MT822: INTRODUCTION TO ALGEBRAIC GEOMETRY

DAWEI CHEN

Contents

1. Algebraic varieties 21.1. Affine varieties 21.2. Projective varieties 21.3. Zariski topology 31.4. Algebraic geometry and analytic geometry 31.5. Singular varieties 31.6. Ideals 41.7. Regular functions and maps 52. Sheaves and cohomology 62.1. The Mittag-Leffler problem 72.2. Sheaves 72.3. Maps of sheaves 82.4. Stalks and germs 102.5. Cohomology of sheaves 113. Vector bundles, line bundles and divisors 163.1. Holomorphic vector bundles 163.2. Vector bundles on a variety and locally free sheaves 183.3. Divisors 193.4. Line bundles 203.5. Sections of a line bundle 224. Algebraic curves 244.1. The Riemann-Roch formula 244.2. The Riemann-Hurwitz formula 254.3. Genus formula of plane curves 274.4. Base point free and very ample line bundles 284.5. Canonical maps 314.6. An informal moduli counting 334.7. Special linear systems 344.8. Weierstrass points 364.9. Plane curves and their duals 415. Chern classes 455.1. Chow ring and rational equivalence 455.2. Formal calculation of Chern classes 475.3. The splitting principle 485.4. Determinantal varieties 50

Date: Spring 2012.During the preparation of the lecture notes the author was partially supported by the NSF

grant DMS-1200329.

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5.5. Degeneracy loci 515.6. Grassmanians and Schubert calculus 526. Algebraic surfaces 546.1. Intersection theory on surfaces 546.2. Blow up 577. Schemes 607.1. The spectrum of a ring 617.2. Zariski topology revisit 617.3. Structure sheaves 627.4. Local rings 637.5. Morphisms 64

1. Algebraic varieties

The main objects in (classical) algebraic geometry are called algebraic varietiesor simply varieties. As a set, a variety consists of the common zero locus of a groupof polynomial equations in an affine or projective space.

1.1. Affine varieties. Let K be a field (e.g. C). Consider the n-dimensionalvector space An over K. Every point of An has affine coordinate (z1, . . . , zn),where zi ∈ K. For K = C, locally we can use standard complex coordinate chartsand identify the affine space An as a smooth n-dimensional complex manifold. Forany m < n, an m-dimensional linear subspace is a subvariety of An (with inducedalgebraic structure), cut out by n −m linearly independent forms, say, zi = 0 fori = 1, . . . , n −m. In general, an affine subvariety in An is the common zero locusof a collection of polynomials f1, . . . , fl ∈ K[z1, . . . , zn].

For beginners, it does not hurt to think of (smooth) varieties as (smooth, com-plex) manifolds. Then a (compact) projective variety can be glued using affinecoordinate charts, where the transition functions are holomorphic.

1.2. Projective varieties. The n-dimensional projective space

Pn ∼= (An+1\0)/K∗,

i.e. (Z0, . . . , Zn) ∼ (Z ′0, . . . , Z′n) if and only if there exists λ 6= 0 such that λZi = Z ′i

for all i. Use the homogeneous coordinates [Z0, . . . , Zn] to denote the correspondingequivalence class. A homogeneous polynomial of degree m on Pn is

F (Z0, . . . , Zn) =∑

i1+···+in=m

ai1···inZi00 · · ·Zinn .

Note that

F (λZ0, . . . , λZn) = λmF (Z0, . . . , Zn),

hence F is not a well-defined function on Pn. Nevertheless, its zero locus is well-defined and called a hypersurface of degree m. In general, a projective subvarietyin Pn is the common zero locus of a collection of homogenous polynomials, orequivalently, the intersection of a collection of hypersurfaces. The degree of thedefining homogeneous polynomial is called the degree of the hypersurface.

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Example 1.1 (P1 and the Riemann Sphere). For n = 1, we have P1 = [Z0, Z1].Let Ui be the open subset defined by Zi 6= 0 for i = 0, 1. Let x = Z1/Z0 andy = Z0/Z1. Then x and y are global (affine) coordinates for U0 and U1, respectively,and Ui ∼= A1. For x 6= 0 and y 6= 0, one can glue the one-punctured Ui’s togethervia the transition function y = 1/x to realize P1 as the Riemann sphere.

Exercise 1.2. For Pn = [Z0, . . . , Zn], let Ui be the open subset defined by Zi 6= 0for i = 0, . . . , n. Show that each Ui is isomorphic to An. How do you glue Ui’stogether to form Pn as a n-dimensional manifold?

Example 1.3 (Conics in P2). The zero locus of Z20−Z1Z2 defines a degree 2 hyper-

surface C, called a (smooth) plane conic, in the projective plane P2 = [Z0, Z1, Z2].

Exercise 1.4. Over the complex number field, can you prove that C is isomorphicto the Riemann sphere (as a complex manifold)?

Exercise 1.5. What is the (projective) dimension of space of degree 2 plane curves?In general, what is the dimension of space of degree k hypersurfaces in Pn?

Exercise 1.6. Prove that a general chosen line in PnC intersects a degree k hyper-surface at k points (counted with multiplicity). This provides another definitionfor the degree of a hypersurface.

1.3. Zariski topology. Let X ⊂ An be an affine variety. Define a basis of opensets by Uf = p ∈ X : f(p) 6= 0 for polynomials f on An. Namely, closedsets are defined by subvarieties of X. If X ⊂ Pn is a projective variety, defineUF = p ∈ X : F (p) 6= 0 for homogeneous polynomials F on Pn. Note that theZariski topology is not Hausdorff.

Exercise 1.7. Show that any two open sets of Pn under the Zariski topologyintersect.

It is convenient to formulate many results in algebraic geometry by the languageof Zariski topology. But when we visualize an algebraic variety, perhaps we still seeits geometry under the classical analytic topology.

1.4. Algebraic geometry and analytic geometry. In PnC, one can identify al-gebraic subvarieties with complex analytic subvarieties (cut out by globally holo-morphic functions). Note that polynomials are holomorphic, hence an algebraicsubvariety in Pn is obviously an analytic subvariety. But some holomorphic (tran-scendental) functions are not polynomials.

Theorem 1.8 (Chow’s Theorem). If X ⊂ PnC is a (closed) complex analytic sub-variety, then X is an algebraic subvariety.

1.5. Singular varieties. In general, algebraic geometers like to study (closed)varieties, and sometimes a variety could be singular, reducible or even non-reduced(as a scheme, explained later). We take these “pathologies” into account, becausethey naturally arise even if we only care about smooth things. Moreover, thissetting provides powerful tools, both conceptually and practically, which are notseen in the category of open, smooth manifolds.

For example, two non-parallel affine lines L1 and L2 in A2 intersects at a uniquepoint. Turn L1 gradually to make it parallel to L2. Then they are disjoint. Itimplies that the intersection number does not behave well (under deformation, say,

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moving the lines) for an open space. Nevertheless, for any two projective lines inP2, the intersection number is always one. This can be made sense even for twoidentical lines.

Now let us consider a family of plane conics Ct given by Z20 − tZ1Z2 = 0, where t

is a base parameter. For t 6= 0, Ct is smooth (prove it) and meets a general line L attwo points. For t = 0, C0 becomes a non-reduced double line supported on Z0 = 0.The intersection number remains to be 2 only if we keep the double structure of C0

instead of the reduced line Z0 = 0.

Remark 1.9. The above example implies that it is often better to characterize avariety by its defining equations than its zero locus.

Exercise 1.10 (Classification of plane conics). Consider the space of degree 2plane curves C, i.e. those cut out by degree 2 homogeneous polynomials in P2 =[Z0, Z1, Z2]. Show that up to projective equivalence, i.e. automorphisms of P2 byPGL(3), C is either a smooth conic, or a pair of lines meeting at one point, or adouble line.

1.6. Ideals. Let X ⊂ An be a variety. Define the ideal of X by

I(X) = f ∈ K[z1, . . . , zn] : f ≡ 0 on X,i.e. functions that are vanishing on X. Conversely, for an ideal I in K[z1, . . . , zn],define an affine variety V (I) by the common zero locus of polynomials in I.

Example 1.11 (Hypersurfaces). By definition, a hypersurface X defined by apolynomial f has ideal I(X) = (f).

Affine varieties are bijective to radical ideals. Recall that I is called radical, iffor a power fk ∈ I we have f ∈ I.

Theorem 1.12 (Hilbert Nullstellensatz). For any ideal I ⊂ K[z1, . . . , zn], we have

I(V (I)) = r(I).

Proof. Denote by X the variety V (I). Take a element f ∈ r(I). Then some powerfk ∈ I, hence fk is vanishing on X, which implies that f is vanishing on X.Therefore we conclude that f ∈ V (X), hence the inclusion r(I) ⊂ I(X). Completeor read the proof for the other direction.

If X ⊂ Pn is a projective variety, define I(X) by homogeneous polynomialsthat are vanishing on X. Conversely let I be a ideal (generated by homogeneouspolynomials) in K[Z0, . . . , Zn]. Define its saturation I by

I = F ∈ K[Z0, . . . , Zn] : (Z0, . . . , Zn)k · F ⊂ I for some k.Saturated ideals play the role of radical ideals that uniquely characterizes subvari-eties in Pn.

Example 1.13. Consider the ideal I = (X2, XY,XZ) ⊂ K[X,Y, Z]. In P2 its zerolocus is the line L defined by X = 0, which is the same as the zero locus of itssaturated ideal I = (X).

Exercise 1.14 (Twisted cubics). Consider the map f : P1 → P3 by

[X,Y ] 7→ [X3, X2Y,XY 2, Y 3].

The image curve R is called a twisted cubic.

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(i) Show that f is an embedding.(ii) Show that for a generally chosen plane H in P3, the intersection locus H ∩R

consists of three points. (The intersection number defines R as a degree three curve,so it is called a twisted cubic).

(iii) Describe the homogeneous ideal I(R).

1.7. Regular functions and maps. Let X ⊂ An be a variety. Define the coordi-nate ring of X to be

A(X) := K[z1, . . . , zn]/I(X).

Essentially we want to define a regular function as the restriction to X of a poly-nomial in K[z1, . . . , zn] modulo those vanishing on X, i.e. an element in A(X).Nevertheless, it is more convenient to first give a local definition.

We say that a function is regular at p ∈ X if in some neighborhood of p it can beexpressed as a quotient f/g, where f, g ∈ K[z1, . . . , zn] and g(p) 6= 0. We say thatthe funtion is regular on X if it is regular at every point of X. If X is a projectivevariety, we can adapt the same definition with the additional assumption that inthe quotient F/G, F and G are homogeneous polynomials of the same degree.

A regular map (also called a morphism) X → An is given by an n-tuple of regularfunctions:

x 7→ (f1(x), . . . , fn(x)).

A regular map X → Y ⊂ An is a map to An with image contained in Y . A mapφ : X → Pn is regular if it is regular locally, i.e. for the standard charts Ui ∼= An,φ−1(Ui)→ Ui is regular, where Ui = [Z0, . . . , Zn] : Zi 6= 0 for 0 ≤ i ≤ n.

Example 1.15. Consider the smooth conic curve C ⊂ P2 given by X2 + Y 2 −Z2.Define φ : C → P1 by

[X,Y, Z] 7→ [X,Z − Y ].

I claim that φ is a regular map.In the affine chart UX = X 6= 0, C is defined by the affine equation z2−y2 = 1,

where y = Y/X and z = Z/X. The map φ is given by

(y, z) 7→ Z − YX

= z − y,

which is regular. In the affine chart Z − Y 6= 0 in P1, make a linear transformationby Y ′ = Y + Z and Z ′ = Z − Y . Then C is given by X2 + Y ′Z ′ = 0 andφ : [X,Y ′, Z ′] 7→ [X,Z ′]. In the affine chart Z ′ 6= 0, C has equation x′2 + y′ = 0where x′ = X/Z ′ and y′ = Y ′/Z ′. The map φ given by (x′, z′) 7→ X/Z ′ = x′ is alsoregular.

In the affine plane Z 6= 0 (over R), the map is given by (x, y) 7→ (1−y)/x, whichcorresponds to projecting points on the unit disk from the north pole (0, 1) to theline y = −1. The north pole just maps to the “infinity” when we compactify theaffine line to P1.

Remark 1.16. Note that at the point [0, 1, 1] the map looks like not well-defined,because X = Z − Y = 0. This suggests that writing down an (n + 1)-tuple ofhomogeneous polynomials of the same degree, we may not immediately see whetheror not it defines a regular map.

What is the relationship between the locally defined regular functions and ele-ments in the coordinate ring?

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Theorem 1.17. The ring of regular functions on X is the coordinate ring A(X).More generally, let Uf = p ∈ X : f(p) 6= 0. Then the ring of regular functionson Uf is the localization A(X)[1/f ].

Proof. Note that K[z1, . . . , zn] is a Noetherian ring, i.e. any ideal is finitely gen-erated. In other words, it satisfies the Descending Chain Condition, i.e. for achain of closed subsets Y1 ⊃ Y2 ⊃ · · · under Zariski topology, for some m we haveYm = Ym+1 = · · · . For any regular function g on Uf , we can find a finite open coverUα of Uf such that in each Uα we can write g = hα/kα, where kα(p) 6= 0 for anyp ∈ Uα. Since kα is nowhere zero in Uα and Uα forms a cover, the common zerolocus of the kα must be contained in the zero locus of f . By Hilbert Nullstellensatz,we have fm ∈ (. . . , kα, . . .) for some m, i.e.

fm =∑α

lαkα.

Consequently we have

fm · g =∑α

lαhα,

i.e. g = (∑α lαhα)/fm ∈ A(X)[1/f ].

Now take f to be a constant function. We have Uf = X and A(X)[1/f ] =A(X).

Corollary 1.18. Any regular function on An must be a polynomial. Any regularfunction on Pn must be a constant.

We say two varieties X and Y are isomorphic if there exist regular maps f : X →Y and g : Y → X inverse to each other. For any two projective subvarieties X,Y ⊂Pn, we say that they are projectively equivalent if there exists an automorphism inPGLn+1(K) of Pn carrying X to Y .

Example 1.19 (Rational normal curves). We have seen conics and twisted cubics,as embeddings of P1 into P2 and P3. This construction can be carried out in general.Consider the map f : P1 → Pn given by

[X,Y ] 7→ [Xn, Xn−1Y, . . . ,XY n−1, Y n].

In affine coordinates, f is given by

t 7→ (t, t2, . . . , tn).

It is easy to see that f is an isomorphism onto its image, which is called a rationalnormal curve in Pn.

Exercise 1.20. Let p1, . . . , pk be k points on a rational normal curve in Pn. Showthat for k ≤ n + 1, the linear span of p1, . . . , pk (i.e. the minimal linear subspacecontaining the pi) is (k − 1)-dimensional.

2. Sheaves and cohomology

Roughly speaking, a sheaf corresponds to a collection of local data on a topo-logical space and cohomology groups of a sheaf contain information about passingfrom local to global. First, let us consider a motivating example.

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2.1. The Mittag-Leffler problem. Let X be a Riemann surface and p ∈ X apoint with local coordinate z. A principal part at p is the polar part of a Laurentseries

∑ni=1 aiz

−i for some n > 0 in a neighborhood of p. Let pi be a collectionof finitely many distinct points on X and fi a fixed principal part for each pi.

Question 2.1. Can we find a global meromorphic function f on X such that it isholomorphic on X\pi and locally around the pi we have f = fi?

A possible solution goes as follows. Let us first introduce some notation. For anopen set U , let O(U) denotes the ring of holomorphic functions on U and M (U)the ring of meromorphic functions.

Consider an open covering U = Uα of X such that each Uα contains at mostone pi. Choose fα ∈ O(Uα) solving the problem locally. On Uαβ = Uα ∩ Uβ , letfαβ = fα − fβ . Note that

fαβ + fβγ + fγα = 0

on Uα ∩ Uβ ∩ Uγ .Suppose there exists a global solution f . Then we have gα = f |Uα − fα ∈ O(Uα)

and fαβ = gβ − gα ∈ O(Uαβ). Conversely, if we have a collection of holomorphicfunctions gα ∈ O(Uα) satisfying gβ − gα = fαβ , then we can define f = gα + fα onUα. Since gα + fα = gβ + fβ on Uαβ , f is globally defined, hence it is a desiredsolution.

Now define two groups

Z1(U,O) =fαβ ∈ O(Uαβ) : fαβ + fβγ + fγα = 0

,

C0(U,O) =gα ∈ O(Uα)

,

and a morphism δ : C0(U,O)→ Z1(U,O) by

gα 7→ fαβ = gβ − gα.

By the above analysis, finding a global solution is equivalent to finding gα ∈ Uαsuch that gβ − gα = fαβ . In other words, the quotient group

H1(U,O) :=Z1(U,O)

δC0(U,O)

is the obstruction to solving this problem in general.

2.2. Sheaves. Let X be a topological space. A sheaf F on X associates to eachopen set U an abelian group F (U), called the sections of F over U , along with arestriction map rV,U : F (V ) → F (U) for any open sets U ⊂ V (for a section σ ∈F (V ), we often write σ|U to denote rV,U (σ)), satisfying the following conditions:

(1) For any open sets U ⊂ V ⊂W , rV,U rW,V = rW,U ;(2) For any open sets U, V and sections α ∈ F (U), β ∈ F (V ), if α|U∩V = β|U∩V ,

then there exists γ ∈ F (U ∪ V ) such that γ|U = α and γ|V = β;(3) For any section γ ∈ F (U ∪ V ), if γ|U = 0 and γ|V = 0, then γ = 0.

Remark 2.2. Note that (2) says that two sections can be glued together if theycoincide on an open subset and (3) implies that such gluing is unique. For manysheaves we will consider, the restriction maps are the obvious ones by restrictingfunctions from bigger subsets to smaller ones.

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Example 2.3. We have the constant sheaf Z on a topological space X, whereZ(U) is the group of constant functions on an open set U ⊂ X. Similarly one candefine constant sheaves Q, R, or K on an algebraic variety defined over a field K.

Let X be a complex manifold and U ⊂ X an open set.(1) Sheaf O of holomorphic functions:

O(U) = holomorphic functions on U.(2) Sheaf O∗ of nonzero holomorphic functions:

O∗(U) = holomorphic functions f on U : f(p) 6= 0 for any p ∈ U.The group structure is given by multiplication.

(3) Sheaf M of meromorphic functions: strictly speaking, a meromorphic func-tion is not a function, even we take ∞ into account. If X is compact, we cannottake a meromorphic function as a global quotient of two holomorphic functions,since any holomorphic function is a constant on X. Instead, we define f ∈M (U)as local quotient of holomorphic functions. Namely, there exists an open coveringUi of U such that on each Ui, f is given by gi/hi for some gi, hi ∈ O(Ui) satisfyinggi/hi = gj/hj , i.e. gihj = gjhi ∈ O(Ui ∩ Uj), hence these local quotients can beglued together over U .

(4) Sheaf M ∗ of meromorphic functions not identically zero: this is definedsimilarly and the group structure is given by multiplication.

Let X be a variety. Replacing holomorphic functions by regular functions, wealso use O to denote the sheaf of regular functions on X and M the sheaf of rationalfunctions. We also call O the structure sheaf of a variety.

2.3. Maps of sheaves. Let E ,F be sheaves on X. A map f : E → F is acollection of group homomorphisms

fU : E (U)→ F (U)such that they commute with the restriction maps, i.e. for open sets U ⊂ V andσ ∈ E (V ) we have

fV (σ)|U = fU (σ|U ).

Define the sheaf of kernel ker(f) as

ker(f)(U) = ker(fU : E (U)→ F (U)).

Exercise 2.4. Prove that in the above definition ker(f) is a sheaf.

Proof. Let U, V be two open subsets of X and α ∈ ker(f)(U), β ∈ ker(f)(V ). Ifα|U∩V = β|U∩V , since α, β are also sections of E , there exists γ ∈ E (U ∪ V ) suchthat γ|U = α and γ|V = β (here we use the gluing property of a sheaf). Moreover,f(γ)|U = f(α) = 0 and f(γ)|V = f(β) = 0 (here we use that f commutes withthe restriction maps), hence f(γ) = 0 ∈ F (U ∪ V ) (here we use the unique gluingproperty of sheaves). It implies that γ ∈ ker(f)(U ∪ V ).

Now suppose for η ∈ ker(f)(U ∪ V ) we have η|U = 0 and η|V = 0. Since η isalso a section of E (U ∪ V ), as a consequence η = 0.

Example 2.5. Let X be a complex manifold. Define the exponential map

exp : O → O∗

by exp(h) = e2π√−1h for any open set U ⊂ X and section h ∈ O(U). It is easy to

see that ker(exp) is the constant sheaf Z.

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The sheaf of cokernel is harder to define. Naively, one would like to definecoker(f)(U) = coker(fU : E (U)→ F (U)), but this is problematic. For instance,consider the exponential map exp : O → O∗ on the punctured plane C\0. Thesection z ∈ O∗(C\0) is not in the image of f , hence it would define a section inthe cokernel. Nevertheless, restricted to any contractible open set U ⊂ C\0, zlies in the image of f . Now cover C\0 by contractible open sets. By the gluingproperty of sheaves, z would be zero everywhere, leading to a contradiction.

Instead, we want a section of coker(f)(U) to be a collection of sections σα ∈F (Uα) for an open covering Uα of U such that for all α, β we have

σα|Uα∩Uβ − σβ |Uα∩Uβ ∈ fUα∩Uβ (E (Uα ∩ Uβ)).

Here the definition still depends on the choice of an open covering. To get rid ofthis, we pass to the direct limit. More precisely, we further identify two collections(Uα, σα) and (Vβ , σβ if for all p ∈ Uα∩Vβ , there exists an open set W satisfyingp ∈W ⊂ Uα ∩ Vβ such that

σα|W − σβ |W ∈ fW (E (W )).

This identification yields an equivalence relation and correspondingly we definecoker(f)(U) as the group of equivalence classes of the above sections.

Exercise 2.6. Prove that in the above definition coker(f) is a sheaf.

Consider a sequence of maps

0 −−−−→ Eα−−−−→ F

β−−−−→ G −−−−→ 0.

We say that it is a short exact sequence if E = ker(β) and G = coker(α). In thiscase we also say that E is a subsheaf of F and G is the quotient sheaf F/E .

If Y ⊂ X is a subspace and F is a sheaf on Y , we can extend F by zero toobtain a sheaf F ′ on X by

F ′(U) = F (U ∩ Y ), if U ∩ Y 6= ∅,F ′(U) = 0, if U ∩ Y = ∅.

The restriction maps are the obvious ones. In general, we will abuse notation andstill use F to denote the extension of F by zero.

Exercise 2.7. Prove that in the above definition F ′ is a sheaf on X.

Example 2.8. Let X be a complex manifold. We have the exact exponentialsequence:

0 −−−−→ Z i−−−−→ Oexp−−−−→ O∗ −−−−→ 0,

where i is the natural inclusion and exp(f) = e2π√−1f for f ∈ O(U).

Exercise 2.9. Prove that the exponential sequence is exact.

Example 2.10. Let X be a complex manifold and Y ⊂ X a submanifold. Definethe ideal sheaf IY/X of Y in X (or simply IY if there is no confusion) by

IY (U) = holomorphic functions on U vanishing on Y ∩ U.We have the exact sequence:

0 −−−−→ IYi−−−−→ OX

r−−−−→ OY −−−−→ 0,

where i is the natural inclusion and r is defined by the natural restriction map.Note that here we treat OY as a sheaf on X by extending OY by zero.

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The same exact sequence holds in the context of varieties, with holomorphicfunctions replaced by regular functions.

Exercise 2.11. Prove that the above sequence is exact.

2.4. Stalks and germs. Let F be a sheaf on a topological space X and p ∈ Xa point. Suppose U, V are two open subsets, both containing p, with two sectionsα ∈ F (U) and β ∈ F (V ). Define an equivalence relation α ∼ β, if there exists anopen subset W satisfying p ∈ W ⊂ U ∩ V such that α|W = β|W . Define the stalkFp as the union of all sections in open neighborhoods of p modulo this equivalencerelation. Namely, Fp is the direct limit

Fp := lim−→U3p

F (U) =( ⊔U3p

F (U))/ ∼ .

Note that Fp is also a group, by adding representatives of two equivalence classes.There is a group homomorphism rU : F (U) → Fp mapping a section α ∈ F (U)to its equivalence class. The image is called the germ of α.

Example 2.12. Let X be a Riemann surface and p ∈ X a point with local co-ordinate z. Then the stalk of the sheaf of holomorphic functions OX at p is theset of Taylor series

∑∞i=0 aiz

i converging in a neighborhood of p. Two holomorphicfunctions have the same germ if and only if their Taylor expansions are the sameat p.

Example 2.13 (Skyscraper sheaf). Let p ∈ X be a point on a topological space X.Define the skyscraper sheaf F at p by F (U) = 0 for p 6∈ U and F (U) = A forp ∈ U , where A is a group (or a ring, a vector space, etc). The restriction maps areeither the identity map A→ A or the zero map. For q 6= p, the stalk Fq = 0. Atp, we have Fp = A. Note that F can also be obtained by extending the constantsheaf A at p by zero to X.

Exercise 2.14. Let X be a Riemann surface and p ∈ X a point. Let Ip be theideal sheaf of p in X parameterizing holomorphic functions vanishing at p. We havethe exact sequence

0 −−−−→ Ipi−−−−→ OX

r−−−−→ Op −−−−→ 0.

Show that the quotient sheaf Op is isomorphic to the skyscraper sheaf with stalk Cat p.

It is more convenient to verify injections and surjections for maps of sheaves bythe language of stalks.

Proposition 2.15. Let φ : E → F be a map for sheaves E and F on a topologicalspace X.

(1) φ is injective if and only if the induced map φp : Ep → Fp is injective for thestalks at every point p.

(2) φ is surjective if and only if the induced map φp : Ep → Fp is surjective forthe stalks at every point p.

(3) φ is an isomorphism if and only if the induced map φp : Ep → Fp is anisomorphism for the stalks at every point p.

Proof. The claim (3) follows from (1) and (2). Let us prove (1) only, and one caneasily find the proof of (2) in many books, e.g. Hartshorne.

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Suppose φ is injective. Take a section σ ∈ E (U) on an open subset U . If φ([σ]) =0 ∈ Fp, there exists a smaller open subset V ⊂ U such that φV (σ) = 0 ∈ F (V ),hence σ|V = 0 ∈ E (V ). Consequently the equivalence class [σ] = 0 ∈ Ep and weconclude that φp is injective.

Conversely, suppose φp is injective for every point p. Take a section σ ∈ E (U).If φ(σ) = 0 ∈ F (U), then for every point p ∈ U , [φ(σ)] = 0 ∈ Fp. Since φp isinjective, it implies that [σ] = 0 ∈ Ep i.e. there exists an open subset Up 3 p suchthat σ|Up = 0 ∈ E (Up). Applying the gluing property to the open covering Up ofU , we conclude that σ = 0 ∈ E (U).

Remark 2.16. The image of φ does not automatically form a sheaf. In general,it is only a presheaf, i.e. without the gluing and unique gluing properties in thedefinition of sheaves. If the sheafification of Im(φ) equals F , we say that φ issurjective. In particular, it does not mean E (U) → F (U) is surjective for everyopen set U . Sometimes one has to pass to a smaller open set in order to obtain asurjection between sections.

Example 2.17. Consider the exponential map exp : O → O∗ on the puncturedplane C\0. This map is surjective, but the section z over the total space does nothave an inverse image. It does have an inverse over any contractible open subset.

2.5. Cohomology of sheaves. Let F be a sheaf on a topological space X. Takea locally closed open covering U = Uα of X. Define the k-th cochain group

Ck(U,F ) :=∏

α0,...,αk

F (Uα0∩ · · · ∩ Uαk).

An element σ of Ck(U,F ) consists of a section σα0,...,αk ∈ F (Uα0∩ · · · ∩ Uαk)

for every Uα0 ∩ · · · ∩ Uαk , satisfying the cochain condition, i.e. skew-symmetric:σα0,...,αk = −σα0,...,αi−1,αi+1,αi,αi+2,...,αk .

Define a coboundary map δ : Ck(U,F )→ Ck+1(U,F ) by

(δσ)α0,...,αk+1=

k+1∑j=0

(−1)jσα0,...,αj ,...,αk+1|Uα0∩···∩Uαk+1

.

Example 2.18. Consider U = U1, U2, U3 as an open covering of X. Take acochain element σ ∈ C0(U,F ), i.e. σ is a collection of a section σi ∈ F (Ui) forevery i. Then we have

(δσ)ij = (σj − σi)|Ui∩Uj ∈ F (Ui ∩ Uj).

Now take τ ∈ C1(U,F ), i.e. τ is a collection of a section τij ∈ F (Ui ∩ Uj) forevery pair i, j. Then we have

(δτ)123 = (τ23 − τ13 + τ12)|U1∩U2∩U3∈ F (U1 ∩ U2 ∩ U3).

A cochain σ ∈ Ck(U,F ) is called a cocyle if δσ = 0. We say that σ is acoboundary if there exists τ ∈ Ck−1(U,F ) such that δτ = σ.

Lemma 2.19. A coboundary is a cocycle, i.e. δ δ = 0.

12

Proof. Let us prove it for the above example. The same idea applies in generalwith messier notation. Under the above setting, we have

((δ δ)σ)123 = (δσ)23 − (δσ)13 + (δσ)12

= (σ3 − σ2)− (σ3 − σ1) + (σ2 − σ1)

= 0 ∈ F (U1 ∩ U2 ∩ U3).

Here we omit the restriction notation, since it is obvious.

Exercise 2.20. Prove in full generality that δ δ = 0.

For the coboundary map δk : Ck(U,F ) → Ck+1(U,F ), define the k-th coho-mology group (respect to U) by

Hk(U,F ) :=ker(δk)

Im(δk−1).

This is well-defined due to the above lemma.

Example 2.21. For k = 0, we have H0(U,F ) = ker(δ0). Take an element σi ∈F (Ui) in this group. Because it is a cocycle, it satisfies

σi|Ui∩Uj = σj |Ui∩Uj ∈ F (Ui ∩ Uj).By the gluing property of sheaves, there exists a global section σ ∈ F (X) suchthat σ|Ui = σi. Conversely, if σ is a global section, then define σi = σ|Ui ∈ F (Ui).In this way we obtain a cocycle in C1(U,F ). From the discussion we see thatH0(U,F ) = F (X), which is independent of the choice of an open covering. HenceH0(U,F ) is called the group of global sections of F and we often denote it byH0(X,F ) or simply H0(F ).

In general, we would like to define cohomology independent of open coverings.Take two open coverings U = Uαα∈I and V = Vββ∈J . We say that U is arefinement of V if for every Uα there exists a Vβ such that Uα ⊂ Vβ and we write itas U < V . Then we also have an index map φ : I → J sending α to β. It inducesa map

ρφ : Ck(V ,F )→ Ck(U,F )

given byρφ(σ)α0,...,αk = σφ(α0),...,φ(αp)|Uα0∩···∩Uαk .

One checks that it commutes with the coboundary map δ, i.e. δ ρφ = ρφ δ.Moreover, it induces a map

ρ : Hk(V ,F )→ Hk(U,F ),

which is independent of the choice of φ. Finally, we define the k-th (Cech) coho-mology group by passing to the direct limit:

Hk(X,F ) := lim−→Hk(U,F ).

The definition involves direct limit, which is inconvenient to use in practice.Nevertheless, we can simplify the situation once the open covering U is fine enough.We say that U = Uii∈I is acyclic respect to F , if for any k > 0 and i1, . . . , il ∈ Iwe have

Hk(Ui1 ∩ · · · ∩ Uil ,F ) = 0.

Theorem 2.22 (Leray’s Theorem). If the open covering U is acyclic respect to F ,then H∗(U,F ) ∼= H∗(X,F ).

13

Example 2.23. Let us compute the cohomology of the structure sheaf O on P1C.

It is clear that H0(P1,O) = C, since any global holomorphic function on a compactcomplex manifold is constant. For higher cohomology, use [X,Y ] to denote thecoordinates of P1. Take the standard open covering U = [X,Y ] : X 6= 0 andV = [X,Y ] : Y 6= 0. It is acyclic respect to the structure sheaf O (morally becauseU ∩ V = C∗ is contractible). Let s = Y/X and t = X/Y as affine coordinates of Uand V , respectively. Suppose h is an element in C1(U, V ,O), i.e. h ∈ O(U ∩ V ).We can write

h =

∞∑i=−∞

aisi.

Now take

f = −∞∑i=0

aisi ∈ O(U),

g =

−1∑i=−∞

aisi =

−1∑i=−∞

ait−i ∈ O(V ).

Then we have (f, g) ∈ C1(U, V ,O) and δ((f, g)) = g − f = h. It implies thatH1(P1,O) = 0. All the other Hk(P1,O) = 0 for k > 1, since there are only twoopen subsets in the covering.

Remark 2.24. In the context of complex manifolds, if Ui’s are contractible, thenU is acyclic respect to the sheaves we will consider. While for varieties, if Ui’s areaffine, then U is acyclic.

Example 2.25. Let Ω denote the sheaf of holomorphic one-forms on a Riemannsurface, i.e. locally a section of Ω can be expressed as f(z)dz, where z is localcoordinate and f(z) a holomorphic function. Let us compute the cohomology of Ωon P1. Take the above open covering. Suppose ω is a global holomorphic one-form.Then on the open chart U , it can be written as( ∞∑

i=0

aisi)ds.

Using the relation s = 1/t and ds = −dt/t2, on V it can be expressed as

−( ∞∑i=0

ait−i−2

)dt,

which is holomorphic if and only if ai = 0 for all i. Hence w is the zero one-formand H0(P1,Ω) = 0. Now take ω ∈ C1(U, V ,Ω), i.e. ω ∈ Ω(U ∩ V ) = Ω(C∗), weexpress it as

ω =( ∞∑i=−∞

aiti)dt.

Note that any α ∈ Ω(U) and β ∈ Ω(V ) can be written as

α =( ∞∑i=0

bisi)ds,

β =( ∞∑i=0

citi)dt.

14

Hence on U ∩ V we have

δ((α, β)) = β − α = −( ∞∑i=0

bit−i−2

)dt+

( ∞∑i=0

citi)dt.

Note that only the term t−1 is missing from the expression. We conclude thatH1(P1,Ω) = a−1t

−1dt ∼= C.

Remark 2.26. If P1 is defined over an algebraically closed field K, replacing holo-morphic functions by regular functions, we haveH0(P1,O) ∼= K andH1(P1,O) = 0.Indeed, we have seen that the coordinate ring A(U ∩V ) is given by K[s, 1/s], hencethe above argument goes word by word. Similarly replacing holomorphic one-formsby regular differentials, i.e. in the expression f(z)dz, f(z) is regular and dz satisfiesthe usual differentiation rules, we have H0(P1,Ω) = 0 and H1(P1,Ω) ∼= K. Ingeneral, the rank of H1(X,O) ∼= H0(X,Ω) (by Serre Duality) is called the genusof a Riemann surface (or an algebraic curve) X.

Exercise 2.27. Let D = p1 + · · · + pn be a collection of n points in P1. We saythat D is an effective divisor of degree n. Define the sheaf O(D) on P1 by

O(D)(U) = f ∈M (U) : f ∈ O(U\p1, . . . , pn) and has at most simple pole at pi.Assume that the standard covering of P1 is acyclic respect to O(D). Use it tocalculate the cohomology groups H∗(P1,O(D)).

As many other homology/cohomology theories, one can associate a long exactsequence of cohomology to a short exact sequence. Suppose we have a short exactsequence of sheaves

0 −−−−→ Eα−−−−→ F

β−−−−→ G −−−−→ 0.

Then α and β induce maps

α : Ck(U,E )→ Ck(U,F ), β : Ck(U,F )→ Ck(U,G ).

Since the coboundary map δ is given by alternating sums of restrictions, α andβ commute with δ, hence they send a cocycle to cocycle and a coboundary tocoboundary. Consequently they induce maps for colomology

α∗ : Hk(X,E )→ Hk(X,F ), β∗ : Hk(X,F )→ Hk(X,G ).

Next we define the coboundary map

δ∗ : Hk(X,G )→ Hk+1(X,E ).

For σ ∈ Ck(U,G ) satisfying δσ = 0, after refining U (still denoted by U) suchthat there exists τ ∈ Ck(U,F ) satisfying β(τ) = σ, because β is surjective. Thenβ(δτ) = δ(β(τ)) = δσ = 0, hence after refining further there exists µ ∈ Ck+1(U,E )satisfying α(µ) = δτ . Note that µ is a cocycle. It is because α(δµ) = δ(α(µ)) =δδ(τ) = 0 and α is injective, hence δµ = 0 and µ ∈ ker(δ). We thus take δ∗σ :=[µ] ∈ Hk+1(X,E ). One checks that this is independent of the choice of τ and µ.

We say that a sequence of maps

· · · −−−−→ An−1αn−1−−−−→ An

αn−−−−→ An+1 −−−−→ · · ·is exact if Im(αn−1) = ker(αn).

Proposition 2.28. The long sequence of cohomology associated to a short exactsequence of sheaves is exact.

15

Proof. We prove it under an extra assumption that there exists an acyclic opencovering U such that for any U = Ui1 ∩ · · · ∩Uik we have the short exact sequence:

0→ E (U)→ F (U)→ G (U)→ 0.

At least for sheaves considered in this course, this assumption always holds. Itfurther implies the following sequence is exact:

0→ Ck(U,E )→ Ck(U,F )→ Ck(U,G )→ 0.

Let us prove that

Hk(U,F )β∗−−−−→ Hk(U,G )

δ∗−−−−→ Hk+1(U,E )

is exact. The other cases are easier.Consider τ ∈ Zk(U,F ). In the definition of δ∗, take σ = β(τ). Then there exists

µ ∈ Ck(U,E ) such that α(µ) = δτ = 0. Then we have µ = 0 since α is injective.Consequently δ∗β∗(τ) = δ∗(σ) = µ = 0, hence δ∗β∗ = 0 and Im(β∗) ⊂ ker(δ∗).

Conversely, suppose δ∗σ = 0 for σ ∈ Zk(U,G ). In the definition of δ∗, it impliesthat µ = 0 ∈ Hk+1(U,E ), hence there exists γ ∈ Ck(U,E ) such that δγ = µ.Since α(µ) = δτ , we have δτ = δα(γ) and τ − α(γ) ∈ Zk(U,F ) is a cocycle.Moreover, β(τ − α(γ)) = β(τ) = σ, hence β∗(τ − α(γ)) = σ. We conclude thatker(δ∗) ⊂ Im(β∗).

Exercise 2.29. Prove in general the cohomology sequence is exact.

Example 2.30. Consider the short exact sequence

0 −−−−→ Ipi−−−−→ OP1

r−−−−→ Op −−−−→ 0.

Its long exact sequence of cohomology is as follows:

0→ H0(Ip)→ H0(OP1)→ H0(Op)→ H1(Ip)→ H1(OP1)→ 0.

The last term is zero because p is a point so it does not have higher cohomology.We have H0(OP1) = K because any global regular function on P1 is constant. Notethat H0(Ip) = 0, because vanishing at p forces such a constant function to bezero. Moreover we have seen that H1(OP1) = 0. Altogether it implies H1(Ip) = 0,because H0(OP1)→ H0(Op) is an isomorphism by evaluating at p.

Exercise 2.31. Let D be an effective divisor of degree n on P1. We have the shortexact sequence

0 −−−−→ IDi−−−−→ OP1

r−−−−→ OD −−−−→ 0.

Use the associated long exact sequence to calculate the cohomology H∗(P1,I (D)).

The cohomology of constant sheaves contains topological information for theunderlying space. Let K be a simplicial complex with the underlying topologicalspace X. Let H∗(K,Z) denote the singular cohomology with coefficient Z.

Proposition 2.32. The singular cohomology of K and the Cech cohomology of theconstant sheaf on X are isomorphic:

H∗(K,Z) ∼= H∗(X,Z).

16

Proof. For each vertex vα of K, associate to it an open set Uα by the interior of theunion of all simplices with vα as a vertex. Then U = Uα is an open covering of X.Note that Uα0 ∩· · ·∩Uαk is nonempty and connected if vα0 , . . . , vαk are the verticesof a k-simplex. Otherwise it is empty. Then an element σ ∈ Z(Uα0

∩ · · · ∩ Uαk) iseither an integer if vα0

, . . . , vαk span a k-simplex or it is zero otherwise.Given a cochain σ ∈ Ck(U,Z), define a simplicial cochain σ′ ∈ Ck(K,Z) by

σ′(〈vα0 , . . . , vαk〉) = σα0,...,αk .

The association σ 7→ σ′ induces an isomorphism Ck(U,Z)∼−→ Ck(K,Z). Moreover,

the coboundary maps satisfy

δσ′(〈vα0, . . . , vαk+1

〉) =

k+1∑i=0

(−1)iσ′(〈vα0, . . . , vαi , . . . , vαk+1

〉)

=

k+1∑i=0

(−1)iσα0,...,αi,...,αk+1

= (δσ)α0,...,αk+1

= (δσ)′(〈vα0, . . . , vαk+1

〉),

so we obtain an isomorphism for the chain complexes C∗(U,Z)∼−→ C∗(K,Z) and

consequently an isomorphism H∗(U,Z)∼−→ H∗(K,Z). Finally we subdivide K such

that U can be arbitrarily fine. Then the claim follows, since H∗(K,Z) does notchange.

3. Vector bundles, line bundles and divisors

3.1. Holomorphic vector bundles. Let k be a positive integer. Consider π :E → X a holomorphic map between complex manifolds, such that for every x ∈ Xthe fiber Ex = π−1(x) is isomorphic to Ck and there exists an open neighborhoodU of x along with an isomorphism

φU : EU = π−1(U)∼−→ U × Ck

mapping Ex to x×Ck which is a linear isomorphism between vector spaces. ThenE is called a holomorphic vector bundle of rank k on X and has a trivialization(U, φU ). If E is of rank 1, we say that E is a line bundle.

We give an equivalent characterization of vector bundles based on transitionfunctions. Note that for two trivializations φU and φV , they induce an isomorphismgUV = φV φ−1

U : (U∩V )×Ck ∼−→ (U∩V )×Ck preserving the vector space structurefiber by fiber. In other words,

gUV ∈ Hom(U ∩ V,GL(Ck))

specifies (holomorphically) a linear isomorphism Ex∼−→ Ex over each base point

x ∈ U ∩V . We call such gUV transition functions with respect to the trivialization(U, φU ). The transition functions satisfy the following identities:

gUV (x) · gV U (x) = I, for all x ∈ U ∩ V,gUV (x) · gVW (x) · gWU (x) = I, for all x ∈ U ∩ V ∩W.

Conversely suppose U = Uα is an open covering of X. Given holomorphic func-tions

gαβ : Uα ∩ Uβ → GL(Ck)

17

satisfying the above equalities, we can construct a vector bundle E by gluing Uα×Cktogether. More precisely,

E =⋃

(Uα × Ck)

as a complex manifold is defined by identifying (x, v) with (x, gαβ(v)) for x ∈ Uα∩Uβand v ∈ Ck and E → X is given by projection to the bases Uα.

Define the dual bundle E∗ of E by taking φ∗U : E∗U → U × Ck∗ ∼= U × Ck foropen subsets U ⊂ X. Each fiber of E∗ is given by Hom(Ex,C) ∼= Ck∗. One checksthat the transition functions of E∗ satisfy g∗αβ(x) = ((gαβ(x))−1)t.

Suppose F is another vector bundle of rank l with transition functions hαβ .Define the direct sum E ⊕ F as a vector bundle of rank k + l with transitionfunctions diag(gαβ(x), hαβ(x)) ∈ GL(Ck+l).

Define the tensor product E ⊗ F as a vector bundle of rank kl with transitionfunctions gαβ(x) ⊗ hαβ(x) ∈ GL(Ck ⊗ Cl). Note that if E is a line bundle, thenE∗⊗E is the trivial line bundle X×C (explained in detail later), since the transitionfunctions are identity everywhere.

Define the wedge product ∧rE as a vector bundle of rank(kr

)with transition

functions ∧rgαβ(x) ∈ GL(∧rCk) for 1 ≤ r ≤ k. In particular, ∧kE is a linebundle with transition functions det(gαβ(x)) ∈ GL(C). Therefore, ∧kE is calledthe determinant line bundle of E and often denoted by det(E).

Let f : X → Y be a holomorphic map and E a vector bundle on Y . We definethe pullback bundle f∗E by setting (f∗E)x = Ef(x)

∼= Ck. If φU : EU → U×Ck is a

trivialization in a neighborhood of f(x), then f∗φU : (f∗E)f−1(U) → f−1(U)× Ckdefines f∗E as a complex manifold and moreover as a vector bundle on X withtransition functions given by f∗φU .

Let F ⊂ E be a collection of l dimensional subspaces Fx ⊂ Ex for every x ∈ Xsuch that there exists a neighborhood U of x and a trivialization φU : EU → U×Cksatisfying

φU |FU : FU → U × Cl ⊂ U × Ck.Then we say that F is a rank l subbundle of E. Up to changing coordinates, thetransition functions of E respect to these trivializations can be written as an uppertriangular matrix

gUV (x) =

(fUV (x) jUV (x)

0 hUV (x)

)with fUV (x) being the transition functions of F . Similarly we define the quotientbundle E/F with (E/F )x = Ex/Fx as well as transition functions given by hUV (x).

Exercise 3.1. Verify that the above definitions give rise to vector bundles withthe claimed transition functions.

A map between vector bundles E,F on X is given by a holomorphic map f :E → F such that f(Ex) ⊂ Fx and fx = f |Ex : Ex → Fx is linear. Note that if f(Ex)has the same rank for every x, then ker(f) and Im(f) are naturally subbundles of Eand F , respectively. We say that E and F are isomorphic if fx is an isomorphismfor every x. A vector bundle is called trivial if it is isomorphic to the direct productX × Ck.

Exercise 3.2. Give an example of a map between vector bundles f : E → F on Xsuch that the rank of Im(f |Ex) is not a constant for all x ∈ X.

18

Exercise 3.3. Let L be a line bundle on X. Prove that L ⊗ L∗ is a trivial linebundle.

Define a section σ as a holomorphic map σ : X → E such that σ(x) ∈ Ex forevery x ∈ X, i.e. π σ is identity. If σ(x) = 0 ∈ Ex, we say that σ is vanishing onx.

Exercise 3.4. Let L be a line bundle on X. Prove that L is trivial if and only ifit possesses a nowhere vanishing section.

Example 3.5 (Holomorphic tangent bundles). Let X be a n-dimensional com-plex manifold. Suppose φU : U → Cn are coordinate charts of X. Define the(holomorphic) tangent bundle TX by setting TX =

⋃Tx with

Tx = C ∂

∂z1, . . . ,

∂

∂zn ∼= Cn

as well as transition functions gUV = J(φV φ−1U ), where J denotes the Jacobian

matrix (∂wi∂zj) for 1 ≤ i, j ≤ n. The dual bundle T ∗X is called the cotangent bundle

of X. The determinant det(T ∗X) is called the canonical line bundle of X.

Remark 3.6. Alternatively, one can define vector bundles on a topological space,a differential manifold and an algebraic variety. The above definitions and proper-ties go through literally after replacing “holomorphic map” by “homomorphism”,“smooth map” or “regular map”.

3.2. Vector bundles on a variety and locally free sheaves. In the context ofvarieties (more rigorously, schemes), there is a one-to-one correspondence betweenisomorphism classes of vector bundles of rank n and isomorphism classes of locallyfree sheaves of rank n on a variety X. Here we briefly explain the idea. The readercan refer to Hartshorne II 5, especially Ex. 5.18 for more details.

Let OX be the structure sheaf of a variety X. Note that OX(U) has a ringstructure (not only a group) for any open set U . A sheaf of OX-modules is a sheafF on X such that for each open set U , the group F (U) is an OX(U)-module. AnOX -module F is called free if it is isomorphic to a direct sum of copies of OX . Itis called locally free if there is an open covering U = Uα such that for each opensubset Uα, F |Uα is a free OX |U -module. The rank of F on U is the number ofcopies of O in the summation. In this course we only consider the rank being finite(such sheaves are objects in the category of coherent sheaves). If X is connected,the rank of F does not vary with the open subsets. In particular, a locally freesheaf of rank 1 is also called an invertible sheaf.

Roughly speaking, if F is locally free of rank n, we can choose a set of n gener-ators x1, . . . , xn for the OX(U)-module F (U). They span an n-dimensional affinespace A[x1, . . . , xn] over U , where A is the coordinate ring of U . By changing to adifferent set of generators over another open subset, one can write down the tran-sition functions, hence it associates to F a vector bundle structure. Conversely ifF is a vector bundle on X, locally we have F |U ∼= U × An with x1, . . . , xn a basis(i.e. n linearly independent sections) of An over U . Then we can associate to F |Uan OX(U)-module of rank n using x1, . . . , xn as generators.

Example 3.7. Let X ⊂ Pn be a (smooth) variety and Y ⊂ X a hypersurface, i.e.Y is cut out (transversely) by a hypersurface F in Pn with X. We have the short

19

exact sequence0→ IY/X → OX → OY → 0.

The ideal sheaf IY/X is an invertible sheaf. Indeed, for an open subset U ⊂ X,IY/X(U) can be expressed as (F |U ) · OX(U), hence is locally free of rank 1. Thesheaf OY (extended to X by zero) is not locally free. For U ∩ Y = ∅, OY (U) = 0and for U ∩ Y 6= ∅, OY (U) is non-zero. Later we will see how to construct a linebundle corresponding to IY/X .

3.3. Divisors. Let X be a variety (complex or algebraic). Suppose Y ⊂ X is anirreducible subvariety of codimension one. We say that Y is an irreducible divisorof X. More precisely, for every p ∈ Y there exists an open neighborhood U ⊂ Xof p such that U ∩ Y is cut out by a (holomorphic or regular) function f . Wecall f a local defining equation for Y near p. A divisor D on X is a formal linearcombination of irreducible divisors:

D =∑

aiYi,

where ai ∈ Z (or Q, R depending on the context). If ai ≥ 0 for all i, we say thatD is effective and denote it by D ≥ 0. The divisors on X form an additive groupDiv(X).

Suppose f is a local defining equation of an irreducible divisor Y ⊂ X on anopen subset U ⊂ X. For another function g on X, locally we can write

g = fa · hsuch that the regular function h is coprime with f in OX(U). We say that a isthe vanishing order of g along Y ∩ U . Note that the vanishing order is locally aconstant, hence is independent of U . We use

ordY (g) = a

to denote the vanishing order of g along Y .For two regular functions g, h on X, we have

ordY (gh) = ordY (g) + ordY (h).

For a meromorphic function f = g/h, we define

ordY (f) = ordY (g)− ordY (h).

If ordY (f) > 0, we say that f has a zero along Y . If ordY (f) < 0, we say that fhas a pole along Y . We also define the divisor associated to f by

(f) =∑Y

ordY (f),

as well as the divisor of zeros

(f)0 =∑Y

ordY (g)

and the divisor of poles

(f)∞ =∑Y

ordY (h).

They satisfy(f) = (f)0 − (f)∞.

If D = (f) is the associated divisor of a global meromorphic function f , D is calleda principal divisor.

20

Let M ∗ be the multiplicative sheaf of (not identically zero) meromorphic func-tions and O∗ the multiplicative sheaf of nowhere vanishing regular functions, whichis a subsheaf of M ∗.

Proposition 3.8. We have a correspondence Div(X) ∼= H0(X,M ∗/O∗).

Proof. Suppose fα represents a global section of M ∗/O∗ with respect to an opencovering U = Uα. Associate to it a divisor Dα = (fα) in Uα. We claim thatDα = Dβ in Uα ∩ Uβ . This is due to

fαfβ∈ O∗(Uα ∩ Uβ),

hence fα and fβ define the same divisor. Consequently Dα defines a globaldivisor. Moreover, if fα and gα define the same divisor, then fα/gα ∈ O∗(Uα),hence fα and gα represent the same section of M ∗/O∗. This shows an injection

H0(X,M ∗/O∗) → Div(X).

Conversely, suppose D =∑aiYi is a divisor on X with ai ∈ Z and Yi effective.

We can choose an open covering U = Uα such that Yi is locally defined bygiα ∈ O(Uα). Consider

fα =∏i

(giα)ai ∈M ∗(Uα).

Then we havefαfβ

=∏i

(giαgiβ

)ai.

Both giα and giβ cut out the same divisor Yi|Uα∩Uβ in Uα ∩Uβ , hence we concludethat

giαgiβ∈ O∗(Uα ∩ Uβ),

fαfβ∈ O∗(Uα ∩ Uβ).

Then fα defines a global section of M ∗/O∗. Finally if D determines the zerosection of M ∗/O∗ (which is 1 since the group structure is multiplicative), it meanslocally fα ∈ O∗(Uα) (after refining the open covering). Then it does not have zerosor poles, hence D|Uα = 0 for each Uα and D is globally zero. This shows the otherinjection

Div(X) → H0(X,M ∗/O∗).

3.4. Line bundles. Recall that a line bundle L on X is a vector bundle of rank1. Equivalently, it is a locally free sheaf of rank 1. Define the Picard group Pic(X)parameterizing isomorphism classes of line bundles on X. The group law is givenby tensor product. We can interpret Pic(X) as a cohomology group.

Proposition 3.9. There is a one-to-one correspondence between the isomorphismclasses of line bundles on X and H1(X,O∗), i.e.

Pic(X) ∼= H1(X,O∗).

Proof. Take an open covering U = Uα of X with respect to the trivialization ofa line bundle L. The transition function

gαβ : (Uα ∩ Uβ)× C→ (Uα ∩ Uβ)× C

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can be regarded as a section of O∗(Uα ∩ Uβ), satisfying

gαβ · gβα = 1,

gαβ · gβγ · gγα = 1.

Therefore, gαβ is a cocycle in C1(U,O∗), hence represents a cohomology class inH1(X,O∗).

Suppose M is another line bundle with transition functions hαβ. If M andL are isomorphic, then L ⊗M∗ is trivial, i.e. gαβ/hαβ are transition functionsof L ⊗ M∗, which has a nowhere vanishing section σ. Suppose on Uα we haveσα : Uα → C∗ as the restriction of σ. Then on Uα ∩ Uβ we have

gαβhαβ· σα = σβ .

Therefore we conclude thatgαβhαβ

=σβσα∈ δC0(U,O∗).

Now we describe another important correspondence between line bundles anddivisors. Suppose D is a divisor on X with local defining equations fα such thatfα ∈M ∗(Uα). Define

gαβ =fβfα.

Then we have gαβ ∈ O∗(Uα∩Uβ). Moreover, gαβ satisfy the assumptions imposedto transition functions, hence they define a line bundle, denoted by L = [D] orL = OX(D). We have a group homomorphism

Div(X)→ Pic(X)

induced by

D +D′ 7→ [D]⊗ [D′].

We say that D and D′ are linearly equivalent, if [D] and [D′] are isomorphic linebundles. We denote linear equivalence by

D ∼ D′.

Proposition 3.10. The associated line bundle [D] is trivial if and only if D is aprincipal divisor, i.e. D = (f) for some f ∈M ∗(X). Two divisors D ∼ D′ if andonly if D −D′ is a principal divisor.

Proof. Suppose D = (f) is the associated divisor of a meromorphic function f onX. Then D has local defining equations fα = f |Uα. The transition functionsassociated to [D] are all equal to 1, hence [D] is a trivial line bundle. Conversely,suppose [D] is trivial. Then it has a nowhere vanishing section σ whose restrictionto Uα is denoted by σα. The transition functions gαβ = fβ/fα defined above satisfy

gαβ · σα = σβ ,

hence we havefασα

=fβσβ∈M ∗(Uα ∩ Uβ).

We can glue fα/σα to form a global function f ∈ M ∗(X). Since σ is nowherevanishing, we obtain that (f) = D.

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Let us summarize, using the short exact sequence

0→ O∗ →M ∗ →M ∗/O∗ → 0.

Recall that

Div(X) ∼= H0(X,M ∗/O∗), Pic(X) ∼= H1(X,O∗).

Then we have the long exact sequence

0→ H0(X,O∗)→ H0(X,M ∗)(·)−→ Div(X)

[·]−→ Pic(X)→ · · ·which encodes all information in the above discussions.

3.5. Sections of a line bundle. Let L be a line bundle on X with transitionfunctions gαβ. A holomorphic section s of L has restriction sα ∈ O(Uα), satisfying

gαβsα = sβ .

Conversely, a collection sα ∈ O∗(Uα) such that sβ/sα = gαβ determines a sectionof L.

Similarly, we define a meromorphic section s to be a collection

sα ∈M (Uα)such that gαβsα = sβ . Suppose t 6≡ 0 is another meromorphic section with collectiontα. We have

sβtβ

=sαtα,

hence the quotient s/t is a global meromorphic function. Conversely, if f is a globalmeromorphic function, then f · sα defines another meromorphic section of L.

For a meromorphic section s 6≡ 0, consider the divisor

(sα)

associated to the local section sα in Uα. Sincesβsα

= gαβ

is nowhere vanishing, (sα) form a global divisor (s) on X. Conversely, suppose Dis a divisor. Consider the construction of the associated line bundle [D]. Supposethe local defining equations of D are given by sα. Then the transition functionsof [D] are gαβ = sβ/sα and consequently the collection sα gives rise to ameromorphic section of [D]. Note that for a section s of L, the divisor (s) iseffective if and only if s is a holomorphic section. We thus obtain the followingresult.

Proposition 3.11. For any section s of L, we have L ∼= [(s)]. A line bundle L isassociated to a divisor D if and only if it has a meromorphic section s such that(s) = D. In particular, L has a holomorphic section if and only if it is associatedto an effective divisor.

Now we treat a line bundle as a locally free sheaf of rank 1 and reinterpret theabove correspondence. Let D be a divisor on X. Define a sheaf OX(D) or simplyO(D) by

O(D)(U) = f ∈M (U) : (f) +D|U ≥ 0.It has a vector space structure since (f) +D|U ≥ 0 and (g) +D|U ≥ 0 implies that(af + bg) +D|U ≥ 0 for any a, b in the base field.

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Proposition 3.12. The space of holomorphic sections of [D] can be identified withH0(X,O(D)).

Proof. A global section s ∈ H0(X,O(D)) is a meromorphic function satisfying

(s) +D ≥ 0.

Suppose D is locally defined by fα. The associated line bundle [D] has transitionfunctions

gαβ =fβfα.

Then the collection s · fα defines a section σ of [D]. Since

(s) + (fα) ≥ 0

in every Uα, σ is a holomorphic section of [D]. Moreover, the associated divisorD′ = (s) + D of the section is linearly equivalent to D, since D′ − D = (s) isprincipal.

Conversely, given a holomorphic section σ of [D], i.e. a collection hα ∈ O(Uα)such that

hβhα

= gαβ =fβfα.

Then hα/fα defines a global meromorphic function h. Since (hα) ≥ 0 in everyUα, we have

(h|Uα) + (fα) = (hα) ≥ 0,

hence (h) +D ≥ 0 globally on X and h ∈ H0(X,O(D)).

Remark 3.13. Replacing X by any open subset U , the proposition implies thatthe sheaf O(D) can be regarded as gathering local holomorphic sections of the linebundle [D]. If D ∼ D′, i.e. D′ − D = (f) for a global meromorphic function f ,then for any g ∈ O(D′)(U), we have

0 ≤ (g) +D′ = (g) + (f) + (D) = (fg) + (D)

restricted to U . So we obtain an isomorphism

O(D′)(U)·f−→ O(D)(U)

for any open subset U , compatible with the sheaf restriction maps. In this sense,the sheaf O(D) and the line bundle [D] have a one-to-one correspondence up to iso-morphism and linear equivalence (assuming that every line bundle can be associatedto a divisor).

Let |D| be the set of effective divisors that are linearly equivalent to D. We call|D| the linear system associated to D.

Proposition 3.14. Let X be compact and D a divisor on X. Then we have

PH0(X,O(D)) = |D|,

i.e. an effective divisor in |D| and a holomorphic section of [D] (up to scalar)determine each other.

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Proof. For anyD′ ∈ |D|, by definitionD′−D = (f) is principal for some f ∈M (X),hence (f) +D = D′ ≥ 0 and f ∈ H0(X,O(D)). Since X is compact, if g is anotherfunction such that D′ −D = (g), then (f/g) = 0, i.e. f/g is holomorphic, hence itis a constant.

Conversely, any f ∈ H0(X,O(D)) defines an effective divisor D′ = (f) + D. If(f) +D = (g) +D, then (f/g) = 0 and f/g is a constant, since X is compact.

Exercise 3.15. Let D =∑aipi be a divisor on P1 with ai ∈ Z and pi ∈ P1. Define

the degree of D by deg(D) =∑ai.

(1) Prove that D ∼ D′ if and only if deg(D) = deg(D′).(2) Calculate the cohomology H∗(P1,O(D)) in terms of deg(D).

4. Algebraic curves

In this section, we apply the techniques developed before to study algebraiccurves.

4.1. The Riemann-Roch formula. Let X be a compact Riemann surface or aclosed algebraic curve. Define its arithmetic genus by

g := h1(OX) = dimCH1(OX).

Theorem 4.1 (Riemann-Roch Formula). Let D be a divisor on X and O(D) theassociated line bundle or locally free sheaf of rank 1. Then we have

h0(O(D))− h1(O(D)) = 1− g + deg(D).

Remark 4.2. Define the (holomorphic) Euler characteristic of a sheaf F by

χ(F ) :=∑i≥0

(−1)ihi(F ).

The Riemann-Roch formula can be written as

χ(O(D))− χ(OX) = deg(D).

Proof. Let us first prove it for effective divisors of degree ≥ 0. Do induction on n.The formula obviously holds for OX . Suppose it is true for deg(D) < n. ConsiderD = p + D′ with D′ an effective divisor of degree n − 1. We have the short exactsequence

0→ O(D′)→ O(D)→ Cp → 0,

where Cp is the skyscraper sheaf with a single stalk C supported at p. The exactnesscan be easily checked. The map O(D′)→ O(D) is an inclusion, since

(f) +D′ ≥ 0

implies that

(f) +D = (f) +D′ + p ≥ 0.

The quotient corresponds to germs of functions f at p such that

(f)|U +D′|U = −p

in arbitrarily small neighborhoods U of p. In other words, if ordp(D′) = m ≥ 0, we

can write

f = z−m−1h(z),

25

where h ∈ O∗(U). So the quotient sheaf is given by C · z−m−1 ∼= C supported atp. Since the associated cohomology sequence is long exact, we have

χ(O(D)) = χ(O(D′)) + 1 = 1− g + (n− 1) + 1 = 1− g + n.

In general, write a divisor D = D1 − D2, where D1 and D2 are both effectivedivisors of degree d1 and d2, respectively, and d1 − d2 = deg(D). By the sametoken, we have the short exact sequence

0→ O(D)→ O(D1)→ Cd2 → 0.

Then we obtain that

χ(O(D)) = χ(O(D1))− d2 = 1− g + d1 − d2 = 1− g + deg(D).

Remark 4.3. Assuming the Serre duality

H1(O(D)) ∼= H0(K ⊗ O(−D)),

where K is the canonical line bundle or the cotangent line bundle or the dualizingsheaf of X, then we can rewrite the Riemann-Roch formula as

h0(L)− h0(K ⊗ L∗) = 1− g + deg(L),

where L is a line bundle on X. Note that K is a degree 2g − 2 line bundle (to bediscussed later). We conclude that

h0(K) = g, h1(K) = h0(O) = 1.

It implies that the space of holomorphic one-forms on a genus g Riemann surfaceis g-dimensional.

4.2. The Riemann-Hurwitz formula. A branched cover π : X → Y betweentwo (compact, connected) Riemann surfaces is a (surjective) regular morphism. Fora general point q ∈ Y , π−1(q) consists of d distinct points. Call d the degree of π.Locally around p 7→ q, if the map is given by

x 7→ y = xm,

where x, y are local coordinates of p, q, respectively, call m the vanishing order ofπ at p and denote it by

ordp(π) = m.

If ordp(π) > 1, we say that p is a ramification point. If π−1(q) contains a ramifica-tion point, then q is called a branch point. Define the pullback

π∗(q) =∑

p∈π−1(q)

ordp(π) · p.

Note that π∗(q) is a degree d effective divisor on X.

Theorem 4.4 (Riemann-Hurwitz Formula). Let π : X → Y be a branched coverbetween two Riemann surfaces. Then we have

KX ∼ π∗KY +∑p∈X

(ordp(π)− 1) · p,

where KX and KY are canonical divisors on X and Y , respectively.

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Proof. Take a one-form ω on Y locally expressed as f(w)dw around a point q =π(p). Suppose the covering at p is given by

z 7→ w = zm,

then we have

π∗(f(w)dw) = mf(zm)zm−1dz.

Namely, the associated divisors satisfy the relation

(π∗ω)|U = (π∗(ω))|U + (ordp(π)− 1) · p

in a local neighborhood U of p. So globally it implies that

(π∗ω) = π∗(ω) +∑p∈X

(ordp(π)− 1) · p.

Since π∗ω is a one-form on X, (π∗ω) is a canonical divisor of X and the claimedformula follows.

We can interpret the (numerical) Riemann-Hurwitz formula from a topologicalviewpoint. Let χ(X) denote the topological Euler characteristic of X. If X is aRiemann surface of genus g, take a triangulation of X and suppose the number ofk-dimensional edges is ck for k = 0, 1, 2. Then we have

χ(X) = c0 − c1 + c2 = 2− 2g.

Proposition 4.5. Let π : X → Y be a degree d branched cover between two Rie-mann surfaces. Then we have

χ(X) = d · χ(Y )−∑p∈X

(ordp(π)− 1).

Proof. Take a triangulation of Y such that every branch point is a vertex. Pull itback as a triangulation of X. Note that it pulls back a face to d faces, an edge tod edges and a vertex v to |π−1(v)| vertices. Note that if

π−1(v) =

k∑i=1

mipi

for distinct points pi, then |π−1(v)| = m. In other words, we have

|π−1(v)| = d−∑

p∈π−1(v)

(ordp(π)− 1).

Then the claimed formula follows right away.

Corollary 4.6 (Numerical Riemann-Hurwitz). Let π : X → Y be a degree dbranched cover between two Riemann surfaces of genus g and h, respectively. Thenwe have

2g − 2 = d(2h− 2) +∑p∈X

(ordp(π)− 1).

In particular, if g < h, such branched covers do not exist.

Corollary 4.7. The canonical line bundle of a genus g Riemann surface X hasdegree equal to 2g − 2.

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Proof. Every Riemann surface X possesses a nontrivial meromorphic function, sayby the Riemann-Roch formula. It induces a branched cover π : X → P1 of degreed. By the Riemann-Hurwitz Formula we know

deg(KX) = d(−2) +∑p∈X

(ordp(π)− 1),

since we have seen that deg(KP1) = −2. By the Numerical Riemann-Hurwitz wehave

2− 2g = 2d−∑p∈X

(ordp(π)− 1).

Then the claim follows immediately.

Exercise 4.8. Let X be a Riemann surface or algebraic curve of genus g. If Xadmits a branched cover of degree 2 to P1, we say that X is a hyperelliptic curve.Prove that every g ≤ 2 curve is hyperelliptic. For g ≥ 2, can you calculate thedimension of the parameter space of genus g hyperelliptic curves?

Remark 4.9. The moduli space of genus g curves has dimension 3g − 3, which isbigger than 2g − 1 for g > 2. Hence a general g > 2 curve is not hyperelliptic.

4.3. Genus formula of plane curves. Suppose F (Z0, Z1, Z2) is a general degreed homogeneous polynomial whose vanishing locus is a plane curve C ⊂ P2. Since Fis general, C is smooth. In other words, the singularities of C locate at the commonzeros of F = 0 and ∂F/∂Zi = 0 for all i, which are empty for a general F .

Theorem 4.10. In the above setting, the genus g of C is given by

g =(d− 1)(d− 2)

2.

Proof. We give two proofs. The first one is more algebraic. Note that all degree dcurves are linearly equivalent in P2. Hence it makes sense to use OP2(d) to denotethe associated line bundle. In particular, O(1) is the associated line bundle of aline L. Then we have the short exact sequence

0→ OP2(−1)→ OP2 → OL → 0.

Tensor it with OP2(1−m). We obtain that

0→ OP2(−m)→ OP2(−(m− 1))→ OP2(1−m)|L → 0.

Since OP2(1−m)|L is the line bundle associated to a degree 1−m divisor on L andL ∼= P1, we conclude that

χ(OP2(−(m− 1)))− χ(OP2(m)) = χ(OP1(1−m)) = 2−m,where we apply the Riemann-Roch formula to P1 in the last equality. Then weobtain that

χ(OP2)− χ(OP2(−d)) =

d∑m=1

(2−m) = −d(d− 3)

2.

Now by the exact sequence

0→ OP2(−d)→ OP2 → OC → 0,

we have

1− g = χ(OC) = −d(d− 3)

2,

28

hence the genus formula follows.The other proof is an application of the Riemann-Hurwitz formula. Without loss

of generality, suppose o = [0, 0, 1] 6∈ C. Let L be the line Z2 = 0 and project C toL from o, i.e.

[Z0, Z1, Z2] 7→ [Z0, Z1].

In affine coordinates x = Z1/Z0 and y = Z2/Z0, this map is given by

(x, y) 7→ x,

i.e. we project C vertically to the x-axis. This yields a degree d branched cover

π : C → L ∼= P1.

A point p is a ramification point of π if and only if there exists a vertical line tangentto C at p, i.e. p is a common zero of F and ∂F/∂Z2. Since F and ∂F/∂Z2 havedegree d and d − 1, respectively, they intersect at d(d − 1) points. By Riemann-Hurwitz, we have

2g − 2 = d(−2) + d(d− 1),

hence the genus formula follows. In order to make sure all the ramifications aresimple, we may choose a general projection direction such that it is different fromthose of the (finitely many) lines with higher tangency to C.

Remark 4.11. In the first proof, indeed we did not use the smoothness of C.So the (arithmetic) genus formula holds for an arbitrary plane curve, even if it issingular, reducible or non-reduced. In the second proof, even if the projection hashigher ramification points, a detailed local study plus Riemann-Hurwitz still givesrise to the desired formula.

4.4. Base point free and very ample line bundles. Let L be a line bundle ora locally free sheaf of rank 1 on X. We say that L has a base point at p ∈ X if pbelongs to the vanishing locus of every regular section of L. If the base locus of Lis empty, then L is called base point free.

For a base point free line bundle L, let σ0, . . . , σn be a basis of the space H0(L)of regular sections. Locally around a point p ∈ X, treat σi as a regular functionand associate to p the point

[σ0(p), . . . , σn(p)] ∈ Pn.This is well-defined, since if we take a different chart, then we get σ′i(p) = gαβσi(p)where gαβ are transition functions of L. Therefore, we obtain a regular map

φL : X → Pn.We can give a more conceptual and coordinate free description of φL. Since L is

base point free, the space of regular sections σ vanishing at p forms a hyperplaneHp ⊂ H0(L) ∼= Cn+1. Then one can define φL(p) = [Hp] ∈ (Pn)∗ in the dualprojective space parameterizing hyperplanes.

Proposition 4.12. In the above setting, there is a one-to-one correspondence be-tween hyperplane sections of X and effective divisors in the linear system |L|.

Proof. This is just a reformulation of the one-to-one correspondence

|L| = PH0(L),

which we proved before. In other words, an effective divisor in |L| uniquely deter-mines a regular section σ =

∑aiσi mod scalar.

29

Example 4.13. If L = O, then we have φO maps X to a single point.

Example 4.14. Let X = P1 and L = O(2p) where p = [0, 1]. Then H0(L) is3-dimensional and we can choose a basis by

1,Y

X,

Y 2

X2.

Recall that around p the sections of L are given by f · (X/Y )2. Hence we obtainthat

φL([X,Y ]) = [X2, XY, Y 2],

which is a smooth conic in P2. The genus formula for a plane curve of degree 2 alsoimplies that the image has g = 0.

Exercise 4.15. A variety X ⊂ Pn is called non-degenerate if it is not contained inany hyperplane.

(1) Show that any non-degenerate smooth rational curve in Pn has degree ≥ n.(2) For d ≥ n ≥ 3, show that there exist non-degenerate smooth degree d rational

curves in Pn.

Example 4.16. Let E be an elliptic curve and L = O(2p). By Riemann-Roch,h0(L) = 2. Moreover, L is base point free. Otherwise if q is a base point, then q hasto be p and there exists another effective divisor p+ r ∈ |2p| such that p+ r ∼ 2p.But this implies r − p is principal and E ∼= P1, leading to a contradiction. NowφL : E → P1 is a branched cover of degree 2. Two points s and t lie in the samefiber if and only if s+ t ∼ 2p.

The above example indicates that φL is not always an embedding. We say thatL is very ample if φL is an embedding and that L is ample if L⊗m is very amplefor some m > 0.

Example 4.17. The line bundle O(d) is very ample on P1 if and only if d > 0.The induced map φ embeds P1 into Pd as a degree d smooth rational curve, i.e. arational normal curve.

Let us give a criterion when L is base point free or very ample on an algebraiccurve.

Proposition 4.18. Let L be a line bundle on a curve X.(1) L is base point free if and only if

h0(L⊗ O(−p)) = h0(L)− 1

for any p ∈ X.(2) L is very ample if and only L is base point free and for any p, q ∈ X (not

necessarily distinct)

h0(L⊗ O(−p− q)) = h0(L⊗ O(−p))− 1 = h0(L⊗ O(−q))− 1.

Proof. Treat L as a locally free sheaf of rank 1. By the short exact sequence

0→ L⊗ O(−p)→ L→ Cp → 0,

we haveh0(L)− 1 ≤ h0(L⊗ O(−p)) ≤ h0(L).

Then L has a base point at p if and only if all sections of L vanish at p, i.e.H0(L⊗ O(−p)) = H0(L). This proves (1).

30

For (2), a very ample line bundle is necessarily base point free by definition.If p 6= q ∈ X have the same image under φL, it is equivalent to saying that thesubspace of sections vanishing at p is the same as the subspace of sections vanishingat q, which is further equivalent to

h0(L⊗ O(−p)) = h0(L⊗ O(−p− q)) = h0(L⊗ O(−q)).

Moreover, φL induces an injection restricted to the tangent space Tp(X) if andonly if there exists a hyperplane such that it cuts out X locally a simple point atp, namely, if and only if there is a section vanishing at p with multiplicity 1, i.e.

h0(L⊗ O(−2p)) < h0(L⊗ O(−p)).

But we have

h0(L⊗ O(−2p)) ≥ h0(L⊗ O(−p))− 1.

Hence (2) follows by combining the two cases.

Remark 4.19. In (2), for p 6= q the condition geometrically means the sectionsof L separate any two points. When p = q, it says that the sections of L separatetangent vectors at p.

Example 4.20. Let E be an elliptic curve, i.e. a torus as a Riemann surface ofgenus 1. Fix a point p ∈ E. The morphism

τ : E → J(E) ∼= Pic0(E)

by τ(q) = [q− p] is an isomorphism. This defines a group law on E with respect top, i.e. q + r = s, where s ∈ E is the unique point satisfying

(q − p) + (r − p) ∼ s− p.

Now consider the linear system |3p| on E. Since

h0(O(3p)) = 3, h0(O(2p)) = 2, h0(O(p)) = 1,

O(3p) is very ample. It induces an embedding of E into P2 as a plane cubic. A linecuts out a divisor of degree 3 in E, say, q + r + s (not necessarily distinct) if andonly if

q + r + s ∼ 3p.

Note that the tangent line L of E at p is a flex line, i.e. the tangency multiplicity(L · E)p = 3. Such p is called a flex point.

Exercise 4.21. Show that there are in total 9 flex points on a smooth plane cubic.

Let V ⊂ |L| be a linear subspace. We say that V is a linear series of L. Thelinear system |L| is also called a complete linear series. The above definitions andproperties go through similarly for the induced map φV .

Exercise 4.22. Write down a linear series of |O(3)| on P1 such that it maps P1

into P2 as a singular plane cubic. How many different types of such singular planecubics can you describe?

31

4.5. Canonical maps. Let K be the canonical line bundle a curve X. If X is P1,deg(K) = −2 and K is not effective. If X is an elliptic curve, then K ∼= O andthe induced map φK is onto a point. From now on, assume that the genus g of Xsatisfies g ≥ 2. Recall that X is called hyperelliptic if it admits a degree 2 branchedcover of P1. Two points p, q ∈ X are called conjugate if they have the same imagein P1. A ramification point of the double cover is called a Weierstrass point of X,i.e. it is self conjugate. By Riemann-Hurwitz, a genus g ≥ 2 hyperelliptic curvepossesses 2g + 2 Weierstrass points.

Lemma 4.23. If X is a hyperelliptic curve of genus ≥ 2, then X admits a uniquedouble cover of P1.

Proof. Otherwise suppose h0(O(p+ q)) = 2 and h0(O(p+ r)) = 2 for q 6= r. Notethat h0(O(p+ q + r)) < 3, since X cannot be a plane cubic by the genus formula.Then we conclude that

H0(O(p+ q)) = H0(O(p+ r)) = H0(O(p+ q + r)),

which implies that both q, r are base points of |p+q+r| and h0(O(p)) = 2, X ∼= P1,leading to a contradiction.

Proposition 4.24. Let X be a curve of genus g ≥ 2. Then the canonical linebundle K is base point free. The induced map

φK : X → Pg−1

is an embedding if and only if X is not hyperelliptic. If X is hyperelliptic, φK is adouble cover of a rational normal curve in Pg−1.

Proof. First, let us show that K is base point free. For any point p ∈ X, byRiemann-Roch we have

h0(K ⊗ O(−p))− h0(O(p)) = 1− g + (2g − 3),

h0(K ⊗ O(−p)) = g − 1 = h0(K)− 1.

Hence K satisfies the criterion of base point freeness.Next, K fails to separate p, q (not necessarily distinct) if and only if

h0(K ⊗ O(−p− q)) = h0(K ⊗ O(−p)) = g − 1,

which is equivalent to, by Riemann-Roch again, that

h0(O(p+ q)) = 2.

In other words, the linear system |p+ q| induces a double cover X → P1.Finally, if X is hyperelliptic of genus ≥ 2, it admits a unique double cover of P1.

By the above analysis, two points p, q have the same image under the canonicalmap if and only if h0(p + q) = 2, i.e. p, q are conjugate. Then the canonical mapis a double cover of a rational curve of degree deg(K)/2 = g − 1 in Pg−1, i.e. arational normal curve. A hyperplane section of φK(X) pulls back to X a divisor

g−1∑i=1

(pi + qi),

where pi, qi are conjugate or pi = qi a Weierstrass point.

Remark 4.25. For a non-hyperelliptic curve X, φK is called the canonical embed-ding of X and its image is called a canonical curve.

32

Example 4.26. Let X be a genus 2 curve. Then h0(K) = 2, hence X is hyper-elliptic and the double cover of P1 is induced by the canonical line bundle, as wehave seen.

Example 4.27. A genus 3 non-hyperelliptic curve admits a canonical embeddingto P2 as a plane quartic. An effective canonical divisor corresponds to a line sectionof the quartic. By the genus formula, any smooth plane quartic also has genusequal to 3. Moreover, a smooth plane quartic X gives rise to a line bundle L ofdegree 4 on X by restricting OP2(1). By Riemann-Roch, h0(K ⊗ L∗) ≥ 1, butdeg(K ⊗ L∗) = 0, hence L = K. So any plane quartic is a canonical embedding ofa genus 3 non-hyperelliptic curve.

Example 4.28. Let X be a genus 4 non-hyperellipic curve. Then its canonicalembedding is a degree 6 curve in P3. Let OP3(1) denote the line bundle on P3

associated to a hyperplane class. Its restriction to X is the canonical line bundleKX . We have the exact sequence

0→ IC ⊗ OP3(2)→ OP3(2)→ OC(2)→ 0.

Since h0(OP3(2)) = 10 > h0(OC(2)) = 9, we conclude that C is contained in aquadric surface Q in P3. Indeed Q is unique, because otherwise deg(C) ≤ 4.

If Q is smooth, it is isomorphic to P1 × P1. Up to projective equivalence, it hascoordinates

[XZ,XW,Y Z, Y W ],

i.e. its equation isZ0Z3 − Z1Z2 = 0.

Note thatZ0 − aZ1 = Z2 − aZ3 = 0

defines a family A of lines in Q parameterized by the value of a. Any two lines inA are disjoint. Similarly,

Z0 − bZ2 = Z1 − bZ3 = 0

defines another family B of lines in Q parameterized by b, and any two lines in Bare disjoint. Moreover, a line L1 in A and a line L2 in B intersect at a unique point[ab, b, a, 1]. Hence they span a plane H in P3. Suppose H · X = D1 + D2, whereDi is a divisor in Li. Since Li varies in Q in a P1 family, we have h0(OX(Di)) ≥ 2for i = 1, 2. Moreover, because D1 + D2 ∼ KX , we have deg(D1) + deg(D2) = 6.Since X is not hyperelliptic, we conclude that deg(Di) = 3 for i = 1, 2. This canalso be seen since X is a complete intersection of Q with a cubic surface, hence itsclass in Q is linearly equivalent to 3L1 + 3L2. Therefore, we obtain two distinctline bundles of degree 3 given by OX(D1) and OX(D2) such that h0(OX(Di)) = 2.In other words, X admits two triple covers of P1. This can be seen by projectingX along the direction of Li for i = 1, 2.

If Q is singular, since X is non-degenerate, Q is a quadric cone whose equationis given by

Z20 − Z1Z2 = 0.

It has a unique singular point v = [0, 0, 0, 1] as the vertex and Q can be viewed asthe cone over a plane conic C defined by the same equation. Every line L containedin Q passes through v, hence C parameterizes a P1-family of lines. Similarly, onechecks that L ·X = 3 and X admits a unique triple cover of P1 by projecting fromv to C.

33

Exercise 4.29. Suppose g ≥ 3. Let X ⊂ Pg−1 be a non-degenerate smooth genusg curve of degree 2g − 2. Show that X is a canonical curve.

Exercise 4.30. Let X be a genus 4 non-hyperelliptic curve. If D = p+ q + r is adegree 3 effective divisor on X such that h0(O(D)) = 2, show that p, q and r arecollinear in the canonical embedding of X in P3.

Exercise 4.31. A hyperelliptic curve X of genus g can be explicitly written as thelocus of (z, w) satisfying

w2 = (z − a1) · · · (z − a2g+2).

One can treat (w, z) 7→ z as the double cover of P1 which is branched at a1, . . . , a2g+2.Prove that a basis for the space H0(K) of holomorphic one-forms on X is

dz

w, zdz

w, . . . , zg−1 dz

w.

4.6. An informal moduli counting. In this section we present some (heuristic)properties about moduli of curves. We want to find a “moduli space”Mg parame-terizing isomorphism classes of genus g curves and study its geometry. If g = 0, wejust get a pointed space representing the unique isomorphism class of P1. For g = 1,the j-invariant characterizes elliptic curves, so their parameter space should be of1-dimensional. Alternatively, we can take the space of smooth plane cubics moduloPGL(3), which gives rise to a 1-dimensional parameter space of elliptic curves. Forhigher genus, let us first consider the moduli space of hyperelliptic curves.

Proposition 4.32. The space of genus g ≥ 2 hyperelliptic curves has dimensionequal to 2g − 1.

Proof. If X is hyperelliptic of genus g ≥ 2, it admits a unique double cover of P1

with 2g + 2 branch points. Conversely, fix 2g + 2 points on P1. There exists aunique genus g, connected double cover of P1 branched at the fixed points, by theHurwitz monodromy description, i.e. the monodromy around each branch point isthe unique simple transposition (12) ∈ S2. Therefore, up to Aut(P1) ∼= PGL(2) wehave a one-to-one correspondence between hyperelliptic curves and (2g+2)-pointedP1. Hence the dimension of the space of hyperelliptic curve is (2g+ 2)− 3 = 2g− 1for g ≥ 2. Along the way we also see how to construct the moduli space, by

(Sym2g+2(P1)−∆)/PGL(2),

where ∆ is the big diagonal of the symmetric product.

Example 4.33. Every genus 2 curve is hyperelliptic. In particular, the modulispace M2 is 3-dimensional.

Example 4.34. Every smooth plane quartic corresponds to a canonical curve ofgenus 3. Consequently the moduli space of (non-hyperelliptic) genus 3 curves can beidentified with (an open subset of) P14/PGL(3). In particular, it is 6-dimensional.On the other hand, the space of hyperelliptic curves of genus 3 is 5-dimensional,hence a general genus 3 curve is not hyperelliptic.

In general, one approach to count the dimension of moduli of genus g curves isby using branched covers of P1 with sufficiently high degree.

Proposition 4.35. The moduli space Mg of genus g ≥ 2 curves has dimensionequal to 3g − 3. In particular, a general genus g ≥ 3 curve is not hyperelliptic.

34

Proof. Fix d > 2g − 2 and 2g − 2 + 2d distinct points on P1 and consider degreed, connected, genus g simply branched covers of P1 branching at the fixed points.Due to the monodromy description, there exist finitely many (not necessarily one)such covers. Conversely, to construct such a cover on a genus g curve X, we canfirst take a degree d effective divisor D which maps to∞. Then the map is inducedby a meromorphic function f on X such that

(f) +D ≥ 0,

i.e. f is a section of O(D). Since D is non-special, we have

h0(O(D)) = 1− g + d.

Putting everything together, the space Mg depends on

2g − 2 + 2d− [d+ (1− g + d)] = 3g − 3

parameters. Here a scalar acting on f matters, since if f = g/h and [g, h] is theinduced map, it is different from the map [λg, h], where λ is a constant.

For g ≥ 3, we have

3g − 3 > 2g − 1,

hence a general genus g ≥ 3 curve is non-hyperelliptic.

Remark 4.36. Note that the naive dimension count fails for g = 0, 1. For g = 0,Aut(P1) is 3-dimensional, hence the choice of the pole divisor D depends on d− 3parameters only. While for g = 1, due to the translation of an elliptic curve, itdepends on d − 1 parameters. Taking this correction into account, we recover theright moduli for both cases. Of course here we implicitly used the assumption thatthe automorphism group of a genus g ≥ 2 curve is finite (discussed later).

4.7. Special linear systems. Let D = p1+· · ·+pd be an effective divisor of degreed on a genus g curve X. Recall that the linear system |D| can be identified withPH0(O(D)) parameterizing effective divisors linearly equivalent to D. Suppose asa projective space

r = dim |D| = h0(O(D))− 1.

By Riemann-Roch, we have

dim |K ⊗ O(−D)| = r + g − d− 1.

Note that |K ⊗ O(−D)| is the linear system of canonical divisors that contain D.By the canonical map

φK : X → Pg−1,

it says that the space of hyperplanes of Pg−1 that contain φK(p1), . . . , φK(pd) is(r + g − d− 1)-dimensional. In other words, the linear span of φK(p1), . . . , φK(pd)is a

(g − 2)− (r + g − d− 1) = (d− 1)− rdimensional subspace in Pg−1. Since we expect d points to span a (d−1)-dimensionallinear subspace, geometrically it says that φK(D) fails to impose

r = dim |D|

independent conditions. We summarize the discussion as a geometric version of theRiemann-Roch formula.

35

Theorem 4.37 (Geometric Riemann-Roch). In the above setting, let φK(D) bethe linear span of the image of D under the canonical map. Then we have

dim |D| = deg(D)− 1− dimφK(D).

Remark 4.38. Even if D contains multiple point, the formulation still holds, bytaking the multiplicity into account. Say, if D contains 2p, then 2p span the tangentline at p. If D contains 3p, then 3p spans an osculating 2-plane at p.

Example 4.39. Let us revisit the canonical embedding of a genus 4 non-hyperellipticcurve X in P3. Recall that X is contained in a unique quadric surface Q and weshowed that X admits a triple cover of P1 corresponding to a family of lines in Q.Conversely, if D = p + q + r induces a triple cover of P1, i.e. if dim |D| = 1, byGeometric Riemann-Roch, we have

dimφK(D) = 3− 1− 1 = 1,

i.e. p, q and r are collinear in a line L in P3. Because L ·Q = 2 unless L is containedin Q, any triple cover of P1 on X corresponds to a family of lines in Q. We haveseen that if Q is smooth, there are two such families of lines, i.e. X admits twodistinct triple covers of P1, while if Q is singular, such a triple cover is unique.

Let us study in detail the dimension of a linear system.

Lemma 4.40. Let D be a divisor on a curve X. Then dim |D| ≥ k if and only iffor every k points p1, . . . , pk ∈ X there exists an effective divisor in |D| containingall of them.

Proof. Since∑ki=1 pi varies in a k-dimensional family, then dim |D| ≥ k is obvious.

Alternatively, we may also prove it by induction. Suppose it holds for ≤ k. Assumefor every p1, . . . , pk+1, there exists D′ ∈ |D| containing all of them. Then weconclude that dim |D − p| ≥ k for any p ∈ X. Since |D| cannot contain entirely Xin its base locus, choose a point p not in the base locus of |D|. Consequently wehave

dim |D| = dim |D − p|+ 1 ≥ k + 1.

Now suppose dim |D| ≥ k. Then we have

h0(O(D −k∑i=1

pi)) ≥ h0(O(D))− k ≥ 1.

It implies that there exists a non-zero meromorphic function f such that

(f) +D −k∑i=1

pi ≥ 0,

hence (f) +D = D′ is an effective divisor in |D| containing p1, . . . , pk.

Corollary 4.41. For any two effective divisors D1 and D2 on X, we have

dim |D1|+ dim |D2| ≤ dim |D1 +D2|.

Proof. Suppose dim |Di| = ki for i = 1, 2. Take any k1 + k2 points

p1, . . . , pk1 , q1, . . . , qk2

36

in X. By the above lemma, there exist D′1 ∈ |D1| and D′2 ∈ |D2| such that D′1contains all the pi and D′2 contains all the qj . Then D′1 +D′2 ∈ |D1 +D2| containsall the pi, qj , hence we obtain

dim |D1 +D2| ≥ k1 + k2

by the lemma again.

Note that if h0(K ⊗ (−D)) = 0, then Riemann-Roch determines that

h0(O(D)) = 1− g + deg(D).

Some subtlety may occur if

h0(K ⊗ (−D)) 6= 0

and we call such a divisor D a special divisor and the associated linear system |D|a special linear system. By Riemann-Roch, any D with deg(D) > 2g − 2 is non-special. By Geometric Riemann-Roch, D is non-special if and only if the linearspan of φK(D) is the entire space Pg−1.

Theorem 4.42 (Clifford’s Theorem). Let D be a special effective divisor on X.Then we have

dim |D| ≤ 1

2· deg(D).

Proof. Since D is special, there exists an effective divisor D′ such that D+D′ ∼ K,hence we have

dim |D|+ dim |D′| ≤ dim |K| = g − 1.

By Riemann-Roch, we have

dim |D| − dim |D′| = 1− g + deg(D).

The desired inequality follows by combining the two relations.

Remark 4.43. Indeed, the equality holds only if D = 0, D = K or X is hyper-elliptic. If D = 0 or D = K, one easily checks that the equality holds. If X ishyperelliptic, we may choose D = p+q where p, q are conjugate and dim |p+q| = 1.To prove that these are the only possibilities, we need the uniform position theo-rem regarding a general hyperplane section of a non-degenerate space curve, cf.[Griffiths-Harris, p. 249].

Exercise 4.44. Let X be a genus ≥ 2 hyperelliptic curve. For 0 < 2k ≤ g, findan effective divisor D of degree 2k on X such that dim |D| = k. Classify all suchdivisors up to linear equivalence.

4.8. Weierstrass points. We want to generalize the concept of Weierstrass pointson a hyperelliptic curve to an arbitrary curve. Let X be a curve of genus g ≥ 2 andp ∈ X a point. Set N = n ∈ Z : n > 0 and define Hp ⊂ N by

Hp = n : ∃f ∈M (X) such that (f)∞ = np.

Note that if (f)∞ = np and (h)∞ = mp, then (fh)∞ = (m+ n)p. We say that Hp

is the Weierstrass semigroup of p. Define Gp ⊂ N as the complement of Hp, i.e.

Gp = n : @f ∈M (X) such that (f)∞ = np.

We say that Gp is the Weierstrass gap sequence of p.

37

Lemma 4.45. We have n ∈ Hp if and only if

h0(O(np)) = h0(O((n− 1)p)) + 1

and n ∈ Gp if and only if

h0(O(np)) = h0(O((n− 1)p)).

In other words, n ∈ Gp (resp. Hp) if and only if p is (resp. not) a base point of thelinear system |np|. Moreover, Gp is a subset of 1, . . . , 2g − 1 with cardinality g.

Proof. By the exact sequence

0→ O((n− 1)p)→ O(np)→ Cp → 0,

we conclude that

h0(O((n− 1)p)) ≤ h0(O(np)) ≤ h0(O((n− 1)p)) + 1.

The right hand side equality holds if and only if p is not a base point of |np|, namely,if and only if there exists f ∈M (X) such that (f) +np ≥ 0 but (f) + (n− 1)p 6≥ 0,which is equivalent to saying that (f)∞ = np, i.e. if and only if n ∈ Hp.

For any n ≥ 2g, by Riemann-Roch we have

h0(np) = h0((n− 1)p) + 1.

Hence we conclude that Gp ⊂ 1, . . . , 2g − 1. Moreover, we have

g − 1 = h0(O((2g − 1)p))− h0(OX)

=

2g−1∑n=0

(h0(O(np))− h0(O((n− 1)p))

)So there are g− 1 elements of 1, . . . , 2g− 1 belonging to Hp. In other words, thecardinality of Gp is

(2g − 1)− (g − 1) = g.

Lemma 4.46. We have n ∈ Gp if and only if there exists a section of the canonicalline bundle, i.e. a holomorphic differential ω, such that ordp(ω) = n − 1. As aconsequence we conclude again that Gp is a subset of 1, . . . , 2g − 1 of cardinalityg.

Proof. By Riemann-Roch, h0(O(np)) = h0(O((n− 1)p)) if and only if

h0(K ⊗ O(−(n− 1)p)) = h0(K ⊗ O(−np)) + 1.

Note that H0(K ⊗ O(−mp)) parameterizes holomorphic one-forms ω such that(ω) ≥ mp. So the above equality holds if and only if there exists ω such that(ω) ≥ (n− 1)p but (ω) 6≥ np, namely, if and only if ordp(ω) = (n− 1)p.

Since h0(K) = g, one can choose a basis ω1, . . . , ωg such that ordp(ωi) = ai and

a1 < a2 < · · · < ag.

Then we obtain that

Gp = a1 + 1, a2 + 1, . . . , ag + 1.

Because deg(K) = 2g − 2, we have 0 ≥ ai ≤ 2g − 2 for all i, hence Gp is a subsetof 1, . . . , 2g − 1.

38

We say that p is a Weierstrass point of X if Gp 6= 1, 2, . . . , g and p is a normalWeierstrass point if Gp = 1, 2, . . . , g − 1, g + 1. Define the weight of p by

w(p) =( ∑n∈Gp

n)− (1 + 2 + · · ·+ g)

=( ∑n∈Gp

n)− g(g + 1)

2.

Then p is a Weierstrass point if and only if w(p) > 0 and p is a normal Weierstrasspoint if and only if w(p) = 1.

Now we introduce some basic properties of a semigroup of N. In general, ifH ⊂ N is a semigroup whose complement G = N−H consists of g elements, definethe weight of H by

w(H) =(∑n∈G

n)− g(g + 1)

2.

Lemma 4.47. In the above setting, we have

w(H) ≤ (g − 1)g

2.

The equality holds if and only if 2 ∈ H.

Proof. Let H = a1, a2, . . . , such that a1 < a2 < · · · . Suppose b ∈ G is thesmallest element such that b > an. Then for 1 ≤ i ≤ n, we have

b− ai ∈ G, b− ai ≤ an.

The two disjoint sets

b− an, b− an−1, . . . , b− a1, a1, . . . , an

are both contained in 1, 2, . . . , an. Hence we conclude that

an ≥ n+ n = 2n.

Now suppose G = b1, . . . , bg. Then H can be expressed as

1, . . . , b1 − 1; b1 + 1, . . . , b2 − 1; b3 + 1, . . . , b4 − 1; . . ..

Using the inequality, we have

bk − 1 = abk−k ≥ 2(bk − k),

hence we conclude that bk ≤ 2k − 1.Now by definition, we have

w(H) =

g∑k=1

(bk − k)

≤g∑k=1

(k − 1)

=(g − 1)g

2.

Moreover, the equality holds if and only if bk = 2k − 1 for 1 ≤ k ≤ g, henceG = 1, 3, . . . , 2g − 1 and 2 ∈ H.

39

Proposition 4.48. Let X be a curve of genus g ≥ 2. Then we have∑p∈X

w(p) = (g − 1)g(g + 1).

In particular, X has finitely many Weierstrass points and the number of distinctWeierstrass points is at least 2g + 2. This minimum number occurs if and only ifX is hyperelliptic.

Proof. Choose a basis ω1, . . . , ωg of H0(X,K) and write ωi = fi(z)dz for i =1, . . . , g in terms of a local coordinate z around p. Consider the Wronskian

W (z) = det

f1(z) · · · fg(z)f ′1(z) · · · f ′g(z)

......

f(g−1)1 (z) · · · f

(g−1)g (z)

It is a basic fact that for linearly independent functions f1, . . . , fg, W (z) is non-zero.

Suppose z is another coordinate around p such that ωi = fi(z)dz. Let

ψ =dz

dz∈ O∗(U ∩ U).

For f = ψf , we have

df

dz= ψ

df

dz

dz

dz+dψ

dzf

= ψ2 df

dz+dψ

dzf

and in general

dnf

dzn= ψn+1 d

nf

dzn+ · · ·

for higher derivatives. Denote by

N = 1 + 2 + · · ·+ g = g(g + 1)/2.

Then we conclude that

W (z) =(dzdz

)NW (z).

For instance, if g = 2, we have

det

(f1 f2

f ′1 f ′2

)= det

(ψf1 ψf2

ψ2f ′1 + ψ′f1 ψ2f ′2 + ψ′f2

)= ψ1+2 det

(f1 f2

f ′1 f ′2

).

Consequently it implies that W (z)(dz)N defines a global section of H0(X,K⊗N ).

40

Moreover, from the expression of W (z) we have

ordp(W (z)) =( g∑i=1

ai

)− (0 + 1 + · · ·+ (g − 1))

=( g∑i=1

(ai + 1))− (1 + 2 + · · ·+ g)

=( ∑n∈Gp

n)− g(g + 1)/2

= w(p).

Since deg(K⊗N ) = (g − 1)g(g + 1), the desired formula follows right away.By definition, a Weierstrass point has strictly positive weight, so X has finitely

many Weierstrass points. Since w(p) ≤ (g − 1)g/2, we conclude that X has atleast 2g+ 2 distinct Weierstrass points. If this minimum number occurs, then eachWeierstrass point p satisfies 2 ∈ Hp, i.e. X admits a double cover of P1 with p asa ramification point, hence X is hyperelliptic.

Exercise 4.49. Let X be a hyperelliptic curve of genus ≥ 2 with a double coverπ : X → P1 and p ∈ X a point. Prove directly that

(1) p is a ramification point of π if and only if Gp = 1, 3, . . . , 2g − 1;(2) p is not a ramification point of π if and only if Gp = 1, 2, . . . , g.Hint: use either the canonical map of X or an explicit basis of H0(KX).

Exercise 4.50. Determine all possible Weierstrass gap sequences that may occurfor a point on a genus 3 curve.

As an application, we have the following result.

Proposition 4.51. The automorphism group of a genus g ≥ 2 curve is finite.

Proof. Let X be a genus g curve. Since g ≥ 2, X has finitely many Weierstrasspoints. In particular, an automorphism τ of X sends a Weierstrass point to aWeierstrass point.

If X is hyperelliptic, there are 2g + 2 Weierstrass points, corresponding to the2g+2 ramification points of the associated double cover π : X → P1. Since the spaceof canonical divisors containing p is the same of that containing τ(p), τ factorizesthrough the canonical map. Modulo the hyperelliptic involution exchanging thetwo sheets of φ, τ is induced by an automorphism of P1 that sends the 2g + 2branch points to themselves. In other words, Aut(X) is a (Z/2)-extension of theautomorphism group of a (2g+2)-pointed P1. Since 2g+2 > 3, such automorphismsof P1 are of finitely many, because any automorphism of P1 fixing three points mustbe the identity.

Suppose X is non-hyperelliptic. It is sufficient to show that if an automorphismτ fixes all Weierstrass points of X, then τ is the identity. Suppose τ is not identityand fixes all Weierstrass points of X. Take general points p1, . . . , pg+1 on X. Thenthere exists f ∈M (X) such that

(f)∞ = p1 + · · ·+ pg+1.

Consider h = f − τ∗f ∈ M (X). Since pi’s are general, h is not identically zero.Since deg(h) ≤ g+ 1, h can have at most g+ 1 zeros and g+ 1 poles. If p is a fixedpoint of τ , then p is either a zero or a pole of h. So the number of fixed points of τ

41

is bounded above by 2g+ 2. But X has more than 2g+ 2 Weierstrass points, sinceit is not hyperelliptic, leading to a contradiction.

Remark 4.52. We used the fact that for general p1, . . . , pg+1 on X, the linearsystem |p1 + · · · + pg+1| has dimension ≥ 1 and is base point free. The dimensionbound follows directly from Riemann-Roch. The base point freeness follows if wecan show that dim |p1 + · · ·+pg| ≥ 1 only happens for special collections of pi’s. ByRiemann-Roch again, this is equivalent to saying that there is a hyperplane in Pg−1

that contains p1, . . . , pg in the canonical image of X. We just need to take generalpi’s such that any g of them span the total space Pg−1 other than a hyperplane.

4.9. Plane curves and their duals. Let C ⊂ P2 be a smooth degree d planecurve defined by a homogeneous polynomial F . For every point p ∈ C, its tangentline Tp(C) is a point in the dual plane P2∗. Varying p in C, the union of all tangentlines of C forms a curve C∗ in P2∗. We say that C∗ is the dual curve of C. At apoint [Z0, Z1, Z2] ∈ C, the tangent line equation is given by

2∑i=0

∂F

∂ZiZi = 0,

i.e. C∗ has coordinates [∂F/∂Z0, ∂F/∂Z1, ∂F/∂Z2].

Example 4.53. Let Q be a smooth plane conic defined by

XZ − Y 2 = 0.

At a point [X,Y, Z], suppose the tangent line of Q is given by AX + BY + CZ.One checks that the coefficients A, B and C satisfy

4AC −B2 = 0.

In other words, the dual curve Q∗ is a smooth plane conic in the dual plane.

Proposition 4.54. Let C be a smooth degree d plane curve. Then its dual curveC∗ in P2∗ has degree equal to d(d− 1).

Proof. Let p ∈ P2 be a point. All lines passing through p form a line in the dualplane P2∗. Conversely, every line in P2∗ arises in this way. In other words, we havethe incidence correspondence variety

I = ([X,Y, Z], [A,B,C]) | AX +BY + CZ = 0 ⊂ P2 × P2∗

parameterizing the pairs (p, q), where the line representing by q (resp. p) passesthrough p (resp. q).

Let Lp be a line in the dual plane P2∗ parameterizing all lines in P2 passingthrough p. Note that q lies in the intersection of Lp and C∗ if and only if thereexists a tangent line Lq of C such that Lq contains p. Namely, the degree of C∗ isequal to the number of tangent lines of X through p.

Fix a line L0 in P2 and project C from p to L0. We obtain a degree d branchedcover π : C → L0

∼= P1. If p is a general point, we can make sure that the tangentlines of X through p are simple, i.e. π has only simple ramification points. ByRiemann-Hurwitz, the number of ramification points of π is

2g − 2 + 2d = d(d− 1).

42

Since each simple ramification point gives rise to a tangent line through p and viceversa, we conclude that

deg(C∗) = d(d− 1).

Remark 4.55. If C has isolated singularities, the dual curve can still be definedby the closure of the union of its tangent lines at smooth points.

Proposition 4.56. The dual of the dual is the original curve, i.e. C∗∗ = C.

Proof. Let p0 ∈ C be a regular point. The tangent line p∗0 ∈ P2∗ to C at p0 isthe limit of the secant lines p0p as p approaches p0. Similarly, p∗∗0 is the limit ofthe secant lines p∗p∗0 as p∗ approaches p∗0. Note that the dual of the secant linep∗p∗0 is the intersection point of the two lines p∗ and p∗0 in P2. In other words,p∗∗0 ∈ P2∗∗ = P2 is the limit of the intersection point of the tangent lines to C at pand p0 as p approaches p0. It is clear that this limit is just p0.

Remark 4.57. The above two propositions seem to contradict each other. Indeed,nothing is wrong, because the dual curve C∗ is not always smooth, even if C issmooth. For instance, if C has a bitangent line L, i.e. L is tangent to C at twodistinct points p and q, then C∗ will self-intersect at a node [L] ∈ P2∗. If C has aflex line L, i.e. L is tangent to C at p with tangency multiplicity ≥ 3, then C∗ willhave a cusp at [L] ∈ P2∗. Hence it makes sense to discuss these basic plane curvesingularities.

Let C be a plane curve and C∗ its dual. Suppose the defining equations of Cand C∗ are f and f∗, respectively. We say that C has traditional singularities ifevery point p ∈ C is one of the following:

1. A regular point. Namely, p is a smooth point of C and the tangent line to Cat p is a smooth point of C∗. In local coordinates, f can be expressed as

[1, z + · · · , z2 + · · · ],

i.e. f = y − x2 modulo the ideal (y2, xy, x3).2. An ordinary flex. Namely, p is a smooth point of C and the tangent line to

C at p has contact of order three. Locally f can be expressed as

[1, z + · · · , z3 + · · · ],

i.e. f = y − x3 modulo (y2, xy, x4). The tangent lines around p have equation

y − (z3 + · · · )x− (z + · · · )

=∂y

∂x

=3z2 + · · ·1 + · · ·

.

Writing it as AX +BY + CZ, the dual curve C∗ has coordinates

[−3z2 + · · · , 1 + · · · , 2z3 + · · · ],

i.e. it has a singularity of type y2 − x3 = 0, which is a cusp.3. A cusp. Locally f is expressed as

[1, z2 + · · · , z3 + · · · ],

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i.e. f = y2 − x3 modulo (y3, xy, x4). The tangent lines around p have equation

y − (z3 + · · · )x− (z2 + · · · )

=∂y

∂x

=3z2 + · · ·2z + · · ·

.

Writing it as AX +BY + CZ, the dual curve C∗ has coordinates

[−3z2 + · · · , 2z + · · · , z4 + · · · ] = [−3z + · · · , 2 + · · · , z3 + · · · ],

i.e. C∗ has a flex line.

Remark 4.58. We see that cusps and flexes correspond to each other between acurve and its dual.

4. A bitangent. Namely, the tangent line at p is also tangent to X at anotherpoint q with both contact orders equal to two.

5. A node (or ordinary double point). Namely, X has two branches intersectingtransversely at p. Locally f has expression f = xy modulo (x2, y2).

Remark 4.59. Geometrically it is clear that nodes and bitangents correspond toeach other between a curve and its dual.

Let C be a plane curve and C∗ its dual with traditional singularities. Let g bethe geometric genus of C, i.e. the genus of its normalization C. Let d be the degreeof C, b the number of bitangents, f the number of flexes, κ the number of cuspsand δ the number of nodes. We also write g∗, d∗ etc to denote the correspondingquantities for C∗. Based on the above discussion, we have

b = δ∗, δ = b∗,

κ = f∗, f = κ∗.

Proposition 4.60. We have the following classical Plucker formulas:

g =(d− 1)(d− 2)

2− δ − κ,

d∗ = d(d− 1)− 2δ − 3κ,

g∗ =(d∗ − 1)(d∗ − 2)

2− b− f,

d = d∗(d∗ − 1)− 2b− 3f.

Moreover, we have g = g∗.

Proof. By assumption C may have nodes and cusps. Each singularity makes thegeometric genus drop by one compared to the arithmetic genus. Hence the geomet-ric genus of C equals

g =(d− 1)(d− 2)

2− δ − κ.

Recall how we determine the degree of C∗ for a smooth plane curve C. Weproject C from a general point p to a line L. Then the degree of C∗ is equal to thenumber of ramification points of the map. In case when C has nodes and cusps, theprojection π : C → L induces a branched cover π : C → L. If the projection goesthrough a cusp, the inverse image of the cusp is a simple ramification point of π.

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If the projection passes a node, we see two distinct points in C, both unramified.By Riemann-Hurwitz we conclude that

2g − 2 + 2d = d∗ + κ.

Hence we obtain thatd∗ = d(d− 1)− 2δ − 3κ.

The other two identities follow from the duality relation. Since C and C∗ areisomorphic away from finitely many points, they have the same normalization bythe Riemann extension theorem, hence g = g∗.

Remark 4.61. Let π : C → C be the normalization map. Then we have an exactsequence

0→ OC → π∗OC → F → 0,

where F has 0-dimensional support at the singularities of C. For example, if Chas a cusp, locally

OC = k[x, y]/(y2 − x3), OC = k[t].

The map on the left side is given by

x 7→ t2, y 7→ t3,

hence the cokernel is one-dimensional spanned by t.The cohomology of π∗OC is the same as that of OC . Hence we have

g(C) = 1− χ(OC)

= 1− χ(OC)− h0(F )

= pa(C)− h0(F ).

Note the the structure of F is determined by the local type of the singularities.Since a plane rational cubic with a node or a cusp has arithmetic genus equal toone but geometric genus equal to zero, we conclude that a node or a cusp makesthe genus of a plane curve drop by one.

Example 4.62. Let C be a smooth plane cubic. Then we have

d = 3, g = 1, δ = 0, κ = 0.

Moreover, by the degree reason C does not possess a bitangent, hence b = 0. Bythe Plucker formulas, we conclude that d∗ = 6, hence

3f = d∗(d∗ − 1)− d = 27

and f = 9. We have seen that the nine flexes correspond to the nine torsion pointsin the Jacobian of an elliptic curve.

Now suppose C is a plane rational nodal cubic. Then we have

d = 3, g = 0, δ = 1, κ = 0, b = 0.

We conclude thatd∗ = 4, f = 3,

i.e. the number of flexes of C drops to three.If C is a plane rational cuspidal cubic, we have

d = 3, g = 1, δ = 0, κ = 1, b = 0.

We conclude thatd∗ = 3, f = 1,

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i.e. C has a unique flex at the cusp.

Example 4.63. Let C be a smooth plane quartic such that its dual has traditionalsingularities. Then we have

d = 4, g = 3, δ = 0, κ = 0.

By the Plucker formulas, we have

d∗ = 12,

b+ f = 52,

2b+ 3f = 128,

hence we obtain that

b = 28, f = 24.

In other words, C possesses 28 bitangents and 24 flexes. Note that the 24 flexescorrespond to the 24 simple Weierstrass points on a general curve of genus three.

Exercise 4.64. Let C be a smooth degree d plane curve whose dual curve hastraditional singularities. Calculate the number of bitangents and flexes of C.

5. Chern classes

Chern classes are objects associated to vector bundles. They can be defined invarious ways. Here our approach is to treat them as elements in the Chow ring ofthe base variety.

5.1. Chow ring and rational equivalence. Let X be an n-dimensional variety,smooth or with mild singularities. Let Zd(X) be the free abelian group generated byirreducible subvarieties of codimension d. Elements of Zd(X) are called d-cocyclesor (n− d)-cycles.

Let Y ⊂ X be a (k + 1)-dimensional subvariety. For a k-cycle Z, if there existsf ∈M (Y ) such that (f) = Z, we say that Z is a principal k-cycle. Define rationalequivalence ∼ between two k-cycles if their difference is a principal k-cycle. Define

Ad(X) = Zd(X)/ ∼

as the dth Chow group of X and use [Z] to denote the equivalence class of Z.

Remark 5.1. Geometrically, if two k-cycles Z and W are rationally equivalent,then f induces a map Y → P1. We can construct an incidence correspondence

Y = (y, t) | f(y) = t ⊂ X × P1.

In other words, Y is a (k + 1)-cycle of X × P1 such that the fibers of Y → P1 overtwo points are Z and W , respectively. Hence rational equivalence is analogous tocobordism in topology.

Proposition 5.2. Let X be an n-dimensional irreducible variety.(1) A0(X) ∼= Z is generated by [X].(2) Ad(X) = 0 for d > n.(3) If X is smooth, then A1(X) ∼= Pic(X).(4) There exists a degree map An(X)→ Z.

46

Proof. The first three trivially hold by definition. For (4), note that An(X) isgenerated by rational equivalence classes of points. Since the degree of a principaldivisor on a curve is zero, the degree map Zn(X) → Z factorizes through An(X).

Remark 5.3. Rational equivalence could be subtle even for 0-dimensional cycles.Two points are rationally equivalent if and only if there exists a chain of rationalcurves (not necessarily smooth) connecting them. For instance, any two points onP1 are rationally equivalent. But any two points on a positive genus curve are notrationally equivalent.

Example 5.4. One can show that any two k-dimensional linear subspaces of Pnare rationally equivalent. Moreover, any two hypersurfaces of the same degree arerationally equivalent in Pn. In general, let Y be a subvariety of Pn of degree k.Take a general point p and project Y from p to a hyperplane Pn−1. We can choosep such that the projection π : Y → Pn−1 is generically finite, i.e. for a general pointq in the image of π, the inverse of q consists of finitely many points. Suppose p hascoordinate [1, 0, . . . , 0] and the hyperplane is given by X0 = 0. Then π is given by

[X0, . . . , Xn] 7→ [0, X1, . . . , Xn].

It implies that [X0, . . . , Xn] ∈ Y varies in a P1 family [tX0, X1, . . . , Xn] ∈ Yt to itsprojection image as t varies from 1 to 0. Then we conclude that Y is rationallyequivalent to deg(π) · π(Y ).

Let [H] be the hyperplane class of Pn. By induction we thus know that Ad(Pn) =Z generated by [Hd], where [Hd] is the class of a linear subspace of codimensiond. This notation makes sense because we can take d hyperplanes H1, . . . ,Hd suchthat they cut out a linear subspace of codimension d. Moreover, [Hi] = [H] for alli. This example suggests that the union of Ad(X) for all d possesses a graded ringstructure induced by intersection between cycles.

Exercise 5.5. Prove directly that two linear subspaces of Pn of the same dimensionare rationally equivalent.

Let V and W be two irreducible subvarieties. If

codim(V ∩W ) = codim(V ) + codim(W ),

we say that V and W intersect properly. This induces an intersection product

Ad(X)×Ae(X)→ Ad+e(X)

by [V ] · [W ] = [V ∩W ] with appropriate multiplicities on each component of theintersection. Define the Chow ring of X as

A∗(X) =

n⊕d=0

Ad(X).

This is a graded ring with unit 1 = [X].

Remark 5.6. The fact that the intersection product is well-defined is based onChow’s moving lemma. It says that if Y, Z are two cycles, then there exists Y ′

rationally equivalent to Y such that Y ′ and Z intersect properly. Moreover, if Y ′′

is another such cycle, then Y ′ ∩ Z and Y ′′ ∩ Z are rationally equivalent.

47

Example 5.7. Let H be the hyperplane class of Pn. Then we have

A∗(Pn) ∼= Z[H]/Hn+1.

Note that the Chow ring and the cohomology ring of Pn coincide over C. But ingeneral this is not the case.

5.2. Formal calculation of Chern classes. Let E be a vector bundle of rank mon X. Assign to E an element ck(E) ∈ Ak(X) for 1 ≤ k ≤ m. Call ck(E) the kthChern class of E and define the total Chern class by

c(E) = 1 + c1(E) + · · ·+ cm(E).

We want these assignments to satisfy the following properties:(1) If L is a line bundle associated to a divisor D, then c1(L) = [D].(2) If there exists an exact sequence 0→ E → F → G→ 0, then

c(F ) = c(E)c(G).

(3) If f : Y → X is a map and E a vector bundle on X, then c(f∗E) = f∗c(E).The actual definition of Chern classes will be introduced later by degeneracy

loci. Here we do some calculation first.

Example 5.8. Let Ei = O(di) be a line bundle of degree di on Pn for i = 1, 2.Then we have

c(E1 ⊕ E2) = c(E1)c(E2)

= (1 + d1H)(1 + d2H)

= 1 + (d1 + d2)H + d1d2H2.

Hence we conclude that

c1(E1 ⊕ E2) = (d1 + d2)H,

c2(E1 ⊕ E2) = d1d2H2.

Example 5.9. Let us compute the Chern class of the tangent bundle on Pn. Let Sbe the tautological line bundle such that the fiber S|[L] represents the 1-dimensional

subspace L ⊂ Cn+1. Then S∗ is the universal line bundle such that the fiber S∗|[L] is

the space of linear functionals Hom(L,C). Take a non-zero linear map φ : Cn+1 → Cand define Φ([L], v) = φ(v), where v is a vector in L. Then Φ can be regarded asa section of S∗. Moreover, Φ is the zero map at [L] if and only if L ⊂ ker(φ), i.e.the zero locus of Φ in Pn is a hyperplane given by the projectivization of ker(φ). Itimplies

S∗ = OPn(1), S = OPn(−1).

We have an exact sequence

0→ S → V → Q→ 0,

where V = Cn+1 × X is the trivial bundle of rank n + 1, S → V is the naturalinclusion L ⊂ Cn+1 fiber by fiber and Q is the quotient bundle with fiber Q|[L] =

Cn+1/L. Tensor it by S∗ and apply the identification S = OPn(−1). We thusobtain

0→ OPn →n+1⊕i=1

OPn(1)→ S∗ ⊗Q→ 0.

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Note that the fiber of the tangent bundle TPn |[L] is canonically isomorphic to

Hom(L,Cn+1/L). More precisely, given φ ∈ Hom(L,Cn+1) and v ∈ L, one candefine an arc v(t) = v+ tφ(v) ∈ Cn+1. The part φ(v) modulo L is a tangent vectorat [L]. Conversely, one can lift a tangent vector as an equivalence class of arcsthrough v ∈ L and deduce φ accordingly. Globally it implies that

TPn ∼= S∗ ⊗Q.

Hence we obtain the following Euler sequence

0→ OPn →n+1⊕i=1

OPn(1)→ TPn → 0.

The total Chern class of TPn is

c(TPn) = c(OPn(1))n+1

= (1 +H)n+1

= 1 + (n+ 1)H +

(n+ 1

2

)H2 + · · ·+

(n+ 1

n

)Hn.

Note that for n = 2, c(TP2) = 1 + 3H + 3H2 is not a product of two linear forms ofH. Hence TP2 cannot be an extension of two line bundles on P2, i.e. there do notexist two line bundles L and M such that they fit in the exact sequence

0→ L→ TP2 →M → 0.

5.3. The splitting principle. In an ideal world, a vector bundle would split as adirect sum of line bundles. Then we could calculate its Chern class based on thatof line bundles. Well, in real world there exist tons of non-splitting vector bundles.But for computing the Chern classes only, one can still pretend that a vector bundlesplits and carry out the calculation formally.

To illustrate what this means, consider a vector bundle E of rank n. Suppose Esplits as a direct sum of line bundles

E = L1 ⊕ · · · ⊕ Ln

and let ai = c1(Li). Then we have

c(E) =

n∏i=1

(1 + ai)

= 1 +( n∑i=1

ai

)+(∑i6=j

aiaj

)+ · · ·

hence we conclude that

c1(E) =

n∑i=1

ai, c2(E) =∑i6=j

aiaj , . . .

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Suppose L is a line bundle with first Chern class c1(L) = b. Then

c(E ⊗ L) = c((L1 ⊗ L)⊕ · · · ⊕ (Ln ⊗ L))

= (1 + a1 + b) · · · (1 + an + b)

= 1 +( n∑i=1

ai + nb)

+

(∑i 6=j

aiaj +

(n

2

)b2 + (n− 1)

( n∑i=1

ai

)b

)+ · · ·

= 1 + (c1(E) + nc1(L)) +

(c2(E) +

(n

2

)c21(L) + (n− 1)c1(E)c1(L)

)+ · · ·

Hence we conclude that

c1(E ⊗ L) = c1(E) + nc1(L),

c2(E ⊗ L) = c2(E) +

(n

2

)c21(L) + (n− 1)c1(E)c1(L)

and etc.The splitting principle says that even if E is not decomposable as a direct sum

of line bundles, the above calculation still holds. In other words, one can write

c(E) =

n∏i=1

(1 + ai),

where ai’s are formal roots of the Chern polynomial

ct(E) = 1 + c1(E)t+ · · · cn(E)tn.

We call a1, . . . , an the Chern roots of E. If E splits as a direct sum of line bundles,then ai’s are just the first Chern classes of the summands, hence belong to A1(X).In general, ai may not correspond to a divisor class. But the Chern classes of Eare represented by symmetric functions of ai’s, hence eventually one can get rid ofai’s and express the final result of this kind of calculation by the Chern classes ofthe original vector bundles.

Let us use the splitting principle to study the Chern class of the canonical linebundle on a variety.

Proposition 5.10. Let K be the canonical line bundle and T the tangent bundleon a smooth variety X. Then we have

K ∼ −c1(T ).

Proof. If T splits as⊕n

i=1 Li, thenn∧T =

n⊗i=1

Li.

Hence we conclude that

c1

( n∧T)

=

n∑i=1

c1(Li)

= c1(T ).

Therefore, the associated divisor of K is the dual, i.e. −c1(T ). The splittingprinciple ensures that it holds regardless of whether or not T is decomposable.

Corollary 5.11. On Pn we have K = O(−n− 1).

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Proof. We have seen thatc(TPn) = (1 +H)n+1.

In particular, we have c1(TPn) = (n+ 1)H. Then the claim follows from the aboveproposition.

Finally let us say a word about why the splitting principle works. Roughlyspeaking, for a vector bundle E on X, one can always find f : X → X suchthat f∗E is split and A∗(X) → A∗(X) is injective. Then one can carry out the

calculation in A∗(X) regarded as a subring of A∗(X).

Exercise 5.12. Let E and F be vector bundles of rank two on X. Use their Chernclasses to calculate the Chern classes of E∗, Sym2(E),

∧2E, E ⊕ F and E ⊗ F .

Exercise 5.13. Let X be an k-dimensional smooth variety embedded in Pn. Sup-pose the tangent bundle of X is trivial. Prove that n ≥ 2k and give an example ofsuch X when the equality holds.

Hint: use the exact sequence 0 → TX → TPn → N → 0, where N is the normalbundle of X, and consider their Chern classes.

5.4. Determinantal varieties. Let M = M(m,n) be the space of m×n matrices.As a variety, M is isomorphic to Amn. For 0 ≤ k ≤ min(m,n), denote by Mk =Mk(m,n) the locus of matrices of rank at most k, that is, cut out by all the(k+ 1)× (k+ 1) minors. We say that Mk is the kth generic determinantal variety.

Proposition 5.14. Mk is an irreducible subvariety of M of codimension (m −k)(n− k).

Proof. Define the incidence correspondence

Mk = (A,W ) | A ·W = 0 ⊂M ×G(n− k, n).

Then Mk admits two projections p1 and p2 to M and G(n − k, n), respectively.The map p1 is onto Mk and generically one to one. In other words, p1 has positivedimensional fiber over A if and only if the rank of A is at most k− 1. On the otherhand, for a fixed W ∈ G(n − k, n), the fiber of p2 over W is isomorphic to Amk.Hence we conclude that

dimMk = dim Mk

= dimG(n− k, n) +mk

= (m+ n)k − k2.

Therefore, the codimension of Mk in M is equal to

mn− (m+ n)k + k2 = (m− k)(n− k).

Moreover, Mk is a vector bundle on G(n−k, n), hence is irreducible. So is Mk.

Remark 5.15. As a vector bundle, Mk is smooth, but the contraction Mk couldbe singular. With a more detailed study of the tangent space, one can show thatthe singular locus of Mk is exactly Mk−1.

Determinantal varieties are useful to the study of maps between vector bundles.Let E and F be two vector bundles on X of rank n and m, respectively. Supposeφ : E → F is a bundle map. Choose a local trivialization of E and F . Then φ locallycorresponds to an m× n matrix, whose entries are holomorphic functions. Denote

51

by Φk the locus in X where the rank of φ is at most k. Then Φk is the inverseimage of the generic determinantal variety Mk, hence it has expected codimension(m− k)(n− k) in X.

5.5. Degeneracy loci. Let E be a vector bundle of rank n on a projective varietyX. Moreover, suppose E is globally generated, i.e. for any p ∈ X, the fiber E|pis fully spanned by its global sections. We will define Chern classes for such E.In general, one can twist a vector bundle by a very ample line bundle to obtain aglobally generated vector bundle, define its Chern classes and then twist back tothe original bundle.

For 1 ≤ i ≤ minn, dimX, let s1, . . . , si be general sections of E. At a pointp ∈ X, we expect these sections to be linearly independent. Hence we denote byDi the degeneracy locus in X where s1, . . . , si are linearly dependent. Equivalently,Di is the locus of p in X such that

s1(p) ∧ · · · ∧ si(p) = 0.

Note that Di possesses a determinantal structure. Take a trivialization of E ina local neighborhood U and let e1, . . . , en be sections that generate E|U . Then onecan write

sj =

n∑l=1

ajlel

for j = 1, . . . , i, where ajl ∈ O(U). Then Di|U is defined by all the i× i minors ofthe i×n matrix (ajl). Adapting the dimension of the corresponding determinantalvariety, we conclude that the (expected) codimension of Di in X is n − i + 1. Wethus define the Chern class

cn−i+1(E) := [Di] ∈ An−i+1(X).

Remark 5.16. If the sections s1, . . . , si are general, then Di has the right codi-mension as expected. Moreover, different choices of general sections give rise torationally equivalent degeneracy loci.

The definition by degeneracy loci satisfies the formal properties of Chern classes.For instance, if E = L1⊕L2 such that Li is a very ample line bundle, take a generalsection si of Li such that (si) = Si is an effective divisor. Then D1 is the locuswhere (s1, s2) is zero, i.e.

c2(E) = [D1]

= [S1] · [S2]

= c1(L1)c1(L2).

Moreover, take (s1, 0) and (0, s2) as two sections of E. Then D2 is the locus whereeither s1 or s2 is zero, i.e.

c1(E) = [D2]

= [S1] + [S2]

= c1(L1) + c1(L2).

Hence we recover that c(E) = c(L1)c(L2) as required by the formal properties ofChern classes.

Let us consider some applications along this circle of ideas.

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Example 5.17 (Bezout’s Theorem). Consider the vector bundle E = O(a)⊕O(b)on P2 with a, b > 0. We know

c(E) = 1 + (a+ b)L+ abL2,

where L is the line class of P2. Let F,G be two general homogenous polynomials inthree variables of degree a and b, respectively. Then (F,G) is a general section of E.The degeneracy locus D1 associated to (F,G) consists of points that are commonzeros of F and G. In other words, let C and D be the corresponding plane curvesdefined by F and G, respectively. Then we conclude that

#(C ∩D) = deg(c2(E))

= ab.

5.6. Grassmanians and Schubert calculus. As the dual projective space pa-rameterizes hyperplanes, one can consider the Grassmanian G(k, n) parameterizingk-dimensional linear subspaces of Pn, or equivalently (k + 1)-dimensional linearsubspaces of Cn+1. Recall that dimG(k, n) = (k+ 1)(n− k). As we showed beforefor Pn, there is a similar exact sequence on G(k, n):

0→ S → V → Q→ 0,

where S is the tautological bundle of rank k + 1 whose fiber over [W ] ∈ G(k, n)represents the subspace W , V is the trivial bundle of rank n + 1 and Q is thequotient bundle of rank n − k. The tangent bundle TG is isomorphic to S∗ ⊗ Q,hence it fits in the Euler exact sequence

0→(k+1)2⊕

O →n+1⊕

S∗ → TG → 0.

In order to compute the Chern class of TG, we need to know the Chow ring ofG(k, n). In general, the Chow ring of a Grassmanian is generated by its subvari-eties parameterizing k-dimensional linear subspaces satisfying certain interpolationconditions. Below we will use an example to illustrate the idea.

Consider G(1, 3) parameterizing lines in P3. This is a 4-dimensional variety. Fixa flag (p ∈ L ⊂ Λ), where p is a point, L is a line and Λ is a plane in P3. Definethe Schubert cycles of G(1, 3) as follows:σL: the locus of lines that intersect L;σp: the locus of lines that passes through p;σΛ: the locus of lines that are contained in Λ.One can perturb the flag configuration and the resulting cycles are rationally

equivalent to the original ones. So those σ• are elements of A∗(G(1, 3)).

Exercise 5.18. Show that the codimensions of σL, σp and σΛ in G(1, 3) are 1, 2and 2, respectively.

Since σL ∈ A1(G(1, 3)), let us compute σ2L in A2(G(1, 3)). Modulo rational

equivalence, we can take two lines L1, L2 and compute σL1 ∩ σL2 instead. Indeed,we specialize L1, L2 such that they intersect at one point p. Then they also span aplane Λ. In this special configuration, [L0] ∈ σL1

∩σL2means either L0 is contained

in Λ or L0 passes through p. It implies that

σ2L = σp + σΛ

in A∗(G(1, 3)). Of course one has check that it does not cause higher multiplicitiesduring the specialization. But it can be verified by a further study of the tangent

53

space of Schubert cycles. In general, the study of intersection products betweenSchubert cycles is called Schubert calculus.

Exercise 5.19. Carry out the following calculation in A∗(G(1, 3)).(1) Show that σpσΛ = 0, σ2

p = 1 and σ2Λ = 1.

(2) Calculate the degree of σ4L in A4(G(1, 3)) ∼= Z. In other words, calculate the

number of lines that intersect four general lines in P3.

Back to the exact sequence 0→ S → V → Q→ 0 applied to G(1, 3). We knowthat S∗|[W ] = Hom(W,C), where W ⊂ C4 is a 2-dimensional linear subspace. Letus use the degeneracy loci to calculate the Chern classes of S∗.

Take two general linear maps φi : C4 → C for i = 1, 2. They induce two generalsections Φi of S∗ by Φi([W ])(w) = φi(w) for w ∈ W , i = 1, 2. By definition, thevanishing locus of Φ1 (or Φ2) is c2(S∗). Note that Φ1 is the zero map on W if andonly if W ⊂ ker(φ1), that is the line [W ] is contained in the plane Λ obtained byprojectivizing ker(φ1). We thus obtain that

c2(S∗) = σΛ.

Similarly the locus where Φ1 and Φ2 fail to be linearly independent is c1(S∗). Thishappens if and only if W ∩ (ker(φ1) ∩ ker(φ2)) 6= 0, i.e. [W ] intersects the line Lcut out by the projectivization of ker(φ1) ∩ ker(φ2). We thus obtain that

c1(S∗) = σL.

In total, we conclude that

c(S∗) = 1 + σL + σΛ.

Then we have

c(S) = 1− σL + σΛ,

c(Q) =1

c(S)= 1 + σL + σp,

hence one can readily calculate c(TG) = c(S∗ ⊗Q).

Example 5.20 (Lines on a cubic surface). Let X be a smooth cubic surface in P3

defined by a polynomial F . Note that Sym3 S∗ is a rank four vector bundle whosefiber is H0(L,O(3)), the space of cubic forms on a line L in P3. The upshot isthat F can be regarded as a section of Sym3 S∗, i.e. over [L] it specifies a cubicform F |L ∈ H0(L,O(3)). Moreover, L is contained in X if and only if F |L = 0.In other words, the locus of lines lying on X as a cycle in the Grassmannian hasclass c4(Sym3 S∗) ∈ A4(G(1, 3)). Using the Chern classes of S∗, we can show thatc4(Sym3 S∗) = 27. Modulo some multiplicity check, it implies that a smooth cubicsurface contains 27 lines.

Exercise 5.21. Do the explicit calculation to show that c4(Sym3 S∗) = 27.

Exercise 5.22. Use Chern classes to explain that a degree d ≥ 4 surface in P3 isexpected to contain no lines.

Exercise 5.23. Show that the expected number of lines on a quintic 3-fold (i.e. adegree 5 hypersurface in P4) is 2875.

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6. Algebraic surfaces

Throughout this section X will be a smooth projective surface. Recall thatDiv(X) is the group of all divisors on X whose quotient modulo linear equivalenceis the Picard group Pic(X) of isomorphism classes of line bundles. A curve on Xis also a divisor.

6.1. Intersection theory on surfaces. Let C,D be two curves on X and p apoint in their intersection. We say that C and D meet transversally at p if thedefining equations f, g of C,D generate the maximal ideal mp = (x, y) in the localring, where x, y are the local coordinates at p. We also use the notation OC(D) todenote the line bundle OX(D)⊗ OC on C.

Exercise 6.1. Show that if C and D meet transversally at p, then C,D are bothnonsingular at p.

If C and D meet transversally at r distinct points in total, it is natural to definethe intersection number C.D = r. We want to generalize this intersection pairingto any two divisors in Div(X).

Theorem 6.2. There is a unique intersection pairing Div(X) × Div(X) → Z,denoted by C.D for two divisors C,D such that

(1) if C and D meet transversally everywhere, then C.D = #(C ∩D),(2) it is symmetric: C.D = D.C,(3) it is additive: (C1 + C2). D = C1. D + C2. D, and(4) if C1 ∼ C2, then C1. D = C2. D.

The proof of the theorem is decomposed into several steps.

Lemma 6.3. Suppose two curves C and D meet transversally. Then

#(C ∩D) = deg(OC(D)).

Proof. Note that OX(D) ⊗ OC is a line bundle on C whose associated divisor isC ∩D. Since the intersection is transverse everywhere, this divisor thus consists of#(C ∩D) points, hence its degree equals the degree of the line bundle.

Proof of Theorem 6.2. First let us show the uniqueness. Since X is projective, letH be a fixed ample divisor. For any two divisors C,D, we can find n 0 such thatC +nH, D+nH and nH are all very ample. Using certain results of Bertini type,one can choose C ′ ∈ |C +nH|, D′ ∈ |D+nH|, E′ ∈ |nH| and F ′ ∈ |nH| such thatthey are all smooth, D′ transversal to C ′, E′ transversal to D′ and F ′ transversalto C ′, E′. Since C ∼ C ′ − E′ and D ∼ D′ − F ′, by the properties we have

C.D = #(C ′ ∩D′)−#(C ′ ∩ F ′)−#(E′ ∩D′) + #(E′ ∩ F ′).It implies that C.D is determined by the natural intersection numbers.

For the existence, first we define for two very ample divisors C,D by settingC.D = #(C ′∩D′) where C ′ ∈ |C| and D′ ∈ |D| such that C ′, D′ meet transversally.Since D and D′ (resp. C and C ′) represent the same line bundle, by Lemma 6.3the above setting does not depend on the choices of C ′ and D′. Now we have awell-defined pairing between very ample divisor classes. To define it for any twodivisors C,D, we can write C ∼ C ′ − E′ and D ∼ D′ − F ′ such that C ′, D′, E′, F ′

are all very ample. Then C.D follows from the intersection pairing between thosevery ample divisors. One can check that if we choose different expressions for C and

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D, then the resulting pairings are the same. Moreover, it satisfies all the desiredproperties.

It would be handy to define the intersection number locally at a point. SupposeC and D are two curves defined by f and g, respectively, without a common com-ponent. Define the intersection multiplicity (C.D)p to be the length of Op,X/(f, g),i.e. the dimension of the k-vector space k[[x, y]]/(f, g).

Example 6.4. Let f = y and g = y − x2 over C. Then k[[x, y]]/(f, g) ∼= k[[x]]/x2

is generated by 1 and x as a k-vector space, hence the length is two.

Proposition 6.5. If C and D do not have a common component, then

C.D =∑

p∈C∩D(C.D)p.

Proof. Recall the exact sequence

0→ OC(−D)→ OC → OC∩D → 0,

where (as a scheme) C ∩D is defined by (f, g) locally, namely the structure sheafat p is Op,X/(f, g). Then we have

h0(OC∩D) = χ(OC)− χ(OC(−D)).

The left hand side equals∑p∈C∩D(C.D)p by definition. The right hand side only

depends on the linear equivalence classes of C and D, hence we can replace themby differences of smooth curves that are transversal to each other appropriately.Therefore, it reduces to deg(OC(D)) = C.D as in the definition of the pairing.

Exercise 6.6. Consider two plane curves Q : Y Z−X2 = 0 and C : Y Z2−X3 = 0.Calculate the intersection multiplicity of Q and C at every intersection point andconfirm that they satisfy Proposition 6.5.

Example 6.7 (Self-intersection). Even if C is nonsingular, we cannot directly seehow to define C2 = C.C. Instead, we use C2 = deg(OC(C)), where OC(C) is thenormal bundle of C in X, often denoted by NC/X or NC .

Let us treat normal bundles in a more general setting. Suppose X is a smoothcomplex manifold and D ⊂ X a smooth submanifold of codimension k. Denote byND/X or simply ND the normal bundle of D in X as

ND := (TX |D)/TD.

Similarly, let N∗D be the conormal bundle that fits in the exact sequence

0→ N∗D → ΩX |D → ΩD → 0,

where Ω is the cotangent bundle, i.e. the sheaf of differentials. Both ND and N∗Dare of rank k.

Proposition 6.8 (Adjunction formula I). Suppose D is a smooth, effective divisorin X. Then we have

ND ∼= OX(D)|D, N∗D∼= OX(−D)|D.

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Proof. Suppose D is defined by (Uα, fα). Then the line bundle OX(D) has transi-tion functions gαβ = fα/fβ . Since fα ≡ 0 on D ∩ Uα, the differential dfα defines alocal section of N∗D on Uα. Moreover,

dfα = gαβ · dfβon D∩Uαβ , hence the union (Uα, dfα) form a global section of N∗D⊗OX(D), whichis nowhere zero, for otherwise D would be singular at the vanishing locus. It impliesthat N∗D ⊗ OX(D) is a trivial line bundle.

Proposition 6.9 (Adjunction formula II). In the above setting, suppose D is adivisor and let K be the canonical line bundle of X. Then

KD ∼ (KX ⊗ OX(D))|D.

Proof. By the defining sequence of N∗D and Propostion 6.8, we have

c1(OD(−D)) = c1(N∗D) = c1(KX |D)− c1(KD),

hence KD ∼ (KX ⊗ OX(D))|D.

Example 6.10 (Canonical line bundle of a hypersurface). Let X be a degree ksmooth hypersurface in Pn. Then we have

KX ∼ (KPn ⊗ OPn(k))|X = OX(k − n− 1).

Now we specialize to the surface case.

Proposition 6.11 (Adjunction formula for surfaces). If C is a smooth curve ofgenus g on X and if KX is the canonical divisor of X, then

2g − 2 = C. (C +KX).

Proof. We know thatKC ∼ (KX ⊗ O(C))|C .

Take the degree of both sides. The left hand side is 2g − 2 and the right hand sideis C. (C +KX). Hence the desired formula follows.

Example 6.12 (Plane curves). Suppose D and E are plane curves of degree d ande, respectively. Then D.E = de as we saw in Bezout’s theorem. Moreover, if C isa smooth plane curve of degree d, then by Proposition 6.11 we recover that

g =(d− 1)(d− 2)

2.

Example 6.13 (Quadric surfaces). Let Q be a smooth quadric surface in P3. Onecan show that Pic(Q) ∼= Z ⊕ Z, generated by lines in the two families of rulings.Denote the generators by f1 = (1, 0) and f2 = (0, 1). We have

f21 = f2

2 = 0, f1. f2 = 1,

because two lines in the same family are disjoint and two lines of opposite familiesmeet transversally at a point. A hyperplane section of Q has class (1, 1) and thecanonical line bundle K of Q has class (−2,−2) by adjunction.

Let C be a smooth curve of type (a, b). Then C +K has class (a− 2, b− 2). ByProposition 6.11, we have

2g − 2 = (a, b). (a− 2, b− 2) = 2ab− 2a− 2b,

g = (a− 1)(b− 1).

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In particular, if g = 0 and if C has degree d in P3, we have (a, b) = (1, d − 1) or(d− 1, 1).

Recall the Riemann-Roch formula on curves. It says that χ(L)−χ(C) = deg(L)for a line bundle L on a curve C. In other words, for fixed degree the holomorphicEuler characteristic of L only depends on the topology of C. An analogous resultholds for surfaces (as well as for higher dimensional varieties).

Theorem 6.14 (Riemann-Roch for surfaces). Let D be a divisor on a surface X.Then we have

χ(OX(D))− χ(OX) =1

2D. (D −KX).

Proof. Since the result only depends on the linear equivalence classes, we can writeD ∼ C − E for two nonsingular curves C and E. By the exact sequences

0→ OX(C − E)→ OX(C)→ OE(C)→ 0,

0→ OX → OX(C)→ OC(C)→ 0,

we obtain that

χ(OX(C − E))− χ(OX) = χ(OC(C))− χ(OE(C)).

Moreover, by Riemann-Roch on curves we have

χ(OC(C)) = 1− gC + C2,

χ(OE(C)) = 1− gE + C.E.

Finally, by Proposition 6.11 we know

gC =1

2C. (C +KX) + 1,

gE =1

2E. (E +KX) + 1.

Combining the above, we thus obtain the desired formula.

Exercise 6.15 (Bertini). Let X ⊂ Pn be a smooth projective variety. Prove thata general hyperplane cuts out a nonsingular subvariety of X. Use this to show thatif D is a very ample divisor on X, then a general element in the linear system |D|is a smooth divisor.

Exercise 6.16. Suppose that X is a smooth surface of degree d in P3 and supposeit contains a line L. Prove that L2 = 2− d on X.

Exercise 6.17. Let X be a smooth surface of degree d in P3. Calculate the self-intersection K2

X .

6.2. Blow up. Blow up is a general procedure which applies to a subspace of avariety. To illustrate the idea, here we restrict to blowing up a point on a smoothsurface S. Since the operation is local, we use affine coordinates x, y for S = A2 oran open disk and suppose that p = (0, 0) is the origin.

Consider all lines passing through p. We get a one-dimensional family of linesthat are distinguished by the direction y/x if x 6= 0 or x/y if y 6= 0. Use [u, v] to

be the coordinates of P1. Define the blow up of S at p by S as follows:

S := (x, y, [u, v]) | xv = yu ⊂ S × P1.

Note that S admits a contraction π to S by forgetting the coordinates [u, v].

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Proposition 6.18. In the above setting, S is a smooth surface, S\π−1(p) → S\pis an isomorphism and π−1(p) ∼= P1.

Proof. For any q 6= p, i.e. q has coordinate (x, y) such that x, y are not bothzero, without loss of generality suppose that x 6= 0. Then v/u = y/x is uniquelydetermined by q, hence π−1(q) is a single point. Indeed one can use x, y to be the

local coordinates of S nearby π−1(q), hence S is smooth away from π−1(p) and

S\π−1(p) ∼= S\p.At p, since x = y = 0, there is no constraint to [u, v]. We thus obtain that

π−1(p) ∼= P1.Finally for a point q = (0, 0, [u, v]) ∈ π−1(p), without loss of generality suppose

that u 6= 0 and v/u = t. Then a local coordinate system nearby q can be given by

x, t, namely, (x, tx, [1, t]). Therefore, S is also smooth along π−1(p).

Denote by E = π−1(p) ∼= P1 the exceptional curve of the blow up. Let C ⊂ S

be an effective curve. Denote by C the closure of π−1(C\p) and call it the proper

transform of C. If C does not contain p, then C is just π−1(C) and π : C → C isan isomorphism. The purpose of doing blow up is to separate tangent directions ofcurves that pass through p.

Example 6.19. Let L be a line passing through p. Suppose (a, b) is a point onL not equal to p. Then a, b cannot be both zero. Suppose a is not zero and lett = b/a. Then π−1(L\p) parameterize (x, y, [u, v]) such that y/x = v/u = t. In

other words, the equations that define L are y − tx = 0 and v − tu = 0. Note that

L ∩ E is a single point (0, 0, [1, t]) = (0, 0, [a, b]). Therefore, the proper transforms

of lines through p form a disjoint union of lines in S, each of which meets E at adistinct point corresponding to the direction of the line.

Remark 6.20. The proper transform of a curve is different from the total inverse

image under π, if the curve contains p. For instance, π−1(L) = L ∪ E is reducible,consisting of the proper transform as well as the exceptional curve.

Example 6.21. Consider a plane nodal cubic C : y2 − x2(x + 1) = 0. Blow up

p = (0, 0) and consider the proper transform of C. In A2, let v/u = b, then π−1(C)is defined by y−bx = 0 and x(x+1−b2) = 0, i.e. it consists of the exceptional curve

E along with the proper transform C defined by uy−xv = 0 and (x+1)u2−v2 = 0.

Note that C meets E at two points q1 = (0, 0, [1, 1]) and q2 = (0, 0, [1,−1]), whereq1, q2 correspond to the tangent directions of the two branches of C at the node p.

Moreover, C is smooth at both q1 and q2. This example illustrates that blow up isa useful tool for resolving singularities.

Exercise 6.22. Consider a plane cuspidal curve C : y2 − x3 = 0. Blow up the

origin and write down explicitly the equations that define the proper transform C

of C. Moreover, find the intersection of C and the exceptional curve E.

Now let us study the intersection paring associated to the blown up surface.

Let X be a smooth, projective surface and p ∈ X a point. Denote by X theblow up of X at p and E the exceptional curve. We have seen that E ∼= P1

parameterizing tangent directions at p. Pulling back a line bundle induces a natural

map π∗ : Pic(X)→ Pic(X). On the other hand, if D is an effective curve on X, π∗D

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is effective on X, which induces a projection π∗ : Pic(X)→ Pic(X). In particular,π∗E = 0, which can also be seen from Proposition 6.23 (5) below.

Proposition 6.23. In the above setting, we have the following results:(1) E2 = −1;

(2) Pic(X) ∼= Pic(X)⊕ Z;(3) (π∗C). (π∗D) = C.D for any C,D ∈ Pic(X);(4) (π∗C). E = 0 for any C ∈ Pic(X);

(5) (π∗C). D = C. (π∗D) for any C ∈ Pic(X) and D ∈ Pic(X);(6) KX = π∗KX + E and K2

X= K2

X − 1.

Proof. For (1), since E ∼= P1 parameterizes all linear directions at p, the normalbundle NE/X is isomorphic to the tautological bundle OP1(−1). By the adjunction

formula, we have E2 = deg(NE/X) = −1.For (2), we have the exact sequence

Z→ Pic(X)→ Pic(X)→ 0,

where the far left term is generated by [E]. Since (nE)2 = −n2 6= 0, the far left mapis an injection. Moreover, since a line bundle or a divisor is uniquely determined

modulo any codimension two locus, we know Pic(X) ∼= Pic(X\p) ∼= Pic(X\E),hence π∗ splits the exact sequence.

For (3), we can write C ∼ C1 − C2 and D ∼ D1 − D2 such that Ci, Dj arevery ample, hence can be chosen to not contain p. Then the claim follows since

Ci. Dj = π∗(Ci). π∗(Dj) in X\p = X\E.

For (4), as in (3) we write C ∼ C1−C2 such that Ci does not meet p, then π∗Ciis disjoint with E.

For (5), the same idea applies.Finally for (6), suppose KX = π∗KX +nE for certain n ∈ Z. By adjunction, we

have

2gE − 2 = (KX + E). E

= (π∗KX). E − (n+ 1)

= −n− 1.

Since gE = 0, we read off n = 1 and the self-intersection of KX follows.

Remark 6.24. We may also verify (6) via a local calculation, as we did for proving

the Riemann-Hurwitz formula. Suppose in an open subset U of X we have u 6= 0 andlet t = v/u. Then y = tx and v = tu, hence we can use x, t as the local coordinatesin U . Suppose ω is a section of KX with a local expression ω = f(x, y)dx ∧ dy.Then π∗ω gives rise to a section of KX . Restricted to U it can be written as

fdx ∧ (tdx+ xdt) = (xf)dx ∧ dt.Then the associated divisor of π∗ω consists of the zero of x, i.e. the exceptionalcurve E as well as the associated divisor of the pullback of f , i.e. π∗KX .

Next, we define the multiplicity of a curve at a point. Suppose that a curve Cin a smooth surface is locally defined by f(x, y) such that at the origin p we havef = fr modulo (x, y)r+1, where fr is a homogenous polynomial of degree r in x, y.In other words, r is the largest integer such that f ∈ (x, y)r. Then we say that themultiplicity of C at p is r and denote it by µp(C) = r.

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Example 6.25. It is clear that µp(C) = 1 if and only if C is nonsingular at p. Fora plane nodal cubic C : y2 − x2(x+ 1) = 0 and p = (0, 0), we have µp(C) = 2. Fora plane cuspidal cubic C : y2 − x3 = 0, we also have µp(C) = 2. For three distinctlines in A2 meeting at the origin, say C : y3 − x3 = 0, we have µp(C) = 3. Ingeneral, µp(C) measures how singular C is at p.

Proposition 6.26. Let C be an effective curve in a smooth surface X with multi-

plicity µp(C) = r at a point p. Let π : X → X be the map associated to the blow

up of X at p, E the exceptional curve and C the proper transform of C. Then

π∗C = C + rE.

Proof. Write f = fr + g in local coordinates x, y in X such that g ∈ (x, y)r+1.

Consider the open subset U of X where u 6= 0 and let t = v/u. Then v = ut andy = xt, hence x, t give rise to a local coordinate system of U . Pulling back f , wehave

π∗f = fr(x, xt) + g(x, xt) = xr(fr(1, t) + xh(x, t)),

where h(x, t) = g(x, xt)/xr+1 is holomorphic and fr(1, t)+xh(x, t) does not containE in its vanishing locus. From this expression, we see that the associated divisor ofπ∗f , i.e. π∗C consists of two parts, one coming from the zeros of xr, i.e. r copiesof E and the other coming from the proper transform of C.

Corollary 6.27. In the above setting, we have C. E = r.

Proof. By Propositions 6.23 and 6.26, we have

C. E = (π∗C − rE). E

= 0− r · (−1)

= r.

Exercise 6.28. Let C,D be two effective curves on a surface meeting at a point p.

Blow up p and denote by C, D the proper transforms of C,D, respectively. Show

that C. D = C.D − µp(C) · µp(D).

Exercise 6.29. Consider the plane nodal cubic C : Y 2Z −X2(X + Z) = 0 in P2.

Blow up the origin to obtain a surface P2 and denote by C the proper transform of

C. Calculate C2 and KP2 . C. Then use the adjunction formula to verify that C is

isomorphic to P1, which is the normalization of C.

7. Schemes

Recall that a variety, roughly speaking, is the vanishing locus of a family ofpolynomials. There is certain ambiguity, say, x2 and x both cut out the line x = 0in A2, but x2 has vanishing order two along the line. In other words, x is a nilpotentin the ring of functions K[x, y]/(x2) on the “double line” x2 = 0. The language ofschemes allows us to enlarge the category of varieties such that we can deal with“vanishing locus with multiplicity” in a formal way.

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7.1. The spectrum of a ring. Let R be a commutative ring. Define the spectrumSpecR as the set of all (proper) prime ideals of R. We sometimes use [p] to denotea point in SpecR corresponding to a prime ideal p. Note that the zero ideal (0) isprime if and only if R is an integral domain.

Example 7.1. If R is Z, then SpecZ consists of ideals (p) where p = 0 or p is aprime number. If R = C[x]/x2, every element of type a + bx with a 6= 0 is a unit,hence the only prime ideal of R is (x).

Each element f ∈ R gives rise to a “function” in the following sense. For a pointx = [p] ∈ SpecR, write κ(x) or κ(p) to denote the quotient field of the integraldomain R/p, i.e. the field of fractions a/b for a, b ∈ R/p and b 6= 0. We call κ(x)the residue field at x. Then we can define f(x) ∈ κ(x) by the image of f under thecanonical map R→ R/p→ κ(x).

Example 7.2. Consider SpecZ. The “function” 10 at the point (7) and at thepoint (5) has value 3 and 0, by taking the remainder modulo 7 and 5, respectively.

Example 7.3. Consider the ring C[x] and let p(x) be a polynomial. For a ∈ C,(x− a) is a prime ideal (indeed maximal). It is easy to see that κ((x− a)) ∼= C andthe function p(x) at (x− a) ∈ SpecC[x] has value equal to p(a). In addition, (0) isthe only prime ideal not of type (x− a). Hence, SpecC[x] as a set compared to A1

is almost the same, except with one additional point corresponding to (0).

More generally, if R is the coordinate ring of regular functions on a variety Vdefined over an algebraically closed field K and if p is the maximal ideal of a pointx ∈ V (i.e. all functions vanishing at x), then κ(x) = K and f(x) is the value off ∈ R at x.

Example 7.4. Consider SpecC[x]/x2. It consists of a unique point (x) with residuefield C. Then 0 and x are both vanishing at (x), hence as functions they are thesame. In other words, f ∈ R may not be determined by its function values onSpecR.

7.2. Zariski topology revisit. We want to make X = SpecR a topological space.For every subset S ⊂ R, define

V (S) = [p] ∈ SpecR | p ⊃ S.It is easy to see that the definition is equivalent to

V (S) = x ∈ X | f(x) = 0 for all f ∈ S.Think of f as a function and x as a point. Then it is natural to use such V (S)as closed sets for the space SpecR and V (S) is just the vanishing locus of all“functions” in S. The resulting topology is called the Zariski topology. Since⋂α V (Sα) = V (

⋃α Sα), the Zariski topology is well-defined. Moreover, if I is the

ideal generated by S, then V (I) = V (S).The open sets of SpecR are given by the complements of V (S). In particular, for

every f ∈ R we have a distinguished open set Xf = X − V (f) = |SpecRf |, whereRf is the localization of R by adjoining 1/f and the last equality is based on thecorrespondence [p] 7→ [pRf ] for a prime ideal p not containing f . The distinguishedopen sets provide a base for the Zariski topology:

SpecR− V (S) = SpecR−⋂f∈S

V (f) =⋃f∈S

(SpecR)f .

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Remark 7.5. The distinguished open set makes sense even if f is nilpotent, i.e.the multiplicative set fn contains 0. In that case, by the definition of localizationwe have Rf = 0 and SpecRf = ∅. On the other hand, since f is nilpotent, everyprime ideal in R contains f , hence Xf = SpecR− V (f) = ∅.

Proposition 7.6. The closure of a point [p] consists of all [q] such that q ⊃ p. Inparticular, [p] is a closed point if and only if p is maximal.

Proof. The smallest closed set containing [p] is V (p) by definition. Therefore,V (p) = [p] if and only if no other prime ideal contains p, namely, if and only ifp is maximal, since every proper ideal of R is contained in a maximal ideal.

If X is an affine variety defined over an algebraically closed field, then the closedpoints of X correspond precisely to maximal ideals of its ring of functions. Theclosed points in the closure of the point [p] are precisely the points of X in thesubvariety determined by p.

Remark 7.7. Usually SpecR is not a Hausdorff space. For instance, if R is an in-tegral domain, then (0) is a prime ideal and is contained in every prime ideal, henceits closure V (0) = SpecR is the total space. Due to this reason [(0)] is called thegeneric point. In general, every irreducible component of an affine scheme possessesa generic point corresponding to the minimal prime defining the component.

Example 7.8. Consider R = C[x, y]. We want to compare SpecR with the affineplane A2. For a, b ∈ C, the ideal (x − a, y − b) is maximal and corresponds to theclosed point (a, b) in A2. Let f ∈ R be an irreducible polynomial. Then (x−a, y−b)contains (f) if and only if f(a, b) = 0. Geometrically it says that the closed pointsin the closure of [(f)] form the curve in A2 defined by the zero locus of f .

7.3. Structure sheaves. The last ingredient to define a scheme is an additionalstructure on the topological space X = SpecR, analogous to differential structureson smooth manifolds or complex structures on complex manifolds. It is called thestructure sheaf or sheaf of regular functions on X and denoted by OX . Let f ∈ R bea non-zero element. We first define on the distinguished open set Xf = |SpecRf |that

OX(Xf ) = Rf .

If Xg ⊂ Xf , then V (f) ⊂ V (g), hence g ∈ rad(f), i.e. certain power gn is a multipleof f . Then the restriction map Rf → Rg can be defined via the localization mapRf → Rfg = Rg. Modulo some detail in commutative algebra, the sections on thedistinguished open sets Xf over all f ∈ R uniquely determine OX as a sheaf. Inparticular, take f = 1 and then Xf = X, hence OX(X) = R, i.e. elements of R are“regular functions” on X as we expect. Similarly, sections of OX(U) for an open(dense) subset U are “rational functions” on X.

Now we are ready to define schemes in general. A scheme is simply a topologicalspace X together with a sheaf OX such that (X,OX) is locally affine, i.e. X can becovered by open sets Uα such that there exists a ring Rα satisfying Uα ∼= |SpecR|and OX |Uα ∼= OSpecRα . If there is no confusion, we simply use X to denote thescheme.

Let us mention some important notions and leave to the reader to study theirproperties in detail. We deal with affine schemes only, but an appropriate gluinggeneralizes to arbitrary schemes.

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A closed subscheme of an affine scheme X = SpecR is the spectrum of a quotientring of R. Therefore, closed subschemes of X correspond one to one with the idealsin R.

Let SpecR/I and SpecR/J be two closed subschemes. Define their union bySpecR/(I ∩ J) and their intersection by SpecR/(I + J).

Let Rred be the quotient of R modulo the ideal of nilpotent elements. DefineXred = SpecRred to be the reduced scheme associated to X. Note that as topologicalspaces |X| and |Xred| are identical. If X = Xred as schemes, we say that X isreduced.

If |X| is not a union of two properly contained closed sets, then X is calledirreducible.

Example 7.9. The double line corresponding to the ideal (x2) is a non-reducedclosed subscheme of the affine plane A2 = SpecC[x, y]. The cuspidal cubic Ccorresponding to the ideal (y2−x3) is a reduced closed subscheme of A2. Moreover,C is irreducible. On the other hand, consider a closed subscheme D correspondingto the ideal (y2−x2). Then (y2−x2) = (y−x)∩ (y+x), hence D is reducible andgeometrically it is the union of two lines defined by y − x = 0 and y + x = 0.

7.4. Local rings. Let x ∈ X be a point on a scheme X. Define the local ring ofX at x by

OX,x := lim−→x∈U

OX(U),

i.e. the germ of the structure sheaf at x.Since the definition is local, we can assume that X = SpecR is affine and x = [p]

corresponds to a prime ideal p in R. Next we restrict to the distinguished open setsSpecRf = X − V (f) containing x, i.e. f(x) 6= 0 or equivalently f /∈ p. Thus weobtain that

OX,x := lim−→f /∈p

Rf = Rp

as well as its (unique) maximal ideal

mX,x := pRp.

The dimension of X at x, denoted by dim(X,x) is the dimension of OX,x, whichis defined as the supremum of lengths of chains (i.e. the number of strict inclusions)of prime ideals in OX,x. The dimension of X is defined by the supremum of theselocal dimensions.

The Zariski cotangent space to X at x is mX,x/m2X,x, regarded as a vector space

over the residue field κ(x) = OX,x/mX,x. The dual of this vector space is calledthe Zariski tangent space at x. Let us explain this definition when X is a complexalgebraic variety. In that case the tangent space at x is just the vector spaceof derivations from the ring of germs of analytic functions at x to C, mX,x isthe ideal of germs of regular functions vanishing at x and C is the residue field.Therefore, the tangent space at x (in the traditional sense) can be identified withHomC(mX,x/m

2X,x,C) = (mX,x/m

2X,x)∗.

If the dimension of the Zariski tangent space is the same as dim(X,x), we saythat X is nonsingular at x. Otherwise we say that X is singular at x.

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Example 7.10. Consider R = C[x1, . . . , xn]. At the origin x = [(x1, . . . , xn)] ∈An = SpecR, we have

OAn,x = R(x1,...,xn) = f/g | g(0, . . . , 0) 6= 0,mAn,x = f/g | f(0, . . . , 0) = 0 and g(0, . . . , 0) 6= 0.

Then mAn,x/m2An,x is a n-dimensional vector space over C with a basis given by

x1, . . . , xn.

Example 7.11. Consider R = C[x]/x2 and let X = SpecR. At x = [(x)] theunique prime ideal in R, we have

OX,x = R(x) = R,

mX,x = (x),

mX,x/m2X,x = (x).

Thus we have dim(X,x) = 0, since there is no strict inclusion of two prime ideals inR. On the other hand, the Zariski tangent space is one-dimensional at x. Hence weconclude that X is a zero-dimensional singular scheme. Indeed, X can be regardedas a “double point” with support Xred = SpecC = [(0)].

Example 7.12. Consider R = C[x, y]/(y2− x3), C = SpecR and let p = [(x, y)] ∈C. Geometrically p is the cusp on the cuspidal cubic C in the affine plane. Wehave

OC,p = f/g | g(0, 0) 6= 0,mC,p = f/g | f(0, 0) = 0 and g(0, 0) 6= 0.

Then mC,p/m2C,p = (x, y)/(x, y)2 as a vector space over C is generated by x and y,

hence has dimension two. On the other hand, note that (x) and (y) are not primeideals in R, because y2 = x3, and any other function not in (x, y) is a unit in OC,p,hence dim(C, p) = 1 by the inclusion 0 ⊂ (x, y). As a result C is singular at p.

7.5. Morphisms. Morally speaking, a morphism from one scheme to another islocally a homomorphism of their local rings. The rigorous definition is a bit techni-cal, so we simplify the situation by considering morphisms between affine schemesand leave to the reader to “patch them together”.

Let X = SpecS and Y = SpecR be two affine schemes. Then a morphismψ : X → Y is the same as a morphism φ : R→ S via the relation ψ([p]) = [φ−1(p)].Therefore, we have the following correspondence.

Corollary 7.13. The category of affine schemes is equivalent to the category ofcommutative rings, with reversed arrows.

Example 7.14. Consider the morphism ψ : C = SpecC[x, y]/(x − y2) → D =SpecC[x] induced by φ : C[x] → C[x, y]/(x − y2) such that φ(x) = x. Thenψ([(x− a2, y − a)]) = [φ−1(x− a2, y − a)] = [(x− a2)]. But ψ([(x− a2, y + a)]) hasthe same image. Geometrically this is the vertical projection C → D by forgettingthe y-coordinate, which is a double cover between two affine lines. For a 6= 0, theclosed point (x− a2) in D has two inverse images (x− a2, y− a) and (x− a2, y+ a)in C. At the origin (x) in D, we see a branch point, since it has only one inverseimage given by (x, y) in C.