . . , dr, 0, . . . , 0} for some P and Q.By the change of basis theorem the new basis is related by α′

i = die′i for

i = 1. . . . , r, where ei = e′iP−1. Thus L = 〈d1e′1, . . . , dre′r〉. Note that the

isomorphism φ : Zq → Zq, α→αP−1, induces a K-algebra isomorphism

ϕ : S′ = K[t±11 , . . . , t±1

q ] −→ S′ (tαϕ−→ tφ(α)).


Consider the ideal I ′ = (tα − ρ(α)|α ∈ L) ⊂ S′. If α ∈ L, then we canwrite α =

∑ri=1 aiα

′i for some a1, . . . , ar in Z and tα − ρ(α) maps under ϕ

to ta1d11 · · · tardrr −ρ(α′1)a1 · · · ρ(α′

r)ar . Therefore, setting hi = tdii −ρ(α′

i) fori = 1, . . . , r, it follows that ϕ(I ′) is equal to (h1, . . . , hr). This expressionfor ϕ(I ′) is useful to study the structure of Iρ(L) as is seen below.

(i): Let K be the algebraic closure of K and write ρ(α′i) = γdii for

some γi in K. If Iρ(L) contains a monomial, then ϕ(I ′) contains a Laurent

monomial since Iρ(L) ⊂ I ′, a contradiction because tdii − γdii = 0 if ti = γi.

(ii): As h1, . . . , hr form a regular sequence, we get ht(ϕ(I ′)) = r. SinceS′/I ′ is isomorphic to S′/ϕ(I ′), by Exercise 8.2.31, it follows that the Krulldimension of S/Iρ(L) is equal to q − r.

(iii): Let G = {g1, . . . , gs} be a Grobner basis for Iρ(L) with respect to aterm order ≺. For simplicity we set i = 1. Assume that there is 0 �= h ∈ Ssuch that t1h ∈ Iρ(L). By the division algorithm (see Theorem 3.3.6) wecan write h = h1g1+· · ·+hsgs+f , where no term in f is divisible by in≺(gi)for all i; that is, the distinct terms ta1 , . . . , tar occurring in f are standardmonomials. To show that t1 is not a zero divisor of S/Iρ(L), it suffices toprove that f = 0. Assume that f �= 0. Let ∼ be the congruence given bya ∼ b if a − b ∈ L. Since Iρ(L) cannot contain monomials (see part (i)),we can decompose t1f into simple components with respect to ∼ and applyLemma 8.1.8 to derive that (e1 + ai)− (e1 + aj) ∈ L for some i �= j. Hencetai − ρ(ai − aj)taj belongs to Iρ(L) and either tai or taj is not standard, acontradiction.

(iv): If Zq/L is torsion-free, then di = 1 for i = 1, . . . , r. Thus ϕ(I ′) isprime because hi = ti − ρ(α′

i) for all i. Hence I′ is also prime. By part (iii)

it follows that I ′ ∩ S = Iρ(L). Thus Iρ(L) is prime as well. �

Definition 8.2.3 The saturation of a lattice L ⊂ Zq, denoted by Ls, is thelattice consisting of all α ∈ Zq such that ηα ∈ L for some η ∈ Z \ {0}.

From the proof of Theorem 8.2.2 we obtain the following expression forthe saturation (cf. Lemma 8.2.5).

Corollary 8.2.4 If L is a lattice in Zq, then Ls = Z{e′1, . . . , e′r}

Proof. Since α′i = die

′i for i = 1, . . . , r, we need only show that Ls is

contained in Z{e′1, . . . , e′r}. Take α ∈ Ls, then α =∑ri=1(aidi/n)e

′i with

n, ai ∈ Z;n �= 0. Using that the set {e′1, . . . , e′q} is a Z-basis for Zq we obtainthat all the coefficients aidi/n are in Z, thus α is in Z{e′1, . . . , e′r}. �

The next result gives another method to compute the saturation.

Lemma 8.2.5 If L is a lattice of rank r in Zq, then there is an integralmatrix A of size (q − r) × q and rank(A) = q − r such that L ⊂ kerZ(A),with equality if and only if Zq/L is torsion-free. Furthermore kerZ(A) is thesaturation of L.

318 Chapter 8

Proof. We may assume that L = Zα1 ⊕ · · · ⊕ Zαr, where α1, . . . , αq is abasis of the vector space Qq. Consider the hyperplane Hi of Qq generatedby α1, . . . , αi, . . . , αq. Note that the subspace of Qq generated by α1, . . . , αris equal to Hr+1 ∩ · · · ∩Hq. There is a normal vector wi ∈ Zq such that

Hi = {α ∈ Qq| 〈α,wi〉 = 0}.

It is not hard to see that the matrix A with rows wr+1, . . . , wq is the matrixwith the required conditions because by construction αi ∈ Hj for i �= j andconsequently wr+1, . . . , wq are linearly independent. In particular we havethe equality rank(L) = rank(kerZ(A)). Since kerZ(A)/L is a finite group itfollows that Ls, the saturation of L, is equal to kerZ(A). �

Lemma 8.2.6 Let L be a lattice in Zq. If f = ta−λtb ∈ S with 0 �= λ ∈ K,then f ∈ Iρ(L) if and only if a− b ∈ L and λ = ρ(a− b).

Proof. Assume that f ∈ Iρ(L). Consider the congruence in Nq given bya ∼L b if a − b ∈ L. By Theorem 8.2.2, Iρ(L) contains no monomials.Hence, by Lemma 8.1.8, f is simple, i.e., a− b ∈ L. As ta − ρ(a− b)tb is inIρ(L), we get λ = ρ(a− b). The converse is clear. �

Proposition 8.2.7 Let Iρ(L) ⊂ S be a lattice ideal and let ≺ be a termorder. Then the elements in the reduced Grobner basis of Iρ(L) are of theform ta+ − ρ(a)ta− with a ∈ L.

Proof. Let G be a finite set of generators of Iρ(L) consisting of polynomialsof the form λ1t

a − λ2tb with λi ∈ K∗ for i = 1, 2 and let f, g ∈ G. As Iρ(L)contains no monomials (see Theorem 8.2.2), the S-polynomial S(f, g) andthe remainder of S(f, g) with respect to G are both also of this form. Then itfollows that the output of the Buchberger’s algorithm (see Theorem 3.3.12)is a Grobner basis of Iρ(L) consisting of polynomials of the same form.Hence, by Lemma 8.2.6 and Theorem 8.2.2(iii), the elements in the reducedGrobner basis of Iρ(L) are of the form ta+ − ρ(a)ta− with a ∈ L. �

Theorem 8.2.8 If I is a non-pure binomial ideal of S, then I is a latticeideal of the form Iρ(L) for some lattice L and some partial character ρ ifand only if I � S and ti is a non-zero divisor of S/I for all i.

Proof. ⇒) This part follows from items (i) and (iii) of Theorem 8.2.2.⇐) Let L be the set of a ∈ Zq such that ta+ −λta− ∈ I for some λ ∈ K∗

and let a, b be two points in L. Then tb+ − λ1tb− ∈ I for some λ1 in K∗.We claim that L is a lattice. Since ta− − λ−1ta+ ∈ I, we get −a ∈ L.Notice that ta++b+ − λλ1ta−+b− is in I. Factoring the gcd of ta++b+ andta−+b− , we get a+ b ∈ L. Given a ∈ L there is a unique λ ∈ K∗ such thatta+ − λta− ∈ I. Setting ρ(a) = λ, we get a partial character. Thus I is thelattice ideal Iρ(L). This part was adapted from [134]. �


Let L ⊂ Zq be a lattice. We associate to L an equivalence relation onthe monomials in S = K[t1, . . . , tq], by t

α ∼L tβ if α− β ∈ L.

Lemma 8.2.9 Let Iρ(L) be a lattice ideal and let I be an ideal generatedby a subset {tai−ρ(ai− bi)tbi}ri=1 of Iρ(L). If G = Z{ai− bi}ri=1 and f ∈ I,then every simple component of f with respect to ∼G also belongs to I.

Proof. It follows from Lemma 8.1.8 because the relation, a ∼G b if a−b ∈ Gdefines a congruence in Nq and tai − ρ(ai − bi)tbi is simple for all i. �

Lemma 8.2.10 Let a, αi, b, βi be in Nq for i = 1, . . . , r, let a− b be in thelattice L = Z{α1−β1, . . . , αr−βr} and let ρ : L → K∗ be a partial character.Then there is tδ ∈ S such that

tδ(ta − ρ(a− b)tb) ∈ (tα1 − ρ(α1 − β1)tβ1 , . . . , tαr − ρ(αr − βr)tβr).

Proof. We set f = ta−ρ(a−b)tb and gi = tαi−ρ(αi−βi)tβi for i = 1, . . . , r.There are integers n1, . . . , nr such that

(ta/tb)− ρ(a− b) =(tα1/tβ1

)n1 · · ·(tαr/tβr

)nr − ρ(a− b). (∗)

We may assume that ni ≥ 0 for all i by replacing, if necessary, tαi/tβi by itsinverse. Writing tαi/tβi = ((tαi/tβi)−ρ(αi−βi))+ρ(αi−βi) and using thebinomial theorem, from Eq. (∗), it follows that tδf ∈ (g1, . . . , gr) for somemonomial tδ. �

Lemma 8.2.11 A lattice L ⊂ Zq is generated by a1, . . . , am if and only if

Iρ(L) = ((ta+1 − ρ(a1)ta

−1 , . . . , ta

+m − ρ(am)ta

−m) : (t1 · · · tq)∞).

Proof. ⇒) We set I = (ta+1 − ρ(a1)ta

−1 , . . . , ta

+m − ρ(am)ta

−m). “⊂”: Take

α ∈ L. By Lemma 8.2.10 there is tδ ∈ S such that tδ(tα+ − ρ(α)tα−

) ∈ I.Thus tα

+ − ρ(α)tα− ∈ (I : (t1 · · · tq)∞). “⊃”: This inclusion follows readilyfrom Theorem 8.2.8.⇐) This part follows from Lemma 8.2.9 �

The next result can be used to compute the presentation ideal of a sub-ring generated by rational functions.

Proposition 8.2.12 Let F = {f1/g1, . . . , fq/gq} ⊂ K(x) be a finite set ofrational functions, with fi, gi ∈ K[x] = K[x1, . . . , xn] and gi �= 0 for all i,and let ϕ be the homomorphism of K-algebras

ϕ : S = K[t1, . . . , tq] −→ K[F ], induced by ϕ(ti) = fi/gi.

Then ker(ϕ) = (g1t1− f1, . . . , gqtq − fq, yg1 · · · gq − 1)∩S, where y is a newvariable. If gi = 1 for all i, then ker(ϕ) = (t1 − f1, . . . , tq − fq) ∩ S.

320 Chapter 8

Proof. We set J = (g1t1 − f1, . . . , gqtq − fq, yg1 · · · gq − 1) and hi = fi/gifor i = 1, . . . , q. Let us show the first equality. The second equality willfollow by similar arguments. To show that ker(ϕ) ⊂ J ∩ S take f ∈ ker(ϕ).Making ti = (ti − hi) + hi and using the binomial theorem we can write:

f = f(t1, . . . , tq) =∑γ aγ(t1 − h1)γ1 · · · (tq − hq)γq +

∑β aβh

β1

1 · · ·hβqq ,

with aγ ∈ K[F ] for all γ = (γ1, . . . , γq) and all γ’s in Nq \ {0}. As

f(h1, . . . , hq) = 0, we get∑

β aβhβ1

1 · · ·hβqq = 0. Therefore multiplying

the equality above by an appropriate positive power of g1 · · · gq we get

(g1 · · · gq)sf =∑

γ bγ(g1t1 − f1)γ1 · · · (gqtq − fq)γq , (8.2)

where bγ is polynomial in K[x] for all γ. Making z = yg1 · · · gq − 1 we getg1 · · · gq = (z + 1)/y. Thus from Eq. (8.2), we obtain that (z + 1)sf ∈ J .Thus f ∈ J ∩ S. Conversely let f ∈ J ∩ S. Then

f = f(t1, . . . , tq) = a1(g1t1 − f1) + · · ·+ aq(gqtq − fq) + b(g1 · · · gqy − 1).

Hence f(h1, . . . , hq) = b′(g1 · · · gqy − 1). The polynomial f is independentof y. Making y = 1/g1 · · · gq, we get f(h1, . . . , hq) = 0, i.e., f ∈ ker(ϕ). �

Let us illustrate how to compute presentation ideals using Grobner basesand elimination of variables.

Example 8.2.13 Let F = {xixj | 1 ≤ i < j ≤ 4} and let ϕ : K[t] → K[F ]be the map of K-algebras induced by ϕ(tij) = xixj , where t is the set ofvariables {tij | 1 ≤ i < j ≤ 4}. By Proposition 8.2.12 one has

PF = ker(ϕ) = L ∩K[tij ], L = (tij − xixj | 1 ≤ i < j ≤ 4).

Using Macaulay2 [199] we get that the reduced Grobner basis of L withrespect to the elimination ordering in the first variables x1, . . . , x4 is:

t14t23 − t12t34, t13t24 − t12t34, x24t23 − t24t34, x24t13 − t14t34,x24t12 − t14t24, x3t24 − x4t23, x3t14 − x4t13, x3x4 − t34,x23t12 − t13t23, x2t34 − x4t23, x2t14 − x4t12, x2t13 − x3t12,x2x4 − t24, x2x3 − t23, x1t34 − x4t13, x1t24 − x4t12,x1t23 − x3t12, x1x4 − t14, x1x3 − t13, x1x2 − t12,x3t12t34 − x4t13t23.

Therefore PF = (t14t23 − t12t34, t13t24 − t12t34) by Theorem 3.3.20.

The ring of Laurent polynomials in the variables x1, . . . , xn, denoted byK[x±1] or K[x±1

1 , . . . , x±1n ], is generated as a K-vector space by the set of

Laurent monomials , i.e., by the set of all xa with a ∈ Zn.


Definition 8.2.14 Let A = {v1, . . . , vq} be a set of points in Zn. Themonomial subring generated or spanned by F = {xv1 , . . . , xvq} is:

K[F ] :=⋂D∈F

D,

where F is the family of all subrings D of R such that K ∪ F ⊂ D.

The elements of K[F ] have the form∑ca (x

v1)a1 · · · (xvq )aq with ca ∈ K

and all but a finite number of ca’s are zero. As a K-vector space K[F ] isgenerated by the set of monomials of the form xa, with a in the semigroupNA generated by A. That is K[F ] = K[xa| a ∈ NA]. This means that K[F ]coincides with K[NA], the semigroup ring of the semigroup NA (see [180]).

Definition 8.2.15 An ideal P ⊂ S is called a toric ideal if there is afinite set F = {xv1 , . . . , xvq} in K[x±1] such that P is the kernel of theepimorphism of K-algebras:

ϕ : S = K[t1, . . . , tq] −→ K[F ] −→ 0, induced by ϕ(ti) = xvi .

The ideal P is called the toric ideal of K[F ] and is denoted by PF .

Proposition 8.2.16 [400, p. 32] If F = {xv1 , . . . , xvq} is a set of Laurentmonomials in K[x±1], then the corresponding toric ideal is given by

PF = (xv−1 t1 − xv

+1 , . . . , xv

−q tq − xv

+q , yx1 · · ·xn − 1) ∩ S.

Proof. It follows adapting the proof of Proposition 8.2.12. �

This result gives a method to compute toric ideals. Another methodcan be found in [109]. In [38] some algorithms are devised, that in manyrespects improve existing algorithms for the computation of toric ideals.

Lemma 8.2.17 Let B = K[y1, . . . , yn, t1, . . . , tq] be a polynomial ring overa field K. If I is a binomial ideal of B, then the reduced Grobner basisof I with respect to any term order ≺ consists of binomials. FurthermoreI ∩K[t1, . . . , tq] is a binomial ideal.

Proof. Let B be a finite set of generators of I consisting of binomials andlet f, g ∈ B. Since the S-polynomial S(f, g) is again a binomial and theremainder of S(f, g) with respect to B is also a binomial, the output of theBuchberger’s algorithm (Theorem 3.3.12) is a Grobner basis of I consistingof binomials. Hence the reduced Grobner basis of I consists of binomials.

If ≺ is the lex order y1 " · · · " yn " t1 " · · · " tq and K[t] is the ringK[t1, . . . , tq], then by elimination theory (see Theorem 3.3.20) G ∩K[t] is aGrobner basis of I ∩K[t]. Hence I ∩K[t] is a binomial ideal. �

322 Chapter 8

Corollary 8.2.18 The reduced Grobner basis of any toric ideal P ⊂ Sconsists of binomials of the form ta+ − ta− with respect to any term order.

Proof. By Proposition 8.2.16 and Lemma 8.2.17, P is a binomial ideal.Therefore, by Theorem 8.2.8, P is a lattice ideal because P is prime. Hencethe result follows at once from Proposition 8.2.7. �

Definition 8.2.19 Let F = {xv1 , . . . , xvq} be a set of Laurent monomials.The associated matrix of the monomial subring K[F ], denoted by A, is thematrix whose columns are the vectors v1, . . . , vq.

Corollary 8.2.20 If A is the associated matrix of a monomial subringK[F ], then the toric ideal PF is the lattice ideal of kerZ(A).

Proof. By Corollary 8.2.18, PF is generated by binomials of the formta+ − ta− . Since a binomial ta+ − ta− is in PF if and only if A(a+) = A(a−),we get that PF is generated by the set of all ta+ − ta− such that Aa = 0,i.e., PF is the lattice ideal of kerZ(A). �

Corollary 8.2.21 [216] Let K[F ] be a Laurent monomial subring and letA be its associated matrix. Then dimK[F ] = rank(A).

Proof. The toric ideal P of K[F ] is the lattice ideal of L = kerZ(A) byCorollary 8.2.20. Since the rank of L is q − rank(A), using Theorem 8.2.2,we get that dimK[F ] = rank(A). �

Toric ideals form a special class of lattice ideals.

Theorem 8.2.22 Let L ⊂ Zq be a lattice. The following are equivalent :

(a) I(L) is a toric ideal.

(b) I(L) is a prime ideal.

(c) Zq/L is torsion-free.

(d) L = kerZ(A) for some integral matrix A.

Proof. (a) ⇒ (b): If I(L) is a toric ideal, then S/I(L) � K[F ] for somefinite set of Laurent monomials ofK[x±1]. SinceK[F ] is an integral domain,the ideal I(L) is prime.

(b) ⇒ (c): Assume nα ∈ L for some integer n > 0 and α ∈ Zq. First weassume that either p = char(K) = 0 or that p �= 0 is relatively prime to n.

Since tnα+ − tnα−

is in I(L) and n · 1K �= 0, from the equation

xn − yn = (x− y)(xn−1 + xn−2y + · · ·+ xyn−2 + yn−1)

with x = tα+

and y = tα−we get tα

+ − tα− ∈ I(L). Thus α ∈ L. If p > 0and gcd(p, n) = p, we write n = prm, where p and m are relatively prime.

Since tmα+ − tmα− ∈ I(L), by the previous argument we obtain α ∈ L.


(c) ⇒ (d): Let A be as in Lemma 8.2.5. Thus kerZ(A)/L is a finitegroup, which must be zero because it is torsion-free.

(d) ⇒ (a): Let v1, . . . , vq be the columns of A. If F = {xv1 , . . . , xvq}, byCorollary 8.2.20, PF is the lattice ideal of L = kerZ(A). �

Given a non-pure binomial g = tα−λtβ , with λ ∈ K∗, we set g = α−β.If B ⊂ Zq, we denote by 〈B〉 or ZB the subgroup of Zq generated by B.

Lemma 8.2.23 Let L ⊂ Zq be a lattice. If g1, . . . , gr is a set of non-purebinomials that generate Iρ(L), then L = Z{g1, . . . , gr}. In particular if I isa lattice ideal, there are unique L and ρ such that I = Iρ(L).

Proof. Consider the lattice G = Z{g1, . . . , gr}. First we show the inclusion

L ⊂ G. Take 0 �= a ∈ L. Then f = ta+ − ρ(a)ta− is in Iρ(L) = (g1, . . . , gr).

By Lemma 8.2.9 any simple component of f , with respect to ∼G , is also in

(g1, . . . , gr). Since ta+ and ta

−are not in Iρ(L), then f simple with respect

to ∼G , i.e., a = a+ − a− ∈ G. Thus L ⊂ G. To show the other inclusionnotice that, by Lemma 8.2.6, gi ∈ L for all i, i.e., G ⊂ L. �

Lemma 8.2.24 Let I be an ideal generated by a set of binomials g1, . . . , grof S and let h1, . . . , hs be a set of binomials of S that generate rad(I). If

char(K) = 0, then Z{g1, . . . , gr} = Z{h1, . . . , hs}.

Proof. We begin by writing the binomials gi and hi as gi = tαi − tβi andhi = tγi − tδi . We set L1 = Z{g1, . . . , gr} and L2 = Z{h1, . . . , hs}.

“⊂” Since gi ∈ rad(I), then by Lemma 8.2.9, any simple component ofgi with respect to L2 belongs to rad(I). Therefore αi ∼L2 βi otherwiserad(I) would contain tαi , which is impossible. This proves that L1 ⊂ L2.

“⊃” Since hi = tγi − tδi is in rad(I), then hpm

i is in I for m 0 and foran arbitrary odd prime number p. We claim that tp

mγi ∼L1 tpmδi . Consider

the following equality which is obtained from the binomial theorem

hpm

i =

pm∑k=0

(−1)k(pm

k

)(tγi)p

m−k(tδi)k .

If tpmγi and tp

mδi are not in the same simple component of hpm

i with respectto L1, then there is C ⊂ {1, . . . , pm − 1}, C �= ∅, such that the polynomial

f = tpmγi +

∑k∈C

(−1)k(pm

k

)(tγi)p

m−k(tδi)k

is a simple component of hpm

i with respect to L1. By Lemma 8.2.9, f ∈ I.Hence, since char(K) = 0, we get

f(1, . . . , 1) = 0 = 1 +∑k∈C

(−1)k(pm

k

),

324 Chapter 8

a contradiction to(pm

k

)≡ 0 mod(p) for 1 ≤ k ≤ pm− 1; see Exercise 8.3.43.

Therefore tpmγi ∼L1 t

pmδi . Thus pm(γi − δi) ∈ L1. If we pick another oddprime number q �= p and t 0, repeating the previous argument, we getqt(γi − δi) ∈ L1, and hence γi − δi ∈ L1, as required. �

Theorem 8.2.25 Let I be a binomial ideal and let I = q1 ∩ · · · ∩ qm be anirredundant primary decomposition of I. If q1, . . . , qm are binomial idealsand char(K) = 0, then rad(I) = I.

Proof. There are binomials g1, . . . , gr and h1, . . . , hs that generate I andrad(I), respectively. Say gi = tαi − tβi and hi = tγi − tδi . Since rad(I) isequal to rad(q1) ∩ · · · ∩ rad(qm), it suffices to prove the case m = 1. Thus,let us assume that I is primary and show that hi = tγi− tδi belongs to I forall i = 1, . . . , s. By Lemma 8.2.24, we can write γi − δi =

∑rj=1 ηj(αj − βj)

with ηj ∈ Z for all j, and by Lemma 8.2.10 there is a monomial tγ suchthat tγhi ∈ I. If hi /∈ I, then (tγ)� ∈ I for some � ≥ 1 because I is primary,but this is impossible. Thus hi ∈ I, as required. �

Corollary 8.2.26 Let I be a binomial ideal without embedded primes. Ifchar(K) = 0 and rad(I) is prime, then rad(I) = I and I is a prime ideal.

Proof. Set p = rad(I). Since I has no embedded primes, by Exercise 2.1.48,one has Ass(S/I) = {p}. Thus I is a p-primary ideal and I is a radical idealby Theorem 8.2.25. Thus I = p and I is a prime ideal. �

Theorem 8.2.27 Let I(L) be the lattice ideal of L and let p = char(K).

(a) If p = 0, then rad(I(L)) = I(L).

(b) If p �= 0, then rad(I(L)) = (ta − tb| pr(a− b) ∈ L for some r ∈ N).

Proof. (a) Assume tra− trb ∈ I(L) for r ≥ s0 > 0. By Theorem 8.1.12 andExercise 8.1.13 it suffices to prove that a − b ∈ L. Note that the relation“α ∼ β if and only if α − β ∈ L” is a congruence in Nq, thus r(a − b) ∈ Lfor r ≥ s0 and consequently a− b ∈ L.

(b) It follows from Theorem 8.1.10 and Lemma 8.1.8. �

Definition 8.2.28 An ideal I ⊂ S is called a binomial set-theoretic com-plete intersection if there are binomials f1, . . . , fr in I such that rad (I) isequal to rad (f1, . . . , fr), where r = ht (I).

Theorem 8.2.29 [293] Let I(L) ⊂ S be a lattice ideal of height r and letI be an ideal of S generated by binomials g1, . . . , gr. If rad(I(L)) = rad(I)and char(K) = 0, then I(L) = I.


Proof. By Theorem 8.2.27 the ideal I(L) is radical. Let I = q1 ∩ · · · ∩ qsbe a primary decomposition of I. Since I is an ideal of height r generatedby r elements, by the unmixedness theorem I has no embedded primes; seeTheorem 2.3.27. Hence rad(qi) = pi is a minimal prime of both I and I(L).Take a binomial h in I(L). By Lemma 8.2.24 and Lemma 8.2.10, there is amonomial tα so that tαh ∈ I. Thus tαh ∈ qi for all i. It suffices to provethat h belongs to qi for all i. If h /∈ qi for some i, then (tα)� ∈ qi andconsequently pi must contain a variable, a contradiction to Theorem 8.2.8because none of the variables of S is a zero divisor of S/I(L). �

Exercises

8.2.30 If L is a lattice of Zq and Iρ(L) ⊂ S is a lattice ideal. Prove that

Iρ(L) = ({ta − ρ(a)| a ∈ L}) ∩ S

and ({ta − ρ(a)| a ∈ L}) = Iρ(L)S′, where S′ = K[t±1 , . . . , t±q ].

8.2.31 Let T be the set of monomials of S. If L = Iρ(L) is a lattice idealof S and p ∈ AssS(S/L), then T ∩ p = ∅. Prove that ht(L) is equal toht(T−1(L)) and that T−1(L) = ({ta − ρ(a)| a ∈ L}) ⊂ K[t±1 , . . . , t

±q ].

8.2.32 Let I be a binomial ideal of S = K[t1, . . . , tq]. Then

(a) I is a lattice ideal of S if and only if I = (I : (t1 · · · tq)∞).

(b) If I = (ta1 − tb1 , . . . , tas − tbs), then (I : (t1 · · · tq)∞) = I(L), where Lis the lattice L = Z{ai − bi}si=1 in Zq generated by {ai − bi}si=1.

8.2.33 Let S = K[t1] be a polynomial ring in one variable. Prove that thelattice ideals of S are the ideals of the form (tm1 − 1), where m ≥ 0.

8.2.34 Let I ⊂ K[t1, . . . , tq] be a binomial ideal generated by tc+i − tc−i ,

i = 1, . . . ,m, and let L ⊂ Zq be the lattice generated by c1, . . . , cs. Provethat I(L) = IK[t±1] ∩ S.

8.2.35 Let K ⊂ F be a field extension, let S = K[t] and B = F [t] bepolynomial rings and let I be an ideal of S. Then the following hold. (a)IB∩S = I. (b) If ∩ri=1qi is a primary decomposition of IB, then ∩ri=1(qi∩S)is a primary decomposition of I such that rad(qi ∩ S) = rad(qi) ∩ S. (c) IfI is the lattice ideal of L in S, then IB is the lattice ideal of L in B.

8.2.36 Let L be a lattice in Zq and let ρ : L → K∗ be a partial character.If Iρ(L) is prime and K is algebraically closed, then Zq/L is torsion-free.

8.2.37 Let L be the lattice 2Z and let ρ : L → Q∗ be the partial charactergiven by ρ(2p) = (−1)p. If K = Q, prove that Iρ(L) = (t21 + 1).

326 Chapter 8

8.3 Monomial subrings and toric ideals

Let R = K[x] = K[x1, . . . , xn] = ⊕i≥0Ri be a polynomial ring over a fieldK with the standard grading. As usual, x denotes the set of indeterminatesx1, . . . , xn of the ring R. In this section we focus on monomial subringsgenerated by monomials in the polynomial ring R.

Let F = {f1, . . . , fq} be a finite set of monomials in R with fi = xvi andfi �= 1 for all i. The set F has a corresponding set of vectors in Nn:

F = {xv1 , . . . , xvq} ←→ A = {v1, . . . , vq}.

The monomial subring K[F ] is a graded subring of R with the gradinggiven by K[F ]i = K[F ]∩Ri. There is a graded epimorphism of K-algebras

ϕ : S = K[t1, . . . , tq] −→ K[F ] −→ 0, induced by ϕ(ti) = fi,

where S = ⊕i≥0Si is a positively graded polynomial ring with the gradinginduced by setting deg(ti) = deg(xvi) = |vi|. We also denote S by K[t].

The kernel of ϕ, denoted by PF , is the so called toric ideal of K[F ] withrespect to f1, . . . , fq. We also denote the toric ideal of K[F ] by IA. In thiscase we say that IA is the toric ideal of A.

Let A be the n× q matrix with column vectors v1, . . . , vq. This matrixis called the associated matrix of K[F ]. Closely related to the map ϕ is thehomomorphism ψ : Zq −→ Zn determined by the matrix A in the standardbases of Zq and Zn. Indeed, we have ϕ(tα) = xψ(α) for all α ∈ Nq.

Recall that given a binomial g = tα − tβ , we set g = α− β.

Proposition 8.3.1 If g1, . . . , gr is a set of binomials that generate the toricideal PF , then g1, . . . , gr generate kerZ(A).

Proof. By Corollary 8.2.20, PF is the lattice ideal of kerZ(A). Thus theresult follows from Lemma 8.2.23. �

Definition 8.3.2 A binomial tα− tβ ∈ PF is called primitive if there is noother binomial tγ − tδ ∈ PF such that tγ divides tα and tδ divides tβ .

Lemma 8.3.3 If f is a binomial in the reduced Grobner basis of PF withrespect to some term order ≺, then f is a primitive binomial.

Proof. Let G be the reduced Grobner basis of PF . One may assumef = tα+ − tα− and α+ " α−. Assume g = tβ+ − tβ− �= f is a binomial in PFsuch that tβ+ divides tα+ and tβ− divides tα− . Note β+ �= α+ and β− �= α−,otherwise f = g. If β+ " β−, then tβ+ ∈ in(G \ {f}) and consequentlytα+ ∈ in(G \ {f}), a contradiction. If β− " β+, then t

β− ∈ in(G \ {f}) andconsequently tα− ∈ in(G \ {f}), a contradiction. �

According to a result of [400], the set GPF of all primitive binomials ina toric ideal PF is finite and is called a Graver basis of PF .


Definition 8.3.4 The universal Grobner basis of a toric ideal PF is theunion of all reduced Grobner basis of PF . It is denoted by UPF .

Definition 8.3.5 If α is a circuit of ker(A) we call the binomial tα+ − tα−

a circuit of PF . The set of circuits of PF is denoted by CPF .

Proposition 8.3.6 [400] CPF ⊂ UPF ⊂ GPF .

Proof. The second inclusion follows from Lemma 8.3.3. To show the firstinclusion take a circuit f = tα+ − tα− of the toric ideal PF . We may assumethat f = tα1

1 · · · tαrr − t

−αr+1

r+1 · · · t−αss , α = (αi) = α+ − α−. We reorder the

set of variables t as t\{t1, . . . , ts}, t1, . . . , ts. Consider the lex order: ta " tbif the first non-zero entry of a− b is positive. Let G be the reduced Grobnerbasis of PF with respect to " and let g = tβ+−tβ− be a binomial in G so thatin(g) = tβ+ divides in(f). Clearly supp(tβ+) ⊂ supp(tα+). By the choice ofterm order it follows that supp(tβ−) ⊂ supp(tα). Thus supp(β) ⊂ supp(α).Since α is a circuit we get equality. Thus using Lemma 1.9.3 we obtain thatβ = λα for some λ ∈ Q, consequently α = β because the non-zero entriesof α (resp. β) are relatively prime. �

Corollary 8.3.7 If in every circuit ta+ − ta− of PF both monomials ta+

and ta− are square-free, then CPF = UPF = GPF .

Proof. By Proposition 8.3.6 it suffices to show the inclusion GPF ⊂ CPF .Let f = tβ+ − tβ− ∈ GPF , i.e., f is primitive. By Lemma 1.9.5 there is acircuit α, in harmony with β, whose support is contained in the supportof β. Since tα+ and tα− are square-free, we conclude that tα+ (resp. tα−)divides tβ+ (resp. tβ−). Hence f = tα+ − tα− because f is primitive. �

Corollary 8.3.8 If A is t-unimodular, then CPF = UPF = GPF .

Proof. By Corollary 1.9.11 in every circuit ta+ − ta− of PF both terms ta+

and ta−are square-free. Thus the equality follows from Corollary 8.3.7. �

Definition 8.3.9 A universal Grobner basis of a polynomial ideal I is afinite set U ⊂ I which is a Grobner basis of I with respect to all term orders.

Proposition 8.3.10 GPF is a universal Grobner basis of PF .

Proof. It follows from Proposition 8.3.6. �

Theorem 8.3.11 The following are equivalent:

(a) In every circuit ta+ − ta− of PF both monomials are square-free.

(b) Every initial monomial ideal of PF is square-free.

328 Chapter 8

Proof. That (a) implies (b) follows from Corollary 8.3.7. Conversely letf be a circuit of PF . We may assume that f = tarr · · · tass − t

as+1

s+1 · · · taqq .

Consider the lex order " with t1 " · · · " tq and consider the weight vectorω = e1 + · · ·+ er−1 ∈ Nq. We define the term order: a "ω b if and only if

ω · a > ω · b or ω · a = ω · b and a " b. Then there is g = tb+ − tb− , in the

reduced Grobner basis of PF , such that tb+

is square-free, in(g) = tb+

, and

tb+

divides ta+

. Hence supp(tb+

) is contained in supp(ta+

). Consequently,

by definition of "ω, we have that supp(tb−) is contained in {tr, . . . , tq}.

Since a is a circuit we get that a and b have the same support. Thus, byLemma 1.9.3, a = λb for some rational number λ. Using that a (resp. b) has

relatively prime non-zero entries we obtain that a = ±b and tb+ = ta+

, i.e.,ta

+

is square-free. A similar argument shows that ta−is also square-free. �

Corollary 8.3.12 [17, Theorem 4] If char(K) = 0 and PF is a binomialset theoretic complete intersection, then PF is a complete intersection.

Proof. Set r = dimS/PF . By hypothesis, there are g1, . . . , gq−r binomialsof S such that rad(g1, . . . , gq−r) = PF . Since the ideal (g1, . . . , gq−r) is C–M(see Proposition 2.3.24), the result follows from Corollary 8.2.26. �

Lemma 8.3.13 [134] Let R be a ring and let x1, . . . , xn ∈ R. If I is anideal of R, then the radical of I satisfies

rad(I) = rad(I : (x1 · · ·xn)∞) ∩ rad(I, x1) ∩ · · · ∩ rad(I, xn).

Proof. The inclusion “⊂” is clear. To show the inclusion“⊃” take a primep ⊃ I, it suffices to show that p contains one of the ideals on the right-handside. If rad(I : (x1 · · ·xn)∞) �⊂ p, there is f ∈ R \ p such that f(x1 · · ·xn)sis in I for some s ≥ 1. Hence xi ∈ p for some i and p ⊃ rad(I, xi). �

Proposition 8.3.14 Let p be a prime ideal of a ring R and let x1, . . . , xnbe a sequence in R \ p. If I ⊂ p is an ideal, then rad(I) = p if and only ifp = rad(I : (x1 · · ·xn)∞) and rad(I, xi) = rad(p, xi) for all i.

Proof. If J is any ideal of R and y1, . . . , yn ∈ R, then by Lemma 8.3.13rad(J) is equal to rad(J : (y1 · · · yn)∞)∩ rad(J, y1)∩ · · · ∩ rad(J, yn). Hencethe result follows by applying this equality to I and p, together with thefact that x1 · · ·xn is regular on R/p. �

Proposition 8.3.15 [139] Let I be an ideal generated by a set of binomialsg1, . . . , gr in the toric ideal PF . If char(K) = p �= 0 (resp. char(K) = 0)and L is the lattice Z{gi}ri=1, then the following conditions are equivalent :

(a1) PF = rad(I : (t1 · · · tq)∞).

(a2) puker(ψ) ⊂ L for some u ∈ N (resp. ker(ψ) = L).


Proof. (a1) ⇒ (a2): First we consider the case char(K) = p �= 0. Settingz = t1 · · · tq, by hypothesis, there exists u ≥ 0 such that fp

u ∈ (I : z∞) forall f ∈ PF . Let γ ∈ ker(ψ). Write γ = α−β with α, β ∈ Nq and g = tα−tβ.Thus g ∈ PF and g = α − β = γ. The lattice L defines a congruence inNq given by a ∼L b if a − b ∈ L. We know that tδgp

u ∈ I for some δ ∈ Nq

and using that p = char(K), one has tδgpu

= tαpu+δ − tβpu+δ. Since the

simple components of tδgpu

with respect to L belong to I and hence to PF(see Lemma 8.2.9), and since tδtαp

u

does not belong to I because it doesnot belong to PF , it follows that t

δgpu

is simple, i.e., δ + puγ ∼L puβ + δ.Hence puα− puβ ∈ L. This shows the inclusion puker(ψ) ⊂ L.

Next consider the case char(K) = 0. By Lemma 8.2.11 one has theequality (I : z) = I(L). Thus PF = rad(I(L)). As PF is the lattice ideal ofker(ψ) (Corollary 8.2.20), by Lemmas 8.2.23 and 8.2.24 we get L = ker(ψ).

(a1)⇐ (a2): It suffices to prove the inclusion PF ⊂ rad(I : z∞), the otherinclusion is clear. Assume char(K) = p > 0. Let f = tc − te be a binomialin PF . By (a2) and Lemma 8.2.10, we get that tγ(tp

uc − tpue) = tγfpu

isin I for some monomial tγ . Thus f is in rad(I : z∞). If char(K) = 0, thesame argument works if we replace pu by 1. �

Theorem 8.3.16 [139] Let I ⊂ PF be an ideal generated by a finite setof binomials g1, . . . , gr. If char(K) = p �= 0 (resp. char(K) = 0), thenrad(I) = PF if and only if

(a) puker(ψ) ⊂ Z{gi}ri=1 for some u ∈ N (resp. ker(ψ) = Z{gi}ri=1), and

(b) rad(I, ti) = rad(PF , ti) for all i.

Proof. It is a consequence of Propositions 8.3.14 and 8.3.15. �

Next we give a result that holds over a field of arbitrary characteristic.It is a consequence of the proofs of Theorem 8.2.25 and Corollary 8.2.26.

Proposition 8.3.17 Let I ⊂ S be an ideal, and let {gi}ri=1, {hi}si=1 be setsof binomials that generate I, rad(I), respectively. If Z{gi}ri=1 is equal to

Z{hi}si=1, I has no embedded primes, and rad(I) is prime, then I = rad(I).

Corollary 8.3.18 A toric ideal PF of height r is a complete intersection ifand only if there are binomials g1, . . . , gr in PF such that

(a) ker(ψ) = Z{g1, . . . , gr}, and(b) (g1, . . . , gr, ti) = (PF , ti) for all i.

Proof. ⇒) It follows readily from Proposition 8.3.1.⇐) By Theorem 8.3.16 we get rad(g1, . . . , gr) = PF . Notice that the

ideal I = (g1, . . . , gr) is a complete intersection because r = ht(PF ) and Iis quasi-homogeneous. Hence applying Proposition 8.3.17 yields I = PF . �

Given a subset I ⊂ S we denote its zero set in AqK by V (I), and givena subset X ⊂ AqK we denote its vanishing ideal in S by I(X).

330 Chapter 8

Corollary 8.3.19 Let I be an ideal generated by binomials g1, . . . , gr inPF . If char(K) = p �= 0 (resp. char(K) = 0) and pm ker(ψ) ⊂ Z{gi}ri=1 forsome m ∈ N (resp. ker(ψ) = Z{gi}ri=1), then V (I) ⊂ V (PF ) ∪ V (t1 · · · tq).

Proof. By Proposition 8.3.15, there is a monomial tδ and an integer Nsuch that tδPNF ⊂ I. It follows that

V (I) ⊂ V (PNF ) ∪ V (tδ) ⊂ V (PF ) ∪ V (t1 · · · tq). �

Lemma 8.3.20 Let I be a binomial ideal of S such that V (I, ti) = {0} forall i. If p is a prime ideal containing (I, tm) for some 1 ≤ m ≤ q, thenp = (t1, . . . , tq).

Proof. Let h1, . . . , hr be a set of binomials that generate I. For simplicityof notation assume that m = 1. We may assume that t1, . . . , tk are in p andti /∈ p for i > k. If ti ∈ supp(hj) for some 1 ≤ i ≤ k, say hj = taj − tbjand ti ∈ supp(taj ), then tbj ∈ p and there is 1 ≤ � ≤ k such that t� is in thesupport of tbj . Thus, hj ⊂ (t1, . . . , tk). Hence, for each 1 ≤ j ≤ r, either

(i) supp(hj) ∩ {t1, . . . , tk} = ∅ or (ii) hj ∈ (t1, . . . , tk).

Consider the point c = (ci) ∈ AqK , with ci = 0 for i ≤ k and ci = 1 fori > k. If (i) occurs, then hj(c) = (taj − tbj )(c) = 1 − 1 = 0. If (ii) occurs,then hj(c) = (taj − tbj )(c) = 0− 0 = 0. Clearly the polynomial t1 vanishesat c. Hence, c ∈ V (I, t1) = {0}. Therefore, k = q. Thus, p contains all thevariables of S, i.e., p = (t1, . . . , tq). �

Corollary 8.3.21 [139] Let I be an ideal generated by binomials g1, . . . , grin PF . If F = {xd11 , . . . , x

dq1 } and char(K) = p �= 0 (resp. char(K) = 0),

then rad(I) = PF if and only if

(a) pmker(ψ) ⊂ Z{gi}ri=1 for some m ∈ N (resp. ker(ψ) = Z{gi}ri=1)

(b) V (g1, . . . , gr, ti) = {0}, for all i.

Proof. ⇒) By Theorem 8.3.16 condition (a) holds and rad(I, ti) is equal torad(PF , ti) for all i. On the other hand rad(PF , ti) = (t1, . . . , tq), becausethe height of PF is q − 1 and ti is regular on S/PF . Hence V (I, ti) = {0}.⇐) By Proposition 8.3.15 one has PF = rad(I : z∞), where z = t1 · · · tq.

Hence there is monomial tδ and an integer N such that tδPNF ⊂ I.To prove rad(I) = PF , it suffices to show that every prime ideal p

containing I also contains PF . If p contains no variables, then the inclusionstδPNF ⊂ I ⊂ p imply that PF ⊂ p, as desired. If p contains at least onevariable, then p contains all the variables by Lemma 8.3.20 and consequentlycontains PF , as desired. �


Proposition 8.3.22 Let I ⊂ S be a graded binomial ideal. The followinghold. (a) If V (I, ti) = {0} for all i, then ht(I) = q − 1. (b) If I is a latticeideal and ht(I) = q − 1, then V (I, ti) = {0} for all i.

Proof. (a) As I is graded, all associated prime ideal of S/I are graded.Thus, all associated prime ideals of S/I are contained in m = (t1, . . . , tq).If ht(I) = q, then m would be the only associated prime of S/I, that is, mis the radical of I, a contradiction because I cannot contain a power of tifor any i. Thus, ht(I) ≤ q − 1. On the other hand, by Lemma 8.3.20, theideal (I, tq) has height q. Hence, q = ht(I, tq) ≤ ht(I) + 1 (here we use thefact that I is graded). Altogether, we get ht(I) = q − 1.

(b) Let L be the lattice that defines I and g1, . . . , gr a set of homogeneousbinomials that generate I. By Lemma 8.2.23, we get L = 〈g1, . . . , gr〉.Notice that q − 1 = ht(I) = rank(L). Given two integers 1 ≤ i, k ≤ q, thevector space Qq is generated by ek, g1, . . . , gr. Hence, as L is homogeneouswith respect to ω = (ω1, . . . , ωq), there are positive integers ri and rk suchthat riei − rkek ∈ L and riωi − rkωk = 0. By Lemma 8.2.10, there is tδ

such that tδ(trii − trkk ) is in I. Hence, by Theorem 8.2.8, trii − t

rkk is in I.

Therefore, V (I, ti) = {0} for all i. �

Lemma 8.3.23 Let I ⊂ S be a graded binomial ideal. If V (I, ti) = {0} forall i and I is a complete intersection, then I is a lattice ideal.

Proof. By Proposition 8.3.22(a), the height of I is q−1. It suffices to provethat ti is a non-zero divisor of S/I for all i (see Theorem 8.2.8). If ti is azero divisor of S/I for some i, there is an associated prime ideal p of S/Icontaining (I, ti). Hence, using Lemma 8.3.20, we get that p = m, a contra-diction because I is a complete intersection of height q−1 and all associatedprime ideals of I have height equal to q − 1 (see Proposition 2.3.24). �

Theorem 8.3.24 [293] If L ⊂ S is a graded lattice ideal and V (L, ti) = {0}for all i, then L is a complete intersection if and only if there are homoge-neous binomials h1, . . . , hq−1 in L satisfying the following conditions :

(i) L = 〈h1, . . . , hq−1〉, where L is the lattice that defines L.

(ii) V (h1, . . . , hq−1, ti) = {0} for all i.

(iii) hi = ta+i − ta−i for i = 1, . . . , q − 1.

Proof. As L is graded, By Proposition 8.3.22, the height of L is q − 1.⇒) By Lemma 3.1.29 and Exercise 3.3.26, L is generated by homoge-

neous binomials h1, . . . , hq−1. Then, by Lemma 8.2.23 and Theorem 8.2.8,(i) and (iii) hold. From the equality (L, ti) = (h1, . . . , hq−1, ti), it followsreadily that (ii) holds.⇐) We set I = (h1, . . . , hq−1). By hypothesis I ⊂ L. Thus, we need only

show the inclusion L ⊂ I. Let g1, . . . , gm be a generating set of L consisting

332 Chapter 8

of binomials, then gi ∈ L for all i. Using condition (i) and Lemma 8.2.10,for each i there is a monomial tγi such that tγigi ∈ I. Hence, tγL ⊂ I,where tγ is equal to tγ1 · · · tγm . By (ii) and Proposition 8.3.22, the heightof I is q − 1. This means that I is a complete intersection. As tγL ⊂ I, toshow the inclusion L ⊂ I, it suffices to notice that by (ii), Lemma 8.3.23and Theorem 8.2.8 ti is a non-zero divisor of S/I for all i. �

Remark 8.3.25 The result remains valid if we remove condition (iii), i.e.,condition (iii) is redundant. In both implications of the theorem the seth1, . . . , hq−1 is shown to generate L.

Definition 8.3.26 The circuit ideal of the matrix A is given by

I = ({tα+ − tα− |α is a circuit of ker(A)}).

The next lemma is useful in induction arguments.

Lemma 8.3.27 Let A′ be the matrix obtained from A by removing its ithcolumn. Then the inclusion map j : Zq−1 → Zq given by

(a1, . . . , ai−1, ai+1, . . . , aq)j−→ (a1, . . . , ai−1, 0, ai+1, . . . , aq)

induces a bijection between the circuits of ker(A′) and the circuits of ker(A)whose ith entry is equal to zero.

Proof. Since the inclusion map j takes ker(A′) into ker(A), the resultfollows readily. �

Proposition 8.3.28 [134] If I is the circuit ideal of A, then rad (I) = PF .

Proof. We proceed by induction on q, the number of columns of A. If q = 1or q = 2, then PF is a principal ideal and the result is clear. Accordingto Theorems 8.3.16 and 1.9.10 it suffices to prove that rad (I, ti) is equalto rad (PF , ti) for all i. For simplicity of notation we assume i = 1. Let

γ1, . . . , γr be the circuits of ker(A). We set fi = tγ+i −tγ−

i for i = 1, . . . , r andf0 = fr+1 = 0. If p is a prime ideal containing (I, t1), then after relabelingt1, . . . , tq and f1, . . . , fr there are t1, . . . , ts ∈ p and 0 ≤ k ≤ r such that (i)(fk+1, . . . , fr) ⊂ (t1, . . . , ts) and (ii) ti /∈ ∪kj=1supp(fj) for i = 1, . . . , s.

We claim that p contains (PF , t1). Take f = tα+ − tα−

in PF . Settingα = α+ − α−, by Theorem 1.9.6, we can write pα =

∑ri=1 niγi for some

p ∈ N+ and n1, . . . , nr ∈ N, such that for each i either ni = 0 or ni > 0 andγi is in harmony with α and supp(γi) ⊂ supp(α).

Case (I): If ti ∈ supp(f) for some 1 ≤ i ≤ s, then ti ∈ supp(fj) for somej such that nj > 0. Thus by (ii) one has j ≥ k + 1 and fj ∈ (t1, . . . , ts).From the equality pα =

∑ri=1 niγi it follows that f ∈ (t1, . . . , ts) ⊂ p.


Case (II): If ti /∈ supp(f) for 1 ≤ i ≤ s. Let ϕ′ (resp. ψ′) be therestriction of ϕ (resp. ψ) to K[ts+1, . . . , tq] (resp. Zes+1 + · · · + Zeq). Weset F ′ = {xvs+1 , . . . , xvq}. Note that γ1, . . . , γk are the circuits of ker(ψ′)and P ′

F ′ = K[ts+1, . . . , tq]∩PF is the toric ideal ofK[F ′]; see Exercise 8.3.33and Lemma 8.3.27. Hence by induction one has

rad (I ′) = P ′F ′ = K[ts+1, . . . , tq] ∩ PF ,

where I ′ is a circuit ideal generated by f1, . . . , fk. Since f ∈ P ′F ′ , then there

is m ≥ 1 such that fm ∈ I ′ ⊂ p. Thus f ∈ p.Altogether p contains (PF , t1). As rad (PF , t1) is the intersection of the

minimal primes of (PF , t1) one concludes rad (PF , t1) ⊂ rad (I, t1). Sincethe other inclusion is clear one has equality. �

Exercises

8.3.29 If the matrix A is t-unimodular, then the reduced Grobner basis ofPF with respect to any term order consists of square-free binomials, i.e., ofbinomials ta

+ − ta− with ta+

and ta−square-free.

8.3.30 Let B be a set of binomials of S = K[t1, . . . , tq] and let f be abinomial. Prove that dividing f by B produces a residue h = tγ − tδ whichis independent of the field K as long as we follow identical steps in thedivision process when changing fields.

8.3.31 Let F be a finite set of monomials in a polynomial ring Q[x1, . . . , xn]and let P be the toric ideal of Q[F ]. Consider the toric ideal PF of K[F ]over a field K. If G is a Grobner basis of P consisting of binomials withrespect to a fixed term order, prove that G is also a Grobner basis for PF .

8.3.32 Use the previous exercise to draw some conclusion if P is a toricideal minimally generated by a Grobner basis over a field K.

8.3.33 Let F = {f1, . . . , fq} be a set of monomials of a polynomial ring Rover a field K. Consider the homomorphism

ϕ : S = K[t1, . . . , tq] −→ K[F ] −→ 0, induced by ϕ(ti) = fi.

Prove that ker(ϕ)∩K[t1, . . . , tr] is a binomial ideal which is equal to ker(ϕ′),where ϕ′ is the restriction of ϕ to K[t1, . . . , tr].

8.3.34 If I is an ideal of S = K[t1, . . . , tq] and rad(I) = PF , prove thatrad(IS′) = PFS

′, where S′ = K[t±11 , . . . , t±1

q ].

8.3.35 Let A be an n × q integral matrix. If L is a subgroup of kerZ(A),prove that T(Zq/L) = T(kerZ(A)/L), where T denotes torsion.

334 Chapter 8

8.3.36 Let A be an n × q integral matrix with non-zero columns. If L isa subgroup of kerZ(A) and p ≥ 2 a prime number, then pukerZ(A) ⊂ L forsome u ∈ N if and only if kerZ(A)/L is a finite p-group or kerZ(A)/L = (0).

8.3.37 Let d1, . . . , dq be a sequence of relatively prime integers and ψ thelinear map from Zq to Z induced by ψ(ei) = di. If L is a subgroup of ker(ψ),then T(Zq/L) ⊂ ker(ψ)/L, with equality if ker(ψ)/L is a finite group.

8.3.38 Let A be an n×q integer matrix. If L is a subgroup of kerZ(A), thenL = kerZ(A) if and only if Zq/L is torsion-free of rank equal to rank(A).

8.3.39 Let P be the toric ideal of the subring K[x61, x81, x

91] and let I be the

ideal generated by g1, g2, where g1 = t31 − t23, g2 = t92 − t83. Then

(a) P = (g1, t32 − t1t23) for any field K, (b) Z3/〈g1, g2〉 � Z× Z3,

(c) rad(I) = P if char(K) = 3, and rad(I) �= P if char(K) = 2,

(d) Γ = {(x6, x8, x9)|x ∈ K} = {(0, 0, 0), (1, 1, 1)} = V (I), if K = Z2.

8.3.40 Let d1, . . . , dn be a sequence of relatively prime integers and let ψbe the linear map from Zn to Z induced by ψ(ei) = di, then the set

L =

{dj

gcd(di, dj)ei −

digcd(di, dj)

ej

∣∣∣∣ 1 ≤ i < j ≤ n}

is a generating set for ker(ψ), where ei is the ith unit vector.

8.3.41 Let d1, . . . , dn be a sequence of non-zero integers and let ψ be thelinear map from Zn to Z induced by ψ(ei) = di. Prove that the set

L =

{(djd

)ei −

(did

)ej

∣∣∣∣ 1 ≤ i < j ≤ n}

is a generating set for ker(ψ), where d = gcd(d1, . . . , dn).

8.3.42 Let k ≥ 1 be an integer and let p be a prime number. Considerthe expansion of k in base p: k =

∑m1

i=0 aipi, where am1 �= 0, the ai’s are

integers and 0 ≤ ai ≤ p− 1 for all i. If pu1 is the largest power of p dividingk! (the factorial of k), prove the equalities

u1 =

m1∑i=1

ai

(pi − 1

p− 1

)and u1 =

pm − 1

p− 1if k = pm.

8.3.43 If p > 1 is a prime number and m ≥ 1 is an integer. Prove that pdivides the binomial coefficient

(pm

k

)for 1 ≤ k ≤ pm − 1.


8.4 Toric varieties

In this section we use linear algebra methods to characterize when a toricset is an affine toric variety in terms of the existence of certain roots in thebase field and a vanishing condition.

First we fix some notation. Let K be a field and let A be an n × qmatrix with entries in N and with non-zero columns v1, . . . , vq. The toricset parameterized by F = {xv1 , . . . , xvq} ⊂ K[x] is the set given by

Γ ={(av111 · · · avn1

n , . . . , av1q1 · · ·avnq

n ) ∈ AqK∣∣ a1, . . . , an ∈ K} ,

where vi = (v1i, . . . , vni) for all i. Similarly one can define the toric set Γparameterized by xv1 , . . . , xvq , with v1, . . . , vq in Zn, as the set of all points(av111 · · · avn1

n , . . . , av1q1 · · ·avnq

n ) in AqK that are well-defined for some ai’s inK. As usual the toric ideal of K[F ] is denoted by PF . Recall that PF is aprime ideal of the polynomial ring S = K[t1, . . . , tq] over the field K.

There are unimodular integral square matrices U = (uij) and Q = (qij)of orders n and q, respectively, such that

D = UAQ = diag(λ1, . . . , λs, 0, . . . , 0),

where s is the rank of A and λ1, . . . , λs are the invariant factors of A; thatis, λi divides λi+1 and λi > 0 for all i (see Theorem 1.2.2). For use belowwe set U−1 = (fij) and Q

−1 = (bij). In what follows ei will denote the ithunit vector in Zq.

Lemma 8.4.1 kerZ(A) = Zqs+1 ⊕ · · · ⊕ Zqq, where qi is the ith column ofthe matrix Q.

Proof. Let x ∈ Zq. Make the change of variables y = Q−1x. As D = UAQ,it follows that Ax = 0 if and only if Dy = 0. Set y = (y1, . . . , yq).

First note that qi ∈ kerZ(A) for i ≥ s+ 1, because DQ−1qi = Dei. If xis in kerZ(A), then λiyi = 0 for i = 1, . . . , s. Thus x = Qy =

∑qi=s+1 yiqi.

To complete the proof observe that the columns of Q are a basis for Zq. �

Proposition 8.4.2 kerZ(A) = Z{wj − ej}qj=1, where wj =∑si=1 bijqi and

qi is the ith column of Q.

Proof. Note that ej =∑qi=1 bijqi for all j, because QQ

−1 = I. Hence

wj =

s∑i=1

bijqi = ej −q∑

i=s+1

bijqi (j = 1, 2, . . . , q).

336 Chapter 8

Then, using Lemma 8.4.1, we obtain wj − ej ∈ kerZ(A) for j = 1, . . . , q. ByLemma 8.4.1, it suffices to observe that from the equality above one has

q∑j=1

qjk(ej − wj) =

q∑j=1

qjk

(q∑

i=s+1

bijqi

)=

q∑i=s+1

qi

⎛⎝ q∑j=1

qjkbij

⎞⎠=

q∑i=s+1

qiδik = qk,

for all k ≥ s+ 1, where δik = 1 if i = k and δik = 0 otherwise. �

Definition 8.4.3 An affine toric variety V is defined as the zero set of atoric ideal, that is, V = V (PF ) for some toric ideal PF .

Theorem 8.4.4 [354] Let Γ ⊂ Kq be the toric set parameterized by F .Then Γ = V (PF ) if and only if the following two conditions hold :

(a) If (ai) ∈ V (PF ) and ai �= 0 ∀i, then aq1i1 · · ·aqqiq has a λi-root in K for

i = 1, . . . , s, where λ1, . . . , λs are the invariant factors of A,

(b) V (PF , ti) ⊂ Γ for i = 1, . . . , q.

Proof. ⇐) One invariably has Γ ⊂ V (PF ). To prove the other inclusiontake a point a = (a1, . . . , aq) in V (PF ), by condition (b) one may assumeai �= 0 for all i. Thus using (a) there are x′1, . . . , x

′s in K such that

(x′i)λi = aq1i1 · · ·aqqiq = aqi (i = 1, . . . , s). (8.3)

For convenience of notation we extend the definition of x′i by putting x′i = 1for i = s+ 1, . . . , n and x′ = (x′1, . . . , x

′n). Set

xj = (x′1)u1j · · · (x′n)unj (j = 1, . . . , n). (8.4)

We claim that xvk = xv1k1 · · ·xvnkn = ak for k = 1, . . . , q. Setting U−1 = (fij)

and comparing columns in the equality U−1D = AQ one has:

λifi =

q∑j=1

qjivj (i = 1, 2, . . . , s), (8.5)

where fi = (f1i, . . . , fni) and vj = (v1j , . . . , vnj) denote the ith and jthcolumns of U−1 and A. Comparing columns in A = (U−1D)Q−1 we get:

vk =

s∑j=1

λjbjkfj (k = 1, 2, . . . , q), (8.6)


where Q−1 = (bij). Using UU−1 = I and Eq.( 8.4) we rapidly conclude:

xfk = x′k (k = 1, . . . , n). (8.7)

From Proposition 8.4.2 we derive Awj = Aej = vj for j = 1, . . . , q, where

wj =

s∑i=1

bijqi =

(s∑�=1

q1�b�j , . . . ,

s∑�=1

qn�b�j

)(j = 1, . . . , q). (8.8)

Hence Aw+j = A(ej + w−

j ), that is, tw+

j − tej+w−j belongs to the toric ideal

PF . Using that a ∈ V (PF ) yields aw+

j = aej+w−j , thus

awj = aej = aj (j = 1, . . . , q). (8.9)

Therefore putting everything together:

xvk8.6= x

∑sj=1 λjbjkfj = (xf1 )λ1b1k · · · (xfs )λsbsk 8.7

= (x′1)λ1b1k · · · (x′s)λsbsk

8.3= (aq1)b1k · · · (aqs)bsk = aq1b1k+···+qsbsk 8.8

= awk8.9= ak

for k = 1, . . . , q. Thus a ∈ Γ, as required.⇒) It is clear that (b) holds because V (PF , ti) ⊂ V (PF ). To prove (a)

take (ai) ∈ V (PF ) with ai �= 0 for all i, then by definition of Γ there arex1, . . . , xn in K such that aj = xvj for all j. Therefore by Eq. (8.5) one has:

(xfi)λi = xλifi = xq1iv1 · · ·xqqivq = aq1i1 · · · aqqiq . �

The proof of Theorem 8.4.4 works for arbitrary Laurent monomials.Another classification, that uses polyhedral geometry, of the affine toricvarieties that are parameterized by Laurent monomials is given in [272].

Corollary 8.4.5 V (PF ) ⊂ Γ ∪ V (t1 · · · tq) if K is algebraically closed.

Proof. Let a = (ai) ∈ V (PF ) such that ai �= 0 for all i. Since K isalgebraically closed condition (a) above holds. Therefore one may proceedas in the first part of the proof of Theorem 8.4.4 to get a ∈ Γ. �

Next we present another consequence that can be used to prove thatmonomial curves over arbitrary fields are affine toric varieties.

Corollary 8.4.6 If the columns of A generate Zn as Z-module, then theequality Γ = V (PF ) holds if and only if V (PF , ti) ⊂ Γ for all i.

Proof. Since Zv1 + · · ·+Zvq = Zn, one has λi = 1 for all i, thus condition(a) holds. Therefore Γ is an affine toric variety if and only if (b) holds. �

338 Chapter 8

Corollary 8.4.7 If K is algebraically closed, then Γ = V (PF ) if and onlyif V (PF , ti) ⊂ Γ for all i.

Proof. If K is algebraically closed, then (a) is satisfied, thus Γ is a toricvariety if and only if V (PF , ti) ⊂ Γ for all i. �

Proposition 8.4.8 [139] If A has only one row and v1, . . . , vq are relativelyprime positive integers, then Γ = V (PF ) over any field K.

Proof. As Z = Zv1 + · · · + Zvq by Corollary 8.4.6 it suffices to showV (PF , ti) ⊂ Γ. Let a ∈ V (PF , ti) since all the binomials t

vji − t

vij vanish on

a one obtains a = 0 and a ∈ Γ. �

A natural question is whether a toric set Γ can be a variety but not atoric variety (see Exercise 8.4.18):

Example 8.4.9 Let K be the field Z3 and let A be the matrix (2, 4). Then

Γ = {(0, 0)} ∪ {(1, 1)} = V (t1 − t2, t22 − t2).

On the other hand PF = (t1 − t22) and (1, 2) ∈ V (PF ). Thus Γ �= V (PF ).

Affine toric varieties are fully parameterized by monomials when K isalgebraically closed or when the variety is normal.

Theorem 8.4.10 [271, 272] Let F = {xv1 , . . . , xvq} be a set of Laurentmonomials such that K is algebraically closed or K[F ] is normal. Thenthere exists a toric set ΓG parameterized by G = {xε1 , . . . , xεq} with εi ∈ Zn

such that V (PG) = ΓG and PF = PG, where PG is the toric ideal of K[G].

Theorem 8.4.11 (Combinatorial Nullstellensatz [7]) Let S = K[t1, . . . , tq]be a polynomial ring over a field K, let f ∈ S, and let c = (ci) ∈ Nq.Suppose that the coefficient of tc in f is non-zero and deg (f) = c1+ · · ·+cq.If A1, . . . , Aq are subsets of K, with |Ai| > ci for all i, then there area1 ∈ A1, . . . , aq ∈ Aq such that f (a1, . . . , aq) �= 0.

Lemma 8.4.12 Let f be a polynomial in K[x1, . . . , xn]. If K is an infinitefield and f(a) = 0 for all a ∈ Kn, then f is the zero polynomial.

Proof. It follows readily from Theorem 8.4.11. �

Corollary 8.4.13 Let R = K[x] be a polynomial ring over a field K and letX ⊂ Kq be a set parameterized by ti = fi(x), where F = {f1, . . . , fq} ⊂ R.If K is infinite and I(X) is the vanishing ideal of X, then

(a) I(X) is equal to PF , the presentation ideal of K[F ], and

(b) the Zariski closure of X is equal to V (PF ).


Proof. (a): As PF = (t1−f1, . . . , tq−fq)∩K[t1, . . . , tq], one has PF ⊂ I(X).Conversely take f ∈ I(X) and consider the revlex ordering such that ti > xjfor all i, j. By the division algorithm f(t1, . . . , tq) =

∑i gi(ti−fi)+r, where

r is a polynomial in R. Hence for any a ∈ Kn one has

f(f1(a), . . . , fq(a)) =∑i gi(a, x)(fi(a)− fi) + r = 0.

Thus r(a) = 0 for any a ∈ Kn. Therefore r is the zero polynomial becauseK is an infinite field (see Lemma 8.4.12).

(b): By part (a) and Proposition 3.2.3 we get X = V (PF ). �

Affine normal toric varieties Let A = {v1, . . . , vq} ⊂ Zn be a pointconfiguration such that R+A ∩ ZA = NA. Hence the monomial subring

K[xv1 , . . . , xvq ] ⊂ K[x±11 , . . . , x±1

n ]

is normal (Corollary 9.1.3). Pick a Z-basis B = {w1, . . . , wr} of ZA, wherer = rank(ZA), and consider the linear map T : ZA −→ Zr, wi �→ ei. For1 ≤ i ≤ q we can write vi = u1iw1 + · · · + uriwr. Notice that T (vi) = ui,where ui = (u1i, . . . , uri). Setting H = {u1, . . . , uq}, we have ZH = Zr.

Lemma 8.4.14 Zr ∩ R+H = NH.

Proof. The equality follows readily by observing that T can be extendedto an isomorphism from QA onto Qr. �

Let P and P1 be the toric ideals of K[xv1 , . . . , xvq ] and K[xu1 , . . . , xuq ],respectively.

Lemma 8.4.15 P = P1 and K[xv1 , . . . , xvq ] � K[xu1 , . . . , xuq ].

Proof. Let K[t1, . . . , tq] be a polynomial ring over the field K. Considerthe following ring maps

K[t1, . . . , tq]ϕ1 ��

ϕ

��

K[xu1 , . . . , xuq ]

K[xv1 , . . . , xvq ]

ϕ1

��

induced by ϕ1(ti) = xui and ϕ(ti) = xvi . Since the ringsK[xv1 , . . . , xvq ] andK[xu1 , . . . , xuq ] have dimension r it suffices to prove that ker(ϕ) ⊂ ker(ϕ1)and to observe that P = ker(ϕ), P1 = ker(ϕ1). Let f = ta11 · · · t

aqq −tc11 · · · t

cqq

be a binomial in ker(ϕ). Then

(xv1)a1 · · · (xvq )aq = (xv1)c1 · · · (xvq )cq

340 Chapter 8

and∑q

i=1(ai − ci)vi = 0. Applying T we get∑q

i=1(ai − ci)ui = 0. Thus

(xu1 )a1 · · · (xuq )aq = (xu1 )c1 · · · (xuq )cq .

This proves that f ∈ ker(ϕ1), as required. �

Lemma 8.4.16 The map φ : (K∗)r → (K∗)q ∩ V (P ), x �→ (xu1 , . . . , xuq ),is bijective, where K∗ = K \ {0}.

Proof. As ei ∈ ZH = Zr for 1 ≤ i ≤ r, there is αi ∈ NH such thatei + αi = βi and βi ∈ NH. Then the monomial xα = xα1 · · ·xαr satisfiesthat xα ∈ NH and xix

α = xγi ∈ NH. Thus we can write

xα = (xu1)r1 · · · (xuq )rq ; xγi = (xu1 )si1 · · · (xuq )siq ; (ri ∈ N; sij ∈ N).

Setting hi(t1, . . . , tq) = tr11 · · · trqq and h0(t1, . . . , tq) = tsi11 · · · t

siqq , we have

xihi(xu1 , . . . , xuq ) = h0(x

u1 , . . . , xuq ) (8.10)

for 1 ≤ i ≤ r. If φ(x1, . . . , xr) = φ(y1, . . . , yr), then xui = yui for all i.Thus, from Eq. (8.10), xi = yi for all i. This proves that φ is injective.

To prove that φ is onto take a = (ai) ∈ V (P ) ∩ (K∗)q. Let Au bethe matrix with column vectors u1, . . . , uq. There are qij in Z such thatej =

∑qi=1 qijui for j = 1, . . . , r. Thus Au(qj) = ej for j = 1, . . . , r, where

qj = (q1j , . . . , qqj). Hence, setting wi =∑rj=1 ujiqj for i = 1, . . . , q, we get

Au(wi) = ui = Au(ei) (i = 1, . . . , q)

Hence Aw+i = A(ej+w

−i ), that is, t

w+i −tei+w−

i is in the toric ideal P . Using

that a ∈ V (P ) yields aw+i = aei+w

−i , thus awi = aei = ai for i = 1, . . . , q.

Setting xj = aqj = aq1j1 · · ·a

qqjq for j = 1, . . . , r, we get

xui = xu1i1 · · ·xuri

r = (aq1)u1i · · · (aqr )uri = awi = ai

for i = 1, . . . , q. Thus a is in the image of φ, as required. �

Theorem 8.4.17 If K is algebraically closed, then

(a) V (P ) contains the torus (K∗)r as a dense open subset, and

(b) there is an action of the torus (K∗)r on the toric variety V (P ):

(K∗)r × V (P ) −→ V (P )

(x, (y1, . . . , yq)) �−→ (xu1y1, . . . , xuqyq)

Proof. The set Γ∗ := im(φ) = (K∗)r ∩ V (P ) is an open subset of V (P )because the set V (t1 · · · tq) is closed in the Zariski topology of Kq, i.e., theset (K∗)q = Kq \ V (t1 · · · tq) is open in Kq. By Hilbert Nullstellensatz (seeTheorem 3.2.10) V (P ) is irreducible. Hence Γ∗ is dense in V (P ). Thus (a)follows from Lemma 8.4.16. Part (b) is left as an exercise. �


Exercises

8.4.18 Let Γ be a toric set parameterized by F . If K is an infinite fieldand Γ = V (J) for some ideal J , then J ⊂ PF and Γ = V (PF ).

8.4.19 (Rational quartic curve in P3) LetK be an algebraically closed fieldand let Γ be the curve parameterized by F = {x41, x31x2, x1x32, x42}. Thenrad (t2t3 − t1t4, t33 − t2t24, t32 − t21t3) = PF and Γ is an affine toric variety.

8.4.20 Prove that Corollaries 8.4.5 and 8.4.7 are valid assuming condition(a) of Theorem 8.4.4, instead of assuming K algebraically closed.

8.4.21 Let S = K[t1, t2, t3, t4] be a polynomial ring over a field K. Then

V (t1t3 − t2t4) = {(x1x2, x2x3, x3x4, x1x4)| x1, . . . , x4 ∈ K}.

8.4.22 (Segre variety) Let m,n be two positive integers. Prove that

Γ = {(aibj) ∈ Kmn| ai ∈ K, bj ∈ K ∀ i, j} ⊂ AmnK

is an affine toric variety over any field K.

8.4.23 (Veronese variety) Let d be a positive integer and let A be the setof k-partitions of d consisting of all a ∈ Nk such that |a| = d. If K isan algebraically closed field and F = {xa| a ∈ A}, then the toric set Γparameterized by F is an affine toric variety.

8.4.24 Let A = {v1, . . . , vq} ⊂ Nn and let K be a field. Prove thatK[xv1 , . . . , xvq ] is the coordinate ring of an affine toric variety, in the senseof [176], if and only if R+A∩ Zn = NA and rank(A) = n.

8.4.25 Let A = {v1, . . . , vq} ⊂ Zn and let K be a field. Prove that ifK[F ] = K[xv1 , . . . , xvq ] is normal and rank(A) = n, thenK[F ] is isomorphicto the coordinate ring of an affine toric variety, in the sense of [176].

8.4.26 An affine variety X ⊂ Kq is called a set-theoretic complete intersec-tion if I(X) is a set-theoretic complete intersection. If K is algebraicallyclosed, then X is a set-theoretic complete intersection if and only if X canbe defined by q− dim(X) polynomials in q variables with coefficients in K.

8.4.27 Let K = Z2 and let X ⊂ K3 be a space curve given parametricallyby t1 = x71, t2 = x81, t3 = x91. If F = {x71, x81, x91}, prove that PF �= I(X)and PF = (t51 − t2t33, t22 − t1t3, t41t2 − t43).

8.4.28 [272] Let F = {x21x3, x42x23, x1x2x3} and let K = R be the field ofreal numbers. Prove that V (PF ) is never fully parameterized by Laurentmonomials in the sense of Theorem 8.4.10.

342 Chapter 8

8.5 Affine Hilbert functions

In this section, we introduce the notion of degree via affine Hilbert functionsand show additivity of the degree. For lattice ideals the degree turns out tobe independent of the base field and the partial character.

Let S = K[t1, . . . , tq] be a polynomial ring over a field K and let I bean ideal of S. The vector space of polynomials in S (resp. I) of degree atmost i is denoted by S≤i (resp. I≤i). The functions

HaI (i) = dimK(S≤i/I≤i) and HI(i) = Ha

I (i)−HaI (i− 1)

are called the affine Hilbert function and the Hilbert function of S/I. Letu = ts+1 be a new variable and let Ih ⊂ S[u] be the homogenization of I,where S[u] is given the standard grading.

Lemma 8.5.1 If m = (t1, . . . , tq) is the irrelevant maximal ideal of S andI ⊂ S is a graded ideal of dimension zero, then

�Sm(Sm/Im) = �S(S/I) = �S/I(S/I) = dimK(S/I).

Proof. Let m be the image of m in the quotient ring S = S/I and let r bethe least positive integer such that mr = (0). From the ascending chain

(0) = mr ⊂ mr−1 ⊂ · · · ⊂ m2 ⊂ m ⊂ S

and using that S/m = K, we get

�(S) =

r−1∑i=0

�S(mi/mi+1) =

r−1∑i=0

�S/m(mi/mi+1) =

r−1∑i=0

dimK(mi/mi+1).

As the last summation is equal to dimK(S), we get that �(S) is equal todimK(S). The equality �Sm(Sm/Im) = �S(S/I) follows readily by localizingat the maximal ideal m. �

Remark 8.5.2 If S = ⊕∞i=0Si has the standard grading and I ⊂ S is a

graded ideal, thenHaI (i) =

∑ik=0 dimK(Sk/Ik)

where Ik = I ∩ Sk. Thus, one has HI(i) = dimK(Si/Ii) for all i. There areisomorphisms of K-vector spaces

S/mi+1S � S/(mi+1 + I) � S≤i/I≤i,

where S = S/I and m = (t1, . . . , tq). By Lemma 8.5.1, one has

�S(S/(mi+1 + I)) = dimK(S/(mi+1 + I)).

Thus, HaI is the Samuel or Hilbert–Samuel function of S with respect to m

(see Section 4.4 and [413, Definition B.3.1]).


Definition 8.5.3 The graded reverse lexicographical order (GRevLex forshort) on the monomials of S is defined as tb " ta if deg(tb) > deg(ta) ordeg(tb) = deg(ta) and the last non-zero entry of b− a is negative.

Lemma 8.5.4 Let I be an ideal of S. Then the following hold :

(a) dim(S[u]/Ih) = dim(S/I) + 1.

(b) HaI (i) = HIh(i) for i ≥ 0.

Proof. (a): Let ≺ be the GRevLex order and let G be a Grobner basisof I. Then {gh| g ∈ G} is a Grobner basis of Ih and in≺(g) = in≺(g

h)for g ∈ G (see Proposition 3.4.2). Since dim(S/I) = dim(S/in≺(I)) anddim(S[u]/I) = dim(S[u]/in≺(I

h)), the equality follows.(b): Fix i ≥ 0. The mapping S[u]i → S≤i induced by mapping u �→ 1 is

a K-linear surjection. Consider the induced composite K-linear surjectionS[u]i → S≤i → S≤i/I≤i. An easy check shows that this has kernel Ihi .Hence, we have an isomorphism of K-vector spaces

S[u]i/Ihi � S≤i/I≤i ⇒ Ha

I (i) = HIh(i). �

The degree of affine algebras and lattice ideals Let d be the Krulldimension of S/I. By Lemma 8.5.4 and Proposition 5.1.6, there are unique

polynomials haI (t) =∑d

i=0 aiti ∈ Q[t] and hI(t) =

∑d−1i=0 cit

i ∈ Q[t] ofdegrees d and d−1, respectively, such that haI (i) = Ha

I (i) and hI(i) = HI(i)for i 0. By convention, the zero polynomial has degree −1. Notice thatd! ad = (d− 1)! cd−1 for d ≥ 1. If d = 0, then Ha

I (i) = dimK(S/I) for i 0.

Definition 8.5.5 The integer d! ad, denoted by deg(S/I) or e(S/I), iscalled the degree or multiplicity of S/I.

Proposition 8.5.6 deg(S/I) = deg(S[u]/Ih).

Proof. From Lemma 8.5.4(a), dim(S[u]/Ih) is equal to dim(S/I) + 1.Hence, the equality follows from Lemma 8.5.4(b). �

Lemma 8.5.7 Let F be a field extension of K, let B = F [t1, . . . , tq] be apolynomial ring, and let I be an ideal of S, then deg(S/I) = deg(B/IB).

Proof. Let G be a Grobner basis of I w.r.t the GRevLex order. Thanks toBuchberger’s criterion, G is a Grobner basis for IB. By Proposition 3.4.2,Ih and IBh are both generated by Gh = {gh| g ∈ G}, and Gh is a Grobnerbasis for IBh. Hence to show the equality deg(S/I) = deg(B/IB) noticethat, by Lemma 8.5.4, one has

HaI (i) = HIh(i) = dimK(S[u]/Ih)i =

dimF (B[u]/IBh)i = HIBh(i) = HaIB(i) for i ≥ 0. �

344 Chapter 8

For a lattice ideal Iρ(L) the degree depends only on the lattice L andis independent of the base field K and the character ρ. In Section 9.4 weshow effective methods to compute the degree of any lattice ideal.

Proposition 8.5.8 Let ρ : L → K∗ be a partial character of a lattice L inZq and let IQ(L) be the lattice ideal over the field Q. Then

deg(S/Iρ(L)) = deg(S/I(L)) = deg(Q[t1, . . . , tq]/IQ(L)).Proof. First we show the equality on the left. The ideal Iρ(L) contains nomonomials (Theorem 8.2.2). Let G = {g1, . . . , gm} be the reduced Grobner

basis of I(L) w.r.t the GRevLex order ". We can write gi = ta+i − ta−i with

in≺(gi) = ta+i . We set hi = ta

+i − ρ(ai)ta

−i and H = {h1, . . . , hm}. Next,

we show that H is a Grobner basis of Iρ(L). Let f �= 0 be any non-purebinomial of Iρ(L). We claim that f reduces to zero w.r.t H. By the DivisionAlgorithm, f =

∑mi=1 fihi + g, where in≺(f) " in≺(fihi) for all i, and g is

a non-pure binomial in Iρ(L) such that none of the terms of g is divisible

by any of the terms ta+1 , . . . , ta

+m . Then we can write

g = μ(ta − λtb) = μtδ(tu+ − λtu−),

with μ, λ ∈ K∗ and a − b = u+ − u−. As tu+ − λtu− is in Iρ(L), byLemma 8.2.6, u = u+ − u− is in L and λ = ρ(u). If g �= 0, we obtain that

tu+ − tu− , being in I(L), has one of its terms in (ta+1 , . . . , ta

+m) = in≺(I(L)),

a contradiction. Thus, g must be zero, i.e., f reduces to zero w.r.t H. Thisproves the claim. In particular, Iρ(L) is generated by H. Note that theS-polynomial of hi and hj is a binomial; thus, by the claim, it reduces tozero with respect to H. Therefore by Buchberger’s criterion, H is a Grobnerbasis of Iρ(L). Hence, S/I(L) and S/Iρ(L) have the same degree.

Now we show the equality on the right. Let" be the GRevLex order on Sand SQ = Q[t1, . . . , tq], and on the extensions S[u] and SQ[u]. Let GQ be thereduced Grobner basis of IQ(L). We set I = I(L) and IQ = IQ(L). Noticethat Z ⊂ K if char(K) = 0 and Zp ⊂ K if p = char(K) > 0. In the secondcase, one has a map Z �→ K, 1 �→ 1K . Hence SZ = Z[t1, . . . , tq] embeds intoS if char(K) = 0 and SZ maps into S if p = char(K) > 0. If G denotes theimage of GQ under either of these two maps, by Buchberger’s criterion, itis seen that G is a Grobner basis of I. Hence, by Proposition 3.4.2, GhQ and

Gh are Grobner basis of IhQ and Ih, where Gh is the set of fh with f ∈ G.Thus, the standard monomials of SQ[u]/I

hQ and S[u]/Ih are the “same.”

Therefore, these two rings have the same Hilbert function and the samedegree. Thus, by Proposition 8.5.6, the result follows. �

Proposition 8.5.9 (Additivity of the degree) If I is an ideal of S andI = q1 ∩ q2 ∩ · · · ∩ qm is an irredundant primary decomposition, then

deg(S/I) =∑

ht(qi)=ht(I)

deg(S/qi).


Proof. We may assume that p1, . . . , pr are the associated primes of I ofheight ht(I) and that rad(qi) = pi for i = 1, . . . , r. The proof is by inductionon m. We set J = ∩mi=2qi. There is an exact sequence of K-vector spaces

0→ S≤i/(q1 ∩ J)≤iϕ→ S≤i/(q1)≤i ⊕ S/(J)≤i

φ→ S≤i/(q1 + J)≤i → 0,

where ϕ(f) = (f,−f) and φ(f 1, f2) = f1 + f2. Hence

Haq1(i) +Ha

J (i) = HaI +Ha

q1+J(i). (8.11)

As the decomposition of I is irredundant, by Exercise 8.5.10, one hasdim(S/(q1 + J)) < dim(S/J). If r = 1, then dim(S/J) < dim(S/I).Hence, from Eq. (8.11), we get that deg(S/I) = deg(S/q1). If r > 1,then dim(S/J) = dim(S/I) = dim(S/q1). Hence, from Eq. (8.11), we getthat deg(S/I) is deg(S/q1)+deg(S/J). Therefore, by induction, we get therequired formula. �

Exercises

8.5.10 Let I ⊂ S be an ideal and let I = q1∩q2∩· · ·∩qm be an irredundantprimary decomposition. If J = ∩mi=2qi, then dim(S/(q1 + J)) < dim(S/J).

8.5.11 Let I ⊂ S be an ideal. If q is a primary component of I of heightht(I), then qh is a primary component of Ih of height ht(Ih).

8.6 Vanishing ideals over finite fields

Let K = Fq be a finite field with q elements and X a subset of a projectivespace Ps−1 over K. Let S = K[t1, . . . , ts] be a polynomial ring over thefinite field K with the standard grading S = ⊕∞

d=0Sd and let I(X) be thevanishing ideal of X which is the ideal of S generated by the homogeneouspolynomials of S that vanish at all points of X .

The following family arises in algebraic coding theory [348].

Proposition 8.6.1 [330] I(X) is a lattice ideal if and only if

X = {[(xv1 , . . . , xvs)]|xi ∈ K∗ = K \ {0} ∀ i} ⊂ Ps−1, (8.12)

for some monomials xvi := xvi11 · · ·xvinn , i = 1, . . . , s.

In what follows we assume that X is a subset of Ps−1 parameterized bymonomials as in Eq. (8.12).

The following formula allows us to compute the algebraic invariants ofI(X) (degree, regularity, Hilbert function) using Macaulay2 [199].

346 Chapter 8

Theorem 8.6.2 [348] I(X) = ({ti − xviz}si=1 ∪ {xq−1i − 1}ni=1) ∩ S.

Lemma 8.6.3 The degree of S/I(X) is |X |.

Proof. For [P ] ∈ X , let I[P ] be its vanishing ideal. Since I(X) = ∩[P ]∈XI[P ]

and since deg(S/I[P ]) = 1 (see Exercise 8.6.10), the lemma follows from theadditivity of the degree (see Proposition 8.5.9). �

In this context it is usual to denote the Hilbert function of S/I(X)simply by HX .

Proposition 8.6.4 There is an integer r ≥ 0 such that

1 = HX(0) < HX(1) < · · · < HX(r − 1) < HX(d) = |X | for d ≥ r.

Proof. As S/I(X) is Cohen–Macaulay of dimension 1, by Theorem 2.2.4,its Hilbert polynomial has degree 0. Hence, by Lemma 8.6.3, HX(d) = |X |for d 0. Consequently the result follows readily from Theorem 5.6.4. �

Recall from Chapter 5 that the integer r is the index of regularity ofS/I(X) and that by Theorem 6.4.1 this number is the so-called Castelnuovo–Mumford regularity of S/I(X). For these reasons r is simply called theregularity and is denoted by reg(S/I(X)).

Definition 8.6.5 Let X = {[P1], . . . , [Pm]} and let f0(t1 . . . , ts) = td1,where d ≥ 1. The linear map of K-vector spaces:

evd : Sd → K |X|, f �→(f(P1)

f0(P1), . . . ,

f(Pm)

f0(Pm)

)is called an evaluation map. Notice that ker(evd) = I(X)d. The image ofevd, denoted by CX(d), is a linear code and CX(d) is called a parameterizedcode of order d.

The algebraic invariants of S/I(X) are closely related to the so-calledbasic parameters of parameterized codes [120, 348].

Definition 8.6.6 The basic parameters of the linear code CX(d) are:

(b1) dimK CX(d) = HX(d), the dimension,

(b2) |X | = deg(S/I(X)), the length,

(b3) δX(d) = min{‖v‖ : 0 �= v ∈ CX(d)}, the minimum distance, where‖v‖ is the number of non-zero entries of v.

Lemma 8.6.7 δX(d) = 1 for d ≥ reg(S/I(X)).


Proof. As CX(d) is a linear subspace of K |X| of dimension equal to |X | ford ≥ reg(S/I(X)), we get the equality CX(d) = K |X|. Thus δX(d) is equalto 1 for d ≥ reg(S/I(X)). �

A good linear code should have large |X | together with dimK CX(d)/|X |and δX(d)/|X | as large as possible. From Lemma 8.6.7 we conclude thatthe potentially good codes CX(d) can occur only if 1 ≤ d < reg(S/I(X)).

Problem 8.6.8 Find explicit formulas, in terms of the numerical datan, s, q, d, and the combinatorics of xv1 , . . . , xvs , for the basic parameters:

(a) HX(d), (b) deg(S/I(X)), (c) δX(d), (d) reg(S/I(X)).

Formulas for (a)-(d) are known when X is a projective torus [367] andwhenX is parameterized by xd11 , . . . , x

dss , 1, where di ∈ N+ for al i [192, 292].

Example 8.6.9 If X = {[(x901 , x362 , x203 , 1)]|xi ∈ F∗181 for i = 1, 2, 3}, then

reg(S/I(X)) = 13. The basic parameters of the family {CX(d)}d≥1 are:

d 1 2 3 4 5 6 7 8 9 10 11 12 13|X | 90 90 90 90 90 90 90 90 90 90 90 90 90

HX(d) 4 9 16 25 35 45 55 65 74 81 86 89 90δX(d) 45 36 27 18 9 8 7 6 5 4 3 2 1

Exercises

8.6.10 Let X be a finite subset of a projective space Ps−1 over a field K.Let [P ] be a point in X , with P = (α1, . . . , αs) and αk �= 0 for some k, andlet I[P ] be the ideal generated by the homogeneous polynomials of S thatvanish at [P ]. Prove that I[P ] is a prime ideal of height s− 1,

I[P ] = ({αkti − αitk| k �= i ∈ {1, . . . , s}),

deg(S/I[P ]) = 1 and that I(X) =⋂

[Q]∈X I[Q].

8.7 Semigroup rings of numerical semigroups

Let S �= (0) be a semigroup of (N,+), that is, S is a subset of N whichis closed under addition and 0 ∈ S. Note that there exists a sequenced1 < · · · < dq of positive integers such that

S = d1N+ · · ·+ dqN.

If gcd(d1, . . . , dq) = 1, S is called a numerical semigroup [360]. Let K[x]be a polynomial ring in one variable over a field K, the semigroup ring ofS, denoted by K[S], is equal to the monomial subring K[xd1 , . . . , xdq ].

348 Chapter 8

Lemma 8.7.1 If S is a numerical semigroup, then c+ N ⊂ S for some c.

Proof. There are integers r1, . . . , rq such that 1 = r1d1 + · · · + rqdq. Theinteger c = (|r1|d1 + · · ·+ |rq|dq)dq satisfies the required property. �

Definition 8.7.2 Let S be a numerical semigroup. The Frobenius numberof S, denoted by g(S), is the largest integer not in S. The semigroup S iscalled symmetric if g(S)− z ∈ S for all z /∈ S.

Proposition 8.7.3 A numerical semigroup S is symmetric if and only if

(g(S) + 1)/2 = |N \ S|.

Proof. We set c = g(S). Consider the map ϕ : N\S → [0, c]∩N, z �→ c−z.Note that S is symmetric if and only if ϕ(N\S) = S ∩ [0, c). Because of thedecomposition [0, c] = (N \S)∪ (S ∩ [0, c)), one obtains that S is symmetricif and only if c+ 1 = 2|N \ S|. �

Proposition 8.7.4 [173] Let S be a numerical semigroup. If

T (S) = {y ∈ N \ S | y + s ∈ S, ∀s ∈ S, s > 0},

then S is symmetric if and only if T (S) = {g(S)}.

Proof. ⇒) We set c = g(S). Let y ∈ T (S) and assume y < c, then0 < c− y ∈ S and y + (c− y) = c is in S, which is impossible, hence y = c.⇐) Let z /∈ S, z > 0, one must show c − z ∈ S. If c − z is not in S,

choose the least positive integer z such that c−z is not in S. Since c−z �= c,by definition of T (S) there is s ∈ S, s > 0 with (c− z) + s not in S, hencez − s > 0, which contradicts the choice of z. �

The following nice result is due to Froberg. It gives an expression forthe Cohen–Macaulay type for the semigroup ring of a numerical semigroup.

Theorem 8.7.5 (Froberg [173]) Let S be a numerical semigroup. If

T (S) = {y ∈ N \ S | y + s ∈ S, ∀s ∈ S, s > 0},

then |T (S)| is equal to the Cohen–Macaulay type of K[S].

Proof. We set D = K[S]. Let d1 < · · · < dq be a minimal generating set ofS with d1, . . . , dq relatively prime and s ∈ S, s > 0. Consider the mapping

ϕ : T (S)→ Soc(D/xsD), y �→ xy+s + xsD.

The type of K[S] is, by definition, equal to dimK(Soc(D/xsD)) (see thelast paragraph of Section 2.3). Thus it is enough to show that the image


of T (S) is a K-vector space basis of Soc(D/xsD), because ϕ is injective.Note that Soc(D/xsD) has a K-basis consisting of monomials of the formxw + xsD, with w − s /∈ S and xdi(xw + xsD) = xsD for all i, writedi + w = s+ si, where si is in S.

We claim w− s ∈ T (S). It follows that (w− s)+ s′ is in S for all s′ ∈ S,s′ > 0. Observe that w−s > 0; otherwise from w−s = s1−d1 = s2−d2 wederive s1 = 0 and d2 = s2+d1, hence d2 = μd1, which is impossible becaused1 and d2 are part of a minimal generating set of S. Hence w − s ∈ T (S)and ϕ(w − s) = w. To finish note that ϕ(T (S)) is linearly independent. �

Theorem 8.7.6 [287] Let S be a numerical semigroup. Then the ring K[S]is Gorenstein if and only if S is symmetric.

Proof. It follows from Proposition 8.7.4 and Theorem 8.7.5. �

Proposition 8.7.7 Let S be a numerical semigroup. If K[S] ⊂ K[x] hasthe induced grading, then the Frobenius number of S is equal to the degree,as a rational function, of the Hilbert series of K[S].

Proof. Let c = g(S) be the Frobenius number of S. Since K[S]i = Kxi ifi ∈ S and K[S]i = (0) otherwise, one has

F (K[S], z) =∑i∈S

zi = f(z) +

∞∑i=c+1

zi = f(z) +1

(1− z) −c∑i=0

zi

where f(z) is a polynomial with coefficients in {0, 1} of degree at most c−1.Hence deg(F (K[S], z)) = c. �

Arithmetical symmetric semigroups Let S be a numerical semigroupof N. The problem of Frobenius consists in determining g(S), the Frobeniusnumber of S. This problem can be explained as follows. Given a sufficientsupply of coins of various denominations find the largest amount that cannotbe formed with these coins. The problem of computing g(S) has beenexamined by several authors [173, 359, 365, 374]. In [364, 365] the famousformula of Scarf and Shallcross [368], for the Frobenius number, is expressedin algebraic terms using homological algebra and lattice theory.

Lemma 8.7.8 Let u, v ∈ N+. If gcd(u, v) = 1 and S = uN + vN, theng(S) = uv − u− v.

Proof. We set c = uv − u − v. It is easy to see that c is not in S. Hence,to show that c = g(S), it suffices to show that c + i ∈ S for i ∈ N+. Onecan write 1 = μ1u+ μ2v for some μ1, μ2 in Z, hence

c+ i = (v − 1 + iμ1)u+ (−1 + iμ2)v.

350 Chapter 8

By the division algorithm we can write c+ i = b1u+ b2v, where 0 ≤ b2 < u.From the identity

c− b1u− b2v = (−b1 − 1)u+ (u − 1− b2)v = −i /∈ S,

one concludes b1 ≥ 0 and consequently c+ i ∈ S. �

Lemma 8.7.9 Let q ≥ 3 and v ≥ 1 be integers and let S be the semigroup

S = d1N+ · · ·+ dqN,

where di = d1 + (i − 1)v and d1 = r + k(q − 1) for some k, r ∈ N. Ifgcd(d1, v) = 1 and 2 ≤ r ≤ q, then g(S) is equal to (k + v)d1 − v.

Proof. Setting c = (k + v)d1 − v, we will show that c + i ∈ S for i ≥ 1.One may assume 1 ≤ i ≤ d1. Notice

c+ i = (k + 1)d1 + λ+ i,

where λ = vd1 − d1 − v. Consider U = d1N + vN. By Lemma 8.7.8, λ isequal to g(U). Hence λ+ i = ad1 + bv, for some a, b in N. One has

vd1 + i = (a+ 1)d1 + (b + 1)v.

Hence b ≤ (q− 1)(k+ a+1) and we can write b = j(k+ a+1)+ �, for some1 ≤ � ≤ k + a+ 1 and 0 ≤ j ≤ q − 2. Therefore the equality

c+ i = (k + a+ 1)d1 + bv = (k + a+ 1− �)dj+1 + �dj+2,

gives c+ i in S. It remains to show c /∈ S. If c ∈ S, then c =∑q

i=1 aidi, forsome ai’s in N. Setting s =

∑qi=1 ai, note that one can rewrite c as:

c = (k + 1)d1 + λ = a1d1 + · · ·+ aqdq = sd1 + pv,

for some p ∈ N. Since λ /∈ U , one derives s ≤ k and c ≤ sdq ≤ kdq, whichis a contradiction since c = kdq + (r − 1)v. �

Theorem 8.7.10 [148, 265] Let r, k, q, v ∈ N and let

d1 = r + k(q − 1) and di = d1 + (i − 1)v

be an arithmetical sequence. If 2 ≤ r ≤ q �= 2 and gcd(d1, v) = 1, thenr = 2 if and only if the semigroup S = d1N+ · · ·+ dqN is symmetric.

Proof. ⇒) Let d1 = 2 + (q − 1)k. Set c = (k + v)d1 − v = kdq + vand U = d1N + vN. Assume z /∈ S. First consider the case z ∈ U , writez = rd1 + sv, r, s ∈ N. By the Euclidean algorithm and Lemma 8.7.9one can write z = adq + bv, where a, b are integers so that 0 ≤ a ≤ k


and 1 ≤ b < d1. Writing a = k − i, where 0 ≤ i ≤ k, we claim that1 ≤ b ≤ (q − 1)i + 1. If b > (q − 1)i + 1, then z = x1d1 + x2v, withx1 = k+ v− i and x2 = b− (q− 1)i− 2. Since 0 ≤ x2 ≤ (q− 1)x1, it is nothard to show that z ∈ S, which is impossible. Hence 1 ≤ b ≤ (q − 1)i + 1.Set y1 = i and y2 = (q − 1)i − b + 1. Since 0 ≤ y2 ≤ (q − 1)y1 from theequality c − z = y1d1 + y2b, one gets c − z ∈ S. Now consider the casez /∈ U . One can write z = ad1 + bv, with 1 ≤ b < d1 and a < 0. Setw1 = k+ 1 and w2 = d1 − b− 1. Using c− z = w1d1 − (1 + a)d1 +w2v and0 ≤ w2 ≤ (q − 1)w1, gives c− z ∈ S. Therefore S is symmetric.⇐) Assume that S is symmetric and keep the same notations as above.

As v /∈ S one has c − v ∈ S and c − v =∑qi=1 bidi, where bi ∈ N for all i.

Set a =∑q

i=1 bi and notice that

c− v = (k + 1)d1 + λ− v = ad1 + q1v,

for some q1 ∈ N. Therefore a ≤ k, otherwise one obtains λ ∈ U . From theinequality c− v ≤ adq ≤ kdq, it follows that r = 2. �

Exercises

8.7.11 Let S be a numerical semigroup of N. Prove that there is a uniqueminimal set of generators {d1, . . . , dq} of S with gcd(d1, . . . , dq) = 1.

8.7.12 Let S �= (0) be a semigroup of N. Prove that the following conditionsare equivalent:

(a) S is a numerical semigroup.

(b) c+ N ⊂ S for some c ∈ S.(c) |N \ S| <∞.

8.7.13 Prove that the semigroup S = 4N+ 5N+ 6N is symmetric.

8.7.14 If S = 5N + 6N + 7N + 8N and K is a field, prove that K[S] is aGorenstein ring which is not a complete intersection.

8.7.15 Let S ⊂ N be a numerical semigroup with Frobenius number c andlet K be a field. If D = K[S] ⊂ K[x] has the induced grading, prove thatthe degree of the Hilbert series of D/xc+1D is 2c+ 1.

8.7.16 Let K be a field and let D′ = K[t1, t2]/(td21 − td12 , ta11 ta22 ), where ti

has degree di ∈ N+. If gcd(d1, d2) = 1 and c+ 1 = a1d1 + a2d2, prove thatthe Hilbert series of D′ is

(1− zc+1)(1 − zd1d2)(1− zd1)(1 − zd2) ,

where c is the Frobenius number of the proper semigroup S = d1N + d2N.Note D′ � D/xc+1D, where D = K[xd1 , xd2 ].

352 Chapter 8

8.8 Toric ideals of monomial curves

In this section we study toric ideals of monomial curves. Using binary trees,we characterize when the toric ideal of an affine monomial curve over anarbitrary field is a complete intersection. If the base field has positive char-acteristic, we show that 1-dimensional graded lattice ideals are set theoreticcomplete intersections.

Let S = Nd1 + · · ·+Ndq be a numerical semigroup of N generated by asequence d = {d1, . . . , dq} of relatively prime positive integers.

As usual S = K[t1, . . . , tq] denotes a polynomial ring over a field K. Thekernel of the homomorphism of K-algebras

ϕ : S = K[t1, . . . , tq] −→ K[S] = K[xd1 , . . . , xdq ],

induced by ϕ(ti) = xdi , is called the toric ideal of K[S] and will be denotedby P . The map ϕ is graded if we endow S and K[x] with the gradingsinduced by setting deg(ti) = di and deg(x) = 1.

The corresponding monomial curve in AqK parameterized by ti = xdi ,i = 1, . . . , q, will be denoted by Γ. If q = 3, Γ is called a monomial spacecurve. In this context the ideal P is also called the toric ideal of Γ.

Lemma 8.8.1 P is a binomial graded prime ideal of S of dimension 1 andΓ = V (P ). If K is an infinite field, then I(Γ) = P .

Proof. As K[x] is an integral domain, we get that P is prime. Since K[x]is integral over K[S], by Proposition 2.4.13, we have ht(P ) = q − 1. ByCorollary 8.2.18, the toric ideal P is generated by binomials. Accordingto Proposition 8.4.8, if gcd (d) = 1, Γ is an affine toric variety, that isΓ = V (P ). If K is infinite, by Corollary 8.4.13, we get I(Γ) = P . �

Definition 8.8.2 A binomial fi = tmi

i −∏j �=i t

aijj ∈ P is called critical

with respect to ti or ti-critical if mi is the least positive integer such that

midi ∈∑j �=i

djN.

The notion of a critical binomial was first introduced by Eliahou [135],and later studied in [4]. This notion—and some of the results of thissection—can be extended to binomial and lattice ideals [335].

Definition 8.8.3 A set {f1, . . . , fq} is called a full set of critical binomialsif fi is a ti-critical binomial for all i.

The interest in studying ideals generated by critical binomials comesfrom a famous result of Herzog [216] showing that the toric ideal of anymonomial space curve is generated by a full set of critical binomials (seeTheorem 8.8.8). Herzog’s result is no longer true for toric ideals of monomialcurves in higher dimensions.


Example 8.8.4 [135] Letm be a positive integer. If d1 = m2, d2 = 2m2−1,d3 = 3m2 + m, d4 = 4m2 + m − 1, then μ(P ) ≥ m, where μ(P ) is theminimum number of generators of P .

If S is symmetric and generated non-redundantly by four integers, thenP is minimally generated by 3 or 5 elements [53]; thus even in this case P isnot generated by critical binomials. A main open problem is to determinewhen a given full set of critical binomials generates P . If the toric idealof a monomial curve in A4

K is generated by a full set of critical binomials,produced by any critical-binomial-algorithm, then one can predict how thesegenerators should look [4].

In what follows, for simplicity of notation, fi will stand for a ti-criticalbinomial for i = 1, . . . , q. Note that we do not assume ±f1, . . . ,±fq distinct.

Theorem 8.8.5 [216] Let B = {f1, f2, f3} be a full set of critical binomialsand let I = (B). If f1 = ta11 − ta22 ta33 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 andbi > 0, ci > 0 for all i, then

(a) ai > 0 for all i and a1 = b1 + c1, b2 = a2 + c2, c3 = a3 + b3.

(b) B is a Grobner basis for I w.r.t the revlex order t1 " t2 " t3.(c) The ideal I = (f1, f2, f3) is equal to the toric ideal P of K[S].

Proof. (a) If a2 = 0, then a3 ≥ c3. As f1 + ta3−c33 f3 = ta11 − tc11 tc22 ta3−c33 ,one obtains a contradiction, hence a2 > 0, by a similar argument a3 > 0.We set α = (−a1, a2, a3), β = (b1,−b2, b3), γ = (c1, c2,−c3). Consider thevector (v1, v2, v3) = α + β + γ. We now show that vi ≥ 0 for all i. Notea1 > b1, otherwise using t

b1−a11 tb33 f1+f2 one derives a contradiction. Hence,

using tb33 f1+ ta1−b11 f2, one has a3+b3 ≥ c3 and v3 ≥ 0. Similarly, one shows

that v1, v2 ≥ 0. As v1d1 + v2d2 + v3d3 = 0, we get vi = 0 for all i.(b) It follows from part (a) and Buchberger’s criterion (Theorem 3.3.17).(c) Let f be a binomial in P and assume we order the monomials with

the revlex ordering t1 " t2 " t3. As f1, f2, f3 are binomials, by the divisionalgorithm (Theorem 3.3.6), we can write

f = h1f1 + h2f2 + h3f3 + r, where r = tr11 tr22 t

r33 − t

s11 t

s22 t

s33 ∈ P,

r1, s1 < a1, r2, s2 < b2 and such that none of the monomials occurring in ris divisible by the leading term of f3. It suffices to show that r = 0.

Assume r �= 0. As tr33 or ts33 is a factor of r, one may assume that s3 = 0.If s1 ≤ r1 or s2 ≤ r2 we get a contradiction with the minimality of b2 ora1, respectively; hence s1 > r1 and s2 > r2. From the equality r = tr11 t

r22 h,

where h = tr33 − ts1−r11 ts2−r22 , we get h ∈ P and r3 ≥ c3. Using the identity

h− tr3−c33 f3 = tr3−c33 tc11 tc22 − ts1−r11 ts2−r22

and the minimality of b2 and a1 allows us to conclude s1 − r1 > c1 ands2 − r2 > c2, a contradiction because ts11 t

s22 is not divisible by tc11 t

c22 . �

354 Chapter 8

Proposition 8.8.6 If f1 = ta11 − ta22 = −f2, f3 = tc33 − tc11 tc22 is a full setof critical binomials, then (f1, f3) is equal to the toric ideal P of K[S].

Proof. Let f be a binomial in P and assume we order the terms with therevlex order t3 " t1 " t2. As f1, f3 are binomials, by the division algorithm,we can write f = h1f1 + h3f3 + r, where r = tr11 t

r22 t

r33 − ts11 ts22 ts33 ∈ P , such

that r1, s1 < a1 and r3, s3 < c3.Assume r �= 0. As tr22 or ts22 is a factor of r, one may assume that

s2 = 0. Note s1 > r1 and s3 > r3. From the equality r = tr11 tr33 h, where

h = tr22 − ts1−r11 ts3−r33 , we get h ∈ P and r2 ≥ a2. Therefore the identity

h+ tr2−a22 f1 = ta11 tr2−a22 − ts1−r11 ts3−r33

contradicts the choice of c3 because one has the inequality a1 > s1 − r1. �

Definition 8.8.7 The support of a binomial f = tα − tβ , denoted bysupp(f), is defined as supp(tα) ∪ supp(tβ), where supp(tα) = {ti|αi > 0}.

The next result also holds for graded lattice ideals [335].

Theorem 8.8.8 [216] If {f1, f2, f3} is a full set of critical binomials, thenthe ideal (f1, f2, f3) is equal to the toric ideal P of K[S].

Proof. Let f1 = ta11 − ta22 ta33 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 . ByTheorem 8.8.5(c) one may assume supp(fi) �= {t1, t2, t3} for at least two fi.Hence one may assume a3 = 0 and the cases to consider are the following:(i) b1 = 0, (ii) b3 = 0, (iii) c1 = 0, and (iv) c2 = 0.

(i) Note b3 ≥ c3, if b3 = c3, then f2 is a critical binomial w.r.t t3 and byProposition 8.8.6 one concludes that P is generated by {f1, f2}. If b3 > c3,from the identity f2 + tb3−c33 f3 = tb22 − tb3−c33 tc11 t

c22 , we get c2 = 0. Thus

f ′2 = f2 + tb3−c33 f3 = tb22 − tb3−c33 tc11 and

f ′3 = f3 + tc1−a11 f1 = tc33 − tc1−a11 ta22 .

Applying Theorem 8.8.5(a) to f1, f′2, f

′3 yield c1 = a1. Hence, f3 is t1-

critical. Using Proposition 8.8.6, we get that P is generated by {f2, f3}.(ii) As a2 ≥ b2 and b1 ≥ a1, from f1 − ta2−b22 f2 = ta11 − ta2−b22 tb11 we get

f1 = −f2 and P = (f1, f3) by Proposition 8.8.6.One may now assume b1 > 0 and b3 > 0. (iii) Using tc2−b22 f2 + f3 we

readily see that this case cannot occur. (iv) Using ta2−b22 f2 + f1 we derivea contradiction. �

Binary trees and complete intersections

Definition 8.8.9 A binary tree is a connected directed rooted tree suchthat: (i) two edges leave the root and every other vertex has either degree


1 or 3, (ii) if a vertex has degree 3, then one edge enters the vertex and theother two edges leave the vertex, and (iii) if a vertex has degree 1, then oneedge enters the vertex. The vertices of degree 1 are called terminal .

Proposition 8.8.10 If G is a binary tree with q terminal vertices, then thenumber of non-terminal vertices of G is q − 1.

Proof. It follows by induction on q. �

Definition 8.8.11 A binary tree G is said to be labeled by [q] := {1, . . . , q}if its terminal vertices are labeled by {1}, . . . , {q}. Extending this defini-tion, we will also consider binary trees with q terminal vertices labeled byarbitrary finite subsets of N with q elements.

If G is a binary tree labeled by [q] and v is a non-terminal vertex of G,consider v1 and v2, the two vertices of G such that� v

�� v2

��

��v1

is a subgraph of G, and denote by G1, resp. G2, the subtree of G whoseroot is v1, resp. v2. We denote by �1[v] and �2[v] the two disjoint subsetsof [n] formed by the union of the labels of the terminal vertices of G1 andG2, respectively.

Definition 8.8.12 Let B = {g1, . . . , gq−1} be a set of binomials of S withgi = tαi − tβi , supp(tαi) ∩ supp(tβi) = ∅, and αi �= 0, βi �= 0 for alli = 1, . . . , q− 1, and let G be a binary tree labeled by [q]. We say that G iscompatible with B if, denoting by F the set of non-terminal vertices of G,there is a bijection

B f−→ Fsuch that supp(tαi) ⊂ �1[f(gi)] and supp(tβi) ⊂ �2[f(gi)] for all i ∈ [q − 1].

Example 8.8.13 The following binary tree G labeled by [5] is compatiblewith B = {g1 = t21t

42 − t4t5 , g2 = t1 − t2t3 , g3 = t44 − t25 , g4 = t2 − t73}� g1

��

��

�g2��

��

�g4��

��

� {3}

��

��

�{2}

��

�{1}

��

��

� g3��

��

��

�{4} � {5}

and if v is the root of G, then �1[v] = {1, 2, 3} and �2[v] = {4, 5}.

356 Chapter 8

Theorem 8.8.14 Let B = {g1, . . . , gq−1} be a set of binomials of S suchthat gi = tαi − tβi , supp(tαi) ∩ supp(tβi) = ∅, and αi �= 0, βi �= 0 for alli = 1, . . . , q − 1. Then the following two conditions are equivalent:

(1) V (B, ti) = {0} for all i = 1, . . . , q.

(2) There is a binary tree G labeled by [q] which is compatible with B.

Proof. (1) ⇒ (2): Set V1 := {1}, . . . , Vq := {q} and consider the partitionF1 := {V1, . . . , Vq} of [q]. Let us show that there exist Vq+1, . . . , V2q−1, sub-sets of [q], and F2, . . . ,Fq, partitions of [q], such that, reindexing g1, . . . , gq−1

if necessary, the following assertions hold for all i ∈ [q − 1]:

(a) Vq+i = Vj ∪ Vk for some Vj , Vk ∈ Fi, j �= k.

(b) supp(tαi) ⊂ Vj and supp(tβi) ⊂ Vk.

(c) Fi+1 = (Fi \ {Vj, Vk}) ∪ {Vq+i}.

Consider the digraphG with 2q−1 vertices, denoted by v1, . . . ,v2q−1, wherewe connect vq+i with vj and vk as follows:� vq+i

�� vk

��

��vj

whenever Vq+i = Vj ∪ Vk in (a), it is not hard to see that G is a binary treelabeled by [q]. The root of G is v2q−1, and the set of its non-terminal verticesis F := {vq+1, . . . ,v2q−1}. Moreover, by construction, for i ∈ {1, . . . , q−1},one has that �1[vq+i] = Vj and �2[vq+i] = Vk for Vj and Vk in (a). Hence,by (b), G is compatible with B via the map f : B → F , gi �→ vq+i, and (2)will follow.

Let us first construct Vq+1 and F2 satisfying (a), (b), and (c). We firstclaim that for all i ∈ [q], there exists an element gj ∈ B such that eithersupp(tαj ) ⊂ Vi or supp(tβj ) ⊂ Vi because otherwise, we have that theith unit vector ei of A

qK belongs to V (B, t1, . . . , ti−1, ti+1, . . . , tq) which is

{0} by (1). Since |F1| = q and |B| = q − 1, by the pigeonhole principlethere exists an element in B, say g1, and Vj , Vk ∈ F1 with j �= k, suchthat supp(tα1) ⊂ Vj and supp(tβ1) ⊂ Vk. Setting Vq+1 := Vj ∪ Vk andF2 := (F1 \ {Vj , Vk}) ∪ {Vq+1}, (a), (b), and (c) hold for i = 1.

Assume now that for i ∈ [[2, q−1]], we have constructed Vq+1, . . . , Vq+i−1

and F2, . . .Fi such that (a), (b), and (c) hold, and let us construct Vq+i andFi+1 satisfying (a), (b), and (c).

Observe first that for all j ≤ i−1, supp(gj) is contained in some elementof Fi. Set Bi := B \ {g1, . . . , gi−1}. We claim that for each Vk ∈ Fi, thereexists gj ∈ Bi such that either supp(tαj ) ⊂ Vk or supp(tβj ) ⊂ Vk. Inorder to prove this, we show that if there exists an element in Fi, say Vs =


{i1, . . . , im}, that does not satisfy the claim, then α := ei1+· · ·+eim belongsto V (B), which is a contradiction by (1). Take gj ∈ B. If gj ∈ Bi, thensupp(tαj ) �⊂ Vs and supp(tβj ) �⊂ Vs by definition of Vs, and hence gj(α) = 0.If gj /∈ Bi, i.e., if j ≤ i−1, then supp(gj) is contained in some element of Fi,say Vt. If t = s, i.e., if supp(gj) ⊂ Vs, then gj(α) = 1 − 1 = 0. Otherwise,since Fi is a partition of [q] and Vs, Vt ∈ Fi, one has that Vs ∩ Vt = ∅, andhence supp(gj) ∩ Vs = ∅. Thus, gj(α) = 0, and the claim is proved.

We have proved that for each Vk ∈ Fi, there exists gj ∈ Bi such thateither supp(tαj ) ⊂ Vk or supp(tβj ) ⊂ Vk. Since |Fi| = q − i + 1 and|Bi| = q− i, and using that Fi is a partition of [q], we get by the pigeonholeprinciple that there exists an element in Bi, say gi, and Vj , Vk ∈ Fi suchthat supp(tαi) ⊂ Vj and supp(tβi) ⊂ Vk. Setting Vq+i := Vj ∪ Vk andFi+1 = (Fi \ {Vj , Vk}) ∪ {Vq+i}, the statements (a), (b), and (c) hold, andwe are done.

(2) ⇒ (1): The proof is by induction on q. The result is clear if q = 2.Denoting by v the root of G, we may assume without loss of generality, that�1[v] = [r] and �2[v] = [[r + 1, q]] for some r ∈ {1, . . . , q − 1}. Then, if G1

and G2 are the two connected components of the digraph G \ {v} obtainedfrom G by removing the vertex v, one has that G1 and G2 are binary trees(for convenience we regard an isolated vertex as a binary tree), labeled by[r] and [[r+1, q]], respectively. Reindexing the gi’s if necessary, we may alsoassume that G1 is compatible with B1 := {g2, . . . , gr}, G2 is compatiblewith B2 := {gr+1, . . . , gq−1}, and g1 = tα1 − tβ1 with supp(tα1) ⊂ [r] andsupp(tβ1) ⊂ [[r + 1, q]]. Then, supp(gi) ⊂ [r] if i = 2, . . . , r, and supp(gi) ⊂[[r+1, q]] if i = r+1, . . . , q−1. Moreover, applying the induction hypothesis,one has that {0} = V (B1, ti) ⊂ Kr for i = 1, . . . , r, and {0} = V (B2, ti) ⊂Kq−r for i = r + 1, . . . , q. Fix i ∈ [q] and take a ∈ V (B, ti). The resultwill be proved if we show that a = 0. By symmetry, we may assume that1 ≤ i ≤ r. The vector a = (a1, . . . , aq) can be decomposed as a = b + c,where b = (a1, . . . , ar, 0, . . . , 0). Then b ∈ V (B1, ti), and hence b = 0. Onthe other hand, g1(a) = 0 implies that aj = 0 for some j ∈ {r + 1, . . . , q}.Thus c ∈ V (B2, tj) which is {0}, and hence a = 0, as required. �

Complete intersections Let d = {d1, . . . , dq} be a set of distinct positiveintegers and let P ⊂ S be the corresponding toric ideal defined at thebeginning of the section. The exact sequence

0 −→ ker(ψ) −→ Zqψ−→ Z −→ 0; ei

ψ�−→ di

is related to P as follows. If g = ta − tb is a binomial, then g ∈ P if andonly if g = a− b ∈ ker(ψ).

Definition 8.8.15 Let G be a binary tree labeled by [q], and consider a setof vectors in Zq, W = {w1, . . . , wq−1}. We say that G is compatible with W

if G is compatible with the set of binomials {tw+i − tw−

i ; i = 1, . . . , q − 1}.

358 Chapter 8

Theorem 8.8.16 Let G be a binary tree labeled by [q] and denote by F theset of its non-terminal vertices. The following two conditions are equivalent:

(1) There exist vectors w1, . . . , wq−1 ∈ Zq such that G is compatible withW = {w1, . . . , wq−1}, and ker(ψ) = ZW .

(2) For all v ∈ F ,

gcd(dj ; j ∈ �1[v]) gcd(dj ; j ∈ �2[v])gcd(dj ; j ∈ �1[v] ∪ �2[v])

∈2⋂i=1

N{dj ; j ∈ �i[v]}.

Proof. Let v be the root of G and let G1 and G2 be the two connectedcomponents of the digraph G\{v} obtained from G by removing v. We mayassume that �1[v] = [r] and �2[v] = [[r + 1, q]] for some r ∈ {1, . . . , q − 1}.Then G1 and G2 are binary trees (for convenience we regard an isolatedvertex as a binary tree) labeled by [r] and [[r + 1, q]]. The result is clear ifq = 2, and we will prove both implications by induction on q.

(1)⇒ (2): Reindexing the wi’s if necessary, we may assume that wq−1 isthe element ofW associated to v through the map that makes G compatiblewith W , and that W1 = {w1, . . . , wr−1} and W2 = {wr, . . . , wq−2} are theset of vectors in W such that Gi is compatible with Wi for i = 1, 2.

There is a decomposition Zq = Zr⊕Zq−r, where Zr := Zr×{0}q−r andZq−r := {0}r × Zq−r. Consider the linear maps

ψ1 : Zq −→ Z, ψ1(ei) =

{di if 1 ≤ i ≤ r,0 if r < i ≤ q,

and ψ2 = ψ−ψ1. Let ψ1 (resp. ψ2) be the restriction of ψ1 (resp. ψ2) to Zr

(resp. Zq−r). It is not hard to see that ker(ψ1) = ZW1 and ker(ψ2) = ZW2.Hence, setting

d = gcd(d1, . . . , dq), d′ = gcd(d1, . . . , dr), d′′ = gcd(dr+1, . . . , dq),

d = {d1, . . . , dq}, d′ = {d1, . . . , dr}, d′′ = {dr+1, . . . , dq},

and using induction we need only show (d′d′′)/d ∈ Nd′∩Nd′′. For 1 ≤ j ≤ rand r + 1 ≤ k ≤ q we can write

dkdej −

djdek = λ1jkw1 + · · ·+ λq−1

jk wq−1,

for some λ1kj , . . . , λq−1kj in Z. We can write the last vector in the set W as

wq−1 = (a1, . . . , ar,−ar+1, . . . ,−aq) = w+q−1 − w−

q−1.

Hence, we get −(dj/d)ek = λrjkwr + · · ·+ λq−1jk wq−1. This implies

(dj/d)dk = λq−1jk (ar+1dr+1 + · · ·+ aqdq).


Set h = ar+1dr+1 + · · ·+ aqdq. If we fix k and vary j, we get

gcd ((d1/d)dk, . . . , (dr/d)dk) = μkh (μk ∈ Z) ⇒ dkd′ = μkhd.

Therefore varying k yields

gcd (dr+1d′, . . . , dqd

′) = gcd(μr+1hd, . . . , μqhd) = hdμ (μ ∈ Z).

As a consequence (d′d′′)/d = (hdμ)/d ∈ Nd′′. A symmetric argument gives(d′d′′)/d ∈ Nd′, as required.

(2)⇒ (1): By induction hypothesis there are

W1 = {w1, . . . , wr−1} and W2 = {wr, . . . , wq−2}

such that Gi is compatible with Wi and ker(ψi) = ZWi for i = 1, 2. Theresult will be proved if we give wq−1 ∈ Zq such that supp(w+

q−1) ⊂ [r],

supp(w−q−1) ⊂ [[r + 1, q]], and ker(ψ) = ZW for W =W1 ∪W2 ∪ {wq−1}.

By hypothesis there are where a1, . . . , aq in N such that

d′d′′

d=

gcd(d1, . . . , dr) gcd(dr+1, . . . , dq)

gcd(d1, . . . , dq)=

r∑i=1

aidi =

q∑i=r+1

aidi.

Setting wq−1 := (a1, . . . , ar,−ar+1, . . . ,−aq) and W :=W1 ∪W2 ∪ {wq−1},we get supp(w+

q−1) ⊂ [r] and supp(w−q−1) ⊂ [[r+1, q]]. Thus G is compatible

with W . To complete the proof it remains to prove that ZW = ker(ψ).Clearly ZW ⊂ ker(ψ). To prove the reverse inclusion define

σjk = (dj/d)ek − (dk/d)ej ; j, k ∈ [q].

By Exercise 8.3.41 the set {σjk| j, k ∈ [q]} generates ker(ψ). Thus weneed only show that σjk ∈ ZW for all j, k ∈ [q]. If j, k ∈ [r] or j, k ∈ [[r+1, q]],then σjk ∈ ker(ψ1) ⊂ ZW or σjk ∈ ker(ψ2) ⊂ ZW . Assume j ∈ [r] andk ∈ [[r + 1, q]]. From the equalities

S1 =

r∑i=1

ai

(did′ej −

djd′ei

)=d′′

dej −

djd′

r∑i=1

aiei,

S2 =

q∑i=r+1

ai

(did′′ek −

dkd′′ei

)=d′

dek −

dkd′′

q∑i=r+1

aiei

we conclude

dkd′′S1 −

djd′S2 =

(dkdej −

djdek

)− djdkd′d′′

wq−1.

Since Si ∈ ker(ψi) ⊂ ZW we obtain σjk ∈ ZW , as required. �

Notation Given a binomial g = ta − tb, we set g = a− b.

360 Chapter 8

Theorem 8.8.17 Let K be an arbitrary field and let B = {g1, . . . , gq−1} bea set of binomials in the toric ideal P . Then P = (B) if and only if

(a) ker(ψ) = Z{g1, . . . , gq−1} and

(b) V (g1, . . . , gq−1, ti) = {0} for i = 1, . . . , q.

Proof. If P = (B), then (a) follows at once from Proposition 8.3.1, and(b) follows from Corollary 8.3.21(b). Conversely, if (a) and (b) hold then byCorollary 8.3.21, one has rad(B) = P . Let {h1, . . . , hs} be a set of generatorsof P consisting of binomials. Notice that ker (ψ) = Z{g1, . . . , gr} by (a),

and ker (ψ) = Z{h1, . . . , hs} by Proposition 8.3.1. Thus using that (B) is aquasi homogeneous complete intersection and applying Proposition 8.3.17,(B) is a radical ideal, and hence P = (B). �

Theorem 8.8.18 [33] The toric ideal P is a complete intersection if andonly if there is a binary tree G labeled by [q] such that, for all non-terminalvertex v of G, one has that

gcd(dj , j ∈ �1[v]) gcd(dj , j ∈ �2[v])gcd(dj , j ∈ �1[v] ∪ �2[v])

∈ N{dj , j ∈ �1[v]} ∩N{dj , j ∈ �2[v]}.

Proof. ⇒) There are binomials g1, . . . , gq−1 such that P = (g1, . . . , gq−1).We may assume that gi = tαi − tβi and supp(tαi) ∩ supp(tβi) = ∅ for alli. By Theorem 8.8.17(b) and Theorem 8.8.14 there exists a binary tree Glabeled by [q] which is compatible with {g1, . . . , gq−1}. ThenG is compatiblewith W = {g1, . . . , gr} and ker(ψ) = Z{g1, . . . , gr} (see Theorem 8.8.17(a)).Thus applying Theorem 8.8.16 we obtain the required conditions.⇐) By Theorem 8.8.16, there is W = {w1, . . . , wq−1} ⊂ Zq such that W

is compatible with G and ker(ψ) = ZW . Setting gi := tw+i − tw−

i , one hasthat G is compatible with {g1, . . . , gq−1}. Hence, by Theorem 8.8.14, we getV (g1, . . . , gq−1, ti) = {0} for i = 1, . . . , q. Therefore, by Theorem 8.8.17, wededuce the equality P = (g1, . . . , gq−1). �

There is a characterization of complete intersection semigroups of Ngiven in [107] (see also [158] for a generalization of this description to semi-groups of arbitrary dimension). The complete intersection property of toricideals of monomial curves has been studied in [30] from a computationalpoint of view (showing an efficient algorithm that checks this property). Inthe area of complete intersection toric ideals there are some other papers(see the introductions of [182, 184, 322] and the references therein).

Remark 8.8.19 If P is a complete intersection and G is a binary treelabeled by [q] such that the conditions of Theorem 8.8.18 hold, then

(i) The generators {g1, . . . , gq−1} of P and their degrees D1, . . . , Dq−1 canbe obtained as shown in the proofs of Theorems 8.8.16 and 8.8.18.


(ii) The Frobenius number g(S) of a numerical semigroup S = Nd can beexpressed entirely in terms of {d1, . . . , dq}.

This last assertion is a consequence of the following. Recall that thequasi-homogeneous Hilbert series of S/P is

F (S/P, z) =f(z)

(1 − zd1) · · · (1− zdq)

for some polynomial f ∈ Z[z]; see Theorem 5.1.4. When d1, . . . , dq arerelatively prime, by Proposition 8.7.7 and its proof, one can write

F (S/P, z) =h(z)

1− zfor some polynomial h of degree g(S) + 1. If P is a complete intersection,one can write f(z) = (1 − zD1) · · · (1 − zDq−1) where D1, . . . , Dq−1 are thedegrees of the minimal generators of P (see Exercise 5.1.20), and henceg(S) = D1 + · · ·+Dq−1 − (d1 + · · ·+ dq). Denoting by {v1, . . . ,vq−1} theset of non-terminal vertices of G and using (i), we get :

g(S) =(q−1∑i=1

gcd(dj , j ∈ �1[vi]) gcd(dj , j ∈ �2[vi])gcd(dj , j ∈ �1[vi] ∪ �2[vi])

)−(

q∑i=1

di

).

Example 8.8.20 Let K be a field, and consider d1 = 16, d2 = 27, d3 = 45,and d4 = 56. The corresponding toric ideal P ⊂ K[t1, t2, t3, t4] is a completeintersection because using the following binary tree labeled by [4]��

��{3}

��

��

��{2} ��

�� {4}�{1}

the arithmetical conditions in Theorem 8.8.18 are satisfied:

112 =(16)(56)

gcd(16, 56)∈ 16N ∩ 56N; 2(56)

(1)= 7(16)

135 =(27)(45)

gcd(27, 45)∈ 27N ∩ 45N; 3(45)

(2)= 5(27)

72 =gcd(16, 56) gcd(27, 45)

gcd(16, 27, 45, 56)∈ (16, 56)N ∩ (27, 45)N;

1(16) + 1(56)(3)= 1(27) + 1(45).

Moreover, the equalities (1), (2) and (3) provide, by Remark 8.8.19(i), a setof minimal generators of P :

g1 = t24 − t71, g2 = t33 − t52, g3 = t1t4 − t2t3 .

362 Chapter 8

Finally, by Remark 8.8.19(ii), the Frobenius number of the numerical semi-group S = N{16, 27, 45, 56} is

g(S) = 112 + 135 + 72− (16 + 27 + 45 + 56) = 175 .

Monomial curves in positive characteristic Let K be a field and letP ⊂ S be the toric ideal of a monomial curve. In characteristic zero it is anopen problem whether P is a set theoretic complete intersection. Severalauthors have studied this problem; see [135, 302, 319, 402]. The solution ofthe case n = 3 is treated in [54].

The next result gives a nice family of binomial set theoretic completeintersections. We show this result using a theorem of Katsabekis, Morales,and Thoma [270, Theorem 4.4(2)].

Proposition 8.8.21 [293] If K is a field of positive characteristic and L isa graded lattice ideal of S of dimension 1, then L is a binomial set theoreticcomplete intersection.

Proof. Let L be the homogeneous lattice of Zq such that L = I(L). Noticethat L is a lattice of rank q − 1 because ht(L) = rank(L). Thus, there isan isomorphism of groups ψ : Zq/Ls → Z, where Ls is the saturation of Lconsisting of all a ∈ Zq such that da ∈ L for some 0 �= d ∈ Z. For each1 ≤ i ≤ q, we set ai = ψ(ei + Ls), where ei is the ith unit vector in Zq.Following [270], the multiset A = {a1, . . . , aq} is called the configurationof vectors associated to L. Recall that q − 1 = rank(L). Hence, as L ishomogeneous with respect to d = (d1, . . . , dq), there are positive integers riand rk such that riei − rkek ∈ L and ridi − rkdk = 0. Thus, riai = rkakand ai has the same sign as ak. This means that a1, . . . , aq are all positiveor all negative. It follows that A is a full configuration in the sense of [270,Definition 4.3]. Thus, I(L) is a binomial set theoretic complete intersectionby [270, Theorem 4.4(2)] and its proof. �

Corollary 8.8.22 [319] Let P ⊂ S be the toric ideal of a monomial curve.If char(K) > 0, then P is a binomial set theoretic complete intersection.

Proof. By Lemma 8.8.1 P is a 1-dimensional graded lattice ideal. Thus,the result follows at once from Proposition 8.8.21. �

Remark 8.8.23 Working over an algebraically closed field K of character-istic zero Eliahou [136] proved that the toric ideal P of a monomial curvein AnK is the radical of an ideal generated by n binomials.


Exercises

8.8.24 Let P be the toric ideal of K[xd1 , xd2 , xd3 ], where K is a field andd1, d2, d3 are relatively prime positive integers. Prove that P is a completeintersection if and only if P is Gorenstein.

8.8.25 Let f1 = ta11 − ta22 ta33 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 be a fullset of critical binomials. If bi > 0, ci > 0 for all i, prove the equalitiesd1 = a2b3 + a3b2, d2 = a1b3 + a3b1, d3 = a1b2 − a2b1.

8.8.26 (a) If d1 = 7, d2 = 8, d3 = 9, d4 = 17. Prove that

f1 = t51 − t23t4, f2 = t22 − t1t3, f3 = t43 − t41t2, f4 = t4 − t2t3

is a set of critical binomials that generate a height 3 prime ideal of type 2.

(b) If d1 = 204, d2 = 855, d3 = 1216, d4 = 1260. Prove that

f1 = t471 − t42t33t24, f2 = t82 − t151 t34, f3 = t63 − t191 t42, f4 = t54 − t131 t33

is a set of critical binomials that generate a height 3 prime ideal of type 3.

(c) If d1 = 9, d2 = 12, d3 = 18, d4 = 19. Prove that

f1 = −f3 = t21 − t3, f2 = t32 − t23, f4 = t34 − t1t2t23

is a complete intersection prime ideal.

8.8.27 Let f1 = ta11 − ta22 ta33 , f2 = tb22 − tb11 tb33 , f3 = −tc11 tc22 + tc33 be a fullset of critical binomials and I = (f1, f2, f3). If bi > 0, ci > 0 for all i andS = K[t1, t2, t3] has the grading induced by deg(ti) = di, then the minimalresolution of S/I is given by

0→ S(−d2b2 − d3a3)⊕ S(−d3c3 − d2a2)→S(−d1a1)⊕ S(−d2b2)⊕ S(−d3c3)→ S → S/I → 0

and max{d2a2 + d3c3 − (d1 + d2 + d3), d2b2 + d3a3 − (d1 + d2 + d3)} is theFrobenius number of S = N{d1, d2, d3}.Hint Use Theorem 8.8.5(b) and Proposition 8.7.7.

8.8.28 Let K be a field, let P be the toric ideal of K[td1 , . . . , tdq ] and let

I = ({tdji − tdij | 1 ≤ i < j ≤ q}).

If gcd(d1, . . . , dq) = 1 and Γ is the monomial curve in AqK parameterized byti = xdi , then rad(I) = P and Γ = V (I).

364 Chapter 8

8.8.29 Let K be an algebraically closed field and let X be the curve in AqKgiven parametrically by ti = fi(x), where fi(x) ∈ K[x] \K for i = 1, . . . , q.Use the extension theorem [99, Chapter 3] to prove that V (P ) = X , whereP is the presentation ideal of K[f1(x), . . . , fq(x)].

8.8.30 Prove that part (c) of Theorem 8.8.5 is valid for arbitrary distinctpositive integers d1, d2, d3.

8.8.31 [216] If P is the toric ideal of K[xd1 , xd2 , xd3 ], prove that P is a settheoretic complete intersection.

8.8.32 A rational polyhedral cone C ⊂ Rn is unimodular if there existsΓ = {γ1, . . . , γn} a Z-basis of Zn such that C = R+B for some B ⊂ Γ. Alattice polytope P ⊂ Rn with vertex set A is smooth [61, p. 371] if the coneR+(A− v) is unimodular for any v ∈ A, where A− v = {a− v| a ∈ A}. LetP ⊂ Rn be a lattice polytope of dimension n and let

K[P ] := K[{xat| a ∈ Zn ∩ P}] ⊂ K[x1, . . . , xn, t]

be its polytopal subring, where K[x1, . . . , xn, t] is a polynomial ring over afield K. An open problem posed by Bøgvad is whether for a smooth normalpolytope P the toric ideal of K[P ] is generated by quadrics (see [61, 400]).If n = 1, prove that P is smooth, K[P ] is normal, and deg(K[P ]) = vol(P).

Hint Use Theorem 9.3.5, Lemma 9.3.7, and Theorem 9.3.25.

8.8.33 If P = conv(v1, v1 + (s − 1)) ⊂ R, with v1 ∈ N+ and s ≥ 3 aninteger, prove that the toric ideal P of K[P ] is the ideal I2(M) generatedby the 2× 2 minors of the following 2× (s− 1) generic Hankel matrix

M =

(t1 t2 · · · ts−1

t2 t3 · · · ts

).

Hint The ideal I2(M) is a prime ideal of height s− 2 [128, p. 612].

8.8.34 If P = conv((−1,−1), (1, 0), (0, 1)) ⊂ R2, prove that the polytope Pis not smooth, K[P ] is normal, and the toric ideal P of K[P ] is generatedby t1t2t3 − t34. Then prove that the point [(0, 0, 1, 0)] is a singular point ofthe projective toric variety V (P ) ⊂ P3 defined by P .

Hint Use the Jacobian criterion of [210, 5.8, p. 37].

Chapter 9

Monomial Subrings

This chapter deals with point configurations and their lattice polytopes.We consider various subrings and ideals associated with them, e.g., Ehrhartrings, Rees algebras, homogeneous subrings, lattice and toric ideals, matrixand Laplacian ideals of graphs.

Throughout we shall use the following symbology and terminology whichis consistent with the notation introduced thus far:

K; R field; Laurent polynomial ring K[x±11 , . . . , x±1

n ]A;F A = {v1, . . . , vq} ⊂ Zn; F = {xv ∈ R | v ∈ A}If A ⊂ Nn set R = K[x] = K[x1, . . . , xn], I = (F ) ⊂ RA n× q matrix with column vectors v1, . . . , vqP := conv(A) ⊂ Rn lattice polytopeR[t] the polynomial ring K[x1, . . . , xn, t]A(P) Ehrhart ring K[{xati | a ∈ Zn ∩ iP , i ∈ N}]K[P ] polytopal subring K[{xat | a ∈ P ∩ Zn}]R[Ft] Rees algebra R[xv1t, . . . , xvq t]K[F ] monomial subring K[xv1 , . . . , xvq ]

S polynomial ring K[t] = K[t1, . . . , tq]PF ⊂ S; I(L) ⊂ S toric ideal of K[F ]; lattice ideal of L ⊂ Zq

in≺(PF ) initial ideal of PF with respect to ≺K[Ft] monomial subring K[xv1t, . . . , xvq t]

K[Ft] integral closure of K[Ft].

Following [146, 189, 383], we shall be interested in comparing thesemonomial subrings and their algebraic properties and invariants. One hasthe following inclusions:

K[Ft] ⊂ K[P ] ⊂ A(P) ⊂ R[Ft] and K[Ft] ⊂ A(P).

Using Hilbert functions, polyhedral geometry and Grobner bases, we willstudy normalizations of monomial subrings as well as initial ideals, special

366 Chapter 9

generating sets, primary decompositions and multiplicities of lattice andtoric ideals. The reciprocity law of Ehrhart for integral polytopes and theDanilov–Stanley formula for canonical modules of monomial subrings willbe introduced here. Applications of these results will be given.

9.1 Integral closure of monomial subrings

Here we present a description of the integral closure of a monomial subringusing polyhedral geometry and integer programming techniques.

Normality of monomial subrings If A is an integral domain with fieldof fractions KA, recall that the normalization or integral closure of A is thesubring A consisting of all the elements of KA which are integral over A.

Normal monomial subrings arise in the theory of toric varieties [100].The next result is essentially shown in the book of Fulton [176, pp. 29–30 ].

Theorem 9.1.1 If A ⊂ Zn is a finite set of points and S = ZA ∩ R+A,then the following hold:

(a) K[S] := K[{xa| a ∈ S}] is normal,

(b) K[F ] = K[S], where F = {xa| a ∈ A}.

Proof. (a): By Theorem 1.1.29, there are non-zero vectors a1, . . . , ap in Zn

such that R+A = H+a1 ∩ · · · ∩ H+

ap . We set Si = ZA ∩ H+ai . Since K[S] is

equal to K[S1] ∩ · · · ∩K[Sp] and since the intersection of normal domainsis a normal domain, it suffices to show that K[Si] is normal for all i. IfZA ⊂ Hai , then Si = ZA � Zr. Hence, K[Si] � K[Zr] = K[x±1

1 , . . . , x±1r ]

which is normal by Corollary 4.3.9. Thus, we may assume that ZA �⊂ Hai .Setting r = rank(ZA) and L = ZA ∩ Ha, one has rank(L) = r − 1. Thisfollows by noticing that QL = QA ∩Ha, and using the equality

n = dimQ(QA+Ha) = dimQ(QA) + dimQ(Ha)− dimQ(QA∩Ha)

together with the fact that the ranks of ZA and L are equal to dimQ(QA)and dimQ(QL), respectively. The quotient group H = ZA/L is torsion-freeof rank 1. Thus, H is a free abelian group of rank 1 and we can writeH = Zα for some is 0 �= α ∈ ZA such that 〈α, ai〉 > 0. As a consequence,we get that Si = L ⊕Nα � Zr−1 ⊕ N and K[Si] � K[Zr−1 ⊕ N]. ThereforeSi is normal because K[Zr−1 ⊕N] is equal to K[x±1

1 , . . . , x±1r−1, xr] and this

ring is normal again by Corollary 4.3.9.(b): Since K[F ] ⊂ K[S], taking integral closures and using part (a)

gives the inclusion K[F ] ⊂ K[S]. To show the reverse inclusion, note theequality ZA∩R+A = ZA∩Q+A (see Corollary 1.1.27). A straightforwardcalculation shows that xα is in the field of fractions of K[F ] if α ∈ ZA, andxα is an integral element over K[F ] if α ∈ Q+A. Hence K[S] ⊂ K[F ]. �

Monomial Subrings 367

Theorem 9.1.2 Let A be a finite set of points in Zn. If R+A is a pointedcone, then the integral closure of K[F ] in R = K[x±1] is equal to

K[{xa| a ∈ Zn ∩R+A}].

Proof. We set B = K[{xa| a ∈ Zn ∩ R+A}]. Let K[F ] be the integralclosure of K[F ] in R. To show K[F ] ⊂ B, take z ∈ K[F ]. One canwrite z = d1x

γ1 + · · · + drxγr , where di ∈ K \ {0} for all i and γ1, . . . , γr

distinct non-zero points in Zn. Write γi = niβi, with ni equal to the gcdof the entries of γi (observe that β1, . . . , βr are not necessarily distinct).Consider the cone C spanned by A and {β1, . . . , βr}. It suffices to showthat βi ∈ R+A for all i. Assume on the contrary that C is not equal toR+A. By Theorem 1.1.54 one may assume that R+β1 is a face of C notcontained in R+A. Let Ha be a hyperplane so that Ha ∩ C = R+β1 andC ⊂ H−

a . There is 1 ≤ � ≤ r such that

n� = sup1≤j≤r

{nj| γj = njβ1}.

Since z is integral over K[F ] it satisfies a monic polynomial f of degree mwith coefficients in K[F ]. The monomials that occur in the expansion off(z) as a sum of monomials are of the form xα(xγi1 )mi1 · · · (xγit )mit , wheremij > 0 for all j, m ≥

∑tj=1mij and α ∈ R+A. To derive a contradiction

we claim that the term xmγ� occurs only once in the expansion of f(z) as asum of monomials. Assume the equality

(xγ�)m = xα(xγi1 )mi1 · · · (xγit )mit ,

where mij > 0 for all j, m ≥∑t

j=1mij , and α ∈ R+A. As γ� = n�β1, onehas 〈mγ�, a〉 = 0 and from this equality one rapidly derives

〈α, a〉 = 〈βi1 , a〉 = · · · = 〈βit , a〉 = 0.

Hence α, βii , . . . , βit ∈ Ha ∩ C = R+β1. Note β1 /∈ R+A. Thus α = 0 andβij = β1 for all j. Therefore t = 1 and γ� = γi1 , as claimed. Altogether weget f(z) �= 0, which is impossible.

Now we show the inclusion B ⊂ K[F ]. By Corollary 1.1.27 one has theequality Zn ∩R+A = Zn ∩Q+A. A straightforward calculation shows thatxa is an integral element over K[F ] if a ∈ Q+A ∩ Zn. Hence B ⊂ K[F ]. �

Corollary 9.1.3 (a) K[F ] is normal if and only if NA = ZA ∩R+A.(b) K[F ] = K[xγ1 , . . . , xγr ] for some γ1, . . . , γr in ZA ∩Q+A.(c) K[F ] = xβ1K[F ] + · · ·+ xβmK[F ] for some xβ1 , . . . , xβm in K[F ].

(d) If A is t-unimodular, then K[F ] is normal.

368 Chapter 9

Proof. (a): It suffices to observe the equality K[F ] = K[{xα|α ∈ NA}]and to use the description of K[F ] given in Theorem 9.1.1(b).

(b): It follows from Corollary 1.1.27, Lemma 1.3.2, and Theorem 9.1.1.(c): By part (b) K[F ] is generated, as a K[F ]-algebra, by a finite set of

Laurent monomials. As K[F ] is integral over K[F ], by Corollary 2.4.4, weget that K[F ] is finite over K[F ] and the result follows readily.

(d): It follows from Theorem 1.6.7 and part (a). �

Definition 9.1.4 A decomposition K[F ] =⊕∞

i=0K[F ]i of the K-vectorspaceK[F ] is an admissible grading if K[F ] is a positively gradedK-algebrawith respect to this decomposition and each component K[F ]i has a finiteK-basis consisting of monomials.

Theorem 9.1.5 (Danilov, Stanley [65, Theorem 6.3.5]) Let A ⊂ Zn be afinite set of points and let ri(R+A) be the relative interior of R+A. If K[F ]is normal and R+A is a pointed cone, then the canonical module ωK[F ] ofK[F ], with respect to an admissible grading, can be expressed as

ωK[F ] = ({xa| a ∈ NA ∩ ri(R+A)}). (9.1)

The formula above represents the canonical module of K[F ] as an idealof K[F ] generated by Laurent monomials. For a comprehensive treatmentof the Danilov–Stanley theorem, see [61, 65, 102].

Theorem 9.1.6 (Hochster [248], [65, Theorem 6.3.5]) If A ⊂ Zn is a finiteset of points and K[F ] is normal, then K[F ] is Cohen–Macaulay.

Lemma 9.1.7 Let A ⊂ Zn be a finite set. If K[F ] is a normal positivelygraded K-algebra and ωK[F ] is the canonical module of K[F ], then

a(K[F ]) = −min{ i | (ωK[F ])i �= 0}. (9.2)

Proof. Let PF be the toric ideal of K[F ]. Recall that K[F ] is isomorphic,as a K-algebra, to K[t1, . . . , tq]/PF , where K[t1, . . . , tq] has the gradinginduced by deg(ti) = deg(xvi). By Theorem 9.1.6, the algebra K[F ] isCohen–Macaulay. Therefore by Proposition 5.2.3 the formula follows. �

Theorem 9.1.8 If A ⊂ Nn and K[h] := K[h1, . . . , hd] ↪→ K[F ] is a homo-geneous Noether normalization of K[F ], then K[F ] is a free K[h]-module

whose generators have degree at most∑d

i=1 deg(hi).

Proof. Let R = ⊕i≥0Ri be the standard grading of R and endowK[F ] with

the grading K[F ]i = K[F ]∩Ri. The composite K[h] ↪→ K[F ] ↪→ K[F ] is ahomogeneous Noether normalization. Since K[F ] is Cohen–Macaulay (seeTheorem 9.1.6), by Proposition 3.1.27, we get a direct sum decomposition

K[F ] = K[h]xβ1 ⊕ · · · ⊕K[h]xβm .


Therefore the Hilbert series of K[F ] can be expressed as

F (K[F ], t) =m∑i=1

F (K[h]xβi , t) =

(m∑i=1

tdeg βi

)/d∏i=1

(1− tdeghi).

Applying Theorem 9.1.5 one derives that a(K[F ]) is negative. Using thata(K[F ]) is equal to the degree of F (K[F ], t) as a rational function (seeProposition 5.2.3) one concludes deg xβi ≤

∑mi=1 deg hi. �

Applications of the Danilov–Stanley theorem Next we introducesome techniques to compute the canonical module and the a-invariant of awide class of monomial subrings.

Let A be a finite set of points in Zn. The dual cone of R+A is thepolyhedral cone given by

(R+A)∗ = {x | 〈x, y〉 ≥ 0; ∀ y ∈ R+A}.

A set H ⊂ Zn \ {0} is called an integral basis of (R+A)∗ if (R+A)∗ = R+H.

Theorem 9.1.9 Let c1, . . . , cr be an integral basis of (R+A)∗ and let b =(bi) be the vector given by bi = 0 if R+A ⊂ Hci and bi = −1 if R+A �⊂ Hci .Assume there is x0 ∈ Qn such that 〈x0, vi〉 = 1 for all i. If NA = Zn∩R+Aand B is the matrix with column vectors −c1, . . . ,−cr, then

(a) ωK[F ] = ({xa| a ∈ Zn ∩ {x|xB ≤ b}).

(b) a(K[F ]) = −min {〈x0, x〉| x ∈ Zn ∩ {x|xB ≤ b}}.

Proof. K[F ] is a standard graded K-algebra with the grading induced bydeclaring that a monomial xa ∈ K[F ] has degree i if and only if 〈a, x0〉 = i.We set H = {c1, . . . , cr}. As R+H = (R+A)∗ = H+

v1 ∩ · · · ∩H+vq , by duality

(see Corollary 1.1.30), we have the equality

R+A = H+c1 ∩ · · · ∩H

+cr . (9.3)

Observe that R+A ∩Hci is a proper face if bi = −1 and it is an improperface otherwise. From Eq. (9.3) we get that each facet of R+A has the formR+A∩Hci for some i. The relative boundary of the cone R+A is the unionof its facets (see Theorem 1.1.44). Hence, using that H is an integral basis,we obtain the equality

Zn ∩ (R+A)o = Zn ∩ {x|xB ≤ b}. (9.4)

Now, part (a) follows readily from Eq. (9.1) of Theorem 9.1.5 and Eq. (9.4).Part (b) follows from Eq. (9.2) of Lemma 9.1.7 and part (a). �

370 Chapter 9

Definition 9.1.10 If xa is a monomial of K[x], we set log(xa) = a. Thevector a is called the exponent vector of xa. Given a set F of monomials,the log set of F , denoted log(F ), consists of all log(xa) with xa ∈ F .

Example 9.1.11 If F = {x1, . . . , x4, x1x2x5, x2x3x5, x3x4x5, x1x4x5} andA = log(F ), then A is a Hilbert basis and 〈x0, v〉 = 1 for v ∈ A, wherex0 = (1, 1, 1, 1,−1). An integral basis for (R+A)∗ is given by

{e1, e2, e3, e4, e5, (0, 1, 0, 1,−1), (1, 0, 1, 0,−1)}.

Then, using Theorem 9.1.9, it is easy to verify that ωK[F ] is generated bythe set of all xa such that a = (ai) satisfies the system of linear inequalities:

ai ≥ 1 ∀ i; a1 + a3 − a5 ≥ 1; a2 + a4 − a5 ≥ 1. (∗)

The only vertex of the polyhedron defined by Eq. (∗) is v0 = (1, 1, 1, 1, 1).Thus, by Theorem 9.1.9, the a-invariant of K[F ] is −〈x0, v0〉 = −3.

Let R+A = aff(R+A)∩H−a1 ∩ · · · ∩H−

ar be an irreducible representationof R+A such that ai ∈ Zn and the non-zero entries of ai are relatively primefor all i. The existence of this representation follows from Theorem 1.1.29.

Definition 9.1.12 If C is the integral matrix with rows a1, . . . , ar, thepolyhedron

Q = aff(R+A) ∩ {x ∈ Rn|Cx ≤ −1}is called the shift polyhedron of R+A relative to C.

If Zn ∩ ri(R+A) �= ∅, then Q �= ∅. The shift polyhedron will be used togive a technique to compute the canonical module of K[F ].

Example 9.1.13 A shift polyhedron of a cone is a shift of the cone asdrawn below.

�

�

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10��

��

��

��

�

�C =

(1 22 3

)

R+A = {x| Cx ≤ 0}

� � � ��

Here the shift polyhedron of the cone R+A is Q = {x| Cx ≤ −1}.


Proposition 9.1.14 If Q is the shift polyhedron of R+A with respect to Cand R+A is a pointed cone, then Q is a pointed polyhedron,

Zn ∩ Q = Zn ∩ ri(R+A), (9.5)

and conv(Zn ∩ ri(R+A)) is an integral polyhedron if Zn ∩ ri(R+A) �= ∅. IfQ is integral, then Q = conv(Zn ∩ ri(R+A)).

Proof. It is not hard to see that the lineality spaces of R+A and Q areequal. As R+A is pointed, so is Q by Exercise 1.1.80. To prove Eq. (9.5)it suffices to note that a point α is in ri(R+A) if and only if α ∈ R+Aand 〈α, ai〉 < 0 for i = 1, . . . , r, see Theorem 1.1.44. The third assertionfollows by taking convex hulls in Eq. (9.5) and observing that conv(Zn ∩Q)is the integer hull of Q, which is a polyhedron by Proposition 1.1.65(a) andis integral by Proposition 1.1.65(b). The last assertion follows by takingconvex hulls in Eq. (9.5) and using that Q is integral. �

Theorem 9.1.15 Let K[F ] be a monomial subring such that (a) there is0 �= x0 ∈ Qn such that 〈x0, vi〉 = 1 for all i, (b) NA = Zn ∩ R+A, and (c)the shift polyhedron Q, relative to C, of the cone R+A is integral. Then

(i) K[F ] is a normal standard graded K-algebra whose canonical moduleand a-invariant are given by

ωK[F ] = ({xa| a ∈ Zn ∩ Q}) and a(K[F ]) = −min {〈x0, x〉| x ∈ Q}.

(ii) xβ is a minimal generator of ωK[F ] for any vertex β of Q.

Proof. (i): K[F ] is graded as follows: a monomial xa with a ∈ NA hasdegree i if and only if 〈a, x0〉 = i. Using (a), it is seen that K[F ] is astandard graded K-algebra. That K[F ] is normal follows from (b). Thecanonical module is the ideal of K[F ] given by

ωK[F ](9.1)= ({xa| a ∈ NA ∩ ri(R+A)})(b)= ({xa| a ∈ Zn ∩ ri(R+A)}) (9.6)

(9.5)= ({xa| a ∈ Zn ∩ Q}). (9.7)

To show the formula for the a-invariant observe that a monomial xa ofK[F ] has degree 〈x0, a〉. Thus from Eq. (9.7) and Lemma 9.1.7 we obtain

min {〈x0, x〉| x ∈ Q} ≤ −a(K[F ]).

As the minimum is attained at a vertex β of Q (see Proposition 1.1.41), toshow equality, it suffices to observe that β is integral (see Corollary 1.1.64).

(ii): By condition (c), β is integral. If F is a proper face of R+A, thenβ �∈ F because β is in ri(R+A) (see Theorem 1.1.44). Then, by Eq. (9.6),we get xβ ∈ ωK[F ]. There are c ∈ Qn and b ∈ Q such that

372 Chapter 9

(1) {β} = {x|〈x, c〉 = b} ∩ Q and Q ⊂ {x|〈x, c〉 ≤ b}.

Then, by definition of Q, one has vi + β ∈ Q for all i. Thus

〈vi + β, c〉 = 〈vi, c〉+ b ≤ b⇒ 〈vi, c〉 ≤ 0 (i = 1, . . . , q).

Assume there is α ∈ Q and η1, . . . , ηq in N such that

β = η1v1 + · · ·+ ηqvq + α,

then b = 〈β, c〉 = η1〈v1, c〉 + · · · + ηq〈vq, c〉 + 〈α, c〉 ≤ 〈α, c〉 ≤ b. Hence〈α, c〉 = b and by (1) we get α = β. Thus xβ is a minimal generator of thecanonical module ωK[F ]. �

Exercises

9.1.16 If K[F ] = K[x3, y3, x2y], prove that K[F ] = K[F ][xy2].

9.1.17 Prove that the ringK[F ] is normal if and only if any element α ∈ ZAsatisfying rα ∈ NA for some integer r ≥ 1 belongs to NA.

9.1.18 If S = R+A ∩ Zn, then K[S] = K[{xα|α ∈ S}] is normal.

9.1.19 The shift polyhedron Q of R+A, relative to C, can be written asQ = R+A+ P , i.e., R+A is the characteristic cone of Q.

9.1.20 Let A ⊂ Nn be a finite set. If K[F ] is normal and F ′ is the set ofall xvi ∈ F such that x1 is not in the support of xvi , then K[F ′] is normal.

9.1.21 If there is x0 ∈ Qn such that 〈x0, v〉 = 1 for all v ∈ A, then R+A isa pointed cone.

9.2 Homogeneous monomial subrings

The class of homogeneous monomial subrings will be studied here. Examplesin this class include Rees algebras of ideals generated by monomials of thesame degree and polytopal subrings.

Definition 9.2.1 The subring K[F ] (resp. A) is said to be homogeneousif there is x0 ∈ Qn such that 〈vi, x0〉 = 1 for i = 1, . . . , q.

Proposition 9.2.2 K[F ] is homogeneous if and only if K[F ] is a standardgraded algebra with the grading

K[F ]i =∑|a|=i

K(xv1)a1 · · · (xvq )aq , where a = (a1, . . . , aq) ∈ Nq.


Proof. ⇒) There is a vector x0 ∈ Qn such that 〈vi, x0〉 = 1 for all i. Oneclearly has K[F ]iK[F ]j ⊂ K[F ]i+j for all i, j. Thus, it suffices to proveK[F ] =

⊕∞i=0K[F ]i. First let us prove that K[F ]i ∩K[F ]j = {0} for i �= j.

If this intersection is not zero, one has an equality

(xv1)a1 · · · (xvq )aq = (xv1)b1 · · · (xvq )bq ,

with a = (a1, . . . , aq) ∈ Nq, b = (b1, . . . , bq) ∈ Nq, |a| = i, |b| = j. Hence

i = |a| = 〈a1v1 + · · ·+ aqvq, x0〉 = 〈b1v1 + · · ·+ bqvq, x0〉 = |b| = j,

a contradiction. To finish the proof assume fi1 + · · ·+ fir = 0, with fij inK[F ]ij for all j and i1 < · · · < ir. If fir �= 0, using that the monomialsof R form a K-basis, one has K[F ]ir ∩ K[F ]ij �= {0} for some j < r, acontradiction. Thus fir = 0. By induction fij = 0 for all j, as required.⇐) Set V = affQ(v1, . . . , vq) and r = dim(V ). To begin with we claim

that 0 is not in V . Otherwise if 0 ∈ V , write 0 = λ1v1 + · · · + λqvq withλi ∈ Q for all i and

∑qi=1 λi = 1. There is 0 �= p ∈ Z such that pλi ∈ Z

for all i. One may assume, by reordering the vi, that pλi ≥ 0 for i ≤ s andpλi ≤ 0 for i > s. As K[F ] is graded, from the equality

(xv1 )pλ1 · · · (xvs)pλs = (xvs+1)−pλs+1 · · · (xvq )−pλq ,

one derives p∑qi=1 λi = 0, a contradiction. Thus 0 /∈ V , and consequently

r < n. It is well known that a linear variety V in Qn of dimension r canbe written as an intersection of hyperplanes V =

⋂n−ri=1 H(yi, ci), where

0 �= yi ∈ Qn and ci ∈ Q for all i; see [427, Corollary 1.4.2]. Since 0 /∈ V ,one has cj �= 0 for some j. To finish the proof we set x0 = yj/cj. �

Corollary 9.2.3 PF is a graded ideal with respect to the standard gradingof S if and only if K[F ] is homogeneous.

Proof. As PF is generated by a finite set of binomials (Corollary 8.2.18),the result follows readily from Proposition 9.2.2. �

Proposition 9.2.4 If ψ, ϕ are the maps of K-algebras defined by

K[t1, . . . , tq]ϕ ��

ψ

��

K[F ]

K[Ft]

ϕ��

��

��

tiϕ ��

ψ

��

xvi

xvit

ϕ

��

��

then there is a unique epimorphism ϕ such ϕ = ϕψ. In addition ϕ is anisomorphism if and only if K[F ] and K[Ft] have the same Krull dimension.

374 Chapter 9

Proof. To show the existence of ϕ it suffices to prove ker(ψ) ⊂ ker(ϕ). Thisinclusion follows at once because any binomial in ker(ψ) clearly belongs toker(ϕ), and ker(ψ) is a binomial ideal.

Note that ker(ψ) and ker(ϕ) are both prime ideals. The map ϕ is anisomorphism if and only if ker(ψ) = ker(ϕ); thus to finish the proof we needonly observe that the last equality holds if and only if K[F ] and K[Ft] havethe same dimension. �

Corollary 9.2.5 The map ϕ : K[Ft] → K[F ] is an isomorphism if andonly if K[F ] is homogeneous.

Proof. ⇒) As ker(ϕ) = ker(ψ), by Corollary 9.2.3, K[F ] is homogeneousbecause ker(ψ) is homogeneous in the standard grading of S.

⇐) By Corollary 9.2.3, ker(ϕ) is homogeneous in the standard gradingof S. Note that any homogeneous binomial in ker(ϕ) is also in ker(ψ), thusone has ker(ϕ) = ker(ψ) and ϕ is an isomorphism. �

Remark 9.2.6 The subring K[Ft] is always homogeneous because (vi, 1)lies in the hyperplane xn+1 = 1 for all i. If K[F ] is homogeneous, thereis x0 ∈ Qn such that 〈x0, vi〉 = 1 for all i. Then K[F ] = ⊕i∈NK[F ]i is astandard graded K-algebra with ith graded component defined by

K[F ]i =∑

〈a,x0〉=iK{xa|xa ∈ K[F ]}.

The isomorphism of K-algebras ϕ is degree preserving, that is, ϕ mapsK[Ft]i into K[F ]i for all i ∈ N. In particular e(K[F ]) = e(K[Ft]).

Lemma 9.2.7 There is n0 ∈ N+ such that (xα)n0 ∈ K[F ] for xα ∈ K[F ].

Proof. By Lemma 1.3.2 and Theorem 9.1.1, there are γ1, . . . , γr in thesemigroup ZA∩Q+A such thatK[F ] = K[xγ1 , . . . , xγr ]. For each i there is apositive integer ni such that xniγi ∈ K[F ]. The integer n0 = lcm(n1, . . . , nr)satisfies the required condition, because any xα ∈ K[F ] can be written asxα = xa1γ1 · · ·xarγr , for some a1, . . . , ar ∈ N. �

Lemma 9.2.8 There is xγ such that xγK[F ] ⊂ K[F ].

Proof. By Lemma 1.3.2 and Theorem 9.1.1, there are γ1, . . . , γr in thesemigroup ZA ∩ Q+A such that K[F ] = K[xγ1 , . . . , xγr ]. For each i, wecan write xγi = xβi/xδi , where xβi and xδi are in K[F ]. By Lemma 9.2.7there is 0 �= n0 ∈ N such that xn0α ∈ K[F ] for all xα ∈ K[F ]. We setxγ = xn0δ1 · · ·xn0δr . Take xα in K[F ] and write xα = (xγ1)a1 · · · (xγr )ar ,


where ai ∈ N for all i. By the division algorithm, for each i there are qi, ciin N such that ai = qin0 + ci and 0 ≤ ci < n0. From the equality

xγxα =

r∏i=1

(xn0γi)qi︸︷︷︸is in K[F ]

r∏i=1

(xδi+γi)ci︸︷︷︸is in K[F ]

r∏i=1

(xδi)n0−ci

︸︷︷︸is in K[F ]

and using that δi + γi = βi for all i, we get xγxα ∈ K[F ]. As xα was anarbitrary monomial in K[F ], we get xγK[F ] ⊂ K[F ]. �

The normalized Ehrhart ring of P = conv(A) is the graded algebra

AP =

∞⊕i=0

(AP )i ⊂ R[t],

where the ith component is given by

(AP)i =∑

α∈ZA∩iPKxαti.

To prove that AP is a graded algebra note that because of the convexity ofthe polytope P one has

• (AP )i(AP)j ⊂ (AP )i+j and

• (AP )i ∩ (AP )j = {0} for i �= j.

Proposition 9.2.9 K[Ft] ⊂ AP , with equality if K[F ] is homogeneous.

Proof. We set B′ = {(v1, 1), . . . , (vq, 1)} ⊂ Zn+1. First we show:

ZB′ ∩ R+B′ ⊂ {(α, i)|α ∈ ZA ∩ iP and i ∈ N}. (∗)

Take z = (α, i) in ZB′ ∩R+B′. There are ni’s in Z and λi’s in R+ such that

z = n1(v1, 1) + · · ·+ nq(vq, 1) = λ1(v1, 1) + · · ·+ λq(vq, 1).

Note α = 0 if i = 0, and α/i ∈ P if i ≥ 1. Hence z is in the right-hand sideof Eq. (∗). Thus, by Theorem 9.1.1, we get K[Ft] ⊂ AP .

Assume that K[F ] is homogeneous. Let x0 be a vector in Qn such that〈vj , x0〉 = 1 for all j. By Theorem 9.1.1 it suffices to prove that equalityholds in Eq. (∗). Take z = (α, i) in the right-hand side of Eq. (∗) and write

α = n1v1 + · · ·+ nqvq = i(λ1v1 + · · ·+ λqvq),

where nj ∈ Z, λj ≥ 0, and∑

j λj = 1. Hence 〈α, x0〉 =∑

j nj = i, and

z = (α, i) = n1(v1, 1) + · · ·+ nq(vq, 1)

= iλ1(v1, 1) + · · ·+ iλq(vq , 1).

Thus z ∈ ZB′ ∩ R+B′, as required. �

376 Chapter 9

Corollary 9.2.10 If K[F ] is homogeneous, then K[F ] is normal if andonly if K[Ft] = AP .

Proof. It follows from Corollary 9.2.5 and Proposition 9.2.9. �

The normalized Ehrhart function of P is defined as

E′(i) = dimK(AP )i = |ZA ∩ iP|, i ∈ N.

Proposition 9.2.11 If K[F ] is a homogeneous subring of dimension d,then E′ is a polynomial function of degree d− 1.

Proof. As the normalized Ehrhart ring AP is equal to K[Ft], thanks toCorollary 9.1.3 one has that AP is a finitely generated K-algebra. HenceAP is a finitely generated graded K[Ft]-module.

Using Theorem 2.2.4 and Proposition 2.4.13 one concludes that theHilbert function of AP is a polynomial function of degree d− 1. �

Example 9.2.12 Let F = {x41, x31x2, x1x32, x42} and let P be the convexhull of A = log(F ). As K[F ] is homogeneous, using Proposition 9.2.9 andNormaliz [68], we get that the normalized Ehrhart ring is

AP = K[x41t, x31x2t, x1x

32t, x

42t, x

21x

22t].

The subrings K[F ] and AP have different Hilbert functions, but they havethe same Hilbert polynomial, which is equal to 4t+ 1.

Proposition 9.2.13 If K[F ] is homogeneous, then K[F ] is a positivelygraded K-algebra and its multiplicity is equal to the multiplicity of K[F ].

Proof. We set P = conv(A). As K[F ] is a homogeneous subring, usingProposition 9.2.9, we get K[Ft] = AP . Hence K[F ] =

⊕∞i=0K[F ]i is a

graded K-algebra whose ith component is given by

K[F ]i =∑

α∈ZA∩iPKxα.

Recall that K[F ] is graded as in Proposition 9.2.2. Let h and h be theHilbert functions of K[F ] and K[F ], respectively. Since K[F ] and K[F ]have the same dimension d, one can write

h(i) = a0id−1 + terms of lower degree,

h(i) = c0id−1 + terms of lower degree,

for i 0. One clearly has a0 ≤ c0 because K[F ]i ⊂ K[F ]i for all i. Onthe other hand by Lemma 9.2.8 there is xγ ∈ K[F ] of degree m such thatxγK[F ] ⊂ K[F ]. Thus for each i there is a one-to-one map

K[F ]ixγ

−→ K[F ]i+m.


Hence h(i) ≤ h(i + m) for all i, and consequently c0 ≤ a0. Altogethera0 = c0. Therefore

e(K[F ]) = a0(d− 1)! = c0(d− 1)! = e(K[F ]). �

Corollary 9.2.14 [72] If K[F ] is homogeneous and Cohen–Macaulay, then

a(K[F ]) ≤ a(K[F ]).

Proof. As K[F ] is normal, by Theorem 9.1.6, K[F ] is Cohen–Macaulay.Hence, the inequality follows from Propositions 5.2.8 and 9.2.13. �

Definition 9.2.15 A binomial tα − tβ in K[t1, . . . , tq] is said to have asquare-free term if at least one of its two terms tα, tβ is square-free.

The following result gives a necessary condition for the normality of ahomogeneous subring K[F ] in terms of its toric ideal.

Theorem 9.2.16 [385] Let B be a finite set of binomials in the toric idealPF of K[F ]. If K[F ] is a normal homogeneous subring and PF is minimallygenerated by B, then every element of B has a square-free term.

Proof. Recall that the toric ideal of K[F ] is the kernel of the epimorphism

ϕ : S = K[t1, . . . , tq] −→ K[F ] (tiϕ�−→ xvi)

of K algebras. Since K[F ] is homogeneous, PF is a graded ideal in thestandard grading of S (see Corollary 9.2.3). We set B = {g1, . . . , g�} andfi = xvi . Let g be a binomial in B. We proceed by contradiction assumingthat g has no square free-term. After permuting variables one can write

g = ta11 · · · tarr − tar+1

r+1 · · · tass ,

with ai ≥ 1 for all i and 2 ≤ a1 = max{ai}ri=1 ≤ as = max{ai}si=r+1.From the equality fa11 · · · farr = f

ar+1

r+1 · · · fass , we obtain f1 · · · fr/fs ∈ K[F ]because K[F ] is normal. Hence there exists a binomial h1 in PF of the form

h1 = t1 · · · tr − tstα,

with r = deg(h1) = deg(tα) + 1 < deg(g). Note the equality

g − ta1−11 · · · tar−1

r h1 = tsh2, (9.8)

where h2 is a binomial in PF with deg(h2) < deg(g) or h2 = 0. Writing

hj =∑gi �=g

cijgi (j = 1, 2),

378 Chapter 9

and using Eq. (9.8) we conclude g =∑

gi �=g cigi, ci ∈ S, a contradictionbecause PF is minimally generated by g1, . . . , g�. �

A full converse to Proposition 9.2.16 is not true even if the monomialshave the same positive total degree. Notice however that there is a partialconverse due to Sturmfels (see Theorem 9.6.16).

Example 9.2.17 Let F = {x1x2, x2x3, x3x4, x1x4, x21, x22, x23, x24} ⊂ K[x].Using Macaulay2 [199], one obtains that PF is minimally generated by

B = {t24 − t5t8, t23 − t7t8, t1t3 − t2t4, t22 − t6t7,t21 − t5t6, t3t4t6 − t1t2t8, t2t3t5 − t1t4t7}.

Using Normaliz [68], we obtain K[F ] = K[F ][x1x3, x2x4].

The next result complements [400, Theorem 13.14].

Theorem 9.2.18 If K[F ] is normal and homogeneous, then the toric idealPF is generated by homogeneous binomials of degree at most rank(A).

Proof. Let B = {g1, . . . , g�} be a minimal generating set of PF consistingof binomials. Set r = maxi{deg(gi)}.

Claim (I): If deg(gi) = r for some i and r > rank(A), then gi is alinear combination of two binomials in PF of degree strictly less than r. ByTheorem 9.2.16 we may assume that gi can be written as:

gi = t1 · · · tr − tar+1

r+1 · · · tamm with aj ≥ 1 for all j.

One has v1 + · · ·+ vr = ar+1vr+1 + · · ·+ amvm. Applying Caratheodory’stheorem (see Theorem 1.1.18) to both sides of this equality and permutingvariables if necessary, we obtain an equation:

b1v1 + · · ·+ br1vr1 = cr+1vr+1 + · · ·+ cm1vm1

with bj , ck ∈ N \ {0} for all j, k, r1 ≤ rank (A), m1 − r ≤ rank (A), and

m1 ≤ m. Thus the binomial f = tb11 · · · tbr1r1 − t

cr+1

r+1 · · · tcm1m1 belongs to PF .

Case (A): bi = 1 for all i. We can write gi − (tr1+1 · · · tr)f = tr+1h forsome h. Hence gi is a linear combination of f and h, which in this case havedegree less than r.

Case (B): ci = 1 for all i. Notice that deg(f) = m1 − r ≤ rank(A) < r.Pick tγ so that t

ar+1

r+1 · · · tamm = tγ(tr+1 · · · tm1). We can write gi − tγf = t1hfor some h. Thus gi is a linear combination of f and h, which in this casealso have degree less than r.

Case (C): maxi{ci} ≥ maxi{bi} ≥ 2. After permutation of variables wemay assume cr+1 = maxi{ci} ≥ maxi{bi} = b1 ≥ 2. From the equality

f b11 · · · fbr1r1 = f

cr+1

r+1 · · · fcm1m1 (fi = xvi),


we get f1 · · · fr1/fr+1 ∈ K[F ] because K[F ] is normal. Hence there is h1 inPF of the form h1 = t1 · · · tr1−tr+1t

α, with deg(h1) = deg(tα)+1 < deg(gi).Note that r1 = deg(h1). We can write

gi − tr1+1 · · · trh1 = tr+1h2,

where h2 is a binomial in PF with deg(h2) < deg(g) or h2 = 0. Thus onceagain gi is a linear combination of binomials of degree less than r. Theremaining cases follow by symmetry. This completes the proof of the claim.

Applying Claim (I) to each binomial gi of degree r, it follows that PF isminimally generated by binomials of degree less than r. Thus, by induction,it follows that PF is generated by binomials of degree at most rank (A). �

Exercises

9.2.19 If K[F ] is a homogeneous monomial subring over a field K, use theDanilov–Stanley formula to show a(K[F ]) < 0.

9.2.20 Prove that the following conditions are equivalent:

(a) K[F ] = K[xv1 , . . . , xvq ] is homogeneous.

(b) 0 /∈ affQ(v1, . . . , vq) ⊂ Qn.

(c) dimQ Q{(v1, 1), . . . , (vq, 1)} < dimQ Q{(v1, 1), . . . , (vq, 1), (0, 1)}.(d) dimQ Q{(v1, 1), . . . , (vq , 1)} = dimQ Q{v1, . . . , vq}.

9.2.21 Is the integral closure of a homogeneous subring homogeneous?

9.2.22 Is the monomial subring K[x2, xy, y3] ⊂ K[x, y] homogeneous?

9.2.23 Let K[F ] be a normal homogeneous subring. If rank(A) = 2, thenPF is generated by homogeneous binomials of degree at most 2.

9.2.24 Consider the monomial subring Q[F ] = Q[x31x2, x2x33, x

21x

22, x

31x3].

Prove that Q[F ] is equal to Q[F ][x21x2x3, x1x2x23]. Prove that the toric

ideal of Q[F ] is equal to (t51t2 − t33t34).

9.2.25 If K[F ] is a homogeneous and h (resp. E′) is the Hilbert functionof K[F ] (resp. normalized Ehrhart function of P), then(a) h(i) ≤ E′(i) for all i ≥ 0, and

(b) h(i) = E′(i) for all i ≥ 0 if and only if K[F ] is normal.

9.2.26 If K[F ] is a two-dimensional homogeneous monomial subring, provethat the Hilbert polynomials of K[Ft] and K[Ft] are equal.

9.2.27 Let K[F ] be a Cohen–Macaulay homogeneous monomial subring. Ifdim(K[F ]) = 2 and a(K[F ]) < 0, prove that K[F ] is normal.

380 Chapter 9

9.3 Ehrhart rings

In this section we introduce the Ehrhart ring and the polytopal subring ofa lattice polytope. We study some of its properties.

Let A = {v1, . . . , vq} be a set of points in Zn and let P be the latticepolytope P = conv(A) ⊂ Rn. The set A is called a point configuration.A point configuration A is called normal if NA = ZA ∩ R+A, and it iscalled homogeneous if A lies on an affine hyperplane of Rn not containingthe origin. In [106] by a “point configuration” they mean a finite multisetof vectors in Rn. In [106] they also introduce “vector configurations” and“homogeneous point configurations.”

To link the lattice polytope P and the point configuration A to ringtheory consider a Laurent polynomial ring R = K[x±1

1 , . . . , x±1n ] over a field

K. If A ⊂ Nn, we set R = K[x1, . . . , xn]. There is an isomorphism betweenthe group Zn and the multiplicative group of monomials of R given by

a = (a1, . . . , an)←→ xa = xa11 · · ·xann .

In this correspondence:

A = {v1, . . . , vq} ←→ F = {xv1 , . . . , xvq}.

The monomial subring K[P ] = K[{xat| a ∈ P ∩Zn}] ⊂ R[t] is called thepolytopal subring of P . Notice the equality

K[P ] = K[{xat| (a, 1) ∈ Q+{(v1, 1), . . . , (vq, 1)} ∩ Zn+1}],

which readily implies that K[Ft] ⊂ K[P ] is an integral extension. Thepolytope P is said to be normal if K[P ] is normal [62] (cf. [100, 203]).

Proposition 9.3.1 dimK[P ] = dim(P) + 1.

Proof. We set V = R(v2 − v1) + · · ·+ R(vq − v1). From the equality

aff(P) = aff(v1, . . . , vq) = v1 + V,

we get dim(P) = dimR(V ). Let B be the matrix with rows (v1, 1), . . . , (vq, 1)and let B′ be the matrix with rows (v2 − v1, 0), . . . , (vq − v1, 0). UsingCorollary 8.2.21, we obtain

dimK[Ft] = rank(B) = rank(B′) + 1 = dim(P) + 1.

Since K[Ft] ⊂ K[P ] is an integral extension, by Proposition 2.4.13, we getthat dimK[Ft] is equal to dimK[P ].

Lemma 9.3.2 Let A = {v1, . . . , vd+1} be an affinely independent set in Zn.If P = conv(A) is a unimodular d-simplex, then P ∩Zn = A and K[P ] is apolynomial ring.


Proof. By Corollary 1.2.22 we have P ∩ Zn = A. Hence K[P ] is equal toK[xv1t, . . . , xvd+1t]. Thus K[P ] has dimension d + 1 and by Lemma 3.1.5K[P ] is a polynomial ring. �

Proposition 9.3.3 [62] Let {S,Si} be a family of finitely generated semi-groups of Zn such that S = ∪iSi. If ZS = ZSi and K[Si] is normal for alli, then K[S] is normal.

Proof. By Corollary 9.1.3 it suffices to show that ZS ∩ R+S is containedin S. Take a ∈ ZS ∩R+S. There is an integer s ≥ 1 such that sa ∈ S, thussa ∈ Si for some i. As ZS = ZSi and K[Si] is normal we get a ∈ Si ⊂ S,as required. �

Definition 9.3.4 A collection {Δ�} of unimodular lattice simplices in Rn

is called a unimodular covering of P if P = ∪�Δ� and dim(Δ�) = d for all�, where d is the dimension of P .

Theorem 9.3.5 [62] If P has a unimodular covering, then K[P ] is normal.

Proof. Let {Δ�} be a unimodular covering of P . Consider the semigroupsS and S� of Zn+1 defined as:

S = N{(α, 1)|α ∈ Zn ∩ P} and S� = N{(α, 1)|α ∈ Zn ∩Δ�}.

By Lemma 9.3.2 K[S�] is a polynomial ring. In particular K[S�] is normalfor all �. Hence, by Proposition 9.3.3, we need only show that S = ∪�S�and ZS = ZS�. One clearly has the inclusion ∪�S� ⊂ S, because P = ∪�Δ�.To show the reverse inclusion take 0 �= z ∈ S. Then

z = (β, s) = n1(β1, 1) + · · ·+ nr(βr, 1),

where ni ∈ N and βi ∈ Zn ∩ P for all i. Note β/s ∈ P , where β =∑

i niβiand s =

∑i ni. Hence β/s ∈ Δ� for some �. There is an affinely independent

set Γ = {γ0, . . . , γd} in Zn such that Δ� = conv(Γ). Thus β/s =∑d

i=0 μiγi,for some μi’s satisfying μi ≥ 0 and

∑i μi = 1. Hence

z = (β, s) = sμ0(γ0, 1) + · · ·+ sμd(γd, 1) ∈ R+{(γ0, 1), . . . , (γd, 1)}.

As Δ� is unimodular, by Corollary 1.1.27 and Proposition 1.2.21, one has

z ∈ Z{(γ0, 1), . . . , (γd, 1)}

Hence using thatK[S�] = K[xγ0t, . . . , xγdt] is normal yields z ∈ S�. To showthe equality ZS = ZS� note that ZS and ZS� are free groups of rank d+ 1.This follows from the assumption that P and Δ� have the same dimension.Since Δ� is unimodular, by Proposition 1.2.21, we get that Zn+1/ZS� istorsion-free. Thus, ZS = ZS�. �

Recall that the Ehrhart ring of P is:

A(P) = K[xαti|α ∈ Zn ∩ iP ] ⊂ R[t].

382 Chapter 9

Theorem 9.3.6 The following assertions hold:

(a) A(P) = K[{xαti| (α, i) ∈ Zn+1 ∩ R+B′}], B′ = {(α, 1)|α ∈ Zn ∩ P},

(b) A(P) = K[{xαti| (α, i) ∈ Zn+1 ∩ R+B}], B = {(vi, 1)}ni=1,

(c) A(P) is a finitely generated K-algebra and a normal domain,

(d) K[Ft] ⊂ A(P) is a finitely generated integral extension,

(e) K[Ft] = A(P) ⇔ A(P) is contained in the field of fractions of K[Ft],

(f) dim(K[Ft]) = dim(K[Ft]) = dim(A(P)) = dim(P) + 1,

(g) K[Ft] = A(P) if and only if B = {(vi, 1)}qi=1 is a Hilbert basis.

Proof. (a): “⊂” Take a monomial xαti in A(P), then α ∈ Zn ∩ iP andi ∈ N. Thus we can write (α, i) =

∑qj=1 iλj(vj , 1), with

∑qj=1 λj = 1 and

λj ≥ 0 for all j. This proves that (α, i) is in the semigroup Zn+1 ∩ R+B′.“⊃” Take a monomial xαti with (α, i) in Zn+1 ∩ R+B′. Any vector of

the form (β, 1) with β ∈ Zn ∩ P can be written as (β, 1) =∑j μj(vj , 1) for

some μj ’s in R+. Hence we can write (α, i) =∑qj=1 ρj(vj , 1), with ρj ∈ R+

for all j. Consequently α ∈ Zn ∩ iP . Thus xαti ∈ A(P).(b): This follows using part (a).(c): By Proposition 1.3.6 there is a finite Hilbert basis H ⊂ Zn such

that R+B ∩Zn = NH and R+B = R+H. Thus, by Proposition 1.5.2, NH isnormal. Using part (b) and Corollary 9.1.3, we get that A(P) is normal.

(d): If xati is in A(P), then (xati)s ∈ K[Ft] for some 0 �= s ∈ N.Hence A(P) is integral over K[Ft]. As a result, using (c) together withCorollary 2.4.4, we obtain that A(P) is a finitely generated K[Ft]-module.

(e): This follows directly from (c) and (d).(f): The first two equalities follow from Proposition 2.4.13 because the

extensions K[Ft] ⊂ K[Ft] ⊂ A(P) are integral. The third equality followsfrom Proposition 9.3.1.

(g): This follows from part (a) because K[Ft] = K[NB]. �

Multiplicity of monomial subrings The subrings K[Ft], K[Ft] andA(P ) are graded as follows. We denote the multiplicity by e(·) or deg(·).

Notice that K[Ft] is homogeneous because (vi, 1) lies in the hyperplanexn+1 = 1 for all i. The rings K[Ft] and K[Ft] are graded K-algebras withith graded components given by

K[Ft]i =∑

(a,i)∈NBKxati and K[Ft]i =

∑(a,b)∈ZB∩iQ

Kxatb, (9.9)

respectively, where B = {(v1, 1), . . . , (vq, 1)} and Q = conv(B).


The Ehrhart ring A(P) is a normal finitely generated graded K-algebrawith ith component given by

A(P)i =∑

α∈Zn∩iPKxαti.

Since A(P) is a finitely generated module over the homogeneous ring K[Ft],by Proposition 5.1.6, we obtain that its Hilbert function:

χP(i) := |Zn ∩ iP| = cdid + · · ·+ c1i+ c0 (i 0)

is a polynomial function of degree d = dim(P) such that ci ∈ Q for all i andd!cd is an integer, which is the multiplicity of A(P).

The function χP is the Ehrhart function of P . The Hilbert polynomialof A(P), denoted by EP(x), is called the Ehrhart polynomial of P . Recallthat the relative volume of P is given by

vol(P) = limi→∞

|Zn ∩ iP|id

,

where d = dim(P); see Proposition 1.2.10. Hence vol(P) is equal to cd,the leading coefficient of the Ehrhart function of P . For this reason d!cd iscalled the normalized volume of P .

Altogether one has:

Lemma 9.3.7 e(A(P)) = d!cd = d!vol(P).

By the Hilbert–Serre theorem (Theorem 5.1.4) the Hilbert series of A(P)is a rational function that can be written uniquely as:

F (A(P), x) :=∞∑i=0

|Zn ∩ iP|xi = h0 + h1x+ · · ·+ hsxs

(1− x)d+1,

with h0 + h1 + · · ·+ hs �= 0, hi ∈ Z for all i and d = dim(P). This series iscalled the Ehrhart series of P .

Lemma 9.3.8 (a) a(A(P)) = s− (d+ 1) < 0,

(b) EP(i) = χP(i) for all integers i ≥ 0,

(c) hi ≥ 0 for all i,

(d) h0 + h1 + · · ·+ hs = d!vol(P).

Proof. (a): This follows from Theorem 9.1.5. (b): This follows fromCorollary 5.1.9. (c): By Theorem 9.3.6, A(P) is a normal finitely generatedK-algebra. Hence, by Theorem 9.1.6, A(P) is Cohen–Macaulay. Thereforehi ≥ 0 for all i; see Exercise 5.2.9. (d): This follows from Remark 5.1.7. �

384 Chapter 9

Proposition 9.3.9 The Ehrhart function of P can be expressed as:

EP(i) = e0

(i+ d

d

)− e1

(i+ d− 1

d− 1

)+ · · ·+ (−1)d−1ed−1

(i+ 1

1

)+ (−1)ded,

where ei = h(i)(1)/i! for all i and h(x) = h0 + h1x+ · · ·+ hsxs.

Proof. This follows from Proposition 5.1.6 and its proof. �

Definition 9.3.10 e0, . . . , ed are called the Hilbert coefficients .

Theorem 9.3.11 (Reciprocity law of Ehrhart [127]) If E+P is the function

E+P (i) = |Zn ∩ ri(iP)|, where ri(iP) is the relative interior of iP, then

E+P (i) = (−1)dEP (−i) ∀ i ≥ 1.

Proof. By Exercise 5.2.10 and [394, Proposition 4.2.3], we have

F (ωA(P ), t) = (−1)d−1F (A(P ), t−1) = (−1)d∑i≥1

EP(−i)ti.

Therefore the required identity follows by comparing coefficients. �

Example 9.3.12 To illustrate the reciprocity law consider the polytope P :

�

�

1

2

3

4

5

6

y

0 1 2 3 4 5 6 x��

��

��

��

� ��

��

�

�|Z2 ∩ P| = 16

|Z2 ∩ ∂P| = 6

EP(x) = 12x2 + 3x+ 1

E+P (1) = (−1)2EP(−1) = 10

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Proposition 9.3.13 If EP(x) = cdxd + cd−1x

d−1 + · · · + c1x + c0 is theEhrhart polynomial and F1, . . . , Fs are the facets of P, then cd = vol(P),cd−1 = 1

2

∑si=1 vol(Fi) and c0 = 1.

Proof. By Lemma 9.3.7, cd = vol(P). Since EP(0) = χP(0) = 1, we getc0 = 1. Now we show the formula for cd−1. Notice the equalities

iP = ri(iP) ∪ ∂(iP); ∂(P) = F1 ∪ · · · ∪ Fs,


where ∂(P) is the relative boundary. Using the reciprocity law and theinclusion-exclusion principle [2, p. 38, Formula 2.12] we get

|∂(iP) ∩ Zn| = EP (i)− E+(i) = EP (i)− (−1)dEP(−i)= 2cd−1i

d−1 + terms of lower degree

= EF1(i) + · · ·+ EFs(i) + h(i),

where h is a polynomial of degree at most d− 2. Comparing leading terms,the required formula follows. �

In general some of the coefficients c0, . . . , cd of EP (x) may be negative;see [21, Example 3.22] and [241]. It is unknown whether the coefficients arenonnegative if the vertices of P have {0, 1}-entries.

Theorem 9.3.14 If P ⊂ R2 is a lattice polytope of dimension 2, then theEhrhart polynomial is given by:

EP(x) = area(P)x2 + |Z2 ∩ ∂P|2

x+ 1.

Proof. Let EP(x) = c2x2 + c1x+ c0 be the Ehrhart polynomial of P . By

Proposition 9.3.13, c2 = area(P) and c0 = 1. Writing P = ri(P) ∪ ∂(P),where ∂(P) is the boundary of P , we obtain by the reciprocity law:

|∂(P) ∩ Z2| = |P ∩ Z2| − |ri(P) ∩ Z2|= EP(1)− E+

P (1) = EP(1)− EP(−1)= (c2 + c1 + c0)− (c2 − c1 + c0) = 2c1.

Thus one has c1 = |Z2∩∂(P)|2 , as required. �

Corollary 9.3.15 (Pick’s Formula) If P ⊂ R2 and dim(P) = 2, then

|Z2 ∩ P| = area(P) + |Z2 ∩ ∂P|2

+ 1.

Proof. It follows making x = 1 in the formula of Theorem 9.3.14. �

Lemma 9.3.16 If v1, . . . , vq have {0, 1}-entries, then P ∩ Zn = A.

Proof. Take α ∈ P ∩ Zn. Then we can write α =∑

j λjvj with∑

jλj = 1and λj ≥ 0 for all j. We may assume λj > 0 for all j and v1 �= 0. We setvj = (vj1, . . . , vjn). As the ith entry of v1 is 1 for some i, one has

1 ≤ λ1v1i + λ2v2i + · · ·+ λqvqi ≤ λ1 + · · ·+ λq = 1.

Thus q = 1 and λ1 = 1, i.e., α = v1 and α ∈ A. �

386 Chapter 9

Corollary 9.3.17 If v1, . . . , vq have {0, 1}-entries, then EP (−1) = 0.

Proof. By Lemma 9.3.16 and Theorem 9.3.11, (−1)dEP (−1) = E+P (1) = 0.

Thus EP (−1) = 0, as required. �

Corollary 9.3.18 If xv1 , . . . , xvq are square-free, then K[Ft] = K[P ].


Proposition 9.3.19 If A = {v1, . . . , vq} ⊂ Zn and P = conv(A), then themultiplicities of A(P) and K[Ft] are related by

e(A(P)) = |T (Zn/(v2 − v1, . . . , vq − v1))| e(K[Ft]).

Proof. We set B = {(v1, 1)}qi=1 and Q = conv(B). Note that there is abijective map Zn∩ iP → Zn+1∩ iQ, α �→ (α, i), and d = dim(P) = dim(Q).Thus vol(P) = vol(Q). Since B lies in the affine hyperplane xn+1 = 1,applying Theorem 1.2.14 gives

vol(P) = |T (Zn+1/((v2 − v1, 0), . . . , (vq − v1, 0)))| limi→∞

|ZB ∩ iQ|id

. (9.10)

On the other hand, one has:

e(K[Ft]) = e(K[Ft]) = d! limi→∞

|ZB ∩ iQ|id

. (9.11)

The first equality follows from Proposition 9.2.13, while the second equalityfollows by definition of multiplicity noticing that K[Ft] is graded as inEq. (9.9) and using Theorem 9.3.6(f). Hence the asserted equality followsfrom Eq. (9.10), Eq. (9.11), and observing that the map

T (Zn/(v2 − v1, . . . , vq − v1)) −→ T (Zn+1/((v2 − v1, 0), . . . , (vq − v1, 0)))

given by v �−→ (v, 0) is bijective. �

Lemma 9.3.20 Let P h ⊂ S[u] be the homogenization of P = PF . Then

(a) P h is the toric ideal of K[F ′] := K[z, xv2−v1z, . . . , xvq−v1z, x−v1z],where z is a new variable.

(b) The toric ideal of K[xv1z, . . . , xvqz, z] is the toric ideal P h of K[F ′].

Proof. (a): The toric ideal of K[F ′], denoted by P ′, is the kernel of themap ϕ : S[u] → K[F ′] induced by ti �→ xvi−v1z for i = 1, . . . , q + 1, wherevq+1 = 0 and tq+1 = u. Let G be the reduced Grobner basis of P withrespect to the GRevLex order ≺. First, we show the inclusion P h ⊂ P ′.Take an element f of G. By Proposition 3.4.2, it suffices to show that fh is


in P ′. We can write f = ta+ − ta− with in≺(f) = ta

+

. Thus, |a+| ≥ |a−|and fh = ta

+ − ta−u|a|, where a = a+− a−. We set a = (a1, . . . , aq). Using

0 = a1v1 + · · ·+ aqvq = a2(v2 − v1) + · · ·+ aq(vq − v1) + |a|v1,

we get that fh ∈ P ′, as required. Let G′ be the reduced Grobner basis of P ′

with respect to the GRevLex order. Next we show the inclusion P ′ ⊂ P h.Take an element f ′ of G′. It suffices to show that f ′ is in P h. As f ′ ishomogeneous in the standard grading of S[u], we can write f ′ = tc

+−tc−u|c|with in≺(f

′) = tc+

and c = (c1, . . . , cq). Since f′ ∈ P ′, we get

(zc+1 )(xv2−v1z)c

+2 · · · (xvq−v1z)c+q

= (zc−1 )(xv2−v1z)c

−2 · · · (xvq−v1z)c−q (x−v1z)|c|.

Hence, c2(v2− v1) + · · ·+ cq(vq − v1) = −|c|v1. Thus, c1v1 + · · ·+ cqvq = 0,

that is, the binomial f = tc+ − tc− is in P . As f ′ = fh, we get f ′ ∈ P h.

(b): The mapping that sends z to xv1z induces an isomorphism of K-algebras K[F ′]→ K[F ′′]. So this part is a consequence of (a). �

Recall that the multiplicity of an affine algebra S/I is also denoted bydeg(S/I) (see Definition 8.5.5). Thus, by Lemma 9.3.7, one can convenientlyrestate Proposition 9.3.19 as follows:

Proposition 9.3.21 Let B = {β1, . . . , βm} be a set of points of Zn and letQ = conv(B) be its convex hull. If r = dim(Q), then

|T (Zn/〈β1 − βm, . . . , βm−1 − βm〉)| deg(K[xβ1z, . . . , xβmz]) = r!vol(Q).

Theorem 9.3.22 Let P be the toric ideal of K[F ] = K[xv1 , . . . , xvq ], let Abe the matrix whose columns are v1, . . . , vq and let r be the rank of A. Then

|T (Zn/〈v1, . . . , vq〉)| deg(S/P ) = r!vol(conv(v1, . . . , vq, 0)).

Proof. By Proposition 8.5.6, deg(S/P ) = deg(S[u]/P h). On the otherhand, by Lemma 9.3.20(b), P h is the toric ideal of the monomial subring:

K[F ′′] = K[xv1z, . . . , xvqz, z],

where z is a new variable. Hence S[u]/P h � K[F ′′]. Therefore, settingm = q + 1, βi = vi for i = 1, . . . ,m− 1, βm = 0 and B = {β1, . . . , βm}, theresult follows readily from Proposition 9.3.19. �

Corollary 9.3.23 If v1, . . . , vq are positive integers and P is the toric idealof K[xv1 , . . . , xvq ], then gcd(v1, . . . , vq) deg(S/P ) = max{v1, . . . , vq}.

388 Chapter 9

Proof. We may assume that v1 ≤ · · · ≤ vq. The order of the groupT (Z/Z{v1, . . . , vq}) is equal to gcd(v1, . . . , vq). Then, by Theorem 9.3.22,we get that gcd(v1, . . . , vq) deg(S/P ) is vol([0, vq]) = vq. �

Example 9.3.24 Let P be the toric ideal of the monomial subring

K[H ] = K[x2x−11 , x3x

−12 , x4x

−13 , x1x

−14 , x5x

−12 , x3x

−15 , x4x

−15 ].

We employ the notation of Theorem 9.3.22. The rank of A is 4 and theheight of P is 3. Using Normaliz, we get

4!vol(conv(v1, . . . , v7, 0)) = 11.

As the group Z5/Z{v1, . . . , v7} is torsion- free, by Theorem 9.3.22, we havethat deg(K[t1, . . . , t7]/P ) = 11.

Notation If B is an integral matrix, Δi(B) will denote the greatest commondivisor of all the non-zero i× i minors of B.

Theorem 9.3.25 [146] If A = {v1, . . . , vq} ⊂ Zn and B is the matrix whosecolumns are the vectors in B = {(vi, 1)}qi=1, then the following hold:

(a) e(A(P)) = Δr(B)e(K[Ft]), where r = rank(B).

(b) Δr(B) = 1 if and only if K[Ft] = A(P).(c) K[Ft] = A(P) if and only if R+B ∩ ZB = NB and T (Zn/ZB) = (0).

Proof. (a): It follows at once by successively applying Proposition 9.3.19,Lemma 1.2.20 and Lemma 1.3.17.

(b): ⇒) By Theorem 9.3.6, K[Ft] ⊂ A(P) is an integral extension ofrings and A(P) is normal. Thus K[Ft] ⊂ A(P). For the other inclusion,by Theorem 9.3.6(e), it suffices to prove that A(P) is contained in the fieldof fractions of K[Ft]. Let xαti ∈ A(P)i, that is, α ∈ Zn ∩ iP and i ∈ N.Hence the system of equations By = α′ has a rational solution, where α′ isthe column vector (α, i)�. Thus, the augmented matrix [B α′] has rank r.In general Δr([B α

′]) divides Δr(B), so in this case they are equal becauseΔr(B) = 1. By Theorem 1.6.3, the linear system By = α′ has an integralsolution. Hence xαti is in the field of fractions of K[Ft], as required.

(b): ⇐) By hypothesis K[Ft] = A(P). Hence, taking multiplicities andusing Proposition 9.2.13 and part (a), we get Δr(B) = 1.

(c): By Theorem 9.3.6(g), K[Ft] = A(P) if and only if B = {(vi, 1)}qi=1

is a Hilbert basis. Hence this part follows readily from Corollary 1.3.21because R+B is a pointed cone (see Exercise 9.1.21). �

Corollary 9.3.26 If Δr(B) = 1, then K[Ft] is normal if and only ifa(K[Ft]) < 0 and the Hilbert polynomial of K[Ft] is equal to EP (x).


Proof. ⇒) It follows from Theorems 9.1.5 and 9.3.25.

⇐) Since the a-invariants of K[Ft] and A(P) are both negative, thendimK K[Ft]i is equal to dimK A(P)i for i ∈ N. Hence, noticing thatK[Ft]i ⊂ A(P)i, we getK[Ft]i = A(P)i for i ∈ N. ThereforeK[Ft] = A(P)and consequently K[Ft] is normal. �

Definition 9.3.27 If K[Ft] = A(P) and A is the incidence matrix of aclutter C, we say that C is an Ehrhart clutter. If K[Ft] = A(P), we say thatA is an Ehrhart point configuration.

There are many interesting families of Ehrhart point configurations thatwill be presented later; see Theorems 9.3.29, 9.3.31, and Section 14.8.

Corollary 9.3.28 If the subgroup L = ({vi − vj | i, j = 1, . . . , q}) of Zn has

rank n, then K[Ft] = A(P) if and only if L = Zn.

Proof. It follows from Theorem 9.3.25 and Lemma 1.2.20. �

Theorem 9.3.29 If P has a unimodular covering with support in A, thenK[Ft] = A(P).

Proof. Let Δ = conv(vi1 , . . . , vid+1) be any simplex in the unimodular

covering of P , where d = dim(P). Note that Zn+1/Z{(vij , 1)}d+1j=1 is torsion-

free by Proposition 1.2.21. Hence, by Lemma 1.3.17 and Theorem 9.3.25,we obtain the equality K[Ft] = A(P). To finish the proof note that K[Ft]is normal by Exercise 1.6.10. �

Proposition 9.3.30 [146] If A is a unimodular matrix and the vectorsv1, . . . , vq lie on a hyperplane not containing the origin, then K[Ft] = A(P).

Proof. Let B be the matrix with column vectors (v1, 1), . . . , (vq, 1). ByCorollary 9.2.5, we have K[F ] � K[Ft]. Hence r = rank(A) = rank(B).Using Corollary 9.1.3(d), one obtains that K[Ft] is normal. As Δr(A) = 1,applying Theorem 9.3.25 yields the asserted equality. �

The following result is a useful generalization of [383, Theorem 7.1].

Theorem 9.3.31 [146] If the Rees algebra R[Ft] is a normal domain andv1, . . . , vq lie on a hyperplane

b1x1 + · · ·+ bnxn = 1 (bi > 0 ∀i),

then K[Ft] = A(P).

390 Chapter 9

Proof. The inclusion K[Ft] ⊂ A(P) is clear. To show the reverse inclusiontake xαti ∈ A(P). By Theorem 9.3.6(b), one can write

(α, i) = λ1(v1, 1) + · · ·+ λq(vq, 1) (λj ≥ 0 ∀j ). (9.12)

Hence (α, i) ∈ Zn+1 ∩R+A′ = ZA′ ∩ R+A, where

A′ = {(e1, 0), . . . , (en, 0), (v1, 1), . . . , (vq , 1)}

and ei is the ith unit vector in Rn. By the normality of R[Ft] one has(α, i) ∈ NA′. Thus one can write

(α, i) = m1(e1, 0) + · · ·+mn(en, 0) + n1(v1, 1) + · · ·+ nq(vq, 1), (9.13)

where mi, nj are in N for all i, j. From Eqs. (9.12) and (9.13):

α = m1e1 + · · ·+mnen + n1v1 + · · ·+ nqvq

= λ1v1 + · · ·+ λqvq,

i = n1 + · · ·+ nq.

Taking the inner product of α with b = (b1, . . . , bn) yields

〈α, b〉 = m1b1 + · · ·+mnbn + i

and 〈α, b〉 = λ1〈v1, b〉 + · · · + λq〈vq, b〉 = i. Hence mi = 0 for all i. Sincexαti = (xv1t)n1 · · · (xvq t)nq , we get xαti ∈ K[Ft], as required. �

Proposition 9.3.32 If xv1 , . . . , xvq are monomials of degree k ≥ 2 in thepolynomial ring R = K[x1, . . . , xn] and R[Ft] is normal, then

e(R[Ft]) = n!vol(P ′),

where P ′ = conv(A′) and A′ = {(v1, 1), . . . , (vq, 1), e1, . . . , en}.

Proof. The affine hyperplane x1 + · · · + xn − (k − 1)xn+1 = 1 containsA′. Then R[Ft] is a standard graded K-algebra and dim(P ′) = n. Thus,using Theorem 9.3.25, we get an isomorphism R[Ft] � A(P ′), as gradedK-algebras. By Lemma 9.3.7 the multiplicity of A(P ′) is n!vol(P ′). �

Proposition 9.3.33 [385] If R[Ft] is normal and there is x0 ∈ Nn+ suchthat 〈x0, vi〉 = k, i = 1, . . . , q, for some k ≥ 2, then the following hold:

(a) The torsion subgroup of Zn/ZA is cyclic and its order divides k.

(b) If Zn/ZA is a finite group and x0 = (1, . . . , 1), then Zn/ZA � Zk.


Proof. We set B = {(v1, 1)}qi=1. There is an exact sequence of finite groups

0 −→ T (Zn+1/ZB) ϕ−→ T (Zn/ZA) ψ−→ Zk, (∗)

where ϕ and ψ are given by ϕ((α, b)) = α and ψ(α) = 〈α, x0〉, α ∈ Zn,b ∈ Z. Consider the matrix B whose columns are the vectors in the set B.By Theorems 9.3.31 and 9.3.25 we get Δr(B) = 1, where r is the rank ofZB. Since Δr(B) is the order of T (Zn+1/ZB), part (a) follows using Eq. (∗).To prove (b), note that Zn/ZA is a torsion group and the ith unit vectorei maps into the element 1 under the map ψ. Hence ψ is onto, and ψ givesthe required isomorphism. �

Exercises

9.3.34 Let P be a lattice polytope in Zn. Then K[P ] = A(P) if and only ifkP ∩ Zn = P ∩Zn + · · ·+P ∩Zn (k-times) for any k ∈ N+. If this holds Pis said to satisfy the integer decomposition property (cf. Definition 14.6.3).

9.3.35 Let F = {xv1 , . . . , xvq} ⊂ K[x] be a set of square-free monomials ofdegree k. If P = conv(v1, . . . , vq), then K[P ] � K[F ] as graded K-algebras.

9.3.36 Let P = conv((0, 0), (0, 1), (1, 0), (1, 1)) be the unit square. In thefigure below we show the integral points of 4P . Verify the following:

�

�

1

2

3

4

0 1 2 3 4

� ��

��

��

|Z2 ∩ 4P| = 25

EP(x) = (x+ 1)2

vol(P) = 1

A(P) = K[Ft]

F (A(P), x) = (1 + x)/(1 − x)3

9.3.37 Let A be an integral matrix of size n × n with det(A) �= 0 and letA = {v1, . . . , vn} be the set of columns of A. Consider the nth unit cubeQ = [0, 1]n and the parallelotope

P = {λ1v1 + · · ·+ λnvn| 0 ≤ λi ≤ 1}.

If A is a Hilbert basis, prove that the Ehrhart polynomials of Q and P areequal to (x + 1)n. Prove that the normalized volume of P is equal to n!.

392 Chapter 9

9.3.38 If P = conv((0, 0, 0), (12, 1, 0), (0, 1, 1), (1, 0, 1)), use Normaliz [68]to show that the Ehrhart polynomial of P is EP (x) = 13/6x3+x2−1/6x+1.

9.3.39 Let R = K[x1, . . . , xn] be a polynomial ring. If m = (x1, . . . , xn),prove that e(K[md]), the multiplicity of K[md], is equal to dn−1.

9.3.40 Let P be the convex hull of A = {(3, 1), (1, 3)} in R2, let K be thereal numbers field and let F = {x31x2, x1x32} ⊂ K[x1, x2]. Prove that A is aweakly unimodular covering of P and K[Ft] �= A(P).

9.3.41 If K[F ] is homogeneous and the lattice polytope P has a weaklyunimodular covering with support in A, then K[F ] is normal.

9.3.42 If F = {xv1 , . . . , xvq} is a set of monomials in R = K[x1, . . . , xn]and P ′ is the toric ideal of R[Ft], then

deg(K[x1, . . . , xn, t1, . . . , tq]/P′) = (n+ 1)!vol(P ′),

where P ′ = conv(A′ ∪ {0}) and A′ = {(v1, 1), . . . , (vq, 1), e1, . . . , en}.

9.3.43 If {(v1, 1), . . . , (vq , 1)} ⊂ Nn+1 is a Hilbert basis with |vi| = d for alli and rank(v1, . . . , vq) = n, prove that Zn/Z{v1, . . . , vq} � Zd.

9.4 The degree of lattice and toric ideals

In this section we give formulas to compute the degree of a lattice ideal interms of the torsion of certain factor groups of Zq and in terms of relativevolumes of polytopes. Then we study primary decompositions of latticeideals over an arbitrary field. In this section we will use the Eisenbud–Sturmfels theory of binomial ideals over algebraically closed fields [134].

Let L ⊂ Zq be a lattice and let ei be the ith unit vector in Zq+1. Fora = (ai) ∈ Zq define the value of a as |a| =

∑qi=1 ai, and the homogenization

of a with respect to eq+1 as ah = (a, 0)− |a|eq+1 if |a| ≥ 0 and ah = (−a)hif |a| < 0. The homogenization of L, denoted by Lh, is the lattice of Zq+1

generated by all ah such that a ∈ L.

Lemma 9.4.1 Let I(L) ⊂ S be a lattice ideal. Then the following hold.

(a) I(L)h ⊂ S[u] is a lattice ideal.

(b) If L = Z{b1, . . . , br}, then Lh = Z{bh1 , . . . , bhr}.

(c) I(Lh) = I(L)h.



Theorem 9.4.2 [294, 335] If L ⊂ Zq is a lattice of rank r and Iρ(L) is itslattice ideal with respect to a partial character ρ, then

(a) If r < q, there is an integer matrix A of size (q− r)× q and rank q− rsuch that we have the containment of rank r lattices L ⊂ kerZ(A),with equality if and only if Zq/L is torsion-free.

(b) If r < q and v1, . . . , vq are the columns of A, then

deg(S/Iρ(L)) =|T (Zq/L)|(q − r)!vol(conv(0, v1, . . . , vq))

|T (Zq−r/〈v1, . . . , vq〉)|.

(c) If r = q, then deg(S/Iρ(L)) = |Zq/L|.

Proof. (a): This follows at once from Lemma 8.2.5.(b): By Proposition 8.5.8, we may assume that K is algebraically closed

of characteristic zero and that ρ(a) = 1 for a ∈ L, i.e., Iρ(L) = I(L). LetP be the toric ideal of K[xv1 , . . . , xvq ]. By Theorem 9.3.22, we need onlyshow the equality

deg(S/I(L)) = |T (Zq/L)| deg(S/P ).

Let Ls be the saturation of L. By part (a), Ls is equal to kerZ(A). Weset c = |T (Zq/L)|. Notice that T (Zq/L) = Ls/L. According to [134,Corollaries 2.2 and 2.5], there are distinct partial characters ρ1, . . . , ρc of Ls,extending the trivial character ρ(a) = 1 for a in L, such that the minimalprimary decomposition of I(L) is given by

I(L) = Iρ1(Ls) ∩ · · · ∩ Iρc(Ls),

By part (a), P is a minimal prime of I(L). Thus we may assume ρ1(a) = 1for a in Ls, i.e., P is equal to the lattice ideal Iρ1(Ls). By additivity of thedegree (see Proposition 8.5.9), we get

deg(S/I(L)) = deg(S/Iρ1(Ls)) + · · ·+ deg(S/Iρc(Ls)).

Therefore, it suffices to recall that the degree is independent of the character,i.e., deg(S/P ) = deg(S/Iρk(Ls)) for k = 1, . . . , c (see Proposition 8.5.8).

(c): Let tc+1 − tc

−1 , . . . , tc

+m − tc

−m be a set of generators of I(L). By

Lemma 8.2.23, L is generated by c1, . . . , cm. We may assume that |ci| ≥ 0for all i. Then, by Lemma 9.4.1(b), Lh is generated by ch1 , . . . , c

hm. Let C

and Ch be the matrices with rows c1, . . . , cm and ch1 , . . . , chm, respectively.

Notice that Ch is obtained from C by adding the column vector given byb = (−|c1|, . . . ,−|cm|)�. Since b is a linear combination of the columnsof C, using the fundamental theorem of finitely generated abelian group(Theorem 1.3.16), we get that the groups Zq/L and Zq+1/Lh have the

394 Chapter 9

same torsion. Thus, |Zq/L| is equal to |T (Zq+1/Lh)|. By Proposition 8.5.6,deg(S/I(L)) = deg(S[u]/I(L)h), and by Lemma 9.4.1(c), I(L)h = I(Lh).Altogether it suffices to show that the degree of S[u]/I(Lh) is equal to|T (Zq+1/Lh)|. Since S[u]/I(Lh) has dimension 1, we get q = rank(Lh). Bypart (a), there is a row vector A′ = (v′1, . . . , v

′q+1) such that Lh ⊂ kerZ(A

′).

Since the Q-vector spaces generated by Lh and kerZ(A′) are equal and have

dimension q and since 〈α,1〉 = 0 for all α in Lh, it follows readily, usingorthogonal complements, that v′1 = v′i for all i. Thus, by part (b), therequired equality follows. �

Corollary 9.4.3 If I(L) is a lattice ideal of dimension 1 which is homoge-neous with respect to a positive vector d = (d1, . . . , dq), then

gcd(d1, . . . , dq) deg(S/I(L)) = max{d1, . . . , dq}|T (Zq/L)|.

Proof. Let A be the 1× q matrix (d1, . . . , dq). By hypothesis L ⊂ kerZ(A).The order of T (Z/Z{d1, . . . , dq}) is equal to gcd(d1, . . . , dq). Therefore, byTheorem 9.4.2, we get

deg(S/I(L)) = T (Zq/L)vol(conv(0, d1, . . . , dq))T (Z/Z{d1, . . . , dq}) =

|T (Zq/L)|max{d1, . . . , dq}gcd(d1, . . . , dq)

,

as required. �

Corollary 9.4.4 Let L ⊂ Zq be a lattice of rank q− 1 and let α1, . . . , αq−1

be a Z-basis of L. If αi = (αi,1, . . . , αi,q), for i = 1, . . . , q − 1, and

ni = (−1)i det

⎛⎜⎝

α1,1 . . . α1,i−1 α1,i+1 . . . α1,q

......

......

αq−1,1 . . . αq−1,i−1 αq−1,i+1 . . . αq−1,q

⎞⎟⎠ for 1 ≤ i ≤ q,

then deg(S/I(L)) = max{n1, . . . , nq, 0} −min{n1, . . . , nq, 0}.

Proof. Let B be the (q − 1) × q matrix with rows α1, . . . , αq−1 and letA be the 1 × q matrix (n1, . . . , nq). We set r = q − 1. The order ofT (Zq/L) is equal to gcd(n1, . . . , nq), the gcd of the r × r minors of B.The order of T (Z/Z{n1, . . . , nq}) is also equal to gcd(n1, . . . , nq). Sinceαi ∈ ker(A) for all i, we obtain that L ⊂ ker(A). Hence, by Theorem 9.4.2,we get that deg(S/I(L)) is equal to vol(conv(0, n1, . . . , nq)) which is equalto max{n1, . . . , nq, 0} −min{n1, . . . , nq, 0}. �

Normaliz [68] computes normalized volumes of lattice polytopes usingpolyhedral geometry. Thus we can compute the degree of any lattice idealusing Theorem 9.4.2.


Example 9.4.5 Let K be the field of rational numbers and let L be thelattice generated by

a1 = (2, 1, 1, 1,−1,−1,−1,−2), a2 = (1, 1,−1,−1, 1, 1,−1,−1),a3 = (2,−1, 1,−2, 1,−1, 1,−1), a4 = (5,−5, 0, 0, 0, 0, 0, 0).

We use the notation of Theorem 9.4.2. The lattice L has rank 4 anda1, . . . , a4, e1, e3, e4, e5 form a Q-basis of Q8. In this case we obtain thematrix

A =

⎛⎜⎜⎝4 4 0 0 0 −1 1 60 0 1 0 0 1 0 00 0 0 4 0 7 9 −60 0 0 0 2 −3 −3 2

⎞⎟⎟⎠whose columns are denoted by v1, . . . , v8. Let P be the toric ideal ofK[xv1 , . . . , xv8 ]. Therefore, by Theorem 9.4.2, we get

deg(S/I(L)) = 5(4)!vol(conv(0, v1, . . . , v8))

|T (Z4/〈v1, . . . , v8〉)|=

(5)(200)

8= 125.

The normalized volume of the polytope conv(0, v1, . . . , v8) was computedusing Normaliz [68].

Theorem 9.4.6 [335] Let I(L) be a lattice ideal of S over an arbitrary fieldK of characteristic p, let c be the number of associated primes of I(L), andfor p > 0, let G be the unique largest subgroup of T (Zq/L) whose order isrelatively prime to p. Then

(a) All associated primes of I(L) have height equal to rank(L).

(b) |T (Zq/L)| ≥ c if p = 0 and |G| ≥ c if p > 0, with equality if K isalgebraically closed.

(c) deg(S/I(L)) ≥ |T (Zq/L)| if p = 0 and deg(S/I(L)) ≥ |G| if p > 0.

Proof. Let K be the algebraic closure of K and let S = K[t1, . . . , tq] bethe corresponding polynomial ring with coefficients in K. Thus, we have anintegral extension S ⊂ S of normal domains. We set I = I(L) and I = IS,where the latter is the extension of I to S. The ideal I is the lattice ideal ofL in S (see Exercise 8.2.35(c)). Hence, as K is algebraically closed, by [134,Corollaries 2.2 and 2.5] I has a unique irredundant primary decomposition

I = q1 ∩ · · · ∩ qc1 , (†)

where c1 = |T (Zq/L)| if p = 0 and c1 = |G| if p > 0. Notice that c1 =|T (Zq/L)| if p is relatively prime to |T (Zq/L)|. Furthermore, also by [134,Corollaries 2.2 and 2.5], one has that if pi = rad(qi) for i = 1, . . . , c1,

396 Chapter 9

then p1, . . . , pc1 are the associated primes of I and ht(pi) = rank(L) fori = 1, . . . , c1. Hence, by Exercise 8.2.35(b), one has a primary decomposition

I = (q1 ∩ S) ∩ · · · ∩ (qc1 ∩ S) (‡)

such that rad(qi ∩ S) = rad(qi) ∩ S = pi ∩ S. We set pi = pi ∩ S andqi = qi ∩ S for i = 1, . . . , c1.

(a): Since S is a normal domain and S ⊂ S is an integral extension, weget ht(pi) = ht(pi) = rank(L) for all i (see Proposition 2.4.14).

(b): By Eq. (‡), the associated primes of I are contained in {p1, . . . , pc1}.Thus, c1 ≥ c, which proves the first part. Now, assume thatK = K. By (a),we may assume that p1, . . . , pc are the minimal primes of I. ConsequentlyI has a unique minimal primary decomposition I = Q1 ∩ · · · ∩Qc such thatQi is pi-primary and ht(pi) = rank(L) for i = 1, . . . , c. As I = I, fromEq. (†), we get that c1 = c.

(c): Using Lemma 8.5.7, Eq. (†) that was stated at the beginning of theproof, and the additivity of the degree (Proposition 8.5.9), we get

deg(S/I) = deg(S/I) =∑c1

i=1 deg(S/qi) ≥ c1. �

Exercises

9.4.7 Let L = 〈(2,−1,−1), (−3, 1,−1)〉. Prove that I(L) is a non-gradedlattice ideal of height 2 given by

I(L) = ((t21 − t2t3, t2 − t31t3) : (t1t2t3)∞) = (t21 − t2t3, t1t23 − 1).

Then, apply Corollary 9.4.4 with v1 = −2, v2 = −5 and v3 = 1, to showthat K[t1, t2, t3]/I(L) has degree 6.

9.5 Laplacian matrices and ideals

In this section we use algebraic graph theory to study pure binomial matricesand their matrix ideals. We are interested in relating the combinatorics ofa graph with the algebraic invariants and properties of the Laplacian idealassociated to the Laplacian matrix of the graph.

If d = (d1, . . . , dq) ∈ Nq+, then S = ⊕∞k=0Sk has a grading induced by

setting deg(ti) = di for i = 1, . . . , q. A graded ideal of S = K[t1, . . . , tq] isan ideal which is graded with respect to some vector d in Nq+.

Definition 9.5.1 A matrix ideal is an ideal of the form

I(L) = (ta+i − ta

−i | i = 1, . . . ,m) ⊂ S,

where the ai’s are the columns of an integer matrix L of size q ×m.


Note that I(L) is graded if and only if dL = 0 for some d ∈ Nq+.

Example 9.5.2 If L is the matrix

L =

⎛⎝ 2 −1 −10 1 −10 −1 1

⎞⎠ ,

then I(L) = (t21 − 1, t2 − t1t3, t3 − t1t2).

Proposition 9.5.3 [335] Let I = I(L) be a graded matrix ideal and let Lbe the lattice spanned by the columns of L. Suppose that V (I, ti) = {0} forall i. Then we have the following.

(a) I = I(L) if I is unmixed, or else I = I(L) ∩ q, if I is not unmixed,where q is an m-primary component of I and m = (t1, . . . , tq).

(b) (gcd{di}qi=1) deg(S/I) = maxi{di}|T (Zq/L)|.

Proof. By Lemma 8.2.11, one has (I : (t1 · · · tq)∞) = I(L). By consideringa primary decomposition of I, part (a) follows from Lemma 8.3.20 andProposition 8.3.22. Part (b) follows from Corollary 9.4.3 and part (a). �

Definition 9.5.4 Let ai,j ∈ N, i, j = 1, . . . , q, and let L be a q × q matrixof the following special form:

L =

⎛⎜⎜⎜⎝a1,1 −a1,2 · · · −a1,q−a2,1 a2,2 · · · −a2,q

...... · · ·

...−aq,1 −aq,2 · · · aq,q

⎞⎟⎟⎟⎠ .

L is called a pure binomial matrix (PB matrix) if aj,j > 0 for all j, and foreach row and each column of L at least one off-diagonal entry is non-zero.

Definition 9.5.5 Let B = (bi,j) be a q × q real matrix. The underlyingdigraph GB of B has vertex set {t1, . . . , tq}, with an arc from ti to tj iffbi,j �= 0. If bi,i �= 0, we put a loop at ti. A digraph is strongly connected iffor any two vertices ti and tj there is a directed path form ti to tj and adirect path from tj to ti.

Theorem 9.5.6 (Duality Theorem [335]) Let L be a PB matrix of sizeq× q such that Lc� = 0 for some c ∈ Nq+. If GL is strongly connected, thenrank(L) = q − 1 and there is d in Nq+ such that dL = 0.

Proof. Let L be a PB matrix as in Definition 9.5.4. First we treat the casec = 1 = (1, . . . , 1). Let L be a PB matrix as in Definition 9.5.4. We canwrite L = D−A, where D = diag(a1,1, . . . , aq,q) and A is the matrix whose

398 Chapter 9

i, j entry is ai,j if i �= j and whose diagonal entries are equal to zero. We setδi = ai,i for i = 1, . . . , q. By hypothesis L1� = 0, hence rank(L) ≤ q − 1.There exists a non-zero vector d ∈ Zq such that dL = 0. Therefore

dD = dA = dD(D−1A).

Since L1� = 0, we get that D1� = A1� or equivalently (D−1A)1� = 1�.Thus, as the entries of D−1A are nonnegative, the matrix B := D−1A isstochastic. It is well-known that the spectral radius ρ(B) of a stochasticmatrix B is equal to 1 [28, Theorem 5.3], where ρ(B) is the maximumof the moduli of the eigenvalues of B. As the diagonal entries of B arezero and δi > 0 for all i, the underlying digraph GB of B is equal to thedigraph obtained from GL by removing all loops of G. Since GL is stronglyconnected so is GB , and by the Perron–Frobenius Theorem for nonnegativematrices [190, Theorem 8.8.1, p. 178], ρ(B) = 1 and 1 is a simple eigenvalueof B (i.e., the eigenspace of B relative to ρ(B) = 1 is 1-dimensional), and ifz is an eigenvector for ρ(B) = 1, then no entries of z are zero and all havethe same sign. Applying this to z = dD, we get that di �= 0 for all i and allentries of d have the same sign. Hence, ker(L�) = (d�) for any non-zerovector d such that dL = 0 and L has rank q − 1.

The general case follows by considering the matrix L = Ldiag(c1, . . . , cq),

where c = (c1, . . . , cq). Notice that L1� = 0 because Lc� = 0. �

Lemma 9.5.7 Let L be an integer matrix of size q× q with rows �1, . . . , �qand let adj(L) = (Li,j) be the adjoint matrix of L. Suppose Lc� = 0 forsome c in Nq+. The following hold.

(a) If rank(�1, . . . , �i, . . . , �q) = q − 1 and Li,i ≥ 0 for all i, then dL = 0for some d ∈ Nq+.

(b) If Li,i > 0 for all i, then dL = 0 for some d ∈ Nq+.

(c) If L is a PB matrix and rank(�1, . . . , �i, . . . , �q) = q − 1 for all i, thendL = 0 for some d ∈ Nq+.

Proof. (a): Let Li be the ith column of adj(L). Since �1, . . . , �i, . . . , �qare linearly independent, we get Li �= 0. The vector c generates kerQ(L)because L has rank q − 1 and Lc� = 0. Then, because of the equalityLadj(L) = 0, we can write Li = μic for some μi ∈ Q. Notice that μi > 0because Li,i ≥ 0 and c ∈ Nq+. Hence, all entries of adj(L) are positiveintegers. If d is any row of adj(L), we get dL = 0 because adj(L)L = 0.

(b): For any i, the vectors are �1, . . . , �i, . . . , �q are linearly independentbecause Li,i > 0. Thus this part follows from (a).

(c): Let L be as in Definition 9.5.4. (c1): First we treat the case c = 1.By part (a) it suffices to show that Li,i ≥ 0 for all i. Let Hi,i be the


submatrix of L obtained by eliminating the ith row and ith column. By theGershgorin Circle Theorem, every (possibly complex) eigenvalue λ of Hi,i

lies within at least one of the discs {z ∈ C| ‖z − aj,j‖ ≤ rj}, j �= i, whererj =

∑u�=i,j | − aj,u| ≤ aj,j since L1� = 0 and ai,j ≥ 0 for all i, j. If λ ∈ R,

we get |λ − aj,j | ≤ aj,j , and consequently λ ≥ 0. If λ /∈ R, then since Hi,i

is a real matrix, its conjugate λ must also be an eigenvalue of Hi,i. Sincedet(Hi,i) is the product of the q− 1 (possibly repeated) eigenvalues of Hi,i,we get Li,i ≥ 0. (c2): Now, we treat the general case. Let B be the q × qdiagonal matrix diag(c1, . . . , cq), where c = (c1, . . . , cq), and let L = LB.

Notice that L1� = 0 because Lc� = 0, and L is a PB matrix because Lis a PB matrix. Let (Li,j) be the adjoint matrix of L. Since ci > 0 for alli, by the multilinearity of the determinant, it follows that Li,j �= 0 if and

only if Li,j �= 0. Hence, any set of q − 1 rows of L is linearly independent.

Therefore, applying case (c1) to L, we obtain that there is d ∈ Nq+ such

that dL = 0. Then, dL = 0. �

Theorem 9.5.8 [335] Let L be a PB matrix of size q×q such that Lc� = 0for some c ∈ Nq+. The following conditions are equivalent:

(a) GL is strongly connected.

(b) V (I(L), ti) = {0} ∀ i, where V (I(L), ti) is the zero set of (I(L), ti).

(c) Li,j > 0 ∀ i, j; adj(L) = (Li,j) is the adjoint of L.

Proof. (a) ⇒ (b): Let L be as in Definition 9.5.4. For 1 ≤ k ≤ q, let fk bethe binomial defined by the kth column of L. Set I = I(L) and fix i suchthat 1 ≤ i ≤ q. Clearly {0} is contained in V (I, ti) because I is graded byTheorem 9.5.6. To show the reverse containment, let α = (α1, . . . , αq) be apoint in V (I, ti). Consider the set E(ti) of all arrows (ti, tk) leaving ti. If(ti, tk) ∈ E(ti), i.e., ai,k �= 0, we claim that αk = 0. We can write

fk = tak,k

k −∏j �=k

taj,kj .

Using that fk(α) = 0, we get αak,k

k =∏j �=k α

aj,kj . Since ai,k �= 0 and

using that αi = 0, we obtain that αk = 0, as claimed. Let � be an integerin [q] := {1, . . . , q}. Since the digraph GL is strongly connected, there isa directed path {v1, . . . , vr} joining ti and t�, i.e., v1 = ti, vr = t� and(vj , vj+1) ∈ E(GL) for all j. There is a permutation π of [q] such thatπ(1) = i, π(r) = � and vj = tπ(j) for j = 1, . . . , r. Applying the claimsuccessively for j = 2, . . . , r, we obtain απ(2) = 0, απ(3) = 0, . . . , απ(r) = 0.Thus, α� = 0. This proves that α = 0.

(b)⇒ (a): We proceed by contradiction. Assume that GL is not stronglyconnected. Without loss of generality we may assume that there is no

400 Chapter 9

directed path from t1 to tq. Let W be the set of all vertices ti such thatthere is a directed path from t1 to ti, the vertex t1 being included in W .The set W is non-empty because GL has no sources or sinks by definitionof a PB matrix, and the vertex tq is not in W . Consider the vector α ∈ Kq

defined as αi = 1 if ti /∈ W and αi = 0 if ti ∈W . To derive a contradictionit suffices to show that all binomials of I(L) vanish at α. Let L be as inDefinition 9.5.4 and let fk = t

ak,k

k −∏j �=k t

aj,kj be the binomial defined by

the k-columns of L. If tk ∈ W , there is a directed path P from t1 to tk.Then tj is part of the path P for some j such that aj,k > 0 because the lastarrow of the path P has the form (tj , tk). Thus, since tj ∈W , fk(α) = 0. Iftk /∈W , then tj is not inW for any j such that aj,k > 0, because if aj,k > 0,the pair (tj , tk) is an arc of GL. Thus, fk(α) = 0.

(a) ⇒ (c): By the proof of Lemma 9.5.7(c), one has that Li,i ≥ 0 for alli. By Theorem 9.5.6 and using the proof of Lemma 9.5.7(a), we get thatLi,j > 0 for all i, j.

(c)⇒ (a): We proceed by contradiction. Assume that GL is not stronglyconnected. We may assume that there is no directed path from t1 to tq.Let W be as above and let W c = {ti| ti /∈ W} be its complement. Wecan write W c = {t�1 , . . . , t�r}. Consider the r × r submatrix B obtainedfrom L by fixing rows �1, . . . , �r and columns �1, . . . , �r. Notice that by thearguments aboveW c = ∪tk∈W c(supp(fk)). Hence any column of B extendsto a column of L by adding 0’s only, and consequently since dL = 0 forsome d ∈ Ns, the rows of B are linearly dependent. Hence det(B) = 0. Bypermuting rows and columns, L can be brought to the form

L′ =

(C 0C′ C′′

)where C and C′′ are square matrices of orders r and q − r, respectively,and det(C) = 0. Hence the adjoint of L′ has a zero entry, and so does theadjoint of L, a contradiction. �

Laplacian matrices Let G be a connected graph, where V = {t1, . . . , tq}is the set of vertices, E is the set of edges, and let w be a weight function

w : E → N+, e �→ we.

Let E(ti) be the set of edges incident to ti. The Laplacian matrix L(G)of G is the q × q matrix whose (i, j)-entry L(G)i,j is given by

L(G)i,j :=

⎧⎨⎩∑e∈E(ti)

we if i = j,

−we if i �= j and e = {ti, tj} ∈ E,0 otherwise.

Notice that L(G) is symmetric and 1L(G) = 0.


The notion of a Laplacian matrix can be extended to weighted digraphs;see [95] and the references there. Let G = (V,E,w) be a weighted digraphwithout loops and with vertices t1, . . . , tq, let w(ti, tj) be the weight ofthe directed arc from ti to tj and let A(G) be the adjacency matrix ofG given by A(G)i,j = w(ti, tj). The Laplacian matrix of G is given byL(G) = D+(G)−A(G), where D+(G) is the diagonal matrix with the out-degrees of the vertices of G in the diagonal entries. Note that L(G)1� = 0and that the Laplacian matrix of a digraph may not be symmetric. If ti isa sink, i.e., there is no arc of the form (ti, tj), then the ith row of L(G) iszero. Thus, the rank of L(G) may be much less than q − 1.

Definition 9.5.9 The matrix ideal IG of L(G) is called the Laplacian idealof G. If L is the lattice generated by the columns of L(G), the group

K(G) := T (Zq/L)

is called the critical group or the sandpile group of G.

The structure, as a finite abelian group, of K(G) is only known for afew families of graphs [5, 295].

Theorem 9.5.10 (the Matrix-Tree Theorem [396, Theorem 9.8]) If G isregarded as a multigraph (where each edge e occurs we times), then thenumber of spanning trees of G is the (i, j)-entry of the adjoint matrix ofL(G) for any (i, j).

Remark 9.5.11 The order of T (Zq/L) is the gcd of all (q − 1)-minors ofL(G). Thus the order of K(G) is the number of spanning trees of G.

Proposition 9.5.12 [335] Let G be a connected graph, IG ⊂ S its Lapla-cian ideal, and L the lattice spanned by the columns of L(G). Then:

(a) V (IG, ti) = {0} for all i and rank(L(G)) = q − 1.

(b) deg(S/IG) = deg(S/I(L)) = |K(G)|.

(c) If G = Kq is a complete graph, then deg(S/IG) = qq−2.

(d) Hull(IG) = I(L), where Hull(IG) is the intersection of the isolatedprimary components of IG.

Proof. The underlying digraph GL(G) of the Laplacian matrix L(G) isstrongly connected because L(G) is symmetric and G is connected. Hence,by Theorems 9.5.6 and 9.5.8, (a) holds. Parts (b) and (d) follow from (a)and Proposition 9.5.3. Part (c) follows from (b) and the fact that a completegraph with q vertices has qq−2 spanning trees [396, p. 143]. �

402 Chapter 9

Exercises

9.5.13 Let L be the following PB matrix and adj(L) its adjoint:

L =

⎛⎜⎜⎜⎜⎝4 −1 −1 −1 −1−1 4 −1 −1 −10 −1 2 −1 0−1 −1 −1 4 −1−1 0 0 0 1

⎞⎟⎟⎟⎟⎠ .

Prove that V (I(L), ti) = {0} for all i and that GL is strongly connected.

9.5.14 Let G be a connected graph and let IG ⊂ S be its Laplacian ideal.Prove that the following hold.

(a) If degG(ti) ≥ 3 for all i, then IG is not a lattice ideal.

(b) If degG(ti) ≥ 2 for all i, then IG is not a complete intersection.

9.5.15 Let G be the following weighted digraph and L its Laplacian matrix.Prove that the underlying digraph GL is not strongly connected, the matrixideal I(L�) is graded but I(L) is not.

t1

t2 t3

t43

1

1

114 L = L(G) =

⎛⎜⎜⎝5 −4 0 −10 1 −1 00 −1 1 0−3 0 −1 4

⎞⎟⎟⎠

9.5.16 Let G be the complete graph on 3 vertices. Prove that

L(G) =

⎛⎝ 2 −1 −1−1 2 −1−1 −1 2

⎞⎠ ,

IG = (t21 − t2t3, t22 − t1t3, t23 − t1t2), V (IG, ti) = {0} for all i, IG is not acomplete intersection, is a lattice ideal, is not a prime ideal, has degree 3and dimension 1.

9.5.17 Use the package “Binomials” [266] and the following procedure forMacaulay2 [199] to verify that over C the ideal (t21 − t2t3, t22 − t3t1, t23 − t24)is a complete intersection vanishing ideal and is not a lattice ideal.

load "Binomial.m2"

R=QQ[t1,t2,t3,t4]

I=ideal(t1^2-t2*t3,t2^2-t3*t1,t3^2-t4^2)

binomialAssociatedPrimes I

saturate(I,t1*t2*t3*t4)==I


9.6 Grobner bases and normal subrings

Let R = K[x±11 , . . . , x±1

n ] be a Laurent polynomial ring over a field K andconsider a finite set F = {xvi}qi=1 of distinct monomials with vi �= 0 for alli. There is a homomorphism of K-algebras:

S = K[t1, . . . , tq]ϕ−→ K[F ] (ti

ϕ−→ xvi),

where S is a polynomial ring. The kernel of ϕ, denoted by P := PF , is thetoric ideal of K[F ]. In what follows A will denote the n × q matrix withcolumn vectors v1, . . . , vq and A will denote the set {v1, . . . , vq}.

The main reference for this section is the book of Sturmfels on Grobnerbases and convex polytopes [400].

For the rest of this section we assume that ≺ is a fixed term order forthe set of monomials of K[t1, . . . , tq]. We denote the initial ideal of P byin≺(P ) or simply by in(P ). The Stanley–Reisner complex of the square-freemonomial ideal rad (in≺(P )) will be denoted by Δ.

The following result was pointed out by Sturmfels. It can be shown usingessentially “line shellings” of polytopes (as explained in [438, pp. 240–242]).

Theorem 9.6.1 [106, Theorem 9.5.10] Δ is pure shellable.

Corollary 9.6.2 (Sturmfels) rad(in≺(P )) is a Cohen–Macaulay ideal.

Proof. The simplicial complex Δ is pure shellable by Theorem 9.6.1. HenceΔ is Cohen–Macaulay by Theorem 6.3.23. �

Lemma 9.6.3 If σ = {t1, . . . , tr} is a maximal face of Δ, then v1, . . . , vrare linearly independent.

Proof. If v1, . . . , vr are linearly dependent, then after permutation of thevariables we can write λ1v1 + · · · + λjvj = λj+1vj+1 + · · · + λrvr, λi ∈ N

and λr �= 0. Thus if f = tλ11 · · · t

λj

j − tλj+1

j+1 · · · tλrr , then in(f) ∈ in(P ), a

contradiction because rad(in(P )) ⊂ (tr+1, . . . , tq). �

Definition 9.6.4 [400, p. 4] Let I be an ideal of S and let ω = (ω1, . . . , ωq)be a vector in Rq. If f = λ1t

a1 + · · ·+ λstas , define inω(f), the initial form

of f relative to ω, as the sum of all terms λitai such that 〈ω, ai〉 is maximal.

The ideal generated by {inω(f)| f ∈ I} is denoted by inω(I).

If ω ∈ Nq, then inω(f) is the leading coefficient in the x-variable of theunivariate polynomial h(x) = f(xtω1

1 , . . . , xtωqq ) ∈ S[x].

If ω ≥ 0, the following rule defines a term order ≺ω for S: ta ≺ω tb ifand only if 〈ω, a〉 < 〈ω, b〉, or 〈ω, a〉 = 〈ω, b〉 and ta ≺ tb.

404 Chapter 9

Proposition 9.6.5 [400, Proposition 1.8] Let I be an ideal of S. If ω ≥ 0,then in(inω(I)) = in≺ω (I). In addition if inω(I) is a monomial ideal, theninω(I) = in≺ω(I).

Proposition 9.6.6 Let I be an ideal of S and let G be the reduced Grobnerbasis of I. If ω ∈ Rq+, then in(I) = inω(I) if and only if in(g) = inω(g) forall g ∈ G.

Proof. ⇒) Let g ∈ G. There are tas , . . . , tas distinct monomials in Ssuch that g = λ1t

a1 + · · · + λstas and inω(g) = λ1t

a1 + · · · + λrtar , where

r ≤ s and 0 �= λi ∈ k for all i. Since G is reduced there is j satisfyingin(g) = taj ∈ in(I) and tai /∈ in(I) for i �= j. Using that in(I) is a monomialideal we get tai ∈ in(I) for i = 1, . . . , r. Thus r = 1 and j = 1, that is,in(g) = inω(g).⇐) First let us prove in(I) = in≺ω(I). Clearly in(I) ⊂ inω(I). Hence

taking initial ideals with respect to ≺ in both sides and using the previousproposition we get in(I) ⊂ in≺ω(I). To show the reverse inclusion taketa ∈ in≺ω(I). Assume ta /∈ I, otherwise ta ∈ in(I). Since the standardmonomials with respect to ≺ω are aK-basis for S/I (see Proposition 3.3.13),we have ta+I = (λ1t

a1+I)+ · · ·+(λstas+I), λi ∈ K∗, where tai /∈ in≺ω (I).

Therefore ta ∈ in(I), as required.Let inω(f) with f ∈ I. There are tas , . . . , tas distinct monomials in S

and non-zero scalars λ1, . . . , λs such that f = λ1ta1+· · ·+λrtar+· · ·+λstas ,

where inω(f) = λ1ta1 + · · ·+λrt

ar and ta1 " · · · " tar . Note in≺ω (f) = ta1 ,hence ta1 ∈ in(I). There is g ∈ G such that

λ1ta1 = λ1t

δin(g) = λ1tδinω(g) = λ1inω(t

δg).

Set h = f −λ1tδg. As inω(h) = λ2ta2 + · · ·+ λrt

ar , we can repeat the sameargument to get λit

ai ∈ in(I) for 2 ≤ i ≤ r, that is, inω(f) ∈ in(I). �

The next result shows that in(P ) can be represented by a weight vector.

Proposition 9.6.7 [400, Proposition 1.11] in(P ) = inω(P ) for some non-negative integer weight vector ω ∈ Nq.

A fundamental result linking Grobner basis theory with “regular trian-gulations” of point configurations is the following result of Sturmfels.

Theorem 9.6.8 [399, 400] If in(P ) = inω(P ), then

Δ = {σ| ∃ c ∈ Rn such that 〈vi, c〉 = ωi if ti ∈ σ & 〈vi, c〉 < ωi if ti /∈ σ}.

In what follows ω denotes a nonnegative integer vector that representsthe initial ideal of the toric ideal P , that is, in(P ) = inω(P ).


Lemma 9.6.9 If σ = {t1, . . . , tr} is a face of Δ, then

in(P ) = inω′(P ) = in≺ω′ (P )

for some ω′ = (ω′i) ∈ Rq such that ω′

i = 0 if ti ∈ σ and ω′i > 0 if ti /∈ σ.

Proof. According to Theorem 9.6.8 there is c ∈ Rq such that 〈vi, c〉 = ωiif ti ∈ σ and 〈vi, c〉 < ωi if ti /∈ σ. Let g = ta − tb be an element in thereduced Grobner basis of P with ta " tb. Note inω(g) = ta, otherwise ifinω(g) = tb or inω(g) = g, the using in(P ) = inω(P ) we get that tb is not astandard monomial, a contradiction. Thus 〈ω, a〉 > 〈ω, b〉.

Consider the vector ω′ = (ωi) − (〈vi, c〉). By Proposition 9.6.5 andProposition 9.6.6 we need only show the inequality 〈ω′, a〉 > 〈ω′, b〉. Ifa = (ai) and b = (bi), from a1v1 + · · · + aqvq = b1v1 + · · · + bqvq, we geta1〈c, v1〉+ · · ·+ aq〈c, vq〉 = b1〈c, v1〉+ · · ·+ bq〈c, vq〉. Hence

〈ω′, a〉 − 〈ω′, b〉 = 〈ω, a〉 − 〈ω, b〉 > 0. �

Lemma 9.6.10 If in(P ) is square-free and 0 �= v ∈ NA, then there is σ′ aface of Δ such that v ∈ NA′, where A′ = {vi| ti ∈ σ′}.

Proof. Let {tβ1 − tγ1 , . . . , tβp − tγp} be the reduced Grobner basis of thetoric ideal P , where tβi " tγi for all i. There is λ = (λi) in Nq such that v isequal to λ1v1 + · · ·+ λqvq. Consider the monomial h0 = tλ and its supportσ0 = supp(h). If σ0 ∈ Δ, set σ′ = σ0. Assume σ0 /∈ Δ, then h0 ∈ in(P ) andwe may assume h0 = tδ1tβ1 . Set h1 = tδ1tγ1 and σ1 = supp(h1). Note

fλ := fλ11 · · · fλq

q = f δ1fβ1 = f δ1fγ1 .

Thus one may assume σ1 /∈ Δ, otherwise we set σ′ = σ1. Since h0 " h1 andusing that ≺ is Noetherian, the existence of σ′ follows by induction. �

Lemma 9.6.11 Let A′ = {v1, . . . , vr} and let π be the projection map

π : Zq −→ Zq−r (π(ai) = (ar+1, . . . , aq)).

If L = π(ker(A) ∩ Zq), then Zq−r/L � ZA/ZA′.

Proof. It suffices to note that the linear map Zq−rψ−→ ZA/ZA′ given by

ψ(ar+1, . . . , aq) = ar+1vr+1 + · · · + aqvq + ZA′ is an epimorphism whosekernel is equal to L. �

For use below consider the epimorphism of K-algebras:

S = K[t1, . . . , tq]φ−→ S′ = K[tr+1, . . . , tq]

induced by φ(ti) = 1 for i ≤ r and φ(ti) = ti otherwise. Note that φ isrelated to the map π by φ(ti) = tπ(ei) for all i.

406 Chapter 9

Lemma 9.6.12 Let I be a monomial ideal of S and let p = (tr+1, . . . , tq)be a minimal prime of I, then

(a) Ass(S′/I ′) = {p′}, where I ′ = φ(I) and p′ = p ∩ S′,

(b) dim(S′/I ′) = 0, and

(c) I ′ = p′, if I is generated by square-free monomials.

Proof. (a): Let q′ be an associated prime of S′/I ′ and set q = q′S. Notethat I ⊂ I ′S and q′ is generated by variables in V = {tr+1, . . . , tq}. SinceI ⊂ q, there is a minimal prime qi of I contained in q, thus qi is generatedby variables in V . Hence by the minimality of p we get q = p and q′ = p′.

(b): From (a) the radical of I ′ is equal to the irrelevant maximal idealof S′, thus S′/I ′ is Artinian. Part (c) follows by noticing that I ′ is also asquare-free monomial ideal and applying (a). �

Definition 9.6.13 Let I be a monomial ideal of S and let p be a minimalprime of I. The multiplicity of p in I is: mI(p) = �Sp(Sp/Ip).

Proposition 9.6.14 Let I be a monomial ideal of S and p = (tr+1, . . . , tq)a minimal prime of I. If I ′ = φ(I), then mI(p) = dimK(S′/I ′).

Proof. Since Sp/Ip � (S′)p/(I′)p, the result follows from Lemma 8.5.1 �

Proposition 9.6.15 If L is a lattice in Zq−r of rank q − r, then

dimK(S′/I(L)) = |Zq−r/L|.

Proof. It follows at once from Theorem 9.4.2 (c) because dimK(S′/I(L))is equal to degS′/I(L). �

For use below recall that the map σ �→ (σc) induces a bijection betweenthe maximal faces of Δ and the minimal primes of rad(in(P )), where σc isequal to {ti| ti /∈ σ} (see Proposition 6.3.4).

Theorem 9.6.16 [400] If the initial ideal in(P ) is generated by square-freemonomials, then K[F ] is normal.

Proof. Let α ∈ R+A ∩ ZA. By Corollary 1.1.27, pα ∈ NA for some0 �= p ∈ N. By Lemma 9.6.10 there is σ′, a maximal face of Δ, such that pαis in NA′, where A′ = {vi| ti ∈ σ′}. We may assume that σ′ = {t1, . . . , tr}.Consider the prime ideal p of S generated by {tr+1, . . . , tq}. As p is aminimal prime of the initial ideal in(P ), using Lemma 9.6.12(c), we get

φ(in(P )) = p′ = (tr+1, . . . , tq) ⊂ S′.


By Lemma 9.6.9 we may assume that in(P ) = inω(P ) = in≺ω (P ) for someω = (ωi) ∈ Rq such that ωi = 0 if ti ∈ σ′ and ωi > 0 if ti /∈ σ′. Therefore

p′ = φ(in(P )) = inω(φ(P )) = in≺ω (φ(P )). (9.14)

Indeed the first equality follows from Proposition 9.6.6 and noticing thatinω(φ(P )) ⊂ p′. The second equality follows at once from Proposition 9.6.5because inω(φ(P )) is a monomial ideal by the first equality. If ta − tb is abinomial in the reduced Grobner basis of φ(P ) with respect to ≺ω, then taor tb is a standard monomial. Thus by Eq. (9.14) ta = 1 or tb = 1. Hence

φ(P ) = (tr+1 − 1, . . . , tq − 1) ⊂ S′.

Let L be the lattice π(ker(A) ∩Zq). As a result S′/φ(P ) is Artinian, L hasrank q−r and dimK(S′/φ(P )) = 1. Since φ(P ) = I(L), using Lemma 9.6.11and Proposition 9.6.15 we conclude ZA = ZA′. By Lemma 9.6.3 the set A′

is linearly independent, thus α ∈ NA, as required. This proof was adaptedfrom [400]. �

Proposition 9.6.17 Let I be an ideal of S. The following hold.

(a) [336] If in≺(I) is square-free, then rad(I) = I.

(b) [142, Corollary 6.9] If I is graded and in≺(I) is Cohen–Macaulay (resp.Gorenstein), then I is Cohen–Macaulay (resp. Gorenstein).

Proposition 9.6.18 Let I = I(L) ⊂ S be a standard graded lattice ideal.If the initial ideal in≺(I) is square-free, then I is a prime ideal and S/I isnormal and Cohen–Macaulay.

Proof. By Theorem 9.4.6 and Proposition 9.6.17 all associated prime idealsof I have height r = rank(L) and I is a radical ideal. Then I has anirredundant primary decomposition I = p1 ∩ · · · ∩ pm, where pi is a primeideal of height r for all i. Let Ls be the saturation of L consisting of alla ∈ Zq such that pa ∈ L for some 0 �= p ∈ N and let I(Ls) be its latticeideal. Since rank(L) is equal to rank(Ls), by Theorem 8.2.2, we get that r isalso the height of I(Ls). As Zq/Ls is torsion-free, by Theorem 8.2.22, I(Ls)is a prime toric ideal. Then we may assume that p1 = I(Ls). We claimthat in≺(I) = in≺(I(Ls)). Clearly in≺(I) ⊂ in≺(I(Ls)) because I ⊂ I(Ls).To show the reverse inclusion take any element f in the reduced Grobnerbasis of I(Ls). It suffices to show that in≺(f) ∈ in≺(I). By Lemma 8.3.3,

we can write f = ta+ − ta− for some a = a+ − a− in Ls. We may assume

that in≺(f) = ta+

. There is p ∈ N+ such that pa ∈ L. The binomial

g = tpa+ − tpa

−is in I = I(L) and in≺(g) = tpa

+

. Thus tpa+ ∈ in≺(I)

and since this ideal is square-free we get that ta+ ∈ in≺(I). This proves the

claim. Hence deg(S/I) is deg(S/I(Ls)) because S/I and S/I(Ls) have the

408 Chapter 9

same Hilbert function. Therefore, by additivity of the degree, we get thatm = 1. Consequently, by Theorems 9.6.16 and 9.1.6, S/I is normal andCohen–Macaulay. �

Interesting families of graded ideals with square-free initial ideals aregiven in [196]. For these families the corresponding simplicial complexesassociated to the initial ideals are vertex decomposable and hence shellable.

Regular triangulations of cones Consider the primary decompositionof the radical of in(P ) as a finite intersection of face ideals:

rad (in(P )) = p1 ∩ p2 ∩ · · · ∩ pr,

see Theorem 6.1.4. Recall that, by Proposition 6.3.4, the facets of Δ aregiven by σi = {tj| tj /∈ pi}, or equivalently by Ai = {vj | tj /∈ pi} if oneidentifies ti with vi.

According to [400, Theorem 8.3], the family of cones (resp. family ofsimplices if A is homogeneous)

{R+A1, . . . ,R+Ar} (resp. {conv(A1), . . . , conv(Ar)})

is a regular triangulation of the cone R+A (resp. the polytope conv(A)) inthe sense of [400, pp. 63-64] (resp. in the sense of [438, pp. 129-130]). Thismeans that R+A1, . . . ,R+Ar (resp. conv(A1), . . . , conv(Ar)) are obtainedby projection onto the first n coordinates of the lower facets of

Q′ = R+{(v1, ω1), . . . , (vq, ωq)} (resp. conv((v1, ω1), . . . , (vq , ωq))),

a facet of the cone Q′ is lower if it has a normal vector with negative lastentry, i.e., a lower facet of Q′ has the form

F = {x ∈ Q′| 〈α, x〉 = 0}, 〈α, x〉 ≥ 0 valid for Q′, αn+1 < 0,

see Example 9.6.19. Furthermore all regular triangulations of R+A (resp.conv(A) if A is homogeneous) arise in this way, i.e., the regular triangu-lations of R+A (resp. conv(A) if A is homogeneous) are in one to onecorrespondence with the radicals of the monomial initial ideals of the toricideal P ; see [400, Theorem 8.3] and [106, Theorem 9.4.5]. The simplicialcomplex Δ and also its set of facets {A1, . . . ,Ar} is called a regular trian-gulation of the cone R+A (resp. polytope conv(A) if A is homogeneous).For a thorough study of triangulations we refer to [106] and [438].

The regular triangulation {R+A1, . . . ,R+Ar} is called weakly unimodu-lar (resp. unimodular) if ZAi = ZA for all i (resp. the simplex conv(Ai)is unimodular for all i). If A is homogeneous, this regular triangulation isweakly unimodular if and only if the monomial ideal in(P ) is square-free;see [400, Corollary 8.9].


Example 9.6.19 The facets of the cone generated by

(1, 1, 0, 0, 1), (0, 1, 1, 0, 1), (0, 0, 1, 1, 1), (1, 0, 0, 1, 1),

(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0).

are given by the system of linear inequalities

xi ≥ 0, i = 1, . . . , 5 x1 + x3 − x5 ≥ 0, x2 + x4 − x5 ≥ 0,

Thus the lower facets are defined by the last two inequalities.

Example 9.6.20 Let R = Q[x1, x2, x3, y1, y2, y3] be a polynomial ring withthe default monomial order ≺ in Macaulay2 [199], let F = {f1, . . . , f10} bethe set of monomials

f1 = y1y2y3, f2 = x1x2y1, f3 = x2x3y1, f4 = x1x3y1, f5 = x1x2y2

f6 = x2x3y2, f7 = x1x3y2, f8 = x1x2y3, f9 = x2x3y3, f10 = x1x3y3,

and let A = {v1, . . . , v10} be the set of exponent vectors of f1, . . . , f10. Theinitial ideal of P is in(P ) = (t4t5, t3t5, t4t6, t4t8, t3t8, t4t9, t7t8, t6t8, t7t9).The primary decomposition of the initial ideal in(P ) gives the followingregular triangulation of conv(A):

A1 = {v1, v2, v5, v8, v9, v10}, A2 = {v1, v2, v5, v6, v9, v10},A3 = {v1, v2, v3, v6, v9, v10}, A4 = {v1, v2, v5, v6, v7, v10},A5 = {v1, v2, v3, v6, v7, v10}, A6 = {v1, v2, v3, v4, v7, v10}.

It is easy to verify that ZA = ZAi for all i, i.e., the regular triangulation isweakly unimodular. However notice that this triangulation is not unimod-ular in the sense that the simplex Δi = conv(Ai) has normalized volumeequal to 2 for all i, i.e., vol(Δi) = 2/5! and dim(Δi) = 5 for all i. TheEhrhart function of P = conv(A) is given by

|Z6 ∩ iP| = 1

10i5 +

1

2i4 +

11

6i3 +

7

2i2 +

46

15i+ 1 =

12

5!i5 + · · ·

Thus the normalized volume of P is equal to 12.

The notion of weakly unimodular regular triangulation is related to TDIsystems and to Hilbert bases [252].

Proposition 9.6.21 [252] Let A = {v1, . . . , vq} and let A be the matrixwith column vectors v1, . . . , vq. If ZA = Zn, then the system xA ≤ ω isTDI if and only if Δ is a weakly unimodular regular triangulation of R+A.

Proof. It follows from Theorems 1.3.22 and 9.6.8, and Proposition 1.3.14.We leave the details as an exercise. �

410 Chapter 9

Conjecture 9.6.22 [119] If A is the incidence matrix of a uniform clutterC that satisfies the max-flow min-cut property, then the rational polyhedralcone R+{v1, . . . , vq} has a weakly unimodular regular triangulation.

This conjecture holds for some interesting classes of uniform clutterscoming from combinatorial optimization (see Theorem 14.4.17).

Proposition 9.6.23 [400] If A is a t-unimodular matrix, then the regulartriangulation {R+A1, . . . ,R+Ar} of R+A is weakly unimodular.

Proof. From Proposition 1.6.6, we get ZAi = ZA for all i, that is, thetriangulation is weakly unimodular. �

Exercises

9.6.24 Let A = {v1, . . . , vq} ⊂ Nn be a homogeneous configuration. ThenF = conv(v1, . . . , vs) is a face of conv(A) if and only if F ′ = R+{v1, . . . , vs}is a face of R+A.

9.6.25 Let xA ≤ c be a rational system. Assume that Q = {x|xA ≤ c}is pointed and rank(A) < rank ((v1, c1), . . . (vq, cq)), where the vi’s are thecolumns of A. Prove that the map α �→ (α,−1) gives a bijection betweenthe vertices of Q and the lower facets of the cone spanned by the (vi, ci)’s.

9.7 Toric ideals generated by circuits

In this section we give sufficient conditions for the normality of K[F ] andclassify when the toric ideal of a homogeneous normal monomial subring isgenerated by circuits.

Theorem 9.7.1 [49] If each circuit of ker(A) has a positive or negativepart with entries consisting of 0’s and 1’s, then K[F ] is normal.

Proof. The proof is by induction on q, the number of generators of K[F ].The case q = 1 is clear. Assume q ≥ 2 and that the result holds formonomial subrings with less than q generators. To simplify the notation weset fi = xvi for i = 1, . . . , q. Using Theorem 9.1.1 one has:

K[F ] = K[{xa| a ∈ ZA ∩ R+A}].

Let z = xa ∈ K[F ] be a minimal generator of the integral closure of K[F ].

One can write z = xa = fλ11 · · · f

λqq for some λ1, . . . , λq ∈ Z. There is a

positive integer m such that

zm = fmλ11 · · · fmλq

q = f b11 · · · f bqq , (9.15)


for some b1, . . . , bq ∈ N. First we assume that bi = λi = 0 for some i.Consider the matrix A′ obtained from A by removing its ith column, thenLemma 8.3.27 and the induction hypothesis yield z ∈ K[F \ {fi}]. Hence,we may assume that for each i either λi �= 0 or bi > 0.

If ker(A) = (0), the columns of A are linearly independent and K[F ] isnormal because it is a polynomial ring. Assume ker(A) �= (0) and take acircuit 0 �= α ∈ ker(A), one can write

α = (ei1 + · · ·+ eik)︸︷︷︸α+

− (ε1ej1 + · · ·+ εtejt)︸︷︷︸α−

,

where ei is the ith unit vector, εi ∈ N for all i, and i1, . . . , ik, j1, . . . , jtdistinct. Thus one has the equality

k∑�=1

vi� =

t∑�=1

ε�vj� . (9.16)

If bir = 0 for some 1 ≤ r ≤ k, then using Eq. (9.16) one can rewrite

Eq. (9.15) as zm = fmμ1

1 · · · fmμqq = f b11 · · · f

bqq , where μi ∈ Z for all i and

μir = bir = 0, which by induction yields that z ∈ K[F \{fir}]. It remains toconsider the case bir > 0 for all 1 ≤ r ≤ k. One may assume bi1 ≤ · · · ≤ bik .Using Eqs. (9.15) and (9.16) we obtain

ma(9.15)=

q∑i=1

bivi = bi1

k∑�=1

vi� +∑i�=i1

civi(9.16)= bi1

t∑�=1

ε�vj� +∑i�=i1

civi, (9.17)

where the ci’s are nonnegative integers. Therefore, using Eq. (9.16) and

(9.17), we can rewrite Eq. (9.15) as zm = fmδ11 · · · fmδqq = fd11 · · · fdqq , where

δi ∈ Z, di ∈ N for all i, and di1 = δi1 = 0. Hence by the induction hypothesisone obtains z ∈ K[F \ {fi1}]. �

A non-zero binomial ta − tb is said to have a square-free term if ta

is square-free or tb is square-free. If ta and tb are both not square-freemonomials we say that the binomial ta − tb has nonsquare-free terms.

Definition 9.7.2 A binomial g = ta11 · · · taqq − tb11 · · · t

bqq is called balanced if

the following holds: max{a1, . . . , aq} = max{b1, . . . , bq}. If g is not balancedit is called unbalanced. Let g be an unbalanced binomial of the form:

g = tb11 · · · tbrr − tbr+1

r+1 · · · tbss , bi ≥ 1 ∀i,

where 1 ≤ m1 = max{b1, . . . , br} < max{br+1, . . . , bs} = m2. A connectorof g is a binomial: ti1 · · · tij − t

cj+1

ij+1· · · tcmim , ci ≥ 1 ∀i, with a square-free

term ti1 · · · tij such that {i1, . . . , ij} ⊂ {1, . . . , r} and the intersection of{ij+1, . . . , im} with {r + 1, . . . , s} is non-empty.

412 Chapter 9

Theorem 9.7.3 [308] If K[F ] is homogeneous and normal and PF is itstoric ideal, then the following are equivalent:

(a) PF is generated by a finite set of circuits.

(b) PF is generated by a finite set of circuits with a square-free term.

(c) Every unbalanced circuit of PF has a connector which is a linear com-bination (with coefficients in K[t]) of circuits of PF with a square-freeterm.

Proof. As K[F ] is homogeneous, we get that any binomial ta − tb in PF ishomogeneous with respect to the standard grading of K[t] = K[t1, . . . , tq]induced by setting deg(ti) = 1 for all i, a fact that will be used repeatedlybelow without any further notice.

(a) ⇒ (b): The toric ideal PF is minimally generated by a finite set Bof circuits. Thus, by Theorem 9.2.16, each binomial of B is a circuit with asquare-free term.

(b) ⇒ (c): Let g = tb11 · · · tbrr − tbr+1

r+1 · · · tbss , bi ≥ 1 ∀i, be an unbalancedcircuit of A, where 1 ≤ m1 = max{b1, . . . , br} < max{br+1, . . . , bs} = m2.We may assume m2 = br+1. Then

(xv1 · · ·xvr/xvr+1)m1 ∈ K[F ]. (9.18)

The element xv1 · · ·xvr/xvr+1 is in the field of fractions of K[F ] and byEq. (9.18) it is integral over K[F ]. Hence, by the normality of K[F ], theelement xv1 · · ·xvr/xvr+1 is in K[F ]. Since K[F ] is generated as a K-vectorspace by Laurent monomials of the form xa, with a ∈ NA, it is not hard tosee that there is a monomial tγ such that t1 · · · tr − tr+1t

γ ∈ PF . This is aconnector of g and by hypothesis it is a linear combination of circuits of PFwith a square-free term.

(c)⇒ (a): By Theorem 9.2.16, the toric ideal PF is minimally generatedby a finite set B = {f1, . . . , fm} consisting of binomials with a square-freeterm. We will show, by induction on the degree, that each one of the fi’sis a linear combination of circuits. The degree is taken with respect to thestandard grading of K[t].

Let f be a binomial in B. We may assume that f = t1 · · · tp−tap+1

p+1 · · · ta�� ,

ai ≥ 1 ∀i, � ≤ q. Assume that f is not a circuit. Then by Lemma 1.9.5there is a circuit in PF (permuting variables if necessary) of the form

g = tb11 · · · tbrr − tbp+1

p+1 · · · tbss , bi ≥ 1 ∀i,

with r < p or s < �. Setm1 = max{b1, . . . , br} andm2 = max{bp+1, . . . , bs}.We claim that there exist binomials h and h1 (we allow h = h1 or h = 0)

in PF of degree less than deg(f) = p and a binomial h2 which is a linearcombination of circuits of PF such that f is in the ideal of K[t] generatedby g, h, h1, h2. To prove this we consider the following two cases.


Case (A): r = p and s < �. Then

g = tb11 · · · tbpp − tbp+1

p+1 · · · tbss . (9.19)

Subcase (A1): bi = 1 for i = 1, . . . , p. Then we can write

f − g = tbp+1

p+1 · · · tbss − tap+1

p+1 · · · ta�� = tp+1h,

for some binomial 0 �= h ∈ PF (PF is prime) with deg(h) < deg(f) = p.

Subcase (A2): bi = 1 for i = p+1, . . . , s. Then g = tb11 · · · tbpp −tp+1 · · · ts.

By subcase (A1), we may assume that bi ≥ 2 for some 1 ≤ i ≤ p. Then,on the one hand, by the homogeneity of g, p + 1 ≤

∑pi=1 bi = s − p, so

2p+1 ≤ s. On the other hand, by the homogeneity of f , p ≥ �−p ≥ s−p+1,so 2p− 1 ≥ s. This is a contradiction. So this case cannot occur.

Subcase (A3): bi ≥ 2 for some 1 ≤ i ≤ p, bp+j ≥ 2 for some 1 ≤ j ≤ s−p,and m1 ≥ m2. Then f = t1 · · · tp − tap+1

p+1 · · · ta�� , g = tb11 · · · tbpp − tbp+1

p+1 · · · tbss .For simplicity of notation we may assume that m1 = b1. Using that g ∈ PFand m1 ≥ m2 ≥ 2, we get (xvp+1 · · ·xvs/xv1)m2 ∈ K[F ]. Hence, by thenormality of K[F ], there is tγ such that h1 = tp+1 · · · ts− t1tγ is in PF . Thebinomial h1 is non-zero and has degree less than deg(f) because s − p <� − p ≤ p; the second inequality follows from the homogeneity of f . Lettδ = t

ap+1

p+1 · · · ta�� /tp+1 · · · ts. We have

f + h1tδ = f + (tp+1 · · · ts − t1tγ)tδ = t1 · · · tp − t1tγtδ = t1h, (9.20)

where 0 �= h ∈ PF and deg(h) < p.Subcase (A4): bi ≥ 2 for some 1 ≤ i ≤ p, bp+j ≥ 2 for some 1 ≤ j ≤ s−p,

and m1 < m2. Since g is an unbalanced circuit, by hypothesis g has aconnector

h2 = ti1 · · · tik − tik+1tγ , i1 < · · · < ik,

with ik+1 ∈ {p + 1, . . . , s}, {i1, . . . , ik} ⊂ {1, . . . , p} and such that h2 is alinear combination of circuits of PF . Set t

δ = t1 · · · tp/ti1 · · · tik . If f = h2tδ,

then tδ = 1 and f = h2. If f �= h2tδ, then we can write

f − h2tδ = tik+1tγtδ − tap+1

p+1 · · · ta�� = tik+1h, (9.21)

with 0 �= h ∈ PF and deg(h) < p.Case (B): r < p and s ≤ �. In this case

f = t1 · · · tp − tap+1

p+1 · · · ta�� , g = tb11 · · · tbrr − t

bp+1

p+1 · · · tbss .

Subcase (B1): bi = 1 for i = 1, . . . , r. Then

f − gtr+1 · · · tp = tbp+1

p+1 · · · tbss tr+1 · · · tp − tap+1

p+1 · · · ta�� = tp+1h, (9.22)

414 Chapter 9

where 0 �= h ∈ PF and deg(h) < p.Subcase (B2): bi = 1 for i = p+1, . . . , s. Let tγ = t

ap+1

p+1 · · · ta�� /tp+1 · · · ts.

Then we havef − gtγ = t1 · · · tp − tb11 · · · tbrr tγ = t1h (9.23)

where 0 �= h ∈ PF and deg(h) < p.Subcase (B3): bi ≥ 2 for some 1 ≤ i ≤ r, bj ≥ 2 for some p+ 1 ≤ j ≤ s,

and m1 ≤ m2. We may assume m2 = bp+1. Using that g ∈ PF andthat m2 ≥ m1 ≥ 2, we get (xv1 · · ·xvr/xvp+1)m1 ∈ K[F ]. Hence, by thenormality of K[F ], there is tγ such that h1 = t1 · · · tr − tp+1t

γ is in PF .The binomial h1 is non-zero and has degree less than deg(f) because r < p.Then we have

f − h1tr+1 · · · tp = f − (t1 · · · tr − tp+1tγ)tr+1 · · · tp (9.24)

= tp+1tγtr+1 · · · tp − tap+1

p+1 · · · ta�� = tp+1h,

where 0 �= h ∈ PF and deg(h) < p.Subcase (B4): bi ≥ 2 for some 1 ≤ i ≤ r, bj ≥ 2 for some p+ 1 ≤ j ≤ s,

and m1 > m2. Since g is an unbalanced circuit, by hypothesis g has aconnector

h2 = tik+1· · · tik+t

− tidtγ , ik+1 < · · · < ik+t,

with {ik+1, · · · , ik+t} ⊂ {p+ 1, . . . , s}, id ∈ {1, . . . , r}, and such that h2 isa linear combination of circuits of PF . Set tδ = t

ap+1

p+1 · · · ta�� /tik+1

· · · tik+t.

If f = −h2tδ, then tδ = 1 and f = −h2. If f �= −h2tδ, then we can write

f + h2tδ = t1 · · · tp − tidtγtδ = tidh, (9.25)

with 0 �= h ∈ PF and deg(h) < p. This completes the proof of the claim.We are now ready to show that each fi in B is a linear combination

(with coefficients in K[t]) of circuits. We proceed by induction on deg(fi).Let p = min{deg(fi)| 1 ≤ i ≤ m} be the initial degree of PF . If fi is abinomial in B of degree p, then either fi is a circuit or fi is not a circuitand by the claim fi is a linear combination of circuits (notice that in thiscase h = h1 = 0 because there are no non-zero binomials in PF of degreeless than p). Let d be an integer greater than p and let fk be a binomial ofB of degree d (if any). Assume that each fi of degree less than d is a linearcombination of circuits. If fk is a circuit there is nothing to prove. If fk isnot a circuit, then by the claim (or more precisely by Eqs. (9.19)–(9.25))we can write

fk = λg + μh+ μ1h1 + μ2h2, (9.26)

where λ, μ, μ1, μ2 are monomials, h, h1 are binomials in PF of degree lessthan d = deg(fk), h2 is a linear combination of circuits, and g is a cir-cuit. Since PF is a graded ideal with respect to the standard grading ofK[t1, . . . , tq], we get that h and h1 are linear combinations of binomials in


B of degree less than d. Therefore by Eq. (9.26) and the induction hypoth-esis, we conclude that fk is a linear combination of circuits. Therefore theideal PF is generated by a finite set of circuits. �

The following result will be used below to show a class of toric idealsgenerated by circuits. See Theorem 10.3.14 for an application of this result.

Corollary 9.7.4 Let K[F ] be a homogeneous and normal subring. If eachcircuit of PF with nonsquare-free terms is balanced, then PF is generated bya finite set of circuits with a square-free term.

Proof. The circuits of PF satisfy condition (c) of Theorem 9.7.3. Indeed,let f be an unbalanced circuit of PF . Then f has a square-free term byhypothesis. Thus f is a circuit with a square-free term and it is a connectorof f . Hence the result follows from Theorem 9.7.3. �

The best known examples of toric ideals generated by circuits come fromconfigurations whose matrix A is t-unimodular (see Corollary 8.3.8). Otherinteresting examples of toric ideals generated by circuits are the phylogeneticideals studied in [82]. As noted in [328], these phylogenetic ideals actuallyrepresent the family of cut ideals of cycles.

Corollary 9.7.5 If K[F ] is a homogeneous subring such that each circuitof PF has a square-free term, then K[F ] is normal and PF is generated bya finite set of circuits with a square-free term.

Proof. The normality of K[F ] follows from Theorem 9.7.1. Since thecircuits of PF satisfy condition (c) of Theorem 9.7.3, we get that PF isgenerated by a finite set of circuits with a square-free term. �

Exercises

9.7.6 Let F = {x1x2, x2x3, x3x4, x1x4, x1x3, x21, x22, x23, x24}. Use Macaulay2and the procedure below to show that PF is minimally generated by

t3t4 − t5t9, t4t5 − t3t6, t3t5 − t4t8, t2t5 − t1t8, t24 − t6t9, t25 − t6t8,t1t5 − t2t6, t1t2 − t5t7, t1t3 − t2t4, t22 − t7t8, t23 − t8t9, t21 − t6t7.

Prove that Q[F ] is not normal. Notice that PF is minimally generated by“circuits” with square-free part.

KK=ZZ/31991

R=KK[x_1..x_4,t_1..t_9,MonomialOrder=>Eliminate 4]

I=ideal(t_1-x_1*x_2,t_2-x_2*x_3,t_3-x_3*x_4,t_4-x_1*x_4,

t_5-x_1*x_3,t_6-x_1^2,t_7-x_2^2,t_8-x_3^2,t_9-x_4^2)

J= ideal selectInSubring(1,gens gb I)

mingens J

416 Chapter 9

9.7.7 We set F = {xv1 , . . . , xv7}, where the vi’s are given by:

v1 = (0, 1, 1, 0, 1, 0), v2 = (1, 0, 0, 1, 1, 0), v3 = (1, 0, 1, 0, 0, 1),v4 = (1, 0, 0, 1, 0, 1), v5 = (0, 1, 1, 0, 0, 1), v6 = (0, 1, 0, 1, 1, 0),v7 = (0, 1, 0, 1, 0, 1).

Let A be the matrix with column vectors v1, . . . , v7. Prove that the circuitsof ker(A) are given by the rows of the matrix:⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−1 1 0 −1 1 0 0−1 0 1 −1 0 1 0−1 1 1 −2 0 0 10 −1 1 0 −1 1 01 −1 1 0 −2 0 11 1 −1 0 0 −2 10 1 0 −1 0 −1 11 0 0 0 −1 −1 10 0 −1 1 1 0 −1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Then show that K[F ] is a normal monomial subring using the criterion ofTheorem 9.7.1.

9.7.8 In the proof of Theorem 9.7.3 (from (c) to (a)), show that the subcases(A3), (B1), and (B3) cannot occur.

9.8 Divisor class groups of semigroup rings

In this section we present an algorithm to compute the divisor class groupof a Krull semigroup ring of the form K[NA], where K is a field and NA isa subsemigroup of Nn generated by a finite set of vectors.

Krull semigroup rings In this part the semigroups are commutative,satisfy the cancellation law, have a unit, and the total ring of quotients istorsion-free. The semigroups will be written additively.

Let S be a semigroup and let ∼ be the equivalence relation on S × Sgiven by:

(a, b) ∼ (c, d) if and only if a+ d = c+ b.

We denote each equivalence class [(a, b)] by a − b or simply by a if b = 0.Let 〈S〉 be the set of equivalence classes with the operation

(a− b) + (c− d) = (a+ c)− (b + d).

It follows readily that 〈S〉 is a group which is called the total quotient groupof S or the group of differences of S.


Definition 9.8.1 S is called normal if s ∈ 〈S〉 and ns ∈ S for some n ≥ 1implies s ∈ S. S is called completely normal if s ∈ 〈S〉, a ∈ S and a+ns ∈ Sfor all n ≥ 1 implies s ∈ S.

If S = NA is a semigroup of Nn generated by a finite set A, thenthe group of differences of S is 〈S〉 = ZA and S is normal if and only ifNA = ZA ∩ R+A.

Remark 9.8.2 If S is completely normal, then S is normal. The converseholds if S satisfies the ascending chain condition on ideals, thus in particularwhen S is finitely generated (see [83, 161]).

Definition 9.8.3 A discrete valuation of an abelian group G is a homo-morphism of groups ν : G→ Z. The set {x ∈ G | ν(x) ≥ 0} is the valuationsemigroup of ν. The residue group of ν is ker(ν). We say that ν is normalizedif ν(G) = Z.

Proposition 9.8.4 If ν �= 0 is a valuation of a group G, then the valuationsemigroup of ν is isomorphic to ker(ν)× N.

Proof. There is 0 �= n0 ∈ N such that ν(G) = n0Z and we can writen0 = ν(x0) for some x0 ∈ G. The following mapping gives the requiredisomorphism

ϕ : ker(ν)× N→ {x ∈ G | ν(x) ≥ 0}, (x, n) �→ x+ nx0. �

Definition 9.8.5 S is a Krull semigroup if there is a family (νi)i∈I ofdiscrete valuations of 〈S〉 such that S is the intersection of the valuationsemigroups of the νi’s and for each s ∈ S the set {i ∈ I | νi(s) > 0} is finite.

Notation Let F =⊕

i∈I Zei be the free abelian group with basis {ei | i ∈ I}.For each i ∈ I let πi the projection πi : F → Z given by πi(akek)k∈I = ai.The positive part of F is: F+ = {x ∈ F | πi(x) ≥ 0 ∀ i ∈ I}. Notice thatF+ =

⊕i∈I Nei and if S is a subsemigroup of F, then ZS (the subgroup

generated by S) is equal to 〈S〉 (the group of differences of S).

Proposition 9.8.6 [83] S is a Krull semigroup if and only if S ∼= G× S1,where G is a group and S1 is a subsemigroup of a free group F =

⊕i∈I Zei

such that S1 = 〈S1〉 ∩ F+.

Corollary 9.8.7 If S is a semigroup of F such that S = 〈S〉 ∩ F+, then Sis a Krull semigroup.

Proof. Notice that S � {0} × S and apply Proposition 9.8.6. �

Let S be a semigroup and let A be a ring. The semigroup ring of S withcoefficients in A is denoted by A[S]. The elements of A[S] can be written in

418 Chapter 9

the form∑

s∈S rsXs, rs ∈ A, where rs = 0 except for a finite number, with

addition and multiplication defined as for polynomials. See [180, p. 64] fora formal construction of a semigroup ring.

Theorem 9.8.8 [83] A[S] is a Krull ring if and only if A is a Krull ring andS is a Krull semigroup such that 〈S〉 satisfies the ascending chain conditionon cyclic subgroups.

Corollary 9.8.9 Let A ⊂ Nn be a finite set. Then K[NA] is a Krull ringif and only if NA is a Krull semigroup.

The divisor class group of a Krull semigroup

Definition 9.8.10 A subsemigroup S of F is called reduced in F if theprojection πi|〈S〉 : 〈S〉 → Z is onto for all i ∈ I.

Theorem 9.8.11 [83] Let F =⊕

i∈I Zei be a free abelian group and let Sbe a subsemigroup of F such that S = 〈S〉 ∩ F+. If any of the following twoequivalent conditions hold, then Cl(K[S]) ∼= F/〈S〉.

(a) S is a reduced subsemigroup of F, and for all i, j ∈ I with i �= j, thereexists s ∈ S such that πj(s) > 0 = πi(s).

(b) Ti = {xj + 〈S〉 | i �= j ∈ I} generates F/〈S〉 as a subsemigroup ∀ i ∈ I.

Next we state a particular case of Theorem 9.8.11 that can be used tocompute the divisor class group of an affine semigroup.

Theorem 9.8.12 [83] Let F = Ze1 ⊕ · · · ⊕ Zer be a free abelian group ofrank r and let S = Nw1 + · · · + Nwq be a subsemigroup of F such that thefollowing two conditions hold:

(a) S = ZS ∩ F+, where ZS is the subgroup of F generated by S.

(b) πi(ZS) = Z for i = 1, . . . , r and for each pair i �= j in {1, . . . , r} thereexists w ∈ S such that πi(w) > 0 = πj(w).

Then Cl(K[S]) ∼= F/ZS for any field K.

Divisor class groups of affine semigroups Let A = {v1, . . . , vq} be afinite set of distinct vectors in Nn \ {0} and let NA be the affine semigroupof Nn generated by A. If R = K[x1, . . . , xn] is a polynomial ring over afield K and F = {xv1 , . . . , xvq}, then the elements in the monomial subringK[F ] are expressions of the form:∑

ca (xv1)

a1 · · · (xvq )aq ,


where a = (a1, . . . , aq) ∈ Nq, ca ∈ K and ca = 0 except for a finite number.Therefore K[F ] is generated, as a K-vector space, by the set of all xa suchthat a ∈ NA, i.e., K[F ] is the semigroup ring of NA:

K[F ] = K[NA] = K[{xa| a ∈ NA}].

Theorem 9.8.13 The following conditions are equivalent:

(a) K[F ] is normal. (b) K[NA] is a Krull ring. (c) NA = R+A ∩ ZA.

Proof. Notice that K[F ] is a Noetherian domain because K[F ] is a finitelygenerated K-algebra. Thus, by the Mori–Nagata theorem (Theorem 2.6.2),part (a) is equivalent to part (b). From Corollary 9.1.3, we obtain that part(a) is equivalent to part (c). �

Corollary 9.8.14 NA is a Krull semigroup if and only if NA is equal toR+A ∩ ZA.

Proof. It follows from Theorem 9.8.13 and Corollary 9.8.9. �

Theorem 9.8.15 ZA ∩ R+A is a finitely generated Krull semigroup.

Proof. By the finite basis theorem there are a1, . . . , ak in Qn such that

R+A = H+a1 ∩ · · · ∩H

+ak. (†)

For 1 ≤ i ≤ k consider the valuation νi : ZA → Z given by νi(x) = 〈ai, x〉.We may assume that ai ∈ Zn for all i because H+

a = H+ra for r > 0. Hence

ZA ∩R+A =

k⋂i=1

{x ∈ ZA | νi(x) ≥ 0}.

Hence ZA ∩ R+A is a Krull semigroup because it is the intersection of thevaluation semigroups of ν1, . . . , νk. By Gordan’s lemma the semigroup isfinitely generated. �

Proposition 9.8.16 There exists an irreducible representation:

R+A = RA ∩H+�1∩ · · · ∩H+

�r, (9.27)

with �i ∈ Qn for all i and such that the following conditions are satisfied:

(a) 〈�i, vj〉 ∈ N for all i, j.

(b) Z = 〈�i, v1〉Z+ · · ·+ 〈�i, vq〉Z for all i.

420 Chapter 9

Proof. The decomposition of R+A given in Eq. (†) can be brought tothe form of Eq. (9.27), where �i ∈ Qn for all i and RA is the vector spacegenerated by A. By multiplying each �i by an appropriate positive integer,we can choose the �i’s such that condition (a) holds. Let 1 ≤ i ≤ r be afixed integer. We claim that there exists 1 ≤ j ≤ q such that 〈�i, vj〉 > 0.Otherwise 〈�i, vj〉 = 0 for all j and one has R+A ⊂ H�i . By the irreducibilityof the representation (see Theorem 1.1.44), the set

R+A∩H�i = R+A

is a facet (proper face of maximum dimension), a contradiction. Let m bethe greatest common divisor of the non-zero elements of the set

{〈�i, v1〉, . . . , 〈�i, vq〉}.

We can write 〈�i, vk〉 = λkm. Setting �′i = �i/m, conditions (a) and (b) aresatisfied if we substitute �i by �

′i, this follows from the fact that the non-zero

elements of the set λ1, . . . λq are relatively prime. �

For the rest of this section we assume that R+A has an irreduciblerepresentation satisfying the conditions of Proposition 9.8.16. Consider thehomomorphism of Z-modules ϕ : Zn → Zr given by:

ϕ(v) = 〈�1, v〉e1 + · · ·+ 〈�r, v〉er.

In what follows wi will denote the vector of Nr defined as

wi := ϕ(vi) = 〈�1, vi〉e1 + · · ·+ 〈�r, vi〉er (i = 1, . . . , q),

S will denote the semigroup S := Nw1 + · · ·+ Nwq, and F will denote thefree abelian group F := Ze1 ⊕ · · · ⊕ Zer.

Proposition 9.8.17 NA � S.

Proof. Clearly ϕ induces a homomorphism from NA onto S that we alsodenote by ϕ. Thus it suffices to prove that ϕ is an injective map. Letα = a1v1+ · · ·+aqvq be a vector in ker(ϕ). It follows readily that 〈α, �i〉 = 0for i = 1, . . . , r. Therefore α belongs to the set

R+A ∩H�1 ∩ · · · ∩H�r = {0}

and consequently α = 0. The previous equality follows using that the vector0 is a vertex of the pointed cone R+A, hence {0} is an intersection of facets.From the irreducibility of the representation of the cone one obtains thatits facets have the form R+A∩H�i for i = 1, . . . , r (see Theorem 1.1.44). �

Corollary 9.8.18 K[NA] � K[S].


Proof. By Proposition 9.8.17 we have NA � S. Hence applying [180,Theorem 7.2, p. 69] we obtain the required isomorphism. �

Theorem 9.8.19 If K[NA] is a Krull ring , then

Cl(K[NA]) � F/〈S〉 = (Ze1 ⊕ · · · ⊕ Zer) /(Zw1 + · · ·+ Zwq) .

In particular rank(F/〈S〉) = r − rank(〈S〉).

Proof. Taking into account Theorem 9.8.12 and Corollary 9.8.18 it sufficesto prove that the following two conditions are satisfied:

(a) S = ZS ∩ F+.

(b) πi(ZS) = Z for i = 1, . . . , r and for every i �= j in {1, . . . , r} thereexists w ∈ S such that πi(w) > 0 = πj(w).

First we prove (a). Since 〈�j , vi〉 ∈ N for all i, j, we get that wi ∈ F+

and clearly wi ∈ ZS for all i. Therefore S ⊂ ZS ∩ F+. Conversely letα ∈ ZS ∩ F+. We can write

α = a1w1 + · · ·+ aqwq = η1e1 + · · ·+ ηrer (ai ∈ Z; ηj ∈ N ∀ i, j).

Substituting the wi’s and matching the coefficients of the ei’s yields:

ηi = a1〈�i, v1〉+ · · ·+ aq〈�i, vq〉, i = 1, . . . , r.

Consider the vector β = a1v1 + · · ·+ aqvq ∈ ZA. Clearly ϕ(β) = α becauseϕ(vi) = wi for = 1, . . . , q. Using these equations we obtain

〈β, �i〉 = a1〈�i, v1〉+ · · ·+ aq〈�i, vq〉 = ηi ≥ 0, i = 1, . . . , r.

Therefore β ∈ RA ∩ H+�1∩ · · · ∩ H+

�r= R+A and as a consequence we get

β ∈ ZA∩R+A. By the normality of K[NA] we conclude that β ∈ NA. Thismeans that we can rewrite β as β =

∑qi=1 civi with ci ∈ N for all i. Hence

ϕ(β) =∑qi=1 ciwi = α. Thus α ∈ S, as desired.

Next we prove (b). Since πi(wj) = 〈�i, vj〉 for all i, j, one has

πi(ZS) = πi(Zw1 + · · ·+ Zwq) = Zπi(w1) + · · ·+ Zπi(wq)

= Z〈�i, v1〉+ · · ·+ Z〈�i, vq〉 = Z,

where the last equality is satisfied thanks to the choice of �1, . . . , �r (seeProposition 9.8.16). To prove the second part of condition (b) it suffices toprove that for i �= j there exists wk = ϕ(vk) ∈ S such that

〈�i, vk〉 = πi(wk) > 0 = πj(wk) = 〈�j, vk〉.

By the irreducibility of the representation of R+A, the sets Fi = H�i ∩R+Aand Fj = H�j ∩R+A are distinct facets. By Proposition 1.1.23, Fi is a conegenerated by a subset of A and since Fj �⊂ Fi there exists vk ∈ Fj \ Fi.Therefore 〈�j , vk〉 = 0 and 〈�i, vk〉 > 0, as desired. �

422 Chapter 9

Algorithm To compute the divisor class group of the semigroup ringK[NA] proceed as follows:

(1) First, using Normaliz [68], compute the integral closure of K[NA] inits field of fractions to determine whether K[NA] is a Krull ring.

(2) Second, using PORTA [84], compute an irreducible representation ofR+A and adjust this representation (if necessary) to satisfy conditions(a) and (b) of Proposition 9.8.16.

(3) Third, using (2) determine the vectors w1, . . . , wq. Then, using Maple[80], diagonalize (over the integers) the matrix B whose rows arew1, . . . , wq, to obtain a “diagonal matrix”D = diag{d1, . . . , ds}, whered1, . . . , ds are the invariant factors of Zr/Z{w1, . . . , wq}.

(4) Fourth, use Theorem 9.8.19 together with the fundamental theorem forfinitely generated abelian groups (see Theorem 1.3.16), to obtain:

Cl(K[NA]) � Zr/Z{w1, . . . , wq} � Zt ⊕ Z/Zd1 ⊕ · · · ⊕ Z/Zds,

where t = r − rank(B) = r − s.

Example 9.8.20 Let A = {(a1, . . . , ad) ∈ Nn| a1+ · · ·+an = d} be the setof ordered partitions of an integer d ≥ 2. Assume A = {v1, . . . , vq}. Noticethat RA = Rn and the irreducible representation of R+A is:

R+A = Rn+ = H+e1 ∩ · · · ∩H

+en .

In this case r = n and �i = ei for i = 1, . . . , n. It is easy to see thatϕ(vi) = wi = vi for all i. Thus S = ZA. Let B be the matrix whoserows are w1, . . . , wq. Using elementary operations we obtain that the Smithnormal form of B is diag{1, . . . , 1, d}, where the number of 1’s is n− 1. ByTheorem 9.8.19 we get Cl(K[NA]) � Zn/ZA � Zd.

Exercises

9.8.21 A semigroup (0) �= S of Nn is called full if Nn ∩ZS = S. If S ⊂ Nn

is a full semigroup, prove that S is a finitely generated semigroup.

9.8.22 [248] Let A ⊂ Nn be a finite set. If NA is a normal semigroup of Nn,then there is a full semigroup S ⊂ Nr such that NA � S (as semigroups).In particular K[NA] � K[S] (as rings).

9.8.23 If S ⊂ Nn is a full semigroup, prove that R+S ∩ ZS = NS, that is,S is a normal semigroup.

9.8.24 Let F be the set of square-free monomials of degree d of a polynomialring over a field K and let K[F ] be the monomial subring spanned by F .Find a formula for Cl(K[F ]).

Chapter 10

Monomial Subrings ofGraphs

In this chapter we study monomial subrings associated to graphs and theirtoric ideals. We relate the even closed walks and circuits of the vectormatroid of a graph with Grobner bases. A description of the integral closureof the edge subring of a multigraph will be presented along with a descriptionof the circuits of its toric ideal. The Smith normal form and the invariantfactors of the incidence matrix of a graph are fully determined. As anapplication we compute the multiplicity of edge subrings in terms of relativevolumes. The family of ring graphs is studied here. These graphs arecharacterized in algebraic and combinatorial terms.

Several interesting connections between monomial subrings, polyhedralgeometry and graph theory will occur in this chapter. We study in detailthe irreducible representation of an edge cone of a graph and show someapplications to graph theory, e.g., we show the marriage theorem.

10.1 Edge subrings and ring graphs

Let R = K[x1, . . . , xn] be a polynomial ring over a fieldK with the standardgrading R =

⊕∞i=0 Ri induced by deg(xi) = 1 for all i and let G be a graph

on the vertex set X = {x1, . . . , xn}.

Definition 10.1.1 The edge subring of the graph G, denoted by K[G], isthe K-subalgebra of R given by:

K[G] = K[{xixj |xi is adjacent to xj}] ⊂ R.

424 Chapter 10

To obtain a presentation of the edge subring of G note that K[G] is astandard K-algebra with the normalized grading

K[G]i = K[G] ∩R2i

and consider the set F = {f1, . . . , fq} of all monomials xixj such that xi isadjacent to xj . Since the elements in K[G] are polynomial expressions in Fwith coefficients in K, there is a graded epimorphism of K-algebras

ϕ : S = K[t1, . . . , tq] −→ K[G], ti �−→ fi,

where S is a polynomial ring graded by deg(ti) = 1 for all i. The kernel ofϕ, denoted by P (G), is a graded ideal of S called the toric ideal of K[G]with respect to f1, . . . , fq.

Definition 10.1.2 Let w = {x0, x1, . . . , xr = x0} be an even closed walkof G such that fi = xi−1xi. The binomial tw = t1 · · · tr−1− t2 · · · tr is calledthe binomial associated to w.

Remark 10.1.3 tw ∈ P (G) because f1f3 · · · fr−1 = f2f4 · · · fr.

Definition 10.1.4 A closed walk of even length will be called a monomialwalk . A monomial xixj of R is said to be an edge generator if {xi, xj} isan edge of G.

Notation In what follows Is denotes the set of all non-decreasing sequencesα = (i1, · · · , is) of length s. If a1, . . . , aq is a sequence and α = (i1, · · · , is)is in Is, then we set aα = ai1 · · · ais .

Proposition 10.1.5 [417] If G is a graph and P = P (G) is the toric idealof the edge subring K[G], then

(a) P = ({tw|w is an even closed walk}), and

(b) P = ({tw |w is an even cycle}) if G is bipartite.

Proof. (a): We set B = {tw|w is an even closed walk}. As P is a gradedideal in the standard grading of S = K[t1, . . . , tq], we can write

P = S ·( ∞⋃s=2

Ps

).

One clearly has (B) ⊂ P . To prove the other inclusion we use inductionon s. First recall that P is a binomial ideal by Corollary 8.2.18. If s = 2,it is enough to note that the binomials in P2 come from squares. Assume

Monomial Subrings of Graphs 425

Ps−1 ⊂ (B). To show Ps ⊂ (B), take tα − tβ in Ps, where α = (i1, . . . , is)and β = (j1, . . . , js). Define G1 as the subgraph of G having vertex set

V1 = {xi ∈ V |xi divides fα}

and edge set

E1 = {{x, y} ∈ E(G)| f� = xy for some � ∈ {i1, . . . , is, j1, . . . , js}}.

Note that if fi1 · · · fim = fj1 · · · fjm for some m < s and for some orderingof the generators then tα − tβ ∈ (B); to prove it, notice that tα − tβ can bewritten as

tα − tβ = tim+1 · · · tis(ti1 · · · tim − tj1 · · · tjm)

+tj1 · · · tjm(tim+1 · · · tis − tjm+1 · · · tjs)

and use induction hypothesis. Now we can assume fi1 · · · fim �= fj1 · · · fjmfor m < s and for any re-ordering of the fi’s. Observe that this forces G1

to be connected. Take x0 ∈ V1. Since fα = fβ, it is easy to check thatafter re-ordering we can write fik = x2k−2x2k−1 and fjk = x2k−1x2k, where1 ≤ k ≤ s and x0 = x2s. Therefore the monomial walk w = {x0, . . . , x2n}satisfies tw = tα − tβ and the induction is complete.

(b): Assume that G is bipartite. This part can be proven similarly topart (a) if we notice that in this case the condition fi1 · · · fim �= fj1 · · · fjmfor m < s and for any re-ordering of the fi’s implies that the graph G1 isan even cycle. �

Corollary 10.1.6 If G is a graph and f = tα − tβ is a primitive binomialin P (G), then f = tw for some even closed walk w of G.

Proof. It follows from the proof of Proposition 10.1.5. �

Corollary 10.1.7 If G is a bipartite graph and f = tα − tβ is a primitivebinomial in P (G), then f = tw for some even cycle w of G.


Proposition 10.1.8 If G is a graph and f = tα−tβ is a primitive binomialin P (G), then the entries of α satisfy αi ≤ 2 for all i.

Proof. By Corollary 10.1.6 one can write

f = tw = t1t3 · · · t�−1 − t2t4 . . . t�,

where � is even, and w is an even closed walk in G of length �:

w = {x0, x1, . . . , x�−1, x� = x0}

426 Chapter 10

such that f1f3 · · · f�−1 = f2f4 · · · f� and fi = xi−1xi for all i. Note thatthere may be some repetitions of the variables ti and accordingly somerepetitions in the monomials fi. As usual we are identifying the vertices ofG with variables. It suffices to verify that x0 occurs at most three times(including the end vertices) in the closed walk w. If x0 occurs more thanthree times in w, one can write

w = {x0, x1, . . . , xt1 = x0, xt1+1, . . . , xt1+t2 = x0, xt1+t2+1, . . . , x� = x0},

note that t1 must be odd; otherwise using w1 = {x0, x1, . . . , xt1 = x0} onehas that tw1 = tγ− tδ, where tγ divides t1t3 · · · t�−1 and tδ divides t2t4 . . . t�,a contradiction because f is primitive. By a similar argument t2 must beodd. Thus t1 + t2 is even, to get a contradiction note that the closed walk

w′ = {xt1+t2 = x0, xt1+t2+1, . . . , x� = x0}

is of even length, and apply the previous argument. �

In [352] there are characterizations of when a binomial of a toric ideal ofK[G] is primitive, minimal, indispensable, or fundamental in terms of theeven closed walks.

Lemma 10.1.9 Let G be a graph and let P = P (G) be the toric ideal ofK[G]. If f is a polynomial in any reduced Grobner bases of P , then

(a) f is a primitive binomial and f = tw for some even closed walk w ofthe graph G.

(b) If G is bipartite, then f is primitive and f = tw for some even cyclew of the graph G.

Proof. Let B be a reduced Grobner basis of P and take f ∈ B. UsingCorollary 8.2.18 and the fact that normalized reduced Grobner bases areuniquely determined, we obtain f = tα−tβ. By Lemma 8.3.3 f is primitive,thus by Corollaries 10.1.6 and 10.1.7 f has the required form. �

Proposition 10.1.10 If G is a graph, then the set

{tw| tw is primitive and w is an even closed walk}

is a universal Grobner basis of P (G).


Proposition 10.1.11 If G is a bipartite graph, then the set

{tw|w is an even cycle}

is a universal Grobner basis of P (G).



Definition 10.1.12 A chord of a cycle c of a graph G is any edge of Gjoining two non-adjacent vertices of c. A cycle without chords is calledprimitive.

Proposition 10.1.13 If a is a cycle and ta ∈ (tc| c is a cycle, c �= a), thena has a chord.

Proof. If ta = tα − tβ ∈ (tc| c is a cycle, c �= a), then each term of ta mustbe divisible by a term of tc for some c �= a. This means that tα = tδtα1 andtβ = tγtα2 , where tαi are terms of some binomial of the form tci with ci �= aand ci a cycle. Hence (after permuting variables) x1 · · ·xr = xθ(x1 · · ·xs),where x1, . . . , xr and x1, . . . , xs are the vertices of the cycles a and c1. It isseen that a must have a chord. �

Proposition 10.1.14 If G is a bipartite graph, then P (G) is a prime idealminimally generated by the set {tc| c is a primitive cycle of G}.

Proof. It follows from Proposition 10.1.13. �

Rees algebras of edge ideals We will look closely at the toric ideal ofthe Rees algebra of an edge ideal, and then show a few applications.

Let I = I(G) be the edge ideal of a graph G, let F = {f1, . . . , fq} bethe set of edge generators of I, and let R[It] = R[f1t, . . . , fqt] ⊂ R[t] be theRees algebra of I. There is an epimorphism

φ : R[t1, . . . , tq]→ R[It], induced by ti → fit.

We set J = ker(φ) and B = R[t1, · · · , tq], notice that J = ⊕∞i=1Ji is a graded

ideal (in the ti-variables) of B = ⊕∞i=0Bi. The ideal J is the presentation

ideal or toric ideal of R[It] with respect to F . Recall that the ideal I is saidto be of linear type if J = J1B.

Theorem 10.1.15 [417] Let I = I(G) be the edge ideal of a graph G andlet J be the toric ideal of R[It]. Then Js = B1Js−1 +RPs for s ≥ 2 and

J = BJ1 +B ·( ∞⋃s=2

Ps

),

where Ps = {tα − tβ | fα = fβ , for some α, β ∈ Is}.

Corollary 10.1.16 [417] Let G be a connected graph and let I be its edgeideal. Then I is an ideal of linear type if and only if G is a tree or G has aunique cycle of odd length.

428 Chapter 10

Proof. ⇒) If I is an ideal of linear type, then by Proposition 4.2.1, G is atree or G has a unique cycle of odd length.⇐) Assume that G is either a tree or G has a unique cycle of odd length.

We claim that RPs = (0) for s ≥ 2, we proceed by induction on n. If n ≤ 3this is easily verified, assume the claim true for graphs with less than nvertices. Using induction on s we now show RPs = (0) for s ≥ 2, notice thatRP2 = (0) because G has no squares. Assume RPs−1 = (0) and RPs �= (0),take a non-trivial relation fi1 · · · fis = fj1 · · · fjs . Since a tree has a vertexof degree one, by induction hypothesis it follows that G must be a cycle.Therefore, using induction again, we may write {fi1 , . . . , fis} = {g1, . . . , gs},where gcd(gi, gj) = 1 for i �= j and g1 · · · gk = x1 · · ·xn. Notice that thisequality cannot occur if n is odd, hence RPs = (0) and the proof of theclaim is complete. Hence, by Proposition 10.1.15, I is of linear type. �

Arbitrary square-free monomial ideals of linear type have been studied in[6, 162, 376] using simplicial complexes. These papers show a deep interplaybetween graph theory and the defining equations of Rees algebras. Aninteresting family of square-free monomial ideals of linear type is given in[376]. This family includes edge ideals of totally balanced clutters and facetideals of disjoint simplicial cycles of odd lengths. This family is closed underthe operation of adding M -elements.

Let I ⊂ R be a square-free monomial ideal minimally generated bymonomials f1, . . . , fq. The line graph of I, denoted by L(I), is the graphwhose vertex set is f1, . . . , fq and where {fi, fj} is an edge of L(I) if andonly if fi and fj have at least one variable in common.

Recently Fouli and Lin [162] have shown that if the line graph L(I) is adisjoint union of trees and graphs with a unique cycle of odd length, thenI is of linear type. A generalization of this fact is given in [6].

A dimension formula Let us describe the cycle space of a graph G overthe field F = Z2. We denote the edge set and vertex set of G by

E(G) = {f1, . . . , fq} and X = {x1, . . . , xn},

respectively. Let C0 and C1 denote the vector spaces over F of 0-chainsand 1-chains, respectively. Recall that a 0-chain of G is a formal linearcombination ∑

aixi

of points and a 1-chain is a formal linear combination∑bifi

of edges, where ai ∈ F and bi ∈ F. The boundary operator ∂ is the lineartransformation defined by

∂(fk) = ∂({xi, xj}) = xi + xj .


A cycle vector is a 1-chain of the form t1 + · · ·+ tr where t1, . . . , tr are theedges of a cycle of G. The cycle space Z(G) of G over F is equal to ker(∂).

The vectors in Z(G) can be regarded as a set of edge-disjoint cycles. Acycle basis for G is a basis for Z(G) which consist entirely of cycle vectors,such a basis can be constructed as follows:

Remark 10.1.17 If G is connected, then G has a spanning tree T . Thesubgraph of G consisting of T and any edge of G not in T has exactly onecycle, the collection of all the cycle vectors obtained in this way form a cyclebasis for G. See [208] for details (cf. Exercise 6.2.8). In particular

dimF Z(G) = q − n+ 1.

Definition 10.1.18 If G is a graph, the number dimFZ(G) is called thecycle rank of G and is denoted by rank(G).

Lemma 10.1.19 Let G be a connected graph and Ze(G) the subspace ofZ(G) of all cycle vectors of G with an even number of terms. Then

dimFZe(G) ={q − n+ 1 if G is bipartite, andq − n otherwise.

Proof. Let c1, . . . , cl, cl+1, . . . , cm be a cycle basis for G, where c1, . . . , clare even cycles and cl+1, . . . , cm odd cycles. It is clear that a basis for Ze(G)is given by {c1, . . . , cl, cl+1 + cm, . . . , cm−1 + cm}. �

Proposition 10.1.20 [417] If G is a connected graph, then

ht(P (G)) = dimF Ze(G).

Proof. Let G be a connected graph with q edges and n vertices. Assumethat I(G) is minimally generated by the monomials f1, . . . , fq. There isa spanning tree T of G so that (after re-ordering) I(T ) = (f1, . . . , fn−1).Since I(T ) is an ideal of linear type

dimK[G] = tr.degK K[G] ≥ n− 1.

If G is bipartite notice that fk ∈ k(f1, . . . , fn−1) for k ≥ n; to prove itwe write fk = xz and observe that the graph T ∪ {x, z} has a unique cycleof even length. This shows the equality dimK[G] = n− 1.

If G is not bipartite, then for some fk = xz, k ≥ n, the subgraphT ∪ {x, z} has a unique cycle of odd length. By Corollary 10.1.16 the ideal(f1, . . . , fn−1, fk) is of linear type, hence dimK[G] ≥ n; to show equalityrecall that tr.degK K[G] ≤ n. To finish the proof use Lemma 10.1.19. �

430 Chapter 10

Corollary 10.1.21 If G is a connected graph with n vertices and K[G] isits edge subring, then

dim(K[G]) =

{n if G is not bipartite, andn− 1 otherwise.


Definition 10.1.22 The number of primitive cycles of a graph G, denotedby frank(G), is called the free rank of G.

Proposition 10.1.23 If G is a graph, then Z(G) is generated by cyclevectors of primitive cycles. In particular rank(G) ≤ frank(G).

Proof. Let c1, . . . , cr be a cycle basis for the cycle space of G and letc1, . . . , cr be the corresponding cycles of G. It suffices to notice that if somecj has a chord, we can write cj = c′j +c′′j , where c

′j and c′′j are cycle vectors

of cycles of length smaller than that of cj . �

Corollary 10.1.24 Let G be a graph. The following are equivalent :

(a) rank(G) = frank(G).

(b) The set of cycle vectors of primitive cycles is a basis for Z(G).(c) The set of cycle vectors of primitive cycles is linearly independent.

Proof. (a) ⇒ (b): By Proposition 10.1.23 there is a basis B of Z(G)consisting of cycle vectors of primitive cycles. By hypothesis rank(G) isfrank(G). Thus B is the set of all cycle vectors of primitive cycles and B isa basis. That (b) implies (c) and (c) implies (a) are also easy to prove. �

The family of graphs satisfying the equality rank(G) = frank(G) can beconstructed as is seen later in this section.

Ring graphs In this part we characterize ring graphs in algebraic andcombinatorial terms. The reader is referred to Section 7.1 for the notionsof cutvertex, bridge, and block.

Lemma 10.1.25 Let G be a graph and let G1, . . . , Gr be its blocks. Thenrank(G) = frank(G) if and only if rank(Gi) = frank(Gi) for all i.

Proof. ⇒) Let Gi be any block of G. We may assume |V (Gi)| > 2,otherwise rank(Gi) = frank(Gi) = 0. If c is a primitive cycle of Gi, then bythe maximality condition of a block one has that c is also a primitive cycleof G. Thus by Corollary 10.1.24 the set of cycle vectors of primitive cyclesof Gi is linearly independent and rank(Gi) = frank(Gi).


⇐) Let Bi and B be the set of cycle vectors of primitive cycles of Gi andG, respectively. As ∪ri=1Bi is linearly independent, by Corollary 10.1.24 itsuffices to prove that ∪ri=1Bi = B. In the first part of the proof we havealready observed that ∪ri=1Bi ⊂ B. To prove the equality take any cyclevector c of a primitive cycle c of G. Since c is a 2-connected subgraph, itmust be contained in some block of G, i.e., in some Gi. Thus c is a primitivecycle of Gi, so c is in Bi. �

Definition 10.1.26 Given a graph H , we call a path P an H-path if P isnon-trivial and meets H exactly in its ends.

Theorem 10.1.27 ([111, Proposition 3.1.2], [208, Theorem 5.10]) Let G bea graph. Then the following three conditions are equivalent:

(a) G is 2-connected.

(b) G can be constructed from a cycle by successively adding H-paths tographs H already constructed.

(c) Every pair of vertices of G is joined by at least 2 vertex-disjoint paths.

This result suggests the following more restrictive notion:

Definition 10.1.28 A graph G is called a ring graph if each block of Gwhich is not a bridge can be constructed from a cycle by successively addingH-paths of length at least 2 that meet graphs H already constructed in twoadjacent vertices.

Families of ring graphs include forests and cycles.

Remark 10.1.29 Let G be a 2-connected ring graph and let c be a fixedprimitive cycle of G, then G can be constructed from c by successivelyadding H-paths of length at least 2 that meet graphs H already constructedin two adjacent vertices.

Definition 10.1.30 A graph H is called a subdivision of a graph G if Harises from G by replacing edges by paths. That is H is obtained by iter-atively choosing an edge (u, v), introducing a new vertex w, deleting edge(u, v), and adding edges (u,w) and (w, v).

Example 10.1.31 A complete bipartite graph K2,3 and a subdivision

��

��

�K2,3

��

��

��

��

��

��

��

��

��

� ��

��

��

��

��

��

432 Chapter 10

Lemma 10.1.32 [2, Lemma 7.78, p. 387] Let G be a graph with vertex setV. If G is 2-connected and deg(v) ≥ 3 for all v ∈ V, then G contains asubdivision of K4 as a subgraph.

Lemma 10.1.33 Let G be a graph. If rank(G) = frank(G) and x, y aretwo non-adjacent vertices of G, then there are at most two vertex disjointpaths joining x and y.

Proof. Assume that there are three vertex disjoint paths joining x and y:

P1 = {x, x1, . . . , xr, y}, P2 = {x, z1, . . . , zt, y},P3 = {x, y1, . . . , ys, y},

where r, s, t are greater than or equal to 1. We may assume that the sumof the lengths of the Pi’s is minimal. Consider the cycles

c1 = {x, x1, . . . , xr, y, zt, . . . , z1, x},c2 = {x, z1, . . . , zt, y, ys, . . . , y1, x},c3 = {x, x1, . . . , xr, y, ys, . . . , y1, x}.

Thus we are in the following situation:

�x

�z1 zt�� y

��

�y1

x1 xr

�� ys

c1

c2

c3

Observe that, by the choice of the Pi’s, a chord of the cycle c1 (resp. c2,c3) must join xi and zj (resp. zi and yj , xi and yj) for some i, j. Thereforewe can write

c1 =

n1∑i=1

ai, c2 =

n2∑i=1

bi, c3 =

n3∑i=1

di

where a1, . . . , an1 ,b1, . . . ,bn2 ,d1, . . . ,dn3 are distinct cycle vectors of prim-itive cycles of G and ci is the cycle vector corresponding to ci. Thus fromthe equality c3 = c1 + c2. we get that the set of cycle vectors of primitivecycles is linearly dependent, a contradiction to Corollary 10.1.24. �

Definition 10.1.34 A graph G has the primitive cycle property (PCP) ifany two primitive cycles intersect in at most one edge.

Lemma 10.1.35 Let G be a graph. If rank(G) = frank(G), then G has theprimitive cycle property.


Proof. Let c1, c2 be two distinct primitive cycles. Assume that c1 andc2 intersect in at least two edges. Thus c1 and c2 must intersect in twonon-adjacent vertices. Hence there are two non-adjacent vertices x, y in c1and a path P = {x, x1, . . . , xr, y} of length at least two that intersect c1 inexactly the vertices x, y:

�x

�� y��

�

x1 xr

�� c1

P

This contradicts Lemma 10.1.33. Hence c1 and c2 have at most one edge incommon. �

Lemma 10.1.36 Let G be a graph. If G satisfies PCP and G does notcontain a subdivision of K4 as a subgraph, then for any two non-adjacentvertices x, y there are at most two vertex disjoint paths joining x and y.

Proof. Assume that there are three vertex disjoint paths joining x and y:

P1 = {x, x1, . . . , xr, y}, P2 = {x, z1, . . . , zt, y}, P3 = {x, y1, . . . , ys, y},

where r, s, t are greater than or equal to 1. We may assume that the sumof the lengths of the Pi’s is minimal. Consider the cycles

c1 = {x, x1, . . . , xr, y, zt, . . . , z1, x}, c2 = {x, z1, . . . , zt, y, ys, . . . , y1, x},c3 = {x, x1, . . . , xr, y, ys, . . . , y1, x}.

Thus we are in the following situation:

�x

�z1 zt�� y

��

�y1

x1 xr

�� ys

c1

c2

c3

Observe that, by the choice of the Pi’s, a chord of the cycle c1 (resp. c2, c3)must join xi and zj (resp. zi and yj , xi and yj) for some i, j. Using thatG does not contain a subdivision of K4 as a subgraph, it is seen that thecycles c1 and c3 are primitive. Thus, since c1 and c3 have at least two edgesin common, we obtain that G does not satisfy PCP, a contradiction. �

434 Chapter 10

Lemma 10.1.37 Let G be a graph. If rank(G) = frank(G), then G doesnot contain a subdivision of K4 as a subgraph.

Proof. Assume there is a subgraph H ⊂ G which is a subdivision of K4.If K4 is a subgraph of G, then G has four distinct triangles whose cyclevectors are linearly dependent, a contradiction to Corollary 10.1.24. If K4

is not a subgraph of G, then H is a strict subdivision of K4, i.e., H hasmore than four vertices. It follows that there are two vertices x, y in V (H)which are non-adjacent in G. Notice that x, y can be chosen in K4 beforesubdivision. Therefore there are at least three non-adjacent paths joining xand y, a contradiction to Lemma 10.1.33. �

Theorem 10.1.38 [184] Let G be a graph. The following are equivalent:

(a) G is a ring graph.

(b) rank(G) = frank(G).

(c) G satisfies PCP and does not contain a subdivision of K4 as a subgraph.

Proof. (a)⇒ (b): By induction on the number of vertices it is not hard tosee that any ring graph G satisfies the equality rank(G) = frank(G).

(b) ⇒ (c): It follows at once from Lemmas 10.1.35 and 10.1.37.(c) ⇒ (a): Let G1, . . . , Gr be the blocks of G. The proof is by induc-

tion on the number of vertices of G. If each Gi is either a bridge or anisolated vertex, then G is a forest and consequently a ring graph. Hence byLemma 10.1.25 we may assume that G is 2-connected and that G is not acycle. We claim that G has at least one vertex of degree 2. If deg(v) ≥ 3for all v ∈ V (G), then by Lemma 10.1.32 there is a subgraph H ⊂ G whichis a subdivision of K4, which is impossible. Let v0 ∈ V (G) be a vertexof degree 2 as claimed. By the primitive cycle property there is a uniqueprimitive cycle c = {v0, v1, . . . , vs = v0} of G containing v0. The graphH = G \ {v0} satisfies PCP and does not have a subdivision of K4 as asubgraph. Consequently H is a ring graph. Thus we may assume that c isnot a triangle, otherwise G is a ring graph because it can be obtained byadding the H-path {v2, v0, v1} to H .

Next we claim that if 1 ≤ i < j < k ≤ s− 1, then vi and vk cannot bein the same connected component of H \ {vj}. Otherwise there is a path ofH \ {vj} that joins vi with vk. It follows that there is a path P of H \ {vj}with at least three vertices that joins a vertex of {vj+1, . . . , vs−1} with avertex of {v1, . . . , vj−1} and such that P intersects c exactly in its ends, butthis contradicts Lemma 10.1.36. This proves the claim. In particular vi is acutvertex of H for i = 2, . . . , s− 2 and vi−1, vi+1 are in different connectedcomponents of H \ {vi}. For each 1 ≤ i ≤ s − 2 there is a block Ki of Hsuch that {vi, vi+1} is an edge of Ki. Notice that if 1 ≤ i < j < k ≤ s− 1,then vi, vj , vk cannot lie in some K�. Indeed if the three vertices lie in some


K�, then there is a path P ′ in K� \ {vj} that joins vi and vk. Since P ′ isalso a path in H \ {vj}, we get that vi and vk are in the same connectedcomponent of H \ {vj}, but this contradicts the last claim. In particularV (K�) intersects the cycle c in exactly the vertices v�, v�+1 for 1 ≤ � ≤ s−2.

Observe that at least one of the edges of c not containing v0 is not abridge of H . To show this pick x /∈ c such that {x, vk} is an edge of H .We may assume that vk+1 �= v0 (or vk−1 �= v0). Since G′ = G \ {vk}is connected, there is a path P of G′ joining x and vk+1 (or vk−1). Thisreadily yields a cycle of H containing an edge of c which is not a bridge ofH . Hence at least one of the blocks K1, . . . ,Ks−2, say Ki, contains verticesoutside c.

Next we show that two distinct blocks B1, B2 of H cannot intersectoutside c. We proceed by contradiction assuming that V (B1)∩V (B2) = {z}for some z not in c. Let H1, . . . , Ht be the connected components of H \{z}.Notice that t ≥ 2 because {z} is the intersection of two different blocks ofH . We may assume that {v1, . . . , vs−1} are contained in H1. Consider thesubgraph H ′

1 of G \ {z} obtained from H1 by adding the vertex v0 andthe edges {v0, v1}, {v0, vs−1}. It follows that the connected components ofG \ {z} are H ′

1, H2, . . . , Ht , which is impossible because G is 2-connected.Let Ki be a block of H that contains vertices outside of c for some

1 ≤ i ≤ s − 2. By induction hypothesis Ki is a ring graph. Thus byRemark 10.1.29 we can construct Ki starting with a primitive cycle c1 thatcontains the edge {vi, vi+1}, and then adding appropriate paths. Supposethat P1, . . . ,Pm is the sequence of paths added to c1 to obtain Ki. If weremove the path Pm from G and use the fact that distinct blocks of Hcannot intersect outside c, then again by induction hypothesis we obtain aring graph. It follows that G is a ring graph as well. �

A graph G that can be embedded in the (Euclidean) plane R2 is called aplanar graph. The point set of a plane graph is compact. Therefore exactlyone face of G (a region of R2 \G) is unbounded. It is called the unboundedface of G (outer face, exterior face). A polygonal arc is a subset of R2 whichis the union of a finite number of straight line segments

{p+ λ(q − p) | 0 ≤ λ ≤ 1} (p, q ∈ R2)

and is homeomorphic to the closed unit interval [0, 1].

A ring graph is planar by construction. Thus an immediate consequenceof Theorem 10.1.38 is:

Corollary 10.1.39 Let G be a graph. If rank(G) = frank(G), then G isplanar.

Lemma 10.1.40 [320] If G is a planar graph, then G has a representationin the plane such that all edges are simple polygonal arcs.

436 Chapter 10

Proposition 10.1.41 [111, Proposition 4.2.5] If G is a 2-connected planargraph, then every face of G is bounded by a cycle.

Theorem 10.1.42 (Euler’s formula; see [111, Theorem 4.2.7]) Let G be aconnected planar graph with n vertices, q edges, and r faces. Then

n− q + r = 2

Proposition 10.1.43 (Kuratowski; see [48, p. 14, Theorem 17]) A graph isplanar if and only if it does not contain a subdivision of K5 or a subdivisionof K3,3 as a subgraph.

Definition 10.1.44 A planar graph is outerplanar if it can be embeddedin the plane so that all its vertices lie on some face; it is usual to choosethis face to be the exterior.

Definition 10.1.45 Two graphs H1 and H2 are called homeomorphic ifthere exists a graph G such that both H1 and H2 are subdivisions of G.

Theorem 10.1.46 [208, Theorem 11.10] A graph is outerplanar if and onlyif it has no subgraph homeomorphic to K4 or K2,3 except K4 \ {e}, where eis an edge.

Proposition 10.1.47 If G is outerplanar, then rank(G) = frank(G).

Proof. By Theorem 10.1.38(c) it suffices to prove that G satisfies PCPand G does not contain a subdivision of K4 as a subgraph. If G con-tains a subdivision H of K4 as a subgraph, then G contains a subgraph,namely H , homeomorphic to K4, but this is impossible by Theorem 10.1.46.To finish the proof we now show that G has the PCP property. Letc1 = {x1, x2, . . . , xm = x1} and c2 = {y1, y2, . . . , yn = y1} be two dis-tinct primitive cycles having at least one common edge. We may assumethat xi = yi for i = 1, 2 and x3 �= y3. Notice that y3 /∈ c1 becauseotherwise {y2, y3} = {x2, y3} is a chord of c1. We need only show that{x1, x2} = c1 ∩ c2, because this implies that c1 and c2 cannot have morethan one edge in common. Assume that {x1, x2} � c1 ∩ c2. Let r be theminimum integer such that yr belong to (c1 ∩ c2) \ {x1, x2}. Notice thatyr �= x3 because otherwise {x2, x3} is a chord of c2. Hence c1 together withthe path {x2 = y2, y3, . . . , yr} give a subgraph H of G which is a subdivisionof K2,3, a contradiction to Theorem 10.1.46. �

Proposition 10.1.48 Let G be a bipartite graph and let G1, . . . , Gr be theirblocks, then

K[G] � K[G1]⊗K · · · ⊗K K[Gr] and P (G) = (P (G1), . . . , P (Gr)).


Proof. Since G is bipartite it follows from Exercise 10.1.71. �

Definition 10.1.49 A graph G is called a complete intersection (over K)if the toric ideal P (G) of K[G] is a complete intersection.

The complete intersection property of the edge subring of a bipartitegraph was first studied in [114], and later in [379].

Proposition 10.1.50 [379] If G is a bipartite graph, then G is a completeintersection if and only if rank(G) = frank(G).

Proof. Let G1, . . . , Gr be the blocks of G. By Proposition 10.1.48 G is ac.i. if and only if Gi is a c.i. for all i. On the other hand by Exercise 10.1.72Gi is a c.i. if and only if rank(Gi) = frank(Gi). To finish the proof applyLemma 10.1.25. �

Corollary 10.1.51 If G is a bipartite graph and G is a complete intersec-tion, then G is a planar graph.


Corollary 10.1.52 If G is a bipartite graph, then G is a complete inter-section if and only if G is a ring graph.

Proof. By Proposition 10.1.50 G is a complete intersection if and only ifrank(G) = frank(G) and the result follows from Theorem 10.1.38. �

Theorem 10.1.53 [273] The edge subring K[G] of a bipartite graph G is acomplete intersection if and only if G is planar and any two primitive cyclesof G have at most one common edge.

Proof. ⇒) It follows from Lemma 10.1.35 and Corollary 10.1.51.⇐) See [273, Theorem 3.5]. �

Let us summarize some of the results obtained thus far. The followingimplications hold for any graph G:

outerplanar ⇒ ring graph ⇔ PCP+ contains no, subdivision of K4 ⇒ planar

rank = frank as a subgraph

If G is bipartite, then

ring graph ⇐⇒ complete intersection ⇐⇒ PCP + planar.

If G is not bipartite, the complete intersection property of P (G) is hardto characterize in combinatorial terms. In [29] a polynomial time algorithmis presented that checks whether a given graph is a complete intersection.

438 Chapter 10

Definition 10.1.54 Let g1 = tα1 − tβ1 , . . . , gr = tαr − tβr be homogeneousbinomials of degree at least 2 in the polynomial ring S = K[t1, . . . , tq]. Wesay that B = {g1, . . . , gr} is a foliation if the following conditions hold:

(a) tαi and tβi are square-free for all i,

(b) supp(tαi) ∩ supp(tβi) = ∅ for all i, and

(c) |(∪ji=1Ci) ∩Cj+1| = 1 for 2 ≤ j < r, where Ci = supp(gi) for all i.

Proposition 10.1.55 If B = {g1, . . . , gr} is a foliation, then the binomialideal I = (B) is a complete intersection.

Proof. By the constructive nature of B we can order the variables t1, . . . , tqso that the leading terms of g1, . . . , gr, with respect to the lex order ≺, arerelatively prime. Since

dim(S/I) = dim(S/(in≺(g1), . . . , in≺(gr))),

we obtain that the height of I is equal to r, as required.

Corollary 10.1.56 If G is a 2-connected bipartite graph with at least fourvertices, then the toric ideal P (G) is a complete intersection if and only ifit is generated by a foliation.

Exercises

10.1.57 [376] A clutter C is called of linear type if its edge ideal I(C) is oflinear type. If all connected components of a clutter are of linear type, thenthe clutter is of linear type.

10.1.58 [376] If C is a clutter of linear type, then so are all minors of C.

10.1.59 Let R = Q[x1, . . . , x7] and let I be the ideal of R generated byf1 = x1x2x3, f2 = x2x4x5, f3 = x5x6x7, f4 = x3x6x7. Use Macaulay2 toverify that the toric ideal of R[It] is minimally generated by the binomials

x3t3 − x5t4, x6x7t1 − x1x2t4, x6x7t2 − x2x4t3, x4x5t1 − x1x3t2, x4t1t3 − x1t2t4.

Prove that Theorem 10.1.15 does not extend to uniform square-free mono-mial ideals.

10.1.60 Let G be a graph. Then G is planar (resp. outerplanar) if andonly if each block of G is planar (resp. outerplanar).

10.1.61 Let G be a ring graph. Prove that G is outerplanar if and only ifG does not contain a subdivision of K2,3 as a subgraph.


10.1.62 Let G be a ring graph with n ≥ 3 vertices and q edges. If G is2-connected and has no triangles, then q ≤ 2(n− 2).

10.1.63 Let G be a graph with q edges and P (G) the toric ideal of the edgesubring K[G]. If f is a primitive binomial in P (G) prove:

deg(f) ≤{q if q is even, andq − 1 otherwise.

10.1.64 Let G be a bipartite graph with bipartition (X,Y ). Prove that thelargest degree of a binomial which is part of a minimal set of homogeneousgenerators of P (G) is less than or equal to min(|X |, |Y |).

10.1.65 If G is a connected graph, then G has a spanning tree T , that is,there exists a tree T which is a spanning subgraph of G.

10.1.66 [208] Prove that a graph G is bipartite if and only if every cycle insome cycle basis of G is even.

10.1.67 If G is a connected non-bipartite graph with n vertices, prove thatthere is a connected subgraph H of G with n vertices and n edges and witha unique cycle of odd length.

10.1.68 If G is a connected graph, prove that e is a bridge if and only if eis not on any cycle of G.

10.1.69 Let G be a connected graph and let f1, . . . , fq be the edge genera-tors of K[G]. If fr correspond to a bridge of G and

fa = fa11 · · · faqq = f b11 · · · f bqq = f b

for some a, b ∈ Nq such that supp(a) ∩ supp(b) = ∅ and ar > 0, prove thatar is an even integer.

10.1.70 The graph below is planar but not outerplanar and it satisfiesrank(G) = frank(G) = 3. The regions of this embedding are not boundedby primitive cycles. Thus, the converse of Proposition 10.1.47 fails.�

� ��

��

��

��

10.1.71 Let G be a graph that can be written as G = G1 ∪ G2, whereG1 and G2 are subgraphs with at most one vertex in common. If G1 isbipartite, then P (G) = (P (G1), P (G2)) and K[G] ∼= K[G1]⊗K K[G2].

440 Chapter 10

10.1.72 [379] Let G be a connected bipartite graph with n vertices and qedges. Prove that frank(G) = q − n+ 1 if and only if the toric ideal P (G)of K[G] is a complete intersection.

10.1.73 Let G be a bipartite graph. If G is a subdivision of K5 or a subdi-vision K3,3 prove that the binomials of P (G) that correspond to primitivecycles do not form a regular sequence.

10.1.74 Let G be a bipartite graph. Use Kuratowski’s theorem (see Theo-rem 10.1.43) to prove that if the toric ideal of K[G] is a complete intersec-tion, then G is a planar graph.

10.1.75 If G is a ring graph and H is an induced subgraph of G, then His a ring graph.

10.1.76 Prove that the complete graph K4 has the primitive cycle propertyand rank(G) �= frank(G). Thus, the converse of Lemma 10.1.35 fails.

10.2 Incidence matrices and circuits

Let G be a simple graph with vertex set V = {x1, . . . , xn} and with edgeset E = {f1, . . . , fq}, where every edge fi is an unordered pair of distinctvertices fi = {xij , xik}. The incidence matrix AG = (aij) associated to Gis the n× q matrix defined by

aij =

{1 if xi ∈ fj , and0 if xi /∈ fj .

Notice that each column of AG has exactly two 1’s and the rest of itsentries equal to zero. If fi = {xij , xik} define vi = eij + eik , where ei isthe ith canonical vector in Rn. Thus the columns of AG are precisely thevectors v1, . . . , vq regarded as column vectors. In what follows we denotethe set {v1, . . . , vq} by A.

Example 10.2.1 Consider a triangle G with vertices x1, x2, x3. The inci-dence matrix of G is:

AG =

⎛⎝ 1 0 11 1 00 1 1

⎞⎠ ,

with the vectors v1 = e1 + e2, v2 = e2 + e3, and v3 = e1 + e3 correspondingto the edges f1 = {x1, x2}, f2 = {x2, x3}, and f3 = {x1, x3}.

Proposition 10.2.2 If A is the incidence matrix of a bipartite graph G,then A is totally unimodular.


Proof. By induction on the number of columns of A. Let (V1, V2) be thebipartition of G and let B be a square submatrix of A. If a column of B hasat most one entry equal to 1, then by induction det(B) = 0,±1. Thus wemay assume that all the columns of B have two entries equal to 1. Hence∑

xi∈V1

Fi =∑xi∈V2

Fi = (1, . . . , 1),

where x1, . . . , xn are the vertices of G and F1, . . . , Fr are the rows of B.Therefore the rows of B are linearly dependent and det(B) = 0. �

Theorem 10.2.3 [187] Let G be a simple bipartite graph with n verticesand q edges and let A = (aij) be its incidence matrix. If e1, . . . , en are thefirst n unit vectors in Rn+1 and C is the matrix

C =

⎛⎜⎜⎜⎝a11 . . . a1q e1 · · · en...

......

an1 . . . anq1 . . . 1

⎞⎟⎟⎟⎠obtained from A by adjoining a row of 1’s and the column vectors e1, . . . , en,then C is totally unimodular.

Proof. Suppose that {1, . . . ,m} and {m + 1, . . . , n} is the bipartition ofthe graph G. Let C′ be the matrix obtained by deleting the last n − mcolumns from C. It suffices to show that C′ is totally unimodular. Firstone successively subtracts the rows 1, 2, . . . ,m from the row n+1. Then onereverses the sign in the rows m+1, . . . , n. These elementary row operationsproduce a new matrix C′′. The matrix C′′ is the incidence matrix of adirected graph, namely, consider G as a directed graph, and add one morevertex n+ 1, and add the edges (i, n+ 1) for i = 1, . . . ,m. The matrix C′′,being the incidence matrix of a directed graph, is totally unimodular; seeExercise 1.8.10. As the last m column vectors of C′′ are

e1 − en+1, . . . , em − en+1,

one can successively pivot on the first non-zero entry of ei − en+1 for i =1, . . . ,m and reverse the sign in the rows m + 1, . . . , n to obtain back thematrix C′. Here a pivot on the entry c′st means transforming column t ofC′′ into the sth unit vector by elementary row operations. Since pivotingpreserves total unimodularity [338, Lemma 2.2.20] one derives that C′ istotally unimodular, and hence so is C. In the next specific example wedisplay C′ and C′′. �

442 Chapter 10

Example 10.2.4 To illustrate the constructions in the proof above con-sider the complete bipartite graph G = K2,3. In this case the two Z-rowequivalent matrices C′ and C′′ are⎡⎢⎢⎢⎢⎢⎢⎣

1 1 1 0 0 0 1 00 0 0 1 1 1 0 11 0 0 1 0 0 0 00 1 0 0 1 0 0 00 0 1 0 0 1 0 01 1 1 1 1 1 0 0

⎤⎥⎥⎥⎥⎥⎥⎦ ,

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 1 0 0 0 1 00 0 0 1 1 1 0 1−1 0 0 −1 0 0 0 00 −1 0 0 −1 0 0 00 0 −1 0 0 −1 0 00 0 0 0 0 0 −1 −1

⎤⎥⎥⎥⎥⎥⎥⎦ ,

respectively. Note that pivoting in the last two columns of C′′ amounts toadding row 1 to row 6, and then adding row 2 to row 6.

Remark 10.2.5 Let G be a graph, let AG be the incidence matrix of G,and let B be a square submatrix of AG. In [201] it is shown that eitherdet(B) = 0 or det(B) = ±2k, for some integer k such that 0 ≤ k ≤ τ0, whereτ0 is the maximum number of vertex disjoint odd cycles in G. Moreover forany such value of k there exists a minor equal to ±2k.

The number of bipartite (resp. non-bipartite) connected components ofa graph G will be denoted by c0 (resp. c1). Thus c = c0 + c1 is the totalnumber of components of G.

Lemma 10.2.6 [201] If G is a graph with n vertices and AG is its incidencematrix, then rank(AG) = n− c0.

Proof. Let G1, . . . , Gc be the connected components of G, and ni thenumber of vertices of Gi. After permuting the vertices we may assume thatAG is a “diagonal” matrix

AG = diag(AG1 , . . . , AGc),

where AGi is the incidence matrix of Gi. By Proposition 10.1.20 the rankof AGi is equal to ni − 1 if Gi is bipartite, and is equal to ni otherwise.Hence the rank of AG is equal to n− c0. �

The incidence matrix of G is denoted by AG or simply by A. Regardingthe vi’s as column vectors, we define the matrix B as:

B =

(v1 · · · vq1 · · · 1

).

Observe that rank(B) = rank(AG) because the last row of B is a linearcombination of the first n rows. Recall that Δr(B) denotes the gcd of allnon-zero r-minors of B.

For simplicity we keep the notation introduced above throughout therest of this section. We shall assume, if need be, that the graph G has noisolated vertices.


Lemma 10.2.7 If G is a unicyclic graph with a unique odd cycle and r isthe rank of B, then Δr(B) = 1.

Proof. If G is an odd cycle of length n, then the matrix B′ obtained fromB by deleting any of its first n rows has determinant ±1. The matrix B′ isin fact totally unimodular by Theorem 10.2.3. We proceed by induction. IfG is not an odd cycle, then G has a vertex xj of degree 1. Thus the jth rowof A has exactly one entry equal to 1, say ajk = 1. Consider the matrix Cobtained from B by deleting the jth row and the kth column. Thus usinginduction Δr−1(C) = 1, which gives Δr(B) = 1. �

Lemma 10.2.8 Let G be a graph with connected components G0, . . . , Gm.If G0 is non-bipartite, then there exists a subgraph H of G with connectedcomponents H0, . . . , Hm such that H0 is a spanning unicyclic subgraph ofG0 with a unique odd cycle and Hi is a spanning tree of Gi for all i ≥ 2.

Proof. The existence of H2, . . . , Hm is clear because any connected graphhas a spanning tree. Take a spanning tree T0 of G0. For each line e of G0 notin T0, the graph T0∪{e} has exactly one cycle. According to Remark 10.1.17the set Z(T ) of such cycles form a basis for the cycle space of G0. Since agraph is bipartite if and only if every cycle in some cycle basis is even (seeExercise 10.1.66), there is an edge e of G0 such that T0 ∪ {e} is a unicyclicconnected graph with a unique odd cycle. �

Proposition 10.2.9 If G is a graph with exactly one non-bipartite con-nected component and r = rank(B), then Δr(B) = 1.

Proof. Let G0, . . . , Gm be the components of G, with G0 non-bipartite,and let H be as in Lemma 10.2.8. If A′ is the incidence matrix of H , thenusing that G and H have the same number of non-bipartite components onehas rank(A) = rank(A′). Therefore Δr(B) = 1 by Lemma 10.2.7. �

Proposition 10.2.10 Let G be a non-bipartite graph and let A be its inci-dence matrix. If A has rank r, then Δr(A) = 2Δr(B).

Proof. We set A = {v1, . . . , vq} and A′ = {(v1, 1), . . . , (vq, 1)}. It sufficesto prove that there is an exact sequence of groups

0 −→ T (Zn+1/ZA′)ϕ−→ T (Zn/ZA) ψ−→ Z2 −→ 0.

For β = (a1, . . . , an) ∈ Zn and b ∈ Z, define

ϕ(β, b) = β and ψ(β) = a1 + · · ·+ an.

It is not hard to verify that ϕ is injective and that im(ϕ) = ker(ψ).

444 Chapter 10

To show that ψ is onto consider a non-bipartite component G1 of G. Ifx1, . . . , xs are the vertices of G1 and v′1, v

′2, . . . , v

′m are the columns of the

incidence matrix of G1, then Zs/(v′1, v′2, . . . , v

′m) is a finite group. It follows

that for any 1 ≤ j ≤ s, the element ej is in the torsion subgroup of Zn/ZA,hence ψ(ej) = 1, as required. �

Corollary 10.2.11 [201] If G is a graph and c0 (resp. c1) is the numberof bipartite (resp. non-bipartite) components, then

Zn/ZA � Zn−r × Zc12 = Zc0 × Zc12 ,

where r = n− c0 is the rank of AG.

Proof. Let G1, . . . , Gc be the components of G and let Ai be the set ofvectors in A corresponding to the edges of Gi. Since the vectors in Ai canbe regarded as vector in Zηi , where ηi is the number of vertices in Gi, thereis a canonical isomorphism

Zn/ZA � Zη1/ZA1 ⊕ · · · ⊕ Zηc/ZAc .To finish the proof apply Propositions 10.2.9 and 10.2.10 to derive

Zηi/ZAi �{

Z2 if Gi is non-bipartite,Z if Gi is bipartite.

In the second isomorphism we are using the fact that incidence matrices ofbipartite graphs are totally unimodular; see Proposition 10.2.2. �

Corollary 10.2.12 If c1 is the number of non-bipartite components of Gand r is the rank of B, then

Δr(B) =

{2c1−1 if c1 ≥ 1,1 if c1 = 0.


Definition 10.2.13 A square matrix U , with entries in a commutative ringR, is called unimodular if det(U) is a unit of R.

Theorem 10.2.14 [201] If G is a graph with n vertices and AG is theincidence matrix of G, then there are unimodular integral matrices U , U ′

such that

S = UAGU′ =

(D 00 0

),

where D = diag(1, 1, . . . , 1, 2, 2, . . . , 2), n− c is the number of 1’s and c1 isthe number of 2’s.

Proof. It follows using Corollary 10.2.11 and the fundamental structuretheorem for finitely generated abelian groups. See Theorem 1.3.16. �

The matrix S is called the Smith normal form of AG and the numbersin the diagonal matrix D are the invariant factors of AG.


Circuits of the kernel of an incidence matrix We give a geometricdescription of the elementary integral vectors or circuits of the kernel of theincidence matrix of a graph (see Section 10.3 for the multigraph case).

Notation For α = (α1, . . . , αq) ∈ Nq and f1, . . . , fq in a commutative ringwith identity we set fα = fα1

1 · · · fαqq . The support of fα is defined as the

set supp(fα) = {fi |αi �= 0}.Let G be a graph on the vertex set V with edge set E and incidence

matrix A. We set N = ker(A), the kernel of A in Qq. Note that a vectorα ∈ N ∩ Zq determine the subgraph Gα of G having vertex set

Vα = {x ∈ V |x divide fα+}

and edge set Eα = {{x, y} ∈ E |xy ∈ supp(fα+) ∪ supp(fα−)}, wheref1, . . . , fq are the square-free monomials corresponding to the edges of G,that is, the edge generators of G.

The next result can be generalized to multigraphs; see Section 10.3.

Proposition 10.2.15 [417] Let G be a graph with incidence matrix A andlet N be the kernel of A in Qq. Then a vector 0 �= α ∈ Zq ∩ N is anelementary vector of N if and only if

(i) Gα is an even cycle, or

(ii) Gα is a connected graph consisting of two odd cycles with at most onecommon vertex joined by a path.

Proof. Assume α ∈ Zq ∩N is an elementary vector of N so that Gα is notan even cycle. From the equation fα+ = fα− we obtain a walk in Gα

w = {x0, x1, . . . , xn = x0, xn+1, . . . , x�},

so that x1, . . . , x�−1 are distinct vertices, n is odd and x� is in {x1, . . . , x�−2}.The walk w can be chosen so that xixi+1 ∈ supp(fα+) if i is even andxixi+1 ∈ supp(fα−) otherwise. We claim that x� is not in {x1, . . . , xn−1}.Assume x� = xm for some 1 ≤ m ≤ n− 1. If m and �− n are odd then themonomial walk

w1 = {x0, x1, . . . , xm, x�−1, x�−2, . . . , xn = x0}

yields a relation f δ = fγ with supp(δ − γ) properly contained in supp(α),which is a contradiction. If m is odd and �−m is even then we can considerthe walk

w1 = {xn, xn+1, . . . , x� = xm, xm+1, . . . , xn}and derive a contradiction; the remaining cases are treated similarly. Wemay now assume x� = xm for n ≤ m ≤ �− 2. The cycle

C1 = {xm, . . . , x� = xm}

446 Chapter 10

must be odd, otherwise C1 would give a relation f δ = fγ with supp(δ − γ)a proper subset of supp(α). The walk w gives a relation f δ = fγ with

supp(f δ) ⊂ supp(fα+) and supp(fγ) ⊂ supp(fα−).

Hence supp(δ − γ) is a subset of supp(α), and since the support of α isminimal one has the equality supp(δ − γ) = supp(α). Therefore

Vα = {x1, . . . , x�}

and Gα is as required. The converse follows from Corollary 10.1.16. �

Definition 10.2.16 A circuit of a graph G is either an even cycle or asubgraph consisting of two odd cycles with at most one common vertexjoined by a path.

A family of connected graphs is called matroidal if given any graph G,the subgraphs of G isomorphic to a member of the family are the circuits ofa matroid on the edge set of G. Thus Proposition 10.2.15 implies that thefamily of circuits is matroidal as was shown in [387].

Proposition 10.2.17 If A is the incidence matrix of a connected simplegraph G, then the circuits of the vector matroidM [A] consists of even cycles,two edge disjoint odd cycles meeting at exactly one vertex, and two vertexdisjoint odd cycles joined by an arbitrary path.

Proof. It follows from Proposition 10.2.15 (see Section 1.9). �

Circuits of a graph and Grobner bases Let G be a graph. We studythe toric ideal P (G) of the edge subringK[G] from the view point of Grobnerbases. For general results on Grobner bases of toric ideals a standard refer-ence is [400, Chapter 4].

Proposition 10.2.18 If G is a graph with incidence matrix A and P isthe toric ideal of K[G], then

P = ({tα+ − tα− |α ∈ Zq and Aα = 0}).

Proof. The result is a consequence of Corollary 8.2.20. �

Corollary 10.2.19 Let G be a graph with incidence matrix A. If λ = (λi)is a circuit of N = ker(A), then |λi| ≤ 2 for all i.

Proof. Let λ be a circuit ofN . SinceGλ is an even cycle or two edge disjointcycles joined by a path we obtain a circuit α with supp(α) = supp(λ) andso that |αi| ≤ 2 for all i. Because any two elementary vectors with the samesupport are dependent we obtain λ = α. �


Theorem 10.2.20 Let G be a connected graph with q edges and let P bethe toric ideal of K[G]. Then the total degree of a polynomial in any reducedGrobner basis of P is less than or equal to 2q dimZe(G).

Proof. We set N = {α ∈ Qq |Aα = 0}, where A is the incidence matrixof G. Assume F is the reduced Grobner basis of P and take f = f(t) ∈ F .Using Corollary 8.2.18 and Corollary 10.2.18, together with the fact thatnormalized reduced Grobner bases are uniquely determined, we obtain

f(t) = tα − tβ .

Notice that supp(α) ∩ supp(β) = ∅, for otherwise take i in the intersectionand write f(t) = ti(t

a−tb). Since ta−tb reduces with respect to F it followsthat f(t) reduces with respect to F \ {f}, which is impossible. We can nowwrite f(t) = tα+− tα− for some α ∈ N ∩Zq . By Theorem 1.9.6 we can write

α = a1λ1 + · · ·+ arλr

for some ai ∈ Q+, and for some circuit λi each in harmony with α such thatsupp(λi) ⊂ supp(α) for all i and r ≤ dimN . Hence

α+ = a1(λ1)+ + · · ·+ ar(λr)+ and α− = a1(λ1)− + · · ·+ ar(λr)−.

If ai > 1 for some i then αi+ and αi− are componentwise larger than λi+ andλi−, respectively; note that this is not possible by the previous argument.Hence 0 < ai ≤ 1. Using Corollary 10.2.19 it now follows that the entries ofα satisfy |αi| ≤ 2 dimZe(G) for all i, which gives the asserted inequality. �

Exercises

10.2.21 Let G be a cycle of length n and AG its incidence matrix. Prove

det(AG) =

{±2 if n is odd0 if n is even.

10.2.22 Let U be a square matrix with entries in a commutative ring R.Prove that U is unimodular if and only if U is invertible.

10.2.23 Let G be a graph with incidence matrix A and N = ker(A) ⊂ Qq.Give an example to show that the set of circuits of N does not determinethe presentation ideal of K[G].

10.2.24 Let G be a graph with incidence matrix A and N = ker(A) thekernel of A in Qq. Let λ1, . . . , λs be the circuits of N . Define the elementaryzonotope of N as EN = [0, λ1] + · · ·+ [0, λs]. Prove that EN is a polytope.

448 Chapter 10

10.2.25 If N is the kernel of the incidence matrix of a graph G and P isthe toric ideal of K[G], prove that {tα+ − tα− |α ∈ EN ∩ Zq} is a universalGrobner basis for P .

10.2.26 (I. Gitler) Prove that the complete graph K5 is not a circuit andconsists of two edge disjoint odd cycles of length 5.

10.2.27 (K. Truemper) Consider the matrices

V =

⎛⎜⎜⎝

1 0 0 10 1 0 10 0 1 11 1 1 1

⎞⎟⎟⎠ , A =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 0 1 00 0 1 00 1 0 00 1 0 01 0 0 01 0 0 01 0 0 10 1 0 10 0 1 1

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

, B =

(A

1 1 1 1

).

Note det(V ) = −2. Prove that A is totally unimodular but B is not. ThusTheorem 10.2.3 does not extend to k-hypergraphs with k > 2.

10.3 The integral closure of an edge subring

Let G be a multigraph. In this section we unfold a construction for theintegral closure of K[G] and study the circuits of the toric ideal of K[G].

Let G a multigraph with vertex set X = {x1, . . . , xn}, i.e., G is obtainedfrom a simple graph by allowing multiple edges and loops. Thus the edgesof G have the form {xi, xj}. If e = {xi, xj} is an edge of G, its characteristicvector is given by ve = ei+ ej, where ei is the ith unit vector in Rn. Noticethat if e is a loop, i.e., if i = j, then ve = 2ei. The incidence matrix ofG, denoted by A, is the matrix whose column vectors are the characteristicvectors of the edges and loops of G.

Let v1, . . . , vq be the characteristic vectors of the edges and loops of Gand let R = K[x1, . . . , xn] be a polynomial ring over a field K. In whatfollows A denotes the set {v1, . . . , vq} and F denotes the set {f1, . . . , fq}of all monomials xixj in R such that {xi, xj} is an edge of G. The edgesubring of G is given by

K[G] = K[f1, . . . , fq] ⊂ R.

We may assume that fi = xvi for i = 1, . . . , q. One has the followingdescription of the integral closure of K[G]:

K[G] = K[{xa| a ∈ R+A ∩ ZA}],

where R+A is the cone in Rn generated by A and ZA is the subgroup ofZn generated by A. See Theorem 9.1.1.


Proposition 10.3.1 If G is a bipartite graph, then K[G] is normal andCohen–Macaulay.

Proof. By Proposition 10.2.2 the incidence matrix of a bipartite graph istotally unimodular. Then, by Corollary 1.6.8, we have NA = Zn ∩ R+A.Thus K[G] = K[G], that is K[G] is normal. Hence, by Theorem 9.1.6, K[G]is Cohen–Macaulay. �

Definition 10.3.2 A bow-tie of G is an induced submultigraph w of Gconsisting of two odd cycles with at most one common vertex

Z1 = {z0, z1, . . . , zr = z0} and Z2 = {zs, zs+1, . . . , zt = zs}

joined by a path {zr, . . . , zs}. In this case we set Mw = z1 · · · zrzs+1 · · · zt.We regard a loop of G as an odd cycle.

Remark 10.3.3 A bow-tie w, as above, always defines an even closed walk:

w = {z0, z1, . . . , zr, zr+1, . . . , zs, zs+1, . . . , zt = zs, zs−1, . . . , zr = z0}.

Lemma 10.3.4 If w is a bow-tie of a multigraph G, then Mw is in theintegral closure of K[G].

Proof. Let w be a bow-tie. With the notation above. If fi = zi−1zi, then

z21 · · · z2r = f1 · · · fr, z2s · · · z2t−1 = fs+1 · · · ft,

which together with the identities

Mw =∏i odd

fi∏

i evenr < i ≤ s

f−1i and M2

w = f1 · · · frfs+1 · · · ft

gives Mw ∈ K[G]. �

Remark 10.3.5 Putting the two equations above together one obtains∏i oddr < i ≤ s

f2i

∏i oddi /∈ (r, s]

fi =∏

i evenr < i ≤ s

f2i

∏i eveni /∈ (r, s]

fi. (10.1)

If r = s, that is Z1 and Z2 intersect at a vertex, then∏i odd

fi =∏

i even

fi (10.2)

If s = r + 1, that is Z1 and Z2 are joined by an edge {zr, zs}, then∏i odd

fi = f2s

∏i eveni �= s

fi. (10.3)

450 Chapter 10

Example 10.3.6 A bow-tie formed with a triangle and a pentagon joinedby a path of length 2:

��

�

��

�

�

��

��

��

!!!!

""""

z0 = z5 z7 = z10

z6

z8z1

z4 z9

z2

z3

Example 10.3.7 A bow-tie w formed with two loops (odd cycles of length1) joined by a path of length 2:

��

x1 x3

x2 ��

We can write w as an even closed walk w = {x1, x1, x2, x3, x3, x2, x1}. Inthis example Mw = x1x3.

The cone generated by A is given by R+A = R+v1 + · · · + R+vq. Thiscone will be studied in detail in Section 10.7 when G is a graph.

Definition 10.3.8 A multigraph having just one cycle is called unicyclic.

Theorem 10.3.9 [384] If G is a connected multigraph, then

K[G] = K[{f1, . . . , fq} ∪ {Mw|w is a bow-tie}].

Proof. We set B = {f1, . . . , fq} ∪ {Mw|w is a bow-tie}. According toTheorem 9.1.1 and Corollary 1.1.27 one has:

K[G] = K[{xa| a ∈ ZA ∩ R+A}] = K[{xa| a ∈ ZA ∩Q+A}].

Hence, by Lemma 10.3.4, it suffices to show that K[G] ⊂ K[B]. Let xa be aminimal (irreducible) generator of K[G]. Then a ∈ Q+A and a ∈ ZA. ByCaratheodory’s Theorem (see Theorem 1.1.18), and the irreducibility of xa,we can write

a = λ1v1 + · · ·+ λrvr, (10.4)

where v1, . . . , vr are linearly independent vectors in A and λi ∈ Q+ ∩ (0, 1)for i = 1, . . . , r. Permitting an abuse of notation, we will also denote theedges of G by v1, . . . , vq, and refer to vi as an edge of G.

Let Ga be the multigraph whose edges are v1, . . . , vr and let

(Ga)1, . . . , (Ga)s

be the components of Ga. For each i consider the graph Hi obtained from(Ga)i by removing loops and by replacing multiple edges by single edges.


EachHi is either a tree or a unicyclic graph with a unique odd cycle becausev1, . . . , vr are linearly independent. If Hi is a unicyclic graph with a uniqueodd cycle, then (Ga)i has no loops or multiple edges because v1, . . . , vr arelinearly independent and the rank of the incidence matrix of (Ga)i is equalto the number of vertices of (Ga)i. Thus in this case (Ga)i = Hi. If Hi

is a tree, then (Ga)i is a tree with at most one loop. This follows usingthat the rank of the incidence matrix of (Ga)i is equal to |V ((Ga)i)| − 1 if(Ga)i is a tree. Since 0 < λi < 1, by Eq. (10.4), the multigraph Ga hasno vertices of degree 1. Therefore the connected components of Ga are oddcycles. Hence, as 0 < λi < 1 for all i and the components of Ga are oddcycles, we get that the entries of a are in {0, 1}. The multigraph Ga hasat least two components, i.e., two odd cycles Z1 and Z2, because a ∈ ZA.Thus, by the irreducibility of xa, the components of Ga are Z1 and Z2. Asthe graph is connected, there is a bow-tie w whose odd cycles are Z1 andZ2 and such that xa =Mw. So x

a ∈ B as required. This proof was adaptedfrom [263]. �

There are two special types of bow-ties that can be omitted in the de-scription of the integral closure of K[G].

Proposition 10.3.10 Let w be a bow-tie of G with two odd cycles Z1, Z2

such that either Z1 and Z2 meet at exactly one vertex or Z1 and Z2 aredisjoint and are connected by an edge, then Mw is in K[G].


Definition 10.3.11 A multigraph G has the odd cycle condition if everytwo vertex disjoint odd cycles can be joined by at least one edge.

The odd cycle condition comes from graph theory [175]. This notion wasused to study the normality of edge subrings [242, 383, 384] and the circuitsof a graph [358, 417].

Corollary 10.3.12 Let G be a connected multigraph. Then K[G] is normalif and only if G satisfies the odd cycle condition.

Proof. It follows from Theorem 10.3.9 and Proposition 10.3.10. �

Normality and the circuits of a multigraph Let S = K[t1, . . . , tq]be a polynomial ring over a field K. Recall that the toric ideal of K[G],denoted by P (G), is the kernel of the epimorphism of K-algebras:

S = K[t1, . . . , tq] −→ K[G] induced by ti �→ xvi .

Lemma 10.3.13 If f = ta−tb is a circuit of P (G), then f has a square-freeterm or f has nonsquare-free terms and maxi{ai} = maxi{bi} = 2.

452 Chapter 10

Proof. If G is a simple graph, the result was shown in Corollary 10.2.19.In the case of a multigraph it follows using an identical argument as the onegiven in the proof of this corollary. �

Theorem 10.3.14 Let G be a multigraph. If K[G] is normal, then P (G)is generated by circuits with a square-free term.

Proof. It follows at once applying Lemma 10.3.13 and Corollary 9.7.4. �

The converse of this result is not true (see Exercise 10.3.25).

Definition 10.3.15 A sub-multigraph H of G is called a circuit of G if Hhas one of the following forms:

(a) H is an even cycle.

(b) H consists of two odd cycles intersecting in exactly one vertex; a loopis regarded as an odd cycle of length 1.

(c) H consists of two vertex disjoint odd cycles joined by a path.

The circuits of G are in one-to-one correspondence with the circuits ofP (G) as we now explain. (See Section 10.2 for a detailed discussion whenG is a graph.) Any circuit H of G can be regarded as an even closed walk

w = {w0, w1, . . . , wr, w0},

where r is even, w0, w1, . . . , wr are the vertices of H (we allow repetitions)and {wi, wi+1} is an edge (we allow loops) of G for all i. Then the binomialtw = t1t3 · · · tr−1 − t2t4 · · · tr is in P (G), where fi = wi−1wi and ti maps tofi for all i.

Remark 10.3.16 The circuits of P (G) with a square-free term correspondto the following types of circuits of G: (a) Even cycles, (b) two odd cyclesintersecting in exactly one vertex, (c) two vertex disjoint odd cycles joinedby an edge. Thus if K[G] is normal, by Theorem 10.3.14 we obtain a precisegraph theoretical description of a generating set of circuits for P (G).

Toric ideals of edge subrings of oriented graphs were studied in [184].In this case the toric ideal is also generated by circuits and the circuitscorrespond to the cycles of the graph. In [182] they completely characterizethe graphs that are complete intersections for all orientation.

Exercises

10.3.17 Prove that K[G] is normal for every connected graph G with sixvertices, where K is a field.


10.3.18 Let G be a graph and let G1, . . . , Gr be its connected components.Then K[G] is normal if and only if K[Gi] is normal for each i.

10.3.19 Let G be a graph and let G1, . . . , Gr be the connected componentsof G. If K is a field, prove that K[G] is normal if and only if Gi satisfiesthe odd cycle condition for all i.

10.3.20 Let G1 and G2 be two finite sets of monomials in disjoint sets ofindeterminates. Then K[G1 ∪G2] � K[G1]⊗K K[G2] and

K[G1 ∪G2] � K[G1]⊗K K[G2] � K[G1] K[G2],

is the subring generated by the generators of K[G1] and K[G2].

10.3.21 Prove that the smallest connected graph G such that K[G] is non-normal is the graph consisting of two disjoint triangles joined by a 2-path(see Example 10.5.1).

10.3.22 Le G be a graph and let w be one of the following bow-ties of Gwith cycles Z1 and Z2

•

��

Z1

��

•

•

��

��

�

Z2

��• •

•

��

��

�

Z1

•

Z2• •

��

��

��

�

•

�� •

prove that Mw is in K[G].

10.3.23 If G is a graph and K[G] is normal, then P (G) is generated by thebinomials that correspond to the circuits of the incidence matrix of G.

10.3.24 Let F = {x1x2, x2x3, x3x4, x1x4, x21, x22, x23, x24} ⊂ K[x]. Find allthe bow-ties of the multigraph defined by F . Use Theorem 10.3.9 to showthat K[F ] = K[F ][x1x3, x2x4].

10.3.25 (A. Thoma) Let G be the graph whose edge subring is

K[G] = K[x1x2, x2x3, x1x3, x1x4, x5x4, x5x6, x6x4, x5x7, x6x7].

Prove that the toric ideal is a complete intersection generated by circuitsbut that K[G] is not normal.

454 Chapter 10

10.4 Ehrhart rings of edge polytopes

Let G be a graph on the vertex set V = {x1, . . . , xn}. The edge polytope ofG is the lattice polytope

P = conv(A) ⊂ Rn,

where A = {v1, . . . , vq} is the set of vectors in Nn of the form ei + ej suchthat xi is adjacent to xj . Let R = K[x1, . . . , xn] be a polynomial ring overa field K. Recall that the Ehrhart ring of P is

A(P) = K[{xαti|α ∈ Zn ∩ iP}] ⊂ R[t],

the edge subring of G is the K-subalgebra

K[G] = K[xv1 , . . . , xvq ] ⊂ R,

the polytopal subring is given by

K[P ] = K[xαt|α ∈ Zn ∩ P ] ⊂ R[t],

and the incidence matrix of G, denoted by A, is the matrix with columnvector v1, . . . , vq. In what follows B denotes the matrix

B =

(v1 · · · vq1 · · · 1

).

As was earlier observed rank(B) = rank(A). We shall keep these notationsthroughout the rest of this section. If need be, we assume that the graph Ghas no isolated vertices.

Proposition 10.4.1 dim(P) = n− c0− 1, where c0 denotes the number ofbipartite components of G.

Proof. By Lemma 10.2.6 and Proposition 9.3.1 one has the equalities

dim K[G] = rank(A) = n− c0 and dim K[P ] = dim(P) + 1.

Thanks to Lemma 9.3.16 one has that K[P ] is equal to K[xv1t, . . . , xvq t].Thus K[G] � K[P ] ⊂ A(P). Consequently P has dimension n− c0 − 1. �

Theorem 10.4.2 K[P ] = A(P) if and only if the graph G has at most onenon-bipartite connected component.

Proof. ⇒) If G has two non-bipartite components G1, G2, take

Z1 = {x0, x1, . . . , xr = x0}, Z2 = {xr+1, xr+2 . . . , xr+s = xr+1}


two odd cycles of lengths r, s− 1 in G1, G2, respectively. Setting

β = e1 + · · ·+ er + er+1 + · · ·+ er+s−1,

note the equality

β =m

2

[(er + e1m

+

r−1∑i=1

ei + ei+1

m

)+

(er+s−1 + er+1

m+

r+s−2∑i=r+1

ei + ei+1

m

)]

where m = r + s − 1. Hence β ∈ Zn ∩ pP , where p = m/2, that is, xβtp

belongs to A(P) = K[P ]. Hence (β, p) is in the field of fractions of thepolytopal subring K[P ]. There are integers λ1, . . . , λq such that

e1 + e2 + · · ·+ er+s−1 + pen+1 = λ1(v1, 1) + · · ·+ λq(vq, 1).

Since Z1 and Z2 are in different connected components we obtain an equalitye1 + · · · + er =

∑j λjvj , where the sum is taken over all j such that the

vector vj corresponds to an edge of G1. Setting 1 = (1, . . . , 1) and takinginner products in the last equality one has r = 〈e1 + · · ·+ er,1〉 = 2

∑j λj ,

a contradiction because r is odd.⇐) By Lemma 9.3.16 K[P ] = K[xv1t, . . . , xvq t]. If G is bipartite then,

by Proposition 9.3.30, K[P ] = A(P). If G is not bipartite, the equalityK[P ] = A(P) follows from Theorem 9.3.25 and Proposition 10.2.9. �

Corollary 10.4.3 [242, p. 423] If G is connected, then K[P ] = A(P).

Theorem 10.4.4 The multiplicities of A(P) and K[G] are related by

e(A(P)) = vol(P)d! ={

2c1−1e(K[G]) if G is non-bipartite,e(K[G]) if G is bipartite ,

where d = dim(P) and c1 is the number of non-bipartite components of G.

Proof. By Theorem 9.3.25 one has e(A(P)) = Δr(B)e(K[P ]), where r isthe rank of A. From Corollary 10.2.11 one derives

Δr(A) = |T (Zn/ZA)| = 2c1 .

Thus the result follows readily from Proposition 10.2.10 and using the factthat A is totally unimodular if c1 = 0. �

Lemma 10.4.5 Let G be a connected non-bipartite graph with n verticesand let B = {v1, . . . , vn} be a set of linearly independent columns of A. IfK[G] is normal and 0 �= β ∈ R+B ⊂ Rn, then β can be written as

β = μ1v′1 + · · ·+ μnv

′n (μi ≥ 0),

such that v′1, . . . , v′n are linearly independent columns of A defining a span-

ning subgraph of G which is unicyclic connected and non-bipartite.

456 Chapter 10

Proof. For simplicity we identify the edges of G with the columns of A.There are nonnegative real numbers λ1, . . . , λn such that

β = λ1v1 + · · ·+ λnvn (λi ≥ 0). (10.5)

Consider the subgraph H of G defined by the edges v1, . . . , vn and thesubgraph H ′ of G defined by the set of edges {vi|λi > 0}. Using thatv1, . . . , vn are linearly independent one rapidly derives that H is a spanningsubgraph of G whose connected components are unicyclic and non-bipartite(cf. [400, Lemma 9.5]). Let H1, . . . , Hs be the components of H and Zi theunique odd cycle of Hi, for i = 1, . . . , s. First assume that H ′ has at mostone odd cycle, say Z1. For each cycle Zi with i ≥ 2 there is vji in Zi withλji = 0. Construct a graph K by removing vj1 , . . . , vjs from H and thenadding an edge connecting Z1 with Zi for each i ≥ 2, such constructionis possible thanks to Corollary 10.3.12. This new graph K is a spanningconnected subgraph of G, is non-bipartite and has a unique cycle Z1. Thusin this case Eq. (10.5) is itself the required expression for β. Next we assumethat H ′ has at least two odd cycles Z1, Z2. Let v be an edge joining Z1

with Z2. In this case, using Remark 10.3.5, it is seen that we can rewrite

β = μ1v′1 + · · ·+ μnv

′n (μi ≥ 0),

where v′1, . . . , v′n are linearly independent edges such that the graph H ′′

determined by the set {v′i|μi > 0} has one cycle less than H ′. Thus, byinduction, the result follows. �

For connected graphs the next result was shown in [242].

Theorem 10.4.6 If K[G] is normal, then P has a unimodular coveringwith support in the vertices of P if and only if G has at most one non-bipartite component.

Proof. ⇒) By Theorem 9.3.29K[Ft] = A(P). Hence using Theorem 10.4.4we obtain that the number of non-bipartite components of G is at most 1.⇐) Let G1, . . . , Gc be the connected components of the graph G and let

ηi be the number of vertices of Gi.Assume thatG1 is non-bipartite andG2, . . . , Gc are bipartite. We denote

the set of columns of the incidence matrix of G by A = {v1, . . . , vq}. Toconstruct a unimodular covering of P consider any subgraph H of G withconnected components H1, . . . , Hc such that Hi is a spanning tree of Gifor i > 1 and H1 is a unicyclic connected spanning subgraph of G1 with aunique odd cycle. The edge polytope ΔH of H is a simplex of dimensiond = n−c. From Proposition 10.2.9 and 1.2.21 ΔH is a simplex of normalizedvolume equal to 1.

We claim that the family of all simplices ΔH is a unimodular coveringof P . Let β ∈ P . By Caratheodory’s Theorem (see Theorem 1.1.18) one


can write (after permutation of edges) β =∑m

i=1 λivi with λi ≥ 0, wherem = n−(c−1) and v1, . . . , vm are linearly independent. Note that v1, . . . , vmdefine a subgraph H of G. For each i there is a subgraph Hi of Gi such thatH = H1∪· · ·∪Hc. Since v1, . . . , vm are linearly independent, Hi is bipartitefor i > 1 and hence the number of edges of Hi is at most ηi− 1 for all i > 1.On the other hand H1 has at most η1 edges. Altogether the number of edgesof H is at most η1 +(η2− 1)+ · · ·+(ηc − 1), but since the number of edgesof H is exactly this number one concludes that the number of edges of Hi

(resp. H1) is ηi − 1 (resp. η1) for i > 1 (resp. i = 1). Hence H2, . . . , Hc aretrees. Since G1 is connected and non-bipartite applying Lemma 10.4.5 wecan rewrite β such that H1 is a spanning subgraph of G1 which is connectedunicyclic and non-bipartite. Therefore β ∈ ΔH , as required.

If G1 is bipartite the proof follows using similar arguments as above.Indeed is suffices to make G1 = ∅ and proceed as above. �

Exercise

10.4.7 Let G be an arbitrary graph and let P be its edge polytope. Provethat K[P ] = A(P) if and only if G has at most one non-bipartite connectedcomponent and this component satisfies the odd cycle condition.

10.5 Integral closure of Rees algebras

Let G be a graph with vertex set X = {x1, . . . , xn}, let R = K[x1, . . . , xn]be a polynomial ring over a field K and let I(G) be the edge ideal of G. Asusual we denote the Rees algebra of I(G) by R[I(G)t].

Example 10.5.1 (Hochster’s configuration) Let G be the graph

�x1 �x4 �x5�x6�x7

��

��

x2

x3

��

��

Note that f = (x1x2x3)(x5x6x7) satisfies f ∈ I(G)3 \ I(G)3.

Definition 10.5.2 A Hochster configuration of order k of G consists of twoodd cycles C2r+1 and C2s+1 satisfying the following conditions:

(i) C2r+1 ∩NG(C2s+1) = ∅ and k = r + s+ 1.

(ii) No chord of either C2r+1 or C2s+1 is an edge of G.

458 Chapter 10

Proposition 10.5.3 If G has a Hochster configuration of order k, thenI(G)k �= I(G)k.

Proof. We set I = I(G). The monomial obtained by multiplying all the

variables in the two cycles of the configuration is an element of Ik \ Ik.Indeed, let

δ = z1 · · · z2r+1 and γ = w1 · · ·w2s+1

be the monomials corresponding to the two cycles in the configuration. Bydefinition, we have r + s+ 1 = k. Therefore δγ ∈ Ik−1 \ Ik. On the otherhand δ2 ∈ I2r+1 and γ2 ∈ I2s+1, thus (δγ)2 ∈ I2(r+s+1) = I2k. This shows

that δγ ∈ Ik. �

Definition 10.5.4 The cone C(G), over the graphG, is obtained by addinga new vertex t to G and joining every vertex of G to t.

Proposition 10.5.5 R[I(G)t] is isomorphic to K[C(G)].

Proof. Let R[I(G)t] = K[{x1, . . . , xn, tfi| 1 ≤ i ≤ q}] be the Rees algebraof I(G) = (f1, . . . , fq), where f1, . . . , fq are the monomials corresponding tothe edges of G. Let

K[C(G)] = K[{txi, fj | 1 ≤ i ≤ n, 1 ≤ j ≤ q}]

be the monomial subring of the cone. R[I(G)t] and K[C(G)] are integraldomains of the same dimension by Corollary 8.2.21 and Proposition 10.1.20.It follows that there is an isomorphism

ϕ : R[I(G)t] −→ K[C(G)], induced by ϕ(xi) = txi and ϕ(tfi) = fi.

That is R[I(G)t]ϕ� K[C(G)]. �

Corollary 10.5.6 Let G be a connected graph and I = I(G) its edge ideal.Then K[G] is normal if and only if the Rees algebra R[It] is normal.

Proof. ⇒) Let C(G) be the cone over G. By Proposition 10.5.5 R[I(G)t]is isomorphic to K[C(G)]. Let Mw be a bow-tie of C(G) with edge disjointcycles Z1 and Z2 joined by a path P . It is enough to verify that Mw is inK[C(G)]. If t /∈ Z1 ∪ Z2 ∪ P , then w is a bow-tie of G and Mw ∈ K[G].Assume t ∈ Z1 ∪ Z2, say t ∈ Z1. If Z1 ∩ Z2 �= ∅, then Mw ∈ K[C(G)]. Onthe other hand if Z1 ∩Z2 = ∅, then Mw ∈ K[C(G)] because in this case Z1

and Z2 are joined by the edge {t, z}, where z is any vertex in Z2. It remainsto consider the case t /∈ Z1 ∪ Z2 and t ∈ P , since G is connected there is apath in G joining Z1 with Z2. Therefore Mw = Mw1 for some bow-tie w1

of G and Mw ∈ K[G].⇐) Conversely if R[I(G)t] is normal, then by Proposition 4.3.42 we

obtain that K[G] is normal. �


Example 10.5.7 Let F = {x1x2, x3x4x5, x1x3, x2x4, x2x5, x1x5} and letI = (F ). Using Normaliz [68] we get that R[It] is normal but K[F ] is not.

Proposition 10.5.8 Let G be a graph. Then R[I(G)t] is normal if andonly if either (i) G is bipartite, or (ii) exactly one of the components, sayG1, is non-bipartite and K[G1] is a normal domain.

Proof. It follows adapting the proof of Corollary 10.5.6. �

Corollary 10.5.9 [383] The edge ideal of a graph G is normal if and onlyif G admits no Hochster configurations.

Remark 10.5.10 (a) Let G be a graph consisting of two disjoint triangles.Note that G itself is a Hochster configuration but G is not contained in abow-tie, since the two triangles cannot be joined by a path.

(b) If two disjoint odd cycles Z1, Z2 of a graph G are in the sameconnected component, then the two cycles form a Hochster configuration ifand only if Z1, Z2 cannot be connected by an edge of G and Z1, Z2 haveno chords in G

Integral closure of Rees algebras Consider the endomorphism ϕ ofthe field K(x1, . . . , xn, t) defined by xi �→ xit, t �→ t−2. If I = I(G) is theedge ideal of G, then by Proposition 10.5.5 it induces an isomorphism

R[It]ϕ�−→ K[C(G)], xi �→ txi, xixjt �→ xixj .

We thank Wolmer Vasconcelos for showing us how to get the descriptionof the integral closure of R[It] given below. First we describe the integralclosure of K[C(G)].

If Z1 = {x1, x2, . . . , xr = x1} and Z2 = {z1, z2, . . . , zs = z1} are twoedge disjoint odd cycles in C(G). Then one has

Mw = x1 · · ·xrz1 · · · zs

for some bow-tie w of C(G), this follows readily because C(G) is a connectedgraph. In particularMw is in the integral closure of K[C(G)]. Observe thatif t occurs in Z1 or Z2, then Mw ∈ K[C(G)] because in this case eitherZ1 and Z2 meet at a point, or Z1 and Z2 are joined by an edge in C(G);see Proposition 10.3.10. Thus in order to compute the integral closure ofK[C(G)] the only bow-ties that matter are those defining the set:

B = {Mw|w is a bow-tie in C(G) such that t /∈ supp(Mw)}.

Therefore by Theorem 10.3.9 we have proved:

Proposition 10.5.11 K[C(G)] = K[C(G)][B].

460 Chapter 10

To obtain a description of the integral closure of the Rees algebra of Inote that using ϕ together with the equality

(tx1)Mw = (tz1)

r+12∏i=1

(x2i−1x2i)

s−12∏i=1

(z2iz2i+1),

where xr+1 = x1, we see that Mw is mapped back in the Rees algebra inthe element

Mwtr+s2 = x1x2 · · ·xrz1z2 · · · zst

r+s2 .

If B′ is the set of all monomials of this form, then one obtains:

Proposition 10.5.12 R[It] = R[It][B′].

Exercises

10.5.13 Let G be a graph. Prove that I(G)2 is integrally closed.

10.5.14 If G is a graph consisting of two disjoint triangles, then K[G] isnormal and the Rees algebra of I(G) is not normal.

10.5.15 Let G be a graph. If H is the whisker graph of G or the cone of Gand I(G) is normal, then I(H) is normal.

10.5.16 Let G be a graph and let m = (r + s)/2 be the least positiveinteger such that G has two vertex disjoint odd cycles of lengths r and s. IfI = I(G) is the edge ideal of G, prove that Ii is integrally closed for i < m.

10.5.17 If G is any non-discrete graph with n vertices and C(G) its cone,prove that the edge subring K[C(G)] has dimension n+ 1.

10.5.18 Let G be a simple graph with vertex set X = {x1, . . . , xn} and letI ⊂ R = K[X ] be its edge ideal. If ϕ, ψ are the maps of K-algebras definedby the diagram

R[t1, . . . , tq]ϕ ��

ψ

��

R[It]

K[C(G)]

tiϕ ��

ψ

��

Tfi

fi

xiϕ ��

ψ

��

xi

xti

where C(G) is the cone of G and f1, . . . , fq are the edge generators, thenker(ϕ) = ker(ψ).

10.5.19 Let R = K[x1, x2, x3, y1, y2, y3, y4] and I = I1J3 + I3J1, where Ir(resp. Jr) denote the ideal of R generated by the square-free monomialsof degree r in the xi variables (resp. yi variables). If z = x1x2x3y

21y2y3y4,

then z2 ∈ I4 and z ∈ I2 \ I2.


10.5.20 Let G be a graph. Explain the differences between a circuit of Gand a Hochster configuration of G.

10.5.21 Let G be a bipartite graph and I(G) its edge ideal. Prove thatthe toric ideal of the Rees algebra of I(G) has a universal Grobner basisconsisting of square-free binomials.

10.6 Edge subrings of complete graphs

In this section we treat two special types of edge subrings, attached tocomplete graphs, and compute their Hilbert series and a-invariant.

Edge subrings of bipartite graphs One result that simplifies the studyof an edge subring associated to a complete bipartite graph is the fact thatits toric ideal is a determinantal prime ideal. Let us begin with a classicalformula for determinantal ideals.

Theorem 10.6.1 [73, Theorem 2.5] Let T = (tij) be an m × n matrix ofindeterminates over a field K and Ir(T ) the ideal generated by the r-minorsof T . If m ≤ n and 1 ≤ r ≤ m+ 1, then

height(Ir(T )) = (m− r + 1)(n− r + 1).

Proposition 10.6.2 Let K be a field and let Km,n be the complete bipartitegraph. Then, the toric ideal P of K[Km,n] is generated by the 2× 2 minorsof a generic m× n matrix.

Proof. Let V1 = {x1, . . . , xm} and V2 = {y1, . . . , yn} be the bipartition ofKm,n and let

S = K[{tij| 1 ≤ i ≤ m, 1 ≤ j ≤ n}]be a polynomial ring over a field K in the indeterminates tij and considerthe graded homomorphism of K-algebras

ϕ : S −→ K[Km,n] = K[xiyj ’s], ϕ(tij) = xiyj .

We claim that the kernel of ϕ is generated by the 2 × 2 minors of ageneric m × n matrix. We set T = (tij). It is clear that the ideal I2(T ),generated by the 2×2 minors of T , is contained in P = ker(ϕ). On the otherhand, since the graph is bipartite, one has that the dimension of K[Km,n]is equal to m+ n− 1; see Corollary 10.1.21. Therefore

height(P ) = (mn)− (m+ n− 1) = (m− 1)(n− 1) = height(I2(T )),

the last equality uses Theorem 10.6.1. Since they are both prime ideals, wehave I2(T ) = P . �

By Proposition 10.3.1 the ring K[Km,n] is normal and Cohen–Macaulay.The Gorensteiness of K[Km,n] has been dealt with in great detail in [71].

462 Chapter 10

Proposition 10.6.3 [418] Let S/P be the presentation of the edge subringK[Km,n]. If n ≥ m, then the Hilbert series of S/P is given by

F (S/P, z) =

m−1∑i=0

(m− 1

i

)(n− 1

i

)zi

(1 − z)n+m−1.

Corollary 10.6.4 If n ≥ m, then the a-invariant and the multiplicity ofK[Km,n] are given by:

a(K[Km,n]) = −n and e(K[Km,n]) =(m+ n− 2

m− 1

).

Proof. It follows using Remark 5.1.7. �

Proposition 10.6.5 Let G be a spanning connected subgraph of Km,n. Ifn ≥ m, then the a-invariant of K[G] satisfies

a(K[G]) ≤ a(K[Km,n]) = −n.

Proof. From Corollary 10.6.4 one has a(K[Km,n]) = −n. Hence one mayproceed as in the proof of Proposition 12.4.13 to rapidly derive the assertedinequality. �

An application of Dedekind–Mertens formula If R is a commutativering and f = f(t) ∈ R[t] is a polynomial, say

f = a0 + · · ·+ amtm,

the content of f is the R-ideal (a0, . . . , am). It is denoted by c(f). Givenanother polynomial g, the Gaussian ideal of f and g is the R-ideal

c(fg).

This ideal bears a close relationship to the ideal c(f)c(g), one aspect ofwhich is expressed in the classical lemma of Gauss: If R is a principal idealdomain, then

c(fg) = c(f)c(g).

In general, these two ideals are different but one aspect of their relationshipis given by a formula due to Dedekind–Mertens (see [126] and [334]):

c(fg)c(g)m = c(f)c(g)m+1. (10.6)

We consider the ideal c(fg) in the case when f , g are generic polynomials.It turns out that some aspects of the theory of Cohen–Macaulay rings showup very naturally when we closely examine c(fg).


One path to our analysis, and its applications to Noether normalizations,starts by multiplying both sides of (10.6) by c(f)m; we get

c(fg)[c(f)c(g)]m = c(f)c(g)[c(f)c(g)]m. (10.7)

It will be shown that this last formula is sharp in terms of the exponentm = deg(f).

To make this connection, we recall the notion of a reduction of an ideal.Let R be a ring and I an ideal. A reduction of I is an ideal J ⊂ I suchthat, for some nonnegative integer r, the equality

Ir+1 = JIr

holds. The smallest such integer is the reduction number rJ (I) of I relativeto J , and the reduction number r(I) of I is the smallest reduction numberamong all reductions J of I.

Thus (10.7) says that J = c(fg) is a reduction for I = c(f)c(g), andthat the reduction number is at most min{deg(f), deg(g)}.

The following result solves the question of Noether normalizations formonomial subrings of complete bipartite graphs.

Theorem 10.6.6 [126] Let R = K[x0, . . . , xm, y0, . . . , yn] be a polynomialring over a field K and let hq =

∑i+j=q xiyj. Then

A = K[h0, h1, . . . , hm+n] ↪→ S = K[{xiyj| 0 ≤ i ≤ m, 0 ≤ j ≤ n}]

is a Noether normalization of S.

Proof. Let R[t] be a polynomial ring in a new variable t and f, g ∈ R[t]the generic polynomials

f = f(t) =

m∑i=0

xiti and g = g(t) =

n∑j=0

yjtj .

Set I = c(f)c(g) and J = c(fg). Note that J is a reduction of I by Eq. (10.7)and more precisely JIm = Im+1. Therefore

R[Jt] ↪→ R[It]

is an integral extension; indeed R[It] is generated as an R[Jt]-module by afinite set of elements of t-degree at most m. On the other hand, as I and Jare generated by homogeneous polynomials of the same degree, one has anintegral embedding:

R[Jt]⊗R R/m = K[h0, . . . , hm+n] ↪→ R[It]⊗R/m = K[xiyj’s],

where m = (x0 . . . , xm, y0, . . . , yn). To complete the proof note that thedimension of K[xiyj’s] is equal to n+m+ 1, by Corollary 10.1.21. �

464 Chapter 10

Theorem 10.6.7 [97]Let R = K[x0, . . . , xm, y0, . . . , yn] be a polynomialring over a field K and let f, g ∈ R[t] be the generic polynomials

f =

m∑i=0

xiti and g =

n∑j=0

yjtj

with n ≥ m. If I = c(f)c(g) and J = c(fg), then rJ (I) = m.

Proof. By Theorem 10.6.6 there is a Noether normalization:

A = K[h0, h1, . . . , hm+n] ↪→ S = K[xiyj ’s],

where hq =∑i+j=q xiyj . Set I = (f1, . . . , fs). We note that the ideal

I = (xiyj’s) is the edge ideal associated to a complete bipartite graph Gwhich is the join of two discrete graphs, one with m+1 vertices and anotherwith n+1 vertices. Since K[G] is Cohen–Macaulay (see Corollary 10.3.12),using Proposition 3.1.27 we get

K[G] = Afβ1 ⊕ · · · ⊕Afβk , βi ∈ Ns.

By Corollary 10.6.4 the a-invariant of K[G] is −n− 1, we can compute theHilbert series of K[G] using the decomposition above to get that

max1≤i≤k

{deg(fβi)} = m,

where the degree is taken with respect to the normalized grading.On the other hand assume r = rJ(I) and JI

r = Ir+1, since we alreadyknow the inequality r ≤ m. It suffices to prove r ≥ m. Note

K[G] =∑α∈I

Afα,

where I = {α |α = (v1, . . . , vs) ∈ Ns and |α| =∑s

i=1 vi ≤ r} . If r < m,this rapidly leads to a contradiction because, by Exercise 10.6.14, there isJ ⊂ I such that K[G] can be written as

K[G] =⊕α∈J

Afα. �

Edge subrings of complete graphs Let R = K[x1, . . . , xn] be a ring ofpolynomials over a field K and let Kn be the complete graph on the vertexset X = {x1, . . . , xn}. For i < j, we set fij = xixj . Let

K[Kn] = K[{fij| 1 ≤ i < j ≤ n}]

be the K-subring of R spanned by the fij . Consider the homomorphism

ψ : An = K[{tij| 1 ≤ i < j ≤ n}]→ K[Kn], induced by tij → fij .

The ideal Pn = ker(ψ) is the presentation ideal or toric ideal of K[Kn].


Proposition 10.6.8 [418] If the terms in An are ordered lexicographicallyby t12 ≺ · · · ≺ t1n < t23 ≺ · · · ≺ t2n ≺ · · · ≺ t(n−1)n, then the set

B = {tijtk� − ti�tjk, tiktj� − ti�tjk| 1 ≤ i < j < k < � ≤ n}.

is a minimal generating set for Pn and B is a reduced Grobner basis for Pn.

Example 10.6.9 If K[{tij| 1 ≤ i < j ≤ 5}] has the lex ordering

t45 ≺ t34 ≺ t35 ≺ t23 ≺ t24 ≺ t25 ≺ t12 ≺ t13 ≺ t14 ≺ t15.

Then the reduced Grobner basis for the toric ideal P of K[K5] is equal to:

t35t24 − t45t23, t34t25 − t45t23, t45t23t13 − t34t35t12,t25t13 − t35t12, t35t14 − t45t13, t23t14 − t34t12,t34t15 − t45t13, t23t15 − t35t12, t24t15 − t45t12.t24t13 − t34t12 t25t14 − t45t12.

Theorem 10.6.10 [418] Let Kn be the complete graph on n vertices andAn/Pn be the presentation of K[Kn]. If n ≥ 3, then the Hilbert series Fn(z)of An/Pn satisfies:

(1 − z)nFn(z) = 1 +n(n− 3)

2z +

�n2 �∑

m=2

(n

2m

)zm

Corollary 10.6.11 The multiplicity e(K[Kn]), of the ring K[Kn], is equalto 2n−1 − n.

Theorem 10.6.12 Let G be a connected graph with n vertices and q edges.Then the following are sharp upper bounds for the multiplicity of K[G]:

e(K[G]) ≤

⎧⎨⎩2n−1 − n, if G is non-bipartite,(m+r−2m−1

), if G is a spanning subgraph of Km,r.

Proof. It follows from Corollary 10.6.4 and Corollary 10.6.11. �

It is easy to produce examples where those bounds are very lose. Theextreme cases being the unicyclic connected non-bipartite graphs and thetrees, because in both cases the multiplicity of K[G] is equal to 1. Inour case adding edges to G increases the volume of the edge polytope (seeLemma 9.3.16); hence the multiplicity is an increasing function of the edges.For additional information on the multiplicity of edge subrings of bipartiteplanar graphs, see [189] and the references therein.

466 Chapter 10

Exercises

10.6.13 Let R[t] be a polynomial ring over a ring R. If f = f(t) andg = g(t) are two polynomials in R[t], prove the Dedekind–Mertens formula

c(fg)c(g)m = c(f)c(g)m+1 (m = deg(f)).

10.6.14 Let M be a graded free module over a polynomial ring A suchthat M = ⊕ri=1Afi, where fi is homogeneous for all i. If M is generated byhomogeneous elements g1, . . . , gs, then M = ⊕rj=1Agij , for some i1, . . . , ir.

10.6.15 Let A = K[x1, . . . , xm] and R = K[y1, . . . , yn] be two polynomialrings over a field K, prove that the Segre product

(A0 ⊗K R0)⊕ (A1 ⊗K R1)⊕ · · · ⊂ A⊗K R

is isomorphic to S = K[xiyj ’s] ⊂ K[xi’s, yj ’s].

10.6.16 Let R = K[x1, . . . , xn] be a polynomial ring over a field K andm = (x1, . . . , xn). Prove that the toric ideal of the Rees algebra R[mt] is theideal generated by the 2-minors of the following matrix of indeterminates:

X =

(x1 x2 · · · xnt1 t2 · · · tn

).

10.6.17 Let R = K[x1, . . . , xn] be a polynomial ring over a field K andI = (x1, . . . , xn)

k. Set s = −�n/k� + n, and write n = qk + r, for someq, r ∈ N such that 0 ≤ k < r. By a “counting degrees” argument prove:

xk−11 · · ·xk−1

n ∈ Is if r = 0,

xk−11 · · ·xk−1

q xr−1q+1x

k−1q+2 · · ·xk−1

n ∈ Is if r > 0.

10.6.18 Let R = K[x1, . . . , xn] be a polynomial ring over a field K andm = (x1, . . . , xn). If I = mk and J = (xk1 , . . . , x

kn), then J is a reduction of

I and the reduction number of I relative to J is given by

rJ (I) = −⌈nk

⌉+ n, where �x� is the ceiling of x.

10.6.19 Let S be a standard algebra over an infinite field K and

A = K[h1, . . . , hd] ↪→ S

a Noether normalization of S with hi ∈ S1 for all i. If b1, . . . , bt is a minimalset of homogeneous generators of S as an A-module, prove

(h1, . . . , hd)Sr = Sr+1,

where r = maxi{deg(bi)}, and that r ≥ 0 is the minimum integer whereequality occurs. The reduction number of S, denote by r(S), is the minimumr taken over all Noether normalizations. See [413, Chapter 9] for a study ofseveral degrees of complexity associated to a graded module.


10.6.20 Let S be a standard algebra over an infinite field K and a(S) itsa-invariant. If S is Cohen–Macaulay, then r(S) = a(S) + dim(S).

10.6.21 Let φ : B → S be a graded epimorphism of standard K-algebras,where K is an infinite field. If dim(B) = dim(S) and

A = K[h1, . . . , hd]ı↪→ B

is a Noether normalization of B with hi ∈ S1 for all i, prove that S isintegral over A′ = K[φ(h1), . . . , φ(hd)] and that the composite

A = K[h1, . . . , hd]ı↪→ B

φ−→ S

is a Noether normalization of S.

10.6.22 Let φ : B → S be a graded epimorphism of standard K-algebras,where K is an infinite field. If dim(B) = dim(S), prove r(B) ≥ r(S).

10.7 Edge cones of graphs

In this section we give explicit combinatorial descriptions of the edge coneof a graph. In Chapter 11 these descriptions will be used to compute the a-invariant and the canonical module of an edge subring. The main tools usedto show these descriptions are linear algebra (Farkas’s Lemma, incidencematrices of graphs, Caratheodory’s Theorem), graph theory and polyhedralgeometry (finite basis theorem, facet structure of polyhedra).

First we fix the notation that will be used throughout this section andadapt the terminology and notation of Chapter 1 for the case of edge cones.

Let G be a simple graph with vertex set V = {x1, . . . , xn} and edge setE = {f1, . . . , fq}. Every edge fi is an unordered pair of distinct verticesfi = {xij , xik}. If fi = {xij , xik} define vi = eij + eik , where ei is the ithunit vector in Rn. The incidence matrix of G, denoted by A, is the matrixof size n× q whose columns are precisely the vectors v1, . . . , vq.

We set AG (or simply A if G is understood) equal to the set {v1, . . . , vq}of row vectors of the transpose of the incidence matrix of G. Since virepresents an edge of G, sometimes vi is called an edge or an edge vector .The edge cone of G, denoted by R+A, is the cone generated by A. Noticethat R+A �= (0) if G is not a discrete graph. By Corollary 10.2.11 one has

n− c0(G) = rank (A) = dim R+A,

where c0(G) is the number of bipartite components of G.

Lemma 10.7.1 If xi is not an isolated vertex of the graph G, then the setF = Hei ∩ R+A is a proper face of the edge cone.

468 Chapter 10

Proof. Note F �= ∅ because 0 ∈ F , and R+A ⊂ H+ei . Since xi is not an

isolated vertex R+A �⊂ Hei . �

Let A be an independent set of vertices of G. The supporting hyperplaneof the edge cone defined by ∑

xi∈A

xi =∑

xi∈N(A)

xi

will be denoted by HA , where N(A) is the neighbor set of A. The sum overan empty set is defined to be zero.

Lemma 10.7.2 If A is an independent set of vertices of the graph G andF = R+A∩HA, then either F is a proper face of the edge cone or F = R+A.

Proof. It suffices to prove R+A ⊂ H−A . Take an edge {xj , x�} of G. If

{xj , x�} ∩ A �= ∅, then ej + e� is in HA, else ej + e� is in H−A . �

Definition 10.7.3 The support of a vector β = (βi) ∈ Rn is defined as

supp(β) = {βi |βi �= 0}.

Lemma 10.7.4 [419] Let G1, . . . , Gr be the components of G. If G1 is atree with at least two vertices and G2, . . . , Gr are unicyclic non-bipartitegraphs, then ker(A�

G) = (β) for some β ∈ Rn with supp(β) = {1,−1} suchthat V (G1) = {vi ∈ V |βi = ±1}.

Proof. If G is a tree, using induction on the number of vertices it is nothard to show that ker(A�) = (β), where supp(β) = {1,−1}. On the otherhand if G = Gi for some i ≥ 2, then A� is non-singular and ker(A�) = (0).The general case follows readily by writing A� as a “diagonal” matrix

A� = diag(A�G1, . . . , A�

Gm),

where AGi is the incidence matrix of the graph Gi. Observe the equalityV (G1) = {xi ∈ V |βi = ±1}. �

To determine whether HA defines a facet of R+A consider the subgraphL = L1 ∪ L2, where L1 is the subgraph of G with vertex set and edge set

V (L1) = A ∪N(A) and E(L1) = {z ∈ E| z ∩ A �= ∅}

respectively, and L2 = G[S] is the subgraph of G induced by S = V \V (L1).The vectors in A ∩HA correspond precisely to the edges of L.

Remark 10.7.5 Let F be a facet of R+A defined by the hyperplane Ha.If the rank of A is n, then it is not hard to see that Ha is generated by alinearly independent subset of A. See Proposition 1.1.23.


Theorem 10.7.6 [419] If rank(A) = n, then F is a facet of R+A if andonly if either

(a) F = Hei ∩ R+A for some i, where all the connected components ofG \ {xi} are non-bipartite graphs, or

(b) F = HA ∩ R+A for some independent set A ⊂ V such that L1 isa connected bipartite graph, and the connected components of L2 arenon-bipartite graphs.

Proof. ⇒) Assume that F is a facet of R+A. By Remark 10.7.5 we mayassume that there is a ∈ Rn such that

(i) F = R+A ∩Ha, 〈a, vi〉 ≤ 0 ∀ i ∈ {1, . . . , q}, and(ii) v1, . . . , vn−1 are linearly independent vectors in Ha.

Consider the subgraph D of G whose edges correspond to v1, . . . , vn−1 andits vertex set is the union of the vertices in the edges of D. Let Γ bethe transpose of the incidence matrix of D, and D1, . . . , Dr the connectedcomponents of D. Without loss of generality one may assume

Γ =

⎡⎢⎢⎢⎣Γ1 0 . . . 00 Γ2 . . . 0...

.... . .

...0 0 . . . Γr

⎤⎥⎥⎥⎦ , rank(Γ) = n− 1,

where Γi is the transpose of the incidence matrix of Di. We set ni and qiequal to the number of vertices and edges of Di, respectively. Let Θ be thematrix with rows v1, . . . , vn−1. We consider two cases.

Case (I): Assume Θ has a zero column. Hence it has exactly one zerocolumn and D has vertex set V (D) = V \ {xi}, for some i. This impliesthat

∑ri=1 ni =

∑ri=1 qi = n− 1. Since ni ≤ qi for all i, one obtains ni = qi

for all i. Note that c0(D) = 0, by Lemma 10.2.6. Hence D1, . . . , Dr areunicyclic non-bipartite graphs. As any component of G \ {xi} contains acomponent of D, we are in case (a).

Case (II): Assume that all the columns of Θ are non-zero, that is, D hasvertex set equal to V . Note that in this case

n =

r∑i=1

ni = 1 +

r∑i=1

qi. (10.8)

Since rank (Γ) = n−c0(D) = n−1, we get that D has exactly one bipartitecomponent, say D1. Hence n1 − 1 ≤ q1 and ni ≤ qi for all i ≥ 2, whichtogether with Eq. (10.8) yields q1 = n1 − 1 and ni = qi for all i ≥ 2.Altogether D1 is a tree and D2, . . . , Dr are unicyclic non-bipartite graphs.

470 Chapter 10

By Lemma 10.7.4, ker(Γ) = (β) and supp(β) ⊂ {1,−1} for some β ∈ Rn.We may harmlessly assume a = β, because Ha = Hβ and a ∈ (β). Set

A = {xi ∈ V | ai = 1} and B = {xi ∈ V | ai = −1}.

Note that A �= ∅, because D has no isolated vertices and its vertex set is V .We will show that A is an independent set in G and B = N(A), where theneighbor set of A is taken w.r.t G.

On the contrary if {xi, xj} is an edge of G for some xi, xj in A, thenvk = ei + ej and by (i) we get 〈a, vk〉 ≤ 0, which is impossible because〈a, vk〉 = 2. This proves that A is an independent set.

Next we show N(A) = B. If xi ∈ N(A), then vk = ei + ej for some xjin A, using (i) we obtain 〈a, vk〉 = ai + 1 ≤ 0 and ai = −1, hence xi ∈ B.Conversely if xi ∈ B, since D has no isolated vertices, there is 1 ≤ k ≤ n−1so that vk = ei + ej , for some j, by (ii) we obtain 〈a, vk〉 = −1 + aj = 0,which shows that xj ∈ A and xi ∈ N(A).

From the proof of Lemma 10.7.4 we obtain V (D1) = A∪N(A). Thereforev1, . . . , vn−1 are in HA and Ha = HA. Observe that L1 (see notation beforeRemark 10.7.5) cannot contain odd cycles because A is independent andevery edge of L1 contains a vertex in A. Hence L1 is bipartite, that L1 isconnected follows from the fact that D1 is a spanning tree of L1. Note thatevery connected component of L2 is non-bipartite because it contains someDi, with i ≥ 2.⇐) Assume that F is as in (a). Since the vectors in A∩Hei correspond

precisely to the edges of G \ {xi} one obtains that F is a facet of the edgecone. If F is as in (b), a similar argument shows that F is also a facet. �

Theorem 10.7.7 If G is a connected graph and F is a facet of the edgecone of G, then either

(a) F = R+A ∩ {x ∈ Rn |xi = 0} for some 1 ≤ i ≤ n, or(b) F = R+A ∩HA for some independent set A of G.

Proof. By Theorem 10.7.6 one may assume that G is bipartite and n ≥ 3.Notice that F = R+B, where B = {vi ∈ A | 〈vi, a〉 = 0}. Since dim(F ) isequal to n− 2, one may assume that v1, . . . , vn−2 are linearly independentvectors in F . There is 0 �= a ∈ Rn such that

(i) F = R+A ∩Ha, 〈a, vi〉 = 0 ∀ i ∈ {1, 2, . . . , n− 2}, and(ii) 〈a, vi〉 ≤ 0 for i ∈ {1, 2, . . . , q}.

Let D be the subgraph of G whose edges correspond to v1, . . . , vn−2

and its vertex set is the union of the vertices in the edges of D. Settingk = |V (D)|, by Corollary 10.2.11 one has n− 2 = rank (AD) = k − c0(D),where AD is the incidence matrix of D and c0(D) is the number of bipartite


components of D. Thus 0 ≤ n − k = 2 − c0(D). This shows that eitherc0(D) = 1 and k = n− 1 or c0(D) = 2 and k = n.

Case (I): Assume c0(D) = 1 and k = n− 1. Set V (D) = {x1, . . . , xn−1}.As D is a tree with n− 2 edges and 〈vi, a〉 = 0 for i = 1, . . . , n− 2, applyingLemma 10.7.4, one may assume a = (a1, . . . , an−1, an), where ai = ±1 for1 ≤ i ≤ n− 1.

Subcase(I.a): Assume that an = 0. Set A = {xi ∈ V | ai = 1} andB = {xi ∈ V | ai = −1}. Note that ∅ �= A ⊂ V (D), because D is a tree withat least two vertices. We will show that A is an independent set of G andB = NG(A). If A is not an independent set of G, there is an edge {xi, xj}of G for some xi, xj in A. Thus vk = ei + ej and by (ii) we get 〈a, vk〉 ≤ 0,which is impossible because 〈a, vk〉 = 2. Next we will show NG(A) = B.If xi ∈ N(A), then vk = ei + ej for some xj in A, using (ii) we obtain〈a, vk〉 = ai +1 ≤ 0 and ai = −1, hence xi ∈ B. Conversely if xi ∈ B, sinceD has no isolated vertices, there is 1 ≤ k ≤ n− 2 so that vk = ei + ej, forsome j, by (i) we obtain 〈a, vk〉 = −1 + aj = 0, which shows that xj ∈ Aand xi ∈ N(A). Therefore Ha = HA and F is as in (b).

Subcase(I.b): Assume an > 0. In this case consider a′ = −en. Then:(c) 〈vi, a′〉 = 0 for i = 1, . . . , n− 2.

(d) If vj ∈ R(v1, . . . , vn−2) ⇒ 〈vj , a′〉 = 〈vj , a〉 = 0.

(e) If vj /∈ R(v1, . . . , vn−2), then 〈vj , a′〉 = −1 and 〈vj , a〉 < 0.

To prove (e) assume vj /∈ R(v1, . . . , vn−2), then vj = ek + en, becauseotherwise the “edge” vj added to the treeD form a graph with a unique evencycle, a contradiction. Thus 〈vj , a′〉 = −1. On the other hand 〈vj , a〉 < 0,because if 〈vj , a〉 = 0, then Ha contains the linearly independent vectorsv1, . . . , vn−2, vj , a contradiction to dim(F ) = n− 2.

Thus given 1 ≤ j ≤ q, then 〈vj , a〉 = 0 if and only if 〈vj , a′〉 = 0. Note

R+A ∩Ha′ = R+B′, (10.9)

where B′ = {vi ∈ A | 〈vi, a′〉 = 0}. Hence using F = R+B and Eq. (10.9) weget F = R+A∩Ha′ , as required. The case an < 0 can be treated similarly.

Case (II): Assume c0(D) = 2 and k = n. Let D1 and D2 be the twocomponents of D and set V1 = V (D1) and V2 = V (D2).

Using Lemma 10.7.4 we can relabel the vertices ofD to write a = rb+sc,where 0 �= r ≥ s ≥ 0 are rational numbers and b = (b1, . . . , bm, 0, . . . , 0),c = (0, . . . , 0, cm+1, . . . , cn), where V1 = {x1, . . . , xm}, bi = ±1 for i ≤ m,and ci = ±1 for i > m. Set a′ = b. Note the following:

(f) 〈vi, a′〉 = 0 for i = 1, . . . , n− 2.

(g) If vj ∈ R(v1, . . . , vn−2) ⇒ 〈vj , a′〉 = 〈vj , a〉 = 0.

(h) If vj /∈ R(v1, . . . , vn−2), then 〈vj , a〉 < 0 and 〈vj , a′〉 = −1 .

472 Chapter 10

Let us prove assertion (h). Assume vj /∈ R(v1, . . . , vn−2). Since dim(F )is equal to = n− 2, by (ii) we have 〈vj , a〉 < 0. Observe that if an “edge”vk has vertices in V1 (resp. V2), then 〈vk, a〉 = 0. Indeed if we add the edgevk to the tree D1 (resp. D2) we get a graph with a unique even cycle andthis implies that v1, . . . , vn−2, vk are linearly dependent, that is, 〈vk, a〉 = 0.Thus vj = ei + e� for some xi ∈ V1 and x� ∈ V2. From the inequality

〈vj , a〉 = r〈vj , b〉+ s〈vj , c〉 = rbi + sc� < 0

we obtain bi = −1 = 〈vj , a′〉, as required. In particular note that Ha′ is asupporting hyperplane of R+A because R+A ⊂ H−

a′ . We may now proceedas in case (I) to obtain the equality F = R+A ∩ Ha = R+A ∩ Ha′ . Usingthat D1 has no isolated vertices, one may proceed as in case (I) to prove thatHa′ = HA for the independent set of vertices given by A = {xi | bi = 1}. �

Theorem 10.7.8 If G is a connected graph, then

R+A =

( ⋂A∈F

H−A

)⋂(n⋂i=1

H+ei

),

where F is the family of all the independent sets of vertices of G and H+ei

is the closed halfspace {x ∈ Rn |xi ≥ 0}.

Proof. Set C = R+A. From Theorem 10.7.7 and Corollary 1.1.32 we get

C = aff(C) ∩( ⋂

A∈FH−

A

)⋂(n⋂i=1

H+ei

).

We may assume that G is bipartite, otherwise aff(C) = Rn and there isnothing to prove. If (V1, V2) is the bipartition of G, then aff(R+A) is thehyperplane

∑xi∈V1

xi =∑xi∈V2

xi, because dim(R+A) = n − 1. Since

H−V1∩H−

V2= HV1 = aff(C), we obtain the required equality. �

Studying the bipartite case For connected bipartite graphs we willpresent more precise results about the irreducible representations of edgecones. Let G be a connected bipartite graph with bipartition (V1, V2). Toavoid repetitions we continue using the notation introduced at the beginningof the section.

Proposition 10.7.9 If A is an independent set of G such that A �= Vi fori = 1, 2, then F = R+A∩HA is a proper face of the edge cone.

Proof. Assume N(A) = V2. Take any xi ∈ V1 \A and any xj ∈ V2 adjacentto xi, then ei + ej /∈ HA. This means that R+A � HA. Thus we mayassume N(A) �= Vi for i = 1, 2.


Case (I): N(A) ∩ Vi �= ∅ for i = 1, 2. If the vertices in N(A) ∩ Vi fori = 1, 2 are only adjacent to vertices in A, then pick vertices xi ∈ N(A)∩Viand note that there is no path between x1 and x2, a contradiction. Thusthere must be a vector in the edge cone which is not in HA.

Case (II): A � V1. If the vertices in N(A) are only adjacent to vertices inA. Then a vertex in A cannot be joined by a path to a vertex in V2 \N(A),a contradiction. As before we obtain R+A �⊂ HA. �

Proposition 10.7.10 Let F be the family of all independent sets A of Gsuch that HA ∩ R+AG is a facet. If A is in F and Vi ∩ A �= ∅ for i = 1, 2,then H−

A is redundant in the following expression of the edge cone

R+A = aff(R+A) ∩( ⋂

A∈FH−

A

)⋂(n⋂i=1

H+ei

).

Proof. Set A = {v1, . . . , vq}. One can write A = A1 ∪ A2 with Ai ⊂ Vi fori = 1, 2. There are v1, . . . , vn−2 linearly independent vectors in HA ∩R+A,where n is the number of vertices of G. Consider the subgraph D of Gwhose edges correspond to v1, . . . , vn−2 and its vertex set is the union ofthe vertices in those edges. Note that D cannot be connected. Indeed thereis no edge of D connecting a vertex in NG(A1) with a vertex in NG(A2)because all the vectors v1, . . . , vn−2 satisfy the equation∑

xi∈A

xi =∑

xi∈NG(A)

xi.

Hence D is a spanning subgraph of G with two components D1, D2, whichare trees, such that V (Di) = Ai ∪ NG(Ai), i = 1, 2. Therefore HAi is aproper support hyperplane defining a facet Fi = HAi ∩R+A, that is A1,A2

are in F . Since H−A1∩H−

A2is contained in H−

A the proof is complete. �

Proposition 10.7.11 If A2 � V2, F = HA2 ∩ R+A is a facet of the edgecone of G and A′ = A ∪ {0}, then

H−A2∩ aff(A′) =

{H−

A1∩ aff(A′), where A1 = V1 \N(A2) �= ∅, or

H+ei ∩ aff(A′), for some xi with G \ {xi} connected.

Proof. Let us assume that G has n vertices x1, . . . , xn and that V1 is theset of the first m vertices of G. There are v1, . . . , vn−2 linearly independentvectors in the hyperplane HA2 . Consider the subgraph D of G whose edgescorrespond to v1, . . . , vn−2 and its vertex set is the union of the vertices inthose edges. As G is connected, either D is a tree with n− 1 vertices or Dis a spanning subgraph of G with two connected components.

474 Chapter 10

Assume that D is a tree (the other case is left as an exercise), writeV (D) = V \ {xi} for some i. Note 〈vj , vA2 〉 = −〈vj , ei〉, j = 1, . . . , q, where

vA2 =∑xi∈A2

ei −∑

xi∈N(A2)

ei.

Indeed if the “edge” vj has vertices in V (D), then both sides of the equalityare zero, otherwise write vj = ei + e�. Observe xi /∈ A2 and x� ∈ N(A2)because asHA2 is a facet it cannot contain vj , thus both sides of the equalityare equal to −1. As a consequence since aff(A′) = R(v1, . . . , vn−2, vj) forsome vj = ei + e� we rapidly obtain 〈α, vA2 〉 = −〈α, ei〉, ∀α ∈ aff(A′).Therefore H−

A2∩ aff(A′) = H+

ei ∩ aff(A′), as required. �

From the proof of Proposition 10.7.11 we get:

Lemma 10.7.12 Let F = HA ∩R+A be a facet of R+A with A � V1.

(a) If N(A) = V2, then A = V1\{xi} for some xi ∈ V1 and F = Hei∩R+A.(b) If N(A) � V2, then F = HV2\N(A) ∩R+A and N(V2 \N(A)) = V1 \A.

Proposition 10.7.13 Let A � V1. Then F = HA∩R+A is a facet of R+Aif and only if

(a) G[A ∪N(A)] is connected with vertex set V \ {v} for some v ∈ V1, or(b) G[A ∪ N(A)] and G[(V2 \ N(A)) ∪ (V1 \ A)] are connected and their

union is a spanning subgraph of G.

Moreover any facet has the form F = HA∩R+A for some A � Vi, i ∈ {1, 2}.

Proof. The first statement follows readily from Lemma 10.7.12 and usingpart of the proof of Theorem 10.7.8. The last statement follows combiningTheorem 10.7.8 with Proposition 10.7.10. �

Remark 10.7.14 In Proposition 10.7.13 the case (a) is included in case (b).To see this fact, take N(A) = V2 and note that G[(V2 \ N(A)) ∪ (V1 \ A)]must consist of a point. The condition in case (a) is equivalent to requireG\{v} connected and in this case F = Hei ∩R+A, where v = xi correspondto the unit vector ei.

Lemma 10.7.15 Let F be a facet of R+A. If F = HA∩R+A = HB∩R+Awith A � V1 and B � V1, then A = B.

Proof. It follows from Lemma 10.7.12 and Proposition 10.7.13. �

Putting together the previous results we get the following canonical wayof representing the edge cone. The uniqueness follows from Lemma 10.7.12and Lemma 10.7.15.


Theorem 10.7.16 There is a unique irreducible representation

R+A = aff(R+A) ∩ (∩ri=1H−Ai) ∩ (∩i∈IH

+ei)

such that Ai � V1 for all i and I = {i|xi ∈ V2 is not cut vertex }.

Corollary 10.7.17 There is an irreducible representation

R+A = aff(R+A) ∩ (∩ri=1H−Ai) ∩ (∩si=1H

−Bi)

such that Ai � V1 for all i and Bi � V2 for all i.

Lemma 10.7.18 Zn ∩ R+A = NA. In particular if (β1, . . . , βn) is an in-tegral vector in the edge cone, then

∑ni=1 βi is an even integer.

Proof. Since G is bipartite, by Proposition 10.2.2 its incidence matrix istotally unimodular, the lemma follows from Corollary 1.6.8. �

As an application we recover the marriage theorem (Theorem 7.1.9).For a generalized version of the marriage theorem; see Proposition 13.5.3.Recall that a pairing off of all the vertices of a graph G is called a perfectmatching.

Theorem 10.7.19 (Marriage Theorem) If G is a connected bipartite graph,then G has a perfect matching if and only if |A| ≤ |N(A)| for every inde-pendent set of vertices A of G.

Proof. G has a perfect matching if and only if the vector β = (1, . . . , 1) isin NA. By Lemma 10.7.18 β is in NA if and only if β ∈ R+A. Thus theresult follows from Theorem 10.7.8. �

Exercises

10.7.20 Let G be a graph with n vertices and let A be its incidence matrix.Prove that if rank(A) = n and A is an independent set of G, then the setsF = R+A∩HA and F = Hei ∩ R+A are proper faces of the edge cone.

10.7.21 Let G be a connected bipartite graph and let Q be the polyhedrondefined as the set of x = (xi) ∈ Rn+1 satisfying

xi ≥ 1 for i = 1, . . . , n+ 1, and−xn+1 +

∑vi∈C xi ≥ 1 for any minimal vertex cover C of G.

If G is unmixed and v0 ≥ 3, then a = (1, . . . , 1) and b = (1, . . . , 1, v0 − 1)are the vertices of Q.

476 Chapter 10

10.7.22 [72] Let G = Kn be the complete graph on n vertices and let A bethe set of column vectors of its incidence matrix. Then a vector x ∈ Rn isin R+A if and only if x = (x1, . . . , xn) satisfies

−xi ≤ 0, i = 1, . . . , nxi −

∑j �=i xj ≤ 0, i = 1, . . . , n.

In addition if n ≥ 4, these equations define all the facets of R+A.

10.7.23 If G is a triangle with vertices {x1, x2, x3}, then the edge cone ofG has three facets defined by

x1 ≤ x2 + x3, x2 ≤ x1 + x3, x3 ≤ x1 + x2.

Note that x1 ≥ 0, x2 ≥ 0 and x3 ≥ 0 define proper faces of dimension 1.

10.7.24 If G is an arbitrary graph, then

R+A =

( ⋂A∈F

H−A

)⋂(n⋂i=1

H+ei

),

where F is the set of all the independent sets of vertices of G.

10.7.25 If G1, . . . , Gr are the components of a graph G, then

R+A =

(⋂A

H−A

)⋂(n⋂i=1

H+ei

),

where the first intersection is taken over all the independent sets of verticesof the components Gi of G.

10.7.26 Let G = Km,n be the complete bipartite graph with m ≤ n. IfV1 = {x1, . . . , xm} and V2 = V \ V1 is the bipartition of G, then a vectorz ∈ Rm+n is in R+A if and only if z = (x1, . . . , xm, y1, . . . , yn) satisfies

x1 + · · ·+ xm = y1 + · · ·+ yn,−xi ≤ 0, i = 1, . . . ,m,−yi ≤ 0, i = 1, . . . , n.

In addition if m ≥ 2, the inequalities define all the facets of R+A.

10.7.27 If G = K1,3 is the star with vertices {v, x1, x2, x3} and center x,then the edge cone of G has three facets defined by

xi ≥ 0, (i = 1, 2, 3)

and x = 0 defines a proper face of dimension 1.


10.8 Monomial birational extensions

Let R = K[x1, . . . , xn] be a polynomial ring over a field K. In this sectionwe consider a finite set of distinct monomials F = {xv1 , . . . , xvq} ⊂ R ofdegree d ≥ 2 and consider the integer matrices:

A = (v1, . . . , vq) and B =

(v1 · · · vq1 · · · 1

),

where the vi’s are regarded as column vectors, A is sometimes called thelog-matrix of F . In the sequel let xd denote the set of all monomials ofdegree d in R. Then K[xd] is the dth Veronese subring R(d) of R.

If L is an integer matrix with r rows, we denote by ZL (resp. QL) thesubgroup of Zr (resp. subspace of Qr) generated by the columns of L.

Definition 10.8.1 An extension D1 ⊂ D2 of integral domains is said to bebirational if D1 and D2 have the same field of fractions.

Proposition 10.8.2 Let F1 = {xβ1 , . . . , xβr} be a set monomials of R suchthat F ⊂ F1. Then the ring extension K[F ] ⊂ K[F1] is birational if andonly if ZA = ZL, where L is the matrix with column vectors β1, . . . , βr.


Following [385] an aim here is to study the birationality of the extensionK[F ] ⊂ R(d) in terms of the matrices:

(a) the linear syzygy matrix Syz�(F ) whose columns are the set of vectorsof the form xiej − xke� such that xix

vj = xkxv� ,

(b) the numerical linear syzygy matrix L obtained from Syz�(F ) by makingthe substitution xi = 1 for all i,

(c) the matrix M whose columns are the set of vectors ei − ek ∈ Rn suchthat ei − ek = vj − v�, i.e., M = AL,

(d) the Jacobian matrix Θ(F ) = (∂fi/∂xj).

Notice that the matrices in the first row of the diagram:

Θ(F ) Syz�(F ) M := Θ(F )�Syz�(F )↓ ↓ ↓A L M := AL

specialize to the matrices in the second row by making xi = 1 for all i. Thematrices Syz�(F ) and L have size q× r, while the matricesM and M havesize n× r.

Proposition 10.8.3 [385] If rank(M) = n − 1, then K[F ] ⊂ K[xd] is abirational extension.

478 Chapter 10

Proof. Let a = (ai) ∈ Nn such that |a| =∑i ai = d. It suffices to prove

that a ∈ ZA. Let w1, . . . , wr be the column vectors of the matrix M . Eachwm has the form wm = ei − ek = vj − v� for some i �= k and j �= �. Hencerank(A) = n because v1 /∈ QM . Therefore we can write

λa = λ1w1 + · · ·+ λrwr + μv1 (λ, μ, λi ∈ Z).

Taking inner product with 1 = (1, . . . , 1) yields

λd = λ|a| = λ1|w1|+ · · ·+ λr|wr |+ μ|v1| = μd ⇒ λ = μ

⇒ λ(a− v1) = λ1w1 + · · ·+ λrwr. (10.10)

Consider the digraph D with vertex set X = {x1, . . . , xn} such that (xi, xk)is a directed edge iff wm = ei − ek for some m. The incidence matrix ofD is M , thus M is totally unimodular by Exercise 1.8.10 and consequentlyZn/ZM is torsion-free. Hence from Eq. (10.10) we get a − v1 ∈ ZM anda ∈ ZM + v1 ⊂ ZA, as required. �

Remark 10.8.4 Let D be the digraph in the proof of Proposition 10.8.3.Then according to [190, Theorem 8.3.1] we have

rank(M) = n− c,

where c is the number of connected components of D. In particular M hasrank n− 1 if and only if D is connected.

A generalization of the next result is given in Theorem 10.8.12.

Corollary 10.8.5 Let G be a connected non-bipartite graph with n verticesand let K[G] be its edge subring. Then K[G] ⊂ K[x2] is birational.

Proof. There is a connected subgraph H of G with n vertices and n edgesand with a unique cycle of odd length. This follows using that any connectedgraph has a spanning tree together with the fact that a graph is bipartite ifand only if every cycle in some cycle basis is even; see Exercise 10.1.67. Byinduction on n it is not hard to see that the numerical matrix M associatedto H has rank n−1. ThusK[H ] ⊂ K[x2] is birational by Proposition 10.8.3,which implies the birationality of K[G] ⊂ K[x2]. �

Proposition 10.8.6 [385] K[F ] ⊂ K[xd] is a birational extension if andonly if n = rank(A) and Δn(A) = d.

Proof. Assume the given extension is birational. Then Zlog(xd) = ZA byProposition 10.8.2. By Proposition 9.3.33(b) one has Zn/Zlog(xd) � Zd.Therefore, Δn(A) = d. Conversely, if Δn(A) = d, the surjection

Zn/ZA −→ Zd, a �−→|a|,

is an isomorphism. Since log(xd) maps to zero under this map, one obtainsZA = Zlog(xd), as required. �


Proposition 10.8.7 [385] The following conditions are equivalent

(a) K[F ] ⊂ K[xd] is birational.

(b) Zn/Z({vi − vj |1 ≤ i < j ≤ q}) is torsion-free of rank 1.

(c) n = rank(A) and

Z({vi − vj |1 ≤ i < j ≤ q}) = Z({ei − ej |1 ≤ i < j ≤ n}).

Proof. If n = rank(A), then there is an exact sequence of finite groups

0 −→ T (Zn+1/ZB)ϕ−→ T (Zn/ZA)

ψ−→ Zd −→ 0, (∗)

where the maps ϕ and ψ are given by

ϕ((α, b)) = α (α ∈ Zn, b ∈ Z),ψ(α) = 〈α, x0〉.

Pick a basis {v1, . . . , vn} of the column space of A and notice that

{v1 − vn, v2 − vn, . . . , vn−1 − vn, vn}

is also a basis because |vi| = d for all i. Therefore one has the equality

Q({vi − vj |1 ≤ i < j ≤ q}) = Q({ei − ej|1 ≤ i < j ≤ n}).

The equivalences now follow using Lemma 1.2.20, Proposition 9.3.33, theexact sequence (∗), together with the fact that the group

Zn/Z({ei − ej |1 ≤ i < j ≤ n})

is torsion-free of rank 1, see the proof of Proposition 10.8.3. �

Proposition 10.8.8 [385] If rank(L) = q − 1 and rank(A) = n, thenrank(M) = n− 1. In particular K[F ] ⊂ K[xd] is birational.

Proof. There are linear maps

QrL−→ Qq

A−→ Qn.

Hence we have a linear map

im(L)A1−→ im(AL) = im(M) −→ 0,

where A1 is the restriction of A to im(L). By hypothesis dim(im(A)) = nand dim(im(L)) = q − 1. Hence

q − 1 = dim(im(L)) = dim(ker(A1)) + dim(im(M)),

q − n = dim(ker(A)) ≥ dim(ker(A1)).

Therefore dim(im(M)) ≥ n− 1. Since im(M) is generated by vectors of theform ei − ej it follows that im(M) has rank n simply because e1 is not inthe linear space generated by {ei − ej|1 ≤ i < j ≤ n}. �

480 Chapter 10

Lemma 10.8.9 If m is a non-zero minor of Syz�(F ) of order s for somes ≥ 1, then

m = ±xa11 · · ·xann (ai ∈ N).

Proof. By induction on s. The case s = 1 is clear because the entries ofSyz�(F ) are monomials. Assume s ≥ 2. Let L1 be a submatrix of Syz�(F )of size s× s. We may assume that L1 is obtained using the first s rows andcolumns of Syz�(F ). Let g1, . . . , gq be the rows obtained from Syz�(F ) byfixing the first s columns. Thus L1 is the matrix with rows g1, . . . , gs.

Case (I): If the vector gi has a non-zero entry for some s + 1 ≤ i ≤ q,say the jth entry of gi is non-zero, then expanding the determinant of L1

along the jth column and using the induction hypothesis we get that

det(L1) = ±xa11 · · ·xann (ai ∈ N).

Case (II): Now we assume that gi is the zero vector for s + 1 ≤ i ≤ q,but this implies that the linear syzygy matrix of F ′ = {xv1 , . . . , xvs} hasrank at least s, a contradiction because from the minimal free resolution ofI ′ = (xv1 , . . . , xvs) one has rank(Syz�(F

′)) ≤ s− 1; see Exercise 10.8.17. �

Theorem 10.8.10 [385] If rank(A) = n and rank(Syz�(F )) = q − 1, thenK[F ] ⊂ K[xd] is birational.

Proof. By Lemma 10.8.9 the matrix L obtained from Syz�(F ) by makingxi = 1 for all i has also rank q − 1. Therefore by Propositions 10.8.8 and10.8.3 we get that K[F ] ⊂ K[xd] is birational. �

Corollary 10.8.11 [385] If I = (F ), rank(A) = n, and I has a linearpresentation, then K[F ] ⊂ K[xd] is birational.

Proof. It follows at once from Theorem 10.8.10 because in this case therank of Syz�(F ) is q − 1. �

Monomials of degree two The birational theory of monomials of degreetwo can be fully established. Theorem 10.8.12 gives a complete answer forthe birationality of subrings of multigraphs.

In what follows we assume that deg(xvi ) = 2 for all i. Consider themultigraph G on the vertex set X = {x1, . . . , xn} whose edges (resp. loops)are the pairs {xi, xj} (resp. {x�, x�}) such that xixj ∈ F (resp. x2� ∈ F )for some i �= j. Notice that, in our situation, the log-matrix A of F is theincidence matrix of G and the monomial subring K[F ] is the edge subringK[G] of the multigraph G.


Theorem 10.8.12 K[G] ⊂ K[x2] is a birational extension if and only ifthe following two conditions hold :

(a) G is connected.

(b) G is bipartite and has at least one loop or G is non-bipartite.

Proof. ⇐) Since G is connected, there is a spanning tree T of G containingall the vertices of G; see Exercise 10.1.65.

Case (I): G is a bipartite and xn is a loop of G. We may then regard T asa tree with a loop at xn. Notice that T has exactly n−1 simple edges plus aloop. The incidence matrix AT of T has order n, is non-singular, and we mayassume that the last column of AT has the form (0, 0, . . . , 0, 2)�. Considerthe matrix A′

T obtained from AT by removing the last column. The matrixA′T is totally unimodular because it is the incidence matrix of a bipartite

graph; see Proposition 10.2.2. Therefore det(AT ) = ±2 and rank(A) = n.From Lemma 10.8.6 we obtain that K[T ] ⊂ K[x2] is birational. HenceK[G] ⊂ K[x2] is birational as well.

Case (II): G is non-bipartite. This case follows from Corollary 10.8.5.For use below notice that in Cases (I) and (II) we have that rank(A) = nand Zn/ZA � Z2.⇒) Let G1, . . . , Gr be the connected components of G and let Ai be the

incidence matrix of Gi. Thus we have A = diag(A1, A2, . . . , Ar). Noticethat for each i the multigraph Gi is either non-bipartite or bipartite withat least one loop, for otherwise rank(Ai) < ni for some i, where ni is thenumber of vertices of Gi, a contradiction because n = n1 + · · · + nr andn = rank(A) =

∑j rank(Aj). Therefore by the observation at the end of the

proof of Case (II) we have rank(Ai) = ni and Zni/ZAi � Z2. This meansthat the Smith normal form of Ai is diag(1, . . . , 1, 2,0) and consequently theSmith normal form of A has exactly r entries equal to 2 and the remainingentries equal to 1 and 0. Hence Zn/ZA � Zr2 and |Zn/ZA| = |Δn(A)| = 2r.By the birationality of the extension, using Lemma 10.8.6, we get r = 1,i.e., G is connected, as required. �

Corollary 10.8.13 If G is a connected graph, then K[G] ⊂ K[x2] is abirational extension if and only if G is non-bipartite.

Proof. If G is bipartite, dimK[G] = rank(A) = n−1, hence K[G] ⊂ K[x2]cannot be birational. The converse follows from Theorem 10.8.12. �

Exercises

10.8.14 Let A be an integral matrix of size n× q with equal column sumd. Prove the equalities

In(A) = In

(A

A1 + · · ·+An

)= In

(Ad1

)= dIn

(A1

),

482 Chapter 10

where A1, . . . , An are the rows of A and In(A) is the ideal of Z generatedby the n-minors of A.

Hint d1 = (d, . . . , d) = A1 + · · ·+An.

10.8.15 Let F = {xv1 , . . . , xvq} and G = {xβ1 , . . . , xβr} be two sets ofmonomials of R such that F ⊂ G. Prove that K[F ] ⊂ K[G] is a birationalextension if and only if Zlog(F) = Zlog(G).

10.8.16 Let F be a set of monomials of R = K[x1, . . . , xn] of the samedegree d such that Zn/Zlog(F ) is a finite group. Prove that K[Ft] = A(P )if and only if Zn/Zlog(F ) � Zd.

10.8.17 Let I = (xv1 , . . . , xvq ) be a monomial ideal of a polynomial ring Rover a field K. Prove that if

· · · −→ Rb2ϕ−→ Rq −→ I −→ 0

is the minimal free resolution of I, then rank(ϕ) = q− 1 and the columns ofthe matrix ϕ have the form xaiej − xake� (cf. Theorem 3.3.19 and [137]).

Chapter 11

Edge Subrings andCombinatorialOptimization

Let G be a connected bipartite graph. We present an approach to computethe canonical module of the edge subring K[G] using linear programmingtechniques and study the Gorenstein property and the type of K[G]. Thea-invariant of K[G] will be interpreted in combinatorial optimization termsas the maximum number of edge disjoint directed cuts.

11.1 The canonical module of an edge subring

Let G be a connected bipartite graph on the vertex set V = {x1, . . . , xp}and let R = K[x1, . . . , xp] = ⊕∞

i=0Ri be a polynomial ring over a field Kwith the standard grading induced by deg(xi) = 1. The edge subring of Gis the K-subalgebra

K[G] = K[{xixj |xi is adjacent to xj}] ⊂ R,

we grade K[G] with the normalized grading K[G]i = K[G] ∩ R2i. Recallthat, by Proposition 10.3.1, K[G] is normal and Cohen–Macaulay.

The set of vectors vk = ei + ej ∈ Rp such that xi is adjacent to xj willbe denoted by A = {v1, . . . , vq}. Note that A is the set of column vectorsof the incidence matrix of G. As G is bipartite, by Lemma 10.7.18, we have

NA = Zp ∩R+A. (11.1)

Thus, by a formula of Danilov–Stanley (see Theorem 9.1.5), the canonical

484 Chapter 11

module ωK[G] is the ideal given by

ωK[G] = ({xa| a ∈ NA ∩ ri(R+A)})(11.1)= ({xa| a ∈ Zp ∩ ri(R+A)}) ⊂ K[G], (11.2)

where ri(R+A) is the interior of R+A relative to the affine hull of R+A.Recall, from Section 10.7, that R+A is called the edge cone of G.

11.2 Integrality of the shift polyhedron

In what follows G will denote a connected bipartite graph with p = m+ nvertices and bipartition (V1, V2). We will assume that the vertices in V1 arex1, . . . , xm and the vertices in V2 are xm+1, . . . , xn+m, where 2 ≤ m ≤ n.

Consider the family

F = {A ∪ A′| ∅ �= A � V1;N(A) ⊂ A′ ⊂ V2} ∪ {A′| ∅ �= A′ ⊂ V2},

where N(A) is the neighbor set of A. For each Y = A∪A′ ∈ F we associatethe following vector

βY =∑xi∈A

ei −∑xi∈A′

ei ∈ Rm+n,

note that if A = ∅ the vector βY is a {0,−1}-vector. Let C′ be the matrixwhose rows are the vectors in {βY }Y ∈F . According to Corollary 10.7.17 theedge cone of G can be written as:

R+A = aff(R+A) ∩ {x|C′x ≤ 0}.

The shift polyhedron of the edge cone of G, with respect to C′, is therational polyhedron:

Q = aff(R+A) ∩ {x|C′x ≤ −1},

see Definition 9.1.12. From the finite basis theorem (see Theorem 1.1.33)the shift polyhedron can be written as the sum of a unique cone and apolytope. In our case:

Q = R+A+ conv(β1, . . . , βr),

where β1, . . . , βr are the vertices ofQ. Recall thatQ is an integral polyhedronif Q = conv(Zp ∩ Q). As Q is a pointed polyhedron, it is integral if andonly if β1, . . . , βr are integral vectors; see Corollary 1.1.64.

A reason for introducing the shift polyhedron is that its integral pointsdefine the canonical module of K[G].

Edge Subrings and Combinatorial Optimization 485

Proposition 11.2.1 Zp ∩ Q = Zp ∩ ri(R+A).

Proof. It follows from Eq. (11.2) and Corollary 1.1.45. �

Thus, the shift polyhedron is a bridge which allows us to use combina-torial optimization techniques to study the edge subring K[G].

Theorem 11.2.2 The shift polyhedron Q = aff(R+A) ∩ {x|C′x ≤ −1} isintegral.

Proof. If G is regarded as the digraph with all its arrows leaving the vertexset V2, then it is seen that one has the equality C′A = −B′, where A is theincidence matrix of G and B′ is the one-way cut-incidence matrix of thefamily F ; see Definition 1.8.5. Let b be any vector in Rp, p = m + n, suchthat the following maximum is finite

max{〈x, b〉|x ∈ Q}. (11.3)

According to Theorem 1.1.63 it suffices to prove that the maximum inEq. (11.3) is attained by an integral vector. As Q ⊂ R+A, any vectorx ∈ Q can be written as x = Ax for some x ≥ 0, x ∈ Rq, where q is thenumber of edges of G. Hence

{〈b, x〉|x ∈ Q} = {〈bA, x〉| x ≥ 0;B′x ≥ 1}.

Therefore

max{〈x, b〉|x ∈ Q} = max{〈bA, x〉| x ≥ 0;B′x ≥ 1}. (11.4)

By Corollary 1.8.8 and Lemma 1.8.9 the polyhedron

Q′ = {x ∈ Rq| x ≥ 0;B′x ≥ 1}

is integral. Hence, by Theorem 1.1.63, the maximum in the right-hand sideof Eq. (11.4) is attained by an integral vector x0 ∈ Q′. Consequently themaximum in Eq. (11.3) is attained by the integral vector Ax0 ∈ Q. �

Example 11.2.3 Consider the following bipartite graph G and make G adigraph with “edges” v1, . . . , v6 as shown below.

�V1

V2

v1v3

v2v4 v6

v5

x1 x2 x3

x4 x5 x6

�

��

��

��

��

��

��#

� � �

486 Chapter 11

The family F consists of the subsets that occur in the directed cuts:

δ+({x1, x4, x6}) = {v3, v6}, δ+({x5}) = {v4, v5},δ+({x2, x4, x5}) = {v1, v5}, δ+({x6}) = {v2, v6},δ+({x3, x5, x6}) = {v2, v4}, δ+({x4, x5}) = {v1, v3, v4, v5},

δ+({x1, x2, x4, x5, x6}) = {v5, v6}, δ+({x4, x6}) = {v1, v2, v3, v6},δ+({x1, x3, x4, x5, x6}) = {v3, v4}, δ+({x5, x6}) = {v2, v4, v5, v6},δ+({x2, x3, x4, x5, x6}) = {v1, v2}, δ+({x4, x5, x6}) = {v1, . . . , v6}.

δ+({x4}) = {v1, v3},

As pointed out in the proof of Theorem 11.2.2 the one-way cut-incidencematrix B′ of F is related to the incidence matrix A of G and to the matrixC′ by the equality C′A = −B′. In this example one has:

C′ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 −1 0 −10 1 0 −1 −1 00 0 1 0 −1 −11 1 0 −1 −1 −11 0 1 −1 −1 −10 1 1 −1 −1 −10 0 0 −1 0 00 0 0 0 −1 00 0 0 0 0 −10 0 0 −1 −1 00 0 0 −1 0 −10 0 0 0 −1 −10 0 0 −1 −1 −1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

and B′ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 0 0 11 0 0 0 1 00 1 0 1 0 00 0 0 0 1 10 0 1 1 0 01 1 0 0 0 01 0 1 0 0 00 0 0 1 1 00 1 0 0 0 11 0 1 1 1 01 1 1 0 0 10 1 0 1 1 11 1 1 1 1 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

The polyhedron Q′ = {x|B′x ≥ 1; x ≥ 0} is integral and has two vertices(0, 1, 1, 0, 1, 0) and (1, 0, 0, 1, 0, 1) that map under A onto the vector 1 of Q.

Corollary 11.2.4 If G is a connected bipartite graph and Q is the shiftpolyhedron of the edge cone of G, with respect to C′, then

Q = conv(Zp ∩ ri(R+A)).

Proof. The polyhedron Q is integral by Theorem 11.2.2. Hence the resultfollows from Proposition 9.1.14. �

Exercises

11.2.5 Let T : Rq → Rn be a linear map and let Q′ ⊂ Rq be an integralpolyhedron. If T (Zq) ⊂ Zn, then T (Q′) is an integral polyhedron.

11.2.6 Prove that, in the proof Theorem 11.2.2, the linear transformationx �→ Ax maps Q′ onto Q. Prove that dim(Q′) = q and dim(Q) = m+n−1.Give a shorter proof of Theorem 11.2.2 using Exercise 11.2.5.


11.3 Generators for the canonical module

Let S be a standard graded K-algebra over a field K. The a-invariant of S,denoted by a(S), is the degree, as a rational function, of the Hilbert seriesof S. If S is Cohen–Macaulay and ωS is the canonical module of S, then

a(S) = −min{ i | (ωS)i �= 0},

see Proposition 5.2.3. In our situation, since G is bipartite, S = K[G] is anormal Cohen–Macaulay standard K-algebra. Thus this formula applies.

Theorem 11.3.1 [405] If Q is the shift polyhedron of the edge cone of aconnected bipartite graph G, then the a-invariant of K[G] is given by

a(K[G]) = −min

{|x|2

∣∣∣∣ x ∈ Q} .Proof. Let C′ be the matrix defining the shift polyhedron. Note thatany row of C′ defines a proper face of the edge cone, except the row withthe first m entries equal to 0 and the last n entries equal to −1. Thus fromCorollary 1.1.45 and Theorem 1.1.44 we have Zp∩Q = Zp∩ri(R+A). Hencethe inequality “≤” follows from the Danilov–Stanley formula of Eq. (11.2).By Proposition 1.1.41 the minimum above is attained at a vertex β of Q.Hence it suffices to observe that β has integral entries by Theorem 11.2.2and Corollary 1.1.64. �

Theorem 11.3.2 [405, Proposition 4.2] Let G be a connected bipartitegraph. If G is regarded as the digraph with all its arrows leaving the vertexset V2, then the following three numbers are equal:

(a) −a(K[G]), minus the a-invariant of K[G].

(b) The minimum cardinality of an edge set that contains at least one edgeof each directed cut.

(c) The maximum number of edge disjoint directed cuts.

Proof. Let A be the incidence matrix ofG and let C′ be the matrix definingthe shift polyhedron. As C′A = −B′, then by Theorem 1.8.7 and LP duality(Theorem 1.1.56) one has that the optimum values in the equality

min{〈1, x〉| x ≥ 0;B′x ≥ 1} = max{〈y,1〉| y ≥ 0; yB′ ≤ 1}

are attained by integral vectors. By looking at B′ as a one-way cut-incidencematrix it follows that the two numbers in (b) and (c) are equal. In general,for digraphs, it is known that the numbers in (b) and (c) are equal. See forinstance [281, Theorem 19.10]. Noticing that

〈1, Ax〉/2 = 〈1, x1v1 + · · ·+ xqvq〉/2 = (2x1 + · · ·+ 2xq)/2 = 〈1, x〉,

488 Chapter 11

where v1, . . . , vq are the column vectors of A, from the equalities

min{〈1, x〉/2|x ∈ aff(R+A);C′x ≤ −1} =min{〈1, Ax〉/2| x ≥ 0;B′x ≥ 1} = min{〈1, x〉| x ≥ 0;B′x ≥ 1},

and Theorem 11.3.1 we get that the numbers in (a) and (b) are equal. �

Proposition 11.3.3 If G is a connected bipartite graph and β is a vertexof the shift polyhedron Q, then xβ is a minimal generator of ωK[G].

Proof. By Eq. (11.2), Corollary 1.1.45, and Theorem 11.2.2 we get thatxβ is in ωK[G]. There are c ∈ Qp and b ∈ Q such that

(i) {β} = {x|〈x, c〉 = b} ∩ Q and (ii) Q ⊂ {x|〈x, c〉 ≤ b}.

If v1, . . . , vq are the columns of the incidence matrix of G, then by definitionof Q one has vi + β ∈ Q for all i. Thus

〈vi + β, c〉 = 〈vi, c〉+ b ≤ b⇒ 〈vi, c〉 ≤ 0 (i = 1, . . . , q).

Assume there are α ∈ Q, η1, . . . , ηq ∈ N such that β = (∑q

i=1 ηivi)+α, then

b = 〈β, c〉 = η1〈v1, c〉+ · · ·+ ηq〈vq, c〉+ 〈α, c〉 ≤ 〈α, c〉 ≤ b.

Hence 〈α, c〉 = b and by (i) we get α = β. Thus xβ is a minimal generatorof the canonical module ωK[G]. �

Definition 11.3.4 Let G be a bipartite graph with incidence matrix Aand let C′ be the matrix defining the shift polyhedron. If C′A = −B′, thepolyhedron

Q′ = {x ∈ Rq| x ≥ 0;B′x ≥ 1}

is called the blocking polyhedron.

Theorem 11.3.5 If G is a connected bipartite graph, then Q′ is integral.

Proof. It follows from Corollary 1.8.8 and Lemma 1.8.9. �

Proposition 11.3.6 Let G be a connected bipartite graph and let Q be theshift polyhedron with respect to C′. If xβ is a minimal generator of ωK[G]

and A is the incidence matrix of G, then there is a vertex α of the blockingpolyhedron Q′ such that Aα = β.

Proof. Let v1, . . . , vq be the column vectors of the matrix A and let C′ bethe matrix defining Q. As A is totally unimodular (see Proposition 10.2.2),using Caratheodory’s theorem (see Theorem 1.1.18) and Heger’s theorem(see Theorem 1.6.4), after permuting the vi’s, we can write β =

∑ri=1 ηivi


with ηi ∈ N \ {0} and r ≤ p − 1 ≤ q, where p = m + n. We claim thatηi = 1 for all i. Assume ηi > 1. Take any row v of C′. Observe that 〈v, vk〉is equal to 0 or −1 for any vk. The vector

β′ = η1v1 + · · ·+ ηi−1vi−1 + (ηi − 1)vi + ηi+1vi+1 + · · ·+ ηrvr

satisfies 〈β′, v〉 ≤ −1 because 〈β, v〉 ≤ −1. Thus β′ ∈ Q, a contradictionbecause β′ = β− vi and xβ is minimal. Thus ηi = 1. Notice that the vectorα = e1 + · · · + er satisfies Aα = β, and from C′A = −B′ we get α ∈ Q′.Consider the linear program

min xr+1 + · · ·+ xq

subject to B′x ≥ 1 and x ≥ 0

and notice that 0 is the optimum value of this linear program because αis in Q′. By Theorem 11.3.5 there is an integral vertex γ of Q′ where theminimum is attained. Hence γi = 0 for i > r. By Exercise 11.3.8 the vectorγ has {0, 1}-entries. If γk = 0 for some 1 ≤ k ≤ r, then Aγ =

∑i�=k εivi ∈ Q,

where εi ∈ {0, 1} for all i �= k, a contradiction to the minimality of xβ . Thusα = γ, as required. Observe that the last part of the argument works evenif q = r, which is the case of a tree. �

Theorem 11.3.7 If G is a connected bipartite graph with incidence matrixA and Q′ is the blocking polyhedron, then the canonical module ωK[G] ofK[G] is generated by the set

{xAα| α is a vertex of Q′}.

Proof. It follows by recalling that the blocking polyhedron Q′ is integraland using Proposition 11.3.6. �

Exercise

11.3.8 If B is a {0, 1}-matrix, then any integral vertex of the rationalpolyhedron Q = {x |x ≥ 0;Bx ≥ 1} is a {0, 1}-vector.

11.4 Computing the a-invariant

Let G be a bipartite graph. In order to use the results of Section 11.3 inan efficient way, we introduce other representations of the shift polyhedron.For each independent set of vertices A of G consider the vector

αA =∑xi∈A

ei −∑

xi∈N(A)

ei.

490 Chapter 11

By Corollary 10.7.17, there exists an irreducible representation of the edgecone of G, as an intersection of closed half-spaces, of the form:

R+A = aff(R+A) ∩H−αA1∩ · · · ∩H−

αAr∩H−

−ei1 ∩ · · · ∩H−−eis , (11.5)

where for each i either Ai � V1 or Ai � V2 and none of the half-spaces canbe omitted from the intersection. Let us denote by C the matrix whose rowsare the vectors αA1 , . . . , αAr ,−ei1 , . . . ,−eis and by C′ the matrix definingthe shift polyhedron, as defined in Section 11.2.

The next result says that the shift polyhedrons Q and Q1 with respectto C′ and C, respectively, are equal.

Theorem 11.4.1 If G is a connected bipartite simple graph, then the shiftpolyhedron Q of the edge cone of G, with respect to C′, is given by

Q = aff(R+A) ∩ {x|Cx ≤ −1}.Proof. It follows from Exercises 11.4.7, 11.4.8, and Lemma 10.7.15. �

Remark 11.4.2 If CA = −B and C′A = −B′, where A is the incidencematrix of G, then B and B′ define the same blocking polyhedron.

There are linear programming techniques to convert the description ofa rational polyhedron given by a “finite basis” into an irreducible represen-tation as intersection of closed half-spaces and vice versa. These techniqueshave been converted into very efficient routines in several programming en-vironments; see for instance PORTA [84]. Thus, one can effectively computea generating set for the ideal ωK[G] (see Example 11.4.4 for an illustration).

Remark 11.4.3 To compute the vertices of a shift polyhedron of an edgecone using PORTA we need a “valid” point. Note that if A = {v1, . . . , vq}is the set of column vectors of the incidence matrix of G, then the pointα =

∑qi=1 vi = (deg(x1), . . . , deg(xn+m)) is valid , that is, α ∈ Q:

Example 11.4.4 Consider the following bipartite simple graph G:

��

��

��

��

��

��

��

x1 x9 x5 x11

x7 x2 x10 x3

x8 x6

x4

with bipartition V1 = {x1, . . . , x6} and V2 = {x7, . . . , x11}. The incidencematrix of G, denoted by A, is the transpose of the following matrix. Wewill display the data as input files for PORTA.


CONE_SECTION (columns of A) corresponding monomial

( 1) 1 0 0 0 0 0 1 0 0 0 0 x1x7

( 2) 1 0 0 0 0 0 0 1 0 0 0 x1x8

( 3) 0 1 0 0 0 0 1 0 0 0 0 x2x7

( 4) 0 1 0 0 0 0 0 0 1 0 0 x2x9

( 5) 0 0 1 0 0 0 0 1 0 0 0 x3x8

( 6) 0 0 1 0 0 0 0 0 0 1 0 x3x10

( 7) 0 0 1 0 0 0 0 0 0 0 1 x3x11

( 8) 0 0 0 1 0 0 0 0 0 1 0 x4x10

( 9) 0 0 0 0 1 0 0 0 0 1 0 x5x10

( 10) 0 0 0 0 0 1 0 0 0 1 0 x6x10

( 11) 0 0 0 0 0 1 0 0 0 0 1 x6x11

( 12) 0 0 0 0 1 0 0 0 1 0 0 x5x9

( 13) 0 0 0 1 0 0 0 0 0 0 1 x4x11

( 14) 1 0 0 0 0 0 0 0 1 0 0 x1x9

( 15) 0 1 0 0 0 0 0 0 0 1 0 x2x10

( 16) 0 0 0 0 0 1 1 0 0 0 0 x6x7

( 17) 0 0 0 0 1 0 0 0 0 0 1 x5x11

Applying PORTA to this input file we obtain an irreducible representationof R+A, which immediately yields the following representation of Q.

VALID POINT: 3 3 3 2 3 3 3 2 3 5 4 (see previous Remark)

INEQUALITIES_SECTION

-xi<= -1 (for i=7,...,11)

( 1) x1+x2+x3+x4+x5+x6-x7-x8-x9-x10-x11 == 0

( 1) x1+x3+x4+x5+ x6-x7-x8-x9-x10-x11 <= -1

( 2) x1+x2+x4+x5+ x6-x7-x8-x9-x10-x11 <= -1

( 3) x1+x2+x3+x5+ x6-x7-x8-x9-x10-x11 <= -1

( 4) x1+x2+x3+x4+ x6-x7-x8-x9-x10-x11 <= -1

( 5) x1+x2+x3+x4+x5 -x7-x8-x9-x10-x11 <= -1

( 6) x4 -x10-x11 <= -1

( 7) x2 -x7 -x9 -x10 <= -1

( 8) x4+ x6-x7 -x10-x11 <= -1

( 9) x4+x5 -x9-x10-x11 <= -1

(10) x3+x4 -x8 -x10-x11 <= -1

(11) x1+x2 -x7-x8-x9-x10 <= -1

(12) x3+x4 +x6-x7-x8 -x10-x11 <= -1

(13) x3+x4+x5 -x8-x9-x10-x11 <= -1

(14) x2 +x4+x5+x6-x7 -x9-x10-x11 <= -1

(15) x1 -x7-x8-x9 <= -1

(16) x2+x3+x4+x5+x6-x7-x8-x9-x10-x11 <= -1

Using PORTA we get that the vertices of the shift polyhedron are:

(1) 1 2 1 1 1 1 1 1 1 1 3 (5) 1 1 1 1 1 1 1 1 1 2 1

(2) 2 1 1 1 1 1 1 1 1 1 3 (6) 1 1 1 1 1 1 1 1 2 1 1

(3) 2 1 1 1 1 1 1 1 1 3 1 (7) 1 1 1 1 1 1 2 1 1 1 1

(4) 1 1 1 1 1 1 1 1 1 1 2

492 Chapter 11

From the equality CA = −B (Remark 11.4.2), we get that the correspondingblocking polyhedron is defined by

INEQUALITIES_SECTION

xi>=0 (for i=1,...,17) (11) x10+x11+x15+x17+x6+x7+x9>=1

( 1) x15+x3+x4>=1 (12) x1+x10+x12+x14+x16+x6+x8+x9>=1

( 2) x5+x6+x7>=1 (13) x1+x15+x17+x3+x6+x7+x9>=1

( 3) x13+x8>=1 (14) x10+x11+x14+x15+x4+x6+x7>=1

( 4) x12+x17+x9>=1 (15) x10+x11+x15+x17+x2+x9>=1

( 5) x10+x11+x16>=1 (16) x10+x12+x16+x5+x6+x8+x9>=1

( 6) x1+x16+x3>=1 (17) x1+x15+x17+x2+x3+x9>=1

( 7) x2+x5>=1 (18) x10+x11+x14+x15+x2+x4>=1

( 8) x12+x14+x4>=1 (19) x1+x14+x6+x7>=1

( 9) x10+x15+x6+x8+x9>=1 (20) x12+x16+x3+x4+x5>=1

(10) x11+x13+x17+x7>=1 (21) x1+x14+x2>=1

Using PORTA we get that the blocking polyhedron Q′ has 173 vertices.The distinct images of those vertices under the incidence matrix A are:

1 2 1 1 1 1 1 1 1 1 3 vertex of the shift polyhedron Q

2 1 1 1 1 1 1 1 1 1 3 vertex of Q

2 1 1 1 1 1 1 1 1 3 1 vertex of Q

1 1 1 1 1 1 1 1 1 1 2 vertex of Q

1 1 1 1 1 1 1 1 1 2 1 vertex of Q

1 1 1 1 1 1 1 1 2 1 1 vertex of Q

1 1 1 1 1 1 2 1 1 1 1 vertex of Q

2 1 1 1 1 1 1 1 1 2 2 not a vertex of Q but correspond to a

minimal generator of the canonical module

1 1 1 1 1 2 1 1 2 2 1 this and the following vectors correspond to

1 1 2 1 1 1 1 1 2 2 1 redundant monomials in the canonical module

.....................

By Theorem 11.3.7, ωK[G] is minimally generated by the eight monomialscorresponding to the first eight vectors above. Thus type(K[G]) = 8 andthe rank of the last module in the graded free resolution of K[G] is 8.

Remark 11.4.5 From Theorem 11.3.1 and Theorem 11.4.1 we obtain aneffective method to compute the a-invariant of K[G] that only requires adescription of the shift polyhedron by linear inequalities and to solve a linearprogram. See Example 11.4.6.

Example 11.4.6 Consider the following bipartite graph G:

�x5�x2�x1

�x6�x3�x8�x7

�x4

��

��

bipartition:

V1 = {x1, x2, x3, x4}V2 = {x5, x6, x7, x8}


In order to estimate the a-invariant of G we set up the next linear programusing the following input file for Mathematica [431]

vars:={x1,x2,x3,x4,x5,x6,x7,x8}

f:=(x1+x2+x3+x4+x5+x6+x7+x8)/2

ieq:={x1+x2+x3+x4-x5-x6-x7-x8 == 0, -x1<= -1, -x2<= -1,

-x4<= -1, -x5<= -1, -x7<= -1,-x8 <= -1,x4-x7-x8 <= -1,

-x3-x4+x7+x8 <= -1,x3+x4-x6-x7-x8 <= -1}

ConstrainedMin[f,ieq,vars]

where the set of inequalities was found using PORTA and it comes from anirreducible representation of the edge cone as in Eq. (11.5). The optimalvalue of this linear program is equal to 5 and is attained at the vertex(1, 1, 2, 1, 1, 2, 1, 1). Hence by Theorems 11.3.1 and 11.4.1 we get that thea-invariant of K[G] is equal to −5.

Exercises

11.4.7 [189, Lemma 7.3.2] LetG = G(m,n) be a connected bipartite graph.If Q is the shift polyhedron of G, with respect to C′, and x ∈ Q, then xi ≥ 1for all i = 1, . . . ,m+ n.

11.4.8 [189, Proposition 7.3.3] Let G be a connected bipartite graph. If Qis the shift polyhedron of G, with respect to C′, and x ∈ Q, then for A � V1(resp. A � V2) such that N(A) ⊂ A′ ⊂ V2 (resp. N(A) ⊂ A′ ⊂ V1) one has∑

xi∈A

xi −∑xi∈A′

xi ≤ −1.

11.5 Algebraic invariants of edge subrings

In this section we give upper bounds for the a-invariant of the edge subringK[G] valid for any bipartite connected graphG. We examine the Gorensteinproperty and the type of K[G]. As a tool we use a technique, introduced inSection 10.1, based on decomposing a graph into blocks.

Proposition 11.5.1 If G is a connected bipartite graph with a bipartition(V1, V2) and |V1| = m ≤ |V2| = n, then a(K[G]) ≤ −n.Proof. Let V1 = {x1, . . . , xm}. Take a monomial xβ in the canonicalmodule of K[G], that is, β = (βi) is an integral vector in ri(R+A). Onemay assume m ≥ 2 because if m = 1 the graph G is a star and the result isclear. Thus βi ≥ 1 by Corollary 1.1.45, and

∑mi=1 βi =

∑n+mi=m+1 βi because

the edge cone lies in the hyperplane

m∑i=1

xi =

m+n∑i=m+1

xi.

494 Chapter 11

Hence deg(xβ), the normalized degree of xβ , is∑m+n

i=m+1 βi. Therefore

deg(xβ) ≥ n, which proves the required inequality. �

Corollary 11.5.2 [63] Let Km,n be the complete bipartite graph. If n ≥ m,then a(K[Km,n]) = −n.

Proof. By Exercise 10.7.26 and Proposition 11.5.1 it suffices to observethat the monomial xβ with

β = (n−m+ 1, 1, . . . , 1︸︷︷︸m entries

, 1, . . . , 1︸︷︷︸n entries

)

is in the canonical module of K[G] and has degree equal to n. �

There are 2-connected graphs where the a-invariant does not reach theupper bound of Proposition 11.5.1 (see Exercise 11.5.14).

A formula for the type of a determinantal ring is due to Brennan; see[73, p. 115]. We show a special case of this formula.

Proposition 11.5.3 Let G = Km,n be the complete bipartite graph. Ifn ≥ m, then K[G] is a level algebra and its type is given by:

type(K[G]) =

(n− 1

m− 1

).

Proof. From Exercise 10.7.26 it is seen that the canonical module ofK[G] isminimally generated by the set of monomials of the form xa11 · · ·xamm y1 · · · yn,such that a1 + · · ·+ am = n and ai ≥ 0 for all i. Thus the number of suchpartitions is the type. As the canonical module is generated in a singledegree the algebra K[G] is level. �

Proposition 11.5.4 If G is a bipartite connected graph with bipartition(V1, V2) and |V1| = m ≤ |V2| = n, then a(K[G]) = −m if and only if m = nand 1 = (1, . . . , 1) is in ri(R+A).

Proof. ⇒) By Corollary 1.1.45, the canonical module ωK[G] of the ring

K[G] is generated by monomials xβ such that all the entries of β are positiveintegers. As n = m we derive that x1 = x1x2 · · ·xm+n is in the canonicalmodule and it is the only monomial of normalized degree n in ωK[G].⇐) It follows from Proposition 11.5.1. �

Definition 11.5.5 Let G be a graph. A cycle containing all the vertices ofG is said to be a Hamilton cycle, and a graph containing a Hamilton cycleis said to be Hamiltonian.

Corollary 11.5.6 If G is a Hamiltonian bipartite graph with 2n vertices,then a(K[G]) = −n.


Proof. Let H be a Hamiltonian cycle of G and let 2n be the numberof vertices of G. Note that a(K[H ]) = −n (because K[H ] is the ring ofa hypersurface of degree n) and dim K[H ] = dim K[G]. Hence one hasa(K[H ]) ≤ a(K[G]) and consequently a(K[G]) = −n. �

Corollary 11.5.7 If G = Kn,n is a complete bipartite graph, then the a-invariant of K[G \ {e}] is equal to −n.

Proof. Note that for n ≥ 3 the graph G \ {e} is Hamiltonian. �

Decomposing a bipartite graph into blocks Let G be a connectedbipartite graph. Next we describe in a condensed form a technique thatcan be used to simplify the computation of the a-invariant and the type ofK[G]. It is based in the graph theoretical notion of block (see Section 7.1)and the following result.

Proposition 11.5.8 If B and C are two standard K-algebras, then

a(B ⊗K C) = a(B) + a(C).

If in addition B and C are Cohen–Macaulay, then

type(B ⊗K C) = type(B) · type(C).

Proof. The first equality follows from the proof of Proposition 5.1.11. Forthe second equality, see [171]. �

Proposition 11.5.9 If G1, . . . , Gr are the blocks of a connected bipartitegraph, then

a(K[G]) =

r∑i=1

a(K[Gi]) and type(K[G]) =

r∏i=1

type(K[Gi]).

Proof. Since G is connected and bipartite one has

K[G] � K[G1]⊗K K[G2]⊗K · · · ⊗K K[Gr].

Hence the result follows from Proposition 11.5.8. �

The next theorem is a combinatorial obstruction for an edge subring tobe Gorenstein. Because of Proposition 11.5.9 it suffices to consider the classof Gorenstein subrings of 2-connected bipartite graphs.

Theorem 11.5.10 Let G be a Gorenstein connected bipartite graph withbipartition (V1, V2), then G has a perfect matching and furthermore

|A| < |N(A)| for all A � V1.

496 Chapter 11

Proof. Let xβ be the generator of ωK[G]. For each 1 ≤ i ≤ m + n choosea spanning tree of G \ {xi} and enlarge this to a spanning tree Ti of G.Note that the monomial xγi with γi = (degTi

(x1), . . . , degTi(xn+m)) and

degTi(xi) = 1 is the generator of the canonical module of K[Ti]. Hence

since aff(R+Ai) = aff(R+A), where R+Ai is the edge cone of Ti, we getxγi ∈ ωK[G]. Therefore βi = 1, and consequently β = 1. Noting that β isin the affine space generated by the edge cone of G it follows m = n. Thusfrom Theorem 10.7.16 we obtain |A| < |N(A)| for all A � V1. �

The converse of Theorem 11.5.10 does not hold.

Example 11.5.11 Consider the bipartite graph G:

� ��

� ��

��

��

��

��

$$$$$$��

��

x1 x8

x5

x7

x2

x4

x6

x3

Using the following procedure inMacaulay2 [199], we rapidly obtain thatK[G] has type equal to three. On the other hand 1 = (1, . . . , 1) is in therelative interior of the edge cone. In this example the shift polyhedron hasvertices 1, (2, 1, 1, 1, 1, 1, 1, 2) and (1, 2, 1, 1, 1, 1, 2, 1).

KK=ZZ/31991

R=KK[x_1..x_8,t_1..t_11,MonomialOrder=>Eliminate 8]

I=ideal(t_1-x_1*x_5, t_2-x_1*x_6, t_3-x_1*x_7, t_4-x_2*x_5,

t_5-x_2*x_6, t_6-x_2*x_8, t_7-x_3*x_6, t_8-x_3*x_7,

t_9-x_3*x_8, t_10-x_4*x_7, t_11-x_4*x_8)

J= ideal selectInSubring(1,gens gb I)

M=coker gens J

res M

Corollary 11.5.12 Let G be a bipartite connected graph with bipartition(V1, V2) and |V1| = m ≤ |V2| = n. If G has no cut vertices and K[G] isGorenstein, then G has a perfect matching and a(K[G]) = −n.

Proof. It follows from the proof of Theorem 11.5.10. �

If one of the blocks of a graph G is a bridge e one can say a bit moreabout the relationship between G and G \ {e}.

Proposition 11.5.13 Let G be a graph with a bridge e. If G\{e} = G1∪G2

with G1 and G2 disjoint graphs such that K[Gi] is Gorenstein for i = 1, 2,then K[G] is Gorenstein.


Proof. There is an isomorphism K[G] � (K[G1]⊗KK[G2])[t], where t is avariable. Since the tensor product of Gorenstein rings is Gorenstein and apolynomial ring over a Gorenstein ring is Gorenstein, the result follows. �

Exercises

11.5.14 Consider the following bipartite graph G:

� � � ��

�

�

��

��

��

��

��

��

��

��

x1 x8x2 x7

x6 x3

x5 x4

Using Theorems 11.3.1 and 11.4.1, show that a(K[G]) = −5 and that theshift polyhedron has four vertices.

11.5.15 Prove that the following are the equations of the edge cone of thegraph G shown on the left.

�x1�x2

� x5�x3 �� x4

��

��

��

��

��

−x1 ≤ 0,−x4 ≤ 0,−x5 ≤ 0,x3 ≤ x1 + x2 + x4 + x5,x2 ≤ x1 + x3 + x4 + x5,

x1 + x4 + x5 ≤ x2 + x3.

Show that the a-invariant of K[G] is equal to −4.

11.5.16 Let G be the graph consisting of two squares joined by an edge.Prove that K[G] is a Gorenstein ring, its a-invariant is −5, it has a perfectmatching, and α = (1, . . . , 1) is not in the relative interior of the edge cone ofG. Prove that the monomial corresponding to (1, 1, 2, 1, 1, 2, 1, 1) generatesthe canonical module of K[G].

11.5.17 Let G be a graph with a bridge e. If G \ {e} = G1 ∪ G2 with G1

and G2 disjoint graphs, then

a(G) = a(G1) + a(G2)− 1.

11.5.18 Let G be a graph and let A be the set of column vectors of itsincidence matrix. If K[G] is normal, then the canonical module of K[G] is:

ωK[G] = ({xa| a ∈ NA ∩ ri(R+A)}).

Use this formula to prove the following:

498 Chapter 11

(a) The join G1 ∗G2 of two disjoint graphs G1 and G2 consists of G1∪G2

and all the lines joining G1 with G2. If G1, G2 are connected graphson n vertices and K[G1], K[G2] are normal, then

a(K[G1 ∗G2]) = −n.

(b) Let G be a connected non-bipartite graph on n vertices. If K[G] isnormal, then

a(K[G])− 1 ≤ a(K[C(G)]) ≤ −⌈n+ 1

2

⌉.

(c) Let G1, G2 be two connected non-bipartite graphs. If K[G1] andK[G2] are normal, then

a(K[G1]) + a(K[G2]) ≤ a(K[G1 ∗G2]).

Chapter 12

Normality of ReesAlgebras of MonomialIdeals

Let R = K[x1, . . . , xn] be a polynomial ring over a field K and let I be anideal of R generated by a finite set F = {xv1 , . . . , xvq} of monomials. In thischapter we study the integral closure of I, and the normality and invariantsof R[It], the Rees algebra of I. The normalization of a Rees algebra isexamined using the Danilov–Stanley formula, Caratheodory’s theorem, andHilbert bases of Rees cones.

Interesting classes of normal ideals, such as ideals of Veronese type andpolymatroidal ideals, are introduced in the chapter. The divisor class groupof a normal Rees algebra is computed using polyhedral geometry.

12.1 Integral closure of monomial ideals

Proposition 12.1.1 [275] Let I be a monomial ideal of a polynomial ringR over a field K. Then I, the integral closure of I, is a monomial ideal.

Proposition 12.1.2 Let R = K[x1, . . . , xn] be a polynomial ring over afield K and let I ⊂ R be a monomial ideal. Then

(a) I = (xα|xmα ∈ Im for some m ≥ 1), and

(b) I = (xα|α ∈ conv(log(I)) ∩ Zn).

Proof. (a): If xmα ∈ Im, then xα satisfies a polynomial of the formzm+ am, with am ∈ Im, hence xα is in I. On the other hand if z = xα ∈ I,then there is an equation zr + a1z

r−1 + · · · + ar−1z + ar = 0 with ai ∈ Ii

500 Chapter 12

for all i. Since I is a monomial ideal one obtains zm ∈ Im for some m ≥ 1.Observing that I is a monomial ideal the asserted equality follows.

(b): “⊃”: Let α ∈ conv(log I) ∩ Zn. One can write α =∑q

i=1 μiβi,where μi ∈ Q+,

∑qi=1 μi = 1 and xβi ∈ I. Pick m ∈ N+ such that mμi ∈ N

for all i. Then xmα ∈ Im and xα ∈ I.“⊂”: Since I is generated by monomials for any xα ∈ I there is m ∈ N+

such that xmα ∈ Im. It follows readily that α ∈ conv(log I). �

Example 12.1.3 If R = K[x1, x2, x3] and I = (x31, x2x23), then using the

program Normaliz [68] we get I = (x31, x2x23, x

21x2x3).

A geometric description of the integral closure Let α ∈ Qn+, whereQ+ is the set of nonnegative rational numbers. We define the upper rightcorner or ceiling of α as the vector �α� whose entries are given by

�α�i ={αi if αi ∈ N,(αi)+ 1 if αi /∈ N,

where (αi) stands for the integer part of αi. Accordingly we can definethe ceiling of any vector in Rn or the ceiling of any real number. Letconv(v1, . . . , vq) be the convex hull (over the rationals), that is,

conv(v1, . . . , vq) =

{q∑i=1

λivi

∣∣∣∣∣q∑i=1

λi = 1, λi ∈ Q+

}is the set of all convex combinations of v1, . . . , vq.

Proposition 12.1.4 Let R = K[x1, . . . , xn] be a polynomial ring over afield K. If I ⊂ R is an ideal generated by monomials xv1 , . . . , xvq , then

I =({x�α�

∣∣∣ α ∈ conv(v1, . . . , vq)})

and I is generated by all xa with a ∈ (conv(v1, . . . , vq) + [0, 1)n) ∩Nn.

Proof. “⊃”: Let α =∑qi=i λivi be a convex combination of v1, . . . , vq with

λi ∈ Q+ for all i. Since �α� ≥ α, there is β ∈ Qn+ such that �α� = β + α.Hence there is 0 �= p ∈ N so that pβ ∈ Nn and pλi ∈ N for all i. Therefore

xp�α� = xpβxpα = xpβ (xv1)pλ1 · · · (xvq )pλq ∈ Ip ⇒ x�α� ∈ I.

“⊂”: Let xγ ∈ I, that is, xpγ ∈ Ip for some 0 �= p ∈ N. There arenonnegative integers s1, . . . , sq such that

xpγ = xδ (xv1)s1 · · · (xvq )sq and s1 + · · ·+ sq = p.

Hence γ = (δ/p) +∑q

i=1 (si/p) vi. We set α =∑q

i=1(si/p)vi. By dividingthe entries of δ by p, we can write γ = θ + β + α, where 0 ≤ βi < 1 forall i and θ ∈ Nn. Notice β + α ∈ Nn. It is seen that �α� = β + α. Thusxγ = xθx�α� with α ∈ conv(v1, . . . , vq) as required. �

Normality of Rees Algebras of Monomial Ideals 501

Example 12.1.5 If I = (x31, x42) ⊂ K[x1, x2], by Proposition 12.1.4, I is

generated by the monomials of the points marked in the figure:

�

�

��

1

2

3

4

0 1 2 3

��

��

��

Thus I = (x31, x42, x1x

32, x

21x

22).

Corollary 12.1.6 Let I ⊂ R = K[x1, . . . , xn] be an ideal generated bymonomials of degree at most k. The following hold.

(a) I is generated by monomials of degree at most k + n− 1.

(b) If ht(I) = n, then I is generated by monomials of degree at most k.

Proof. (a) Let G(I) = {xv1 , . . . , xvq} be the minimal set of generators ofI. Take α = (αi) ∈ conv(v1, . . . , vq) and set β = (βi) = �α�. As βi is equalto αi + δi, for some δi in [0, 1), one has deg(xβ) < k + n, as required.

(b) One has (xk1 , . . . , xkn) = (x1, . . . , xn)

k ⊂ I because xki ∈ I for all i.Therefore I does not require minimal generators of degree k + 1. �

Complete and m-full ideals in dimension two Let R = K[x1, x2] bea polynomial ring over an infinite field K, let m = (x1, x2) and let I be anm-primary ideal of R minimally generated by q monomials, q = μ(I), thatare listed lexicographically,

I = (xa11 , xa21 x

bq−1

2 , . . . , xai1 xbq−i+1

2 , . . . , xaq−1

1 xb22 , xb12 )

with a1 > a2 > · · · > aq−1 > aq := 0, b1 > b2 > · · · > bq−1 > bq := 0. Weassume that I �= mq−1.

The ideal I is said to be m-full if (mI : f) = I for some f ∈ m \ m2 orequivalently I is m-full if (I : m) = (I : f) [259, Proposition 14.1.6]. A usefulcharacterization of m-full ideals is the following result.

Theorem 12.1.7 (Rees–Watanabe [425, Theorem 4]) If I is an m-primaryideal of a regular local ring (R,m) of dimension two, then I is m-full if andonly if for all ideals I ⊂ J , μ(J) ≤ μ(I).

502 Chapter 12

The order of I is ord(I) := max{k| I ⊂ mk}. In regular local rings ofdimension two one has μ(I) ≤ ord(I) + 1 [259, Lemma 14.1.3].

Lemma 12.1.8 If μ(I) = ord(I) + 1, then I is m-full.

Proof. Let J be an ideal of R. Note that I ⊂ J implies ord(J) ≤ ord(I).Since μ(J) ≤ ord(J) + 1 ≤ μ(I), I satisfies Theorem 12.1.7. �

The following result classifies all m-full monomial ideals of K[x1, x2].

Theorem 12.1.9 [181] I is m-full if and only if there is 1 ≤ k ≤ q suchthat the following conditions hold

(i) bq−i − bq−i+1 = 1 for 1 ≤ i ≤ k − 1,

(ii) k = q or k < q and bq−k − bq−k+1 ≥ 2,

(iii) ai − ai+1 = 1 for k ≤ i ≤ q − 1.

Proof. ⇒) If I is m-full, ord(I) ≤ q − 1, as otherwise I ⊂ (x1, x2)q, which

has q + 1 minimal generators, which would violate Theorem 12.1.7. Thusord(I) ≤ q − 1. Let xk1x

q−1−k2 be a monomial of I of degree q − 1. This

means that there are at most q − 1 − k elements prior to xk1xq−1−k2 and at

most k elements after. This gives

I = (xa11 , xa21 x2, . . . , x

ak−1

1 xq−2−k2 , xk1x

q−1−k2 , xk−1

1 xbq−k

2 , . . . , x1xb22 , x

b12 ).

By choosing k as small as possible we achieve all three conditions.⇐) By Lemma 12.1.8 we need only show μ(I) ≥ ord(I) + 1. Since

q = μ(I) its suffices to show that I has a monomial of degree q − 1. Noticethat bq−i = i for 1 ≤ i ≤ k−1 and ai = q− i for k ≤ i ≤ q−1. In particular

ak = q − k and bq−k+1 = k − 1. Thus the monomial xq−k1 xk−12 belongs to I

and has degree q − 1. �

If I is complete, then I is m-full [425, Theorem 5]. In polynomial ringsin two variables any complete ideal is normal; more generally one has thefollowing result of Zariski.

Theorem 12.1.10 [436, Appendix 5] If I1, . . . , Ir are complete ideals in aregular local ring R of dimension two, then I1 · · · Ir is complete.

In [101] Crispin Quinonez studied the normality of m-primary monomialideals in K[x1, x2] and established a criterion in terms of certain partialblocks and associated sequences of rational numbers.

Example 12.1.11 The ideal I = (x31, x21x

82, x1x

152 , x

212 ) is m-full and is not

normal. The integral closure of I is I = (x31, x21x

72, x1x

142 , x

212 ).


Linear programming test We give a test to check whether or not agiven monomial lies in the integral closure of a monomial ideal.

Proposition 12.1.12 (Membership test) Let v1, . . . , vq be a set of vectorsin Nn and let A be the n×q matrix whose columns are the vectors v1, . . . , vq.If I is the ideal generated by xv1 , . . . , xvq , then a monomial xb lies in theintegral closure of I if and only if the linear program:

Maximize x1 + · · ·+ xq

Subject to

Ax ≤ b and x ≥ 0

has an optimal value greater than or equal to 1, which is attained at a vertexof the rational polytope P = {x ∈ Rq |Ax ≤ b and x ≥ 0}.

Proof. ⇒) Let xb ∈ I, that is, xpb ∈ Ip for some positive integer p. Thereare nonnegative integers r1, . . . , rq such that xpb = xδ (xv1)

r1 · · · (xvq )rq andr1 + · · ·+ rq = p. Hence the column vector c with entries ci = ri/p satisfiesAc ≤ b and c1+ · · ·+ cq = 1. Thus the linear program has an optimal valuegreater than or equal to 1.⇐) Since the vertices of P have rational entries (see Proposition 1.1.46)

and the maximum of x1 + · · ·+ xq is attained at a vertex of the polytope P(see Proposition 1.1.41), there are c1, . . . , cq in Q+ such that c1+· · ·+cq ≥ 1and c1v1+· · ·+cqvq ≤ b. Hence there are ε1, . . . , εq in Q such that 0 ≤ εi ≤ cifor all i and

∑qi=1 εi = 1. Therefore there is a vector δ ∈ Qn+ such that

b = δ +∑qi=1 εivi. Thus there is an integer p > 0 such that

pb = pδ︸︷︷︸∈Nn

+ pε1︸︷︷︸∈N

v1 + · · ·+ pεq︸︷︷︸∈N

vq ⇒ xb ∈ I. �

Remark 12.1.13 By Theorem 1.1.56 one can also use the dual problem

Minimize b1y1 + · · ·+ bnyn

Subject to

yA ≥ 1 and y ≥ 0

to check whether xb is in I. In this case one has a fixed polyhedron

Q = {y ∈ Rn | yA ≥ 1 and y ≥ 0}

that can be used to test membership of any monomial xb, while in the primalproblem the polytope P depends on b.

Example 12.1.14 Consider the ideal I = (x31, x42, x

53, x

64, x1x2x3x4) and

the exponent vector b = (0, 1, 1, 4). To check whether xb is in I we use thefollowing procedure in Mathematica [431]:

504 Chapter 12

ieq:={3x1+x5<=0,4x2+x5<=1,5x3+x5<=1,6x4+x5<=4}

vars:={x1,x2,x3,x4,x5}

f:=x1+x2+x3+x4+x5

ConstrainedMax[f,ieq,vars]

The answer that Mathematica gives is:

{67/60, {x1 -> 0, x2 -> 1/4, x3 -> 1/5, x4 -> 2/3, x5 -> 0}}

where the first entry is the optimal value and the other entries correspondto a vertex of the polytope P . By Proposition 12.1.12, we get xb ∈ I.

Using PORTA [84] one obtains that the vertices of the polyhedral set Qdescribed in Remark 12.1.13 are the rows of the matrix:

M =

⎡⎢⎢⎣1/3 1/4 1/5 13/601/3 1/4 1/4 1/61/3 3/10 1/5 1/623/60 1/4 1/5 1/6

⎤⎥⎥⎦.This is a “membership test matrix” in the sense that a monomial xb lies

in I if and only if Mb ≥ 1. If xb = x2x3x44, we get:

Mb = (79/60, 7/6, 7/6, 67/60)� ≥ (1, 1, 1, 1) ⇒ xb ∈ I.

Exercises

12.1.15 Let R = K[x1, . . . , xn] be a polynomial ring over a field K and Fa finite set of monomials of the same degree. If I = (F ) is complete, then

conv(log(F )) ∩ Zn = log(F ).

12.1.16 Let R = K[x1, . . . , xn] be a polynomial ring over a field K. Ifm = (x1, . . . , xn), then md is the integral closure of (xd1 , . . . , x

dn).

12.1.17 Let I = (xv1 , . . . , xvq ) be a monomial ideal of a polynomial ring

over a field K. Prove that Ii = (xiv1 , . . . , xivq ) for i ≥ 1.

12.1.18 If I ⊂ K[x1, x2] is an ideal generated by monomials of degree k,then so is I.

12.1.19 Prove that I ∩ J �= I ∩ J , where I = (x51, x32) and J = (x52, x

23).

12.1.20 If I = (x21, x32) ⊂ K[x1, x2] and P = conv((2, 0), (0, 3)), then P∩Z2

is equal to {(2, 0), (0, 3)} and I = I + (x1x22).

12.1.21 If I = (x21, x1x22, x

32) ⊂ K[x1, x2], then I is normal.


12.2 Normality criteria

In this section we present a simple normality criterion which is suitable tostudy the normality of monomial ideals.

Proposition 12.2.1 Let (R,m) be a Noetherian local ring and let I be aproper ideal of R. Then I is normal if and only if

(a) IRp is normal for any prime ideal I ⊂ p �= m, and

(b) Ir ∩ (Ir : m) = Ir for all r ≥ 1.

Proof. ⇒) That (a) is satisfied follows from the fact that integral closuresand powers of ideals commute with localizations. Part (b) is clearly satisfied.⇐) Assume Ir �= Ir for some r ≥ 1. Take p ∈ AssR(M), where M =

Ir/Ir. Since p is in the support of M we get Mp �= (0) and consequently,by (a), we must have p = m. Hence there is an embedding

R/m ↪→M (1 �→ x0),

where x0 ∈ Ir \ Ir and m = ann(x0). From (b) one concludes that x0 ∈ Ir,a contradiction. Thus Ir = Ir for all r ≥ 1. �

Proposition 12.2.2 Let (R,m) be a Noetherian local ring and let I be aproper ideal of R. Then I is integrally closed if and only if IRp is integrallyclosed for any prime ideal I ⊂ p �= m and I ∩ (I : m) = I.


The next result proves that a normal monomial ideal of K[x1, . . . , xn]stays normal if we make any variable, say xn, equal to 1 or 0.

Proposition 12.2.3 Let I = (xv1 , . . . , xvq ) be a normal ideal. If each xvi

is written as xain gi, where gi is a monomial in R′ = K[x1, . . . , xn−1], then

(a) J = (g1, . . . , gq) ⊂ R′ is a normal ideal, and

(b) L = ({gi| ai = 0}) ⊂ R′ is a normal ideal.

Proof. (a) We will show that Jr = Jr for all r ≥ 1. Take a monomial xα

of R′ that belong to Jr. Then (xα)p ∈ Jrp for some p ≥ 1 and we can write

(xα)p = gb11 · · · gbqq xβ ,

where∑qi=1 bi = rp and xβ ∈ R′. Multiplying both sides of the equation

above by xspn with s =∑q

i=1 aibi, we get (xsnxα)p ∈ Irp. Hence xsnx

α ∈ Irand we can write xsnx

α = (xa1n g1)c1 · · · (xaqn gq)cqxγ , where

∑qi=1 ci = r.

Evaluating the last equality at xn = 1 gives that xα ∈ Jr, as required. Part(b) follows using similar arguments. �

506 Chapter 12

Theorem 12.2.4 (Normality criterion) Let I = (xv1 , . . . , xvq ) be a mono-mial ideal of R = K[x1, . . . , xn] and let Ji be the ideal of R generated byall monomials obtained from {xv1 , . . . , xvq} by making xi = 1. Then I isnormal if and only if

(a) Ji is normal for all i (resp. Ip is normal for any prime ideal p �= m),

(b) Ir ∩ (Ir : m) = Ir for all r ≥ 1, where m = (x1, . . . , xn).

Proof. ⇒) Condition (a) follows from Proposition 12.2.3 (resp. that Ip isnormal follows using that integral closures and powers of ideals commutewith localizations; see Proposition 4.3.4). Part (b) is clearly satisfied.⇐) To prove that Ir = Ir set M = Ir/Ir. We proceed by contradiction

assuming Ir � Ir. Take p ∈ AssR(M). There are embeddings

R/pϕ↪→M ↪→ R/Ir.

Thus p ∈ AssR(R/Ir). Consequently p is generated by a subset of

{x1, . . . , xn} and p ⊂ m. Note p = m, otherwise if xi /∈ p for some i, thenby (a) IRp = JiRp is normal and Mp = (0), a contradiction because p isin the support of M . Thus p = m. If x0 = ϕ(1), we get x0 ∈ Ir \ Ir andx0 ∈ (Ir : m), which contradicts (b). �

Proposition 12.2.5 Let R = ⊕∞i=0Ri be an N-graded ring and let I = R≥α

be the ideal of R generated by the elements of degree at least α. If R is adomain, then I is integrally closed.

Proof. For f = fs + fs+1 + · · ·+ fr ∈ R, with fi ∈ Ri for all i and fs �= 0,we set deg(f) = s. As R is a domain one has deg(gh) = deg(g)+deg(h) forg, h ∈ R and gh �= 0. Take 0 �= f ∈ I with s = deg(f). There is an equation

fn + a1fn−1 + · · ·+ an−1f + an = 0 (ai ∈ Ii).

Assume s < α. If aifn−i �= 0, then deg(aif

n−i) ≥ iα+ (n− i)s > ns. Thusone can rewrite the equation above as

fns + (terms of degree greater than ns) = 0,

a contradiction because fs �= 0 and R is graded. Hence deg(f) ≥ α. �

Theorem 12.2.6 [153] Let K be a field and let R = K[x1, . . . , xn] be apositively graded K-algebra. If R is a domain and c is the least commonmultiple of the weights of the xi, then I = R≥nc is a normal ideal. If R isnormal, then the Rees ring R[It] is normal.

Proof. In [153] it is proved that Ip is equal to R≥pnc for all p ≥ 1, wherethe latter is integrally closed. �


Ideals of mixed products Let R = K[x1, . . . , xm, y1, . . . , yn] be a ringof polynomials over a field K. Given k, r, s, t in N such that k + r = s+ t,consider the square-free monomial ideal of mixed products given by

L = IkJr + IsJt,

where Ik (resp. Jr) is the ideal of R generated by the square-free monomialsof degree k in the variables x1, . . . , xm (resp. y1, . . . , yn). The Betti numbersand some of the invariants of L have been studied in [246, 260, 355]. Theinvariants of the symmetric algebra of L are studied in [288].

Next we present a complete classification of the normal ideals of mixedproducts that can be shown using the normality criterion of Theorem 12.2.4.

Theorem 12.2.7 [351] If L �= R, then L is normal if and only if it can bewritten (up to permutation of k, s and r, t) in one of the following forms:

(a) L = IkJr + Ik+1Jr−1, k ≥ 0 and r ≥ 1.

(b) L = IkJr, k ≥ 1 or r ≥ 1.

(c) L = IkJr + IsJt, 0 = k < s = m, or 0 = t < r = n, or k = t = 0, s = 1.

Definition 12.2.8 Let G be a simple graph (resp. digraph) with verticesx1, . . . , xn and let t ≥ 2 be an integer. The path ideal of G, denoted byIt(G), is the ideal of K[x1, . . . xn] generated by all square-free monomialsxi1 · · ·xit such that the xij is adjacent to xij+1 (resp. ej = (xij , xij+1 ) is adirected edge from xij to xij+1 ) for all 1 ≤ j ≤ t− 1.

Path ideals of digraphs were introduced by Conca and De Negri [89]and later studied in [50, 211]. Kubitzke and Olteanu [284] have studied thealgebraic properties and invariants of path ideals of certain types of posets.

Corollary 12.2.9 If G is a complete bipartite graph, then the path idealIt(G) is normal for all t ≥ 2.

Proof. Let x1, . . . , xm, y1, . . . , yn be the vertex set of G, one may assumethat the edges of G are precisely the pairs of the form {xi, yj}. ThereforeIt(G) is IdJd+1 + Id+1Jd if t = 2d + 1 and It(G) is IdJd if t = 2d. Hencethe result follows from Theorem 12.2.7. �

Exercises

12.2.10 Let R = K[x1, . . . , xm, y1, . . . , yn] and L = IsJt, where Is (resp.Jt) denote the ideal of R generated by the square-free monomials of degrees (resp. t) in the xi (resp. yi) variables. Prove that L is normal.

508 Chapter 12

12.2.11 Let R = K[x1, . . . xn] and R[x] be polynomial rings over a fieldK and let Lt = It + xIt−2, where It is the ideal of R generated by thesquare-free monomials of degree t ≥ 2. Then Lt is a normal ideal of R[x].

12.2.12 Let I and J be two ideals of the polynomial rings K[x1, . . . , xm]and K[y1, . . . , yn], respectively. The join of I and J is:

I ∗ J = I + J + (xiyj | 1 ≤ i ≤ m and 1 ≤ j ≤ n).

If I and J are normal ideals generated by square-free monomials of the samedegree t ≥ 2, then their join I ∗ J is normal.

12.2.13 Let X = {x1, . . . , xm} and {y1, . . . , yn} be two sets of variablesover a field K. Let I be a normal ideal of K[X ] generated by square-freemonomials of degree t and let L = I + (X)(Y ). Then L is a normal ideal.

12.2.14 Let R = K[x1, . . . , x8] be a polynomial ring over a field K ofcharacteristic zero and let I be the ideal of R generated by

x1x2x3x4, x5x6x7x8, x1x2x4x5, x3x4x6x8, x1x3x7, x2x7x8.

Prove that I is normal, whereas the ideal J = I + (x1y1, . . . , x8y1) is not.

12.2.15 Let R = K[x1, . . . , x6] be a polynomial ring over a field K. Provethat I = (x3x5x6, x3x4x6, x2x5x6, x2x4x5, x1x3x5, x1x2x6) is not normal.

12.3 Rees cones and polymatroidal ideals

Let R = K[x1, . . . , xn] be a polynomial ring over a field K. In this part weprove that Rees algebras of polymatroidal ideals are normal domains. Inparticular ideals of Veronese type are normal.

Quasi-ideal Rees cones Let I be a monomial ideal of R minimally gen-erated by the set F = {xv1 , . . . , xvq}. The Rees cone of I is the rationalpolyhedral cone in Rn+1, denoted by R+A′ or R+(I), generated by

A′ := {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Rn+1,

where ei is the ith unit vector. Consider the index set

J = {1 ≤ i ≤ n| 〈ei, vj〉 = 0 for some j} ∪ {n+ 1}.

Notice that dimR+A′ = n + 1. By Theorem 1.4.2, the Rees cone has aunique irreducible representation

R+A′ =

(⋂i∈J

H+ei

)⋂(r⋂i=1

H+�i

)(12.1)


such that, for each i, 0 �= �i ∈ Zn+1, the non-zero entries of �i are relativelyprime, the first n entries of �i are in N and the last entry of �i is negative.

Let P be the convex hull of A = {v1, . . . , vq}. Recall that the Ehrhartring of P and the Rees algebra of I are the monomial subrings given by

A(P) = K[xatb| a ∈ bP ∩ Zn] and R[It] = K[x1, . . . , xn, xv1 t, . . . , xvq t]

respectively, where t is a new variable. Using Theorem 9.1.1, it followsreadily that the normalization of R[It] is

R[It] = K[xatb| (a, b) ∈ R+A′ ∩ Zn+1].

Notation For use below we set [n] = {1, . . . , n}.

Theorem 12.3.1 If �k has the form �k = −dken+1 +∑

i∈Akei for some

Ak ⊂ [n] for k = 1, . . . , r, then A(P)[x1, . . . , xn] = R[It].

Proof. “⊂”: This inclusion is clear because A(P) ⊂ R[It].“⊃”: Let xatb = xa11 · · ·xann tb ∈ R[It] be a minimal generator, i.e., (a, b)

cannot be written as a sum of two non-zero integral vectors in the Rees coneR+A′. We may assume ai ≥ 1 for 1 ≤ i ≤ m, ai = 0 for i > m, and b ≥ 1.

Case (I): 〈(a, b), �i〉 > 0 for all i. The vector γ = (a, b) − e1 satisfies〈γ, �i〉 ≥ 0 for all i, that is γ ∈ R+A′. Thus since (a, b) = e1 + γ we derivea contradiction.

Case (II): 〈(a, b), �i〉 = 0 for some i. We may assume

{�i| 〈(a, b), �i〉 = 0} = {�1, . . . , �p}.

Subcase (II.a): ei ∈ H�1 ∩ · · · ∩H�p for some 1 ≤ i ≤ m. It is not hardto verify that the vector γ = (a, b) − ei satisfies 〈γ, �k〉 ≥ 0 for all k. Thusγ ∈ R+A′, a contradiction because (a, b) = ei + γ.

Subcase (II.b): ei /∈ H�1 ∩ · · · ∩H�p for all 1 ≤ i ≤ m. Since the vector(a, b) belongs to the Rees cone it follows that we can write

(a, b) = λ1(v1, 1) + · · ·+ λq(vq, 1) (λi ≥ 0).

Therefore a ∈ bP , i.e., xatb ∈ A(P). �

If �1, . . . , �r satisfy the condition of Theorem 12.3.1 (resp. dk = 1 for allk), we say that the Rees cone of I is quasi-ideal (resp. ideal).

Corollary 12.3.2 If the Rees cone of I is quasi-ideal and K[Ft] = A(P),then the Rees algebra R[It] is normal.

Proof. By Theorem 12.3.1 we get

R[It] = K[Ft][x1, . . . , xn] = A(P)[x1, . . . , xn] = R[It]. �

510 Chapter 12

Polymatroidal sets of monomials Let A = {v1, . . . , vq} ⊂ Nn be theset of bases of a discrete polymatroid of rank d, i.e., |vi| = d for all i andgiven any two a = (ai), c = (ci) in A, if ai > ci for some index i, then thereis an index j with aj < cj such that a−ei+ej is in A. We refer to [221, 227]for the theory of discrete polymatroids.

The set F = {xv1 , . . . , xvq} (resp. the ideal I = (F ) ⊂ R) is called apolymatroidal set of monomials (resp. a polymatroidal ideal).

Lemma 12.3.3 F ′ = {xvi/x1 : x1 occurs in xvi} is also polymatroidal.


Lemma 12.3.4 Let A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)}. If E is a facetof R+A′, then E = R+A′ ∩ Hb, where b = ei for some 1 ≤ i ≤ n + 1 orb = (bi) ∈ Zn+1, bi ∈ {0, 1} for i = 1, . . . , n and bn+1 ≥ −d.

Proof. The proof is by induction on d. The case d = 1 is clear because thematrix whose columns are the vectors in A′ is totally unimodular.

Assume that d ≥ 2. For 1 ≤ k ≤ q, we set fvi = (vi, 1), Bi = supp(vi)and vk = (vk1, . . . , vkn). Let E be a facet of R+A′. There is a unique0 �= b = (bi) in Zn+1 whose non-zero entries are relatively prime such that

(a) E = R+A′ ∩Hb �= R+A′, R+A′ ⊂ H+b , and

(b) there is a linearly independent set A ⊂ Hb ∩ A′ with |A| = n.

If A = {e1, . . . , en} (resp. A ⊂ {fv1 , . . . , fvq}), then b = en+1 (resp. b =(1, . . . , 1,−d)). Notice that bi ≥ 0 for i = 1, . . . , n. Thus we may assumethat A is the set {e1, . . . , es, fv1 , . . . , fvt}, where 1 ≤ s ≤ n − 1, s + t = n,and {1, . . . , s} is the set of all i ∈ [n] such that ei ∈ Hb. We may assumethat 1 ∈ Bk for some 1 ≤ k ≤ q. Indeed if 1 is not in ∪qi=1Bi, then thefacets of R+A′ different from He1 ∩R+A′ are in one-to-one correspondencewith the facets of

R+{e2, . . . , en, fv1 , . . . , fvq}.Assume that 1 /∈ B1. Since vk1 > v11 for some k, by the symmetric exchangeproperty of A [221, Theorem 4.1, p. 241] there is j with vkj < v1j such thatv1 + e1 − ej = vi for some vi in A. Hence

fv1 + e1 = fvi + ej ⇒ 〈fvi , b〉 = −〈ej , b〉 ⇒ 〈fvi , b〉 = 〈ej , b〉 = 0,

i.e., fvi and ej belong to Hb. Thus from the outset we may assume that1 ∈ B1. Next assume that t ≥ 2 and 1 /∈ B2. Since v11 > v21, by thesymmetric exchange property of A there is j with v1j < v2j such thatv2 + e1 − ej = vi for some vi in A. Hence

fv2 + e1 = fvi + ej ⇒ 〈fvi , b〉 = 〈ej , b〉 = 0,


i.e., fvi and ej belong to Hb. Notice that i �= 1. Indeed if i = 1, thenfrom the equality above we get that {fv2 , e1, fv1 , ej} is linearly dependent,a contradiction. Applying the arguments above repeatedly shows that wemay assume that 1 belongs to Bi for i = 1, . . . , t. We may also assume thatv1, . . . , vr is the set of all vectors in A such that 1 ∈ Bi, where t ≤ r. For1 ≤ i ≤ r, we set v′i = vi − e1. By Lemma 12.3.3 the set {v′1, . . . , v′r} isagain the set of basis of a discrete polymatroid of rank d− 1. Consider theRees cone R+A′′ generated by

A′′ = {e1, e2, . . . , en, fv′1 , . . . , fv′r}.

Notice that R+A′′ ⊂ H+b and that {e1, e2, . . . , es, fv′1 , . . . , fv′t} is a linearly

independent set. Thus R+A′′ ∩ Hb is a facet of the cone R+A′′ and theresult follows by induction. �

Theorem 12.3.5 [221] If P = conv(v1, . . . , vq), then K[Ft] = A(P).

Theorem 12.3.6 [422] If I is a polymatroidal ideal, then I is normal.

Proof. By Lemma 12.3.4 and Theorem 12.3.5 R+A′ is quasi-ideal andK[Ft] = A(P). Thus, by Corollary 12.3.2, we get that R[It] is normal. �

Corollary 12.3.7 If I is a polymatroidal ideal, then I has the persistenceproperty.

Proof. By Theorems 12.3.6 and 7.7.3 I has the persistence property. �

Normality of an ideal of Veronese type Let R = K[x1, . . . , xn] be aring of polynomials over an arbitrary field K. Given a sequence of integers(s1, s2, . . . , sn; d) such that 1 ≤ sj ≤ d ≤

∑ni=1 si for all j, we define A as

the set of partitions

A = {(a1, . . . , an) ∈ Zn| a1 + · · ·+ an = d; 0 ≤ ai ≤ si ∀ i},

and F as the set of monomials F = {xa| a ∈ A} = {f1, . . . , fq}.

Definition 12.3.8 The ideal I = (F ) ⊂ R is said to be of Veronese typeof degree d with defining sequence (s1, . . . , sn; d). If si = 1 for all i we callI the dth square-free Veronese ideal , and if si = d for all i we call I the dthVeronese ideal . Similar terminology may be applied to subrings.

The monomial subring K[F ] ⊂ R and its toric ideal have been studiedby Sturmfels in [400]. In loc. cit. it is shown that the toric ideal of K[F ]has a quadratic Grobner basis whose initial ideal is square-free.

Proposition 12.3.9 [147] If I ⊂ R is an ideal of Veronese type, then I isnormal.

512 Chapter 12

Proof. The ideal I is a polymatroidal ideal. Then, the Rees algebra of Iis normal by Theorem 12.3.6, i.e., I is a normal ideal. �

Proposition 12.3.10 [418] If I is the dth square-free Veronese ideal of R,then I is normal.

Proof. The result follows at once from Proposition 12.3.9. �

The basis monomial ring of a matroid LetM be a matroid on a finiteset X = {x1, . . . , xn} and let B be the collection of bases of M . The clutterwith vertex set X and edge set B is called the clutter of bases of M .

Definition 12.3.11 The set of all xi1 · · ·xir ∈ R such that {xi1 , . . . , xir}is in B is denoted by FM . The basis monomial ring ofM is the ring K[FM ].The square-free monomial ideal (FM ) is the basis monomial ideal of M .

It is well known [338] that all bases of a matroid M have the samenumber of elements; this common number is called the rank of the matroid.Thus K[FM ] is a standard graded K-algebra and all monomials of FM havethe same degree; see Section 9.2.

Corollary 12.3.12 If I = (FM ) is the basis monomial ideal of a matroid,then R[It] is normal.


Corollary 12.3.13 [429] K[FM ] is normal.

Proof. As K[FM t] � K[FM ], by Theorems 12.3.5 and 9.3.6, K[FM ] isnormal. �

Definition 12.3.14 The standard graded K-algebra S = K[FM ] is said tobe Koszul if the residue field K has a linear S-free resolution.

A sufficient condition for K[FM ] to be Koszul is that its toric ideal hasa quadratic Grobner basis; see [46, 221]. In [430, Conjecture 12] Whiteconjectured that the toric ideal of K[FM ] is generated by quadrics. Morerecently Blum [46] asked whether K[FM ] is Koszul. An open problem posedby Bøgvad is whether the toric ideal of a smooth projectively normal toricvariety is generated by quadrics (see [61, 400]).

Exercises

12.3.15 Let G = {xu1 , . . . , xus} and let d = maxi{|ui|}. Suppose thatF = {xu1 , . . . , xur} is the set of all xui of degree d. If Q = conv(u1, . . . , us)and K[Gt] = A(Q), then K[Ft] = A(P), where P = conv(u1, . . . , ur).

12.3.16 If B = {{x1, x2}, {x2, x3}, {x3, x4}, {x1, x4}}, prove that B satisfiesthe bases exchange property of matroids.


12.4 Veronese subrings and the a-invariant

In this section we present an explicit generating set for the canonical moduleof a square-free Veronese subring and compute its a-invariant. A sharpupper bound for the a-invariant of the normalization of a subring generatedby square-free monomials of the same degree will be presented, this boundis attained at a square-free Veronese subring.

Let R = K[x1, . . . , xn] = ⊕∞i=0Ri be a polynomial ring over a field K

with the standard grading and let k ∈ N+. An element in Rik is said tohave normalized degree i. The kth Veronese subring of R is given by:

R(k) :=

∞⊕i=0

Rik ⊂ R,

and the kth square-free Veronese subring of R is given by:

K[Vk] := K[{xi1 · · ·xik | 1 ≤ i1 < · · · < ik ≤ n}].

The kth Veronese subring of R is graded by (R(k))i = Rki. Thus R(k)

is a K-algebra generated by all the monomials of R of normalized degree 1and K[Vk] is a graded subring of R(k) with the normalized grading:

K[Vk] =

∞⊕i=0

(K[Vk])i,

where (K[Vk])i = K[Vk] ∩ (R(k))i. In what follows we shall assume thatK[Vk] has the normalized grading.

Let F a finite set of square-free monomials of the same degree k. In thiscase there are embeddings

K[F ] ⊂ K[Vk] ⊂ R(k).

Let us fix some of the notation that will be used throughout the restof the section. Let n ≥ 2k ≥ 4 be two integers (this is not an essentialrestriction; see Remark 12.4.11). We set

V = {ei1 + · · ·+ eik | 1 ≤ i1 < · · · < ik ≤ n},

where e1, . . . , en are the canonical vectors in Rn. The K-subring of Rspanned by the set {xa| a ∈ V} is equal to K[Vk].

Remark 12.4.1 dimK[Vk] = n. This follows from Corollary 8.2.21. Thevector space generated by V is equal to Rn and dim R+V = n.

514 Chapter 12

Lemma 12.4.2 [72] Set N1 = {−e1, . . . ,−en}, N = N1 ∪N2 and

N2 = {−e1 − · · · − ei−1 + (k − 1)ei − ei+1 − · · · − en| 1 ≤ i ≤ n}.

If H is a supporting hyperplane of R+V such that H contains a set of n− 1linearly independent vectors in V, then H = Ha for some a ∈ N .

Remark 12.4.3 The converse of Lemma 12.4.2 is also true because we areassuming n ≥ 2k ≥ 4, and this implies n ≥ k + 2. Note that if n = 3 andk = 2, then the cone R+V has only three facets.

Proposition 12.4.4 [72] A point x = (x1, . . . , xn) ∈ Rn is in R+V if andonly if x is a feasible solution of the system of linear inequalities

−xi ≤ 0, i = 1, . . . , n(k − 1)xi −

∑j �=i xj ≤ 0, i = 1, . . . , n.

Proof. Let R+V = H−b1∩ · · · ∩ H−

bmbe the irreducible representation of

R+V . By Theorem 1.1.44 the set Hbi ∩ R+V is a facet of R+V . Notethat Hbi is generated by a set of linearly independent vectors in V ; seeProposition 1.1.23. Hence, by Lemma 12.4.2, we get Hbi = Ha for somea ∈ N , where N is the set defined in Lemma 12.4.2. �

Lemma 12.4.5 Let a be a vector in C ∩ ri(R+V) and set I = {i | ai ≥ 2}.If |I| ≥ k and i1, . . . , ik are distinct integers in I, then a′ = a−ei1−· · ·−eikalso belongs to NV ∩ ri(R+V).

Proof. Without loss of generality one may assume a1 ≥ a2 ≥ · · · ≥ an,ak ≥ 2 and a′ = a− e1 − · · · − ek. As R+V has dimension n, it is seen thata′ ∈ ri(R+V). By Proposition 12.3.10 the subringK[Vk] is a normal domain,and therefore NV = ZV ∩ R+V . Since a′ ∈ ZV , we conclude a′ ∈ NV . �

Theorem 12.4.6 [72] Let ωK[Vk] be the canonical module of K[Vk] and letB be the set of monomials xa11 · · ·xann satisfying the following conditions:

(a) ai ≥ 1 and (k − 1)ai ≤ −1 +∑

j �=i aj, for all i.

(b)∑ni=1 ai ≡ 0mod (k).

(c) |{ i | ai ≥ 2}| ≤ k − 1.

If n ≥ 2k ≥ 4, then B is a generating set for ωK[Vk].

Proof. By Theorem 9.1.5 we get ωK[Vk] = ({xa| a ∈ NV∩ri(R+V)}). Usingthe arguments of the proof of Lemma 12.4.5 and by a repeated use of thislemma it is enough to prove that B ⊂ ωK[Vk]. Let M ∈ B; we may assume


M = xa11 · · ·xak−1

k−1 xk · · ·xn, where a1 ≥ · · · ≥ ak−1 ≥ 1. The monomial

N = xa11 · · ·xak−1

k−1 can be factored as

N =

k−1∏i=1

Ni, where Ni =

{(x1 · · ·xi)ai−ai+1 , if 1 ≤ i ≤ k − 2(x1 · · ·xk−1)

ak−1 , if i = k − 1.

On the other hand, by (a) and (b), we can write∏ni=k xi = N ′∏k−1

i=1 N′i ,

where deg(N ′i) = (k − i)(ai − ai+1) if 1 ≤ i ≤ k − 2, deg(N ′

i) = ak−1 if

i = k − 1, and deg(N ′) ≡ 0mod (k). Hence M = N ′∏k−1i=1 (NiN

′i) is in

K[Vk], which readily implies M ∈ ωK[Vk]. �

Let S = K[{ti1···ik | 1 ≤ i1 < · · · < ik ≤ n}] be a polynomial ring over afield K with one variable ti1···ik for each monomial xi1 · · ·xik . Here S hasthe standard grading. There is a graded homomorphism of K-algebras

ϕ : S −→ K[Vk], induced by ti1···ikϕ�−→ xi1 · · ·xik ,

the ideal P = ker(ϕ) is the toric ideal of K[Vk]. Thus the Cohen–Macaulaytype of the ring K[Vk], denoted by type(K[Vk]), is the last Betti number inthe minimal free resolution of S/P as an S-module; see Proposition 5.3.4.

Remark 12.4.7 To compute the type of K[Vk] recall that this number isthe minimal number of generators of the canonical module ωK[Vk] of K[Vk];see Corollary 5.3.6. Notice that a monomial xa11 · · ·x

ak−1

k−1 xk · · ·xn is in B if

and only if for all 1 ≤ i ≤ k − 1 one has∑k−1

j=1 aj = mk − n + k − 1 and1 ≤ ai ≤ m− 1, for some m ≥ 2. Therefore n

k ≤ m ≤ n− 2k+2. Hence, byTheorem 12.4.6, the computation of the type of K[Vk] reduces to countingpartitions of positive integers.

Corollary 12.4.8 If k = 2, n ≥ 2k and n is odd, then

ωK[Vk] = ({x1 · · ·xj−1x2ij xj+1 · · ·xn| 1 ≤ j ≤ n, 1 ≤ i ≤ (n− 3)/2}).

Corollary 12.4.9 If k = 2 and n ≥ 2k, then type(K[Vk]) is n(n− 3)/2 ifn is odd, and type(K[Vk]) is (n2 − 4n+ 2)/2 if n is even.

Corollary 12.4.10 [72] If n ≥ 2k ≥ 4, then the a((K[Vk]) = −⌈nk

⌉.

Proof. Set D = K[Vk] and m =⌈nk

⌉. To compute the a-invariant of D

we use the formula of Lemma 9.1.7. It follows from Remark 12.4.7 that thedegree of the generators in least degree of ωD is at leastm. To complete theproof we exhibit some generators of ωD living in degreem. Write n = qk+r,0 ≤ r < k; note q ≥ 2. If r ≥ 1, observe that the monomials

x21 · · ·x2k−rxk−r+1 · · ·xk−1xk · · ·xn and x1 · · ·xn−k+rx2n−k+r+1 · · ·x2n−1x2n

516 Chapter 12

belong to (ωD)m. In particular, D cannot be a Gorenstein ring in this case.If r = 0, then the monomial M = x1 · · ·xn satisfies M ∈ (ωD)m. �

Remark 12.4.11 (Duality) Let 1 ≤ k ≤ n− 1 be an integer and let

Dn,k = K[{xi1 · · ·xik | 1 ≤ i1 < · · · < ik ≤ n}],

be the K-subring of R spanned by the xi1 · · ·xik ’s. Observe that there is agraded isomorphism of K-algebras of degree zero:

ρ : Dn,k −→ Dn,n−k, induced by ρ(xi1 · · ·xik) = xj1 · · ·xjn−k,

where {j1, . . . , jn−k} = {1, . . . , n} \ {i1, . . . , ik}. Thus if n ≤ 2k, then

a(Dn,k) = a(Dn,n−k) = −⌈

n

n− k

⌉.

Because of this duality one may always assume that n ≥ 2k.

Corollary 12.4.12 [108] The subring Dn,k is a Gorenstein ring if and onlyif k ∈ {1, n− 1} or n = 2k.

Proof. By duality one may assume n ≥ 2k ≥ 4. Set D = Dn,k. IfD is Gorenstein, then by the proof of Corollary 12.4.10 we may assumen = qk. If q ≥ 3, then x1 · · ·xn and x31x

22 · · ·x2k−1xk · · ·xn belong to (ωD)q

and (ωD)q+1 respectively, which is impossible. Therefore q = 2, as required.

Conversely assume n = 2k. Let B be as in Theorem 12.4.6. Take amonomial xa in B; it suffices to verify that xa is equal to x1 · · ·xn. Onemay assume that xa = xa11 · · ·x

ak−1

k−1 xk · · ·xn, where ai ≥ ai+1 ≥ 1. By

hypothesis∑k−1

j=1 aj = k(m − 1) − 1 for some m ≥ 2. On the other hand

one has kai ≤ k +∑k−1

j=1 aj for all 1 ≤ i ≤ k − 1. Hence ai ≤ m − 1for all i. Therefore using the previous equality again we rapidly derivek(m−1)−1 ≤ (k−1)(m−1), which yields m ≤ 2. As a consequence ai = 1for all i, as required. �

Theorem 12.4.13 [72] Let F be a finite set of square-free monomials ofdegree k in R such that dim(K[F ]) = n. The following hold.

(i) a(K[F ]) ≤ a(K[Vk]).

(ii) a(K[F ]) ≤ −⌈nk

⌉if n ≥ 2k and a(K[F ]) ≤ −

⌈n

n−k

⌉if n ≤ 2k, n �= k.

(iii) If n ≥ 2k ≥ 4, then K[F ] is generated as a K-algebra by elements ofnormalized degree less than or equal to n−

⌈nk

⌉.


Proof. (i): We set V = {ei1 + · · · + eik | 1 ≤ i1 < · · · < ik ≤ n} andD = K[Vk]. Let C, CF be the semigroups of Nn generated by V , log(F ),respectively. By Proposition 12.3.10 D is normal. Hence, K[F ] ⊂ D. Let

M1 = {xa| a ∈ CF ∩ ri(R+CF )} and M2 = {xa| a ∈ C ∩ ri(R+C)},

where CF = ZCF ∩ R+CF . Notice R+CF = R+CF and aff(R+CF ) =Rn. Therefore the relative interior of R+CF equals its interior in Rn. Forsimilar reasons we have ri(R+C) = (R+C)

o. Hence ri(R+CF ) ⊂ ri(R+C).Altogether we obtain M1 ⊂M2. Let x

b be an element of minimal degree inM1 so that deg(xb) = −a(K[F ]). Set r = −a(K[F ]). Since xb is in M2 andxb ∈ K[F ]r ⊂ Dr, we conclude

−a(D) = min{deg(xa)|a ∈M2} ≤ r = −a(K[F ]).

(ii): It follows from Corollary 12.4.10, part (i), and the duality given inRemark 12.4.11.

(iii): It follows from the proof of part (i) and using part (ii). �

Example 12.4.14 Let K[F ] be the subring of R = K[x1, . . . , x8] spannedby the monomials of R defining the edges of the graph shown below.

� ��

��

�

x1 x2

x3

x4

x5

x6

x7

x8

��

��

��

��

��

��

f1 = x1x2,

f2 = x2x3,

f3 = x1x3,

f4 = x3x4,

f5 = x4x5,

f6 = x5x6,

f7 = x6x7,

f8 = x5x7,

f9 = x7x8,

f10 = x5x8,

f11 = x6x8,

f12 = x1x2x3x5x6x8,

f13 = x1x2x3x6x7x8,

f14 = x1x2x3x5x6x7,

f15 = x1x2x3x5x7x8.

F := {f1, . . . , f11},

K[F ] =

K[F ][f12, f13, f14, f15].

The generators of K[F ] can be computed using the program Normaliz [68].See also [413, Section 7.3]. A Noether normalization for K[F ] is given by

A0 = K[h1, . . . , h8] ↪→ K[F ] ↪→ K[F ],

where h1 = f1, h3 = f8− f11, h5 = f2− f3, h7 = f5− f7− f9− f11, h2 = f6,h4 = f9 − f10, h6 = f3 − f5, h8 = f1 −

∑11i=2 fi.

As K[F ] is Cohen–Macaulay, by Proposition 3.1.27, we get

K[F ] = A01⊕A0f7 ⊕A0f10 ⊕ a0f11⊕A0f

27 ⊕A0f

210 ⊕A0f10f11 ⊕A0f7f11⊕

A0f37 ⊕A0f

27 f11 ⊕A0f

310 ⊕A0f

210f11⊕

A0f12 ⊕A0f13 ⊕A0f14 ⊕A0f15 ⊕A0f310f11.

518 Chapter 12

Therefore the Hilbert series of K[F ] is equal to

H(K[F ], z) =1 + 3z + 4z2 + 8z3 + z4

(1 − z)8 and a(K[F ]) = −4.

Exercises

12.4.15 Let P be the toric ideal of the Veronese subring R(k) or the toricideal of the kth square-free Veronese subring K[Vk]. If k ≥ 2, then P isgenerated by homogeneous binomials of degree two.

12.4.16 [197, 311] Prove the following: (a) R(k) is Gorenstein if and onlyif k divides n. (b) a(K[Vk]) ≤ a(R(k)). (c) If n = 5 and k = 3, thena(K[Vk]) = −3 and a(R(k)) = −2. (d) a(R(k)) = −

⌈nk

⌉.

12.4.17 (a) If n = 2k + 1 ≥ 5, then the canonical module of K[Vk] isgenerated by the set of all xa, a = (a1, . . . , an), such that |{i|ai = 2}| = k−1and |{i|ai = 1}| = n− k + 1.

(b) If h1, . . . , hn is a system of parameters for K[Vk] with each hi a formof degree 1, then K[Vk]/(h1, . . . , hn) is a level algebra, i.e., all the non-zeroelements of its socle have the same degree.

12.4.18 If n = 2k + 1 ≥ 5, then type(K[Vk]) =(nk−1

).

12.4.19 Let F be the set of all monomials xixj such that xi and xj areconnected by an edge of the following graph:

�x1 �x4 �x5�x6�x7

��

��

x2

x3

��

��

The monomial subring A = K[F ] is a graded subring of A, where both ringsare endowed with the normalized grading. Prove that the Hilbert series andHilbert polynomial of A and A are given by:

F (A, t) =1 + t+ t2 + t3 + t4

(1− t)7 , F (A, t) =1 + t+ t2 + 2t3

(1− t)7

ϕ(A, t) =1

144t6 +

1

16t5 +

55

144t4 +

65

48t3 +

47

18t2 +

31

12t+ 1,

ϕ(A, t) =1

144t6 +

17

240t5 +

55

144t4 +

21

16t3 +

47

18t2 +

157

60t+ 1.

In particular e(A) = e(A) = 5, a(A) = −3 and a(A) = −4.


12.5 Normalizations of Rees algebras

Let R = K[x1, . . . , xn] be a polynomial ring over a field K, and let I be amonomial ideal of R generated by xv1 , . . . , xvq . In this section we study thenormality of R[It] and the behavior of the filtration F = {Ii}i≥0.

Proposition 12.5.1 If I = (xv1 , . . . , xvq ) is a monomial ideal of the ringof polynomials R = K[x1, . . . , xn], then R[It] is generated as a K-algebraby monomials of degree in t at most n− 1.

Proof. If H is the minimal integral Hilbert basis of the Rees cone of I,then R[It] = K[NH] and the result follows from Proposition 1.5.4. �

Lemma 12.5.2 Let I be a monomial ideal of R = K[x1, . . . , xn]. If R[It]is generated as a K-algebra by monomials of degree in t at most s and Ii isintegrally closed for i ≤ s, then R[It] is normal.

Proof. Let xαtb be a generator of R[It] with b ≤ s. Note that xα ∈ Ib = Ib.Hence xαtb ∈ R[It], as required. �

Corollary 12.5.3 [345] If I is a monomial ideal of a polynomial ring in nvariables and Ii is integrally closed for i ≤ n− 1, then I is normal.

Proof. It follows from Proposition 12.5.1 and Lemma 12.5.2. �

Theorem 12.5.4 Let I = (xv1 , . . . , xvq ) ⊂ K[x1, . . . , xn] be a monomialideal and let r0 = rank(v1, . . . , vq). The following hold.

(i) Ib = IIb−1 for b ≥ n.

(ii) If A = {v1, . . . , vq} is homogeneous, then Ib = IIb−1 for b ≥ r0.

Proof. We set A′ = {(v1, 1), . . . , (vq, 1), e1, . . . , en}. Assume b ≥ n (resp.

b ≥ r0 and A homogeneous). Notice that we invariably have IIb−1 ⊂Ib. To show the reverse inclusion take xα ∈ Ib. Thus xαtb ∈ R[It] and(α, b) is in the Rees cone R+A′. By the proof of Proposition 1.5.4 and byCaratheodory’s theorem (see Theorem 1.1.18), we can write

(α, b) = λ1(vi1 , 1) + · · ·+ λr(vir , 1) + μ1ej1 + · · ·+ μsejs (λi, μk ∈ Q+),

where {(vi1 , 1), . . . , (vir , 1), ej1 , . . . , ejs} is a set of vectors (resp. linearlyindependent set of vectors) contained in A′ and r ≤ n (resp. r ≤ r0). Sinceb = λ1 + · · ·+ λr, we obtain that λi ≥ 1 for some i. It follows readily thatxα ∈ IIb−1, as required. �

520 Chapter 12

Normalizations and Hilbert functions Let R = K[x1, . . . , xn] be apolynomial ring over a field K with n ≥ 2 and let I be a zero dimensionalmonomial ideal of R.

The integral closure of Ii, denoted by Ii, is again a monomial ideal forall i ≥ 1 (see Proposition 12.1.2). The length of R/Ii will be denoted simply

by �(R/Ii). By Lemma 8.5.1 the length of R/Ii is dimK(R/Ii). We areinterested in computing the Hilbert function

HF(i) := �(R/Ii) = dimK(R/Ii); i ∈ N \ {0}; HF(0) = 0,

of the filtration F = {Ii}∞i=0. The Hilbert function of F is a polynomialfunction of degree n:

HF (i) = cnin + cn−1i

n−1 + · · ·+ c1i+ c0 (i 0),

where c0, . . . , cn ∈ Q and cn �= 0 (see the proof of Proposition 12.5.16).The polynomial ϕF (x) := cnx

n + · · · + c0 is the Hilbert polynomial of F .By the results of [104], one has the equality n!cn = e(I), where e(I) is themultiplicity of the ideal I in the sense of Section 4.4.

There is a unique set of monomials F = {xu1 , . . . , xur} that minimallygenerate I. For 1 ≤ i ≤ n, we may assume that the ith entry of ui is ai andthat all other entries of ui are zero, where a1, . . . , an are positive integers.We set α0 = (1/a1, . . . , 1/an). Let {un+1, . . . , us} be the set of ui such that〈ui, α0〉 < 1, and let {us+1, . . . , ur} be the set of ui such that i > n and〈ui, α0〉 ≥ 1. Consider the convex rational polyhedron

Q = Qn+ + conv(u1, . . . , ur)

= Qn+ + conv(u1, . . . , un, un+1, . . . , us).

The second equality follows from the finite basis theorem and the equality

Qn+ + conv(u1, . . . , un) = {x |x ≥ 0; 〈x, α0〉 ≥ 1}.

Lemma 12.5.5 Ii = ({xa| a ∈ iQ∩ Zn}) for 0 �= i ∈ N.

Proof. Let xv1 , . . . , xvq be a set of generators of I. Let xα ∈ Ii, i.e.,xpα ∈ Iip for some 0 �= p ∈ N. Hence

α/i ∈ conv(v1, . . . , vq) +Qn+ ⊂ conv(u1, . . . , ur) +Qn+ = Q

and α ∈ iQ∩ Zn. Conversely let α ∈ iQ∩ Zn. Hence xpα ∈ (I)ip for some

0 �= p ∈ N, this yields xα ∈ (I)i. Since (I)i ⊂ Ii and the latter is complete

we get xα ∈ Ii. �

Corollary 12.5.6 �(R/Ii) = |Nn \ iQ| for i ≥ 1.


Proof. It follows from Lemmas 8.5.1 and 12.5.5. �

In what follows we set P := conv(u1, . . . , us), S := conv(0, u1, . . . , un)and d = dim(P). The Ehrhart function of P is given by

χP(i) := |Zn ∩ iP|, i ∈ N.

This is a polynomial function of degree d. The Ehrhart polynomial of Pis the unique polynomial EP (x) of degree d such that EP(i) = χP(i) fori 0. Ehrhart functions are studied in Section 9.3.

Proposition 12.5.7 HF(i) = ES(i)− EP(i) for i ∈ N.

Proof. Since EP(0) = ES(0) = 1, we get HF(0) = 0. Assume i ≥ 1.Notice that from the decomposition Q = (Qn+ \ S) ∪ P we get

iQ = (Qn+ \ iS) ∪ iP ,Nn \ iQ = [Nn ∩ (iS)] \ [Nn ∩ (iP)].

Hence, by Corollary 12.5.6, �(R/Ii) = |Nn \ iQ| = ES(i)− EP(i). �

The Hilbert function of F and its Hilbert polynomial are equal for non-negative integer values:

Corollary 12.5.8 HF (i) = cnin + · · ·+ c1i+ c0 for i ∈ N and c0 = 0.

Proof. By Lemma 9.3.8, EP(i) = χP(i) and ES(i) = χS(i) for i ∈ N. Thusthe result follows from Proposition 12.5.7. �

Example 12.5.9 If I is the monomial ideal (x21, x32), then I = I + (x1x

22).

Notice that P = conv((2, 0), (0, 3)) because (1, 2) lies above the hyperplanex1/2 + x2/3 = 1 and S = conv(0, (2, 0), (0, 3)). Using Normaliz [68] we getES(i) = 3i2 + 3i+ 1 and EP (i) = i+ 1. Thus HF (i) = 3i2 + 2i.

Example 12.5.10 If I = (x41, x52, x

63, x1x2x

23), using Normaliz [68] we get:

I = I + (x1x42, x

21x

32, x

31x

22, x

21x

22x3, x

42x

23, x1x

53, x

32x

33, x

21x

33, x

31x2x3,

x22x43, x1x

32x3, x

31x

23, x2x

53).

Notice that P = conv((4, 0, 0), (0, 5, 0), (0, 0, 6), (1, 1, 2)) because (1, 1, 2) isthe only ui lying strictly below the hyperplane x1/4+ x2/5+ x3/6 = 1 andS = conv(0, (4, 0, 0), (0, 5, 0), (0, 0, 6)). Using Normaliz [68] we get

HF (i) = ES(i)− EP(i)

= (1 + 6i+ 19i2 + 20i3)− (1 + (1/6)i+ (3/2)i2 + (13/3)i3)

= (35/6)i+ (35/2)i2 + (47/3)i3.

522 Chapter 12

Remark 12.5.11 For i ≥ n, Ii = IIi−1, see Theorem 12.5.4. Thus we canuse polynomial interpolation and Corollary 12.5.8 to determine c1, . . . , cn.

Example 12.5.12 Consider the ideal of Example 12.5.10. For i = 0, 1, 2, 3the values of HF (i) = �(R/Ii) are 0, 39, 207, 598. To compute the Hilbertpolynomial ϕF (x) of F , using polynomial interpolation, one can use thefollowing command in Maple [80]:

interp([0,1,2,3],[0,39,207,598],x);

to get that ϕF(x) = (47/3)x3 + (35/2)x2 + (35/6)x.

Example 12.5.13 Consider the ideal I = (x101 , x82, x

53) ⊂ R. The values of

HF(i) at i = 0, 1, 2, 3 are 0, 112, 704, 2176. Using interpolation we obtain:

HF (i) = �(R/Ii) =200

3i3 + 40x2 +

16

3i.

Proposition 12.5.14 [424] Let Ii be the ideal obtained from I by makingxi = 0 in the generators of I and let e(Ii) be its multiplicity. Then

2cn−1 ≥n−1∑i=1

e(Ii)

(n− 1)!.

Let e0, e1, . . . , en be the Hilbert coefficients of HF . Recall that we have:

HF(i) = e0

(i+ n− 1

n

)− e1

(i+ n− 2

n− 1

)+ · · ·+ (−1)n−1en−1

(i

1

)+ (−1)nen,

where e0 = e(I) is the multiplicity of I and cn = e0/n!. Notice that en = 0because HF (0) = 0.

Corollary 12.5.15 e0(n− 1)− 2e1 ≥ e(I1) + · · ·+ e(In−1) ≥ n− 1.

Proof. From the equality

cn−1 =1

n!

[e0

(n

2

)− ne1

]=

1

(n− 1)!

[e0(n− 1)

2− e1

]and using Proposition 12.5.14 we obtain the desired inequality. �

Proposition 12.5.16 ei ≥ 0 for all i.

Proof. Since the Rees algebra of F is R[It] and this algebra is normal and

Cohen–Macaulay, we get that the ring grF (R) = ⊕i≥0Ii/Ii+1 is Cohen–Macaulay (see [198, Corollary 2.1, p. 74]). Recall that grF(R) is a finitelygenerated (R/I)-algebra, i.e., grF(R) = (R/I)[z1, . . . , zm], where z1, . . . , zm


are homogeneous elements of positive degree (see [65, Proposition 4.5.4]).The canonical map grI(R) → grF (R) is integral. Thus grF(R) is a posi-tively graded Noetherian module over the standard graded algebra grI(R).Hence, by the Hilbert–Serre theorem, we can write the Hilbert series ofgrF(R) as h(x)/(1− x)n, where h(x) is a polynomial with nonnegative in-teger coefficients because grF(R) is Cohen–Macaulay. To finish the proofnotice that ei = h(i)(1)/i! (see [65, Proposition 4.1.9]). �

Example 12.5.17 Let m = (x1, . . . , xn) be the irrelevant maximal idealand let I = mk be the kth Veronese ideal of R. Then

HF (i) =

(ki+ n− 1

n

)=kn

n!in +

kn−1

(n− 2)!2in−1 + terms of lower degree

because I is normal. Here e0 = e(I) = kn, e1 = (n− 1)(kn − kn−1)/2 and

cn−1 =1

2

n−1∑i=1

e(Ii)

(n− 1)!.

Exercises

12.5.18 Let I = (xv1 , . . . , xvq ) ⊂ R be a monomial ideal and let r0 be therank of A = (v1, . . . , vq). If A = {v1, . . . , vq} is homogeneous and Ii isintegrally closed for i = 1, . . . , r0 − 1, then R[It] is a normal domain.

12.5.19 If R = Q[x1, x2] and I = (x21, x32, x1x2), find a presentation of

grI(R) and find the multiplicity of I using the CoCoA [88] procedure:

Use R::=Q[t[1..3],x[1..2]];

J:=Toric([[2,0,1,1,0],[0,3,1,0,1],[1,1,1,0,0]]);

I:=Ideal(x[1]^2,x[2]^3,x[1]*x[2]);

L:=I+J;

Multiplicity(R/LT(L));

12.5.20 If I = (x21, x32, x1x2) ⊂ Q[x1, x2], use Exercise 12.5.19 to prove:

�(Ii/Ii+1) =

{5i+ 4 if 0 ≤ i �= 2,27 if i = 2.

Prove that the Samuel function of I is given by

χIR(i) := �(R/Ii+1) =5

2i2 +

13

2i+ 4 (i ≥ 0).

12.5.21 If R = Q[x1, x2] and I = (x1x2, x21 + x22), prove that e(I) = 4.

12.5.22 If I = (x2, y3, z5, w6, xy, z2w, xz2, y2z2) ⊂ R = K[x, . . . , w], usethe methods of this section to prove:

�(R/Ii) = (7/2)i4 + (65/6)i3 + (21/2)i2 + (19/6)i; ∀ i ∈ N.

524 Chapter 12

12.6 Rees algebras of Veronese ideals

In this section we compute the a-invariant of the Rees algebra of a Veroneseideal. This will be used to show some degree bounds for the normalizationof a uniform ideal; see Theorem 12.6.7.

The Rees algebra of a square-free Veronese ideal Let K be a fieldand let R = K[x1, . . . , xn] be a polynomial ring with coefficients in K.Given two positive integers k, n with k ≤ n, we define

Vk = {xi1 · · ·xik | 1 ≤ i1 < · · · < ik ≤ n}.

The ideal L := (Vk) is called the kth square-free Veronese ideal of R. Recallthat the Rees algebra of L is given by

R[Lt] = K[x1, . . . , xn, f1t, . . . , f(nk)t] ⊂ R[t],

where t is a new variable and Vk = {f1, . . . , f(nk)}. We can make R[Lt]

a standard K-algebra with the grading induced by setting δ(xi) = 1 andδ(t) = 1− k. In this grading δ(fit) = 1 for all i.

The Rees algebra of L satisfies the hypothesis of Theorem 9.1.5 because:(i) R[Lt] is a normal domain by Proposition 12.3.10, and (ii) R[Lt] is astandard K-algebra.

The next expression for the a-invariant follows from Theorem 9.1.5.

Theorem 12.6.1 [3] If 1 ≤ k < n and n = pk + r, where 0 ≤ r < k, then

a(R[Lt]) =

⎧⎪⎪⎨⎪⎪⎩−(p+ 2) if r ≥ 2,−(p+ 1) if r = 1,−(p+ 1) if r = 0 and k > 1,−p if r = 0 and k = 1.

There is an alternative expression for a Rees algebra of a monomial idealgenerated by monomials of the same degree k.

Proposition 12.6.2 Let F = {xv1 , . . . , xvq} be a set of monomials in R offixed degree k ≥ 1. If x = {x1, . . . , xn}, then R[Ft] � K[F,xt].

Proof. Consider the epimorphisms of K-algebras:

φ : R[t1, . . . , tq]→ R[tF ], induced by φ(ti) = xvit and φ(xi) = xi,ψ : R[t1, . . . , tq]→ K[F,xt], induced by ψ(ti) = xvi and ψ(xi) = xit.

It suffices to notice that ker(φ) = ker(ψ). This equality follows from thefact that ker(φ) and ker(ψ) are binomial ideals (see Corollary 8.2.18). �

Corollary 12.6.3 [72] The a-invariant of the edge subring of a completegraph Kn+1 is given by a(K[Kn+1]) = −

⌈n+12

⌉.

Proof. It follows from Theorem 12.6.1 and Proposition 12.6.2. �


The Rees cone of a Veronese ideal In this part, we refer to Chapter 1for standard terminology and notation on polyhedral cones. Setting

A = {e1, . . . , en, ke1 + en+1, . . . , ken + en+1},A′ = {e1, . . . , en, a1e1 + · · ·+ anen + en+1| ai ∈ N; |a| = k},

where ei is the ith unit vector in Rn+1 and |a| := a1 + · · ·+ an, one has theequality R+A = R+A′. This cone is the so-called Rees cone.

Lemma 12.6.4 The irreducible representation of R+A is:

R+A = H+e1 ∩ · · · ∩H

+en ∩H

+en+1∩H+

a , a = (1, . . . , 1,−k).

Proof. Set N = {e1, . . . , en+1, a}. It suffices to prove that F is a facet ofR+A if and only if F = Hb ∩ R+A for some b ∈ N . Let 1 ≤ i ≤ n + 1.Consider the following sets

{e1, e2, . . . , ei, . . . , en, en+1} and {ke1 + en+1, . . . , ken + en+1},

where ei means to omit ei from the list. Since those sets are linearly inde-pendent it follows that F = Hb∩R+A is a facet for b ∈ N . Conversely let Fbe a facet of R+A. There are linearly independent vectors α1, . . . , αn ∈ Aand 0 �= b = (b1, . . . , bn+1) ∈ Rn+1 such that

(i) F = R+A∩Hb,

(ii) Rα1 + · · ·+ Rαn = Hb, and

(iii) R+A ⊂ H+b .

Since e1, . . . , en are in A, by (iii), one has 〈ei, b〉 = bi ≥ 0 for i = 1, . . . , n.Set B = {α1, . . . , αn} and consider the matrix MB whose rows are thevectors in B. Let us prove that there exists c ∈ N such that Hb = Hc andR+A ⊂ H+

c . Consider the following cases. Case (1): If the ith column ofMB is zero for some 1 ≤ i ≤ n + 1, it suffices to take c = ei. Case (2): IfB = {ke1 + en+1, . . . , ken + en+1}, then take c = a. Case (3): If

B = {ei1 , . . . , eis , kej1 + en+1, . . . , kejt + en+1},

where s, t > 0, s+ t = n, 1 ≤ i1 < · · · < is ≤ n, 1 ≤ j1 < · · · < jt ≤ n, andMB has all its columns different from zero. Since bi1 = 0, using

〈kei1 + en+1, b〉 = bn+1 ≥ 0,〈kej1 + en+1, b〉 = kbj1 + bn+1 = 0,

and bj1 ≥ 0, we obtain bn+1 = 0. Then en+1 ∈ Hb. It follows that Hb isgenerated by the set {e1, e2, . . . , ei, . . . , en, en+1} for some 1 ≤ i ≤ n and wetake c = ei. �

526 Chapter 12

Theorem 12.6.5 Let R[Jt] be the Rees algebra of the kth Veronese idealJ . (a) If 2 ≤ k < n and n = pk + s, where 0 ≤ s < k, then

a(R[Jt]) =

{−(p+ 2) if s ≥ 2,−(p+ 1) if s = 0 or s = 1.

(b) If k ≥ n, then a(R[Jt]) = −2.

Proof. Let M = xaxbtc be a monomial in the canonical module ωR[Jt] of

R[Jt], where xbtc = (f1t) · · · (fct) and fi is a monomial in R of degree k forall i. Note δ(M) = |a|+ c. Recall that log(M) = (a+ b, c) is in the interiorof R+A by Theorem 9.1.5. Therefore, using Lemma 12.6.4, one has c ≥ 1,ai + bi ≥ 1 for all i, and |a|+ |b| ≥ kc+ 1. As |b| = kc, altogether we get:

|a|+ |b| ≥ n and |a| ≥ 1. (12.2)

In particular δ(M) ≥ 2 and a(R[Jt]) ≤ −2. To prove (b) note that byLemma 12.6.4 the monomial m = xk−n+2

1 x2 · · ·xnt is in ωR[Jt] and δ(m) isequal to 2. Hence a(R[Jt]) = −2. To prove (a) there are the following threecases to consider.

Case s ≥ 2: First we show δ(M) ≥ p + 2. If c > p, then δ(M) ≥ p+ 2follows from Eq.(12.2). On the other hand assume c ≤ p. Observe:

k(p− c) + s ≥ (p− c) + 2. (∗)

From Eq.(12.2) one has |a|+ |b| = |a|+ kc ≥ n = kp+ s. Consequently

δ(M) = |a|+ c ≥ k(p− c) + s+ c. (∗∗)

Hence from (∗) and (∗∗) we get δ(M) ≥ p+ 2, as required. Therefore onehas the inequality a(R[Jt]) ≤ −(p+ 2), to show equality we claim that

m = x21x22 · · ·x2k−s+1xk−s+2 · · ·xntp+1

is in ωR[Jt] and δ(m) = p+2. An easy calculation shows that m is in R[Jt]and δ(m) = p+ 2. Finally let us see that m is in ωR[Jt] via Lemma 12.6.4.That the entries of log(m) satisfy yi > 0 for all i is clear. The inequality

y1 + y2 + · · ·+ yn > kyn+1, (∗ ∗ ∗)

after making yi equal to the ith entry of log(m), transforms into

2(k − s+ 1) + (n− (k − s+ 1)) > k(p+ 1)

but the left-hand side is k(p+ 1) + 1, hence log(m) satisfies (∗ ∗ ∗).Case s = 1: If c ≥ p, then clearly |a|+ c ≥ p+ 1, so let us suppose that

c < p, then n+ c(1 − k) ≥ n+ p(1− k) and using Eq. (12.2) we get

|a|+ c = (|a|+ |b| − n) + (n+ c(1− k)) ≥ n+ c(1− k) ≥ p+ 1


so a(R[Jt]) ≤ −(p+ 1). Now observe that the monomial m = x1x2 · · ·xntphas δ(m) = p + 1 and m ∈ R[Jt]. Finally, an argument similar to theprevious case shows that m is in ωR[Jt]. Thus a(R[Jt]) = −(p+ 1).

Case s = 0: First note that p > 1 because n > k. If c ≥ p, then clearly|a|+ c ≥ p+ 1. So let us suppose c < p. Then, using Eq. (12.2), we get

|a| ≥ k(p− c) > (p− c).

Then adding c we derive |a|+ c ≥ p+ 1. Hence a(R[Jt]) ≤ −(p+ 1). Notethat the monomial m = x21x2 · · ·xntp has δ(m) = p + 1 and belongs toωR[Jt]. Thus, one has the required equality. �

Corollary 12.6.6 Let I be an ideal of Veronese type of degree k. If 2 ≤k < n and n = pk + s, where 0 ≤ s < k, then

a(R[It]) =

{−(p+ 2) if s ≥ 2,−(p+ 1) if s = 0 or s = 1.

Proof. Let L and J be the kth square-free Veronese ideal and the kthVeronese ideal, respectively. From Theorems 12.6.1 and 12.6.5 we obtainthat a(R[Lt]) and a(R[Jt]) can be computed using the formula above. SinceR[It] is normal, using the proof of Theorem 12.4.13, we get

a(R[Lt]) ≤ a(R[It]) ≤ a(R[Jt]).

Therefore a(R[It]) is given by the formula above. �

Normalization of a uniform ideal Let R = K[x1, . . . , xn] be a ring ofpolynomials over a fieldK and let I be an ideal of R generated by monomialsxv1 , . . . , xvq of degree k ≥ 2. Ideals of this type are called uniform. In thiscase the Rees algebra of I is a standard graded K-algebra with the gradingδ induced by δ(xi) = 1 and δ(t) = 1− k. Notice that δ(xvi t) = 1 for all i.

Notice that we can embed R[It] in R[Jt], where J is the kth Veroneseideal of R. The next result complements [72, Theorem 3.4] for the class ofuniform ideals.

Theorem 12.6.7 Let R = R[It] be the Rees algebra of I. If 2 ≤ k < n,then the normalization R of I is generated as an R-module by monomialsg ∈ R[t] of degree in t at most n− (n/k).

Proof. We set A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)}. As ZA′ = Zn+1,according to Theorem 9.1.1, the normalization of I can be expressed as:

R = K[{xatb| (a, b) ∈ Zn+1 ∩ R+A′}].

528 Chapter 12

If m = xatb with 0 �= (a, b) ∈ Zn+1 ∩ R+A′, then δ(m) ≥ b. To show thiswrite (a, b) =

∑ni=1 λiei +

∑qi=1 μi(vi, 1) for some λi ≥ 0, μi ≥ 0. Hence

|a| = λ1 + · · ·+ λn + (μ1 + · · ·+ μq)kb = μ1 + · · ·+ μq,

and consequently δ(m) = |a|+(1− k)b = (λ1 + · · ·+λn)+ b ≥ b. ThereforeR is generated as a K-algebra by monomials of positive degree. There is aNoether normalization of R

A = K[z1, . . . , zn+1]ϕ↪→R ψ

↪→R,

where z1, . . . , zn+1 ∈ R1. Notice that the composite A → R is a Noethernormalization of R. By Theorem 9.1.6 R is Cohen–Macaulay. Hence, byProposition 3.1.27, R is a free module over A and one can write

R = Am1 ⊕ · · · ⊕Amp, (12.3)

where mi = xβitbi . Using that the length is additive one has the followingexpression for the Hilbert series

F (R, z) =p∑i=0

zδ(mi)

(1− z)n+1=h0 + h1z + · · ·+ hsz

s

(1− z)n+1,

where hi = |{j | δ(mj) = i}|. Note that a(R), the a-invariant of R, is equalto s− (n+ 1). On the other hand one has

a(R) = −min{i | (ωR)i �= 0},

where ωR is the canonical module of the normalization R. Using the proofof Theorem 12.4.13 together with Theorem 12.6.5 yields

a(R) = s− (n+ 1) ≤ a(R[Jt]) ≤ −⌊nk

⌋− 1

and s ≤ n−(n/k). Altogether ifmi = xβitbi one has bi ≤ δ(mi) ≤ n−(n/k),that is, the degree in t of mi is at most n− (n/k), as required. �

Proposition 12.6.8 Ii = IIi−1 for i ≥ n + 2 + a(R[Jt]). Furthermoreequality holds for i ≥ n− (n/k)+ 1.

Proof. It follows for the proof of Theorem 12.6.7 and using Eq.(12.3). �

Definition 12.6.9 The index of normalization of I, denoted by s(I), is the

smallest integer s such that Ii = IIi−1 for i > s.

Corollary 12.6.10 s(I) ≤ n− (n/k).


Exercises

12.6.11 Let B = K[x1, . . . , xn, t1, . . . , tn] be a polynomial ring, let A bethe following matrix of indeterminates

A =

(x1 x2 · · · xnt1 t2 · · · tn

),

and let I2(A) be the ideal generated by the 2-minors of A. If R is thepolynomial ring K[x1, . . . , xn] and m = (x1, . . . , xn), then the toric ideal ofR[mt] is I2(A) and a(R[mt]) = a(B/I2(A)) = −n.

12.6.12 [72] Let K[F ] ⊂ R be a subring generated by a set F of monomialsof degree k. Prove that K[F ] is generated as a K[F ]-module by elementsg ∈ R of normalized degree at most dim(K[F ])− 1.

12.7 Divisor class group of a Rees algebra

In this section we determine the divisor class group of the Rees algebra ofa normal monomial ideal.

Let R = K[x1, . . . , xn] be a polynomial ring over a fieldK and let I be anideal of R of height g ≥ 2 minimally generated by monomials xv1 , . . . , xvq ofdegree at least two. For technical reasons we shall assume that each variablexi occurs in at least one monomial xvj .

The Rees cone of the point configurationA = {v1, . . . , vq} is the rationalpolyhedral cone in Rn+1, denoted by R+A′, consisting of the nonnegativelinear combinations of the set

A′ := {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Rn+1,

where ei is the ith unit vector. Notice that dim(R+A′) = n + 1. Thusaccording to Corollary 1.1.50 there is a unique irreducible representation

R+A′ = H+e1 ∩ · · · ∩H

+en+1∩H+

�1∩ · · · ∩H+

�r(12.4)

such that 0 �= �i ∈ Zn+1 and the non-zero entries of �i are relatively primefor all i.

Lemma 12.7.1 If �j = (a1, . . . , an,−an+1), then ai ≥ 0 for i = 1, . . . , nand an+1 > 0.

Proof. Clearly ai ≥ 0 because ei ∈ R+A′ for i = 1, . . . , n. Notice thatai > 0 for some 1 ≤ i ≤ n. Assume an+1 ≤ 0. Then

H+e1 ∩ · · · ∩H

+en+1

⊂ H+e1 ∩ · · · ∩H

+en+1∩H+

�j.

530 Chapter 12

To show the inclusion take x = (xi) is in the left-hand side, then xi ≥ 0 forall i and a1x1 + · · ·+ anxn ≥ 0 ≥ an+1xn+1. Thus x ∈ H+

�j. Hence H+

�jcan

be deleted from Eq. (12.4), a contradiction. Thus an+1 > 0. �

Lemma 12.7.2 If 1 ≤ i ≤ r, then the following equality holds

Z = 〈e1, �i〉Z+ · · ·+ 〈en, �i〉Z+ 〈(v1, 1), �i〉Z+ · · ·+ 〈(vq , 1), �i〉Z. (12.5)

Proof. Fix an integer 1 ≤ i ≤ r. Since H�i ∩ R+A′ is a facet of the Reescone there is (vj , 1) such that 〈(vj , 1), �i〉 = 0. Setting vj = (c1, . . . , cn) and�i = (a1, . . . , an,−an+1), we get c1a1 + · · ·+ cnan − an+1 = 0. Notice thatthe right-hand side of Eq. (12.5) is equal to

a1Z+ · · ·+ anZ+ 〈(v1, 1), �i〉Z + · · ·+ 〈(vq, 1), �i〉Z.

Thus we need only show the equality Z = a1Z+ · · ·+anZ. By Lemma 12.7.1an+1 > 0. Thus, from the equality

∑ni=1 ciai = an+1, and using that the

non-zero entries of �j are relatively prime, we get that the non-zero elementsin a1, . . . , an are relatively prime, as desired. �

Next we prove a result about the torsion freeness of the divisor classgroup of R[It]. It shows that the rank of this group can be read off fromthe irreducible representation of the Rees cone given in Eq (12.4).

Proposition 12.7.3 [380, Theorem 1.1] If R[It] is normal, then its divisorclass group Cl(R[It]) is a free abelian group of rank r.

Proof. Set p = n+ 1 + r. Consider the map ϕ : Zn+1 → Zp given by

ϕ(v) = 〈v, e1〉e1 + · · ·+ 〈v, en+1〉en+1 + 〈v, �1〉en+2 + · · ·+ 〈v, �r〉ep

and consider the matrix B whose rows are the vectors in the set

B = ϕ(A′) = {ϕ(e1), . . . , ϕ(en), ϕ(v1, 1), . . . , ϕ(vq , 1)}.

The matrix B is given by:

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

e1 0 �11 · · · �r1e2 0 �12 · · · �r2· · · · · · · · · · · · · · ·en 0 �1n · · · �rnv1 1 〈(v1, 1), �1〉 · · · 〈(v1, 1), �r〉· · · · · · · · · · · · · · ·vq 1 〈(vq , 1), �1〉 · · · 〈(vq, 1), �r〉

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦,

where �i = (�i1, . . . , �i(n+1)). It is not hard to see that the Smith normalform of B is equal to diag(1, . . . , 1, 0, . . . , 0), where the number of 1’s isn+ 1. Hence, using Theorem 9.8.19, we get Cl(R[It]) � Zp/ZB � Zr . �


Exercise

12.7.4 Let I = (x1x2, x2x3, x3x4, x1x4) be the edge ideal of a cycle of length4. Prove that Cl(R[It]) is a free abelian group of rank 2.

12.8 Stochastic matrices and Cremona maps

In this section we discuss a conjecture on the normality of ideals arisingfrom doubly stochastic matrices and study Cremona monomial maps.

Let A = (aij) be a non-singular matrix of order n with entries in N. Thematrix A is called doubly stochastic of degree d if

n∑k=1

aki =n∑k=1

ajk = d ≥ 2 for all 1 ≤ i, j ≤ n.

If∑nk=1 aki = d ≥ 2 for all i, we say that A is a d-stochastic matrix by

columns. The set

A = {v1, . . . , vn}

will denote the set of columns of A and I will denote the monomial ideal ofthe polynomial ring R = K[x1, . . . , xn] generated by xv1 , . . . , xvn .

Proposition 12.8.1 If A = (aij) is a d-stochastic matrix by columns andR[It] is normal, then det(A) = ±d.

Proof. By Proposition 9.3.33 we have H = Zn/(v1, . . . , vn) � Zd. ByLemma 1.3.17 the order of H is | det(A)|. Thus d = | det(A)|. �

Conjecture 12.8.2 The Rees algebra R[It] is normal for every doublystochastic matrix A of degree d with entries in {0, 1} and det(A) = ±d.

To prove this conjecture one could try to use the Birkhoff von Neu-mann theorem: “A matrix is doubly stochastic if and only if it is a convexcombination of permutation matrices multiplied by d” [372, Theorem 8.6].

Proposition 12.8.3 If A = (aij) is a d-stochastic matrix by columns, thendet(A) is a multiple of d.

Proof. Let F1, . . . , Fn be the rows of A and let B be the matrix with rowsF1, . . . , Fn−1,

∑ni=1 Fi. Since det(B) = det(A) and

∑ni=1 Fi = (d, . . . , d),

the assertion follows. �

Proposition 12.8.4 Let A = (aij) be a 2-stochastic matrix by columnswith entries in {0, 1}. If det(A) = ±2, then R[It] is normal.

532 Chapter 12

Proof. Since d = 2, A is the incidence matrix of a simple graph G = (V,E)with vertex set V = {x1, . . . , xn} and edge set E, where {xi, xj} ∈ E if andonly if ei + ej ∈ A. Notice that |V | = |E| because A is a square matrix.

We claim that G is a connected. Let G1, . . . , Gk be the connected com-ponents of G. By re-ordering the vertices of G we can write

A = diag(M1, . . . ,Mk),

where Mi is the incidence matrix of Gi. Let ni be the number of vertices ofGi. By Lemma 10.2.6, rank(Mi) = ni if Gi is non-bipartite and rank(Mi) =ni − 1 if Gi is bipartite. As G is a disjoint union of G1, . . . , Gk, we getn = n1+· · ·+nk. Thus, using that rank(A) is

∑ki=1 rank(Mi), it follows that

Gi is a connected non-bipartite graph with at least ni edges for any i. Sincethe number of edges of G is n we obtain that Mi is a square matrix of orderni for all i. Then det(A) = det(M1) · · · det(Mk). By Proposition 12.8.3 weget det(Mi) = 2mi for 1 ≤ i ≤ k, where mi ∈ Z for all i. Hence k = 1.

Let T = (VT , ET ) be a spanning tree of G (see Exercise 10.1.65). Then|ET | + 1 = |VT | = |V | = |E| and G is obtained from T by adding oneedge. This means that G has a unique odd cycle and some branches. UsingCorollary 10.3.12 we get K[G] = K[{xa| a ∈ A}] is normal and accordingto Corollary 10.5.6 we have that the Rees algebra R[It] is normal. �

Example 12.8.5 Consider the following doubly stochastic matrix

At =

⎡⎢⎢⎢⎢⎢⎢⎣0 0 1 1 1 10 1 0 1 1 11 1 1 0 1 01 1 1 1 0 01 0 0 1 1 11 1 1 0 0 1

⎤⎥⎥⎥⎥⎥⎥⎦whose rows are v1, . . . , v6. Notice that det(A) = ±8 �= ±4. The ideal I isminimally non-normal , that is I is not normal and I ′ is normal for everyproper minor I ′ of I (see Definition 14.2.23 for the notion of minor).

Lemma 12.8.6 If A is a d-stochastic matrix by columns and det(A) = ±d,then A−1(ei − ej) ∈ Zn for all i, j.

Proof. Fixing indices i, j, we can write A−1(ei − ej) =∑n

k=1 λkek forsome λ1, . . . , λn in Q. Notice that A−1(ei) is the ith column of A−1. Since1A = d1, we get 1/d = 1A−1. Therefore |A−1(ei)| = |A−1(ej)| = 1/d and∑

k λk = 0. Then we can write

A−1(ei − ej) =n∑k=2

λk(ei − e1) =⇒ ei − ej =n∑k=2

λk(vk − v1).


Thus there is 0 �= s ∈ N such that s(ei − ej) belong to Z{v1 − vk}nk=2. ByProposition 10.8.7, the group Zn/Z{v1 − vk}nk=2 is free. Then we can writeei − ej =

∑nk=2 ηk(vk − v1), for some ηi’s in Z. Since Z{v1 − vk}nk=2 is also

free (of rank n−1), the vectors v2−v1, . . . , vn−v1 are linearly independent.Thus λk = ηk ∈ Z for all k ≥ 2, hence ultimately A−1(ei − ej) ∈ Zn. �

Theorem 12.8.7 [386] If A is a d-stochastic matrix by columns such thatdet(A) = ±d, then there are unique vectors β1, . . . , βn, γ ∈ Nn such that thefollowing two conditions hold:

(a) Aβi = γ + ei for all i, where βi, γ and ei are column vectors ;

(b) The matrix B with columns β1, . . . , βn has at least one zero entry inevery row.

Moreover, det(B) = ±(|γ|+ 1)/d = ±|βi| for all i.

Proof. First we show the uniqueness. Assume that β′1, . . . , β

′n, γ

′ is aset of vectors in Nn such that: (a′) Aβ′

i = γ′ + ei for all i, and (b′) Thematrix B′ whose column vectors are β′

1, . . . , β′n has at least one zero entry

in every row. Let Δ = (Δi) and Δ′ = (Δ′i) be nonnegative vectors such

that A−1(γ − γ′) = Δ′ −Δ. Then from (a) and (a′) we get

βi−β′i = A−1(γ−γ′) = Δ′−Δ, ∀ i =⇒ βik−β′

ik = Δ′k−Δk, ∀ i, k, (12.6)

where βi = (βi1, . . . , βin) and β′i = (β′

i1, . . . , β′in). It suffices to show that

Δ = Δ′. If Δ′k > Δk for some k; then, by Eq. (12.6), we obtain βik > 0 for

i = 1, . . . , n, which contradicts (b). Similarly if Δ′k < Δk for some k; then,

by Eq. (12.6), we obtain β′ik > 0 for i = 1, . . . , n, which contradicts (b′).

Thus Δk = Δ′k for all k, i.e., Δ = Δ′.

Next we prove the existence of β1, . . . , βn and γ. By Lemma 12.8.6, fori ≥ 2 we can write

0 �= αi = A−1(e1 − ei) = α+i − α−

i

where α+i and α−

i are in Nn. Notice that α+i �= 0 and α−

i �= 0. Indeed thesum of the entries of A−1(ei) is equal to 1/d. Thus |αi| = |α+

i | − |α−i | = 0,

and consequently the positive and negative part of αi are both non-zero fori ≥ 2. The vector α+

i can be written as α+i = (α+

i1, . . . , α+in) for i ≥ 2. For

1 ≤ k ≤ n consider the integers given by mk = max2≤i≤n{α+ik} and set

β1 = (m1, . . . ,mn). Since β1 ≥ α+i , for each i ≥ 2 there is θi ∈ Nn such

that β1 = θi + α+i . Therefore

αi = A−1(e1 − ei) = α+i − α−

i = β1 − (θi + α−i ).

We set βi = θi+α−i for i ≥ 2. Since we have Aβ1− e1 = Aβi− ei for i ≥ 2,

it follows readily that Aβ1 − e1 ≥ 0 (make i = 2 in the last equality and

534 Chapter 12

compare entries). Thus, setting γ := Aβ1 − e1, it follows that β1, . . . , βnand γ satisfy (a). If each row of B has some zero entry the proof of theexistence is complete. If every entry of a row of B is positive we subtractthe vector 1 = (1, . . . , 1) from that row and change γ accordingly so that(a) is still satisfied. Applying this argument repeatedly we get a sequenceβ1, . . . , βn, γ satisfying (a) and (b).

We now prove the last part of the assertion. If βij denotes the j-entryof βi, then Aβi = γ + ei is equivalent to βi1v1 + · · ·+ βinvn = γ + ei. Thus|βi|d = |γ|+ 1. Condition (a) is equivalent to AB = Γ + I, where Γ is thematrix all of whose columns are equal to γ. Since det(B) = ± det(Γ + I)/dit suffices to show that det(Γ + I) = |γ|+ 1. By Exercise 12.8.16, if Γ hasrank at most one and D is the identity, we get det(Γ+ I) = trace(Γ)+1. �

Example 12.8.8 Consider the following matrix A and its inverse:

A =

⎛⎝d d− 1 00 1 d− 10 0 1

⎞⎠ ; A−1 =1

d

⎛⎝1 1− d (d− 1)2

0 d d(1− d)0 0 d

⎞⎠ .

To compute the βi’s and γ we follow the proof of Theorem 12.8.7. Thenβ1 = (2, d, 0), β2 = (1, d+ 1, 0), β3 = (d, 1, 1), γ = (d2 + d− 1, d, 0), and

B =

⎛⎝2 1 dd d+ 1 10 0 1

⎞⎠ .

By subtracting the vector (1, 1, 1) from rows 1 and 2, we get

B′ =

⎛⎝ 1 0 d− 1d− 1 d 00 0 1

⎞⎠ .

The column vectors β′1 = (1, d − 1, 0), β′

2 = (0, d, 0), β′3 = (d − 1, 0, 1),

γ′ = (d2 − d, d− 1, 0) satisfy (a) and (b).

Cremona maps Let R = K[x1, . . . , xn] be a polynomial ring over a fieldK. In what follows we assume that A is a d-stochastic matrix by columnsand that the corresponding set of monomials F = {xv1 , . . . , xvn} ⊂ R haveno non-trivial common factor. We also assume throughout that every xidivides at least one member of F , a harmless condition.

Definition 12.8.9 F defines a rational (monomial) map Pn−1 �� Pn−1

denoted again by F and written as a tuple F = (xv1 , . . . , xvn). F is calleda Cremona map if it admits an inverse rational map with source Pn−1.


A rational monomial map F is defined everywhere if and only if thedefining monomials are pure powers of the variables, in which case it is aCremona map if and only if F = (xσ(1), . . . , xσ(n)) for some permutation σ.

Proposition 12.8.10 [385, Proposition 2.1] F is a Cremona map if andonly if det(A) = ±d.

Thus Conjecture 12.8.2 has the following reformulation: “the rationalmap F : Pn−1 �� Pn−1 is a Cremona map if and only ifR[It] is normal.” Ford = 2 this conjecture was proved in [385]; see Proposition 12.8.4. Cremonamaps defined by monomials of degree d = 2 are thoroughly analyzed andclassified via integer arithmetic and graph combinatorics in [98].

Theorem 12.8.11 If F : Pn−1 �� Pn−1 is a Cremona map then its inverseis also defined by monomials of fixed degree.

Proof. We set fi = xvi for i = 1, . . . , n. By Proposition 12.8.10, A hasdeterminant ±d. Therefore Theorem 12.8.7 implies the existence of an n×nmatrix B such that AB = Γ+I, where Γ is a matrix with repeated column γthroughout. Let g1, . . . , gn denote the monomials defined by the columns ofB and call G the corresponding rational monomial map. The above matrixequality translates into the equality

(f1(g1, . . . , gn), . . . , fn(g1, . . . , gn)) = (xγ · x1, . . . , xγ · xn).

Thus the left-hand side is proportional to the vector (x1, . . . , xn) whichmeans that the composite map F ◦G is the identity map wherever the twoare defined. On the other hand Theorem 12.8.7 also says that G is also aCremona map. Therefore G has to be the inverse of F , as required. Noticethat the proof of Theorem 12.8.7 provides an algorithm to compute B andγ. The input for this algorithm is the matrix A. �

Exercises

12.8.12 If B is a real non-singular matrix such that the sum of the elementsin each row is d �= 0, then the sum of the elements in each row of B−1 isequal to 1/d

12.8.13 Let A = (aij) be a d-stochastic matrix by columns of order n. IfB is the matrix (bij) = (1− aij), then

(n− d) det(A) = (−1)n−1d det(B).

12.8.14 Let A = (aij) be a doubly stochastic matrix of order n with

n∑i=1

aik =

n∑j=1

a�j = d (∀ k, �).

If | det(A)| = d ≥ 1, then gcd{n, d} = 1.

536 Chapter 12

12.8.15 Let F be a Cremona map of the form F = (xa11 xa22 , x

b22 x

b33 , x

c11 x

c33 ).

Prove that, up to permutation of the variables and the monomials, F is oneof the following two kinds:

F = (x1x2, x2x3, x1x3) or F = (xd1, x2xd−13 , xd−1

1 x3).

12.8.16 Let Γ = (γi,j) be an n × n matrix over a commutative ring, andlet D = diag(d1, . . . , dn) be a diagonal matrix. Then

det(Γ +D) = det(Γ)

+∑i

diΔ[n]\{i} +∑

1≤i1<i2≤ndi1di2Δ[n]\{i1,i2}

+ · · ·+∑

1≤i1<···<in−1≤ndi1 · · · din−1Δ[n]\{i1,...,in−1}

+ detD,

where Δ[n]\{i1,...,ik} denotes the principal (n− k)× (n− k)-minor of Γ withrows and columns [n] \ {i1, . . . , ik}. Here [n] = {1, . . . , n}.

Chapter 13

Combinatorics ofSymbolic Rees Algebras ofEdge Ideals of Clutters

In this chapter we give a description—using notions from combinatorialoptimization and polyhedral geometry— of the minimal generators of thesymbolic Rees algebra of the edge ideal of a clutter and show a completegraph theoretical description of the minimal generators of the symbolic Reesalgebra of the ideal of covers of a graph. For a connected non-bipartite graphG whose edge subring is normal, we give conditions for G to have a perfectmatching.

The notion of a indecomposable graph is related to the strong perfectgraph theorem. We give a description—in terms of cliques—of the symbolicRees algebra and the Simis cone of the edge ideal of a perfect graph.

13.1 Vertex covers of clutters

In this section we characterize the notion of a minimal vertex cover of aclutter in algebraic and combinatorial terms and relate the matching numberand the vertex covering number of a clutter to the algebraic invariants ofthe corresponding edge ideal.

Let C be a clutter with vertex set X = {x1, . . . , xn} and let R = K[X ]be a polynomial ring over a field K. We denote the incidence matrix of Cby A and denote the column vectors of A by v1, . . . , vq. The edge ideal ofC, denoted by I(C), is the ideal of R generated by xv1 , . . . , xvq .

Definition 13.1.1 A subset C ⊂ X is a minimal vertex cover of a clutterC if: (c1) every edge of C contains at least one vertex of C, and (c2) there is

538 Chapter 13

no proper subset of C with the first property. If C only satisfies condition(c1), then C is called a vertex cover of C.

Proposition 13.1.2 Let Q(A) = {x|x ≥ 0; xA ≥ 1} be the set coveringpolyhedron of C. The following are equivalent :

(a) p = (x1, . . . , xr) is a minimal prime of I = I(C).(b) C = {x1, . . . , xr} is a minimal vertex cover of C.(c) α = e1 + · · ·+ er is a vertex of Q(A).

Proof. (a) ⇔ (b): This follows at once from Lemma 6.3.37.(b) ⇒ (c): Fix 1 ≤ i ≤ r. To make notation simpler fix i = 1. We

may assume that there is s1 such that xvj = x1mj for j = 1, . . . , s1 andx1 /∈ supp(xvj ) for j > s1. Notice that supp(mk1)∩ (C \ {x1}) = ∅ for some1 ≤ k1 ≤ s1, otherwise C \ {x1} is a vertex cover of C strictly contained inC, a contradiction. Hence for each 1 ≤ i ≤ r there is vki in {v1, . . . , vq} suchthat xvki = ximki and supp(mki) ⊂ {xr+1, . . . , xn}. The vector α is clearlyin Q(A), and since {ei}ni=r+1 ∪ {vk1 , . . . , vkr} is linearly independent, and

〈α, ei〉 = 0 (i = r + 1, . . . , n), 〈α, vki 〉 = 1 (i = 1, . . . , r),

we get that the vector α is a basic feasible solution of the linear systemx ≥ 0;xA ≥ 1. Therefore α is a vertex of Q(A) by Corollary 1.1.49.

(c)⇒ (b): If C′ � C is a vertex cover of C, then α′ =∑

xi∈C′ ei satisfiesα′A ≥ 1 and α′ ≥ 0. Using that α is a basic feasible solution of the linearsystem x ≥ 0;xA ≥ 1, it is not hard to verify that α′ is also a vertex ofQ(A). If V is the vertex set of Q(A), by Theorem 1.1.42, we can write

Q(A) = Rn+ + conv(V ).

Since α = β+α′, for some β ∈ Rn+, we obtain Q(A) = Rn+ +conv(V \ {α}),a contradiction (see Propositions 1.1.36 and 1.1.39). Thus C is a minimalvertex cover, as required. �

Corollary 13.1.3 A vector α ∈ Rn is an integral vertex of Q(A) if andonly if α = ei1 + · · ·+ eis for some minimal vertex cover {xi1 , . . . , xis} of C.

Proof. By Proposition 13.1.2 it suffices to observe (see Exercise 11.3.8)that any integral vertex of Q(A) has entries in {0, 1}. �

Let C be a clutter. A set of edges of C is called independent or a matchingif no two of them have a common vertex. We denote the smallest number ofvertices in any minimal vertex cover of C by α0(C) and the maximum numberof independent edges of C by β1(C). We call α0(C) the vertex coveringnumber and β1(C) the matching number of C. These numbers are relatedto min-max problems (see Exercise 13.1.7).

Combinatorics of Symbolic Rees Algebras of Edge Ideals 539

Definition 13.1.4 If α0(C) = β1(C) we say that the clutter C (resp. theedge ideal I(C)) has the Konig property.

Definition 13.1.5 The monomial grade of I(C), denoted by mgrade(I(C)),is the maximum integer r such that there exists a regular sequence of mono-mials xα1 , . . . , xαr in I(C).

The combinatorial invariants α0(C) and β1(C) can be interpreted interms of algebraic invariants of I(C).

Proposition 13.1.6 ht I(C) = α0(C) and β1(C) = mgrade(I(C)).

Proof. The first equality follows at once from Proposition 6.5.4. Thesecond equality follows readily from Exercise 13.1.10. �

Exercises

13.1.7 Let A be the incidence matrix of a clutter C. Prove that the vertexcovering number and the matching number of C satisfy:

α0(C) ≥ min{〈1, x〉|x ≥ 0;xA ≥ 1}= max{〈y,1〉| y ≥ 0;Ay ≤ 1} ≥ β1(C).

Prove that α0(C) = β1(C) if and only if both sides of the equality haveintegral optimum solutions.

13.1.8 If C is a clutter, prove that the following conditions are equivalent:

(a) α0(C) = β1(C).

(b) x1 · · ·xntg belongs to R[It], where g = ht(I) and I = I(C).

13.1.9 Let f = {f1, . . . , fr} be a sequence of elements of a ring R and letf0 = 0. Prove that f is a regular sequence if and only if

((f1, . . . , fi−1) : fi) = (f1, . . . , fi−1) for all i ≥ 1.

13.1.10 A sequence xα1 , . . . , xαr of monomials of R is a regular sequenceif and only if xαi and xαj have no common variables for all i �= j.

13.1.11 Let C be a clutter and let C be a minimal vertex cover of C. Provethat |C ∩ e| = 1 for some edge e of C.

540 Chapter 13

13.2 Symbolic Rees algebras of edge ideals

Let C be a clutter with vertex set X = {x1, . . . , xn} and edge set E(C), letR = K[X ] be a polynomial ring over a field K, and let I = I(C) be the edgeideal of C. The blowup algebra studied here is the symbolic Rees algebra

Rs(I) = R⊕ I(1)t⊕ · · · ⊕ I(i)ti ⊕ · · · ⊂ R[t]

of I, where t is a new variable and I(i) is the ith symbolic power of I. Themain theorem of this section is a description—in combinatorial optimizationterms—of the minimal set of generators of Rs(I) as a K-algebra.

As usual, we denote by C∨ the clutter whose edges are the minimalvertex covers of C. If C is a subset of X , its characteristic vector is thevector v =

∑xi∈C ei, where ei is the ith unit vector in Rn. Let C1, . . . , Cs

be the minimal vertex covers of C and let uk be the characteristic vector ofCk for 1 ≤ k ≤ s. In our situation, according to Propositions 4.3.24 and6.5.4, the bth symbolic power of I has a simple expression:

I(b) = pb1 ∩ · · · ∩ pbs = ({xa| 〈a, uk〉 ≥ b for k = 1, . . . , s}), (13.1)

where pk is the prime ideal of R generated by Ck. In particular, if b = 1,we obtain the primary decomposition of I because I(1) = I.

Definition 13.2.1 The Simis cone of I = I(C) is the polyhedral cone:

Cn(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

(u1,−1) ∩ · · · ∩H+(us,−1).

The term Simis cone was coined in [147] to recognize the pioneeringwork of Aron Simis on symbolic powers of monomial ideals [377]. TheSimis cone is a finitely generated rational cone (Theorem 1.1.29). Hence,by Theorem 1.3.9, there is a unique minimal finite set of integral vectorsH ⊂ Zn+1 such that Zn+1 ∩ R+H = NH and Cn(I) = R+H (minimalrelative to taking subsets). The set H is called the minimal Hilbert basis ofCn(I). See Section 1.3 for a detailed study of Hilbert bases.

Theorem 13.2.2 H is the set of all integral vectors 0 �= α ∈ Cn(I) suchthat α is not the sum of two other non-zero integral vectors in Cn(I).

Proof. It follows at once from Theorem 1.3.9. �

Theorem 13.2.3 [147] Let H ⊂ Nn+1 be a Hilbert basis of Cn(I). IfK[NH] is the semigroup ring of NH, then Rs(I) = K[NH].

Proof. Recall that K[NH] = K[{xatb| (a, b) ∈ NH}]. Take xatb ∈ Rs(I),that is, xa ∈ pbi for all i. Hence 〈(a, b), (ui,−1)〉 ≥ 0 for all i or equivalently(a, b) ∈ Cn(I). Thus (a, b) ∈ NH and xatb ∈ K[NH]. Conversely takexatb ∈ K[NH], then (a, b) is in Cn(I) and 〈(a, b), (ui,−1)〉 ≥ 0 for all i.Hence xa ∈ pbi for all i and x

a ∈ I(b), as required. �


Corollary 13.2.4 [301] Rs(I) is a finitely generated K-algebra.

Proof. Let Cn(I) be the Simis cone of I and let H ⊂ Nn+1 be a Hilbertbasis of Cn(I). Applying Theorem 13.2.3, we get Rs(I) = K[NH]. Thus,Rs(I) is a finitely generated K-algebra. �

Definition 13.2.5 Let a = (ai) �= 0 be a vector in Nn and let b ∈ N. If a, bsatisfy 〈a, uk〉 ≥ b for k = 1, . . . , s, we say that a is a b-cover of C∨.

Lemma 13.2.6 xatb is in Rs(I) if and only if a is a b-cover of C∨.

Proof. It follows from the description of I(b) given in Eq. (13.1). �

The notion of a b-cover occurs in combinatorial optimization (see forinstance [373, Chapter 77, p. 1378] and the references therein) and algebraiccombinatorics [147, 225].

Definition 13.2.7 A b-cover a of C∨ is called reducible if there exists ani-cover c and a j-cover d of C∨ such that a = c+ d and b = i+ j. If a is notreducible, we call a irreducible.

The irreducible 0 and 1 covers of C∨ are the unit vectors e1, . . . , en andthe characteristic vectors v1, . . . , vq of the edges of C, respectively.

Lemma 13.2.8 A monomial xatb is a minimal generator of Rs(I), as aK-algebra, if and only if a is an irreducible b-cover of C∨.

Proof. It follows from the discussion above, by decomposing any b-coverinto irreducible ones. �

Let S be a set of vertices of a clutter C. The induced subclutter on S,denoted by C[S], is the maximal subclutter of C with vertex set S. Thusthe vertex set of C[S] is S and the edges of C[S] are exactly the edges of Ccontained in S. Notice that C[S] may have isolated vertices, i.e., verticesthat do not belong to any edge of C[S]. If C is a discrete clutter, i.e., all thevertices of C are isolated, we set I(C) = 0 and α0(C) = 0.

Let C be a clutter and let X1, X2 be a partition of its vertex set V (C)into non-empty sets. Clearly, one has the inequality

α0(C) ≥ α0(C[X1]) + α0(C[X2]). (13.2)

If C is a graph and equality occurs, Erdos and Gallai [144] call C a decom-posable graph. This motivates the following similar notion for clutters.

Definition 13.2.9 A clutter C is called decomposable if there are non-empty vertex sets X1, X2 such that X = V (C) is the disjoint union ofX1 and X2, and α0(C) = α0(C[X1]) + α0(C[X2]). If C is not decomposable,it is called indecomposable.

542 Chapter 13

Examples of indecomposable graphs include complete graphs, odd cyclesand complements of odd cycles of length at least five (see Lemma 13.3.1).

Definition 13.2.10 (Schrijver [373]) The duplication of a vertex xi of aclutter C means extending its vertex set X by a new vertex x′i and replacingthe edge set E(C) by

E(C) ∪ {(e \ {xi}) ∪ {x′i}|xi ∈ e ∈ E(C)}.

The deletion of xi, denoted by C\{xi}, is the clutter formed from C by delet-ing the vertex xi and all edges containing xi. A clutter obtained from C by asequence of deletions and duplications of vertices is called a parallelization.

It is not difficult to verify that these two operations commute. If a = (ai)is a vector in Nn, we denote by Ca the clutter obtained from C by successivelydeleting any vertex xi with ai = 0 and duplicating ai − 1 times any vertexxi if ai ≥ 1 (for graphs cf. [191, p. 53]).

Example 13.2.11 Let G be the graph whose only edge is {x1, x2} and leta = (3, 3). We set x1i = xi for i = 1, 2. The parallelization Ga is a completebipartite graph with bipartition V1 = {x11, x21, x31} and V2 = {x12, x22, x32}.Note that xki is a vertex, i.e., k is an index not an exponent.

��x1

x2

G

Graph

� ��

Duplications of x1

�x11 x21 x31

x12��

��

��

G(3,1) �x11 x21 x31

x12 x22 x32��

��

��

��

��

�

��

� � �G(3,3)

Duplications of x1 and x2

Lemma 13.2.12 Let C be a clutter and let A be its incidence matrix. Ifa = (ai) is a vector in Nn, then

β1(Ca) ≤ max{〈y,1〉| y ∈ Nq; Ay ≤ a}.

Proof. We may assume that a = (a1, . . . , am, 0, . . . , 0), where ai ≥ 1 fori = 1, . . . ,m. Recall that for each 1 ≤ i ≤ m the vertex xi is duplicatedai − 1 times and for each i > m the vertex xi is deleted. We denote theduplications of xi by x

2i , . . . , x

aii and set x1i = xi. Thus, we can write

V (Ca) = {x11, . . . , xa11 , . . . , x1i , . . . , xaii , . . . , x1m, . . . , xamm }.

There are f1, . . . , fβ1 independent edges of Ca, where β1 = β1(Ca). Each fihas the form

fk = {xjk1k1 , xjk2k2, . . . , x

jkrkr} (1 ≤ k1 < · · · < kr ≤ m; 1 ≤ jki ≤ aki).


We set gk = {x1k1 , x1k2, . . . , x1kr} = {xk1 , xk2 , . . . , xkr}. By definition of Ca

we get that gk ∈ E(C) for all k. We may re-order the fi’s so that

g1 = g2 = · · · = gs1︸︷︷︸s1

, gs1+1 = · · · = gs2︸︷︷︸s2−s1

, . . . , gsr−1+1 = · · · = gsr︸︷︷︸sr−sr−1

and gs1 , . . . , gsr distinct, where sr = β1. Let vi be the characteristic vectorof gsi . Set y = s1e1+(s2− s1)e2+ · · ·+(sr− sr−1)er. We may assume thatthe incidence matrix A of C has column vector v1, . . . , vq. Then y satisfies〈y,1〉 = β1. For each ki the number of variables of the form x�ki that occurin f1, . . . , fβ1 is at most aki because the fi are pairwise disjoint. Hence foreach ki the number of times that the variable x1ki occurs in g1, . . . , gβ1 is atmost aki . Then

Ay = s1v1 + (s2 − s1)v2 + · · ·+ (sr − sr−1)vr ≤ a. �

Definition 13.2.13 The clutter of minimal vertex covers of C, denoted byb(C) or C∨, is called the blocker of C or the Alexander dual of C.

Lemma 13.2.14 ([116, Lemma 2.15], [373, p. 1385, Eq. (78.6)]) Let C bea clutter and let C∨ be the blocker of C. If a = (ai) ∈ Nn, then

min

{∑xi∈C

ai

∣∣∣∣∣ C ∈ C∨}

= α0(Ca).

Proof. We may assume that a = (a1, . . . , am, am+1, . . . , am1 , 0, . . . , 0),where ai ≥ 2 for i = 1, . . . ,m, ai = 1 for i = m+ 1, . . . ,m1, and ai = 0 fori > m1. Thus for i = 1, . . . ,m the vertex xi is duplicated ai − 1 times. Wedenote the duplications of xi by x

2i , . . . , x

aii and set x1i = xi.

We prove first the inequality “≤”. Let Ca be a minimal vertex cover ofCa with α0 elements, where α0 is equal to α0(Ca). We may assume thatCa ∩ {x1, . . . , xm1} = {x1, . . . , xs}. Note that x1i , . . . , x

aii are in Ca for

i = 1, . . . , s. Indeed since Ca is a minimal vertex cover of Ca, there existsan edge e of Ca such that e ∩Ca = {x1i }. Then (e \ {x1i })∪ {x

ji} is an edge

of Ca for j = 1, . . . , ai. Consequently xji ∈ Ca for j = 1, . . . , ai. Hence

a1 + · · ·+ as ≤ |Ca| = α0. (13.3)

On the other hand the set C′ = {x1, . . . , xs} ∪ {xm1+1, . . . , xn} is a vertexcover of C. Let D be a minimal vertex cover of C contained in C′. Let eDdenote the characteristic vector of D. Then, since ai = 0 for i > m1, usingEq. (13.3) we get

〈a, eD〉 =∑xi∈D

ai =∑

xi∈D∩{x1,...,xs}ai ≤

∑xi∈{x1,...,xs}

ai ≤ α0.

544 Chapter 13

This completes the proof of the inequality “≤”. Next we show the inequality“≥”. Let C be a minimal vertex cover of C. Note that the set

C ′ = ∪xi∈C{x1i , . . . , xaii }

is a vertex cover of Ca. Indeed any edge ea of the clutter Ca is of the formea = {xj1i1 , . . . , x

jrir} for some edge e = {xi1 , . . . , xir} of C and since e is

covered by C, we have that ea is covered by C′. Therefore one has thatα0(Ca) ≤ |C′| =

∑xi∈C ai. As C was an arbitrary vertex cover of C we get

the asserted inequality. �

Theorem 13.2.15 [307] Let C be a clutter with vertex set X = {x1, . . . , xn}and let 0 �= a ∈ Nn, b ∈ N. Then xatb is a minimal generator of Rs(I(C)),as a K-algebra, if and only if Ca is indecomposable and b = α0(Ca).

Proof. We may assume that a = (a1, . . . , am, 0, . . . , 0), where ai ≥ 1 fori = 1, . . . ,m. For each 1 ≤ i ≤ m the vertex xi is duplicated ai − 1 times,and the vertex xi is deleted for each i > m. We denote the duplications ofxi by x

2i , . . . , x

aii and set x1i = xi for 1 ≤ i ≤ m. The vertex set of Ca is

Xa = {x11, . . . , xa11 , . . . , x1i , . . . , xaii , . . . , x1m, . . . , xamm } = Xa1 ∪ · · · ∪Xam ,

where Xai = {x1i , . . . , xaii } for 1 ≤ i ≤ m and Xai ∩Xaj = ∅ for i �= j.⇒) Assume that xatb is a minimal generator of Rs(I(C)). Then, by

Lemma 13.2.8, a is an irreducible b-cover of C∨. First we prove that bis equal to α0(Ca). There is k such that ak �= 0. We may assume thata− ek �= 0. By Lemma 13.2.14 we need only show the equality

b = min

{∑xi∈C

ai

∣∣∣∣∣ C ∈ C∨}.

As a is a b-cover of C∨, the minimum is greater than or equal to b. If theminimum is greater than b, then we can write a = (a−ek)+ek, where a−ekis a b-cover and ek is a 0-cover of C∨, a contradiction.

Next we show that Ca is indecomposable. We proceed by contradiction.Assume that Ca is decomposable. Then there is a partition X1, X2 of Xa

such that α0(Ca) = α0(Ca[X1]) + α0(Ca[X2]). For 1 ≤ i ≤ n, we set

�i = |Xai ∩X1| and pi = |Xai ∩X2|

if 1 ≤ i ≤ m, and �i = pi = 0 if i > m. Consider the vectors � = (�i)and p = (pi). Notice that a has a decomposition a = � + p because onehas a partition Xai = (Xai ∩ X1) ∪ (Xai ∩ X2) for 1 ≤ i ≤ m. To derivea contradiction we now claim that � (resp. p) is an α0(Ca[X1])-cover (resp.α0(Ca[X2])-cover) of C∨. Take an arbitrary C in C∨. The set

Ca =⋃xi∈C{x1i , . . . , xaii } =

⋃xi∈C

Xai


is a vertex cover of Ca. Indeed, if fk is any edge of Ca, then fk has the form

fk = {xjk1k1 , . . . , xjkrkr} (1 ≤ k1 < · · · < kr ≤ m; 1 ≤ jki ≤ aki) (13.4)

for some edge {xk1 , . . . , xkr} of C. Since {xk1 , . . . , xkr} ∩ C �= ∅, we getfk∩Ca �= ∅. Thus Ca is a vertex cover of Ca. Therefore Ca∩X1 and Ca∩X2

are vertex covers of Ca[X1] and Ca[X2], respectively, because E(Ca[Xi]) iscontained in E(Ca) for i = 1, 2. Hence using the partitions

Ca ∩X1 =⋃xi∈C

(Xai ∩X1) and Ca ∩X2 =⋃xi∈C

(Xai ∩X2)

we obtain

α0(Ca[X1]) ≤ |Ca ∩X1| =∑xi∈C

�i and α0(Ca[X2]) ≤ |Ca ∩X2| =∑xi∈C

pi.

This completes the proof of the claim. Consequently a is a reducible b-coverof C∨, where b = α0(Ca), a contradiction to the irreducibility of a.⇐) Assume that Ca is an indecomposable clutter and b = α0(Ca). We set

a = (a1, . . . , an). To show that xatb is a minimal generator of Rs(I(C)) weneed only show that a is an irreducible b-cover of C∨. To begin with, noticethat a is a b-cover of C∨ by Lemma 13.2.14. We proceed by contradictionassuming that there is a decomposition a = � + p, where � = (�i) is a c-cover of C∨, p = (pi) is a d-cover of C∨, and b = c + d. Each Xai can bedecomposed as Xai = X�i ∪ Xpi , where X�i ∩ Xpi = ∅, �i = |X�i|, andpi = |Xpi |. We set

X� = X�1 ∪ · · · ∪X�m and Xp = Xp1 ∪ · · · ∪Xpm .

Then one has a decomposition Xa = X� ∪Xp of the vertex set of Ca. Wenow show that α0(Ca[X�]) ≥ c and α0(Ca[Xp]) ≥ d. By symmetry, it sufficesto prove the first inequality. Take an arbitrary minimal vertex cover C� ofCa[X�]. Then C� ∪ Xp is a vertex cover of Ca because if f is an edge ofCa contained in X�, then f is covered by C�, otherwise f is covered by Xp.Hence there is a minimal vertex cover Ca of Ca such that Ca ⊂ C�∪Xp. Weset Vm = {x1, . . . , xm}. Since C[Vm] is a subclutter of Ca, there is a minimalvertex cover C1 of C[Vm] contained in Ca. Then the set C1 ∪ {xi| i > m} isa vertex cover of C. Therefore there is a minimal vertex cover C of C suchthat C ∩ Vm ⊂ Ca. Altogether one has:

C ∩ Vm ⊂ Ca ⊂ C� ∪Xp ⇒ C ∩ Vm ⊂ Ca ∩ Vm ⊂ (C� ∪Xp) ∩ Vm. (13.5)

We may assume that Ca∩Vm = {x1, . . . , xs}. Next we claim that Xai ⊂ Cafor 1 ≤ i ≤ s. Take an integer i between 1 and s. Since Ca is a minimalvertex cover of Ca, there exists an edge e of Ca such that e ∩ Ca = {x1i }.

546 Chapter 13

Then (e \ {x1i }) ∪ {xji} is an edge of Ca for j = 1, . . . , ai, this follows using

that the edges of Ca are of the form described in Eq. (13.4). Consequentlyxji ∈ Ca for j = 1, . . . , ai. This completes the proof of the claim. Thus onehas X�i ⊂ Xai ⊂ Ca for 1 ≤ i ≤ s. Hence, by Eq. (13.5), and noticing thatX�i ∩ Xp = ∅, we get X�i ⊂ C� for 1 ≤ i ≤ s. So, using that �i = 0 fori > m, we get

α0(Ca[X�]) ≥ |C�| ≥s∑i=1

�i ≥∑

xi∈C∩Vm

�i =∑xi∈C

�i ≥ c.

Therefore α0(Ca[X�]) ≥ c. Similarly α0(Ca[Xp]) ≥ d. Thus

α0(Ca[X�]) + α0(Ca[Xp]) ≥ c+ d = b = α0(Ca),

and consequently, by Eq. (13.2), we have equality. Thus we have shownthat Ca is a decomposable clutter, a contradiction. �

Lemma 13.2.16 If C is an indecomposable clutter with the Konig property,then either C has no edges and has exactly one isolated vertex or C has onlyone edge and no isolated vertices.

Proof. Let f1, . . . , fg be a set of independent edges and let X ′ = ∪gi=1fi,where g = α0(C). Note that g = 0 if C has no edges. Then V (C) has apartition

V (C) = (∪gi=1fi) ∪(∪xi∈V (C)\X′{xi}

).

As C is indecomposable, we get that either g = 0 and V (C) = {xi} for somevertex xi or g = 1 and V (C) = fi for some i. Thus in the second case, as Cis a clutter, we get that C has exactly one edge and no isolated vertices. �

Corollary 13.2.17 Let C be a clutter and let I = I(C) be its edge ideal.Then all indecomposable parallelizations of C satisfy the Konig property ifand only if Ii = I(i) for i ≥ 1.

Proof. ⇒) It suffices to prove that R[It] = Rs(I). Clearly R[It] ⊂ Rs(I).To prove the reverse inclusion take a minimal generator xatb of Rs(I). Ifb = 0, then a = ei for some i and xatb = xi is in R[It]. Assume b ≥ 1.By Theorem 13.2.15 Ca is an indecomposable clutter such that b = α0(Ca).As Ca satisfies the Konig property, using Lemma 13.2.16, it is not hard tosee that b = 1 and that E(Ca) = {e} consists of a single edge e of C, i.e.,xatb = xet, where xe =

∏xi∈e xi. Thus x

atb ∈ R[It].⇐) Since R[It] = Rs(I), by Theorem 13.2.15, we obtain that the only

indecomposable parallelizations are either induced subclutters of C withexactly one edge and no isolated vertices or subclutters consisting of exactlyone isolated vertex. Thus in both cases they satisfy the Konig property. �


Corollary 13.2.18 Let C be a clutter with vertex set X = {x1, . . . , xn} andlet S ⊂ X. Then the induced clutter H = C[S] is indecomposable if and onlyif the monomial (

∏xi∈S xi)t

α0(H) is a minimal generator of Rs(I(C)).

Proof. We set a =∑xi∈S ei. Since Ca = C[S], the result follows from

Theorem 13.2.15. �

Corollary 13.2.19 Let H be the Hilbert basis of Cn(I(C)). Then

H = {(a, b)|xatb is a minimal generator of Rs(I(C))}= {(a, α0(C))| Ca is an indecomposable parallelization of C}= {(a, b)| a is an irreducible b-cover of C∨}.

Proof. The first and third equalities follow readily from Lemma 13.2.8 andTheorem 13.2.2. The second equality follows from Theorem 13.2.15. �

This result allows us to compute all indecomposable parallelizations ofC and all indecomposable induced subclutters of C using Hilbert bases (seealso Exercise 13.2.31).

Computing the Hilbert basis of a Simis cone Let xv1 , . . . , xvq be theminimal set of generators of the edge ideal I(C) of a clutter C and let R+A′

be the Rees cone of A = {v1, . . . , vq}.

Lemma 13.2.20 If �i = (ui,−1) for i = 1, . . . , s, then the halfspace H+�i

occurs in the irreducible representation of R+A′.

Proof. By Corollary 13.1.3 the integral vertices of Q(A) are u1, . . . , us,where A is the incidence matrix of the clutter C. Hence the result followsfrom Theorem 1.4.2. �

Procedure 13.2.21 To determine the Hilbert basis of Cn(I) and the gen-erators of Rs(I(C)) we proceed as follows. Using Normaliz , one computesthe irreducible representation of R+A′ and apply Lemma 13.2.20 to obtainthe set of “inequalities” Γ = {e1, . . . , en+1, �1, . . . , �s} that define Cn(I).Then one can use Γ as input for Normaliz [68] to obtain the Hilbert basis.Then one uses Corollary 13.2.19 to obtain the generators of Rs(I(C)).

This procedure is based on the fact that the program Normaliz [68] isable to compute the minimal Hilbert basis of a pointed cone given by alinear system of inequalities (cf. [189, Procedure 4.6.10]).

Example 13.2.22 Using Procedure 13.2.21 and Normaliz [68], we computethe minimal generators of the symbolic Rees algebra of the edge ideal

I = I(C) = (x1x2, x2x3, x1x3).

The input file for Normaliz is:

548 Chapter 13

3

3

1 1 0

0 1 1

1 0 1

3

Using Normaliz, with mode 3, and Lemma 13.2.20, we obtain that Cn(I) isdefined by the “inequalities” given by the rows of the matrix:

7

4

0 0 1 0

0 0 0 1

1 0 1 -1

1 0 0 0

0 1 1 -1

0 1 0 0

1 1 0 -1

4

Using this input file for Normaliz, with mode 4, to compute the Hilbertbasis of Cn(I), we get:

7 Hilbert basis elements:

0 1 1 1

1 0 1 1

1 1 1 2

0 0 1 0

0 1 0 0

1 0 0 0

1 1 0 1

Thus, by Corollary 13.2.19, the symbolic Rees algebra of the edge ideal I isminimally generated as a K-algebra by x1, x2, x3, It, x1x2x3t

2.

Algebras of vertex covers In this part we will further examine minimalsets of generators of symbolic Rees algebras of edge ideals using polyhedralgeometry and vertex covers.

Let I = I(C) be the edge ideal of a clutter C and let v1, . . . , vq be thecharacteristic vectors of the edges of C. Recall that Rees cone of I, denotedby R+(I) or R+A′, is the rational cone generated by the set

A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Rn+1,

where ei is the ith unit vector. By Proposition 1.1.51 and Theorem 1.4.2,the Rees cone has a unique irreducible representation

R+(I) = H+e1 ∩H

+e2 ∩ · · · ∩H

+en+1∩H+

�1∩H+

�2∩ · · · ∩H+

�r(13.6)


such that, for each k, �k ∈ Zn+1, the non-zero entries of �k are relativelyprime, the first n entries of �k are in N, the last entry of �k is negative, andnone of the closed halfspaces H+

e1 , . . . , H+en+1

, H+�1, . . . , H+

�rcan be omitted

from the intersection.The facets (i.e., the proper faces of maximum dimension or equivalently

the faces of dimension n) of the Rees cone are exactly:

He1 ∩ R+(I), . . . , Hen+1 ∩ R+(I), H�1 ∩R+(I), . . . , H�r ∩ R+(I).

Let C1, . . . , Cs be the minimal vertex covers of C and let u1, . . . , usbe their characteristic vectors, i.e., uk =

∑xi∈Ck

ei for k = 1, . . . , s. ByLemma 13.2.20 we may always assume that

�k = (uk,−1) for 1 ≤ k ≤ s,

and that every facet H�k ∩ R+(I), with k > s, satisfies 〈�k, en+1〉 < −1.Thus one has:

Remark 13.2.23 The primary decomposition of I(C) as well as all theminimal vertex covers of C can be read off from the irreducible representationof the Rees cone R+(I(C)).

The ideal of covers of C, denoted by Ic(C), is the ideal of R generated byxu1 , . . . , xus . We also denote Ic(C) by I(C)∨ and call I(C)∨ the Alexanderdual of I(C). Notice that Ic(C) = I(C∨) = I(C)∨.

The symbolic Rees algebra of Ic(C) can also be interpreted in terms of“b-vertex covers” because Rs(Ic(C)) is the symbolic Rees algebra of the edgeideal of C∨ and (C∨)∨ = C. Let a = (a1, . . . , an) �= 0 be a vector in Nn andlet b ∈ N. Recall that a is a b-vertex cover of C if 〈vi, a〉 ≥ b for i = 1, . . . , q.

Definition 13.2.24 The algebra of vertex covers of I, denoted by Rc(I),is the K-subalgebra of R[t] generated by all monomials xatb such that a isa b-cover of C. This algebra turns out to be equal to Rs(Ic(C)).

The irreducible 0 and 1 covers of C are the unit vector e1, . . . , en andthe vectors u1, . . . , us. The minimal generators of Rc(I), as a K-algebra,correspond to the irreducible covers of C (see Lemma 13.2.8). Notice thefollowing dual descriptions:

I(b) = ({xa| 〈a, ui〉 ≥ b for i = 1, . . . , s}),Ic(C)(b) = ({xa| 〈a, vi〉 ≥ b for i = 1, . . . , q}).

Thus, we have the following duality between symbolic Rees algebras.

Theorem 13.2.25 Rc(I) = Rs(Ic(C)) and Rc(Ic(C)) = Rs(I).

550 Chapter 13

Theorem 13.2.26 Rc(I) is a finitely generated K-algebra.

Proof. It follows from Corollary 13.2.4 because Rc(I) is the symbolic Reesalgebra of the edge ideal Ic(C). �

Lemma 13.2.27 If �k = (ak,−dk) is any of the vectors of Eq. (13.6), whereak ∈ Nn, dk ∈ N+, then ak is an irreducible dk-cover of C and xak tdk is aminimal generator of Rs(Ic(C)).

Proof. We proceed by contradiction and assume that there is a d′k-covera′k and a d′′k-cover a

′′k such that ak = a′k + a′′k and dk = d′k + d′′k . Set

F ′ = H(a′k,−d′k) ∩ R+(I) and F ′′ = H(a′′k ,−d′′k ) ∩ R+(I). Clearly F ′, F ′′

are proper faces of R+(I) and F = R+(I) ∩ H�k = F ′ ∩ F ′′. ApplyingTheorem 1.1.44(d) to F ′ and F ′′ it is seen that F ′ ⊂ F or F ′′ ⊂ F , i.e.,F = F ′ or F = F ′′. We may assume F = F ′. Hence H(a′k,−d′k) = H�k .Taking orthogonal complements we get that (a′k,−d′k) = λ(ak,−dk) forsome λ ∈ Q+, because the orthogonal complement of H�k is generated by�k. Since the non-zero entries of �k are relatively prime, we may assumethat λ ∈ N. Thus d′k = λdk ≥ dk ≥ d′k and λ must be 1. Hence ak = a′kand a′′k must be zero, a contradiction. �

Remark 13.2.28 Let Fn+1 be the facet of the Rees cone R+(I) determinedby the hyperplane Hen+1 . Thus, by Lemma 13.2.27, we have a map ψ:

{Facets of R+(I(C))} \ {Fn+1}ψ−→ Rs(Ic(C))

H�k ∩ R+(I)ψ−→ xak tdk , where �k = (ak,−dk)

Hei ∩ R+(I)ψ−→ xi

whose image is a good approximation for the minimal set of generators ofRs(Ic(C)) as a K-algebra. Likewise, from the facets of R+(Ic(C)), we obtainan approximation for the minimal set of generators of Rs(I(C)).

The next example shows a connected graph G for which the image ofthe map ψ does not generate Rs(Ic(G)).

Example 13.2.29 Consider the following graph G:

�� x2 x4 x6 x8

��

��

x1 x3 x5 x7 x9

��

��

��

��

��

Using Normaliz [68] it is seen that the vector a = (1, 1, 2, 0, 2, 1, 1, 1, 1) is anirreducible 2-cover of G such that the supporting hyperplane H(a,−2) doesnot define a facet of the Rees cone of I(G).


For balanced clutters, the image of the map ψ generates Rs(Ic(C)). Thisfollows from the next result (see Corollary 14.3.11). In particular the imageof the map ψ generates Rs(Ic(C)) when C is a bipartite graph.

Proposition 13.2.30 [183, 185] If C is a balanced clutter, then

Rs(Ic(C)) = R[Ic(C)t].

This result was first shown for bipartite graphs in [183, Corollary 2.6]and later generalized to balanced clutters in [185].

Exercises

13.2.31 Let C be a clutter with vertices x1, . . . , xn and α = (a1, . . . , an, b)a vector in {0, 1}n×N. Then α is in the minimal Hilbert basis of Cn(I(C))if and only if the induced subclutter H = C[{xi| ai = 1}] is indecomposablewith b = α0(H).

13.2.32 Let C be a clutter. Prove that (1, . . . , 1, α0(C)) is in the minimalHilbert basis of Cn(I(C)) if and only if C is indecomposable.

13.3 Blowup algebras in perfect graphs

Harary and Plummer [209] studied indecomposable graphs. To the bestof our knowledge there is no structure theorem for indecomposable graphs.In this section we relate these graphs to the theory of perfect graphs andto symbolic Rees algebras. Then we present some general properties ofindecomposable clutters.

Lemma 13.3.1 Let Cn = {x1, . . . , xn} be a cycle of length n. (a) If n is oddand n ≥ 5, then the complement Cn of Cn is indecomposable, (b) if n is odd,then Cn is indecomposable, and (c) any complete graph is indecomposable.

Proof. (a) Assume that G = Cn is decomposable. Then there are disjointsets X1, X2 such that V (G) = X1∪X2 and α0(G) = α0(G[X1])+α0(G[X2]).Since β0(G) = 2, it is seen that G[Xi] is a complete graph for i = 1, 2. Wemay assume that x1 ∈ X1. Then x2 must be in X2, otherwise {x1, x2} is anedge ofG[X1], a contradiction. By induction it follows that x1, x3, x5, . . . , xnare in X1. Hence, {x1, xn} is an edge of G[X1], a contradiction. (b) Thisis left as an exercise. (c) It follows readily from the fact that the coveringnumber of a complete graph in r vertices is r − 1. �

For graphs, we can use this lemma together with Corollary 13.2.19 andExercise 13.2.31 to locate all induced odd cycles (odd holes) and all inducedcomplements of odd cycles (odd antiholes) of length at least five. Notice that,

552 Chapter 13

by Theorem 13.2.15, odd holes and odd antiholes correspond to minimalgenerators of the symbolic Rees algebra of the edge ideal of the graph.

A graph G is called a Berge graph if and only if G has no odd holes orodd antiholes of length at least five.

Theorem 13.3.2 (Strong perfect graph theorem [85]) A graph G is perfectif and only if G is a Berge graph.

In commutative algebra odd holes occurred for the first time in [377],and later in the description of I(G){2}, the join of an edge ideal of a graphG with itself [381], and in the description of the associated primes of powersof ideals of vertex covers of graphs [164]. The expository paper [163] surveysalgebraic techniques for detecting odd cycles and odd holes in a graph andfor computing the chromatic number of a hypergraph (see Theorem 7.7.19).

Theorem 13.3.3 [164] If G is a graph and p is an associated prime ofIc(G)

2, then p = (xi, xj) for some edge {xi, xj} of G or p = (xi| ∈ S),where S is a set of vertices of G that induces an odd hole.

Example 13.3.4 Let G be the graph below. Computing the generatorsof Rs(I(G)) via Procedure 13.2.21 and Corollary 13.2.19, gives that theindecomposable parallelizations of G are: seven vertices, nine edges, oneinduced triangle, three induced pentagons, and the duplication shown below.

��

��

��

��

x4 x3

x5

x1

x2x6

x7

Decomposable graph G

��

��

��

��

��

x4 x3

x5

x1

x2x6

x7

x′1

Indecomposable graph G(2,1,1,1,1,1,1)

The next result shows that indecomposable graphs occur naturally inthe theory of perfect graphs.

Proposition 13.3.5 [117, Proposition 2.13] A graph G is perfect if andonly if the indecomposable parallelizations of G are exactly the completesubgraphs of G

The following was one of the first deep results in the study of symbolicpowers of edge ideals from the viewpoint of graph theory.

Corollary 13.3.6 [383, Theorem 5.9] Let G be a graph and let I be its edgeideal. Then G is bipartite if and only if Ii = I(i) for i ≥ 1.


Proof. ⇒) If G is a bipartite graph, then any parallelization of G is again abipartite graph. This means that any parallelization of G satisfies the Konigproperty because bipartite graphs satisfy this property; see Theorem 7.1.8.Thus Ii = I(i) for all i by Corollary 13.2.17.⇐) Assume that Ii = I(i) for i ≥ 1. Thanks to Corollary 13.2.17 all

indecomposable induced subgraphs of G have the Konig property. If G isnot bipartite, then G has an induced odd cycle, a contradiction becauseinduced odd cycles are indecomposable by Lemma 13.3.1 and do not satisfythe Konig property. �

Basic properties of indecomposable clutters If e is a edge of a clutterC, we denote by C \ {e} the spanning subclutter of C obtained by deleting eand keeping all the vertices of C.

Definition 13.3.7 A clutter C is called vertex critical (resp. edge critical)if α0(C \ {xi}) < α0(C) (resp. α0(C \ {e}) < α0(C)) for all xi ∈ V (C) (resp.for all e ∈ E(C)).

Lemma 13.3.8 Let xi be a vertex and let e be an edge of a clutter C.(a) If α0(C \ {xi}) < α0(C), then α0(C \ {xi}) = α0(C)− 1.

(b) If α0(C \ {e}) < α0(C), then α0(C \ {e}) = α0(C)− 1.

Proof. Part (a) is left as an exercise (cf. Proposition 7.2.11). Part (b)follows from Proposition 7.2.13. �

Definition 13.3.9 A clutter C is called connected if there is no U ⊂ V (C)such that ∅ � U � V (C) and such that e ⊂ U or e ⊂ V (C) \U for each edgee of C.

Proposition 13.3.10 If a clutter C is indecomposable, then it is connectedand vertex critical.

Proof. Assume that C is disconnected. Then there is a partition X1, X2 ofV (C) such that

E(C) ⊂ E(C[X1]) ∪ E(C[X2]). (13.7)

For i = 1, 2, let Ci be a minimal vertex cover of C[Xi] with α0(C[Xi])vertices. Then, by Eq. (13.7), C1∪C2 is a minimal vertex cover of C. Henceα0(C[X1]) + α0(C[X2]) is greater than or equal to α0(C). So α0(C) is equalto α0(C[X1]) + α0(C[X2]), a contradiction to the indecomposability of C.Thus C is connected.

We now show that α0(C \{xi}) < α0(C) for all i. If α0(C \{xi}) = α0(C),then V (C) = X1 ∪X2, where X1 = V (C) \ {xi} and X2 = {xi}. Note thatC[X1] = C \ {xi}. As α0(C[X1]) = α0(C) and α0(C[X2]) = 0, we contradictthe indecomposability of C. Thus α0(C \ {xi}) < α0(C). �

554 Chapter 13

Proposition 13.3.11 If C is a connected edge critical clutter, then C isindecomposable.

Proof. Assume that C is decomposable. Then there is a partition X1, X2 ofV (C) into non-empty vertex sets such that α0(C) = α0(C[X1]) + α0(C[X2]).Since C is connected, there is an edge e ∈ E(C) intersecting both X1 andX2. Pick a minimal vertex cover C of C \ {e} with less than α0(C) vertices.As E(C[Xi]) is a subset of E(C \ {e}) = E(C) \ {e} for i = 1, 2, we get thatC covers all edges of C[Xi] for i = 1, 2. Hence C must have at least α0(C)vertices, a contradiction. �

Exercises

13.3.12 Let D be a clutter obtained from C by adding a new vertex v andsome new edges containing v and some vertices of V (C). If a = 1 ∈ Nn is anirreducible α0(C)-cover of C∨ such that α0(D) = α0(C) + 1, then a′ = (a, 1)is an irreducible α0(D)-cover of D∨.

13.3.13 [12] If G is a complete graph andH is the graph obtained by takinga cone over a pentagon, then

Rs(I(G)) = K[{xatb|xa is square-free ; deg(xa) = b+ 1}],Rs(I(H)) = R[I(H)t][x1 · · ·x5t3, x1 · · ·x6t4, x1 · · ·x5x26t5].

13.3.14 [209] Prove that any odd cycle is indecomposable.

13.3.15 Using Procedure 13.2.21 and Corollary 13.2.19, show that thegraph G below has exactly 103 indecomposable parallelizations, 92 of whichcorrespond to subgraphs. The only indecomposable parallelization Ga, withai > 0 for all i, is that obtained by duplication of the five outer vertices,i.e., a = (2, 2, 2, 2, 2, 1, 1, 1, 1, 1) and α0(G

a) = 11.

� �

�

� ��

��

%%%%%

&&&&&

��

��

��

'''''

��(((((

��

��

'''''

��

""""

��

(((((

!!!!

)))))

&&&&&

%%%%%

****

*

x4 x3

x1

x9 x8

x10x7

x6x5 x2

Decomposable graph G


13.4 Algebras of vertex covers of graphs

Let G be a graph and let Ic(G) be its ideal of covers. In this section wegive a graph theoretical description of the irreducible b-covers of G, i.e., wedescribe the minimal generators of the symbolic Rees algebra of Ic(G).

Lemma 13.4.1 If a = (a1, . . . , an) ∈ Nn is an irreducible b-cover of G,then 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2 for i = 1, . . . , n.

Proof. Recall that a is a b-cover of G if and only if ai + aj ≥ b for anyedge {xi, xj} of G. If b = 0 or b = 1, then by the irreducibility of a it isseen that either a = ei for some i or a = ei1 + · · · + eir for some minimalvertex cover {xi1 , . . . , xir} of G. Thus we may assume that b ≥ 2.

Case (I): ai ≥ 1 for all i. Clearly 1 = (1, . . . , 1) is a 2-cover. If a−1 �= 0,then a−1 is a b−2 cover and a = 1+(a−1), a contradiction. Hence a = 1.Pick any edge {xi, xj} of G. Since a is a b-cover, we get 2 = ai+aj ≥ b andb must be equal to 2.

Case (II): ai = 0 for some i. We may assume ai = 0 for 1 ≤ i ≤ rand ai ≥ 1 for i > r. Notice that the set S = {x1, . . . , xr} is independentbecause if {xi, xj} is an edge and 1 ≤ i < j ≤ r, then 0 = ai + aj ≥ b,a contradiction. Consider the neighbor set NG(S) of S. We may assumethat NG(S) = {xr+1, . . . , xs}. Observe that ai ≥ b ≥ 2 for i = r + 1, . . . , s,because a is a b-cover. Write

a = (0, . . . , 0, ar+1 − 2, . . . , as − 2, as+1 − 1, . . . , an − 1)+

(0, . . . , 0︸︷︷︸r

, 2, . . . , 2︸︷︷︸s−r

, 1, . . . , 1︸︷︷︸n−s

) = c+ d.

Clearly d is a 2-cover. If c �= 0, using that ai ≥ b ≥ 2 for r + 1 ≤ i ≤ sand ai ≥ 1 for i > s it is not hard to see that c is a (b − 2)-cover. Thisgives a contradiction, because a = c+ d. Hence c = 0. Therefore ai = 2 forr < i ≤ s, ai = 1 for i > s, and b = 2. �

Corollary 13.4.2 Rs(Ic(G)) is generated, as a K-algebra, by monomialsof degree in t at most two and total degree at most 2n.

Proof. Let xatb be a minimal generator of Rs(Ic(G)). Then a = (ai) is anirreducible b-cover of G. By Lemma 13.4.1, we get 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2for all i. If b = 0 or b = 1, we get that the degree of xatb is at most n becausewhen b = 0 or 1 one has a = ei or a =

∑xi∈Ck

ei for some minimal vertexcover Ck of G, respectively. If b = 2, by the proof of Lemma 13.4.1, eithera = 1 or ai = 0 for some i. Thus deg(xa) ≤ 2(n− 1). �

For use below consider the vectors �1, . . . , �r that occur in the irreduciblerepresentation of R+(I(G)) given in Eq. (13.6).

556 Chapter 13

Corollary 13.4.3 If �i = (�i1, . . . , �in,−�i(n+1)), then 0 ≤ �ij ≤ 2 forj = 1, . . . , n and 1 ≤ �i(n+1) ≤ 2.

Proof. It suffices to observe that (�i1, . . . , �in) is an irreducible �i(n+1)-coverof G and to apply Lemma 13.4.1. �

Lemma 13.4.4 a = (1, . . . , 1) is an irreducible 2-cover of G if and only ifG is non-bipartite.

Proof. ⇒) We proceed by contradiction. Assume that G is bipartite.Then G has a bipartition (V1, V2). Set a

′ =∑xi∈V1

ei and a′′ =

∑xi∈V2

ei.Since V1 and V2 are minimal vertex covers of G, we can decompose a asa = a′ + a′′, where a′ and a′′ are 1-covers, which is impossible.⇐) Notice that a cannot be the sum of a 0-cover and a 2-cover. Indeed

if a = a′ + a′′, where a′ is a 0-cover and a′′ is a 1-cover, then a′′ hasan entry ai equal to zero. Pick an edge {xi, xj} incident with xi, then〈a′′, ei + ej〉 ≤ 1, a contradiction. Thus we may assume that a = c + d,where c, d are 1-covers. Let Cr be an odd cycle of G of length r. Notice thatany vertex cover of Cr must contain a pair of adjacent vertices. Hence thevertex covers of G corresponding to c and d must contain a pair of adjacentvertices, a contradiction because c and d are complementary vectors andthe complement of a vertex cover is an independent set. �

Theorem 13.4.5 Let 0 �= a = (ai) ∈ Nn and let G∨ be the clutter ofminimal vertex covers of G. Then the following hold.

(i) If G is bipartite, then a is an irreducible b-cover of G if and only ifb = 0 and a = ei for some 1 ≤ i ≤ n or b = 1 and a =

∑xi∈C ei for

some C ∈ E(G∨).

(ii) If G is non-bipartite, then a is an irreducible b-cover if and only if ahas one of the following forms:

(a) (0-covers) b = 0 and a = ei for some 1 ≤ i ≤ n,(b) (1-covers) b = 1 and a =

∑xi∈C ei for some C ∈ E(G∨),

(c) (2-covers) b = 2 and a = (1, . . . , 1),

(d) (2-covers) b = 2 and up to permutation of vertices

a = (0, . . . , 0︸︷︷︸|A|

, 2, . . . , 2︸︷︷︸|NG(A)|

, 1, . . . , 1)

for some independent set of vertices A �= ∅ of G such that

(d1) NG(A) is not a vertex cover of G and V �= A ∪NG(A),(d2) the induced subgraph G[V \ (A ∪ NG(A))] has no isolated

vertices and is not bipartite.


Proof. (i) ⇒) Since G is bipartite, by Proposition 13.2.30, we have theequality Rs(Ic(G)) = R[Ic(G)t]. Thus the minimal set of generator ofRs(Ic(G)) as a K-algebra is {x1, . . . , xn, xu1t, . . . , xur t}, where u1, . . . , urare the characteristic vectors of the minimal vertex covers of G. As a is anirreducible b-cover of G, xatb is a minimal generator of Rs(Ic(C)). Thereforeeither a = ei for some i and b = 0 or a = ui for some i and b = 1. Theconverse follows readily and is valid for any graph or clutter.

(ii) ⇒) By Lemma 13.4.1, 0 ≤ b ≤ 2 and 0 ≤ ai ≤ 2 for all i. If b = 0 orb = 1, then clearly a has the form indicated in (a) or (b), respectively.

Assume b = 2. If ai ≥ 1 for all i, then ai = 1 for all i, otherwise if ai = 2for some i, then a − ei is a 2-cover and a = ei + (a − ei), a contradiction.Hence a = 1. Thus we may assume that a has the form

a = (0, . . . , 0, 2, . . . , 2, 1, . . . , 1).

We set A = {xi| ai = 0} �= ∅, B = {xi| ai = 2}, and C = V \(A∪B). Observethat A is an independent set because a is a 2-cover and B = NG(A) becausea is irreducible. Hence it is seen that conditions (d1) and (d2) are satisfied.By Lemma 13.4.4, the proof of the converse is straightforward. �

If I ⊂ R is an ideal, the ring Fs(I) = Rs(I)/mRs(I) is called the symbolicspecial fiber of I (cf. Definition 14.2.10). If G is a graph, the algebraicproperties of the ring Fs(I) are studied in [91], when G is bipartite thesymbolic special fiber of Ic(G) turns out to be Koszul [91, 356] and in thiscase, by Proposition 13.2.30, one has Fs(Ic(G)) = R[Ic(G)t]/mR[Ic(G)t].

Exercises

13.4.6 Let G be a graph with vertex set V (G) = {x1, . . . , xn} and let H+α

be any of the closed halfspaces that occur in the irreducible representationof R+(Ic(G)) with α = (a1, . . . , an,−b), ai ∈ N for all i, 0 �= b ∈ N, and thenon-zero entries of α are relatively prime.

(a) Let H be the cone over G. If ai ≥ 1 for all i and

β = (a1, . . . , an, (∑n

i=1 ai)− b,−∑ni=1 ai) = (β1, . . . , βn+1,−βn+2),

then Hβ ∩ R+(Ic(H)) is a facet of R+(Ic(H)) and xβ1

1 · · ·xβn+1

n+1 tβn+2 is a

minimal generator of Rs(I(H)).

(b) Let G0 = G and let Gr be the cone over Gr−1 for r ≥ 1. If α isequal to (1, . . . , 1,−g) and b = n+ (r − 1)(n− g), then

(1, . . . , 1︸︷︷︸n

, n− g, . . . , n− g︸︷︷︸r

)

is an irreducible b-cover of Gr and Rs(I(Gr)) has a generator of degree in tequal to b.

558 Chapter 13

(c) Let G = Cs be an odd cycle of length s = 2k + 1. Then, α0(Cs) isequal to (s+ 1)/2 = k + 1 and

x1 · · ·xsxks+1 · · ·xks+rtrk+k+1

is a minimal generator of Rs(I(Gr)). This proves that the degree in t of theminimal generators of Rs(I(Gr)) is much larger than the number of verticesof the graph Gr [225].

13.5 Edge subrings in perfect matchings

In this section we present membership criteria and a generalized versionof the marriage theorem. For a connected non-bipartite graph whose edgesubring is normal, we give conditions for the existence of a perfect matching.

Let G be a connected graph and let A = {v1, . . . , vq} be the set of allvectors ei + ej such that {xi, xj} is an edge of G. The edge cone of G,denoted by R+A, is defined as the cone generated by A. Below we recoveran explicit combinatorial description of the edge cone (see Section 10.7).

Let A be an independent set of vertices of G. The supporting hyperplaneof the edge cone of G defined by∑

xi∈A

xi −∑

xi∈NG(A)

xi = 0

is denoted by HA. If aA =∑

xi∈A ei −∑xi∈NG(A) ei, then HA = HaA .

Edge cones and their representations by closed halfspaces are a usefultool to study a-invariants of edge subrings [419, 405]; see Chapter 11. Thefollowing result is a prototype of these representations (cf. Theorem 10.7.8and Exercise 10.7.24). As an application we give a direct proof of the nextresult using Rees cones and Theorem 13.4.5.

Corollary 13.5.1 [406, Corollary 2.8] A vector a = (a1, . . . , an) ∈ Rn is inR+A if and only if a satisfies the following system of linear inequalities

ai ≥ 0, i = 1, . . . , n;∑xi∈NG(A) ai −

∑xi∈A ai ≥ 0, for all independent sets A ⊂ V (G).

Proof. We set B = {(v1, 1), . . . , (vq, 1)} and I = I(G). Notice the equality

R+(I) ∩ RB = R+B, (13.8)

where RB is the R-vector space spanned by B. Consider the irreduciblerepresentation of R+(I) given in Eq. (13.6) and write �i = (ai,−di), where0 �= ai ∈ Nn, 0 �= di ∈ N. Next we show the equality:

R+A = RA ∩Rn+ ∩H+(2a1/d1−1) ∩ · · · ∩H

+(2ar/dr−1), (13.9)


where 1 = (1, . . . , 1). Take α ∈ R+A. Clearly α ∈ RA∩ Rn+. We can write

α = λ1v1 + · · ·+ λqvq ⇒ |α| = 2(λ1 + · · ·+ λq) = 2b.

Thus (α, b) = λ1(v1, 1) + · · · + λq(vq, 1), i.e., (α, b) ∈ R+B. Hence, fromEq. (13.8), we get (α, b) ∈ R+(I) and

〈(α, b), (ai,−di)〉 ≥ 0 ⇒ 〈α, ai〉 ≥ bdi = (|α|/2)di = |α|(di/2).

Writing α = (α1, . . . , αn) and ai = (ai1, . . . , ain), the last inequality gives:

α1ai1 + · · ·+ αnain ≥ (α1 + · · ·+ αn)(di/2) ⇒ 〈α, ai − (di/2)1〉 ≥ 0.

Then 〈α, 2ai/di − 1〉 ≥ 0 and α ∈ H+(2ai/di−1) for all i. This proves the

inclusion “⊂” of Eq. (13.9). The other inclusion follows similarly. Now,by Lemma 13.2.27, ai is an irreducible di-cover of G. Therefore, usingTheorem 13.4.5, we readily get the equality

R+A =

( ⋂A∈F

H−A

)⋂(n⋂i=1

H+ei

),

where F is the collection of all the independent sets of vertices of G. Fromthis equality the assertion follows at once. �

Proposition 13.5.2 [419] Let G be a connected non-bipartite graph with n

vertices. If K[G] is a normal domain, then a monomial xβ1

1 · · ·xβnn belongs

to K[G] if and only if the following two conditions hold

(i) β = (β1, . . . , βn) is in the edge cone of G, and

(ii)∑ni=1 βi is an even integer.

Proof. ⇐) LetA = {v1, . . . , vq} be the set of column vector of the incidencematrix of G. Assume that β ∈ R+A and deg(xβ) even. We proceed byinduction on deg(xβ). Using Corollary 10.2.11 one has the isomorphism

Zn/(v1, . . . , vq) � Z2.

Hence 2β ∈ R+A ∩ ZA = NA and one can write

2β = 2

q∑i=1

sivi +

q∑i=1

εivi, si ∈ N and εi ∈ {0, 1},

by induction one may assume∑q

i=1 sivi = 0. Therefore, from the equalityabove, one concludes that the subgraph whose edges are defined by the set{vi| εi = 1} is an edge disjoint union of cycles C1, . . . , Cr. By induction onemay further assume that all the Ci’s are odd cycles. Note r ≥ 2, becausedeg(xb) is even. As G is connected, using Corollary 10.3.12, it follows thatxβ is in K[G].⇒) It follows readily by Corollary 9.1.3 because K[G] is normal. �

560 Chapter 13

Proposition 13.5.3 Let G be a connected graph with n vertices. If n iseven and K[G] is normal of dimension n, then x1 · · ·xn is in K[G] if andonly if |A| ≤ |N(A)| for every independent set of vertices A of G.

Proof. ⇒) LetA = {v1, . . . , vq} be the set of column vector of the incidencematrix of G. Since K[G] is normal: R+A ∩ ZA = NA. Hence the vectora = (a1, . . . , an) = 1 is in R+A. Using Corollary 13.5.1 we get that asatisfies the inequalities:

|A| =∑xi∈A

ai ≤∑

xi∈N(A)

ai = |N(A)|

for every independent set of vertices A of G, as required.⇐) First we use Corollary 13.5.1 to conclude that a is in the edge cone,

then apply Proposition 13.5.2 to get xa ∈ K[G]. �

Corollary 13.5.4 (Marriage theorem) Let G be a graph with an even num-ber of vertices. If G is connected and satisfies the odd cycle condition, thenthe following are equivalent

(a) G has a perfect matching.

(b) |A| ≤ |N(A)| for all A independent set of vertices of G.

Proof. The proof follows using Theorem 7.1.9 and Proposition 13.5.3. �

Exercises

13.5.5 Let G be a connected non-bipartite graph with n vertices and letv1, . . . , vq be the columns of its incidence matrix. If n is even, prove thatthe vector (1, . . . , 1) is in L = Zv1 + · · ·+ Zvq.

13.5.6 Let G be a connected non-bipartite graph with n vertices. If n iseven, prove that the monomial x1 · · ·xn is in the field of fractions of K[G].

13.5.7 Let G be a connected non-bipartite graph with n vertices. If K[G]is a normal domain, then x1 · · ·xn is in K[G] if and only if (1, . . . , 1) is inthe edge cone of G and n is even.

13.5.8 Prove that the following graph does not have a perfect matching byexhibiting an independent set A such that |A| �≤ |N(A)|.

��

��

��

��

��

��

��

��

��

��


13.5.9 Consider the graphG = G1∪G2 with six vertices x1 · · ·x6 consistingof two disjoint triangles:

•

��

��

�

G1

•

G2• •

��

��

��

�

•

�� •Prove that the vector 1 = (1, . . . , 1) belongs to the edge cone of G and thatx1 = x1 · · ·x6 is not in K[G].

13.6 Rees cones and perfect graphs

Let G be a graph and let Ic(G) be its ideal of covers. In this section wecharacterize when G is perfect in terms of the irreducible representation ofthe Rees cone of Ic(G) and show an application to Ehrhart rings.

Let G be a graph with vertex set X = {x1, . . . , xn}. In what follows weshall always assume that G has no isolated vertices. We denote a completesubgraph of G with r vertices by Kr. The empty set is regarded as anindependent set whose characteristic vector is the zero vector.

Theorem 13.6.1 [86, 296] The following statements are equivalent :

(a) G is a perfect graph.

(b) The complement of G is perfect.

(c) The independence polytope of G, i.e., the convex hull of the incidencevectors of the independent sets of G, is given by:{

(ai) ∈ Rn+|∑xi∈Kr

ai ≤ 1; ∀Kr ⊂ G}.

The equivalence between (a) and (b) is due to Lovasz [296] and this iscalled the weak perfect graph theorem. That (b) and (c) are equivalent isdue to Fulkerson and Chvatal [86].

Lemma 13.6.2 Let G be a graph. Then the set

F = {(ai) ∈ Rn+1|∑

xi∈V (Kr)ai = (r − 1)an+1} ∩ R+(Ic(G))

is a facet of R+(Ic(G)), where Kr is a complete subgraph of G.

Proof. If Kr = ∅, then r = 0 and F = Hen+1 ∩ R+(Ic(G)), which is afacet because e1, . . . , en ∈ F . If r = 1, then F = Hei ∩ R+(Ic(G)) for some1 ≤ i ≤ n, which is a facet because ej ∈ F for j /∈ {i, n + 1} and there is

562 Chapter 13

at least one minimal vertex cover of G not containing xi. We may assumethat V (Kr) = {x1, . . . , xr} and r ≥ 2. For each 1 ≤ i ≤ r there is a minimalvertex cover Ci of G not containing xi. Notice that Ci contains V (Kr)\{xi}.Let ui be the characteristic vector of Ci. Since rank(u1, . . . , ur) is r, theset {(u1, 1), . . . , (ur, 1), er+1, . . . , en} is linearly independent and containedin F , i.e., dim(F ) = n. Hence F is a facet of R+(Ic(G)) because thehyperplane that defines F is a supporting hyperplane. �

We regardK0 as the empty set with zero elements. A sum over an emptyset is defined to be 0.

Proposition 13.6.3 Let J = Ic(G) be the ideal of covers of G. Then G isperfect if and only if the following equality holds

R+(J) ={(ai) ∈ Rn+1|

∑xi∈Kr

ai ≥ (r − 1)an+1; ∀Kr ⊂ G}. (13.10)

Moreover this is the irreducible representation of R+(J) if G is perfect.

Proof. ⇒) The inclusion “⊂” is clear because any minimal vertex cover ofG contains at least r − 1 vertices of any Kr. To show the reverse inclusiontake a vector a = (ai) satisfying b = an+1 �= 0 and∑

xi∈Krai ≥ (r − 1)b; ∀Kr ⊂ G ⇒

∑xi∈Kr

(ai/b) ≥ r − 1; ∀Kr ⊂ G.

This implication follows because by making r = 0 we get b > 0. We mayassume that ai ≤ b for all i. Indeed if ai > b for some i, say i = 1, then wecan write a = e1 + (a− e1). From the inequality∑

xi∈Krx1∈Kr

ai = a1 +∑

xi∈Kr−1

ai ≥ a1 + (r − 2)b ≥ 1 + (r − 1)b

it is seen that a − e1 belongs to the right-hand side of Eq. (13.10). Thus,if necessary, we may apply this observation again to a − e1 and so on tillwe get that ai ≤ b for all i. Hence, by Theorem 13.6.1(c), the vectorγ = 1− (a1/b, . . . , an/b) belongs to the independence polytope of G. Thuswe can write

γ = λ1w1 + · · ·+ λsws; (λi ≥ 0;∑

i λi = 1),

where w1, . . . , ws are characteristic vectors of independent sets of G. Hence

γ = λ1(1− u′1) + · · ·+ λs(1− u′s),

where u′1, . . . , u′s are characteristic vectors of vertex covers of G. For each i

we can write u′i = ui+ εi, where ui is the characteristic vector of a minimalvertex cover of G and εi ∈ {0, 1}n. Therefore

1− γ = λ1u′1 + · · ·+ λsu

′s =⇒

a = bλ1(u1, 1) + · · ·+ bλs(us, 1) + bλ1ε1 + · · ·+ bλsεs.


Thus a ∈ R+(J). If b = 0, clearly a ∈ R+(J). Hence we get equalityin Eq. (13.10), as required. The converse follows using similar arguments.To finish the proof notice that, by Lemma 13.6.2, the decomposition ofEq. (13.10) is irreducible. �

Remark 13.6.4 Normaliz [68] determines the irreducible representation ofa Rees cone. Thus Proposition 13.6.3 can be used to check whether a givengraph is perfect.

Definition 13.6.5 The clique clutter of a graph G, denoted by cl(G), isthe clutter on V (G) whose edges are the maximal cliques of G (maximalwith respect to inclusion).

Definition 13.6.6 The incidence matrix A of a clutter is called perfect ifthe polytope defined by the system x ≥ 0; xA ≤ 1 is integral.

Theorem 13.6.7 ([24], [373, Corollary 83.1a(vii)]) Let C be a clutter andlet A be its incidence matrix. Then C is balanced if and only if every sub-matrix of A is perfect.

The vertex-clique matrix of a graph G is the incidence matrix of cl(G),the clique clutter of G.

Theorem 13.6.8 ([296], [86]) Let A be the incidence matrix of a clutter.Then the following are equivalent:

(a) The system x ≥ 0; xA ≤ 1 is TDI.

(b) A is perfect.

(c) A is the vertex-clique matrix of a perfect graph.

Let v1, . . . , vq be a set of points in Nn and let P = conv(v1, . . . , vq). TheEhrhart ring of the lattice polytope P is the K-subring of R[t] given by

A(P) = K[{xatb| a ∈ bP ∩ Zn}].

Corollary 13.6.9 Let A be a perfect matrix with column vectors v1, . . . , vq.If there is x0 ∈ Rn such that all the entries of x0 are positive and 〈vi, x0〉 = 1for all i, then A(P) = K[xv1t, . . . , xvq t].

Proof. The inclusion “⊃” is clear. To show the other inclusion take xatb

in A(P). Then we can write (a, b) =∑qi=1 λi(vi, 1), where λi ≥ 0 for all i.

Hence 〈a, x0〉 = b. By Theorem 13.6.8 the system x ≥ 0; xA ≤ 1 is TDI.Therefore, applying Proposition 1.3.28, we have:

(a, b) = η1(v1, 1) + · · ·+ ηq(vq, 1)− δ1e1 − · · · − δnen (ηi ∈ N; δi ∈ N).

Then b = 〈a, x0〉 = b − δ1〈x0, e1〉 − · · · − δn〈x0, en〉. Using that 〈x0, ei〉 > 0for all i, we conclude that δi = 0 for all i, i.e., xatb ∈ K[xv1t, . . . , xvq t]. �

564 Chapter 13

Exercises

13.6.10 Let A be an n × q integer matrix with column vectors v1, . . . , vq.If the polyhedron {x|x ≥ 0; xA ≤ 1} is integral and R+B ∩ Zn+1 = NB,where B = {(vi, 1)}qi=1 ∪ {−ei}ni=1, then the system x ≥ 0; xA ≤ 1 is TDI.

13.6.11 Let G be a graph and let G be its complement. Then:

(a) Ic(G) = ({xa|X \ supp(xa) is a maximal clique of G}).(b) If G is perfect, then R+(Ic(G)) is equal to{

(ai) ∈ Rn+1|∑xi∈S ai ≥ (|S| − 1)an+1; ∀S independent set of G

}.

13.6.12 Let G be a perfect graph with vertex set X and let β0 be its vertexindependence number. Then there is a partition X1, . . . , Xβ0 of X suchthat Xi is a clique of G for all i. If G is unmixed, then cl(G)∨ has a perfectmatching, where G is the complement of G.

13.7 Perfect graphs and algebras of covers

Let G be a graph with vertex set X = {x1, . . . , xn} and let I = I(G) beits edge ideal. The main purpose of this section is to study the symbolicRees algebra of I and the Simis cone of I when G is a perfect graph, i.e.,we study the algebra of vertex covers of G∨. We show that the cliques ofa perfect graph G completely determine both the Hilbert basis of the Simiscone and the symbolic Rees algebra of I(G).

The Simis cone of I is denoted by Cn(I) (see Definition 13.2.1). LetC1, . . . , Cs be the minimal vertex covers of G. For 1 ≤ k ≤ s let uk be thecharacteristic vector of Ck, i.e., uk =

∑xi∈Ck

ei.If H is the minimal integral Hilbert basis of Cn(I), then Rs(I(G)) is

equal to K[NH], the semigroup ring of NH (see Corollary 13.2.19). Nextwe describe H when G is perfect.

Theorem 13.7.1 Let ω1, . . . , ωp be the characteristic vectors of the non-empty cliques of a perfect graph G. If

H = {(ω1, |ω1| − 1), . . . , (ωp, |ωp| − 1)}.

Then NH = Cn(I) ∩ Zn+1, that is, H is the integral Hilbert basis of Cn(I).

Proof. The inclusion NH ⊂ Cn(I)∩Zn+1 is clear because any clique of sizer intersects any minimal vertex cover in at least r− 1 vertices. Let us showthe reverse inclusion. Let (a, b) be a minimal generator of Cn(I) ∩ Zn+1,where 0 �= a = (ai) ∈ Nn and b ∈ N. Then∑

xi∈Ckai = 〈a, uk〉 ≥ b, (13.11)


for all k. If b = 0 or b = 1, then (a, b) = ei for some i ≤ n or (a, b) =(ei + ej , 1) for some edge {xi, xj}, respectively. In both cases (a, b) ∈ H.Thus we may assume that b ≥ 2 and aj ≥ 1 for some j. Using Eq. (13.11)we obtain∑

xi∈Ck

ai +∑

xi∈X\Ck

ai = |a| ≥ b+∑

xi∈X\Ck

ai = b+ 〈1− uk, a〉, (13.12)

for all k. Set c = |a| − b. Notice that c ≥ 1 because a �= 0. Indeed ifc = 0, from Eq. (13.12) we get

∑xi∈X\Ck

ai = 0 for all k, i.e., a = 0, a

contradiction. Consider the vertex-clique matrix of G′:

A′ = (1− u1 · · ·1− us) ,

where 1− u1, . . . ,1− us are regarded as column vectors. From Eq. (13.12)we get (a/c)A′ ≤ 1. Hence by Theorem 13.6.1(c) we obtain that a/c belongsto conv(ω0, ω1, . . . , ωp), where ω0 = 0, i.e., we can write a/c =

∑pi=0 λiωi,

where λi ≥ 0 for all i and∑

i λi = 1. Thus we can write

(a, c) = cλ0(ω0, 1) + · · ·+ cλp(ωp, 1).

Using Theorem 13.6.8(a) it follows that the subring K[{xωit|0 ≤ i ≤ p}] isnormal. Hence there are η0, . . . , ηp in N such that

(a, c) = η0(ω0, 1) + · · ·+ ηp(ωp, 1).

Thus |a| = η0|ω0|+ · · ·+ηp|ωp| and c = η0+ · · ·+ηp = |a|− b, consequently:

(a, b) = η0(ω0, |ω0| − 1) + η1(ω1, |ω1| − 1) + · · ·+ ηp(ωp, |ωp| − 1).

Notice that there is u� such that 〈a, u�〉 = b; otherwise since aj ≥ 1, byEq. (13.11) the vector (a, b) − ej would be in Cn(I) ∩ Zn+1, contradictingthe minimality of (a, b). Therefore from the equality

0 = 〈(a, b), (u�,−1)〉 = η0 +∑p

i=1 ηi〈(ωi, |ωi| − 1), (u�,−1)〉

we conclude that η0 = 0, i.e., (a, b) ∈ NH, as required. �

Corollary 13.7.2 If G is a perfect graph, then

Rs(I(G)) = K[xatr|xa is square-free ; 〈supp(xa)〉 = Kr+1; 0 ≤ r < n].

Proof. By Theorem 13.2.3, we have the equality Rs(I(G)) = K[NH], thusthe formula follows from Theorem 13.7.1. �

Lemma 13.7.3 Let G be a graph and let 0 �= a = (ai) ∈ Nn. If ai ∈ {0, 1}for all i and G[{xi| ai > 0}] = Kb+1, then a is an irreducible b-cover of G∨.

566 Chapter 13

Proof. By Lemma 13.6.2, the closed halfspace H+(a,−b) must occur in the

irreducible representation of R+(Ic(G)). Hence a is an irreducible b-coverof G∨ by Lemma 13.2.27. �

Corollary 13.7.4 [423] If G is a graph, then

K[xatr|xa square-free ; 〈supp(xa)〉 = Kr+1; 0 ≤ r < n] ⊂ Rs(I(G))

with equality if and only if G is a perfect graph.

Proof. The inclusion follows from Lemma 13.7.3. If G is a perfect graph,then by Corollary 13.7.2 the equality holds. Conversely if the equality holds,then by Lemmas 13.2.27 and 13.6.2 we have

R+(Ic(G)) ={(ai) ∈ Rn+1|

∑xi∈Kr

ai ≥ (r − 1)an+1; ∀Kr ⊂ G}.

Hence an application of Proposition 13.6.3 gives that G is perfect. �

Exercises

13.7.5 Let G be a graph with clique number ω(G) and chromatic numberχ(G). Notice that the chromatic number is given by

χ(G) = min{k| ∃X1, . . . , Xk independent sets of G whose union is X}.

Prove that ω(G) ≤ χ(G).

13.7.6 If G is a perfect graph and C = G∨, show that the image of ψ givenin Remark 13.2.28 generates Rs(Ic(G

∨)).

Hint Use that the irreducible b-covers of G∨ correspond to cliques of thegraph G (see Corollary 13.7.4). Notice that Ic(G

∨) is equal to I(G).

Chapter 14

CombinatorialOptimization and BlowupAlgebras

In this chapter we relate commutative algebra and blowup algebras withcombinatorial optimization to gain insight on these research areas. A maingoal here is to connect algebraic properties of blowup algebras associated toedge ideals with combinatorial and optimization properties of clutters andpolyhedra. A conjecture of Conforti and Cornuejols about packing problemsis examined from an algebraic point of view. We study max-flow min-cutproblems of clutters, packing problems, and integer rounding properties ofsystems of linear inequalities—and their underlying polyhedra—to analyzealgebraic properties of blowup algebras and edge ideals (e.g., normality,normally torsion freeness, Cohen–Macaulayness, unmixedness). Systemswith integer rounding properties and clutters with the max-flow min-cutproperty come from linear optimization problems [372, 373].

The study of algebraic and combinatorial properties of edge ideal ofclutters and hypergraphs is of current interest; see [119, 155, 206, 285, 326]and the references therein. A comprehensive reference for combinatorialoptimization and hypergraph theory is the 3-volume book of Schrijver [373].For a thorough study of clutters—that includes 18 conjectures in the area—from the point of view of combinatorial optimization, see [93].

In this chapter we make use of polyhedral geometry and combinatorialoptimization to study blowup algebras and vice versa. We refer the readerto Chapter 1 for the undefined terminology and notation regarding theseareas. As a handy reference in Section 14.1 we introduce and fix some ofthe notation and definitions that will be used throughout this chapter.

568 Chapter 14

14.1 Blowup algebras of edge ideals

In this section we introduce some of the notation and definitions that will beused throughout this chapter. In particular the irreducible representationof a Rees cone is introduced here.

Let C be a clutter with vertex set X = {x1, . . . , xn} and let R = K[X ]be a polynomial ring over a field K. The set of vertices and edges of C aredenoted by V (C) and E(C), respectively. Let f1, . . . , fq be the edges of Cand let vk =

∑xi∈fk ei be the characteristic vector of fk for 1 ≤ k ≤ q.

Recall that the edge ideal of C, denoted by I = I(C), is the ideal of Rgenerated by all monomials xe =

∏xi∈e xi such that e is an edge of C.

Thus, I is minimally generated by F = {xv1 , . . . , xvq}. In what follows weshall always assume that the height of I is at least 2.

The blowup algebras and monomial algebras studied in this chapter arethe following: (a) Rees algebra

R[It] := R ⊕ It⊕ · · · ⊕ Iiti ⊕ · · · ⊂ R[t],

where t is a new variable, (b) extended Rees algebra

R[It, t−1] := R[It][t−1] ⊂ R[t, t−1],

(c) symbolic Rees algebra

Rs(I) := R+ I(1)t+ I(2)t2 + · · ·+ I(i)ti + · · · ⊂ R[t],

where I(i) is the ith symbolic power of I, (d) associated graded ring

grI(R) := R/I ⊕ I/I2 ⊕ · · · ⊕ Ii/Ii+1 ⊕ · · · � R[It]⊗R (R/I),

with multiplication

(a+ Ii+1)(b + Ij+1) = ab+ Ii+j+1 (a ∈ Ii, b ∈ Ij),

(e) monomial subring

K[F ] = K[xv1 , . . . , xvq ] ⊂ R

spanned by F = {xv1 , . . . , xvq}, (f) homogeneous monomial subring

K[Ft] = K[xv1t, . . . , xvq t] ⊂ R[t]

spanned by Ft, (g) homogeneous monomial subring

K[Ft ∪ {t}] = K[xv1t, . . . , xvq t, t] ⊂ R[t]

spanned by Ft ∪ {t}, (h) homogeneous monomial subring

S = K[xw1t, . . . , xwr t] ⊂ R[t],

Combinatorial Optimization and Blowup Algebras 569

where {w1, . . . , wr} is the set of all α ∈ Nn such that 0 ≤ α ≤ vi and weallow xvi to be an arbitrary monomial, (i) Ehrhart ring

A(P) = K[{xati| a ∈ Zn ∩ iP ; i ∈ N}] ⊂ R[t]

of the lattice polytope P = conv(v1, . . . , vq), and (j) Stanley–Reisner ring

K[ΔC ] = R/I(C).

of the independence complex ΔC . This ring is also called the edge ring of C.The incidence matrix of C will be denoted by A; it is the n× q matrix

with column vectors v1, . . . , vq. We shall always assume that the rows andcolumns of A are different from zero. Consider the set

A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)} ⊂ Nn+1,

where ei is the ith unit vector. The Rees cone of I, will be denoted by R+(I),it is the rational polyhedral cone generated by A′. The closed halfspace H+

eioccurs in the irreducible representation of R+(I) for i = 1, . . . , n + 1 (seeExercise 14.2.34). Thus, by Theorem 1.4.2, there is a unique irreduciblerepresentation

R+(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

α1∩ · · · ∩H+

αr

such that 0 �= αi ∈ Qn+1, 〈αi, en+1〉 = −1 for all i, and none of the closedhalfspaces can be omitted from the intersection.

Let p1, . . . , ps be the minimal primes of the edge ideal I = I(C) and let

Ck = {xi|xi ∈ pk} (k = 1, . . . , s)

be the corresponding minimal vertex covers of C (see Proposition 13.1.2).The set of column vectors of A will be denoted by A = {v1, . . . , vq} andQ(A) will denote the set covering polyhedron

Q(A) := {x|x ≥ 0;xA ≥ 1}.

Notation Let dk be the unique positive integer such that dkαk has relativelyprime integral entries. We set �k = dkαk for k = 1, . . . , r.

Theorem 14.1.1 The irreducible representation of the Rees cone is:

R+(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

�1∩ · · · ∩H+

�r(14.1)

where �k = −en+1 +∑xi∈Ck

ei for 1 ≤ k ≤ s. Moreover all the vertices ofQ(A) are integral if and only if r = s.

570 Chapter 14

Proof. By Proposition 13.1.2 uk =∑xi∈Ck

ei is a vertex of Q(A). Since�k+ en+1 = (uk, 0) for k = 1, . . . , s, by Theorem 1.4.3, the result follows. �

Notation In the sequel we shall always assume that u1, . . . , us are the vectorsgiven by uk =

∑xi∈Ck

ei, �1, . . . , �s are the vectors defined as

�k = −en+1 +∑xi∈Ck

ei,

and that �1, . . . , �r are the vectors of Eq. (14.1). Notice that dk = 1 fork = 1, . . . , s and dk = −〈�k, en+1〉 for 1 ≤ k ≤ r.

The Simis cone of I is the polyhedral cone:

Cn(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

�1∩ · · · ∩H+

�s,

see Definition 13.2.1. The Simis and Rees cones are related to the normalityand torsion freeness of I (Theorem 14.2.2 and Proposition 14.2.3).

14.2 Rees algebras and polyhedral geometry

In this section we describe when the integral closure of a Rees algebra ofan edge ideal is equal to the symbolic Rees algebra in combinatorial andalgebraic terms. If the Rees algebra is normal, we describe the primarydecomposition of the zero ideal of the associated graded ring. We show thatcertain properties of clutters and edge ideals are closed under taking minorsand Alexander duals.

Lemma 14.2.1 If 1 ≤ j ≤ r and 〈�j , en+1〉 = −1, then 1 ≤ j ≤ s. Inparticular r = s if and only if 〈�j, en+1〉 = −1 for i = 1, . . . , r

Proof. Let �j = (a1, . . . , an,−1). Since R+(I) ⊂ H+�j

we get ai ≥ 0 fori = 1, . . . , n. Consider the ideal p of R generated by the set of xi suchthat ai > 0. Note that I ⊂ p because 〈�j , (vi, 1)〉 ≥ 0 for all 1 ≤ i ≤ n.For simplicity of notation assume that �j = (a1, . . . , am, 0, . . . , 0,−1), whereai > 0 for all i. Take an arbitrary vector v ∈ A′ such that v ∈ H�j , then vsatisfies the equation

a1y1 + · · ·+ amym = yn+1.

Observe that v also satisfies the equation y1 + · · · + ym = yn+1. Usingthat H�j contains n linearly independent vectors from A′ we conclude thatH�j = Hb, where b = e1 + · · ·+ em − en+1. Hence �j = b and consequentlyai = 1 for all i. It remains to show that p is a minimal prime of I, and thisfollows from the irreducibility of Eq. (14.1). �


Theorem 14.2.2 R[It] = Rs(I) ⇐⇒ 〈�i, en+1〉 = −1 for i = 1, . . . , r.

Proof. The equality R[It] = Rs(I) holds if and only if Zn+1 ∩ R+(I) isequal to Zn+1 ∩ Cn(I) if and only if the Rees and Simis cones are equal.By the irreducibility of the representations of both cones it follows thatR[It] = Rs(I) if and only if r = s. Now, to finish the proof recall that byLemma 14.2.1, r = s if and only if 〈�i, en+1〉 = −1 for i = 1, . . . , r. �

Proposition 14.2.3 If I is a monomial ideal, then R[It] is normal if andonly if any of the following two equivalent conditions hold:

(a) A′ is a Hilbert basis, i.e., NA′ = Zn+1 ∩ R+(I).

(b) I is normal, i.e., Ii = Ii for all i ≥ 1.

Proof. The Rees algebra of I can be written as

R[It] = K[{x1, . . . , xn, xv1t, . . . , xvq t}]= K[{xatb| (a, b) ∈ NA′}]= R⊕ It⊕ I2t2 ⊕ · · · ⊕ Iiti ⊕ · · · .

By Theorem 9.1.1 and Exercise 4.3.49, the integral closure of R[It] in itsfield of fractions can be expressed as

R[It] = K[{xatb| (a, b) ∈ Zn+1 ∩R+(I)}]= R⊕ It⊕ I2t2 ⊕ · · · ⊕ Iiti ⊕ · · · ,

where Ii is the integral closure of Ii. Thus, R[It] is normal if and only if

A′ is a Hilbert basis if and only if Ii = Ii for all i ≥ 1. �

Let p be a prime ideal of R. In what follows we denote the localizationof R[It] at the multiplicative set R \ p by R[It]p

Lemma 14.2.4 If p ∈ Spec(R), then pR[It]p ∩R = p.


Lemma 14.2.5 If B = R[It] and p is a minimal prime of I, then pBp∩Bis a minimal prime of IB.

Proof. From the equality pBp = p(Rp[pRpt]) = pRp + p2Rpt + · · · we getthat pBp is a prime ideal. Hence its contraction pBp∩B is also a prime ideal.To prove the minimality take P ∈ Spec(B) such that IB ⊂ P ⊂ pBp ∩ B.Contracting to R and using Lemma 14.2.4 we get P ∩ R = p. From thisequality it follows that PBp ∩B = P . Therefore

pBp ∩B ⊂ PBp ∩B = P =⇒ P = pBp ∩B. �

572 Chapter 14

Notation In what follows J(dk)k will denote the ideal of R[It] given by

J(dk)k = ({xatb| 〈(a, b), �k〉 ≥ dk}) ∩R[It] (k = 1, . . . , r)

and Jk will denote the ideal of R[It] given by

Jk = ({xatb| 〈(a, b), �k〉 > 0}) ∩R[It] (k = 1, . . . , r),

where dk = −〈�k, en+1〉. In general the ideal J(dk)k might not be equal to

the dk-th symbolic power of Jk; see [415]. If dk = 1 we have J(1)k = Jk. By

Lemma 14.2.1 we have that dk = 1 if and only if 1 ≤ k ≤ s.

Proposition 14.2.6 J1, . . . , Jr are distinct height one prime ideals of R[It]that contain IR[It] and Jk is equal to pkR[It]pk

∩R[It] for k = 1, . . . , s. IfQ(A) is integral, then

rad(IR[It]) = J1 ∩ · · · ∩ Js.

Proof. IR[It] is clearly contained in Jk for all k by definition of Jk. Toshow that Jk is a prime ideal of height one it suffices to notice that theright-hand side of the isomorphism:

R[It]/Jk � K[{xatb ∈ R[It]| 〈(a, b), �k〉 = 0}]

is an n-dimensional integral domain, because Fk = R+(I)∩H�k is a facet ofthe Rees cone for all k. The prime ideals J1, . . . , Jr are distinct because thedimension of R+(I) is n + 1 and F1, . . . , Fr are distinct facets of the Reescone. Set Pk = pkR[It]pk

∩R[It] for 1 ≤ k ≤ s. By Lemma 14.2.5 this idealis a minimal prime of IR[It] and admits the following description

Pk = pkRpk[pkRpk

t] ∩R[It]= pk + (p2k ∩ I)t+ (p3k ∩ I2)t2 + · · ·+ (pi+1

k ∩ Ii)ti + · · ·

Notice that xa ∈ pb+1k if and only if 〈a,

∑xi∈Ck

ei〉 ≥ b+1. Hence Jk = Pk.

Assume that Q(A) is integral, i.e., r = s. Take xαtb ∈ Jk for all k. UsingEq. (14.1) it is not hard to see that (α, b + 1) ∈ R+(I); that is, xαtb+1 is

in R[It] and xαtb+1 ∈ Ib+1tb+1. It follows that xαtb is a monomial in theradical of IR[It]. This proves the asserted equality. �

The following formulas, pointed out to us by Vasconcelos, show thedifference between the symbolic Rees algebra and the normalization of I.

Rs(I) =

s⋂k=1

R[It]pk∩R[t]; R[It] =

r⋂k=1

R[It]pk∩R[t],

where pk = Jk ∩R for k = 1, . . . , r. These representations are related to theso-called Rees valuations of I; see [414, Chapter 8].


Theorem 14.2.7 [185, 188] The following are equivalent:

(a) Q(A) is integral.

(b) R+(I) = H+e1 ∩ · · · ∩H+

en+1∩H+

�1∩ · · · ∩H+

�s, i.e., r = s.

(c) Rs(I) = R[It].

(d) The minimal primes of IR[It] are of the form pR[It]p ∩ R[It], wherep is a minimal prime of I.

Proof. (a) ⇔ (b): It follows at once from Theorem 14.1.1. (b) ⇔ (c):It follows from Theorem 14.2.2 and Lemma 14.2.1. (a) ⇒ (d): It followsfrom Proposition 14.2.6. (d)⇒ (b): From Proposition 14.2.6, J1, . . . , Jr areminimal primes of IR[It] and Jk is equal to pkR[It]pk

∩R[It] for k = 1, . . . , s.Thus r = s. �

Corollary 14.2.8 If Q(A) has only integral vertices and C has n vertices,then α0(Ca) ≤ n− 1 for all indecomposable parallelizations Ca of C.

Proof. By Theorem 14.2.7 we have the equality R[It] = Rs(I). Take anyindecomposable parallelization Ca of C and consider the monomialm = xatb,where b = α0(Ca). By Theorem 13.2.15 m is a minimal generator of Rs(I).Now, according to Proposition 12.5.1, a minimal generator of R[It] hasdegree in t at most n− 1, i.e., b ≤ n− 1. �

Example 14.2.9 Let I = I(C) = (x1x2x5, x1x3x4, x2x3x6, x4x5x6) be theedge ideal associated to the clutter:

�

�

� ��

��

��

� �

��

��

��

� ��

��

��

��

��

��

� � ��

� ��

� ��

x1 x2

x3

x5

x6

x4

C

This clutter, usually denoted byQ6 in the literature, plays an important rolein combinatorial optimization [93]. Using Normaliz [68], Theorems 1.4.9,14.2.7, and Remark 13.2.23, we obtain that Q(A) is integral,

R[It] � Rs(I) = R[It] = R[It][x1 · · ·x6t2],

574 Chapter 14

R[It] is not normal, the minimal primes of I are

p1 = (x1, x6), p2 = (x2, x4), p3 = (x3, x5),p4 = (x1, x2, x5), p5 = (x1, x3, x4), p6 = (x2, x3, x6), p7 = (x4, x5, x6),

and C∨, the blocker of C, has the max-flow min-cut property.

The dual hypergraph H� of a hypergraph H = (V,E) is the hypergraphwith vertex set E and edges all sets {e ∈ E| v ∈ e} for v ∈ V . So theincidence matrix of H� is the transpose of the incidence matrix of H. Theclutter Q6 is the dual hypergraph of K6, the complete graph on 6 vertices.

Recall that the analytic spread of I, denoted by �(I), is given by

�(I) = dimR[It]/mR[It].

This number satisfies ht(I) ≤ �(I) ≤ dimR [412, Corollary 5.1.4]. Theanalytic spread of a monomial ideal I can be computed in terms of theNewton polyhedron of I [42].

Definition 14.2.10 The ring F(I) = R[It]/mR[It] is called the specialfiber of I.

Corollary 14.2.11 If Q(A) is integral, then �(I) < n.

Proof. By Theorem 14.2.7 we have r = s. If �(I) = n, then the heightof mR[It] is equal to 1. Hence there is a height one prime ideal P of R[It]such that IR[It] ⊂ mR[It] ⊂ P . By Proposition 14.2.6 the ideal P has theform pkR[It]pk

∩R[It], this readily yields a contradiction. �

This corollary also follows directly from [312, Theorem 3].

Proposition 14.2.12 If C is a uniform clutter, then �(I) = rank(A).

Proof. Since deg(xvi ) = d for all i. There are isomorphisms

R[It]/mR[It] � K[xv1t, . . . , xvq t] � K[xv1 , . . . , xvq ].

Thus, by Corollary 8.2.21, we get that �(I) is the rank of A. �

Theorem 14.2.13 infi{depth(R/Ii)} ≤ dim(R)− �(I) with equality if theassociated graded ring grI(R) is Cohen–Macaulay.

The inequality is due to Burch [76]. If grI(R) is C–M, the equalitycomes from [131]. By a result of Brodmann [55], depthR/Ik is constantfor k 0. Brodmann improved the Burch inequality by showing that theconstant value is bounded by dim(R)− �(I). For a study of the initial andlimit behavior of the numerical function f(k) = depthR/Ik, see [223].


Lemma 14.2.14 If xv1 = x1xv′1 , . . . , xvp = x1x

v′p and x1 /∈ supp(xvi) for

i > p, where x1 is a variable in Ck for some 1 ≤ k ≤ s, then there is xv′j

such that supp(xv′j ) ∩ Ck = ∅.

Proof. If supp(xv′j ) ∩ Ck �= ∅ for all j, then Ck \ {x1} is a vertex cover of

C, a contradiction because Ck is a minimal vertex cover. �

Proposition 14.2.15 If 1 ≤ i ≤ s, then Ji is the Ji-primary component ofIR[It].

Proof. We set P = Ji. It suffices to show the equality R[It]∩IR[It]P = P .In general the left-hand side is contained in the right-hand side. To showthe reverse inclusion we may assume that P can be written as

P = (x1, . . . , xm, xv1t, . . . , xvp t)R[It],

where p = (x1, . . . , xm) is a minimal prime of I. Set C = {x1, . . . , xm}.Case (I): Consider xk with 1 ≤ k ≤ m. By Lemma 14.2.14 there is j

such that xvj = xkxα and supp(xα) ∩ C = ∅. Thus since xα is not in P

(because of the second condition) we obtain xk ∈ R[It] ∩ IR[It]P .Case (II): Now, consider xvk t with 1 ≤ k ≤ p. Since

〈(vk, 1), e1 + · · ·+ em − en+1〉 ≥ 1,

the monomial xvk contains at least two variables in C. Thus we may assumethat x1, x2 are in the support of xvk . Again by Lemma 14.2.14 there are j, �such that xvj = x1x

α, xv� = x2xγ , and the support of xα and xγ disjoint

from C. Hence the monomial xvkxα+γt belongs to I2t and xα+γ is not inP . Writing xvk t = (xvkxα+γ t)/xα+γ , we get xvk t ∈ R[It] ∩ IR[It]P . �

Lemma 14.2.16 rad(J(dk)k ) = Jk for 1 ≤ k ≤ r.

Proof. By construction one has rad(J(dk)k ) ⊂ Jk. The reverse inclusion

follows by noticing that if xatb ∈ Jk, then (xatb)dk ∈ J (dk)k . �

Theorem 14.2.17 [258, Theorem 1.11] If the height of I ≥ 2, then thefollowing are equivalent:

(i) grI(R) is torsion-free over R/I.

(ii) grI(R) is reduced.

(iii) R[It] is normal and Cl(R[It]), the divisor class group of R[It], is afree abelian group whose rank is the number of minimal primes of I.

Proposition 14.2.18 ([70], [188]) If R[It] is normal, then

IR[It] = J(d1)1 ∩ J (d2)

2 ∩ · · · ∩ J (dr)r .

576 Chapter 14

Proof. “⊂”: Assume xαtb ∈ IR[It]. Since xα ∈ Ib+1 we readily obtain(α, b + 1) ∈ NA′. In particular we get (α, b + 1) ∈ R+(I). Therefore

0 ≤ 〈(α, b + 1), �k〉 = 〈(α, b), �k〉 − dk

and consequently xαtb ∈ J (dk) for 1 ≤ k ≤ r.“⊃”: Assume xαtb ∈ J (dk) for all k. Since the element (α, b + 1) is in

R+(I) ∩ Zn+1 and using that R[It] is normal yields (α, b + 1) ∈ NA′. Itfollows that xαtb ∈ Ib+1tb ⊂ IR[It]. �

Since the programNormaliz [68] computes the irreducible representationof the Rees cone and the integral closure of R[It], the following result is aneffective criterion for the reducedness of the associated graded ring.

Theorem 14.2.19 ([258], [147]) The following are equivalent :

(a) R[It] = Rs(I).

(b) grI(R) is reduced.

(c) R[It] is normal and 〈�i, en+1〉 = −1 for i = 1, . . . , r.

Proof. (a) ⇒ (b) By Corollary 4.3.26 the Rees algebra R[It] is normal.Thus using Theorem 14.2.7 we obtain that r = s and that the minimalprimes of IR[It] are J1, . . . , Js. Hence from Proposition 14.2.18 we have

IR[It] = J(d1)1 ∩ J (d2)

2 ∩ · · · ∩ J (ds)s .

Therefore IR[It] is a radical ideal because di = 1 for i = 1, . . . , s. SincegrI(R) � R[It]/IR[It] we get that grI(R) is reduced.

(b) ⇒ (c) By Theorem 14.2.17, the ring R[It] is normal and Cl(R[It]),the divisor class group of R[It], is a free abelian group whose rank is thenumber of minimal primes of I, i.e., the rank of Cl(R[It]) is equal to s. Onthe other hand by Proposition 12.7.3, the rank of Cl(R[It]) is equal to r.Thus s = r and 〈�i, en+1〉 = −1 for i = 1, . . . , r.

(c) ⇒ (a) By Lemma 14.2.1, r = s. Hence, by Theorem 14.2.7, we getR[It] = Rs(I). Thus R[It] = Rs(I). �

Example 14.2.20 Let I = (x1x5, x2x4, x3x4x5, x1x2x3). Using Normaliz[68] with the input file:

4

5

1 0 0 0 1

0 1 0 1 0

0 0 1 1 1

1 1 1 0 0

3


we get the output file:

9 generators of integral closure of Rees algebra:

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

1 0 0 0 1 1

0 1 0 1 0 1

0 0 1 1 1 1

1 1 1 0 0 1

10 support hyperplanes:

0 0 1 1 1 -1

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 0 0 1

0 0 1 0 0 0

1 0 0 1 0 -1

0 0 0 1 0 0

0 0 0 0 1 0

0 1 0 0 1 -1

1 1 1 0 0 -1

The first block of the output file shows the list of generators of R[It]. ThusR[It] is normal. The second block gives the irreducible representation ofthe Rees cone of I. This means that all the �i’s that occur in the irreduciblerepresentation of the Rees cone R+(I) (see Eq. (14.1)) have its last entryequal to −1. Thus, by Theorem 14.2.19, the ring grI(R) is reduced.

Proposition 14.2.21 If R[It] = Rs(I), then R[Ic(C)t] = Rs(Ic(C)).

Proof. We set J = Ic(C). Using Theorem 14.2.7 one has

R+(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

�1∩ · · · ∩H+

�s,

where �i = (ui,−1) for all i. Then, by Corollary 1.1.30, we get:(n⋂i=1

H+ei

)⋂(q⋂i=1

H+(vi,1)

)=

(n+1∑i=1

R+ei

)+

(s∑i=1

R+(ui,−1)). (14.2)

By Theorem 14.2.7 we need only show the equality

R+(J) =

(n+1⋂i=1

H+ei

)⋂(q⋂i=1

H+(vi,−1)

).

578 Chapter 14

Clearly the left-hand side is contained in the right-hand side. To show thereverse inclusion take (α, b) in the right-hand side. Then

0 ≤ 〈(α, b), (vi,−1)〉 = 〈(α,−b), (vi, 1)〉

for all i. Hence using Eq. (14.2) we can write

(α,−b) =s∑i=1

λi(ui,−1) +n+1∑i=1

μiei (λi;μj ≥ 0). (14.3)

Since λ1 + · · ·+ λs − b = μn+1 ≥ 0, there are λ′1, . . . , λ′s such that

0 ≤ λ′i ≤ λi; b = λ′1 + · · ·+ λ′s; μn+1 = (λ1 − λ′1) + · · ·+ (λs − λ′s).

Therefore using Eq. (14.3) we conclude

(α, b) =

s∑i=1

λ′i(ui, 1) +n∑i=1

μiei +

s∑i=1

(λi − λ′i)ui ∈ R+(J). �

Corollary 14.2.22 [93, Theorem 1.17] If Q(A) is integral and A∨ is theincidence matrix of C∨, then Q(A∨) is integral.

Proof. It follows at once from Theorem 14.2.7 and Proposition 14.2.21. �

The following notion of minor of an edge ideal was introduced in [188].It is inspired by the notion of a minor in combinatorial optimization [93]and is consistent with the terminology of Section 6.5 (Definition 6.5.3).

Definition 14.2.23 A minor of I is an ideal (0) � I ′ � R obtained from Iby making any sequence of variables equal to 1 or 0 in F = {xv1 , . . . , xvq}.The ideal I is considered itself a minor. A minor of C is a clutter C′ thatcorresponds to a minor (0) � I ′ � R of I. The clutter C is itself a minor. IfC′ is a minor of C and E(C′) �= E(C), we call C′ a proper minor.

If I ′ ⊂ R is a minor of I, notice that C′ is obtained from I ′ by consideringG(I ′), the unique set of square-free monomials that minimally generate I ′.Indeed if G(I ′) = {xα1 , . . . , xαm}, then I ′ defines a clutter C′ (resp. matrixA′) whose vertices are the variables of R and whose edges (resp. columns)are the supports of the monomials xαi (resp. the column vectors αi). We canalso take as vertex set of C′, the set of variables that occur in xα1 , . . . , xαm .This choice of vertex set is useful in induction arguments, where one needsto reduce the number of variables.

Proposition 14.2.24 If Ii = I(i) for some i ≥ 2 and J = I ′ is a minor ofI, then J i = J (i).


Proof. Assume that J is the minor obtained from I by making x1 = 0.Take xα ∈ J (i). Then xα ∈ I(i) = Ii because J ⊂ I. Thus xα ∈ Ii. Sincex1 /∈ supp(xα) it follows that xα ∈ J i. This proves J (i) ⊂ J i. The otherinclusion is clear because J (i) is integrally closed and J i ⊂ J (i).

Assume that J is the minor obtained from I by making x1 = 1. Takexα ∈ J (i). Notice that xi1x

α ∈ I(i) = Ii. Indeed if x1 ∈ pk, then xi1 ∈ pik, and

if x1 /∈ pk, then J ⊂ pk and xα ∈ pik. Thus x

i1xα ∈ I(i). Hence (xi1xα)p ∈ Iip

for some p ∈ N+. Since x1 /∈ supp(xα) it follows that xα ∈ J i. �

Corollary 14.2.25 (a) If Rs(I) = R[It], then Rs(I′) = R[I ′t] for any

minor I ′ of I. (b) The integrality of Q(A) is preserved under taking minorsand parallelizations.

Proof. (a): It follows from Proposition 14.2.24. (b): This part follows frompart (a), Theorem 14.2.7, and [373, p. 1390]. �

Proposition 14.2.26 Let C∨ be the blocker of C. If R[It] is equal to Rs(I)and |C ∩B| ≤ 2 for C ∈ C and B ∈ C∨, then R[It] is normal.

Proof. Let xatb = xa11 · · ·xann tb ∈ R[It] be a minimal generator, that is(a, b) cannot be written as a sum of two non-zero integral vectors in theRees cone R+(I). We may assume ai ≥ 1 for 1 ≤ i ≤ m, ai = 0 for i > m,and b ≥ 1.

Case (I): 〈(a, b), �i〉 > 0 for all i. The vector γ = (a, b) − e1 satisfies〈γ, �i〉 ≥ 0 for all i, that is γ ∈ R+(I). Thus since (a, b) = e1 + γ we derivea contradiction.

Case (II): 〈(a, b), �i〉 = 0 for some i. We may assume

{�i| 〈(a, b), �i〉 = 0} = {�1, . . . , �p}.

Subcase (II.a): ei ∈ H�1 ∩ · · · ∩H�p for some 1 ≤ i ≤ m. It is not hardto verify that the vector γ = (a, b)− ei satisfies 〈γ, �k〉 ≥ 0 for all 1 ≤ k ≤ s.Thus γ ∈ R+(I), a contradiction because (a, b) = ei + γ.

Subcase (II.b): ei /∈ H�1 ∩ · · · ∩H�p for all 1 ≤ i ≤ m. Since the vector(a, b) belongs to the Rees cone it follows that we can write

(a, b) = λ1(v1, 1) + · · ·+ λq(vq, 1) (λi ≥ 0). (∗)

By the choice of xatb we may assume 0 < λ1 < 1. Set γ = (a, b) − (v1, 1)and notice that by Eq. (∗) this vector has nonnegative entries. We claimthat γ is in the Rees cone. Since by hypothesis one has that �j = uj (seeTheorem 14.2.7) and 0 ≤ 〈(v1, 1), �j〉 ≤ 1 for all j we readily obtain

〈γ, �k〉 ={〈(a, b), �k〉 − 〈(v1, 1), �k〉 = 0 if 1 ≤ k ≤ p,〈(a, b), �k〉 − 〈(v1, 1), �k〉 ≥ 0 otherwise.

Thus γ ∈ R+(I) and (a, b) = (v1, 1)+γ. As a result γ = 0 and (a, b) ∈ R[It],as desired. �

580 Chapter 14

Normality criteria and Rees cones For each proper face G of therational polyhedral cone R+A we set

AG = {vi | vi ∈ G} and FG = {xvi | vi ∈ G}.

If G is a proper face of R+A, there is a supporting hyperplane Ha in Rn

such that G = R+A ∩Ha �= ∅, R+A �⊂ Ha, and R+A ⊂ H+a . Notice that

AG = {vi ∈ A | 〈vi, a〉 = 0} �= ∅ and FG = {xvi | 〈vi, a〉 = 0} �= ∅.

Lemma 14.2.27 If NA is a normal semigroup (resp. A is a Hilbert basis),then NAG is a normal semigroup (resp. AG is a Hilbert basis) for eachproper face G of R+A.

Proof. Assume that NA is normal. Let α ∈ R+AG ∩ZAG . We may assumewithout loss of generality that 〈vi, a〉 = 0 for 1 ≤ i ≤ p and 〈vi, a〉 > 0 forp+ 1 ≤ i ≤ q. Using that α ∈ R+A ∩ ZA = NA we can write:

α = λ1v1 + · · ·+ λqvq = a1v1 + · · ·+ apvp,

where λi ∈ N for 1 ≤ i ≤ q and ai ∈ R+ for 1 ≤ i ≤ p. Taking innerproducts we get the equality:

0 = a1〈v1, a〉+ · · ·+ ap〈vp, a〉 = λp+1〈vp+1, a〉+ · · ·+ λq〈vq, a〉.

Since the λi’s are nonnegative and 〈vi, a〉 > 0 for p + 1 ≤ i ≤ q, one hasλp+1 = · · · = λq = 0. Therefore α ∈ NAG . This shows that NAG is normal.The other assertion is shown similarly. �

Lemma 14.2.28 If K[F ] is normal, then K[FG ] is normal for any properface G of R+A.

Proof. Notice K[F ] = K[NA] and K[FG ] = K[NAG ]. By Corollary 9.1.3K[F ] is normal if and only if NA is normal. Thanks to Lemma 14.2.27 thesemigroup is NAG is normal. Therefore K[FG ] is normal. �

The converse of Lemma 14.2.27 fails as the next example shows.

Example 14.2.29 If A = {(1, 1, 0), (1, 0, 1), (0, 1, 1)} and 1 = (1, 1, 1),then 1 is in R+A ∩ Z3 and 1 �∈ NA. Thus A is not a Hilbert basis. ByProposition 1.1.23 any proper face G of R+A is a cone generated by a propersubset of A. It follows that AG is a Hilbert basis.

Proposition 14.2.30 If K[NA′G ] is normal for every proper face G of the

Rees cone R+(I) and there is a fixed integer 1 ≤ i ≤ n such that the ithentry of �j is either 0 or 1 for all 1 ≤ j ≤ r, then R[It] is normal.


Proof. Recall that R[It] = K[NA′]. Let (α, b) be an element in the minimalintegral Hilbert basis of the affine semigroup NA′, where α ∈ Nn and b ∈ N.It suffices to prove that (α, b) belongs to NA′.

Case (I): Assume (α, b) ∈ rb(R+(I)). By Theorem 1.1.44(b) the relativeboundary of R+(I) is given by:

rb(R+(I)) =

(n+1⋃i=1

R+(I) ∩Hei

)∪(

r⋃i=1

R+(I) ∩H�i

).

Hence (α, b) is in some facet of the Rees cone and by hypothesis it followsreadily that (α, b) is in NA′.

Case (II): If (α, b) ∈ ri(R+(I)), then 〈(α, b), �j〉 > 0 for 1 ≤ j ≤ r and〈(α, b), ek〉 > 0 for 1 ≤ k ≤ n + 1. For simplicity of notation we mayassume i = 1. Write (α, b) = (α1, . . . , αn, b). Notice (α, b) = β + e1, whereβ = (α1 − 1, α2, . . . , αn, b). Therefore

〈β, �j〉 = 〈(α, b), �j〉 − 〈e1, �j〉 ≥ 0

for all j, that is, β ∈ NA′. Since e1 ∈ NA′, we obtain a contradiction. �

Theorem 14.2.31 If Q(A) is integral, then R[It] is normal if and only ifK[NA′

G ] is normal for any facet G of R+(I).

Proof. By Theorem 14.1.1 r = s and the vectors �1, . . . , �r have their firstn entries in {0, 1} because �i = ui for i = 1, . . . , s. Hence the result followsfrom Lemma 14.2.28 and Proposition 14.2.30. �

Exercises

14.2.32 Prove that grI(R) is reduced if and only if Cn(I) ∩ Zn+1 = NA′.

14.2.33 Consider a sequence 1 ≤ i1 < · · · < is ≤ n of integers. Prove thatthe following two conditions are equivalent:

(a) ei1 + · · ·+ eis is a vertex of Q(A).

(b) F = R+(I)∩H(a,−1) is a facet of the Rees cone for some vector a = (ai)in Zn such that {i| ai �= 0} = {i1, . . . , is}.

14.2.34 If height of I ≥ 2, prove the following assertions:

(a) dim(R+(I)) = n+ 1.

(b) Hei ∩ R+(I) is a facet of R+(I) for i = 1, . . . , n+ 1.

(c) H+e1 , . . . , H

+en+1

occur in the irreducible representation of R+(I).

582 Chapter 14

14.2.35 Let A be a matrix with entries in Nn and let A′ be the matrixobtained from A by adjoining a row of 1’s. Prove that the vertices of Q(A)and Q(A′) are related by

vert(Q(A′)) = {en+1} ∪ {(β, 0)|β ∈ vert(Q(A))}.

In particular Q(A) is integral if and only if Q(A′) is integral.

14.2.36 Let C be a uniform clutter with n vertices and let A be its incidencematrix. If A has rank n and I = I(C), then grI(R) is not reduced.

14.2.37 Let I = (xv1 , . . . , xv4), where v1, . . . , v4 denote the column vectorsof the matrix:

A =

⎛⎜⎜⎝2 1 2 19 6 8 40 0 1 11 6 1 2

⎞⎟⎟⎠ .

Prove that R[It] is not normal and det(A) = ±1.

14.2.38 Let L1, L2 be monomial ideals with disjoint sets of variables. IfL1, L2 are generated by monomials of degrees d1 and d2, respectively, then�(L1 + L2) = �(L1) + �(L2), where �(Li) is the analytic spread of Li.

14.2.39 Let I = I(Q6) = (x1x2x5, x1x3x4, x2x3x6, x4x5x6) be the edgeideal of the clutter Q6. Use Macaulay2 to show the equality

I2 = (I2, x1x2 · · ·x6).

Prove that I3/I I2 has four generators and prove that I(i) = II(i−1) fori ≥ �(I) = 4, where �(I) is the analytic spread of I.

14.2.40 Let I = (x6x7x8, x2x4x7, x1x2x6, x1x3x8, x1x3x5) be the edge idealof a clutter C. Prove that R[It] � R[It] = Rs(I), ht(I) = 3 and C satisfiesthe Konig property.

14.2.41 Let I = (xv1 , xv2 , xv3 , xv4 , xv5), where the vi’s are:

v1 = (1, 1, 1, 1, 0, 0, 0, 0, 0, 0), v2 = (1, 0, 0, 0, 1, 1, 1, 0, 0, 0),

v3 = (0, 1, 0, 0, 1, 0, 0, 1, 1, 0), v4 = (0, 0, 1, 0, 0, 1, 0, 1, 0, 1),

v5 = (0, 0, 0, 1, 0, 0, 1, 0, 1, 1).

Prove that the Rees algebra R[It] is not normal, whereas for every properface G of the Rees cone K[NA′

G ] is normal. See Proposition 14.2.30.


14.2.42 Consider the matrix A whose transpose is:

A� =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 1 0 0 1 0 01 0 0 0 1 0 0 1 01 0 0 0 0 1 0 0 10 1 0 1 0 0 0 0 10 1 0 0 1 0 0 1 00 1 0 0 0 1 1 0 00 0 1 0 0 1 1 0 00 0 1 0 1 0 0 1 00 0 1 1 0 0 0 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦This is a doubly stochastic singular matrix of rank 6. Prove that Q(A) isintegral, R[It] is not normal, A(P ) = K[Ft] and K[Ft] �= K[Ft].

14.3 Packing problems and blowup algebras

In this section, we present several characterizations of the max-flow min-cut property of a clutter in algebraic and combinatorial terms and provethat certain properties are closed under taking minors and parallelizations.We study packing problems and analyze the Conforti–Cornuejols conjectureabout the packing property [90].

The following basic result about the nilpotent elements of the associatedgraded ring holds for arbitrary monomial ideals.

Lemma 14.3.1 If I is a monomial ideal of R, then the nilradical of theassociated graded ring of I is given by

nil(grI(R)) = ({g ∈ Ii/Ii+1| gs ∈ Isi+1; i ≥ 0; s ≥ 1})= ({xα ∈ Ii/Ii+1|xsα ∈ Isi+1; i ≥ 0; s ≥ 1}).

Proof. The first equality holds because the radical of any graded algebrais generated by homogeneous elements.

To prove the second equality take an equivalence class 0 �= g ∈ Ii/Ii+1

such that gs ∈ Iis+1. There are non-zero scalars λ1, . . . , λr in the field Kand monomials m1, . . . ,mr in Ii such that

g = λ1m1 + · · ·+ λrmr

and m1 ≺ · · · ≺ mr, where ≺ is the lexicographical ordering. We mayassume that mj /∈ Ii+1 for all j. Since ms

r is the leading term of gs andsince Iis+1 is a monomial ideal, we get ms

r ∈ Iis+1. Thus mr is in thenilradical of grI(R). By induction it follows that mj is in nil(grI(R)) for allj, as required. �

584 Chapter 14

Proposition 14.3.2 If grI(R) is reduced (resp. R[It] is normal) and I ′ isa minor of I, then grI′(R) is reduced (resp. R[I ′t] is normal).

Proof. Assume that grI(R) is reduced. Notice that we need only show thatgrI′(R) is reduced when I ′ is a minor obtained from I by making x1 = 0or x1 = 1. Using Lemma 14.3.1 both cases are quite easy to prove. Weleave the details as an exercise. If R[It] is normal, then R[I ′t] is normal byProposition 12.2.3. �

Theorem 14.3.3 Let C be a clutter and let C′ be a parallelization of C. IfI(C) is normal, then I(C′) is normal.

Proof. By Proposition 14.3.2 normality is closed under taking minors.Thus we need only show that the normality of I = I(C) is preserved whenwe duplicate a vertex of C. Let V (C) = {x2, . . . , xn} be the vertex set of Cand let C′ be the duplication of the vertex x2. We denote the duplicationof x2 by x1. We may assume that

I = I(C) = (x2xw1 , . . . , x2x

wr , xwr+1 , . . . , xwq ),

where xwi ∈ K[x3, . . . , xn] for all i. We must show that the ideal

I(C′) = I + (x1xw1 , . . . , x1x

wr )

is normal. Consider the sets B0 = {(0, 1, w1, 1), . . . , (0, 1, wr, 1)},

B = {e2, . . . , en} ∪ B0 ∪ {(0, 0, wr+1, 1), . . . , (0, 0, wq, 1)},B′ = B ∪ {e1, (1, 0, w1, 1), . . . , (1, 0, wr, 1)}.

As I is normal, by Proposition 14.2.3, we have Zn+1 ∩ R+B = NB andwe need only show the equality Zn+1 ∩ R+(I) = NB′. It suffices to showthe inclusion “⊂”. Take an integral vector w = (a, b, c, d) in R+(I), wherea, b, d ∈ Z and c ∈ Zn−2. Then

w = (a, b, c, d) =

r∑i=1

αi(0, 1, wi, 1) +

q∑i=r+1

αi(0, 0, wi, 1)

+

r∑i=1

βi(1, 0, wi, 1) +

n∑i=1

γiei

for some αi, βi, γi in R+. Comparing entries it follows that

(0, a+ b, c, d) =

r∑i=1

(αi + βi)(0, 1, wi, 1) +

q∑i=r+1

αi(0, 0, wi, 1)

+(γ1 + γ2)e2 +n∑i=3

γiei,


that is, the vector (0, a+ b, c, d) is in Zn+1 ∩R+B = NB. Thus there are λi,μi in N such that.

(0, a+ b, c, d) =

r∑i=1

μi(0, 1, wi, 1) +

q∑i=r+1

μi(0, 0, wi, 1) +

n∑i=2

λiei.

Comparing entries we obtain the equalities a+ b = μ1 + · · ·+ μr + λ2,

c = μ1w1 + · · ·+ μqwq + λ3e3 + · · ·+ λnen,

d = μ1 + · · ·+ μq.

Case (I): b ≤∑r

i=1 μi. If b < μ1, we set b = μ′1, μ

′1 < μ1, and define

μ′′1 = μ1 − μ′

1. Otherwise pick s ≥ 2 such that

μ1 + · · ·+ μs−1 ≤ b ≤ μ1 + · · ·+ μs.

Then b = μ1 + · · ·+ μs−1 + μ′s, where μ

′s ≤ μs. Set μ′′

s = μs − μ′s. Hence

a+ b = μ1 + · · ·+ μr + λ2 = a+ μ1 + · · ·+ μs−1 + μ′s,

a = μs + · · ·+ μr + λ2 − μ′s = μs+1 + · · ·+ μr + μ′′

s + λ2,

w =

s−1∑i=1

μi(0, 1, wi, 1) + μ′s(0, 1, ws, 1) +

q∑i=r+1

μi(0, 0, wi, 1)

+μ′′s (1, 0, ws, 1) +

r∑i=s+1

μi(1, 0, wi, 1) + λ2e1 +

n∑i=3

λiei,

that is, w = (a, b, c, d) ∈ NB′.Case (II): b >

∑ri=1 μi. Then b =

∑ri=1 μi + λ′2. Since

a+ b = μ1 + · · ·+ μr + λ2 = a+ μ1 + · · ·+ μr + λ′2

we get a = λ2 − λ′2. In particular λ2 ≥ λ′2. Then

w =

r∑i=1

μi(0, 1, wi, 1) +

q∑i=r+1

μi(0, 0, wi, 1) + ae1 + λ′2e2 +n∑i=3

λiei,

that is, w = (a, b, c, d) ∈ NB′. �

Definition 14.3.4 The clutter C satisfies the max-flow min-cut (MFMC)property if both sides of the LP-duality equation

min{〈α, x〉|x ≥ 0;xA ≥ 1} = max{〈y,1〉| y ≥ 0;Ay ≤ α} (14.4)

have integral optimum solutions x and y for each nonnegative integral vectorα. The system x ≥ 0;xA ≥ 1 is called totally dual integral (TDI) if themaximum in Eq. (14.4) has an integral optimum solution y for each integralvector α with finite maximum.

586 Chapter 14

Definition 14.3.5 A clutter C is called Mengerian if β1(Ca) = α0(Ca) forall a ∈ Nn, i.e., C is Mengerian if all its parallelizations have the Konigproperty.

Theorem 14.3.6 [147, 188, 258, 373] Let C be a clutter and let A be itsincidence matrix. The following are equivalent:

(i) grI(R) is reduced, where I = I(C) is the edge ideal of C.(ii) R[It] is normal and Q(A) is an integral polyhedron.

(iii) x ≥ 0; xA ≥ 1 is a TDI system.

(iv) C has the max-flow min-cut property.

(v) Ii = I(i) for i ≥ 1.

(vi) I is normally torsion-free, i.e., Ass(R/Ii) ⊂ Ass(R/I) for i ≥ 1.

(vii) C is Mengerian, i.e., β1(Ca) = α0(Ca) for all a ∈ Nn.

Proof. By Theorem 14.2.19, grI(R) is reduced if and only if R[It] is normaland the vectors �1, . . . , �r occurring in Eq. (14.1) are integral. Thus (i) and(ii) are equivalent by Theorem 1.4.3.

(ii)⇒(iii): Let α be a vector such that Eq. (14.4) has a finite maximumequal to b. We may assume b > 0, otherwise the maximum in Eq. (14.4)is attained at y = 0. Note α ≥ 0. As Q(A) is integral, b ∈ N and there is

0 �= s ∈ N such that xsα ∈ Isb. Hence xα ∈ Ib = Ib and this implies thatthe maximum in Eq. (14.4) has an integral optimum solution y.

(iii)⇒ (i): Let xα ∈ Ii/Ii+1 be a nilpotent element and let m = xα;that is, m is a monomial such that: (i) m ∈ Ii/Ii+1 and (ii) ms ∈ Iis+1, forsome 0 �= s ∈ N. By Lemma 14.3.1 it suffices to prove that m ∈ Ii+1. From(ii) there are a1, . . . , aq in N and δ ∈ Nn such that

xsα = (xv1 )a1 · · · (xvq )aqxδ and∑q

j=1aj = is+ 1.

The rational vector y0 = (a1/s, . . . , aq/s) satisfies Ay0 ≤ α, y0 ≥ 0 and thesum of its entries is equal to z0. Hence the linear program

Maximize f(y) =∑q

i=1 yi

Subject to y ≥ 0;Ay ≤ α

has an optimal value greater than or equal to z0 = i + 1/s. Note that thepolyhedron Q = {y|Ay ≤ α; y ≥ 0} is bounded. Indeed any y = (y1, . . . , yq)in Q must satisfy yj ≤ max{α1, . . . , αq} for all j, where αj is the jth entryof α. Since the system x ≥ 0;xA ≤ 1 is TDI, the optimum value of thelinear program above is attained by an integral vector b = (b1, . . . , bq). Thusb1 + · · ·+ bq ≥ i+ 1. As b ∈ Q, we can write

xα = (xv1)b1 · · · (xvq )bqxγ ,


for some γ ∈ Nn. This proves m ∈ Ii+1, as required.(iii)⇔ (iv): This is Proposition 1.4.8. (i)⇔ (v): This was shown in

Theorem 14.2.19. (v) ⇔ (vi): This was shown in Proposition 4.3.29.(iv) ⇒ (vii): Let A′ be the incidence matrix of C′ = Ca. We have shown

that (ii) is equivalent to (iv). Thus I(C) is normal and Q(A) is integral.By Theorem 14.3.3 the ideal I(C′) is normal, and since the integrality ofQ(A) is closed under minors and duplications (see Corollary 14.2.25), weget that Q(A′) is also integral. Thus, using once again that (ii) is equivalentto (iv), we get that C′ satisfies the max-flow min-cut property. Thereforethe LP-duality equation

min{〈1, x〉|x ≥ 0;xA′ ≥ 1} = max{〈y,1〉| y ≥ 0;A′y ≤ 1}

has optimum integral solutions x, y. Observe that the left-hand side of thisequality is α0(C′) and the right-hand side is β1(C′).

(vii) ⇒ (iv): Conversely if Ca has the Konig property for all a ∈ Nn,then by Lemmas 13.2.12 and 13.2.14 both sides of the LP-duality equation

min{〈a, x〉|x ≥ 0;xA ≥ 1} = max{〈y,1〉| y ≥ 0;Ay ≤ a}

have integral optimum solutions x and y for each nonnegative integral vectora, i.e., C has the max-flow min-cut property. �

Corollary 14.3.7 Let C′ be a parallelization or a minor of a clutter C. IfC has the max-flow min-cut property, then so does C′.

Proof. The max-flow min-cut property of a clutter is closed under takingminors; this follows from Proposition 14.3.2 and Theorem 14.3.6. Thatthe max-flow min-cut property is closed under parallelizations follows fromTheorem 14.3.6. �

Corollary 14.3.8 If a clutter C has the max-flow min-cut property, thenall parallelizations and minors of C have the Konig property.

Proof. Assume that the clutter C has the max-flow min-cut property. ByCorollary 14.3.7 it suffices to prove that C has the Konig property. Let Abe the incidence matrix of C. By hypothesis the LP-duality equation

min{〈1, x〉|x ≥ 0;xA ≥ 1} = max{〈y,1〉| y ≥ 0;Ay ≤ 1}

has optimum integral solutions x, y. To complete the proof notice that theleft-hand side of this equality is α0(C) and the right-hand side is β1(C). �

Definition 14.3.9 A clutter C is called dyadic if |C ∩ B| ≤ 2 for C ∈ Cand B ∈ C∨.

588 Chapter 14

Corollary 14.3.10 [94, Theorem 1.3] If Q(A) is integral and C is dyadic,then x ≥ 0; xA ≥ 1 is a TDI system.

Proof. By Proposition 14.2.26 the Rees algebra R[It] is normal. Thus theproof follows from Theorem 14.3.6. �

Corollary 14.3.11 [185] If C is a balanced clutter, then

(i) R[I(C)t] = Rs(I(C)) and R[Ic(C)t] = Rs(Ic(C)).(ii) grI(C)(R) is reduced and all minors of C satisfy the Konig property.

Proof. By [373, Corollary 83.1a(iv), p. 1441], we get that both C and C∨are Mengerian. Thus the result follows at once from Theorem 14.3.6. �

Corollary 14.3.12 [156, Theorem 5.3] If C is the clutter of facets of asimplicial forest, then C has the Konig property.

Proof. We set I = I(C). By Theorem 6.5.17, C is totally balanced. Hence,by Corollary 14.3.11, the associated graded ring grI(R) is reduced and allminors of C satisfy the Konig property. �

Remark 14.3.13 Let C be the clutter of facets of an unmixed simplicialforest. By Theorem 6.5.17, C has no special cycles. As C is balanced, ithas the Konig property (Corollary 14.3.11), and by Lemma 6.5.7 it has aperfect matching of Konig type. Thus, Theorem 6.5.18 holds for C [156].

Corollary 14.3.14 If A is totally unimodular, then grI(R) is reduced.

Proof. If A is totally unimodular, then A is balanced. Thus we can applyCorollary 14.3.11. �

Corollary 14.3.15 [383, Theorem 5.9] If C is a bipartite graph and I is itsedge-ideal, then I is normally torsion-free and R[It] is normal.

Proof. By Proposition 10.2.2, the incidence matrix of a bipartite graphis totally unimodular. Thus the result follows from Corollary 14.3.14 andTheorem 14.3.6. �

Definition 14.3.16 A clutter C is said to satisfy the packing property (PPfor short) if α0(C′) = β1(C′) for any minor C′ of C; that is, all minors of Csatisfy the Konig property.

Theorem 14.3.17 (A. Lehman [290], [93, Theorem 1.8]) If a clutter C hasthe packing property, then Q(A) is integral.

The converse of this result is not true. A famous example is the clutterQ6 of Example 14.2.9. It does not pack and Q(A) is integral.


Corollary 14.3.18 [93] If a clutter C has the max-flow min-cut property,then C has the packing property.

Proof. It follows at once from Corollary 14.3.8. �

Conforti and Cornuejols [90] conjectured that the converse is also true;see also [93, Conjecture 1.6].

Conjecture 14.3.19 (Packing problem, Conforti–Cornuejols) If a clutterC has the packing property, then C has the max-flow min-cut property.

The following is an equivalent version of this conjecture [93].

Conjecture 14.3.20 (Duplication Conjecture, [93, Conjecture 4.24]) If Chas the packing property and C′ is the clutter obtained from C by duplicatinga vertex xi, then C′ has the Konig property.

The Conforti–Cornuejols conjecture can be studied via blowup algebras.By Theorem 14.3.6, the following is an algebraic version of this conjecture.

Conjecture 14.3.21 If α0(C′) = β1(C′) for all minors C′ of C, then thering grI(R) is reduced.

Using Theorems 14.3.17 and 14.3.6 this conjecture reduces to:

Conjecture 14.3.22 If α0(C′) = β1(C′) for all minors C′ of C, then R[It]is normal.

Proposition 14.3.23 Let Ji be the ideal obtained from I by making xi = 1in F . If Q(A) is integral, then I is normal if and only if Ji is normal fori = 1, . . . , n and depth(R/Ik) ≥ 1 for all k ≥ 1.

Proof. ⇒) The normality of an edge ideal is closed under taking minors(Proposition 12.2.3), hence Ji is normal for all i. Since R[It] is normal, weget that R[It] is C–M thanks to Theorem 9.1.6. Then by Theorem 4.3.19the ring grI(R) is C–M. Hence using Theorem 14.2.13 and Corollary 14.2.11we get that depth(R/Ii) ≥ 1 for all i.⇐) It follows readily adapting the arguments given in the proof of the

normality criterion presented in Theorem 12.2.4. �

By Proposition 14.3.23, we get that Conjecture 14.3.21 also reduces to:

Conjecture 14.3.24 If α0(C′) = β1(C′) for any minor C′ of C, then

depth(R/Ii) ≥ 1 for all i ≥ 1.

Definition 14.3.25 A clutter C is called minimally non-packing if it doesnot pack but all its proper minors do.

590 Chapter 14

The next conjecture implies Conjecture 14.3.19 [94].

Conjecture 14.3.26 [94] If C is a minimally non-packing clutter and thepolyhedron Q(A) is integral, then α0(C) = 2.

The next result essentially says that, for uniform clutters, it suffices toprove Conjecture 14.3.19 for Cohen–Macaulay clutters.

Theorem 14.3.27 [116] Let C be a d-uniform clutter. If C satisfies PP(resp. max-flow min-cut), then there is a d-uniform Cohen–Macaulay clutterC′ satisfying PP (resp. max-flow min-cut) such that C is a minor of C′.

Theorem 14.3.28 If R[It] = Rs(I) and K[Ft] = A(P), then R[It] isnormal.

Proof. By Theorem 14.2.7 the Rees cone of I is quasi-ideal. Thus byCorollary 12.3.2 the Rees algebra of I is normal. � �

Proposition 14.3.29 If C is a d-uniform clutter such that Ib = I(b) for allb and d ≥ 2, then Ib is generated by monomials of degree bd for b ≥ 1.

Proof. The monomial ideal Ib has a unique minimal set of generatorsconsisting of monomials. Take xa in this minimal set. Notice that (a, b)is in R+(I). Thus we may proceed as in the proof of Theorem 12.3.1 toobtain that (a, b) is in the cone generated by {(v1, 1), . . . , (vq, 1)}. Thisyields deg(xa) = bd. �

Proposition 14.3.30 [185] If C is a d-uniform clutter and d ≥ 2, thenIi = I(i) for all i ≥ 1 if and only if Q(A) is integral and K[Ft] = A(P).

Proof. ⇒) By Theorem 14.2.7 the polyhedron Q(A) is integral. SinceI(i) is integrally closed (see Corollary 4.3.26), we get that R[It] is normal.Therefore applying Theorem 9.3.31, we obtain K[Ft] = A(P). Here thehypothesis on the degrees of xvi is essential.⇐) By Theorem 14.2.7 Ii = I(i) for all i, thus applying Theorem 14.3.28

gives that R[It] is normal and we get the required equality. In this part thehypothesis on the degrees of xvi is not needed. �

For uniform clutters, using Proposition 14.3.30 and Theorem 14.3.17, weobtain another algebraic version of the Conforti–Cornuejols conjecture:

Conjecture 14.3.31 If C is a uniform clutter with the packing property,then one has the equality K[Ft] = A(P).

Corollary 14.3.32 Let C be a uniform clutter and let A be its incidencematrix. If the polyhedra

{x|x ≥ 0; xA ≤ 1} and {x|x ≥ 0; xA ≥ 1}

are integral, then C has the max-flow min-cut property.


Proof. By Proposition 14.3.30, the clutter C has the max-flow min-cutproperty if and only if Q(A) is integral and K[Ft] = A(P). Thus, the resultfollows from Corollary 13.6.9. �

Notation For an integral matrix B �= (0), the greatest common divisor ofall the non-zero r × r subdeterminants of B will be denoted by Δr(B).

Corollary 14.3.33 Let B be the matrix obtained from A by adjoining a rowof 1’s. If xv1 , . . . , xvq have degree d ≥ 2 and C has the max-flow min-cutproperty, then Δr(B) = 1 with r = rank(B).

Proof. By Theorem 14.3.6 and Proposition 14.3.30, we obtain the equalityA(P) = K[Ft]. Hence a direct application of Theorem 9.3.25 gives thatΔr(B) is equal to 1. �

Definition 14.3.34 Let M be a matroid on X and let C be the clutter ofbases of M . The edge ideal I(C) is called the basis monomial ideal of M .

Proposition 14.3.35 Let C be the clutter of bases of a matroid M on X.If C satisfies the packing property, then grI(C)(R) is reduced.

Proof. Since Q(A) is integral (Theorem 14.3.17) and R[I(C)t] is normal(Corollary 12.3.12), we get that grI(C)(R) is reduced by Theorem 14.3.6. �

Definition 14.3.36 Let X1, . . . , Xd be a partition of X . The matroidMwhose collection of bases is

{{y1, . . . , yd}| yi ∈ Xi for i = 1, . . . , d}

is called the transversal matroid defined by X1, . . . , Xd.

Proposition 14.3.37 Let X1, . . . , Xd be a partition of X and letM be thetransversal matroid on X defined by X1, . . . , Xd. If I is the basis monomialideal of M, then grI(R) is reduced.

Proof. Let C be the clutter of bases ofM and let A be the incidence matrixof C. Then I = I(C). Notice that the edges of C∨ are X1, . . . , Xd, thereforethe incidence matrix of C∨ is totally unimodular. Hence C∨ is a balancedclutter and (C∨)∨ = C. Thus, by Corollary 14.3.11, grI(R) is reduced. �

Another property which is closed under taking minors is being the basismonomial ideal of a matroid.

Proposition 14.3.38 Let I be the basis monomial ideal of a matroid Mon X of rank d. If I ′ is a minor of I, then I ′ is again the basis monomialideal of a matroid M ′.

592 Chapter 14

Proof. Let I ′ be the minor obtained from I by making xn = 1 and letG(I ′) be the unique minimal generating set of I ′ consisting of monomials.

First we prove that I ′ is generated by monomials of degree d− 1. Takef ∈ G(I ′). We proceed by contradiction. Assume that f = x1x2 · · ·xd withxn /∈ supp(f). Pick a monomial in I of the form g = xi1xi2 · · ·xid−1

xn.If supp(g) \ {xn} ⊂ {x1, . . . , xd} we obtain a contradiction because of theminimality of G(I ′). Thus we may assume xi1 /∈ supp(f). By the exchangeproperty (Theorem 1.9.14) we may assume that the monomial

g1 = x1xi2xi3 · · ·xid−1xn

belongs to I. If supp(g1) \ {xn} ⊂ {x1, . . . , xd} we obtain a contradiction.Thus we may assume xi2 /∈ supp(f). Again by the exchange propertywe may assume that g2 = x1x2xi3xi4 · · ·xid−1

xn is in I. Repeating thisprocedure we get x1x2 · · ·xd−1xn ∈ I, a contradiction. Thus I ′ is generatedby monomials of degree d− 1, as claimed.

We set G(I ′) = {xα1 , . . . , xαr}, B′i = supp(xαi ), and Bi = B′

i ∪ {xn}.The collection of bases {B1, . . . , Br} satisfies the exchange property, henceso does the collection B′ = {B′

1, . . . , B′r}. Thus, by Theorem 1.9.14, there is

a matroidM ′ whose collection of bases is B′. Hence I ′ is the basis monomialideal of M ′, as required. The case xn = 0 is also not hard to show. �

Proposition 14.3.39 If C is a graph, then the following are equivalent:

(a) grI(R) is reduced.

(b) C is bipartite.

(c) Q(A) is integral.

(d) C has the packing property.

Proof. (a) ⇒ (d): By Theorem 14.3.6, C has the max-flow min-cut prop-erty. Then, by Corollary 14.3.18, C has the packing property. (d) ⇒ (b):It suffices to show that C has no induced odd cycles (Proposition 7.1.2). IfCr = {x1, . . . , xr} is an induced cycle, then deleting all vertices outside Crwe obtain that Cr is a minor of C. Thus Cr has the Konig property. Itfollows that r must be even. (b) ⇒ (c): By Proposition 10.2.2, the matrixA is totally unimodular. Then by a result of Hoffman and Kruskal (seeTheorem 1.5.6), Q(A) is integral. (c) ⇒ (a): Since any graph is a dyadicclutter, using Corollary 14.3.10, we get that x ≥ 0; xA ≥ 1 is a TDI system.Thus by Theorem 14.3.6, the ring grI(R) is reduced. �

Definition 14.3.40 A clutter is binary if its edges and its minimal vertexcovers intersect in an odd number of vertices.

Theorem 14.3.41 [375] A binary clutter C has the max-flow min-cut prop-erty if and only if Q6 is not a minor of C.


Corollary 14.3.42 If C is a binary clutter with the packing property, thenC has the max-flow min-cut property.

Proof. Any clutter with the packing property cannot have Q6 as a minorbecause Q6 does not satisfy the Konig property. Thus C has the max-flowmin-cut property by Theorem 14.3.41. �

Lemma 14.3.43 If I is an unmixed ideal and C satisfies the Konig prop-erty, then x1 = x1x2 · · ·xn belongs to the subring K[xv1 , . . . , xvq ].

Proof. We may assume x1 = xv1 · · ·xvgxδ for some xδ, where g is the heightof I. If δ �= 0 pick xn ∈ supp(xδ). Since the variable xn occurs in somemonomial minimal generator of I, there is a minimal prime p containingxn. Thus using that xv1 , . . . , xvg have disjoint support we conclude that pcontains at least g + 1 variables, a contradiction. �

Proposition 14.3.44 Let Ii = I ∩ K[X \ {xi}]. If I is an unmixed idealsuch that the following conditions hold

(a1) Q(A) is integral,

(a2) Ii is normal for i = 1, . . . , n, and

(a3) C has the Konig property,

then R[It] is normal.

Proof. Take xatb = xa11 · · ·xann tb ∈ R[It] a minimal generator; that is, xatb

cannot be written as a product of two non-constant monomials of R[It]. Bycondition (a2) we may assume ai ≥ 1 for all i. Notice that x1 · · ·xntg isalways in R[It] if Q(A) is integral, where g is the height of I. We claim thatb ≤ g. If b > g, consider the decomposition

xatb = (x1 · · ·xntg)(xa1−11 · · ·xan−1

n tb−g).

To derive a contradiction consider the irreducible representation of the Reescone R+(I) (see Eq. (14.1)). Observe that using condition (a1) gives∑

i∈Ck

ai ≥ b (k = 1, . . . , s)

because (a, b) ∈ R+(I). Now since I is unmixed we get∑i∈Ck

(ai − 1) ≥ b− g (k = 1, . . . , s),

and consequently xa1−11 · · ·xan−1

n tb−g ∈ R[It], a contradiction to the choiceof xatb. Thus b ≤ g. Using condition (a3) we get x1 · · ·xntg ∈ Igtg ⊂ R[It],which readily implies xatb ∈ R[It]. �

594 Chapter 14

Corollary 14.3.45 Let C be an unmixed clutter with the packing propertyand let I = I(C) be its edge ideal. If the ideal Ii = I ∩K[X \{xi}] is normalfor all i, then C satisfies the max-flow min-cut property.

Proof. By Theorem 14.3.17, the set covering polyhedron Q(A) is integral.Then R[It] is normal by Proposition 14.3.44. Using Theorem 14.3.6, we getthat C satisfies the max-flow min-cut property. �

Corollary 14.3.46 [204] If a minimal counterexample C to the Conforti–Cornuejols conjecture exists, then C cannot be unmixed.

Proof. Assume that C is unmixed. Let I = I(C) be the edge ideal of C,let Ii = I ∩ K[X \ {xi}], and let Ci be the clutter that corresponds to Ii.Then Ci satisfies the max-flow min-cut property, and by Theorem 14.3.6 weget that Ii is normal. Hence by Corollary 14.3.45, the clutter C has themax-flow min-cut property, a contradiction. �

Proposition 14.3.47 Let Y ⊂ X and let IY = I ∩ K[Y ]. If IY has theKonig property for all Y and R[It] is generated as a K-algebra by monomialsof the form xatb, with xa square-free, then R[It] is normal.

Proof. Take xatb a generator of R[It], with xa square-free. By inductionwe may assume xatb = x1 · · ·xntb. Hence, since (1, . . . , 1, b) is in the Reescone, we get |Ck| ≥ b for k = 1, . . . , s. In particular g = ht(I) ≥ b. AsI has the Konig property, we get that the monomial x1 · · ·xn is in Ig andconsequently xatb ∈ R[It]. �

Proposition 14.3.48 Let Ii = I ∩K[X \ {xi}]. If Ii is a normal ideal fori = 1, . . . , n and

H�1 ∩H�2 ∩ · · · ∩H�r ∩ Rn+1+ �= (0), (14.5)

then R[It] is normal.

Proof. Let xatb = xa11 · · ·xann tb ∈ R[It] be a minimal generator. It sufficesto prove that 0 ≤ b ≤ 1 because this readily implies that xa is either avariable or a monomial in F . Assume b ≥ 2. Since Ii is normal we mayassume that ai ≥ 1 for all i. As each variable occurs in at least one monomialof F , using Eq. (14.5) it follows that there is (vk, 1) such that 〈(vk, 1), �i〉 = 0for i = 1, . . . , r. Therefore

〈(a− vk, b− 1), �i〉 ≥ 0 (i = 1, . . . , r).

Thus (a, b)− (vk, 1) ∈ R+(I), a contradiction to the choice of xatb. �


Rees algebras and their a-invariants In this part we compute somea-invariants in terms of combinatorial invariants of graphs and clutters, andexhibit some families of Gorenstein Rees algebras.

Theorem 14.3.49 Let C be a d-uniform clutter and let a(R[It]) be the a-invariant of R[It] with respect to the grading of R[It] induced by deg(xi) = 1and deg(t) = −(d− 1). If grI(R) is reduced, then

a(R[It]) ≥ − [n− (d− 1)(α0(C)− 1)] ,

with equality if I is unmixed.

Proof. By assigning deg(xi) = 1 and deg(t) = −(d−1), the Rees ring R[It]becomes a standard gradedK-algebra. By Theorem 14.3.6 the Rees algebraR[It] is a normal domain. Then according to a formula of Danilov–Stanley(see Theorem 9.1.5) its canonical module is the ideal of R[It] given by

ωR[It] = ({xa11 · · ·xann tan+1| a = (ai) ∈ (R+(I))o ∩ Zn+1}).

Set α0 = α0(C). By Theorem 14.3.6, the Rees algebra of I is normal andQ(A) is integral. The ring R[It] is Cohen–Macaulay by Theorem 9.1.6.Then by Proposition 5.2.3, the a-invariant can be expressed as

a(R[It]) = −min{ i | (ωR[It])i �= 0}.

Using Eq. (14.1) it is seen that (1, . . . , 1, α0 − 1) ∈ (R+(I))o. Thus the

inequality follows by computing the degree of x1 · · ·xntα0−1.Assume that I is unmixed. Take any monomial xatb = xa11 · · ·xann tb in

the ideal ωR[It], that is, (a, b) ∈ (R+(I))o. By Theorem 14.2.7 the vector

(a, b) has positive entries and satisfies

−b+∑

xi∈Ckai ≥ 1 (k = 1, . . . , s). (14.6)

If α0 ≥ b+ 1 we readily obtain the inequality

deg(xatb) = a1 + · · ·+ an − b(d− 1) ≥ n− (d− 1)(α0 − 1). (14.7)

Now assume α0 ≤ b. Using the normality of R[It] and Eqs. (14.1) and (14.6)it follows that the monomial

m = xa1−11 · · ·xan−1

n tb−α0+1

belongs to R[It]. Since xatb = mx1 · · ·xntα0−1, the inequality (14.7) alsoholds in this case. Altogether we conclude the desired equality. �

Corollary 14.3.50 [183] Let C be a d-uniform clutter. If I is unmixed withα0(C) = 2 and grI(R) is reduced, then R[It] is a Gorenstein ring and

a(R[It]) = −(n− d+ 1).

596 Chapter 14

Proof. From the proof of Theorem 14.3.49 it follows that the monomialx1 · · ·xnt generates the canonical module. �

Remark 14.3.51 If α0(C) ≥ 3, then R[It] is not a Gorenstein ring becausex1 · · ·xntα0−1 and x1 · · ·xnt are distinct minimal generators of ωR[It]. Thisholds in a more general setting (see Proposition 4.3.39).

Corollary 14.3.52 [183] Let J = Ic(C) be the ideal of covers of C. If C isa bipartite graph and I = I(C) is unmixed, then R[Jt] is Gorenstein and

a(R[Jt]) = −(n− α0(C) + 1).

Proof. Notice that R[Jt] has the grading induced by assigning deg(xi) = 1and deg(t) = 1 − α0(C). Thus the formula follows from Corollary 14.3.50once we recall that grJ(R) is reduced according to Corollary 14.3.11. �

Corollary 14.3.53 Let C be a bipartite graph. Then the Rees cone R+(I)is the intersection of the closed halfspaces given by the linear inequalities

xi ≥ 0 i = 1, . . . , n+ 1,−xn+1 +

∑vi∈C xi ≥ 0 C is a minimal vertex cover of C,

and none of those halfspaces can be omitted from the intersection.

Proof. Let A be the incidence matrix of C. By Proposition 10.2.2, thematrix A is totally unimodular. Then the vertices of Q(A) are integral byTheorem 1.5.6. Hence the result follows from Theorem 14.1.1. �

Corollary 14.3.54 Let C be a bipartite graph. Then F is a facet of theRees cone R+(I) if and only if

(a) F = R+(I) ∩Hei for some 1 ≤ i ≤ n+ 1, or

(b) F = R+(I) ∩ {x ∈ Rn+1| − xn+1 +∑vi∈C xi = 0} for some minimal

vertex cover C of C.

Proof. The result follows from Theorem 1.1.44 and Corollary 14.3.53. �

Theorem 14.3.55 [187] Let C be a bipartite graph. If a(R[It]) is the a-invariant of R[It] with respect to the grading of R[It] induced by deg(xi) = 1and deg(t) = −1, then

a(R[It]) = −(β0(C) + 1).

Proof. By Corollary 14.3.15, the Rees algebra R[It] is a normal domain.Clearly R[It] is a standard graded K-algebra with the grading induced by


deg(xi) = 1 and deg(t) = −1. Then according to a formula of Danilov–Stanley (see Theorem 9.1.5) the canonical module of R[It] is the ideal ofR[It] given by

ωR[It] = ({xa11 · · ·xann tan+1| a = (ai) ∈ (R+(I))o ∩ Zn+1}),

where (R+(I))o is the topological interior of the Rees cone. The ring R[It]

is Cohen–Macaulay by Theorem 9.1.6. Therefore the a-invariant can beexpressed as

a(R[It]) = −min{ i | (ωR[It])i �= 0},see Theorem 5.2.3. Let a = (ai) be an arbitrary vector in (R+(I))

o ∩Zn+1.By Corollary 14.3.53 a satisfies ai ≥ 1 for 1 ≤ i ≤ n+ 1 and

−an+1 +∑

vi∈C ai ≥ 1

for any minimal vertex cover C of C. Let C be a vertex cover of C withα0(C) elements and let A = V \ C. Note β0(C) = |A|. Hence if m denotesthe monomial xa11 · · ·xann tan+1, then

deg(m) = a1 + · · ·+ an − an+1

=∑vi∈A

ai +∑vi∈C

ai − an+1 ≥ β0(C) + 1.

This proves the inequality a(R[It]) ≤ −(β0(C) + 1). On the other handusing Corollary 14.3.53 and the assumption α0(C) ≥ 2 we get that themonomial m1 = x1 · · ·xntα0−1 is in ωR[It] and has degree β0(C) + 1. Thusa(R[It]) ≥ −(β0(C) + 1). �

Corollary 14.3.56 If C is a bipartite graph, then type(R[It]) ≥ α0(C)− 1.

Proof. By Corollary 14.3.53 it is seen that the monomials x1 · · ·xnti, withi = 1, . . . , α0(C)− 1, are distinct minimal generators of ωR[It]. �

Exercises

14.3.57 Prove that Q6 is a binary non-dyadic clutter.

14.3.58 If a clutter C has the packing property and I = I(C), then I2 = I2.

14.3.59 Let v1 and v2 be two vectors in Rn with entries in {0, 1} and letI = (xv1 , xv2) ⊂ R. Prove that the Rees algebra R[It] is normal.

14.3.60 Prove that the ideal I = (x21, x22) ⊂ K[x1, x2] is not normal.

14.3.61 Let I = (x1x4x7, x3x5x8, x2x6x9, x1x4x8, x3x4x7, x1x5x7) be theedge ideal of a clutter C and let A be its incidence matrix. Prove that (a)grI(R) is reduced, and (b) A is not totally unimodular.

598 Chapter 14

14.3.62 Let I and I ′ be monomial ideals of R and R′ = K[x2, . . . , xn],respectively. If I is square-free and I = x1I

′, prove that grI(R) is reducedif and only if grI′(R

′) is reduced.

14.3.63 If C is a minimally non-packing clutter, then C is either ideal orminimally non-ideal. Recall that C is called ideal if Q(A) is integral, whereA is the incidence matrix of C, and that C is called minimally non-ideal ifC is not ideal but all its proper minors are ideal [94].

14.3.64 If G is an odd cycle, then G does not satisfy the Konig property.

14.3.65 Let C be a clutter with vertex set X = {x1, . . . , xn}. Consider theclutter C′ with vertex set X ∪{y1, . . . , yn} and whose edges are the edges ofC together with {xi, yi} for i = 1, . . . , n. Prove that C satisfies the max-flowmin-cut property if and only if C′ does.

14.3.66 Let C be a clutter with vertex set X = {x1, . . . , xn} such that allits edges have d ≥ 2 vertices. Construct a clutter C′ satisfying the followingproperties: (a) all the edges of C′ have d vertices. (b) C′ is unmixed. (c) Csatisfies the max-flow min-cut property if and only if C′ does.

14.3.67 Prove that condition (a3) of Proposition 14.3.44 is not needed whenI is generated by monomials of the same degree (cf. Corollary 14.4.7).

14.3.68 Let R = K[x1, . . . , xn] and R[z1, . . . , z�] be polynomial rings overa field K. If I is an ideal of R generated by square-free monomials and I isnormal, then J = (I, x1z1 · · · z�) is a normal ideal of R[z1, . . . , z�].

14.3.69 If grI(R) is reduced and C is d-uniform, then any square submatrixB of A of order m such that the sum of the entries in any row or column ofB is equal to d satisfies m ≡ 0mod (d) and det(B) = 0.

14.3.70 Let xa1 , . . . , xar be a set of generators of I and let B be the matrixwhose columns are a1, . . . , ar. Then both sides of the equation

min{〈1, x〉|x ≥ 0;xA ≥ 1} = max{〈y,1〉| y ≥ 0;Ay ≤ 1}

have integral optimum solutions if and only if both sides of the equation

min{〈1, x〉|x ≥ 0;xB ≥ 1} = max{〈y,1〉| y ≥ 0;By ≤ 1}

have integral optimum solutions. If any of the two systems have integraloptimum solutions x, y their optimal values are equal.


14.4 Uniform ideal clutters

A clutter is called ideal if its set covering polyhedron is integral. In thissection we study the combinatorial structure of uniform ideal clutters . Themax-flow min-cut property of clutters is studied here from a linear algebraperspective. If a uniform clutter has the max-flow min-cut property, weprove that its incidence matrix is equivalent over Z to an “identity matrix”and that its column vectors form a Hilbert basis. The structure of normallytorsion-free Cohen–Macaulay edge ideals of height two is also studied here.

Lemma 14.4.1 [185] If C is a d-uniform clutter and Q(A) is integral, thenthere exists a minimal vertex cover of C intersecting every edge of C inexactly one vertex.

Proof. Let B be the incidence matrix of C∨. Using Corollary 14.2.22,we get that Q(B) is an integral polyhedron. We proceed by contradiction.Assume that for each column αk of B there exists vik in {v1, . . . , vq} suchthat vikB ≥ 1 + ek. Consider the vector α = vi1 + · · · + vis . From theinequality

αB ≥ (s+ 1, . . . , s+ 1)

we obtain that α/(s+1) ∈ Q(B). Notice thatQ(B) = Rn++conv(v1, . . . , vq);see Theorem 1.1.42. Thus, we can write

α/(s+ 1) = λ1v1 + · · ·+ λqvq + μ1e1 + · · ·+ μnen (λi, μj ≥ 0;∑λi = 1).

Therefore, taking inner product with 1, we get |α| = sd ≥ (s + 1)d, acontradiction. �

Definition 14.4.2 A graph G is called strongly perfect if every inducedsubgraph H of G has a maximal independent set of vertices A such that|A ∩K| = 1 for any maximal clique K of H .

Bipartite and chordal graphs are strongly perfect. If A is the vertex-clique matrix of G, then G being strongly perfect implies that the cliquepolytope of G, {x|x ≥ 0; xA ≤ 1}, has a vertex that intersects everymaximal clique. In this sense, uniform ideal clutters can be thought of asbeing analogous to strongly perfect graphs.

Proposition 14.4.3 Let C be a d-uniform clutter. If R[It] = Rs(I) and Iis unmixed, then H�1 ∩ · · · ∩H�r ∩ Rn+1

+ �= (0).

Proof. By Proposition 14.2.21 one has R[Ic(C)t] = Rs(Ic(C)). Thus, byLemma 14.4.1, there is vk such that |supp(vk) ∩ Ci| = 1 for i = 1, . . . , r.This means that (vk, 1) is in the intersection of H�1 , . . . , H�r . �

600 Chapter 14

Proposition 14.4.4 If C is a d-uniform clutter whose covering polyhedronQ(A) is integral, then there are X1, . . . , Xd mutually disjoint minimal vertexcovers of C such that X = ∪di=1Xi.

Proof. By induction on d. By Lemma 14.4.1 there is a minimal vertexcover X1 of C such that |supp(xvi ) ∩ X1| = 1 for all i. Consider the idealI ′ obtained from I by making xi = 1 for xi ∈ X1. Let C′ be the cluttercorresponding to I ′ and let A′ be the incidence matrix of C′. The integralityof Q(A) is preserved under taking minors (Corollary 14.2.25), so Q(A′) isintegral. Then C′ is a (d − 1)-uniform clutter with Q(A′) integral andV (C′) = X \ X1. Therefore by induction hypothesis there are X2, . . . , Xd

pairwise disjoint minimal vertex covers of C′ such thatX\X1 = X2∪· · ·∪Xd.To complete the proof we now show that X2, . . . , Xd are minimal vertex

covers of C. If e is an edge of C and 2 ≤ k ≤ d, then e∩X1 = {xi} for somei. Since e \ {xi} is an edge of C′, we get (e \ {xi}) ∩Xk �= ∅. Hence Xk isa vertex cover of C. Furthermore if x ∈ Xk, then by the minimality of Xk

there is an edge e′ of C′ disjoint from Xk \ {x}. Since e = e′ ∪ {y} is anedge of C for some y ∈ X1, we obtain that e is an edge of C disjoint fromXk \ {x}. Therefore Xk is a minimal vertex cover of C, as required. �

Corollary 14.4.5 If C is a d-uniform clutter and Q(A) is integral, thenthere is a partition X1, . . . , Xd of X such that C is a subclutter of the clutterof bases of the transversal matroid M defined by X1, . . . , Xd.

Proof. It suffices to notice that, by Proposition 14.4.4, any edge of Cintersects Xi in exactly one vertex for any i. �

Theorem 14.4.6 [205] Let C be a d-uniform clutter. Assume that I(C) isnormally torsion-free of height two. Then C is Cohen–Macaulay if and onlyif (i) C is unmixed, and (ii) there is a partition

X1 = {x11, x12}, . . . , Xd = {xd1, xd2}

of X and a perfect matching e1 = {x11, . . . , xd1}, e2 = {x12, . . . , xd2} of C sothat all edges of C have the form {x1i1 , . . . , x

did} for some non-decreasing

sequence 1 ≤ i1 ≤ · · · ≤ id ≤ 2.

Proof. ⇒) By Corollary 3.1.17 Cohen–Macaulay rings are unmixed, so (i)holds true. We shall prove (ii) by induction on d.

We claim that C has a free vertex. By Theorem 14.3.6 Q(A) is integral.Hence, by Proposition 14.4.4, there are minimal vertex covers Z1, . . . , Zdof C such that Z1, . . . , Zd partition X and |Zi ∩ e| = 1 for all e ∈ E(C)and i = 1, . . . , d. Since C is unmixed, one has |Zi| = 2 for all i. ByTheorem 14.3.6, C has the Konig property. Then, by Lemma 6.5.7, C hasa perfect matching of Konig type, i.e., there are edges e1, e2 of C such that


e1 ∩ e2 = ∅ and e1 ∪ e2 = X . We may assume that e1 = {x1, . . . , xd},e2 = {y1, . . . , yd}, and Zi = {xi, yi} for all i. Any minimal vertex cover ofC has the form C = {x, y} for some x ∈ e1 and y ∈ e2. Let G = C∨ be theblocker of C, which, in this case, is a graph. The graph G is bipartite withbipartition e1, e2. As R/I(C) is Cohen–Macaulay, I(G) = I(C∨) has a linearresolution (Theorem 6.3.41). Then, by Theorem 7.6.7, the complementgraph G′ of G is chordal. By Lemma 7.5.4, G′ has a simplicial vertex z.We may assume that z = xk for some k. The induced subgraphs G′[e1]and G′[e2] of G

′ are complete graphs on d vertices. Next we prove thatxk /∈ NG′(e2) for any k. If xk ∈ NG′(e2) for some k, then {xk, y�} is anedge of G′ for some �. Then y� would have to be adjacent in G′ to any xiin e1, in particular {x�, y�} ∈ E(G′), a contradiction. Thus {xk, yi} ∈ E(G)for all i. Note that yk is a free vertex of C. Indeed let e be any edge of Ccontaining yk, then xk is not in e because |e∩Zk| = 1. Hence since {xk, yi}is a vertex cover of C for any i we get that yi ∈ e for any i, i.e., e = e2.

Let I ′ be the edge ideal obtained from I(C) by making xk = 1 and yk = 1and let C′ be the clutter on V ′ = X\{xk, yk} such that I ′ = I(C′). The idealI ′ is Cohen–Macaulay of height two, normally torsion-free, and is generatedby monomials of degree d−1. Therefore, by the induction hypothesis, thereis a partition X2 = {x21, x22}, . . . , Xd = {xd1, xd2} of V ′ such that all edges ofC′ have the form {x2i2 , . . . , x

did} for some 1 ≤ i1 ≤ · · · ≤ id ≤ 2. To complete

the proof we set x11 = xk, x12 = yk, and X

1 = {x11, x12}.⇐) It is left as an exercise. Here, the assumption that I(C) is normally

torsion-free is not needed. �

Corollary 14.4.7 Let C be a d-uniform clutter. If R[It] = Rs(I) and I isunmixed, then C and C∨ have the Konig property.

Proof. According to Proposition 14.2.21 one has R[Ic(C)] = Rs(Ic(C)).Thus, by duality (Theorem 6.3.39), we need only show that C has the Konigproperty, and this follows readily from Proposition 14.4.4. �

Theorem 14.4.8 Let C be a d-uniform clutter and let Ii = I∩K[X \{xi}].If I is unmixed and Q(A) is integral, then grI(R) is reduced if and only ifIi is normal for i = 1, . . . , n.


Proposition 14.4.9 If C is a d-uniform clutter with a perfect matchingand Q(A) is integral, then C has the Konig property and is vertex critical.

Proof. By Theorem 1.1.42, we can write Q(A) = Rn+ + conv(u1, . . . , us),where the ui’s are the characteristic vectors of the minimal vertex covers ofC. As 1/d ∈ Q(A), using this equality, we get

1/d = δ + λ1u1 + · · ·+ λsus; (δ ∈ Rn+; λi ≥ 0;∑

i λi = 1).

602 Chapter 14

Therefore n ≥ gd, where g = α0(C). By hypothesis there are mutuallydisjoint edges f1, . . . , fr such that X = f1 ∪ · · · ∪ fr and n = rd. Thusn = rd ≥ gd and r ≥ g. On the other hand g = α0(C) ≥ β1(C) ≥ r. Thusr = g and n = gd. Hence C has the Konig property. We now prove thatC is vertex critical. By Proposition 14.4.4 there are X1, . . . , Xd mutuallydisjoint minimal vertex covers of C such that X = ∪di=1Xi. Hence

n = gd = |X1|+ · · ·+ |Xd|.

As |Xi| ≥ g for all i, we get |Xi| = g for all i. It follows easily that Cis vertex critical. Indeed notice that each vertex xi belongs to a minimalvertex cover of C with g vertices. Hence α0(G \ xi) < α0(G). �

Proposition 14.4.10 Let C be a d-uniform clutter with a perfect matchingf1, . . . , fr. If Q(A) is integral, then r = α0(C) and there are X1, . . . , Xd

disjoint minimal vertex covers of C of size α0(C) such that X = ∪di=1Xi.


Theorem 14.4.11 [119] If C is a d-uniform clutter with the max-flow min-cut property, then

(a) Δr(A) = 1, where r = rank(A).

(b) NA = R+A ∩ Zn, where A = {v1, . . . , vq}.(c) A diagonalizes over Z to an identity matrix.

Proof. (a): Let B be the matrix with column vectors (v1, 1), . . . , (vq, 1).Since the clutter is uniform, the last row vector of B, i.e., the vector 1, isa Q-linear combination of the first n rows of B. Thus A and B have thesame rank. By Proposition 14.3.30 we obtain K[Ft] = A(P). In particularK[Ft] = A(P) because A(P) is normal. Hence, by Theorem 9.3.25, we haveΔr(B) = 1. Recall that Δr(A) = 1 if and only if A is equivalent over Zto an “identity matrix.” We can regard A as a matrix of size (n + 1) × qby adding a row of zeros. Thus it suffices to prove that B is equivalentto A over Z. Thanks to Proposition 14.4.4, there are X1, . . . , Xd mutuallydisjoint minimal vertex covers of C such that X = ∪di=1Xi and

|supp(xvi ) ∩Xk| = 1 ∀ i, k.

By permuting the variables we may assume that X1 is equal to {x1, . . . , xr}.Hence the last row of B, which is the vector 1, is the sum of the first |X1|rows of B, i.e., B is equivalent to A over Z.

(b): It suffices to prove the inclusion “⊃”. Let a be an integral vector inR+A. Then a = λ1v1 + · · ·+λqvq, λi ≥ 0 for all i. Thus, setting b =

∑i λi,

one has |a| = bd. We claim that (a, �b�) belongs to R+(I). Let u1, . . . , us


be the characteristic vectors of the minimal vertex covers of C. Since Q(A)is integral, by Theorem 14.1.1, we can write

R+(I) = H+e1 ∩H

+e2 ∩ · · · ∩H

+en+1∩H+

�1∩H+

�2∩ · · · ∩H+

�s, (14.8)

where �i = (ui,−1) for 1 ≤ i ≤ s. Notice that (a, b) ∈ R+(I), thus usingEq. (14.8) we get that 〈a, ui〉 ≥ b for all i. Hence 〈a, ui〉 ≥ �b� for all ibecause 〈a, ui〉 is an integer for all i. Using Eq. (14.8) again we get that(a, �b�) ∈ R+(I), as claimed. By Theorem 14.3.6 the Rees algebra R[It]is normal. Then, applying Proposition 14.2.3, we obtain that (a, �b�) is inNA′. There are nonnegative integers η1, . . . , ηq and ρ1, . . . , ρn such that

(a, �b�) = η1(v1, 1) + · · ·+ ηq(vq, 1) + ρ1e1 + · · ·+ ρnen.

Hence it is seen that |a| = �b�d +∑i ρi = bd and consequently ρi = 0 for

all i and b = �b�. It follows at once that a ∈ NA as required.(c): By part (a) one has Δr(A) = 1. Thus the invariant factors of A are

all equal to 1 (see Theorem 1.3.15), i.e., the Smith normal canonical formof A is an identity matrix. �

This result and the Conforti–Cornuejols conjecture suggest the followingweaker conjecture.

Conjecture 14.4.12 [185] If C is a uniform clutter with the packing prop-erty, then either of the following equivalent conditions hold:

(a) Zn/ZA is a free group.

(b) A diagonalizes over Z to an identity matrix.

This conjecture will be proved later for d-uniform clutters with a perfectmatching and α0(C) = 2 (see Theorem 14.4.15).

Corollary 14.4.13 Let C be a uniform clutter. Then C satisfies the max-flow min-cut property if and only if Q(A) is integral and NA = R+A∩ Zn.

Proof. ⇒) It follows from Theorems 14.4.11 and 14.3.6.⇐) By Proposition 14.3.30 we need only show K[Ft] = A(P). The

inclusion “⊂” is clear. To show the inclusion “⊃” take xatb ∈ A(P). Thena ∈ bP ∩ Zn. From NA = R+A∩ Zn it is seen that a =

∑qi=1 ηivi for some

ηi ∈ N such that∑

i ηi = b. Thus xatb ∈ K[Ft], as required. �

Corollary 14.4.14 Let C be a d-uniform clutter. If C has the max-flowmin-cut property, then C has a perfect matching if and only if n = dα0(C).

Proof. ⇒) As Q(A) is integral, from the proof of Proposition 14.4.9 weobtain the equality n = dα0(C).

604 Chapter 14

⇐) We set g = α0(C). Let B be the incidence matrix of C∨. As Q(A) isintegral, by Corollary 14.2.22, we get that Q(B) is an integral polyhedron.Consequently, by Theorem 1.1.42, we can write

Q(B) = Rn+ + conv(v1, . . . , vq),

Therefore, since the rational vector 1/g is in Q(B), we can write

1/g = δ + μ1v1 + · · ·+ μqvq (δ ∈ Rn+; μi ≥ 0; μ1 + · · ·+ μq = 1).

Hence n/g = |δ|+(∑qi=1 μi)d = |δ|+d. Since n = dg, we obtain that δ = 0.

Thus the vector 1 is in R+A∩Zn. By Theorem 14.4.11(b) this intersectionis equal to NA. Then we can write 1 = η1v1+ · · ·+ηqvq, for some η1, . . . , ηqin N. Hence it is readily seen that C has a perfect matching. �

The next result gives some support to Conjecture 14.3.19.

Theorem 14.4.15 [119] Let C be a d-uniform clutter with a perfect match-ing such that C has the packing property and α0(C) = 2. Then:

(a) A diagonalizes over Z to an “identity matrix.”

(b) If K[Ft] is normal, then C has the max-flow min-cut property.

(c) If A is linearly independent, then C has the max-flow min-cut property.

Proof. (a): By Lehman theorem Q(A) is integral; see Theorem 14.3.17.Thus by Proposition 14.4.10 there is a perfect matching f1, f2 of X withX = f1 ∪ f2 and there is a partition X1, . . . , Xd of X such that Xi is aminimal vertex cover of C for all i, |Xi| = 2 for all i, and

|supp(xvi ) ∩Xk| = 1 ∀ i, k. (14.9)

Thus we may assume that Xi = {x2i−1, x2i} for i = 1, . . . , d. Notice thatn = 2d because X = f1 ∪ f2.

By induction on r = rank(A). Since 1 is the sum of the first two rows ofA it suffices to prove that 1 is the only invariant factor of A� or equivalentlythat the Smith normal form of A� is the “identity.” Let w1, . . . , w2d be thecolumns of A� and let Vi be the linear space generated by w1, . . . , w2i. Fork odd, one has wk + wk+1 = 1. Hence if k is odd and we remove columnswk and wk+1 from A� we obtain a submatrix whose rank is greater than orequal to r − 1. Thus after permuting columns we may assume

dim(Vi) =

{i + 1 if 1 ≤ i ≤ r − 1,r if r ≤ i. (14.10)

Let J be the monomial ideal defined by the rows of [w1, . . . , w2(r−1)]. Then aminimal set of generators of J consists of monomials of degree r−1 and will


still satisfy condition in Eq. (14.9) with d replaced by r− 1. Furthermore Jsatisfies the packing property because it is a minor of I. If [w1, . . . , w2(r−1)]

diagonalizes (over the integers) to the identity matrix so does A�. Thereforewe may assume d = r − 1 and I = J .

Let B be the matrix [w1, . . . , w2(d−1)] and let I ′ be the monomial idealdefined by the rows of B; that is, I ′ is obtained from I making x2d−1 = 1and x2d = 1. Notice that by induction B diagonalizes to an identity matrix[Ir−1,0] of order r − 1. Since I has the Konig property we may permuterows and columns and assume that the matrix A� is written as:

10 10 10 · · · 10 10 ←01 01 01 · · · 01 01© © © · · · © 10 ←...

......

......

© © © · · · © 10 ←© © © · · · © 01...

......

...© © © · · · © 01︸︷︷︸

B

where either a pair 1 0 or 0 1 must occur in the places marked with a circleand such that the number of 1′s in the last column is greater than or equalto the number of 1′s in any other column. Consider the matrix C obtainedfrom A by removing the rows whose penultimate entry is equal to 1 (theseare marked above with an arrow) and removing the last column. Let K bethe monomial ideal defined by the rows of C; that is, K is obtained from Iby making x2d−1 = 0 and x2d = 1. By the choice of the last column andbecause of Eq. (14.10) it is seen that K has height two. Since K is a minorof I it has the Konig property. As a consequence using row operations A�

can be brought to the form:

10 10 10 · · · 10 1010 10 10 · · · 10 01© © © · · · © 10...

......

...© © © · · · © 10© © © · · · © 01...

......

...© © © · · · © 01︸︷︷︸

B1

606 Chapter 14

where B1 has rank r − 1 and diagonalizes to an identity. Therefore it isreadily seen that this matrix reduces to

Ir−1 0 0 · · · 0 0 00 0 0 · · · 0 1 −10 0 0 · · · 0 a1 b1...

......

......

0 0 0 · · · 0 ar br

To finish the proof observe that by rank considerations (see Eq. (14.10))this matrix reduces to [Ir,0], as required.

(b): By part (a) we have Δr(A) = 1. Consider the matrix B with columnvectors (v1, 1), . . . , (vq, 1). Since |vi| = d for all i, one has Δr(B) = 1 (seeExercise 14.4.23). Then by Theorem 9.3.25 this condition is equivalent tothe equalityK[Ft] = A(P ). HenceK[Ft] = A(P) becauseK[Ft] is a normaldomain. By Lehman theorem (Theorem 14.3.17), Q(A) is integral. Thus Csatisfies max-flow min-cut by Proposition 14.3.30 and Theorem 14.3.6.

(c): xv1 t, . . . , xvq t are algebraically independent and K[Ft] is normalbecause it is a polynomial ring. Hence the result follows from (b). �

Recall that I is called minimally non-normal if I is not normal and allits proper minors are normal.

Theorem 14.4.16 Let C be a d-uniform clutter and {Xi}di=1 a partition ofX such that Xi = {x2i−1, x2i} is a minimal vertex cover of C for all i. Then

(a) rank(A) ≤ d+ 1.

(b) If C is a minimal vertex cover of C, then 2 ≤ |C| ≤ d.(c) If C satisfies the Konig property and there is a minimal vertex cover C

of C with |C| = d ≥ 3, then rank(A) = d+ 1.

(d) If I = I(C) is minimally non-normal and C has the packing property,then rank(A) = d+ 1.

Proof. (a): For each odd integer k the sum of rows k and k + 1 of thematrix A is equal to 1. Thus the rank of A is bounded by d+ 1.

(b): By the pigeon hole principle, any C ∈ C∨ satisfies 2 ≤ |C| ≤ d.(c): Notice that C contains exactly one vertex of each Xj because Xj �⊂

C. Thus we may assume C = {x1, . . . , x2d−1}. Consider xα := x2 · · ·x2dand notice that xkx

α ∈ I for each xk ∈ C because xkxα is in every minimal

prime of I. Writing xk = x2i−1 with 1 ≤ i ≤ d we conclude that themonomial xαi = x2 · · ·x2(i−1)x2i−1x2(i+1) · · ·x2d is a minimal generator ofI. Thus we may assume xαi = xvi for i = 1, . . . , d. The vector 1 belongsto QA because C has the Konig property. It follows that the matrix withrows v1, . . . , vd,1 has rank d+ 1, as required.


(d): Let xαtb be a minimal generator of R[It] not in R[It] and let m =xα. Using Proposition 14.3.29, one has deg(m) = bd. By hypothesis m is

the only associated prime of N = Ib/Ib. Hence, by Corollary 2.1.29, wehave

rad(ann(N)) =⋂

p∈Ass(N)

p = m = (x1, . . . , xn)

and mr ⊂ ann(N) for some r > 0. Thus for i odd we can write

xrixα = (xv1 )a1 · · · (xvq )aqxδ,

where a1 + · · ·+ aq = b and deg(xδ) = r. If we write xδ = xs1i xs2i+1x

γ withxi, xi+1 not in the support of xγ , making xj = 1 for j /∈ {i, i+ 1}, it is nothard to see that r = s1 + s2 and γ = 0. Thus we get an equation:

xs2i xα = (xv1)a1 · · · (xvq )aqxs2i+1

with s2 > 0. Using a similar argument we obtain an equation:

xw1

i+1xα = (xv1)b1 · · · (xvq )bqxw1

i with w1 > 0.

Hence, xs2+w1

i+1 (xv1)a1 · · · (xvq )aq = xs2+w1

i (xv1)b1 · · · (xvq )bq . As Zn/ZA istorsion-free (Theorem 14.4.15), we get ei − ei+1 ∈ ZA for i odd. Finally toconclude that rank(A) = d+ 1 notice that 1 ∈ ZA. �

The next theorem gives an interesting class of uniform clutters, comingfrom combinatorial optimization, that satisfy Conjecture 9.6.22.

Theorem 14.4.17 If A is a balanced matrix and C is a d-uniform clutter,then any regular triangulation of the cone R+A is weakly unimodular.

Proof. Let A1, . . . ,Am be the elements of a regular triangulation of R+A.Then dimR+Ai = dimR+A and Ai is linearly independent for all i. LetCi be the subclutter of C whose edges correspond to the vectors in Ai, andlet Ai be the incidence matrix of Ci. As Ai is a balanced matrix, usingCorollary 14.3.11 and Theorem 14.3.6, we get that the clutter Ci has themax-flow min-cut property. Hence by Theorem 14.4.11 one has Δr(Ai) = 1,where r is the rank of A. Thus the invariant factors of Ai are all equal to 1(see Theorem 1.3.15). Therefore by the fundamental structure theorem forfinitely generated abelian groups (see Theorem 1.3.16) the quotient groupZn/ZAi is torsion-free for all i. Notice that

dimR+A = rankZA and dimR+Ai = rankZAi

for all i. Since r is equal to dimR+A. It follows that the quotient groupZA/ZAi is torsion-free and has rank 0 for all i, consequently ZA = ZAi forall i, i.e., the triangulation is weakly unimodular. �

608 Chapter 14

Definition 14.4.18 Let K[Ft] = K[xv1t, . . . , xvq t] be a homogeneous sub-ring with the standard grading induced by deg(xatb) = b. Let

K[t1, . . . , tq]/IA � K[Ft], ti �→ xvit,

be a presentation of K[Ft]. The regularity of K[Ft], denoted reg(K[Ft]), isdefined as the regularity of K[t1, . . . , tq]/IA as an K[t1, . . . , tq]-module.

Theorem 14.4.19 Let C be a d-uniform unmixed clutter with g = α0(C).If C has the max-flow min-cut property, then K[Ft] = A(P), the a-invariantof A(P) is bounded from above by −g and reg(A(P)) ≤ (d− 1)(g − 1).

Proof. Setting B = {(vi, 1)}qi=1 and A′ = B ∪{ei}ni=1, one has the equality

R+B = RB ∩ R+A′. (14.11)

By Theorem 14.3.6, R[I(C)t] is normal and Q(A) is integral. Hence, usingTheorem 9.3.31, we obtain K[Ft] = A(P) and A(P) becomes a standardgraded K-algebra. By Theorems 14.1.1 and 14.2.7, we obtain the equality

R+(I) = H+e1 ∩ · · · ∩H

+en+1∩H+

(u1,−1) ∩ · · · ∩H+(us,−1), (14.12)

where u1, . . . , us are the characteristic vectors of the minimal vertex coversof C. Hence, by Proposition 14.4.4, there are X1, . . . , Xd mutually disjointminimal vertex covers of C of size g such that X = ∪di=1Xi. Notice that|Xi ∩ f | = 1 for 1 ≤ i ≤ d and f ∈ E(C). We may assume that Xi = Ci for1 ≤ i ≤ d. Therefore, using Eqs. (14.11) and (14.12), we get

R+B = RB ∩H+e1 ∩ · · · ∩H

+en ∩H

+en+1∩(⋂i∈I

H+(ui,−1)

), (14.13)

where i ∈ I if and only ifH+(ui,−1) defines a proper face of R+B. The Ehrhart

ring A(P) is Cohen–Macaulay (Theorem 9.1.6). Then, by Theorem 9.1.5and Eq. (14.13), the canonical module of A(P) is the ideal given by

ωA(P) = ((xt)a| a ∈ RB; ai ≥ 1 ∀ i; 〈(uk,−1), a〉 ≥ 1 for k ∈ I), (14.14)

where (xt)a = xa11 · · ·xann tan+1. The a-invariant of A(P) is given by

a(A(P)) = −min{ i | (ωA(P))i �= 0}, (14.15)

see Proposition 5.2.3. Take an arbitrary monomial xatb = xa11 · · ·xann tb inthe ideal ωA(P). By Eqs. (14.13) and (14.14), the vector (a, b) is in R+Band ai ≥ 1 for all i. Thus we can write (a, b) =

∑qi=1 λi(vi, 1) with λi ≥ 0

for all i. Since 〈vi, uk〉 = 1 for all i, k, we obtain

g = |uk| ≤∑xi∈Ck

ai = 〈a, uk〉 =q∑i=1

λi〈vi, uk〉 =q∑i=1

λi = b


for 1 ≤ k ≤ d. This means that deg(xatb) ≥ g. Thus −a(A(P)) ≥ g, asrequired. Next we show that reg(A(P)) ≤ (d − 1)(g − 1). Since A(P) isCohen–Macaulay, by Theorem 6.4.1, we have

reg(A(P)) = dim(A(P)) + a(A(P)) ≤ dim(A(P))− g. (14.16)

Using that 〈vi, uk〉 = 1 for all i, k, by induction on d, it is seen thatrank(A) ≤ g+(d−1)(g−1). Thus, using the fact that dim(A(P)) = rank(A)and Eq. (14.16), we get reg(A(P)) ≤ (d− 1)(g − 1). �

Exercises

14.4.20 Let C be the clutter with vertex set X = {x1, . . . , x9} and edges:

f1 = {x1, x2}, f2 = {x3, x4, x5, x6}, f3 = {x7, x8, x9},f4 = {x1, x3}, f5 = {x2, x4}, f6 = {x5, x7}, f7 = {x6, x8}.

Prove the following: the incidence matrix of C is a balanced matrix, Q(A)is integral, |C ∩ fi| ≥ 2 for any C ∈ C∨ and for any i, C has a perfectmatching, α0(C \ {x9}) = α0(C) = 4, C is not vertex critical, and theuniformity hypothesis is essential in Propositions 14.4.4 and 14.4.9.

14.4.21 Let A be the incidence matrix of a cycle of length 3 and let B bethe matrix obtained from A by adjoining the row 1. Prove that det(A) = 2and Δ3(B) = 1. In particular A and B are not equivalent over Z.

14.4.22 Consider the clutter C whose incidence matrix is⎡⎢⎢⎢⎢⎣1 0 0 10 1 0 10 0 1 10 1 1 01 0 1 0

⎤⎥⎥⎥⎥⎦ .Prove that C satisfies max-flow min-cut, A is not equivalent over Z to anidentity matrix, and the columns of A do not form a Hilbert basis for thecone they generate. Thus the uniformity hypothesis is essential in the twostatements of Theorem 14.4.11.

14.4.23 Let A = {v1, . . . , vq} ⊂ Zn. If |vi| = d �= 0 for all i and Zn/ZA isa torsion-free Z-module, then Zn+1/Z{(v1, 1)), . . . , (vq, 1)} is torsion-free.

14.4.24 Let G be an unmixed bipartite graph, let A∨ = {u1, . . . , us} be theset of column vectors of the incidence matrix of G∨, and let P∨ = conv(A∨).Then K[xu1t, . . . , xus t] = A(P∨) and reg(A(P∨)) ≤ (|V (G)|/2)− 1.

610 Chapter 14

14.4.25 If we do not require that |vi| = d for all i in Theorem 14.4.17, givean example to show that this theorem is false even if K[F ] is homogeneous.

14.4.26 [119] Let {u1, . . . , ur} be the set of all characteristic vectors of thecollection of bases B of a transversal matroidM. Then the polyhedral coneR+{u1, . . . , ur} has a weakly unimodular regular triangulation.

14.5 Clique clutters of comparability graphs

In this section we prove that the clique clutter of a comparability graphsatisfies the max-flow min-cut property. We also present some classicalcombinatorial results on posets, e.g., Dilworth decomposition theorem anda min-max theorem of Menger.

Let P = (X,≺) be a partially ordered set (poset for short) on the finitevertex set X = {x1, . . . , xn} and let G be its comparability graph. Recallthat the vertex set of G is X and the edge set of G is the set of all unorderedpairs {xi, xj} such that xi and xj are comparable. A clique of G is a subsetof the set of vertices that induces a complete subgraph.

Lemma 14.5.1 If G1 (resp. cl(G)1) is the graph (resp. clutter) obtainedfrom G (resp. cl(G)) by duplicating the vertex x1, then cl(G)1 = cl(G1).

Proof. Let y1 be the duplication of x1. We now prove the inclusionE(cl(G)1) ⊂ E(cl(G1)). The other inclusion follows readily using similararguments. Take e ∈ E(cl(G)1).

Case (i): Assume y1 /∈ e. Then e ∈ E(cl(G)). Clearly e is a clique ofG1. If e /∈ E(cl(G1)), then e can be extended to a maximal clique of G1.Hence e ∪ {y1} must be a clique of G1. Note that x1 /∈ e because {x1, y1}is not an edge of G1. Then e ∪ {x1} is a clique of G, a contradiction.

Case (ii): Assume y1 ∈ e. Then there is f ∈ E(cl(G)), with x1 ∈ f , suchthat e = (f \ {x1}) ∪ {y1}. Since {x, x1} ∈ E(G) for any x in f \ {x1}, onehas that {x, y1} ∈ E(G1) for any x in f \ {x1}. Then e is a clique of G1.If e is not a maximal clique of G1, there is x /∈ e which is adjacent in G toany vertex of f \ {x1} and x is adjacent to y1 in G1. In particular x �= x1.Then x is adjacent in G to x1 and consequently x is adjacent in G to anyvertex of f , a contradiction because f is a maximal clique of G. �

Let D be a digraph; that is, D consists of a finite set V (D) of verticesand a set E(D) of ordered pairs of distinct vertices called edges. Let A, Bbe two sets of vertices of D. For use below recall that a (directed) pathof D is called an A–B path if it runs from a vertex in A to a vertex in B.A common vertex of A and B is also an A–B path. A set C of verticesis called an A–B disconnecting set if C intersects each A–B path. Menger[316] gave a min-max theorem for the maximum number of disjoint A–Bpaths in an undirected graph. This theorem also holds for digraphs.


Theorem 14.5.2 (Menger’s theorem, see [373, Theorem 9.1]) Let D be adigraph and let A, B be two subsets of V (D). Then the maximum num-ber of vertex-disjoint A–B paths is equal to the minimum size of an A–Bdisconnecting vertex set.

Proof. Let k be the minimum size of an A–B disconnecting vertex set andlet � be the maximum number of vertex-disjoint A–B paths. Clearly � ≤ k.Equality will be shown by induction on E(D), the number of edges of D.The case E(D) = ∅ is trivial. Pick an edge a = (u, v) of D. If each A–Bdisconnecting vertex set of D \ a has size at least k, then inductively thereare k vertex-disjoint A–B paths in D \ a, hence in D. Thus we may assumethat there exists an A–B disconnecting vertex set C of D \ a of size lessthan or equal to k − 1. Then C ∪ {u} and C ∪ {v} are A–B disconnectingvertex sets of D of size k.

Now, each A–(C ∪ {u}) disconnecting vertex set A of D \ a has size atleast k, as A is an A–B disconnecting vertex set of D. Indeed, if P is anA–B path of D, then P intersects C ∪ {u}, and hence P contains an A–(C ∪ {u}) path. So P intersects A. Therefore by induction D \ a containsk vertex-disjoint A–(C ∪ {u}) paths P0, . . . ,Pk−1. Similarly D \ a containsk vertex-disjoint (C ∪ {v})–B paths Q0, . . . ,Qk−1. If pi is the last vertexof Pi, we may assume—by reducing the size of Pi if necessary—that anyother vertex of Pi is not in C ∪ {u}. Similarly if qi is the first vertex ofQi, we may assume that all other vertices of Qi are not in C ∪ {v}. Since|C| = k − 1, we have

P0a0��

uQ0

v��

b0Piai��

ciQi

ci��

bi

where C = {c1, . . . , ck−1}, ai ∈ A, bi ∈ B for i = 0, . . . , k − 1. Any pathin the first collection intersects any path in the second collection only in C,since otherwise D \ a contains an A–B path avoiding C. Therefore we canpairwise concatenate the paths to obtain k vertex-disjoint A–B paths. Thisproof was adapted from [195]. �

Theorem 14.5.3 [118] Let P = (X,≺) be a poset on the vertex set X andlet G be its comparability graph. If C = cl(G) is the clutter of maximalcliques of G, then C satisfies the max-flow min-cut property.

Proof. We can regard P as a transitive digraph without cycles of lengthtwo with vertex set X and edge set E(P ), i.e., the edges of P are orderedpairs (a, b) of distinct vertices with a ≺ b such that:

(i) (a, b) ∈ E(P ) and (b, c) ∈ E(P ) ⇒ (a, c) ∈ E(P ), and(ii) (a, b) ∈ E(P ) ⇒ (b, a) /∈ E(P ).

612 Chapter 14

Note that by (i) P is acyclic; that is, it has no directed cycles. We setX = {x1, . . . , xn}. Let x1 be a vertex of P and let y1 be a new vertex.Consider the digraph P 1 with vertex set X1 = X ∪ {y1} and edge set

E(P 1) = E(P ) ∪ {(y1, x)| (x1, x) ∈ E(P )} ∪ {(x, y1)| (x, x1) ∈ E(P )}.

The digraph P 1 is transitive. Indeed let (a, b) and (b, c) be two edges ofP 1. If y1 /∈ {a, b, c}, then (a, c) ∈ E(P ) ⊂ E(P 1) because P is transitive.If y1 = a, then (x1, b) and (b, c) are in E(P ). Hence (x1, c) ∈ E(P ) and(y1, c) ∈ E(P 1). The cases y1 = b and y1 = c are treated similarly. ThusP 1 defines a poset P 1 = (V (P 1),≺1). The comparability graph H of P 1 isprecisely the graph G1 obtained from G by duplicating the vertex x1 by thevertex y1. From Lemma 14.5.1 we get that cl(G)1 = cl(G1), where cl(G)1 isthe clutter obtained from cl(G) by duplicating the vertex x1 by the vertexy1. Altogether we obtain that the clutter cl(G)1 is the clique clutter of thecomparability graph G1 of the poset P 1.

By Theorem 14.3.6(vii) it suffices to prove that cl(G)w has the Konigproperty for all w ∈ Nn. Since duplications commute with deletions wemay assume that w = (w1, . . . , wr , 0, . . . , 0), where wi ≥ 1 for i = 1, . . . , r.Consider the clutter C1 obtained from cl(G) by duplicating wi − 1 timesthe vertex xi for i = 1, . . . , r. We denote the vertex set of C1 by X1. Bysuccessively applying the fact that cl(G)1 = cl(G1), we conclude that thereis a poset P1 with comparability graph G1 and vertex set X1 such thatC1 = cl(G1). As before we regard P1 as a transitive acyclic digraph.

Let A and B be the set of minimal and maximal elements of the poset P1,i.e., the elements of A and B are the sources and sinks of P1, respectively.We set S = {xr+1, . . . , xn}. Consider the digraph D whose vertex set isV (D) = X1 \ S and whose edge set is defined as follows. A pair (x, y) inV (D)×V (D) is in E(D) if and only if (x, y) ∈ E(P1) and there is no vertexz in X1 with x ≺ z ≺ y. Notice that D is a sub-digraph of P1 which is notnecessarily the digraph of a poset. We set A1 = A\S and B1 = B \S. Notethat Cw = C1 \S, the clutter obtained from C1 by removing all vertices of Sand all edges sharing a vertex with S. If every edge of C1 intersects S, thenE(Cw) = ∅ and there is nothing to prove. Thus we may assume that there isa maximal clique K of G1 disjoint form S. Note that by the maximality ofK and by the transitivity of P1 we get that K contains at least one sourceand one sink of P1, i.e., A1 �= ∅ and B1 �= ∅ (see argument below).

The maximal cliques of G1 not containing any vertex of S correspondexactly to the A1–B1 paths of D. Indeed let c = {v1, . . . , vs} be a maximalclique of G1 disjoint from S. Consider the sub-poset Pc of P1 inducedby c. Note that Pc is a tournament, i.e., Pc is an oriented graph (no-cycles of length two) such that any two vertices of Pc are comparable. ByExercise 7.1.21 any tournament has a Hamiltonian path, i.e., a spanning


oriented path. Therefore we may assume that

v1 ≺ v2 ≺ · · · ≺ vs−1 ≺ vs

By the maximality of c we get that v1 is a source of P1, vs is a sink ofP1, and (vi, vi+1) is an edge of D for i = 1, . . . , s− 1. Thus c is an A1–B1

path of D, as required. Conversely let c = {v1, . . . , vs} be an A1–B1 pathof D. Clearly c is a clique of P1 because P1 is a poset. Assume that c isnot a maximal clique of G1. Then there is a vertex v ∈ X1 \ c such that vis related to every vertex of c. Since v1, vs are a source and a sink of P1,respectively, we get v1 ≺ v ≺ vs. We claim that vi ≺ v for i = 1, . . . , s.By induction assume that vi ≺ v for some 1 ≤ i < s. If v ≺ vi+1, thenvi ≺ v ≺ vi+1, a contradiction to the fact that (vi, vi+1) is an edge of D.Thus vi+1 ≺ v. Making i = s we get that vs ≺ v, a contradiction. Thisproves that c is a maximal clique of G1. Therefore, since the maximal cliquesof G1 not containing any vertex in S are exactly the edges of Cw = C1 \ S,by Menger’s theorem (see Theorem 14.5.2) we obtain that Cw satisfies theKonig property. �

Corollary 14.5.4 Let P = (X,≺) be a poset on the vertex X and let Gbe its comparability graph. Then the maximum number of disjoint maximalcliques of G is equal to the minimum size of a set intersecting all maximalcliques of G.

Proof. By Theorem 14.5.3, the clique clutter of G satisfies the max-flowmin-cut property. Then by Theorem 14.3.6, the clique clutter of G satisfiesthe Konig property. �

Let P = (X,≺) be a poset with vertex set X . A subset L of X is calleda chain of P if x ≺ y or y ≺ x for all x �= y in L. A set A of X is called ananti-chain of P if x and y are not related for all x �= y in A.

Proposition 14.5.5 Let P = (X,≺) be a poset with vertex set X. Thenthe maximum size of a chain of P equals the minimum number of disjointanti-chains into which X can be decomposed (covered).

Proof. Let L be a chain of maximum size and let m = |L|. Clearlymax ≤ min, because if X1, . . . , Xk is a decomposition of X into disjointanti-chains, then |L∩Xi| ≤ i for all i, so |L| ≤ k. Let x ∈ X and let h(x) bethe size of the longest chain of P with end vertex x. Since any two verticesx, y with h(x) = h(y) are incomparable, we get a decomposition

X = {x|h(x) = 1} ∪ {x|h(x) = 2} ∪ · · · ∪ {x|h(x) = m}

into at most m disjoint anti-chains, so min ≤ max. �

614 Chapter 14

Corollary 14.5.6 If G is the comparability graph of a poset P = (X,≺),then G is perfect.

Proof. Let ω(G) and χ(G) be the clique number and the chromatic numberof G, respectively. By Proposition 14.5.5, we have ω(G) = χ(G). As theclass of comparability graphs is closed under taking induced subgraphs, weget that G is a perfect graph. �

Theorem 14.5.7 (Dilworth decomposition theorem) Let P = (X,≺) bea poset on X. Then the maximum size of an anti-chain is equal to theminimum number of disjoint chains into which X can be decomposed.

Proof. Let G be the comparability graph of P . By Corollary 14.5.6, G isperfect. Hence by the weak perfect graph theorem (see Theorem 13.6.1),the complement G of G is also perfect. In particular ω(G) = χ(G) and theresult follows readily. �

Recall that the vertex-clique matrix of a graph G is the incidence matrixof the clique clutter of G.

Corollary 14.5.8 Let G be a comparability graph and let A be the vertex-clique matrix of G. Then the following polytopes are integral :

P(A) = {x|x ≥ 0; xA ≤ 1} and Q(A) = {x|x ≥ 0; xA ≥ 1}

Proof. G is a perfect graph by Corollary 14.5.6. Hence the polytope P(A)is equal to the independence polytope of G (Theorem 13.6.1). In particularP(A) is integral. By Theorem 14.5.3 the clique clutter cl(G) has the max-flow min-cut property. Thus applying Theorem 14.3.6, we get that Q(A)has only integral vertices. �

Conjecture 14.5.9 Let G be a perfect graph and let A be the vertex-cliquematrix of G. If the polyhedron Q(A) = {x|x ≥ 0;xA ≥ 1} is integral, thenthe system x ≥ 0;xA ≥ 1 is TDI.

The conjecture holds when the clique clutter of a perfect graph G isuniform (see Corollary 14.3.32).

Corollary 14.5.10 If G is a comparability graph and cl(G) is its cliqueclutter, then the edge ideal of cl(G) is normally torsion-free and normal.

Proof. It follows from Theorems 14.5.3 and 14.3.6. �

Definition 14.5.11 Let d ≥ 2, g ≥ 2 be two integers and let

X1 = {x11, . . . , x1g}, X2 = {x21, . . . , x2g}, . . . , Xd = {xd1, . . . , xdg}

be disjoint sets of variables. A clutter C with vertex set X = X1 ∪ · · · ∪Xd

and edge set E(C) = {{x1i1 , x2i2 , . . . , x

did}| 1 ≤ i1 ≤ i2 ≤ · · · ≤ id ≤ g} is

called a complete admissible d-uniform clutter (cf. Definition 6.6.4).


Complete admissible uniform clutters were first studied in [160] and morerecently in [205, 325] (see Section 6.6).

Theorem 14.5.12 If C is a complete admissible d-uniform clutter, then itsedge ideal I(C) is normally torsion-free and normal.

Proof. Let P = (X,≺) be the poset with vertex set X and partial ordergiven by x�k ≺ xmp if and only if 1 ≤ � < m ≤ d and 1 ≤ k ≤ p ≤ g. Wedenote the comparability graph of P by G. We claim that E(C) = E(cl(G)),where cl(G) is the clique clutter of G. Let f = {x1i1 , x

2i2, . . . , xdid} be an edge

of C, i.e., we have 1 ≤ i1 ≤ i2 ≤ · · · ≤ id ≤ g. Clearly f is a clique of G.If f is not maximal, then there is a vertex x�k not in f which is adjacent inG to every vertex of f . In particular x�k must be comparable to x�i� , whichis impossible. Thus f is an edge of cl(G). Conversely let f be an edgeof cl(G). We can write f = {xk1i1 , x

k2i2, . . . , xksis }, where k1 < · · · < ks and

i1 ≤ · · · ≤ is. By the maximality of f we get that s = d and ki = i fori = 1, . . . , d. Thus f is an edge of C. Hence by Corollary 14.5.10 we obtainthat I(C) is normally torsion-free and normal. �

Exercises

14.5.13 Let G be a graph. Let G1 = G \ x1 (resp. cl(G)1 = cl(G) \ x1)be the graph (resp. clutter) obtained from G (resp. cl(G)) by deleting thevertex x1. If G is a cycle of length three, prove that cl(G)1 �= cl(G1).

14.5.14 Let cl(G) be the clique clutter of a perfect graph G. If cl(G) isd-uniform and satisfies the packing property, prove that cl(G) satisfies themax-flow min-cut property.

14.6 Duality and integer rounding problems

Throughout this section we keep the notation of Section 14.1 but allow I tobe a monomial ideal and A to be a matrix with entries in N.

In this section we characterize the normality of a monomial ideal interms of integer rounding properties of linear systems and show a dualitytheorem for monomial subrings. We show that the Rees algebra of the idealof covers of a perfect graph is normal.

Let I be a monomial ideal of R generated by xv1 , . . . , xvq , and let A bethe n × q matrix with column vectors v1, . . . , vq. In what follows considerthe homogeneous monomial subring

S = K[xw1t, . . . , xwr t] ⊂ R[t],

where {w1, . . . , wr} is the set of all α ∈ Nn such that 0 ≤ α ≤ vi for some i.

616 Chapter 14

Definition 14.6.1 Given a polyhedron Q in Rn, its blocking polyhedron,denoted by B(Q), is defined as:

B(Q) := {z ∈ Rn| z ≥ 0; 〈z, x〉 ≥ 1 for all x in Q}.

Lemma 14.6.2 If Q = Q(A), then B(Q) = Rn+ + conv(v1, . . . , vq).

Proof. “⊂”: Take z in B(Q), then 〈z, x〉 ≥ 1 for all x ∈ Q and z ≥ 0. Letu1, . . . , ur be the vertex set of Q. In particular 〈z, ui〉 ≥ 1 for all i. Then〈(z, 1), (ui,−1)〉 ≥ 0 for all i. From Theorem 1.4.2, we get that (z, 1) is inthe cone generated by A′. Thus z is in Rn+ + conv(v1, . . . , vq).

“⊃′′: This inclusion is clear. �

Definition 14.6.3 A rational polyhedron Q has the integer decompositionproperty if for each natural number k and for each integer vector a in kQ,a is the sum of k integer vectors in Q, where kQ is equal to {ka| a ∈ Q}.

Theorem 14.6.4 I is normal if and only if the blocking polyhedron B(Q)of Q = Q(A) has the integer decomposition property and all minimal integervectors of B(Q) are columns of A (minimal with respect to ≤).

Proof. By Lemma 14.6.2 we get B(Q) ∩ Qn = Qn+ + convQ(v1, . . . , vq),because B(Q) is a rational polyhedron. From this equality we readily obtain

Ik = ({xa| a ∈ kB(Q) ∩ Zn}) for 0 �= k ∈ N. (14.17)

⇒) Assume that I is normal, i.e., Ik = Ik for k ≥ 1. Let a be anintegral vector in kB(Q). Then, by Eq. (14.17), xa ∈ Ik and a is the sumof k integral vectors in B(Q); that is, B(Q) has the integer decompositionproperty. Take a minimal integer vector a in B(Q). Then xa ∈ I = I andwe can write a = δ + vi for some vi and for some δ ∈ Nn. Thus a = vi bythe minimality of a.⇐) Assume that B(Q) has the integer decomposition property and all

minimal integer vectors of B(Q) are columns of A. Take xa ∈ Ik, i.e., ais an integral vector of kB(Q). Hence a is the sum of k integral vectorsα1, . . . , αk in B(Q). Since any minimal vector of B(Q) is a column of A wemay assume that αi = ci + vi for i = 1, . . . , k. Hence xa ∈ Ik. �

Theorem 14.6.5 If I = I and Q = Q(A), then I is normal if and only ifthe blocking polyhedron B(Q) has the integer decomposition property.

Proof. ⇒) If I is normal, by Theorem 14.6.4, the blocking polyhedronB(Q) has the integer decomposition property.

⇐) Take xa ∈ Ik. By Eq. (14.17), a is an integral vector of kB(Q).Hence a is the sum of k integral vectors α1, . . . , αk in B(Q). By Eq. (14.17),with k = 1, we get that α1, . . . , αk are in I = I. Hence xa ∈ Ik. �


Corollary 14.6.6 If I is the edge ideal of a clutter, then I is normal if andonly if the blocking polyhedron B(Q) has the integer decomposition property.

Proof. Recall that I is an intersection of prime ideals. Thus it is seen thatI = I and the result follows from Theorem 14.6.5. �

Definition 14.6.7 Let A be a matrix with entries in N. The linear systemx ≥ 0;xA ≥ 1 has the integer rounding property if

max{〈y,1〉|Ay ≤ w; y ∈ Nq} = (max{〈y,1〉| y ≥ 0;Ay ≤ w}) (14.18)

for each integral vector w for which the right-hand side is finite.

Systems with the integer rounding property have been widely studied;see [372, Chapter 22], [373, Chapter 5], and the references therein.

Theorem 14.6.8 ([19], [373, p. 82]) The system x ≥ 0;xA ≥ 1 has theinteger rounding property if and only if the blocking polyhedron B(Q) ofQ = Q(A) has the integer decomposition property and all minimal integervectors of B(Q) are columns of A (minimal with respect to ≤).

Corollary 14.6.9 Let I = (xv1 , . . . , xvq ) be a monomial ideal and let A bethe matrix with column vectors v1, . . . , vq. Then I is a normal ideal if andonly if the system xA ≥ 1;x ≥ 0 has the integer rounding property.

Proof. According to Theorem 14.6.8, the system xA ≥ 1;x ≥ 0 has theinteger rounding property if and only if the blocking polyhedron B(Q) ofQ = Q(A) has the integer decomposition property and all minimal integervectors of B(Q) are columns of A (minimal with respect to ≤). Thus theresult follows at once from Theorem 14.6.4. �

This corollary was first observed by N. V. Trung when I is the edge idealof a hypergraph.

Lemma 14.6.10 The following equation holds :

conv(w1, . . . , wr) = Rn+ ∩ (conv(w1, . . . , wr) + R+{−e1, . . . ,−en}).

Proof. The inclusion “⊂” is clear. To show the inclusion “⊃” take a vectorz such that z ≥ 0 and

z = λ1w1 + · · ·+ λrwr − δ1e1 − · · · − δnen, (14.19)

where λi ≥ 0, λ1 + · · · + λr = 1, and δj ≥ 0 for all i, j. Consider thevector z′ = λ1w1 + · · · + λrwr − δ1e1. We set T = conv(w1, . . . , wr) andwi = (wi1, . . . , win). We claim that z′ is in T . We may assume that δ1 > 0,

618 Chapter 14

λi > 0 for all i, and that the first entry wi1 of wi is positive for 1 ≤ i ≤ s andis equal to zero for i > s. From Eq. (14.19) we get λ1w11+ · · ·+λsws1 ≥ δ1.

Case (I): λ1w11 ≥ δ1. Then we can write

z′ =δ1w11

(w1 − w11e1) +

(λ1 −

δ1w11

)w1 + λ2w2 + · · ·+ λrwr.

Notice that w1 − w11e1 is again in {w1, . . . , wr}. Thus z′ is a convex com-bination of w1, . . . , wr, i.e., z

′ ∈ T .Case (II): λ1w11 < δ1. Let m be the largest integer less than or equal to

s such that λ1w11 + · · ·+λm−1w(m−1)1 < δ1 ≤ λ1w11 + · · ·+λmwm1. Then

z′ =m−1∑i=1

λi(wi − wi1e1) +[δ1wm1

−(m−1∑i=1

λiwi1wm1

)](wm − wm1e1) +[

λm −δ1wm1

+

(m−1∑i=1

λiwi1wm1

)]wm +

r∑i=m+1

λiwi.

Notice that wi −wi1e1 is again in {w1, . . . , wr} for i = 1, . . . ,m. Thus z′ isa convex combination of w1, . . . , wr, i.e., z

′ ∈ T . This proves the claim. Wecan apply the argument above to any entry of z or z′, thus we obtain thatz′ − δ2e2 ∈ T . Thus by induction we obtain that z ∈ T , as required. �

Definition 14.6.11 [372, p. 117] Let P be a rational polyhedron in Rn.The antiblocking polyhedron of P is defined as:

T (P) := {z| z ≥ 0; 〈z, x〉 ≤ 1 for all x ∈ P}.

Lemma 14.6.12 If P = {x|x ≥ 0; xA ≤ 1}, then

T (P) = conv(w1, . . . , wr).

Proof. One has the equality P = {z| z ≥ 0; 〈z, wi〉 ≤ 1 ∀i} because for eachwi there is vj such that wi ≤ vj . Hence, by Corollary 1.1.34, we can write

P = {z| z ≥ 0; 〈z, wi〉 ≤ 1 ∀i} = conv(β0, β1, . . . , βm) (14.20)

for some β1, . . . , βm in Qn+ and β0 = 0. From Eq. (14.20) we get

{z| z ≥ 0; 〈z, βi〉 ≤ 1 ∀i} = T (P). (14.21)

Using Eq. (14.20) and noticing that 〈βi, wj〉 ≤ 1 for all i, j, we get

Rn+ ∩ (conv(β0, . . . , βm) + R+{−e1, . . . ,−en}) = {z| z ≥ 0; 〈z, wi〉 ≤ 1 ∀i}.

Hence using this equality and [372, Theorem 9.4] we obtain

Rn+ ∩ (conv(w1, . . . , wr) + R+{−e1, . . . ,−en}) ={z| z ≥ 0; 〈z, βi〉 ≤ 1 ∀i}. (14.22)


Therefore, by Lemma 14.6.10 together with Eqs. (14.21) and (14.22), weconclude that T (P) is equal to conv(w1, . . . , wr), as required. �

Definition 14.6.13 Let A be a matrix with entries in N. The linear systemx ≥ 0;xA ≤ 1 has the integer rounding property if

�min{〈y,1〉| y ≥ 0; Ay ≥ a}� = min{〈y,1〉|Ay ≥ a; y ∈ Nq}

for each integral vector a for which min{〈y,1〉| y ≥ 0; Ay ≥ a} is finite.

Remark 14.6.14 Let A be a matrix with entries in N. Then the linearsystem x ≥ 0;xA ≤ 1 has the integer rounding property if and only if

�min{〈y,1〉| y ≥ 0; Ay ≥ a}� = min{〈y,1〉|Ay ≥ a; y ∈ Nq}

for each vector a ∈ Nn for which min{〈y,1〉| y ≥ 0; Ay ≥ a} is finite. Thisfollows decomposing an integral vector a as a = a+ − a− and noticing thatfor y ≥ 0 we have that Ay ≥ a if and only if Ay ≥ a+

Theorem 14.6.15 ([19], [373, p. 82]) If P = {x|x ≥ 0; xA ≤ 1}, then thesystem xA ≤ 1;x ≥ 0 has the integer rounding property if and only if T (P)has the integer decomposition property and all maximal integer vectors ofT (P) are columns of A (maximal with respect to ≤).

Theorem 14.6.16 The system x ≥ 0; xA ≤ 1 has the integer roundingproperty if and only if the subring K[xw1t, . . . , xwr t] is normal.

Proof. Let P = {x|x ≥ 0; xA ≤ 1} and let T (P) be its antiblockingpolyhedron. According to Lemma 14.6.12 one has

T (P) = conv(w1, . . . , wr). (14.23)

Setting B = {(wi, 1)}ri=1, by Theorem 14.6.15 and Corollary 9.1.3, itsuffices to prove that the equality R+B ∩ ZB = NB holds true if and onlyif T (P) has the integer decomposition property and all maximal integervectors of T (P) are columns of A.

Assume that R+B ∩ ZB = NB. Let b be a natural number and let a bean integer vector in bT (P). Then using Eq. (14.23) it is seen that (a, b) is inR+B. In our situation one has ZB = Zn+1. Hence (a, b) ∈ NB and a is thesum of b integer vectors in T (P). Thus T (P) has the integer decompositionproperty. Assume that a is a maximal integer vector of T (P). It is nothard to see that (a, 1) is in R+B, i.e., (a, 1) ∈ NB. Thus (a, 1) is a linearcombination of vectors in B with coefficients in N. Hence (a, 1) is equal to(wj , 1) for some j. There exists vi such that a = wj ≤ vi. Therefore by themaximality of a, we get a = vi for some i. Thus a is a column of A.

620 Chapter 14

Conversely assume that T (P) has the integer decomposition propertyand that all maximal integer vectors of T (P) are columns of A. Let (a, b) bean integral vector in R+B with a ∈ Nn and b ∈ N. Hence, using Eq. (14.23),we get a ∈ bT (P). Thus a = α1 + · · · + αb, where αi is an integral vectorof T (P) for all i. Since each αi is less than or equal to a maximal integervector of T (P), we get that αi ∈ {w1, . . . , wr}. Then (a, b) ∈ NB. �

Corollary 14.6.17 Let A be the incidence matrix of a uniform clutter. Ifeither linear system x ≥ 0;xA ≤ 1 or x ≥ 0;xA ≥ 1 has the integerrounding property and P = conv(v1, . . . , vq), then

K[xv1t, . . . , xvq t] = A(P).

Proof. In general the inclusion “⊂” holds. Assume that x ≥ 0;xA ≤ 1has the integer rounding property and that every edge of C has d elements.Let w1, . . . , wr be the set of all α ∈ Nn such that α ≤ vi for some i. Thenby Theorem 14.6.16 the subring K[xw1t, . . . , xwr t] is normal. Using thatv1, . . . , vq is the set of wi with |wi| = d, it is not hard to see that A(P) iscontained in K[xv1t, . . . , xvq t].

Assume that x ≥ 0;xA ≥ 1 has the integer rounding property. LetI = I(C) be the edge ideal of C and let R[It] be its Rees algebra. ByCorollary 14.6.9, R[It] is a normal domain. Since the clutter C is uniformthe required equality follows at once from Theorem 9.3.31. �

Corollary 14.6.18 If G is a perfect unmixed graph and v1, . . . , vq are thecharacteristic vectors of the maximal stable sets of G, then the subringK[xv1t, . . . , xvq t] is normal.

Proof. The minimal vertex covers of G are exactly the complements of themaximal stable sets of G. Thus |vi| = d for all i, where d = dim(R/I(G)).On the other hand the maximal stable sets of G are exactly the maximalcliques of G. Thus, by Theorem 13.6.8 and Corollary 14.6.17, the subringK[xv1t, . . . , xvq t] is an Ehrhart ring, and consequently it is normal. �

Proposition 14.6.19 Let I = (xv1 , . . . , xvq ) be a monomial ideal and letv∗i = 1− vi. Then R[It] is normal if and only if the set

Γ = {−e1, . . . ,−en, (v∗1 , 1), . . . , (v∗q , 1)}

is a Hilbert basis.

Proof. Let A′ = {e1, . . . , en, (v1, 1), . . . , (vq, 1)}. Assume that R[It] isnormal. Then A′ is a Hilbert basis. Let (a, b) be an integral vector in R+Γ,with a ∈ Zn and b ∈ Z. Then we can write

(a, b) = μ1(−e1) + · · ·+ μn(−en) + λ1(v∗1 , 1) + · · ·+ λq(v

∗q , 1),


where μi ≥ 0 and λj ≥ 0 for all i, j. Therefore

−(a,−b) + b1 = μ1e1 + · · ·+ μnen + λ1(v1, 1) + · · ·+ λq(vq, 1),

where 1 = e1 + · · ·+ en. As A′ is a Hilbert basis we can write

−(a,−b) + b1 = μ′1e1 + · · ·+ μ′

nen + λ′1(v1, 1) + · · ·+ λ′q(vq, 1),

where μ′i ∈ N and λ′j ∈ N for all i, j. Thus (a, b) ∈ NΓ. This proves that Γ

is a Hilbert basis. The converse can be shown using similar arguments. �

Definition 14.6.20 Let A = (aij) be a matrix with entries in {0, 1}. Itsdual is the matrix A∗ = (a∗ij), where a

∗ij = 1− aij.

The following duality is valid for incidence matrices of clutters. It willbe used later to establish a duality theorem for monomial subrings.

Theorem 14.6.21 If A is the incidence matrix of a clutter and A∗ is itsdual matrix, then x ≥ 0;xA ≥ 1 has the integer rounding property if andonly if x ≥ 0;xA∗ ≤ 1 has the integer rounding property.

Proof. We set Q = {x|x ≥ 0;xA ≥ 1} and P∗ = {x|x ≥ 0;xA∗ ≤ 1}. Ifv∗i = 1− vi for all i, then A∗ is the matrix with column vectors v∗1 , . . . , v

∗q .

Let w∗1 , . . . , w

∗s be the set of all α ∈ Nn such that α ≤ v∗i for some i. Then,

using Lemmas 14.6.2 and 14.6.12, we obtain

B(Q) = Rn+ + conv(v1, . . . , vq) and T (P∗) = conv(w∗1 , . . . , w

∗s).

⇒) Thanks to Theorem 14.6.15 we need only show that T (P∗) has theinteger decomposition property and all maximal integer vectors of T (P∗)are columns of A∗. Let b ∈ N+ and let a be an integer vector in bT (P∗).Then we can write a = b(λ1w

∗1 + · · · + λsw

∗s ) with

∑i λi = 1 and λi ≥ 0.

For each 1 ≤ i ≤ s there is v∗ji in {v∗1 , . . . , v∗q} such that w∗i ≤ v∗ji . Thus for

each i we can write 1− w∗i = vji + δi, where δi ∈ Nn. Therefore

1− a/b = λ1(vj1 + δ1) + · · ·+ λs(vjs + δs).

This means that 1− a/b ∈ B(Q), i.e., b1− a is an integer vector in bB(Q).Hence by Theorem 14.6.8 we can write b1 − a = α1 + · · · + αb for someα1, . . . , αb integer vectors in B(Q), and for each αi there is vki in {v1, . . . , vq}such that vki ≤ αi. Thus αi = vki + εi for some εi ∈ Nn and consequently:

a = (1− vk1) + · · ·+ (1− vkb)− c = v∗k1 + · · ·+ v∗kb − c,

where c = (c1, . . . , cn) ∈ Nn. Notice that v∗k1 + · · ·+ v∗kb ≥ c because a ≥ 0.If c1 ≥ 1, then the first entry of v∗ki is non-zero for some i and we can write

a = v∗k1 + · · ·+ v∗ki−1+ (v∗ki − e1) + v∗ki+1

+ · · ·+ v∗kb − (c− e1).

622 Chapter 14

Since v∗ki − e1 is again in {w∗1 , . . . , w

∗s}, we can apply this argument

recursively to obtain that a is the sum of b integer vectors in {w∗1 , . . . , w

∗s}.

This proves that T (P∗) has the integer decomposition property. Let a bea maximal integer vector of T (P∗). As the vectors w∗

1 , . . . , w∗s have entries

in {0, 1}, we get T (P∗) ∩ Zn = {w∗1 , . . . , w

∗s}. Then a = w∗

i for some i. Asw∗i ≤ v∗j for some j, we get that a = v∗j , i.e., a is a column of A∗.⇐) Thanks to Corollary 14.6.9, the system x ≥ 0;xA ≥ 1 has the integer

rounding property if and only ifR[It] is normal. Thus by Proposition 14.6.19we need only show that the set Γ = {−e1, . . . ,−en, (v∗1 , 1), . . . , (v∗q , 1)} is aHilbert basis. Let (a, b) be an integral vector in R+Γ, with a ∈ Zn andb ∈ Z. Then we can write

(a, b) = μ1(−e1) + · · ·+ μn(−en) + λ1(v∗1 , 1) + · · ·+ λq(v

∗q , 1),

where μi ≥ 0, λj ≥ 0 for all i, j. Hence A∗λ ≥ a, where λ = (λi). Byhypothesis the system x ≥ 0;xA∗ ≤ 1 has the integer rounding property.Then one has

b ≥ �min{〈y,1〉| y ≥ 0; A∗y ≥ a}� = min{〈y,1〉|A∗y ≥ a; y ∈ Nq} = 〈y0,1〉

for some y0 = (yi) ∈ Nq such that |y0| = 〈y0,1〉 ≤ b and a ≤ A∗y0. Then

a = y1v∗1 + · · ·+ yqv

∗q − δ1e1 − · · · − δnen,

where δ1, . . . , δn are in N. Hence we can write

(a, b) = y1(v∗1 , 1)+ · · ·+yq−1(v

∗q−1, 1)+(yq+b−|y0|)(v∗q , 1)−(b−|y0|)v∗q −δ,

where δ = (δi). As the entries of A∗ are in N, the vector −v∗q can be written

as a nonnegative integer combination of −e1, . . . ,−en. Thus (a, b) ∈ NΓ.This proves that Γ is a Hilbert basis. �

The following result is a duality theorem for monomial subrings.

Theorem 14.6.22 [52] Let A be the incidence matrix of a clutter and letv∗i be the vector 1 − vi. If w∗

1 , . . . , w∗s is the set of all α ∈ Nn such that

α ≤ v∗i for some i, then the following conditions are equivalent:

(a) R[It] is normal, where I = (xv1 , . . . , xvq ).

(b) S∗ = K[xw∗1 t, . . . , xw

∗s t] is normal.

(c) {−e1, . . . ,−en, (v∗1 , 1), . . . , (v∗q , 1)} is a Hilbert basis.

(d) x ≥ 0;xA ≥ 1 has the integer rounding property.

(e) x ≥ 0;xA∗ ≤ 1 has the integer rounding property.

Proof. (a) ⇔ (c): This was shown in Proposition 14.6.19. (a) ⇔ (d):This was shown in Corollary 14.6.9. (b) ⇔ (e): This part was shown inTheorem 14.6.16. (d) ⇔ (e): This follows from Theorem 14.6.21. �


Definition 14.6.23 Let I be the edge ideal of a clutter C. The dual of I,denoted by I∗, is the ideal of R generated by all monomials x1 · · ·xn/xesuch that e is an edge of C, where xe =

∏xi∈e xi.

Corollary 14.6.24 Let C be a clutter and let A be its incidence matrix. IfP = {x|x ≥ 0;xA ≤ 1} is an integral polytope and I = I(C), then R[I∗t]and S = K[xw1t, . . . , xwr t] are normal.

Proof. Since P has only integral vertices, by a theorem of Lovasz (seeTheorem 13.6.8) the system x ≥ 0;xA ≤ 1 is totally dual integral, i.e., theminimum in the LP-duality equation

max{〈a, x〉|x ≥ 0;xA ≤ 1} = min{〈y,1〉| y ≥ 0;Ay ≥ a} (14.24)

has an integral optimum solution y for each integral vector a with finiteminimum. In particular the system x ≥ 0;xA ≤ 1 satisfies the integerrounding property. Therefore R[I∗t] and K[xw1t, . . . , xwr t] are normal byTheorem 14.6.22. �

Corollary 14.6.25 [423] If G is a perfect graph, then R[Ic(G)t] is normal.

Proof. Let C be the clique clutter of G, the complement of G, and let Abe its incidence matrix. The graph G is perfect by the weak perfect graphtheorem (see Theorem 13.6.1). Hence, by Theorem 13.6.8, the polytopeP(A) = {x|x ≥ 0; xA ≤ 1} is integral. If I is the edge ideal of C, byCorollary 14.6.24, the ideal I∗ is normal. To complete the proof notice thatI∗ is equal to Ic(G). �

Corollary 14.6.26 [165, Corollary 5.11] If Ic(G) is the ideal of covers ofa perfect graph G, then Ass(R/Ic(G)

k) form an ascending chain.

Proof. We set J = Ic(G). By Corollary 14.6.25, R[Jt] is normal. Thus

Jk = Jk for all k, so by Theorem 7.7.3, J has the persistence property. �

Proposition 14.6.27 Let C be a balanced clutter and let I be its edge ideal.Then the Rees algebra R[I∗t] of I∗ is normal.

Proof. Let A be the incidence matrix of C. This matrix is balanced byhypothesis. By Theorem 13.6.7 A is balanced if and only if every submatrixof A is perfect. Thus, the result follows from Corollary 14.6.24. �

Corollary 14.6.28 If G is a perfect and unmixed graph, then R[Ic(G)t] isa Gorenstein standard graded K-algebra.

Proof. We set g = ht(I(G)) and J = Ic(G). By assigning deg(xi) = 1and deg(t) = −(g − 1), the Rees algebra R[Jt] becomes a standard graded

624 Chapter 14

K-algebra. The ring R[Jt] is normal by Theorem 14.6.25. Then, accordingto Theorem 9.1.5, its canonical module is the ideal of R[Jt] given by

ωR[Jt] = ({xa11 · · ·xann tan+1| a = (ai) ∈ R+(J)o ∩ Zn+1}).

By Theorem 9.1.6 the ring R[Jt] is Cohen–Macaulay. Using Eq. (13.10)of Proposition 13.6.3 it is seen that x1 · · ·xnt belongs to ωR[Jt]. Take any

monomial xatb = xa11 · · ·xann tb in the ideal ωR[Jt], that is (a, b) ∈ R+(J)o.

Hence the vector (a, b) has positive integer entries and satisfies∑xi∈Kr

ai ≥ (r − 1)b+ 1 (14.25)

for every complete subgraph Kr of G. If b = 1, clearly xatb is a multiple ofx1 · · ·xnt. Now assume b ≥ 2. Using the normality of R[Jt] and Eqs. (13.10)and (14.25) it follows that the monomial m = xa1−1

1 · · ·xan−1n tb−1 belongs

to R[Jt]. Since xatb = mx1 · · ·xnt, we obtain that ωR[Jt] is generated byx1 · · ·xnt and thus R[Jt] is a Gorenstein ring. �

Corollary 14.6.29 Let B1, . . . , Bq be the collection of basis of a matroidM with vertex set X and let v1, . . . , vq be their characteristic vectors. If Ais the matrix with column vectors v1, . . . , vq, then all systems

x ≥ 0;xA ≥ 1, x ≥ 0;xA∗ ≥ 1, x ≥ 0;xA ≤ 1, x ≥ 0;xA∗ ≤ 1

have the integer rounding property.

Proof. Consider the basis monomial ideal I = (xv1 , . . . , xvq ) of the matroidM . By [338, Theorem 2.1.1], the collection of basis of the dual matroid M∗

of M is given by X \ B1, . . . , X \ Bq. Now, the basis monomial ideal of amatroid is normal (Corollary 12.3.12). Thus the result follows at once fromthe duality of Theorem 14.6.22. �

Linear systems of the form xA ≤ 1 We now turn our attention tostudy the integer rounding property of linear systems of inequalities of theform xA ≤ 1 (see Definition 1.3.23).

Proposition 14.6.30 Let v1, . . . , vq be the column vectors of a nonnegativeinteger matrix A and let A(P) be the Ehrhart ring of P = conv(0, v1, . . . , vq).Then the system xA ≤ 1 has the integer rounding property if and only if

K[xv1t, . . . , xvq t, t] = A(P).

Proof. By Theorem 1.3.24, we have that the system xA ≤ 1 has the integerrounding property if and only if the set B = {(v1, 1), . . . , (vq, 1), (0, 1)} is anintegral Hilbert basis. Thus the proposition follows by noticing the equality

A(P) = K[{xatb|(a, b) ∈ R+B ∩ Zn+1}]

and the inclusion K[xv1t, . . . , xvq t, t] ⊂ A(P). �


Theorem 14.6.31 Let A = {v1, . . . , vq} be the set of column vectors of amatrix A with entries in N. If the system xA ≤ 1 has the integer roundingproperty, then

(a) K[F ] is normal, where F = {xv1 , . . . , xvq}, and(b) Zn/ZA is a torsion-free group.

The converse holds if |vi| = d for all i.

Proof. If xA ≤ 1 has the integer rounding property, then (a) and (b) followfrom Corollary 1.3.32. Conversely assume that |vi| = d for all i and that(a) and (b) hold. We need only show that B = {(vi, 1)}qi=1 ∪ {en+1} is aHilbert basis. Let (a, b) be an integral vector in R+B, where a ∈ Nn andb ∈ N. Then we can write

(a, b) = λ1(v1, 1) + · · ·+ λq(vq, 1) + μ(0, 1), (14.26)

for some λ1, . . . , λq, μ in Q+. Hence using (b) gives that a is in R+A∩ ZA.Hence xa ∈ K[F ] = K[F ], i.e., a ∈ NA. Then we can write a =

∑qi=1 ηivi

for some η1, . . . , ηq in Nn. Since |vi| = d for all i, one has∑i λi =

∑i ηi.

Therefore using Eq. (14.26), we get μ ∈ N. Therefore we get

(a, b) = η1(v1, 1) + · · ·+ ηq(vq, 1) + μ(0, 1) ⇒ (a, b) ∈ NB. �

Corollary 14.6.32 Let A be the incidence matrix of a connected graph G.Then the system xA ≤ 1 has the integer rounding property if and only if Gis a bipartite graph.

Proof. ⇒) Let A = {v1, . . . , vq} be the set of columns of A. If G isnot bipartite, then according to Corollary 10.2.11 one has Zn/ZA � Z2, acontradiction to Theorem 14.6.31(b).⇐) By Corollary 10.2.11 and Proposition 10.3.1 the ring K[xv1 , . . . , xvq ]

is normal and Zn/ZA � Z. Thus, by Theorem 14.6.31, the system xA ≤ 1has the integer rounding property, as required. �

Integer rounding properties in simple graphs Let G be a simplegraph with vertex set X = {x1, . . . , xn}, let v1, . . . , vq be the column vectorsof the incidence matrix A of G, let R = K[x1, . . . , xn] be a polynomial ringover a field K and let I = I(G) be the edge ideal of G. Recall that theextended Rees algebra of I is the subring

R[It, t−1] := R[It][t−1] ⊂ R[t, t−1],

where R[It] is the Rees algebra of I.

Lemma 14.6.33 R[It, t−1] � K[t, x1t, . . . , xnt, xv1t, . . . , xvq t].

626 Chapter 14

Proof. We set S = K[t, x1t, . . . , xnt, xv1t, . . . , xvq t]. Note that S and

R[It, t−1] are integral domains of the same dimension. This follows fromthe dimension formula given in Corollary 8.2.21. Thus it suffices to provethat there is an epimorphism ψ : S → R[It, t−1] of K-algebras.

Let u0, u1, . . . , un, t1, . . . , tq be a new set of variables and let ϕ, ψ be themaps of K-algebras defined by the diagram

K[u0, u1, . . . , un, t1, . . . , tq]ψ ��

ϕ

��

R[It, t−1]

S

ψ��

u0ϕ�−→ t,

uiϕ�−→ xit,

tiϕ�−→ xvit,

u0ψ�−→ t−1,

uiψ�−→ xi,

tiψ�−→ xvit.

As ker(ϕ) is a binomial ideal (Corollary 8.2.18) we get ker(ϕ) ⊂ ker(ψ).Hence there is an epimorphism ψ of K-algebras that makes the diagramcommutative, i.e., ψ = ψϕ. �

Theorem 14.6.34 Let G be a connected graph and let A be its incidencematrix. Then K[G] := K[xv1 , . . . , xvq ] is normal if and only if the systemx ≥ 0;xA ≤ 1 has the integer rounding property.

Proof. Let I = I(G) be the edge ideal ofG. By Corollary 10.5.6 the subringK[G] is normal if and only if R[It] is normal. Using Theorem 4.3.17, we getthat R[It] is normal if and only if R[It, t−1] is normal. By Lemma 14.6.33,R[It, t−1] is normal if and only if the subring

S = K[t, x1t, . . . , xnt, xv1t, . . . , xvq t]

is normal. Finally applying Theorem 14.6.16 we get that S is normal if andonly if the system x ≥ 0;xA ≤ 1 has the integer rounding property. �

Theorem 14.6.35 Let G be a connected graph and let A be its incidencematrix. Then the system x ≥ 0;xA ≤ 1 has the integer rounding propertyif and only if any of the following equivalent conditions hold.

(a) x ≥ 0;xA ≥ 1 is a system with the integer rounding property.

(b) R[It] is a normal domain, where I = I(G) is the edge ideal of G.

(c) K[xv1t, . . . , xvq t] is normal, where v1, . . . , vq are the columns of A.

(d) K[t, x1t, . . . , xnt, xv1t, . . . , xvq t] is normal.

Proof. According to Corollary 14.6.9, the system x ≥ 0;xA ≥ 1 has theinteger rounding property if and only if the Rees algebra R[It] is normal.Thus the result follows from the proof of Theorem 14.6.34. �

Corollary 14.6.36 Let G be a graph and let I = I(G) be its edge ideal.Then R[It] is normal if and only if R[I∗t] is normal.


Proof. Recall that: (i) the Rees algebra R[It] is normal if and only ifthe extended Rees algebra R[It, t−1] is normal (Theorem 4.3.17), and (ii)R[It, t−1] is isomorphic to

K[t, x1t, . . . , xnt, xv1t, . . . , xvq t],

where I is the edge ideal of G (Lemma 14.6.33). Then the result followsapplying Theorem 14.6.22. �

The next example shows that Corollary 14.6.36 does not extend to ar-bitrary uniform clutters.

Example 14.6.37 Consider the clutter C whose incidence matrix A is thetranspose of the matrix:⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 1 1 0 1 1 1 1 10 0 1 0 1 1 1 1 1 10 1 1 0 0 1 1 1 1 11 1 0 0 0 1 1 1 1 10 1 1 0 1 0 1 1 1 11 1 1 1 1 0 0 1 1 01 1 1 1 1 0 0 1 0 11 1 1 1 1 0 1 1 0 01 1 1 1 1 1 1 0 0 01 1 1 1 0 0 1 1 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

Let I = I(C) be the edge ideal of C. Note that C is uniform. UsingNormaliz [68] it is seen that R[It] is normal and that R[I∗t] is not normal.

Proposition 14.6.38 Let G be a graph without isolated vertices and let Gbe its complement. Then I(G)∨ = I(G)∗ if and only if G is triangle free.

Proof. ⇒) Assume that G has a triangle C3 = {x1, x2, x3}. We may assumen ≥ 4. Notice that C = {x4, . . . , xn} is a vertex cover of G, i.e., x4 · · ·xnbelongs to I(G)∨ and consequently it belongs to I(G)∗, a contradictionbecause I(G)∗ is generated by monomials of degree n− 2.⇐) Let xa = x1 · · ·xr be a minimal generator of I(G)∨. Then the set

C = {x1, . . . , xr} is a minimal vertex cover of G. Hence X \C is a maximalcomplete subgraph of G. Thus by hypothesis X \ C is an edge of G, i.e.,xa ∈ I(G)∗. This proves the inclusion “⊂”. Conversely, let xa be a minimalgenerator of I(G)∗. There is an edge {x1, x2} of G such that xa = x3 · · ·xn.Every edge of G must intersect C = {x3, . . . , xn}, i.e., xa ∈ I(G)∨. �

Corollary 14.6.39 Let G be a triangle free graph without isolated vertices.Then R[I(G)t] is normal if and only if R[I(G)∨t] is normal.


628 Chapter 14

Example 14.6.40 Let G be the graph consisting of two vertex disjoint oddcycles of length 5 and let G be its complement. By Proposition 10.5.8 andCorollary 14.6.39 the Rees algebras R[I(G)t] and R[I(G)∨t] are not normal.

Exercises

14.6.41 Let G be a perfect graph with vertex set X = {x1, . . . , xn} and letS be the subring generated by all xat such that supp(xa) is a clique of G.Prove that S is normal.

14.6.42 Let G be a bipartite graph without isolated vertices and let G beits complement. Then I(G)∨ = I(G)∗.

14.6.43 If I ⊂ R is a Cohen–Macaulay square-free monomial ideal of heighttwo, then R[It] is normal.

14.6.44 Let A be the vertex-clique matrix of a graph G and let v1, . . . , vqbe the column vectors of A. Consider the ideal I∗ = (xv

∗1 , . . . , xv

∗q ), where

v∗i = 1− vi for all i. Prove the duality: I∗ is the ideal of covers of G.

14.6.45 If G is a pentagon, then the Rees algebra of Ic(G) is normal andG is not a perfect graph.

14.6.46 Let A be the incidence matrix of a clutter C. If C is uniform andhas the max-flow min-cut property, then the system xA ≤ 1 has the integerrounding property.

14.7 Canonical modules and integer rounding

Here we give a description of the canonical module and the a-invariant forsubrings arising from systems with the integer rounding property.

Let A be a matrix of size n×q with entries in N such that A has non-zerorows and non-zero columns. Let v1, . . . , vq be the columns of A. For usebelow consider the set w1, . . . , wr of all α ∈ Nn such that α ≤ vi for somei. Let R = K[x1, . . . , xn] be a polynomial ring over a field K and let

S = K[xw1t, . . . , xwr t] ⊂ R[t]

be the subring of R[t] generated by xw1t, . . . , xwr t, where t is a new variable.The ring S is a standard graded K-algebra such that deg(xatb) = b for

any monomial xatb of S. If S is normal, then according to Theorem 9.1.5the canonical module of S is the ideal given by

ωS = ({xatb| (a, b) ∈ NB ∩ (R+B)o}), (14.27)


where B = {(w1, 1), . . . , (wr , 1)} and (R+B)o is the interior of R+B relativeto aff(R+B), the affine hull of R+B. In our case aff(R+B) = Rn+1.

Let {β0, β1, . . . , βm} ⊂ Qn+ be the set of vertices of the polytope

P = {x|x ≥ 0;xA ≤ 1},

where β0 = 0, and let β1, . . . , βp be the set of all maximal elements ofβ0, β1, . . . , βm (maximal with respect to ≤).

The following lemma is not hard to prove.

Lemma 14.7.1 For each 1 ≤ i ≤ p there is a unique positive integer δisuch that the non-zero entries of (−δiβi, δi) are relatively prime.

Notation In what follows {β1, . . . , βp} is the set of maximal elements of{β0, . . . , βm} and δ1, . . . , δp are the unique positive integers in Lemma 14.7.1.

Theorem 14.7.2 If the system x ≥ 0;xA ≤ 1 has the integer roundingproperty, then S is normal, the canonical module of S is given by

ωS =

(xatb

∣∣∣∣ (a, b)( −δ1β1 · · · −δpβp e1 · · · enδ1 · · · δp 0 · · · 0

)≥ 1

),

(14.28)and the a-invariant of S is equal to −maxi{�1/δi + |βi|�}.

Proof. Note that in Eq. (14.28) we regard (−δiβi, δi) and ej as columnvectors for all i, j. The normality of S follows from Theorem 14.6.16. Recallthat we have the following duality:

P = {x|x ≥ 0; 〈x,wi〉 ≤ 1 ∀i} = conv(β0, β1, . . . , βm),

T (P) = {x|x ≥ 0; 〈x, βi〉 ≤ 1∀i} = conv(w1, . . . , wr),

see Lemma 14.6.12. Therefore, by the maximality of β1, . . . , βp, we obtain

conv(w1, . . . , wr) = {x|x ≥ 0; 〈x, βi〉 ≤ 1, ∀ i = 1, . . . , p}.

Hence, setting B = {(wi, 1)}ri=1 and noticing ZB = Zn+1, it is seen that

R+B = H+e1 ∩ · · · ∩H

+en ∩H

+(−δ1β1,δ1)

∩ · · · ∩H+(−δpβp,δp)

. (14.29)

Notice that Hei ∩ R+B and H(−δjβj ,δj) ∩ R+B are proper faces of R+B forany i, j. Hence from Eq. (14.29) we get that a vector (a, b), with a ∈ Zn,b ∈ Z, is in the relative interior of R+B if and only if the entries of a arepositive and 〈(a, b), (−δiβi, δi)〉 ≥ 1 for all i. Thus the required expressionfor ωS follows using the normality of S and Eq. (14.27).

It remains to prove the formula for a(S), the a-invariant of S. Considerthe vector (1, b0), where b0 = maxi{�1/δi + |βi|�}. Using Eq. (14.28), it is

630 Chapter 14

not hard to see (by direct substitution of (1, b0)), that the monomial x1tb0 isin ωS . Thus, from Eq. (9.2) of Lemma 9.1.7, we get a(S) ≥ −b0. Converselyif xatb is in ωS , then again from Eq. (14.28) we get 〈(−δiβi, δi), (a, b)〉 ≥ 1for all i and ai ≥ 1 for all i, where a = (ai). Hence

bδi ≥ 1 + δi〈a, βi〉 ≥ 1 + δi〈1, βi〉 = 1 + δi|βi|.

Since b is an integer we obtain b ≥ �1/δi + |βi|� for all i. Therefore b ≥ b0,i.e., deg(xatb) = b ≥ b0. As xatb was an arbitrary monomial in ωS , by theformula for the a-invariant of S given in Eq. (9.2) of Lemma 9.1.7, we obtainthat a(S) ≤ −b0. Altogether one has a(S) = −b0, as required. �

Theorem 14.7.3 Assume that the system x ≥ 0; xA ≤ 1 has the integerrounding property. If S is Gorenstein and c0 = max{|βi| : 1 ≤ i ≤ p} is aninteger, then |βk| = c0 for each 1 ≤ k ≤ p such that βk has integer entries.

Proof. We proceed by contradiction. Assume that |βk| < c0 for someinteger 1 ≤ k ≤ p such that βk is integral. We may assume that βk is(1, . . . , 1, 0, . . . , 0) and |βk| = s. From Eq. (14.29) xβkts−1 cannot be in Sbecause (βk, s − 1) does not belong to H+

(−δkβk,δk). Consider xatb, where

a = βk + 1, b = b0 + s − 1 and b0 = −a(S). We claim that xatb is inωS . By Theorem 14.7.2 it suffices to show that 〈(a, b), (−δjβj , δj)〉 ≥ 1 for1 ≤ j ≤ p. Thus we need only show that 〈(a, b), (−βj , 1)〉 > 0 for 1 ≤ j ≤ p.From the proof of Theorem 14.7.2, it is seen that −a(S) = maxi{(|βi|)}+1.Hence we get b0 = c0 + 1. One has the following equality

〈(a, b), (−βj , 1)〉 = −|βj| − 〈βk, βj〉+ c0 + s.

Set βj = (βj1 , . . . , βjn). From Eq. (14.29) we get that the entries of each βjare less than or equal to 1. Case (I): If βji < 1 for some 1 ≤ i ≤ s, thens − 〈βk, βj〉 > 0 and c0 ≥ |βj |. Case (II): βji = 1 for 1 ≤ i ≤ s. Thenβj ≥ βk. Thus by the maximality of βk we obtain βj = βk. In both caseswe obtain 〈(a, b), (−βj , 1)〉 > 0, as required. Hence the monomial xatb isin ωS . Since S is Gorenstein and ωS is generated by x1tb0 , we obtain thatxatb is a multiple of x1tb0 , i.e., xβk ts−1 must be in S, a contradiction. �

Theorem 14.7.4 Assume that x ≥ 0;xA ≤ 1 has the integer roundingproperty. If −a(S) = 1/δi + |βi| for i = 1, . . . , p, then S is Gorenstein.

Proof. We set b0 = −a(S) and B = {(w1, 1), . . . , (wr , 1)}. The ring S isnormal by Theorem 14.6.16. Since the monomial x1tb0 = x1 · · ·xntb0 is inωS , we need only show that ωS = (x1tb0). Take xatb ∈ ωS . It suffices toprove that xa−1tb−b0 is in S. Using Theorem 14.6.16, one has the equalityR+B ∩ Zn+1 = NB. Thus we need only show that (a− 1, b− b0) is in R+B.


From Eq. (14.29), the proof reduces to showing that (a − 1, b − b0) is inH+

(−βi,1)for i = 1, . . . , p. As (a, b) ∈ ωS , from Theorem 14.7.2, we get

〈(a, b), (−δiβi, δi)〉 = −〈a, δiβi〉+ bδi ≥ 1 =⇒ −〈a, βi〉 ≥ −b+ 1/δi

for i = 1, . . . , p. Therefore

〈(a−1, b−b0), (−βi, 1)〉 = −〈a, βi〉+|βi|+b−b0 ≥ −b+1/δi+|βi|+b−b0 = 0

for all i, as required. �

Corollary 14.7.5 If P = {x|x ≥ 0;xA ≤ 1} is an integral polytope, thenS is Gorenstein if and only if a(S) = −(|βi|+ 1) for i = 1, . . . , p.

Proof. Notice that if P is integral, then βi has entries in {0, 1} for 1 ≤ i ≤ pand consequently δi = 1 for 1 ≤ i ≤ p. Thus the result follows fromTheorems 14.7.3 and 14.7.4. �

Conjecture 14.7.6 If A is the incidence matrix of a connected graph andthe system x ≥ 0; xA ≤ 1 has the integer rounding property, then S isGorenstein if and only if −a(S) = 1/δi + |βi| for i = 1, . . . , p.

The answer to this conjecture is positive if A is the incidence matrix ofa bipartite graph because in this case P is an integral polytope and we mayapply Corollary 14.7.5.

For use below we consider the empty set as a clique whose vertex set isempty. Note that supp(xa) = ∅ if and only if a = 0.

Theorem 14.7.7 Let G be a perfect graph and let S = K[xω1t, . . . , xωr t] bethe subring generated by all square-free monomials xat such that supp(xa)is a clique of G. Then the canonical module of S is given by

ωS =

({xatb

∣∣∣∣ (a, b)( −β1 · · · −βs e1 · · · en1 · · · 1 0 · · · 0

)≥ 1

}),

where β1, . . . , βs are the characteristic vectors of the maximal independentsets of G, and the a-invariant of S is equal to −(maxi{|βi|}+ 1).

Proof. Let v1, . . . , vq be the set of characteristic vectors of the maximalcliques of G. Note that w1, . . . , wr is the set of all α ∈ Nn such that α ≤ vifor some i. Since G is a perfect graph, by Theorem 13.6.1, we have

P = {x|x ≥ 0;xA ≤ 1} = conv(β0, β1, . . . , βp),

where β0 = 0 and β1, . . . , βp are the characteristic vectors of the indepen-dent sets of G. We may assume that β1, . . . , βs correspond to the maximalindependent sets of G. Furthermore, since P has only integral vertices, by

632 Chapter 14

a result of Lovasz (see Theorem 13.6.8) the system x ≥ 0;xA ≤ 1 is totallydual integral, i.e., the minimum in the LP-duality equation

max{〈α, x〉|x ≥ 0;xA ≤ 1} = min{〈y,1〉| y ≥ 0;Ay ≥ α} (14.30)

has an integral optimum solution y for each integral vector α with finite min-imum. In particular the system x ≥ 0;xA ≤ 1 satisfies the integer roundingproperty. Therefore the result follows readily from Theorem 14.7.2. �

Proposition 14.7.8 Let G be a perfect graph, let w1, . . . , wr be the char-acteristic vectors of the cliques of G, and let β1, . . . , βs be the characteristicvectors of the maximal independent sets of G. Then the set

Γ = {(−β1, 1), . . . , (−βs, 1), e1, . . . , en}

is a Hilbert basis of (R+B)∗, where B = {(w1, 1), . . . , (wr, 1)}.

Proof. The characteristic vector of the empty set is set to be equal to zero.As G is perfect we have

conv(w1, . . . , wr) = {x|x ≥ 0;x(β1 · · ·βs) ≤ 1}.

Therefore it is seen that

R+(w1, 1)+ · · ·+R+(wr , 1) = H+e1 ∩· · ·∩H

+en ∩R+(−β1, 1)∩· · ·R+(−βs, 1).

Thus, by duality (see Corollary 1.1.30), we obtain that (R+B)∗ = R+Γ.Using that the system x ≥ 0; x(β1, . . . , βs) ≤ 1 is TDI it follows thatZn ∩ R+Γ = NΓ, i.e., Γ is a Hilbert basis, as required. �

Corollary 14.7.9 Let G be a connected bipartite graph and let I = I(G)be its edge ideal. Then the extended Rees algebra R[It, t−1] is a Gorensteinstandard K-algebra if and only if G is unmixed.

Proof. Let ωS be the canonical module of S = R[It, t−1]. As S is Cohen–Macaulay, recall that S is Gorenstein if and only if ωS is a principal ideal.Since G is a perfect graph. The result follows using Lemma 14.6.33 togetherwith the description of the canonical module given in Theorem 14.7.7. �

Exercise

14.7.10 If A is the incidence matrix of a graph, then δi = 1 or δi = 1/2 foreach i (see Lemma 14.7.1).


14.8 Clique clutters of Meyniel graphs

In this section it is shown that clique clutters of Meynel graphs are Ehrhartin the sense of [306] (see Definition 9.3.27).

Recall that a Meyniel graph is a simple graph in which every odd cycleof length at least five has at least two chords.

Theorem 14.8.1 [247] A graph G is Meyniel if and only if for each inducedsubgraph H and for each vertex v of H, there exists an independent set inH containing v and intersecting all maximal cliques of H (maximal withrespect to inclusion).

Theorem 14.8.2 Let G be a Meyniel graph and let A = {v1, . . . , vq} be theset of columns of the vertex-clique matrix of G. If A(P) is the Ehrhart ringof P = conv(v1, . . . , vq), then K[xv1t, . . . , xvq t] = A(P).

Proof. The inclusion “⊂” is clear. To prove the other inclusion takexatb ∈ A(P), i.e., a ∈ bP ∩ Zn and b ∈ N. Using that P is the convex hullof the vi’s, it is not hard to see that we can write

(a, b) = λ1(v1, 1) + · · ·+ λq(vq, 1), λi ≥ 0 ∀ i. (14.31)

Let A be the vertex-clique matrix of G. Any Meyniel graph is perfect[373, Theorem 66.6]. Thus, by Theorem 13.6.8, the system x ≥ 0;xA ≤ 1is TDI. Therefore, by Proposition 1.3.28, we get that the set

H = {(v1, 1), . . . , (vq , 1),−e1, . . . ,−en}

is a Hilbert basis, i.e., R+H∩Zn+1 = NH. As (a, b) is an integral vector inR+H, we can write

(a, b) = η1(v1, 1) + · · ·+ ηq(vq, 1)− μ1e1 − · · · − μnen, (14.32)

ηi ∈ N, μj ∈ N for all i, j. By Theorem 14.8.1, for each xk in V (G) ={x1, . . . , xn} there exists an independent set Bk of G containing xk andintersecting all maximal cliques of G. We set βk =

∑xi∈Bk

ei for 1 ≤ k ≤ n.Notice that a clique of G and an independent set of G can meet in at mostone vertex. Then β1, . . . , βn are vectors in {0, 1}n such that 〈vj , βi〉 = 1 and〈ei, βi〉 = 1 for all i, j. Hence, using Eqs. (14.31) and (14.32), we obtain

〈a, βi〉 = λ1〈v1, βi〉+ · · ·+ λq〈vq , βi〉 = b, ∀ i.〈a, βi〉 = η1〈v1, βi〉+ · · ·+ ηq〈vq, βi〉 − μ1〈e1, βi〉 − · · · − μn〈en, βi〉

= b− μ1〈e1, βi〉 − · · · − μn〈en, βi〉, ∀ i.

Therefore for i = 1, . . . , n, we get∑n

j=1 μj〈ej , βi〉 = 0. Since 〈ei, βi〉 = 1for all i, we get μi = 0 for all i. Hence the vector (a, b) belongs to thesemigroup NA. Thus xatb ∈ K[xv1t, . . . , xvq t]. �

634 Chapter 14

Corollary 14.8.3 Let G be a Meyniel graph, let A be the vertex-cliquematrix of G, and let v1, . . . , vq be the column vectors of A. The followingconditions are equivalent:

(i) x ≥ 0;xA ≥ 1 is a TDI system.

(ii) Ii = I(i) for i ≥ 1, where I = (xv1 , . . . , xvq ).

(iii) Q(A) = {x|x ≥ 0;xA ≥ 1} is an integral polyhedron.

Proof. (i) ⇒ (ii) and (ii) ⇒ (iii): Follow from Theorem 14.3.6. (iii) ⇒ (i):By Theorem 14.8.2 we get K[xv1t, . . . , xvq t] = A(P). As Q(A) is integral, adirect application of Proposition 14.3.30 and Theorem 14.3.6 gives that thesystem x ≥ 0;xA ≥ 1 is TDI. �

Exercises

14.8.4 Let v1, . . . , v6 be the characteristic vectors of the maximal cliquesof the graph below and let P be its convex hull.

��

��

G

� �

� ��

��

��

��

Use Normaliz [68] to show that K[xv1t, . . . , xv6t] � A(P). Show that theincidence matrix of cl(G), the clique clutter of G, is totally unimodular byshowing that the incidence matrix of cl(G) is the transpose of the incidencematrix of the complete bipartite graph K2,4.

14.8.5 Consider the following graph G:

��

��

��

��

��

��

��

��

� ��

��

��

��

��

��

��

��

Prove that the edge ideal I = I(cl(G)) of the clique clutter of G is notnormal. This graph is chordal, hence perfect. Thus edge ideals of cliqueclutters of perfect graphs are in general not normal.

Appendix

Graph Diagrams

For convenience we will display some Cohen–Macaulay graphs and unmixedgraphs with small number of vertices.

A.1 Cohen–Macaulay graphs

The complete list of Cohen–Macaulay connected graphs with at most sixvertices is:

• • •

��

��

��• •

• •

• •

•

��

��

��•

• •

•

��

��

��• •

• •

•

��

��

��• •

• •

• •

•

��

��

��• •

• •

•

��

��

�

��

•

•

•

��

��

��• •

•

��

��•

636 Appendix

•

��

��

�•

��

• •

• •

•

��

��

�•

��

• •

• •

• •

• •

• •

•

��

��

��• •

•

��

•

•

•

��

��

��• •

•

��

•

•

•

��

��

�

��

��•

�� •

•

•

•

•

��

��

��• ••

��

•

•

•

• •

•

��

��

�

��

•

•

•

��

��

�

��

��• •

• •

•

•

��

��

�

��

��• •

•

•

��

•

•

��

��

�

��

��• •

• •

��

•

•

��

��

�

��

��• •

• •

��

•

Graph Diagrams 637

•

��

��

��•

��

��

�� •

• •

��

•

��

•

��

��

�

��

��•

��

��

�� •

• •

��

•

•

��

��

�

��

��•

�� •

•

•

•

•

��

��

�

��

��•

�� •

•

•

•

•

��

��

��•

��

��

�� •

•

•

��

•

��

•

��

��

�

��

��• •

• •

��

•

•

��

��

�

��

��• •

• •

��

•

•

��

��

�

��

��• •

��

��

��

�� •

•

��

•

638 Appendix

A.2 Unmixed graphs

Next we display the set of all unmixed non-Cohen–Macaulay connectedgraphs with at most six vertices.

• •

• •

•

��

��

��• •

• •

��

•

��

��

�

• •

•

•

��

•

•

��

��

�

��

��•

�� •

•

•

��

•

•

��

��

�

��

��• •

• •

��

•

•

��

��

�

��

��•

��

��

��

�� •

•

•

��

•

•

��

��

�

��

��

��

•

�� •

• •

•

•

��

��

�

��

��•

�� •

•

•

��

•

Bibliography

[1] W. W. Adams and P. Loustaunau, An Introduction to Grobner Bases ,Graduate Studies in Mathematics 3, American Mathematical Society,Providence, RI, 1994.

[2] M. Aigner, Combinatorial Theory, Springer, 1997.

[3] A. Alcantar, Rees algebras of square-free Veronese ideals and theira-invariants, Discrete Math. 302 (2005), 7–21.

[4] A. Alcantar and R. H. Villarreal, Critical binomials of monomialcurves, Comm. Algebra 22 (1994), 3037–3052.

[5] C. Alfaro and C. E. Valencia, On the sandpile group of the cone of agraph, Linear Algebra Appl. 436 (2012), no. 5, 1154–1176.

[6] A. Alilooee and S. Faridi, When is a squarefree monomial ideal oflinear type? Preprint, 2013, arXiv:1309.1771.

[7] N. Alon, Combinatorial Nullstellensatz, Combin. Probab. Comput. 8(1999), no. 1-2, 7–29.

[8] R. P. Anstee, Hypergraphs with no special cycles, Combinatorica 3(1983), no. 2, 141–146.

[9] M. F. Atiyah and I. G. Macdonald, Introduction to Commutative Al-gebra, Addison-Wesley, Reading, MA, 1969.

[10] D. Avis and K. Fukuda, A pivoting algorithm for convex hulls andvertex enumeration of arrangements and polyhedra, Discrete Comput.Geom. 8 (1992), no. 3, 295–313.

[11] D. Avis and K. Fukuda, Reverse search for enumeration, DiscreteAppl. Math. 65 (1996), no. 1-3, 21–46.

[12] C. Bahiano, Symbolic powers of edge ideals, J. Algebra 273 (2004),no. 2, 517–537.

640 BIBLIOGRAPHY

[13] A. Banerjee, The regularity of powers of edge ideals, J. AlgebraicCombin., to appear. DOI: 10.1007/s10801-014-0537-2.

[14] J. Bang-Jensen and G. Gutin, Digraphs. Theory, Algorithms and Ap-plications , Springer Monographs in Mathematics, Springer, 2006.

[15] M. Barile, On the arithmetical rank of the edge ideals of forests,Comm. Algebra 36 (2008), no. 12, 4678–4703.

[16] M. Barile, D. Kiani, F. Mohammadi, and S. Yassemi, Arithmeticalrank of the cyclic and bicyclic graphs, J. Algebra Appl. 11 (2012), no.2, Article Number: 1250039.

[17] M. Barile, M. Morales, and A. Thoma, Set-theoretic complete inter-sections on binomials, Proc. Amer. Math. Soc. 130 (2002), 1893–1903.

[18] H. Bass, On the ubiquity of Gorenstein rings, Math. Z. 82 (1963),8–28.

[19] S. Baum and L. E. Trotter, Integer rounding for polymatroid andbranching optimization problems, SIAM J. Algebraic Discrete Meth-ods 2 (1981), no. 4, 416–425.

[20] D. Bayer and M. Stillman, Computation of Hilbert functions, J. Sym-bolic Comput. 14 (1992), 31–50.

[21] M. Beck and S. Robins, Computing the Continuous Discretely,Springer, New York, 2007.

[22] E. Becker, R. Grobe, and M. Niermann, Radicals of binomial ideals,J. Pure Appl. Algebra 117 (1997), 41–79.

[23] M. Beintema, A note on Artinian Gorenstein algebras defined bymonomials, Rocky Mountain J. Math. 23 (1993), 1–3.

[24] C. Berge, Balanced matrices, Math. Programming 2 (1972), 19–31.

[25] C. Berge, Graphs and Hypergraphs, North-Holland Mathematical Li-brary, Vol. 6, North-Holland Publishing Co., Amsterdam-London;American Elsevier Publishing Co., Inc., New York, 1976.

[26] C. Berge, Some Common Properties for Regularizable Graphs, Edge-critical Graphs and B-graphs, North-Holland Math. Stud. 60, North-Holland, Amsterdam, 1982, pp. 31–44.

[27] C. Berge, Hypergraphs Combinatorics of Finite Sets , MathematicalLibrary 45, North-Holland, 1989.

BIBLIOGRAPHY 641

[28] A. Berman and R. J. Plemmons, Nonnegative Matrices in the Math-ematical Sciences, Classics in Applied Mathematics 9, Society for In-dustrial and Applied Mathematics (SIAM), Philadelphia, PA, 1994.

[29] I. Bermejo, I. Garcıa-Marco, and E. Reyes, Graphs and complete in-tersection toric ideals, J. Algebra Appl., to appear.

[30] I. Bermejo, I. Garcıa-Marco, and J. J. Salazar-Gonzalez, An algorithmfor checking whether the toric ideal of an affine monomial curve is acomplete intersection, J. Symbolic Comput. (42) (2007), 971–991.

[31] I. Bermejo and P. Gimenez, On Castelnuovo–Mumford regularity ofprojective curves, Proc. Amer. Math. Soc. 128 (2000), 1293–1299.

[32] I. Bermejo and P. Gimenez, Saturation and Castelnuovo–Mumfordregularity, J. Algebra 303 (2006), no. 2, 592–617.

[33] I. Bermejo, P. Gimenez, E. Reyes, and R. H. Villarreal, Completeintersections in affine monomial curves, Bol. Soc. Mat. Mexicana (3)11 (2005), 191–203.

[34] I. Bermejo, P. Gimenez, and A. Simis, Polar syzygies in characteristiczero: the monomial case, J. Pure Appl. Algebra 213 (2009), 1–21.

[35] D. Bertsimas and J. N. Tsitsiklis, Introduction to Linear Optimiza-tion, Athena Scientific, Massachusetts, 1997.

[36] A. Bhat, J. Biermann, and A. Van Tuyl, Generalized cover ideals andthe persistence property, J. Pure Appl. Algebra 218 (2014), no. 9,1683–1695.

[37] A. Bigatti, Computation of Hilbert–Poincare series, J. Pure Appl.Algebra 119 (1997), 237–253.

[38] A. Bigatti, R. La Scala, and L. Robbiano, Computing toric ideals, J.Symbolic Comput. 27 (1999), 351–365.

[39] A. Bigatti and L. Robbiano, Borel sets and sectional matrices, Annalsof Combinatorics 1 (1997), 197–213.

[40] N. Biggs, Algebraic Graph Theory, Cambridge University Press, Cam-bridge, 1993.

[41] L. J. Billera, Polyhedral theory and commutative algebra, in Math-ematical Programming: The State of the Art (A. Bachem, M.Grotschel, and B. Korte, Eds.), Springer-Verlag, 1983, pp. 57–77.

[42] C. Bivia-Ausina, The analytic spread of monomial ideals, Comm. Al-gebra 31 (2003), no. 7, 3487–3496.

642 BIBLIOGRAPHY

[43] A. Bjorner, Topological Methods, Handbook of Combinatorics, Vol. 1,2, 1819–1872, Elsevier, Amsterdam, 1995.

[44] A. Bjorner and M. Wachs, Shellable nonpure complexes and posets I,Trans. Amer. Math. Soc. 348 (1996), no. 4, 1299–1327.

[45] A. Bjorner and M. Wachs, Shellable nonpure complexes and posetsII, Trans. Amer. Math. Soc. 349 (1997), no. 10, 3945–3975.

[46] S. Blum, Base-sortable matroids and Koszulness of semigroup rings,European J. Combin. 22 (2001), no. 7, 937–951.

[47] M. Boij and J. Soderberg, Betti numbers of graded modules andthe multiplicity conjecture in the non-Cohen–Macaulay case, AlgebraNumber Theory 6 (2012), no. 3, 437–454.

[48] B. Bollobas, Modern Graph Theory, Graduate Texts in Mathematics184, Springer-Verlag, New York, 1998.

[49] V. Bonanzinga, C. Escobar, and R. H. Villarreal, On the normalityof Rees algebras associated to totally unimodular matrices, ResultsMath. 41, 3/4, (2002), 258–264.

[50] R. R. Bouchat, H. T. Ha, and A. O’Keefe, Path ideals of rootedtrees and their graded Betti numbers, J. Combin. Theory Ser. A 118(2011), no. 8, 2411–2425.

[51] P. Bouwknegt, q-identities and affinized projective varieties I.Quadratic monomial ideals, Commun. Math. Phys. 210 (2000), no.3, 641–661.

[52] J. P. Brennan, L. A. Dupont, and R. H. Villarreal, Duality, a-invariants and canonical modules of rings arising from linear opti-mization problems, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 51(2008), no. 4, 279–305.

[53] H. Bresinsky, Symmetric semigroups of integers generated by 4 ele-ments, Manuscripta Math. 17 (1975), 205–219.

[54] H. Bresinsky, Monomial space curves in A3 as set-theoretic completeintersections, Proc. Amer. Math. Soc. 75 (1979), 23–24.

[55] M. Brodmann, The asymptotic nature of the analytic spread, Math.Proc. Cambridge Philos. Soc. 86 (1979), 35–39.

[56] M. Brodmann, Asymptotic stability of Ass(M/InM), Proc. Amer.Math. Soc. 74 (1979), 16–18.

BIBLIOGRAPHY 643

[57] A. Brøndsted, Introduction to Convex Polytopes , Graduate Texts inMathematics 90, Springer-Verlag, 1983.

[58] H. Bruggesser and P. Mani, Shellable decompositions of cells andspheres, Math. Scand. 29 (1971), 197–205.

[59] P. Brumatti, A. Simis, and W. V. Vasconcelos, Normal Rees algebras,J. Algebra 112 (1988), 26–48.

[60] W. Bruns and J. Gubeladze, Normality and covering properties ofaffine semigroups, J. reine angew. Math. 510 (1999), 161–178.

[61] W. Bruns and J. Gubeladze, Polytopes, Rings, and K-Theory,Springer Monographs in Mathematics, Springer, Dordrecht, 2009.

[62] W. Bruns, J. Gubeladze, and N. V. Trung, Normal polytopes, tri-angulations, and Koszul algebras, J. reine angew. Math. 481 (1997),123–160.

[63] W. Bruns and J. Herzog, On the computation of a-invariants, Manus-cripta Math. 77 (1992), 201–213.

[64] W. Bruns and J. Herzog, On multigraded resolutions, Math. Proc.Cambridge Philos. Soc. 118 (1995), 245–257.

[65] W. Bruns and J. Herzog, Cohen–Macaulay Rings , Cambridge Univer-sity Press, Cambridge, Revised Edition, 1998.

[66] W. Bruns and T. Hibi, Stanley–Reisner rings with pure resolutions.Comm. Algebra 21 (1995), 1201–1217.

[67] W. Bruns and T. Hibi, Cohen–Macaulay partially ordered sets withpure resolutions, European J. Combin. 19 (1998), 779–785.

[68] W. Bruns, B. Ichim, T. Romer, and C. Soger: Normaliz . Algorithmsfor rational cones and affine monoids.Available from http://www.math.uos.de/normaliz.

[69] W. Bruns and R. Koch, Computing the integral closure of an affinesemigroup, Univ. Iagel. Acta Math. 39 (2001), 59–70.

[70] W. Bruns and G. Restuccia, Canonical modules of Rees algebras, J.Pure Appl. Algebra 201 (2005), 189–203.

[71] W. Bruns, A. Simis, and N. V. Trung, Blow-up of straightening-closedideals in ordinal Hodge algebras, Trans. Amer. Math. Soc. 326 (1991),507–528.

644 BIBLIOGRAPHY

[72] W. Bruns, W. V. Vasconcelos, and R. H. Villarreal, Degree bounds inmonomial subrings, Illinois J. Math. 41 (1997), 341–353.

[73] W. Bruns and U. Vetter, Determinantal Rings , Lecture Notes inMathematics 1327, Springer-Verlag, 1988.

[74] B. Buchberger, An algorithmic method in polynomial ideal theory,in Recent Trends in Mathematical Systems Theory (N.K. Bose, Ed.),Reidel, Dordrecht, 1985, 184–232.

[75] D. Buchsbaum and D. Eisenbud, Algebra structures for finite freeresolutions, and some structure theorems for ideals of codimension 3,Amer. J. Math. 99 (1977), 447–485.

[76] L. Burch, Codimension and analytic spread, Proc. Camb. Phil. Soc.72 (1972), 369–373.

[77] A. H. Busch, F. F. Dragan, and R. Sritharan, New Min-Max Theo-rems for Weakly Chordal and Dually Chordal Graphs, Lecture Notesin Computer Science 6509, Springer-Verlag, 207–218, 2010.

[78] G. Carra Ferro and D. Ferrarello, Cohen–Macaulay graphs arisingfrom digraphs. Preprint, 2007, arXiv:0703417.

[79] M. Cavaliere, M. Rossi, and G. Valla, On Short Graded Algebras (Proc.Workshop in Commutative Algebra, Salvador, Brazil), Lecture Notesin Mathematics 1430, Springer-Verlag, Berlin, 1990, pp. 21–31.

[80] B.W. Char, K.G. Geddes, G.H. Gonnet, and S.M. Watt, Maple VLanguage Reference Manual, Springer-Verlag, Berlin, 1991.

[81] J. Chen, S. Morey, and A. Sung, The stable set of associated primesof the ideal of a graph, Rocky Mountain J. Math. 32 (2002), 71–89.

[82] J. Chifman and S. Petrovic, Toric ideals of phylogenetic invariantsfor the general group-based model on claw trees, Lecture Notes inComputer Science 4545, Springer-Verlag, 307–321, 2007.

[83] L. Chouinard, Krull semigroups and divisor class groups, Canad. J.Math. 33 (1981), 1459–1468.

[84] T. Christof, revised by A. Lobel and M. Stoer, PORTA: A PolyhedronRepresentation Transformation Algorithm, 1997.http://www.zib.de/Optimization/Software/Porta.

[85] M. Chudnovsky, N. Robertson, P. Seymour, and R. Thomas, Thestrong perfect graph theorem, Ann. of Math. (2) 164 (2006), no. 1,51–229.

BIBLIOGRAPHY 645

[86] V. Chvatal, On certain polytopes associated with graphs, J. Combi-natorial Theory Ser. B 18 (1975), 138–154.

[87] V. Chvatal, Linear Programming, W.H. Freeman and Company, NewYork, 1983.

[88] CoCoATeam, CoCoA: a system for doing Computations in Commu-tative Algebra. Available at http://cocoa.dima.unige.it.

[89] A. Conca and E. De Negri, M -sequences, graph ideals, and ladderideals of linear type, J. Algebra 211 (1999), 599–624.

[90] M. Conforti and G. Cornuejols, Clutters that pack and the max-flowmin-cut property, The Fourth Bellairs Workshop on CombinatorialOptimization (W. R. Pulleyblank, F. B. Shepherd, Eds.), 1993.

[91] A. Constantinescu and M. Varbaro, Koszulness, Krull dimension, andother properties of graph-related algebras, J. Algebraic Combin. 34(2011), no. 3, 375–400.

[92] S. M. Cooper, R. J. D. Embree, H. T. Ha, and A. H. Hoefel, Symbolicpowers of monomial ideals. Preprint, 2014, arXiv:1309.5082v3.

[93] G. Cornuejols, Combinatorial Optimization: Packing and Covering ,CBMS-NSF Regional Conference Series in Applied Mathematics 74,SIAM (2001).

[94] G. Cornuejols, B. Guenin, and F. Margot, The packing property,Math. Programming 89 (2000), no. 1, Ser. A, 113–126.

[95] H. Corrales and C. E. Valencia, On the critical ideals of graphs, LinearAlgebra Appl. 439 (2013), no. 12, 3870–3892.

[96] A. Corso and U. Nagel, Monomial and toric ideals associated to ferrersgraphs, Trans. Amer. Math. Soc. 361 (2009), no. 3, 1371–1395.

[97] A. Corso, W. V. Vasconcelos, and R. H. Villarreal, Generic Gaussianideals, J. Pure Appl. Algebra 125 (1998), 117–127.

[98] B. Costa and A. Simis, Cremona maps defined by monomials, J. PureAppl. Algebra 216 (2012), no. 1, 202–215.

[99] D. Cox, J. Little, and D. O’Shea, Ideals, Varieties, and Algorithms ,Springer-Verlag, 1992.

[100] D. Cox, J. Little, and H. Schenck, Toric Varieties, Graduate Studiesin Mathematics 124, American Mathematical Society, Providence, RI,2011.

646 BIBLIOGRAPHY

[101] V. Crispin Quinonez, Integral closure and other operations on mono-mial ideals, J. Commut. Algebra 2 (2010), no. 3, 359–386.

[102] V. I. Danilov, The geometry of toric varieties, Russian Math. Surveys33 (1978), 97–154.

[103] H. Dao, C. Huneke, and J. Schweig, Bounds on the regularity and pro-jective dimension of ideals associated to graphs, J. Algebraic Combin.38 (2013), no. 1, 37–55.

[104] D. Delfino, A. Taylor, W. V. Vasconcelos, R. H. Villarreal, and N.Weininger, Monomial ideals and the computation of multiplicities,Commutative Ring Theory and Applications (Fez, 2001), pp. 87–106,Lecture Notes in Pure and Appl. Math. 231, Dekker, New York, 2003.

[105] J. A. De Loera, R. Hemmecke and M. Koppe, Algebraic and Geomet-ric Ideas in the Theory of Discrete Optimization, MOS-SIAM Serieson Optimization 14, Society for Industrial and Applied Mathematics(SIAM), Philadelphia, PA, 2013.

[106] J. A. De Loera, J. Rambau, and F. Santos, Triangulations, Algorithmsand Computation in Mathematics 25, Springer, 2010.

[107] C. Delorme, Sous-monoides d’intersection complete de N, Ann. Sci.Ecole Norm. Sup. 9 (1976), 145–154.

[108] E. De Negri and T. Hibi, Gorenstein algebras of Veronese type, J.Algebra 193 (1997), 629–639.

[109] F. Di Biase and R. Urbanke, An algorithm to calculate the kernel ofcertain polynomial ring homomorphism, Experiment. Math. 4 (1995),227–234.

[110] A. Dickenstein and E. A. Tobis, Independent sets from an algebraicperspective, Internat. J. Algebra Comput. 22 (2012), no. 2, 83–98.

[111] R. Diestel, Graph Theory, Graduate Texts in Mathematics 173,Springer-Verlag, New York, 1997.

[112] G. A. Dirac, On rigid circuit graphs, Abh. Math. Sem. Univ. Hamburg38 (1961), 71–76.

[113] A. Dochtermann and A. Engstrom, Algebraic properties of edge idealsvia combinatorial topology, Electron. J. Combin. 16 (2009), no. 2, R2.

[114] L. Doering and T. Gunston, Algebras arising from planar bipartitegraphs, Comm. Algebra 24 (1996), 3589–3598.

BIBLIOGRAPHY 647

[115] A. Dress, A new algebraic criterion for shellability, Beitrage AlgebraGeom. 34 (1993), 45–55.

[116] L. A. Dupont, E. Reyes, and R. H. Villarreal, Cohen–Macaulay clut-ters with combinatorial optimization properties and parallelizationsof normal edge ideals, Sao Paulo J. Math. Sci. 3 (2009), no. 1, 61–75.

[117] L. A. Dupont and R. H. Villarreal, Symbolic Rees algebras, vertexcovers and irreducible representations of Rees cones, Algebra DiscreteMath. 10 (2010), no. 2, 64–86.

[118] L. A. Dupont and R. H. Villarreal, Edge ideals of clique clutters ofcomparability graphs and the normality of monomial ideals, Math.Scand. 106 (2010), no. 1, 88–98.

[119] L. A. Dupont and R. H. Villarreal, Algebraic and combinatorial prop-erties of ideals and algebras of uniform clutters of TDI systems, J.Comb. Optim. 21 (2011), no. 3, 269–292.

[120] I. M. Duursma, C. Renterıa, and H. Tapia-Recillas, Reed–Muller codeson complete intersections, Appl. Algebra Engrg. Comm. Comput. 11(2001), no. 6, 455–462.

[121] A. M. Duval, Algebraic shifting and sequentially Cohen–Macaulaysimplicial complexes, Electron. J. Combin. 3 (1996), no. 1, R21.

[122] J. A. Eagon and M. Hochster, R-sequences and indeterminates, Quart.J. Math. Oxford 25 (1974), 61–71.

[123] J. A. Eagon and D.G. Northcott, Ideals defined by matrices and acertain complex associated with them, Proc. Royal Soc. 269 (1962),188–204.

[124] J. A. Eagon and V. Reiner, Resolutions of Stanley–Reisner rings andAlexander duality, J. Pure Appl. Algebra 130 (1998), 265–275.

[125] J. Edmonds and R. Giles, A min-max relation for submodular func-tions on graphs, Ann. of Discrete Math. 1, North-Holland, Amster-dam, 1977, pp. 185–204.

[126] H. Edwards, Divisor Theory, Birkhauser, Boston, 1990.

[127] E. Ehrhart, Demonstration de la loi de reciprocite du polyedre ra-tionnel, C. R. Acad. Sci. Paris Ser. A 265 (1967), 91–94.

[128] D. Eisenbud, Commutative Algebra with a View toward Algebraic Ge-ometry, Graduate Texts in Mathematics 150, Springer-Verlag, 1995.

648 BIBLIOGRAPHY

[129] D. Eisenbud, The Geometry of Syzygies: A Second Course in Com-mutative Algebra and Algebraic Geometry, Graduate Texts in Mathe-matics 229, Springer, New York, 2005.

[130] D. Eisenbud and S. Goto, Linear free resolutions and minimal multi-plicity, J. Algebra 88 (1984), 89–133.

[131] D. Eisenbud and C. Huneke, Cohen–Macaulay Rees algebras and theirspecialization, J. Algebra 81 (1983), 202–224.

[132] D. Eisenbud and B. Mazur, Evolutions, symbolic squares, and Fittingideals, J. reine angew. Math. 488 (1997), 189–201.

[133] D. Eisenbud and F.-O. Schreyer, Betti numbers of graded modulesand cohomology of vector bundles, J. Amer. Math. Soc. 22 (2009),no. 3, 859–888.

[134] D. Eisenbud and B. Sturmfels, Binomial ideals, Duke Math. J. 84(1996), 1–45.

[135] S. Eliahou, Courbes monomiales et algebre de Rees symbolique, PhDthesis, Universite de Geneve, 1983.

[136] S. Eliahou, Ideaux de definition des courbes monomiales, LectureNotes in Mathematics 1092, Springer-Verlag, 1984, pp. 229–240.

[137] S. Eliahou, Minimal syzygies of monomial ideals and Grobner bases,Reporte Tecnico No. 6, CINVESTAV–IPN, 1989.

[138] S. Eliahou and R. H. Villarreal, The second Betti number of an edgeideal, Aportaciones Matematicas, Serie Comunicaciones 25 (1999),Soc. Mat. Mex., pp. 115–119.

[139] S. Eliahou and R. H. Villarreal, On systems of binomials in the idealof a toric variety, Proc. Amer. Math. Soc. 130 (2002), 345–351.

[140] J. Elias, L. Robbiano, and G. Valla, Numbers of generators of perfectideals, Nagoya Math. J. 123 (1991), 39–76.

[141] E. Emtander, Betti numbers of hypergraphs, Comm. Algebra 37(2009), no. 5, 1545–1571.

[142] V. Ene and J. Herzog, Grobner Bases in Commutative Algebra, Grad-uate Studies in Mathematics 130, American Mathematical Society,Providence, RI, 2012.

[143] V. Ene, O. Olteanu, and N. Terai, Arithmetical rank of lexsegmentedge ideals, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 53 (2010),no. 4, 315–327.

BIBLIOGRAPHY 649

[144] P. Erdos and T. Gallai, On the minimal number of vertices represent-ing the edges of a graph, Magyar Tud. Akad. Mat. Kutato Int. Kozl.6 (1961), 181–203.

[145] P. Erdos, A. Hajnal, and J.W. Moon, A problem in graph theory,Amer. Math. Monthly 71 (1964), 1107–1110.

[146] C. Escobar, J. Martınez-Bernal, and R. H. Villarreal, Relative volumesand minors in monomial subrings, Linear Algebra Appl. 374 (2003),275–290.

[147] C. Escobar, R. H. Villarreal, and Y. Yoshino, Torsion freeness andnormality of blowup rings of monomial ideals, Commutative Algebra,Lect. Notes Pure Appl. Math. 244, Chapman & Hall/CRC, Boca Ra-ton, FL, 2006, pp. 69–84.

[148] M. Estrada and A. Lopez, A note on symmetric semigroups and al-most arithmetic sequences, Comm. Algebra 22 (1994), 3903–3905.

[149] M. Estrada and R. H. Villarreal, Cohen–Macaulay bipartite graphs,Arch. Math. 68 (1997), 124–128.

[150] E. G. Evans and P. Griffith, Syzygies , London Math. Soc., LectureNote Series 106, Cambridge University Press, Cambridge, 1985.

[151] G. Ewald, Combinatorial Convexity and Algebraic Geometry , Gradu-ate Texts in Mathematics 168, Springer-Verlag, New York, 1996.

[152] M. Farber, Characterizations of strongly chordal graphs, DiscreteMath. 43 (1983), no. 2-3, 173–189.

[153] S. Faridi, Normal ideals of graded rings, Comm. Algebra 28 (2000),1971–1977.

[154] S. Faridi, The facet ideal of a simplicial complex, Manuscripta Math.109 (2002), 159–174.

[155] S. Faridi, Simplicial trees are sequentially Cohen–Macaulay, J. PureAppl. Algebra 190 (2004), 121–136.

[156] S. Faridi, Cohen–Macaulay properties of square-free monomial ideals,J. Combin. Theory Ser. A 109 (2005), no. 2, 299–329.

[157] S. Faridi, Monomial Ideals via Square-free Monomial Ideals, LectureNotes in Pure and Applied Math. 244, Taylor & Francis, Philadelphia,2005, pp. 85–114.

650 BIBLIOGRAPHY

[158] K. Fischer, W. Morris, and J. Shapiro, Affine semigroup rings thatare complete intersections, Proc. Amer. Math. Soc. 125 (1997), 3137–3145.

[159] H. Flaschka and L. Haine, Torus orbits on G/P, Pacific J. Math. 149(1991), 251–292.

[160] G. Fløystad and J. E. Vatne, (Bi-)Cohen–Macaulay simplicial com-plexes and their associated coherent sheaves, Comm. Algebra 33(2005), no. 9, 3121–3136.

[161] R. M. Fossum, The Divisor Class Group of a Krull Domain, Springer-Verlag, 1973.

[162] L. Fouli and K.-N. Lin, Rees algebras of square-free monomial ideals,J. Commut. Algebra, to appear.

[163] C. Francisco, H. T. Ha, and J. Mermin, Powers of square-free mono-mial ideals and combinatorics, in Commutative Algebra (I. Peeva Ed.),373–392, Springer, New York, 2013.

[164] C. A. Francisco, H. T. Ha, and A. Van Tuyl, Associated primes ofmonomial ideals and odd holes in graphs, J. Algebraic Combin. 32(2010), no. 2, 287–301.

[165] C. Francisco, H.T. Ha, and A. Van Tuyl, Colorings of hypergraphs,perfect graphs, and associated primes of powers of monomial ideals,J. Algebra 331 (2011), no. 1, 224–242.

[166] C. Francisco, A. Hoefel, and A. Van Tuyl, EdgeIdeals: a package for(hyper)graphs, J. Softw. Algebra Geom. 1 (2009), 1–4.

[167] C. A. Francisco and A. Van Tuyl, Sequentially Cohen–Macaulay edgeideals, Proc. Amer. Math. Soc. 135 (2007), 2327–2337.

[168] P. Frankl, Extremal set systems, in Handbook of Combinatorics II,Elsevier, 1995, pp. 1293–1329.

[169] R. Froberg, A study of graded extremal rings and of monomial rings,Math. Scand. 51 (1982), 22–34.

[170] R. Froberg, Rings with monomial relations having linear resolutions,J. Pure Appl. Algebra 38 (1985), 235–241.

[171] R. Froberg, A note on the Stanley–Reisner ring of a join and of asuspension, Manuscripta Math. 60 (1988), 89–91.

[172] R. Froberg, On Stanley–Reisner rings, in Topics in Algebra, Part 2.Polish Scientific Publishers, 1990, pp. 57–70.

BIBLIOGRAPHY 651

[173] R. Froberg, The Frobenius number of some semigroups, Comm. Alge-bra 22 (1994), 6021–6024.

[174] R. Froberg and D. Laksov, Compressed Algebras, Lecture Notes inMathematics 1092 (1984), Springer-Verlag, pp. 121–151.

[175] D. R. Fulkerson, A. J. Hoffman, and M.H. McAndrew, Some proper-ties of graphs with multiple edges, Canad. J. Math. 17 (1965), 166–177.

[176] W. Fulton, Introduction to Toric Varieties , Princeton UniversityPress, 1993.

[177] E. Gawrilow, and M. Joswig, polymake: a framework for analyzingconvex polytopes. Polytopes—combinatorics and computation (Ober-wolfach, 1997), 43–73, DMV Sem., 29, Birkhauser, Basel, 2000.

[178] A. V. Geramita, M. Kreuzer, and L. Robbiano, Cayley–Bacharachschemes and their canonical modules, Trans. Amer. Math. Soc. 339(1993), no. 1, 163–189.

[179] A. M. H. Gerards and A. Sebo, Total dual integrality implies localstrong unimodularity, Math. Programming 38 (1987), no. 1, 69–73.

[180] R. Gilmer, Commutative Semigroup Rings , Chicago Lectures inMath., Univ. of Chicago Press, Chicago, 1984.

[181] P. Gimenez, A. Simis, W. V. Vasconcelos, and R. H. Villarreal, Oncomplete monomial ideals, J. Commut. Algebra, to appear.

[182] I. Gitler, E. Reyes, and J. A. Vega, Complete intersection toric idealsof oriented graphs and chorded-theta subgraphs, J. Algebraic Combin.38 (2013), no. 3, 721–744.

[183] I. Gitler, E. Reyes, and R. H. Villarreal, Blowup algebras of ideals ofvertex covers of bipartite graphs, Contemp. Math. 376 (2005), 273–279.

[184] I. Gitler, E. Reyes, and R. H. Villarreal, Ring graphs and completeintersection toric ideals, Discrete Math. 310 (2010), no. 3, 430–441.

[185] I. Gitler, E. Reyes, and R. H. Villarreal, Blowup algebras of square–free monomial ideals and some links to combinatorial optimizationproblems, Rocky Mountain J. Math. 39 (2009), no. 1, 71–102.

[186] I. Gitler and C. Valencia, Bounds for invariants of edge-rings, Comm.Algebra 33 (2005), 1603–1616.

652 BIBLIOGRAPHY

[187] I. Gitler, C. Valencia, and R. H. Villarreal, A note on the Rees algebraof a bipartite graph, J. Pure Appl. Algebra 201 (2005), 17–24.

[188] I. Gitler, C. Valencia, and R. H. Villarreal, A note on Rees algebrasand the MFMC property, Beitrage Algebra Geom. 48 (2007), no. 1,141–150.

[189] I. Gitler and R. H. Villarreal, Graphs, Rings and Polyhedra, Aporta-ciones Mat. Textos, 35, Soc. Mat. Mexicana, Mexico, 2011.

[190] C. Godsil and G. Royle, Algebraic Graph Theory, Graduate Texts inMathematics 207, Springer, 2001.

[191] M. C. Golumbic, Algorithmic Graph Theory and Perfect Graphs , Sec-ond Edition, Annals of Discrete Mathematics 57, Elsevier ScienceB.V., Amsterdam, 2004.

[192] M. Gonzalez-Sarabia, C. Renterıa, and A. J. Sanchez, Minimum dis-tance of some evaluation codes, Appl. Algebra Engrg. Comm. Comput.24 (2013), no. 2, 95–106.

[193] A. Goodarzi, On the Hilbert series of monomial ideals, J. Combin.Theory Ser. A 120 (2013), no. 2, 315–317.

[194] P. Gordan, Uber die Auflosung linearer Gleichungen mit reellen Co-efficienten, Math. Ann. 6 (1873), no. 1, 23–28.

[195] F. Goring, Short proof of Menger’s theorem, Discrete Math. 219(2000), no. 1-3, 295–296.

[196] E. Gorla, J. C. Migliore, and U. Nagel, Grobner bases via linkage, J.Algebra 384 (2013), 110–134.

[197] S. Goto, The Veronesean subrings of Gorenstein rings, J. Math. KyotoUniv. 16 (1976), 51–55.

[198] S. Goto and K. Nishida, The Cohen–Macaulay and Gorenstein Reesalgebras associated to filtrations, Mem. Amer. Math. Soc. 110 (1994),no. 526, 1–134.

[199] D.R. Grayson and M.E. Stillman, Macaulay2, a software system forresearch in algebraic geometry.Available at http://www.math.uiuc.edu/Macaulay2/.

[200] G. M. Greuel and G. Pfister, A Singular Introduction to CommutativeAlgebra, 2nd extended edition, Springer, Berlin, 2008.

BIBLIOGRAPHY 653

[201] J. Grossman, D. M. Kulkarni, and I. Schochetman, On the minors ofan incidence matrix and its Smith normal form, Linear Algebra Appl.218 (1995), 213–224.

[202] A. Grothendieck, Local Cohomology, Lecture Notes in Mathematics41, Springer-Verlag, Berlin, 1967.

[203] H. T. Ha and K.-N. Lin, Normal 0-1 polytopes. Preprint, 2013,arXiv:1309.4807.

[204] H. T. Ha and S. Morey, Embedded associated primes of powers ofsquare-free monomial ideals, J. Pure Appl. Algebra 214 (2010), no. 4,301–308.

[205] H. T. Ha, S. Morey, and R. H. Villarreal, Cohen–Macaulay admissibleclutters, J. Commut. Algebra 1 (2009), no. 3, 463–480.

[206] H. T. Ha and A. Van Tuyl, Monomial ideals, edge ideals of hyper-graphs, and their graded Betti numbers, J. Algebraic Combin. 27(2008), 215–245.

[207] H. Haghighi, N. Terai, S. Yassemi, and R. Zaare-Nahandi, Sequen-tially Sr simplicial complexes and sequentially S2 graphs, Proc. Amer.Math. Soc. 139 (2011), no. 6, 1993–2005.

[208] F. Harary, Graph Theory, Addison-Wesley, Reading, MA, 1972.

[209] F. Harary and M. D. Plummer, On indecomposable graphs, Canad.J. Math. 19 (1967), 800–809.

[210] R. Hartshorne, Algebraic Geometry, Springer-Verlag, New York, 1977.

[211] J. He and A. Van Tuyl, Algebraic properties of the path ideal of atree, Comm. Algebra 38 (2010), no. 5, 1725–1742.

[212] B. Hedman, The maximum number of cliques in dense graphs, Dis-crete Math. 54 (1985), 161–166.

[213] W. Heinzer, A. Mirbagheri, and L.J. Ratliff, Jr., Parametric decom-position of monomial ideals II, J. Algebra 187 (1997), 120–149.

[214] W. Heinzer, L.J. Ratliff, Jr., and K. Shah, Parametric decompositionof monomial ideals I, Houston J. Math. 21 (1995), 29–52.

[215] R. Hemmecke, On the computation of Hilbert bases of cones, Math-ematical software (Beijing, 2002), 307–317, World Sci. Publ., RiverEdge, NJ, 2002.

654 BIBLIOGRAPHY

[216] J. Herzog, Generators and relations of abelian semigroups and semi-group rings, Manuscripta Math. 3 (1970), 175–193.

[217] J. Herzog, Ein Cohen–Macaulay–Kriteriummit Anwendungen auf denKonormalenmodul und den Differentialmodul, Math. Z. 163 (1978),149–162.

[218] J. Herzog, Alexander duality in commutative algebra and combina-torics, Algebra Colloq. 11 (2004), no. 1, 21–30.

[219] J. Herzog, A generalization of the Taylor complex construction,Comm. Algebra 35 (2007), no. 5, 1747–1756.

[220] J. Herzog and T. Hibi, Componentwise linear ideals, Nagoya Math. J.153 (1999), 141–153.

[221] J. Herzog and T. Hibi, Discrete polymatroids, J. Algebraic Combin.16 (2002), no. 3, 239–268.

[222] J. Herzog and T. Hibi, Distributive lattices, bipartite graphs andAlexander duality, J. Algebraic Combin. 22 (2005), no. 3, 289–302.

[223] J. Herzog and T. Hibi, The depth of powers of an ideal, J. Algebra291 (2005), 534–550.

[224] J. Herzog and T. Hibi, Monomial Ideals, Graduate Texts in Mathe-matics 260, Springer-Verlag, 2011.

[225] J. Herzog, T. Hibi, and N. V. Trung, Symbolic powers of monomialideals and vertex cover algebras, Adv. Math. 210 (2007), 304–322.

[226] J. Herzog, T. Hibi, N. V. Trung, and X. Zheng, Standard gradedvertex cover algebras, cycles and leaves, Trans. Amer. Math. Soc.360 (2008), 6231–6249.

[227] J. Herzog, T. Hibi, and M. Vladoiu, Ideals of fiber type and polyma-troids, Osaka J. Math. 42 (2005), 1–23.

[228] J. Herzog, T. Hibi, and X. Zheng, Dirac’s theorem on chordal graphsand Alexander duality. European J. Combin. 25 (2004), no. 7, 949–960.

[229] J. Herzog, T. Hibi, and X. Zheng, Monomial ideals whose powers havea linear resolution, Math. Scand. 95 (2004), 23–32.

[230] J. Herzog and M. Kuhl, On the Betti numbers of finite pure and linearresolutions, Comm. Algebra 12 (1984), 1627–1646.

BIBLIOGRAPHY 655

[231] J. Herzog and D. Popescu, Finite filtrations of modules and shellablemulticomplexes, Manuscripta Math. 121 (2006), no. 3, 385–410.

[232] J. Herzog, A. Rauf, and M. Vladoiu, The stable set of associated primeideals of a polymatroidal ideal, J. Algebraic Combin. 37 (2013), no.2, 289–312.

[233] J. Herzog, A. Simis, and W. V. Vasconcelos, Koszul Homology andBlowing-Up Rings, Lecture Notes in Pure and Applied Math. 84, Mar-cel Dekker, New York, 1983, pp. 79–169.

[234] J. Herzog, A. Simis, and W. V. Vasconcelos, On the arithmetic andhomology of algebras of linear type, Trans. Amer. Math. Soc. 283(1984), 661–683.

[235] J. Herzog, A. Simis, and W. V. Vasconcelos, On the canonical moduleof the Rees algebra and the associated graded ring of an ideal, J.Algebra 105 (1987), 285–302.

[236] J. Herzog, A. Simis, and W. V. Vasconcelos, Arithmetic of normalRees algebras, J. Algebra 143 (1991), 269–294.

[237] J. Herzog and H. Srinivasan, Bounds for multiplicities, Trans. Amer.Math. Soc. 350 (1998), 2879–2902.

[238] J. Herzog, W. V. Vasconcelos, and R. H. Villarreal, Ideals with slidingdepth, Nagoya Math. J. 99 (1985), 159–172.

[239] T. Hibi, Union and glueing of a family of Cohen–Macaulay partiallyordered sets, Nagoya Math. J. 107 (1987), 91–119.

[240] T. Hibi, Algebraic Combinatorics on Convex Polytopes , Carslaw Pub-lications, Australia, 1992.

[241] T. Hibi, A. Higashitani, A. Tsuchiya, and K. Yoshida Ehrhart poly-nomials with negative coefficients. Preprint, 2013, arXiv:1312.7049.

[242] T. Hibi and H. Ohsugi, Normal polytopes arising from finite graphs,J. Algebra 207 (1998), 409–426.

[243] T. Hibi and N. Terai, Betti numbers of minimal free resolutionsof Stanley–Reisner rings, Semigroups, formal languages and com-binatorial on words (Japanese) (Kyoto 1994), SurikaisekikenkyushoKokyuroku 910 (1995), 98–107.

[244] T. Hibi and N. Terai, Alexander duality theorem and second Bettinumbers of Stanley–Reisner rings, Adv. Math. 124 (1996), 332–333.

656 BIBLIOGRAPHY

[245] P.J. Hilton and U. Stammbach, A Course in Homological Algebra,Graduate Texts in Mathematics 4, Springer-Verlag, 1971.

[246] L. T. Hoa and N. D. Tam, On some invariants of a mixed product ofideals, Arch. Math. 94 (2010), no. 4, 327–337.

[247] C. T. Hoang, On a conjecture of Meyniel, J. Combin. Theory Ser. B42 (1987), no. 3, 302–312.

[248] M. Hochster, Rings of invariants of tori, Cohen–Macaulay rings gener-ated by monomials, and polytopes, Ann. of Math. 96 (1972), 318–337.

[249] M. Hochster, Criteria for equality of ordinary and symbolic powers ofprime ideals, Math. Z. 133 (1973), 53–65.

[250] M. Hochster, Cohen–Macaulay rings, combinatorics and simplicialcomplexes, in Ring Theory II , Lecture Notes in Pure and AppliedMath. 26, Marcel Dekker, New York, 1977, pp. 171–223.

[251] S. Hosten and J. Shapiro, Primary decomposition of lattice basis ide-als, J. Symbolic Comput. 29 (2000), no. 4-5, 625–639.

[252] S. Hosten and R. R. Thomas, Gomory integer programs, Math. Pro-gramming 96 (2003), no. 2, Ser. B, 271–292.

[253] C. Huneke, Linkage and the Koszul homology of ideals, Amer. J.Math. 104 (1982), 1043–1062.

[254] C. Huneke, On the associated graded ring of an ideal, Illinois J. Math.26 (1982), 121–137.

[255] C. Huneke, Hyman Bass and ubiquity: Gorenstein rings, Contemp.Math. 243 (1999), 55–78.

[256] C. Huneke and M. Miller, A note on the multiplicity of Cohen–Macaulay algebras with pure resolutions, Canad. J. Math. 37 (1985),1149–1162.

[257] C. Huneke and J. Ribbe, Symbolic squares in regular local rings,Math.Z. 229 (1998), 31–44.

[258] C. Huneke, A. Simis, and W. V. Vasconcelos, Reduced normal conesare domains, Contemp. Math. 88 (1989), 95–101.

[259] C. Huneke and I. Swanson, Integral Closure of Ideals Rings, and Mod-ules, London Math. Soc., Lecture Note Series 336, Cambridge Uni-versity Press, Cambridge, 2006.

BIBLIOGRAPHY 657

[260] C. Ionescu and G. Rinaldo, Some algebraic invariants related to mixedproduct ideals, Arch. Math. 91 (2008), 20–30.

[261] N. Jacobson, Basic Algebra I , Second Edition, W. H. Freeman andCompany, New York, 1996.

[262] N. Jacobson, Basic Algebra II , W. H. Freeman and Company, NewYork, 1989.

[263] P. M. Johnson, A weighted graph problem from commutative algebra.Preprint, 2007, arXiv:0704.3696.

[264] T. Jozefiak, Ideals generated by minors of a symmetric matrix, Com-ment. Math. Helvetici 53 (1978), 595–607.

[265] L. Juan, Some results about symmetric semigroups, Comm. Algebra21 (1993), 3637–3645.

[266] T. Kahle, Decompositions of binomial ideals, J. Softw. Algebra Geom.4 (2012), 1–5.

[267] T. Kahle and E. Miller, Decompositions of commutative monoid con-gruences and binomial ideals, Algebra Number Theory, to appear.

[268] T. Kaiser, M. Stehlık, and R. Skrekovski, Replication in critical graphsand the persistence of monomial ideals, J. Combin. Theory Ser. A 123(2014), no. 1, 239–251.

[269] G. Kalai and R. Meshulam, Unions and intersections of Leray com-plexes, J. Combin. Theory Ser. A 113 (2006), 1586–1592.

[270] A. Katsabekis, M. Morales, and A. Thoma, Binomial generation ofthe radical of a lattice ideal, J. Algebra 324 (2010), no. 6, 1334–1346.

[271] A. Katsabekis and A. Thoma, Toric sets and orbits on toric varieties,J. Pure Appl. Algebra 181 (2003), 75–83.

[272] A. Katsabekis and A. Thoma, Parametrizations of toric varieties overany field, J. Algebra 308 (2007), no. 2, 751–763.

[273] M. Katzman, Bipartite graphs whose edge algebras are complete in-tersections, J. Algebra 220 (1999), 519–530.

[274] M. Katzman, Characteristic-independence of Betti numbers of graphideals, J. Combin. Theory Ser. A 113 (2006), no. 3, 435–454.

[275] G. Kempf, F. Knudsen, D. Mumford, and B. Saint-Donat, ToroidalEmbeddings I, Lecture Notes in Mathematics 339, Springer, 1973.

658 BIBLIOGRAPHY

[276] K. Kimura, Non-vanishingness of Betti numbers of edge ideals, in:Harmony of Grobner Bases and the Modern Industrial Society, WorldScientific, 2012, pp. 153–168.

[277] K. Kimura, Non-vanishingness of Betti numbers of edge ideals andcomplete bipartite subgraphs. Preprint, 2013, arXiv:1306.1333.

[278] K. Kimura and N. Terai, Binomial arithmetical rank of edge ideals offorests, Proc. Amer. Math. Soc. 141 (2013), no. 6, 1925–1932.

[279] K. Kimura, N. Terai, and K. Yoshida, Arithmetical rank of squarefreemonomial ideals of small arithmetic degree, J. Algebraic Combin. 29(2009), 389–404.

[280] S. Kleiman and B. Ulrich, Gorenstein algebras, symmetric matrices,self-linked ideals, and symbolic powers, Trans. Amer. Math. Soc. 349(1997), 4973–5000.

[281] B. Korte and J. Vygen, Combinatorial Optimization Theory and Al-gorithms, Algorithms and Combinatorics 21, Springer, 2000.

[282] M. Kreuzer and L. Robbiano, Computational Commutative Algebra 2,Springer-Verlag, Berlin, 2005.

[283] J. Kruskal, The number of simplices in a complex, in MathematicalOptimization Techniques (R. Bellman, Ed.), University of CaliforniaPress, 1963, pp. 251–278.

[284] M. Kubitzke and A. Olteanu, Algebraic properties of classes of pathideals of posets, J. Pure Appl. Algebra 218 (2014), no. 6, 1012–1033.

[285] M. Kummini, Regularity, depth and arithmetic rank of bipartite edgeideals, J. Algebraic Combin. 30 (2009), no. 4, 429–445.

[286] J. P. S. Kung, Basis-exchange properties, Theory of matroids , 62–75,Encyclopedia Math. Appl. 26, Cambridge Univ. Press, 1986.

[287] E. Kunz, The value-semigroup of a one dimensional Gorenstein ring,Proc. Amer. Math. Soc. 25 (1970), 748–751.

[288] M. La Barbiera and G. Restuccia, Mixed product ideals generated bys-sequences, Algebra Colloq. 18 (2011), no. 4, 553–570.

[289] M. D. Larsen, Multiplicative Theory of Ideals , Pure and AppliedMathematics 43, Academic Press, 1971.

[290] A. Lehman, On the width-length inequality and degenerate projectiveplanes, in Polyhedral Combinatorics (W. Cook and P. Seymour Eds.)DIMACS Series in Discrete Mathematics and Theoretical ComputerScience 1, Amer. Math. Soc., 1990, pp. 101–105.

BIBLIOGRAPHY 659

[291] M. Llabres and F. Rossello, A new family of metrics for biopolymercontact structures, Comput. Biol. Chem. 28 (2004), no. 1, 21–37.

[292] H. H. Lopez, C. Renterıa, and R. H. Villarreal, Affine cartesian codes,Des. Codes Cryptogr. 71 (2014), no. 1, 5–19.

[293] H. H. Lopez and R. H. Villarreal, Complete intersections in binomialand lattice ideals, Internat. J. Algebra Comput. 23 (2013), no. 6,1419–1429.

[294] H. H. Lopez and R. H. Villarreal, Computing the degree of a latticeideal of dimension one, J. Symbolic Comput. 65 (2014), 15–28.

[295] D. Lorenzini, Smith normal form and Laplacians, J. Combin. TheorySer. B 98 (2008), 1271–1300.

[296] L. Lovasz, Normal hypergraphs and the perfect graph conjecture, Dis-crete Math. 2 (1972), no. 3, 253–267.

[297] L. Lovasz and M. D. Plummer, Matching Theory, Annals of DiscreteMathematics 29, Elsevier Science B.V., Amsterdam, 1986.

[298] C. L. Lucchesi and D. H. Younger, A minimax theorem for directedgraphs, J. London Math. Soc. (2) 17 (1978), no. 3, 369–374.

[299] G. Lyubeznik, Set Theoretic Intersections and Monomial Ideals , PhDthesis, Columbia University, 1984.

[300] G. Lyubeznik, The minimal non-Cohen–Macaulay monomial ideals,J. Pure Appl. Algebra 51 (1988), 261–266.

[301] G. Lyubeznik, On the arithmetical rank of monomial ideals, J. Algebra112 (1988), 86–89.

[302] G. Lyubeznik, A survey of problems and results on the number ofdefining equations, in Commutative Algebra (M. Hochster et. al.,Eds.), MSRI Publications 15, Springer-Verlag, 1989, 375–390.

[303] M. Mahmoudi, A. Mousivand, M. Crupi, G. Rinaldo, N. Terai, and S.Yassemi, Vertex decomposability and regularity of very well-coveredgraphs, J. Pure Appl. Algebra 215 (2011), no. 10, 2473–2480.

[304] W. H. Marlow, Mathematics for Operation Research, New York, Wi-ley, 1978.

[305] J. Martınez-Bernal, S. Morey, and R. H. Villarreal, Associated primesof powers of edge ideals, Collect. Math. 63 (2012), no. 3, 361–374.

660 BIBLIOGRAPHY

[306] J. Martınez-Bernal, E. O’Shea, and R. H. Villarreal, Ehrhart clutters:regularity and max-flow min-cut, Electron. J. Combin. 17 (2010), no.1, R52.

[307] J. Martınez-Bernal, C. Renterıa, and R. H. Villarreal, Combinatoricsof symbolic Rees algebras of edge ideals of clutters, Contemp. Math.555 (2011), 151–164.

[308] J. Martınez-Bernal and R. H. Villarreal, Toric ideals generated bycircuits, Algebra Colloq. 19 (2012), no. 4, 665–672.

[309] H. Matsumura, Commutative Algebra, Benjamin-Cummings, Read-ing, MA, 1980.

[310] H. Matsumura, Commutative Ring Theory, Cambridge Studies in Ad-vanced Mathematics 8, Cambridge University Press, 1986.

[311] T. Matsuoka, On an invariant of Veronesean rings, Proc. Japan Acad.50 (1974), 287–291.

[312] S. McAdam, Asymptotic prime divisors and analytic spreads, Proc.Amer. Math. Soc. 80 (1980), 555–559.

[313] S. McAdam, Asymptotic Prime Divisors, Lecture Notes in Mathemat-ics 103, Springer–Verlag, New York, 1983.

[314] P. McMullen, The maximum numbers of faces of a convex polytope,Mathematika 17 (1970), 179–184.

[315] P. McMullen and G. Shephard, Convex Polytopes and the UpperBound Conjecture, London Mathematical Society, Lect. Note Ser. 3,Cambridge University Press, 1971.

[316] K. Menger, Zur allgemeinen Kurventheorie, Fundamenta Mathemati-cae 10 (1927), 96–115.

[317] E. Miller and B. Sturmfels, Combinatorial Commutative Algebra,Graduate Texts in Mathematics 227, Springer, 2004.

[318] M. Miller and R. H. Villarreal, A note on generators of least degreein Gorenstein ideals, Proc. Amer. Math. Soc. 124 (1996), 377–382.

[319] T. T. Moh, Set-theoretic complete intersections, Proc. Amer. Math.Soc. 94 (1985), 217–220.

[320] B. Mohar and C. Thomassen, Graphs on Surfaces , Johns HopkinsUniversity Press, Baltimore, Maryland, 1997.

BIBLIOGRAPHY 661

[321] S. Moradi and D. Kiani, Bounds for the regularity of edge ideal ofvertex decomposable and shellable graphs, Bull. Iranian Math. Soc.36 (2010), 267–277.

[322] M. Morales and A. Thoma, Complete intersection lattice ideals, J.Algebra 284 (2005), 755–770.

[323] S. Morey, Stability of associated primes and equality of ordinary andsymbolic powers of ideals, Comm. Algebra 27 (1999), 3221–3231.

[324] S. Morey, Depths of powers of the edge ideal of a tree, Comm. Algebra38 (2010), no. 11, 4042–4055.

[325] S. Morey, E. Reyes, and R. H. Villarreal, Cohen–Macaulay, shellableand unmixed clutters with a perfect matching of Konig type, J. PureAppl. Algebra 212 (2008), no. 7, 1770–1786.

[326] S. Morey and R. H. Villarreal, Edge ideals: algebraic and combi-natorial properties, in Progress in Commutative Algebra, Combina-torics and Homology, Vol. 1 (C. Francisco, L. C. Klingler, S. Sather-Wagstaff, and J. C. Vassilev, Eds.), De Gruyter, Berlin, 2012, pp.85–126.

[327] T. S. Motzkin, Comonotone curves and polyhedra, Abstract, Bull.Amer. Math. Soc. 63 (1957), 35.

[328] U. Nagel and S. Petrovic, Properties of cut ideals associated to ringgraphs, J. Commut. Algebra 1 (2009), no. 3, 547–565.

[329] W. Narkiewicz, Elementary and Analytic Theory of Algebraic Num-bers , Second Edition, Springer-Verlag, Berlin; PWN—Polish ScientificPublishers, Warsaw, 1990.

[330] J. Neves, M. Vaz Pinto, and R. H. Villarreal, Regularity and algebraicproperties of certain lattice ideals, Bull. Braz. Math. Soc. (N.S.) 45(2014), no. 4, 777–806.

[331] E. Nevo, Regularity of edge ideals of C4-free graphs via the topologyof the lcm-lattice, J. Combin. Theory Ser. A. 118 (2011), no. 2, 491–501.

[332] M. Newman, Integral Matrices , Pure and Applied Mathematics 45,Academic Press, New York, 1972.

[333] K. Nishida, Powers of ideals in Cohen–Macaulay rings, J. Math. Soc.Japan 48 (1996), 409–426.

[334] D. G. Northcott, A generalization of a theorem on the contents ofpolynomials, Proc. Camb. Phil. Soc. 55 (1959), 282–288.

662 BIBLIOGRAPHY

[335] L. O’Carroll, F. Planas-Vilanova, and R. H. Villarreal, Degree andalgebraic properties of lattice and matrix ideals, SIAM J. DiscreteMath. 28 (2014), no. 1, 394–427.

[336] M. Ohtani, Graphs and ideals generated by some 2-minors, Comm.Algebra 39 (2011), no. 3, 905–917.

[337] I. Ojeda and R. Peidra, Cellular binomial ideals. Primary decomposi-tion of binomial ideals, J. Symbolic Comput. 30 (2000), 383–400.

[338] J. Oxley, Matroid Theory, Oxford University Press, Oxford, 1992.

[339] C. Peskine and L. Szpiro, Liaison des varietes algebriques, Invent.Math. 26 (1974), 271–302.

[340] M. D. Plummer, Some covering concepts in graphs, J. CombinatorialTheory 8 (1970), 91–98.

[341] J. L. Ramırez Alfonsın, The Diophantine Frobenius Problem, OxfordLecture Series in Mathematics and its Applications, 30, Oxford Uni-versity Press, Oxford, 2005.

[342] L. J. Ratliff, Jr., On prime divisors of In, n large, Michigan Math. J.23 (1976), no. 4, 337–352.

[343] L. J. Ratliff, Jr., On asymptotic prime divisors, Pacific J. Math. 111(1984), no. 2, 395–413.

[344] G. Ravindra, Well-covered graphs, J. Combinatorics InformationSyst. Sci. 2 (1977), no. 1, 20–21.

[345] L. Reid, L. G. Roberts, and M. A. Vitulli, Some results on normalhomogeneous ideals, Comm. Algebra 31 (2003), no. 9, 4485–4506.

[346] G. Reisner, Cohen–Macaulay quotients of polynomial rings, Adv.Math. 21 (1976), 31–49.

[347] P. Renteln, The Hilbert series of the face ring of a flag complex, GraphsCombin. 18 (2002), no. 3, 605–619.

[348] C. Renterıa, A. Simis, and R. H. Villarreal, Algebraic methods for pa-rameterized codes and invariants of vanishing ideals over finite fields,Finite Fields Appl. 17 (2011), no. 1, 81–104.

[349] C. Renterıa and H. Tapia, Reed–Muller codes: an ideal theory ap-proach, Comm. Algebra 25 (1997), 401–413.

[350] C. Renterıa and R. H. Villarreal, Koszul homology of Cohen–Macaulay rings with linear resolutions, Proc. Amer. Math. Soc. 115(1992), 51–58.

BIBLIOGRAPHY 663

[351] G. Restuccia and R. H. Villarreal, On the normality of monomialideals of mixed products, Comm. Algebra 29 (2001), 3571–3580.

[352] E. Reyes, C. Tatakis, and A. Thoma, Minimal generators of toricideals of graphs, Adv. in Appl. Math. 48 (2012), no. 1, 64–78.

[353] E. Reyes and J. Toledo, On the strong persistence property for mono-mial ideals, preprint, 2013.

[354] E. Reyes, R. H. Villarreal, and L. Zarate, A note on affine toric vari-eties, Linear Algebra Appl. 318 (2000), 173–179.

[355] G. Rinaldo, Betti numbers of mixed product ideals, Arch. Math. 91(2008), no. 5, 416–426.

[356] G. Rinaldo, Koszulness of vertex cover algebras of bipartite graphs,Comm. Algebra 39 (2011), no. 7, 2249–2259.

[357] R. T. Rockafellar, The elementary vectors of a subspace of RN , inCombinatorial Mathematics and its Applications , Proc. Chapel HillConf., Univ. North Carolina Press, 1969, pp. 104–127.

[358] R. T. Rockafellar, Convex Analysis , Princeton University Press, 1970.

[359] O. J. Rodseth, On a linear Diophantine problem of Frobenius, J. reineangew. Math. 301 (1978), 171–178.

[360] J. C. Rosales and P. A. Garcıa-Sanchez, Numerical Semigroups , De-velopments in Mathematics 20, Springer, New York, 2009.

[361] M. E. Rossi and G. Valla, Multiplicity and t-isomultiple ideals, NagoyaMath. J. 110 (1988), 81–111.

[362] M. Roth and A. Van Tuyl, On the linear strand of an edge ideal,Comm. Algebra 35 (2007), no. 3, 821–832.

[363] J. J. Rotman, An Introduction to Homological Algebra, AcademicPress, 1979.

[364] H. Sabzrou and F. Rahmati, Matrices defining Gorenstein lattice ide-als, Rocky Mountain J. Math. 35 (2005), no. 3, 1029–1042.

[365] H. Sabzrou and F. Rahmati, The Frobenius number and a-invariant,Rocky Mountain J. Math. 36 (2006), no. 6, 2021–2026.

[366] J. Sally, Cohen–Macaulay rings of maximal embedding dimension, J.Algebra 56 (1979), 168–183.

664 BIBLIOGRAPHY

[367] E. Sarmiento, M. Vaz Pinto, and R. H. Villarreal, The minimum dis-tance of parameterized codes on projective tori, Appl. Algebra Engrg.Comm. Comput. 22 (2011), no. 4, 249–264.

[368] H. Scarf and D. Shallcross, The Frobenius problem and maximal lat-tice free bodies, Math. Oper. Res. 18 (1993), 511–515.

[369] H. Schenck, Computational Algebraic Geometry, London Math. Soc.,Student Texts 58, Cambridge University Press, Cambridge, 2003.

[370] P. Schenzel, Uber die freien auflosungen extremaler Cohen–Macaulay-ringe, J. Algebra 64 (1980), 93–101.

[371] A. Schrijver, On total dual integrality, Linear Algebra Appl. 38 (1981),27–32.

[372] A. Schrijver, Theory of Linear and Integer Programming , John Wiley& Sons, New York, 1986.

[373] A. Schrijver,Combinatorial Optimization, Algorithms and Combina-torics 24, Springer-Verlag, Berlin, 2003.

[374] E. Selmer, On the linear diophantine problem of Frobenius, Journalfur Mathematik, band 293/294 (1977), 1–17.

[375] P. D. Seymour, The matroids with the max-flow min-cut property, J.Combinatorial Theory Ser. B 23 (1977), 189–222.

[376] Y. H. Shen, On a class of squarefree monomial ideals of linear type,J. Commut. Algebra, to appear.

[377] A. Simis, Combinatoria Algebrica, XVIII Coloquio Brasileiro deMatematica, IMPA, 1991 (Apendice. Palimpsesto 2: Potencias sim-bolicas, 2.1).

[378] A. Simis, Effective computation of symbolic powers by Jacobian ma-trices, Comm. Algebra 24 (1996), 3561–3565.

[379] A. Simis, On the Jacobian module associated to a graph, Proc. Amer.Math. Soc. 126 (1998), 989–997.

[380] A. Simis and N.V. Trung, The divisor class group of ordinary andsymbolic blow-ups, Math. Z. 198 (1988), 479–491.

[381] A. Simis and B. Ulrich, On the ideal of an embedded join, J. Algebra226 (2000), no. 1, 1–14.

[382] A. Simis andW. V. Vasconcelos, The syzygies of the conormal module,Amer. J. Math. 103 (1981), 203–224.

BIBLIOGRAPHY 665

[383] A. Simis, W. V. Vasconcelos, and R. H. Villarreal, On the ideal theoryof graphs, J. Algebra 167 (1994), 389–416.

[384] A. Simis, W. V. Vasconcelos, and R. H. Villarreal, The integral closureof subrings associated to graphs, J. Algebra 199 (1998), 281–289.

[385] A. Simis and R. H. Villarreal, Constraints for the normality of mono-mial subrings and birationality, Proc. Amer. Math. Soc. 131 (2003),2043–2048.

[386] A. Simis and R. H. Villarreal, Combinatorics of Cremona monomialmaps, Math. Comp. 81 (2012), no. 279, 1857–1867.

[387] J. M. S. Simoes-Pereira, On matroids on edge sets of graphs withconnected subgraphs as circuits II, Discrete Math. 12 (1975), 55–78.

[388] D. E. Smith, On the Cohen–Macaulay property in commutative alge-bra and simplicial topology, Pacific J. Math. 141 (1990), 165–196.

[389] A. Soleyman Jahan, and X. Zheng, Pretty clean monomial ideals andlinear quotients. Preprint, 2007, arXiv:0707.2914.

[390] E. Sperner, Ein satz uber untermengen einer endlichen menge, Math.Z. 27 (1928), 544–548.

[391] R. Stanley, The upper bound conjecture and Cohen–Macaulay rings,Stud. Appl. Math. 54 (1975), 135–142.

[392] R. Stanley, Hilbert functions of graded algebras, Adv. Math. 28(1978), 57–83.

[393] R. Stanley, The number of faces of simplicial polytopes and spheres,Ann. New York Acad. Sci. 440 (1985), pp. 212–223.

[394] R. Stanley, Enumerative Combinatorics I , Wadsworth-Brooks/Cole,Monterey, California, 1986.

[395] R. Stanley, Combinatorics and Commutative Algebra, BirkhauserBoston, 2nd ed., 1996.

[396] R. Stanley, Algebraic Combinatorics , Springer, New York, 2013.

[397] I. Stewart and D. Tall, Algebraic Number Theory, Chapman and HallMathematics Series, 1979.

[398] B. Sturmfels, Grobner bases and Stanley decompositions of determi-nantal rings, Math. Z. 205 (1990), 137–144.

[399] B. Sturmfels, Grobner bases of toric varieties, Tohoku Math. J. 43(1991), 249–261.

666 BIBLIOGRAPHY

[400] B. Sturmfels, Grobner Bases and Convex Polytopes , University Lec-ture Series 8, American Mathematical Society, Rhode Island, 1996.

[401] N. Terai, Alexander duality theorem and Stanley–Reisner rings,Surikaisekikenkyusho Kokyuroku 1078 (1999), 174–184.

[402] A. Thoma, On the set-theoretic complete intersection problem formonomial curves in An and Pn, J. Pure Appl. Algebra 104 (1995),333–344.

[403] B. Toft, Colouring, stable sets and perfect graphs, in Handbook ofCombinatorics I, Elsevier, 1995, pp. 233–288.

[404] B. Ulrich, Gorenstein rings and modules with high numbers of gener-ators, Math. Z. 188 (1984), 23–32.

[405] C. Valencia and R. H. Villarreal, Canonical modules of certain edgesubrings, European J. Combin. 24 (2003), no. 5, 471–487.

[406] C. Valencia and R. H. Villarreal, Explicit representations of the edgecone of a graph, Int. Journal Contemp. Math. Sciences 1 (2006), 53-66.

[407] A. Van Tuyl, Sequentially Cohen–Macaulay bipartite graphs: vertexdecomposability and regularity, Arch. Math. 93 (2009), 451–459.

[408] A. Van Tuyl, A beginner’s guide to edge and cover ideals, inMonomialIdeals, Computations and Applications, Lecture Notes in Mathematics2083, Springer, 2013, pp. 63–94.

[409] A. Van Tuyl and R. H. Villarreal, Shellable graphs and sequentiallyCohen–Macaulay bipartite graphs, J. Combin. Theory Ser. A 115(2008), no.5, 799-814.

[410] W. V. Vasconcelos, Divisor Theory in Module Categories , Mathemat-ics Studies 14, North Holland, 1974.

[411] W. V. Vasconcelos, On the equations of Rees algebras, J. reine angew.Math. 418 (1991), 189–218.

[412] W. V. Vasconcelos, Arithmetic of Blowup Algebras , London Math.Soc., Lecture Note Series 195, Cambridge University Press, Cam-bridge, 1994.

[413] W. V. Vasconcelos, Computational Methods in Commutative Algebraand Algebraic Geometry, Springer-Verlag, 1998.

[414] W. V. Vasconcelos, Integral Closure, Springer Monographs in Mathe-matics, Springer-Verlag, New York, 2005.

BIBLIOGRAPHY 667

[415] R. H. Villarreal, Rees algebras and Koszul homology, J. Algebra 119(1988), 83–104.

[416] R. H. Villarreal, Cohen–Macaulay graphs, Manuscripta Math. 66(1990), 277–293.

[417] R. H. Villarreal, Rees algebras of edge ideals, Comm. Algebra 23(1995), 3513–3524.

[418] R. H. Villarreal, Normality of subrings generated by square free mono-mials, J. Pure Appl. Algebra 113 (1996), 91–106.

[419] R. H. Villarreal, On the equations of the edge cone of a graph andsome applications, Manuscripta Math. 97 (1998), 309–317.

[420] R. H. Villarreal, Monomial algebras and polyhedral geometry, inHandbook of Algebra, Vol. 3 (M. Hazewinkel Ed.), Elsevier, Ams-terdam, 2003, pp. 257–314.

[421] R. H. Villarreal, Unmixed bipartite graphs, Rev. Colombiana Mat. 41(2007), no. 2, 393–395.

[422] R. H. Villarreal, Rees cones and monomial rings of matroids, LinearAlgebra Appl. 428 (2008), 2933–2940.

[423] R. H. Villarreal, Rees algebras and polyhedral cones of ideals of vertexcovers of perfect graphs, J. Algebraic Combin. 27 (2008), 293-305.

[424] R. H. Villarreal, Normalization of monomial ideals and Hilbert func-tions, Proc. Amer. Math. Soc. 136 (2008), 1933–1943.

[425] J. Watanabe, m-full ideals, Nagoya Math. J. 106 (1987), 101–111.

[426] A. Watkins, Hilbert Functions of Face Rings Arising from Graphs ,M.S. Thesis, Rutgers University, 1992.

[427] R. Webster, Convexity, Oxford University Press, Oxford, 1994.

[428] C. Weibel, An Introduction to Homological Algebra, Cambridge Uni-versity Press, Cambridge, 1994.

[429] N. L. White, The basis monomial ring of a matroid, Adv. Math. 24(1977), 292–297.

[430] N. L. White, A unique exchange property for bases, Linear AlgebraAppl. 31 (1980), 81–91.

[431] S. Wolfram, Mathematica. A System for Doing Mathematics by Com-puter , Addison Wesley, 1989.

668 BIBLIOGRAPHY

[432] R. Woodroofe, Vertex decomposable graphs and obstructions toshellability, Proc. Amer. Math. Soc. 137 (2009), 3235–3246.

[433] R. Woodroofe, Chordal and sequentially Cohen–Macaulay clutters,Electron. J. Combin. 18 (2011), no. 1, R208.

[434] R. Woodroofe, Matchings, coverings, and Castelnuovo–Mumford reg-ularity, J. Commut. Algebra 6 (2014), no. 2, 287–304.

[435] R. Zaare-Nahandi, Cohen–Macaulayness of bipartite graphs, revisited,Bull. Malays. Math. Sci. Soc. (2), to appear.

[436] O. Zariski and P. Samuel, Commutative Algebra, Vol. II, Van Nos-trand, Princeton, 1960.

[437] X. Zheng, Resolutions of facet ideals, Comm. Algebra 32 (2004), no.6, 2301–2324.

[438] G. M. Ziegler, Lectures on Polytopes , Graduate Texts in Mathematics152, Springer-Verlag, 1994.

This page intentionally left blankThis page intentionally left blank

Monomial Algebras, Second Edition presents algebraic, combina-torial, and computational methods for studying monomial algebras and their ideals, including Stanley–Reisner rings, monomial subrings, Ehrhart rings, and blowup algebras. It emphasizes square-free mono-mials and the corresponding graphs, clutters, or hypergraphs.

New to the Second Edition• Four new chapters that focus on the algebraic properties of

blowup algebras in combinatorial optimization problems of clut-ters and hypergraphs

• Two new chapters that explore the algebraic and combinatorial properties of the edge ideal of clutters and hypergraphs

• Full revisions of existing chapters to provide an up-to-date ac-count of the subject

Bringing together several areas of pure and applied mathematics, this book shows how monomial algebras are related to polyhedral geom-etry, combinatorial optimization, and combinatorics of hypergraphs. It directly links the algebraic properties of monomial algebras to com-binatorial structures (such as simplicial complexes, posets, digraphs, graphs, and clutters) and linear optimization problems.

Features• Presents computational and combinatorial methods in commuta-

tive algebra • Shows how to solve a variety of problems of monomial algebras• Covers various affine and graded rings, including Cohen–Macau-

lay, complete intersection, and normal• Examines their basic algebraic invariants, such as multiplicity,

Betti numbers, projective dimension, and Hilbert polynomial• Contains more than 550 exercises and over 50 examples, many of

which illustrate the use of computer algebra systems

Mathematics


Rafael H. Villarreal

Villarreal

Second EditionM

onomial Algebras

K23008

w w w . c r c p r e s s . c o m

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICSMONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

K23008_cover.indd 1 2/13/15 12:23 PM

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS...

Documents

Transcript of MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS...