Homotopy-theory techniques in commutative algebrassatherw/DOCS/CPHColloq2007print… · The...

Free ResolutionsThe Koszul Complex

An ApplicationSummary

Homotopy-theory techniques in commutativealgebra

Sean Sather-Wagstaff

Department of Mathematical SciencesKent State University

09 January 2007Departmental Colloquium

Joint with Lars W. ChristensenarXiv: math.AC/0612301

Sean Sather-Wagstaff Homotopy-theory techniques in commutative algebra



Outline

1 Free ResolutionsRings and ModulesExamplesResolutions of Modules

2 The Koszul ComplexConstruction of the Koszul ComplexDifferential Graded Algebra Structure on the KoszulComplex

3 An ApplicationSemidualizing ComplexesDescent of Semidualizing Complexes




Rings and ModulesExamplesResolutions of Modules

Set-Up for the Talk

Throughout this talk, let R be a commutative ring with identity.Examples include:

The rings of integers Z and p-adic integers Zp

A field k like Q, R, C, or Z/pZPolynomial rings A[x1, . . . , xn] with coefficients in acommutative ring A with identityRings of formal power series A[[x1, . . . , xn]]

Quotient rings A[x1, . . . , xn]/I and A[[x1, . . . , xn]]/JR-modules are objects that can be “acted upon” by the ring R.Modules are like vector spaces, but more interesting.





Examples of Modules

If k is a field, then M is a k -module if and only if it is ak -vector space.M is a Z-module if and only if it is an abelian group.If I is an ideal of R, then I is an R-module and so is thequotient R/I.

In a sense, modules unify the notions of “vector space,”“abelian group,” “ideal,” and “quotient by an ideal.”

Consider the polynomial ring R = C[x1, . . . , xn].If H is a Hilbert space and T1, . . . ,Tn are pairwisecommuting operators on H, then H is an R-module

f ξ = f (T1, . . . ,Tn)ξ.





“Modules are like vector spaces, but more interesting.”Justification of “more interesting”

An R-module is free if it has a basis. If F is free and has a finitebasis e1, . . . ,er then F ∼= Rr and r is the rank of F .

Most R-modules are not free.

Example. If n ≥ 2, then the Z-module Z/nZ does not have abasis since nb = 0 for all b ∈ Z/nZ: linear independence fails.

More generally, if I is an ideal of R and (0) ( I ( R, then theR-module R/I does not have a basis.

In fact, R is a field if and only if every R-module has a basis.

This is a good thing! The complexity of the modules shouldmirror the complexity of the ring.





Finitely Generated Modules over a PID

Let R be a PID and M be a finitely generated R-module.M is finitely generated, so there is a surjection τ0 : Rr � M.Ker(τ0) is a submodule of the free module Rr .Since R is a PID, there is an isomorphism Ker(τ0)

∼= Rs.Hence, there is an exact sequence

0→ Rs ι−→ Rr τ0−→ M → 0

meaning that the kernel of each map equals the image ofthe preceding map:ι is injective, τ0 is surjective, and Ker(τ0) = Im(ι).The map ι : Rs → Rt is given by a matrix A. Linear algebra!





Resolutions: Use Linear Algebra to Study Modules

R is Noetherian and M is a finitely generated R-module.For example, R = A[x1, . . . , xn]/I or R = A[[x1, . . . , xn]]/J whereA = k or A = Z or A = Zp

There exists a surjection τ0 : Rr0 � M.R is Noetherian, so Ker(τ0) is finitely generated.If Ker(τ0) is free, then stop.If Ker(τ0) is not free, then repeat.There exists a surjection τ1 : Rr1 � Ker(τ0).

The composition Rr1τ1→ Ker(τ0)→ Rr0 is given by a matrix.

Repeating the process yields a free resolution of M

· · · A3−→ Rr2A2−→ Rr1

A1−→ Rr0τ0−→ M → 0

which is an exact sequence that may or may not terminate.Sean Sather-Wagstaff Homotopy-theory techniques in commutative algebra




A Computation of a Resolution

Example. R = A[x , y ] and I = (x , y)R and M = R/I.τ0 : R → R/I is the canonical surjection, and Ker(τ0) = I.A surjection τ1 : R2 → I is given by

τ1([

fg])

= f · x + g · y .

The relation yx − xy = 0 implies

τ1([ y−x])

= yx − xy = 0 =⇒[ y−x]∈ Ker(τ1).

For all f ∈ R we have f[ y−x]∈ Ker(τ1), and furthermore

Ker(τ1) = R[ y−x] ∼= R.

So Ker(τ1) is free! This gives a free resolution

0→ R

h y−x

i−−−→ R2 [ x y ]−−−→ R

τ0−→ R/I → 0





Example: Resolutions of R/I

R = A[x1, . . . , xn] or R = A[[x1, . . . , xn]] and I = (x1, . . . , xn)R.n = 1:

0→ R1 [ x1 ]−−→ R1 τ0−→ R/I → 0

n = 2:

0→ R1

h x2−x1

i−−−−→ R2 [ x1 x2 ]−−−−→ R1 τ0−→ R/I → 0

n = 3:

0→ R1

» x3−x2x1

–−−−−→ R3

" x2 x3 0−x1 0 x3

0 −x1 −x2

#−−−−−−−−−−→ R3 [ x1 x2 x3 ]−−−−−−→ R1 τ0−→ R/I → 0





Example: Resolutions of R/I

R = A[x1, . . . , xn] or R = A[[x1, . . . , xn]] and I = (x1, . . . , xn)R.n = 1:

0→ R1 [ x1 ]−−→ R1 τ0−→ R/I → 0

n = 2:

0→ R1

h x2−x1

i−−−−→ R2 [ x1 x2 ]−−−−→ R1 τ0−→ R/I → 0

n = 3:

0→ R1

» x3−x2x1

–−−−−→ R3

" x2 x3 0−x1 0 x3

0 −x1 −x2

#−−−−−−−−−−→ R3 [ x1 x2 x3 ]−−−−−−→ R1 τ0−→ R/I → 0

Binomial coefficients!




Construction of the Koszul ComplexDifferential Graded Algebra Structure on the Koszul Complex

Exterior PowersHow the Binomial Coefficients Arise

Fix an integer n ≥ 1 and consider the free module Rn.Let e1, . . . ,en ∈ Rn be the standard basis.For each integer d the d th exterior power of Rn is the freeR-module ∧dRn ∼= R(n

d) with basis

{ei1∧ei2∧· · ·∧eid | 1 ≤ i1 < i2 < · · · < id ≤ n}.

∧dRn d = 4 d = 3 d = 2 d = 1 d = 0 d = −1n = 1 0 R1 R1 0n = 2 0 R1 R2 R1 0n = 3 0 R1 R3 R3 R1 0basis {ei1∧ei2∧ei3} {ei1∧ei2} {ei} {1}

Let x1, . . . , xn ∈ R and set I = (x1, . . . , xn)R.Sean Sather-Wagstaff Homotopy-theory techniques in commutative algebra




The Augmented Koszul Complex

Define homomorphisms ∂Kd : ∧d Rn → ∧d−1Rn for d = 1, . . . ,n.

0 - R(nn)

∂Kn- R( n

n−1)∂K

n−1- · · ·∂K

2- R(n1)

∂K1- R(n

0)τ0- R/I - 0

deg n deg n − 1 deg 1 deg 0 deg −1

∂K1 (ei) = xi Rn → R

∂K2 (ei1∧ei2) = xi1ei2 − xi2ei1 R(n

2) → Rn

∂Kd (ei1∧ei2∧· · ·∧eid ) =

d∑j=1

(−1)j+1xij ei1∧ei2∧· · ·∧eij∧· · ·∧eid

∂Kd−1 ◦ ∂K

d = 0 for d = 2, . . . ,n so this is a chain complex.

Ker(τ0) = I = Im(∂K1 )

If R = A[x1, . . . , xn], this is the free resolution of R/I.Sean Sather-Wagstaff Homotopy-theory techniques in commutative algebra




The Wedge Product

The Koszul complex K R = K R(x1, . . . , xn) is the chain complex

0∂K

n+1−−−→ R(nn)

∂Kn−→ R( n

n−1)∂K

n−1−−−→ · · ·∂K

2−→ R(n1)

∂K1−→ R(n

0)∂K

0−→ 0

The wedge product provides a product on the Koszul complex.

(∧dRn)× (∧d ′Rn)→ ∧d+d ′Rn (v ,w) 7→ v∧w =: vw

This gives rise to elements of the form

(ei1∧· · ·∧eid )(ej1∧· · ·∧ejd′ ) = ei1∧· · ·∧eid ∧ ej1∧· · ·∧ejd′

which are not necessarily basis elements because thesubscripts may not be in strictly ascending order.Use the following two relations:

ei1∧· · ·∧eij∧eij+1∧· · ·∧eid+d′= −ei1∧· · ·∧eij+1∧eij∧· · ·∧eid+d′

ei1∧· · ·∧eij∧eij∧· · ·∧eid+d′= 0





Differential Graded AlgebrasThe Homotopy-Theoretic Methods

Basic properties. For v ∈ ∧dRn and w ∈ ∧d ′Rn

wv = (−1)dd ′vwv2 = 0 when d is oddLeibniz Rule: ∂K

d+d ′(vw) = ∂Kd (v)w − (−1)dv∂K

d ′(w)

This gives the Koszul complex the structure of a differentialgraded commutative algebra or DG algebra for short.

The ring R is a DG algebra concentrated in degree 0.The natural map R → K R is a DG algebra homomorphism.R 0 - R - 0

K R

ι?

0 - R(nn) - · · · - R(n

1)?

- R(n0)

∼=?

- 0?





Differential Graded Algebras (cont.)Compatibility with Ring Homomorphisms

A ring homomorphism ϕ : R → S induces a DG algebrahomomorphism K R(x1, . . . , xn)→ K S(ϕ(x1), . . . , ϕ(xn))

K R 0 - R(nn) - · · · - R(n

1) - R(n0) - 0

K S

Kϕ

?

0 - S(nn)

ϕ(nn)

?

- · · · - S(n1)

ϕ(n1)

?

- S(n0)

ϕ(n0)

?

- 0

The maps ι and Kϕ make the following diagram commute.

Rι- K R

S

ϕ? ι

- K S

Kϕ

?




Semidualizing ComplexesDescent of Semidualizing Complexes

The Homothety Homomorphism

Let C be a chain complex over R.The i th homology module of C is the R-moduleHi(C) = Ker(∂C

i )/ Im(∂Ci+1).

The endomorphism complex of C is denoted End(C).For each r ∈ R there is a commutative diagram

· · ·∂C

j+1- Cj∂C

j- Cj−1∂C

j−1- · · ·

· · ·∂C

j+1- Cj

r ·? ∂C

j- Cj−1

r ·? ∂C

j−1- · · ·

The homothety C r ·−→ C is in Ker(∂

End(C)0

)⊆ End(C)0.

The map R → H0(End(C)) given by r 7→ (C r ·−→ C) is anR-module homomorphism.





Semidualizing Complexes

A chain complex C over R is semidualizing ifeach Ci is a free R-module of finite rank,Ci = 0 for each i < 0,Hi(C) = 0 for i � 0,R ∼= H0(End(C)), andHi(End(C)) = 0 for each i 6= 0.

Example. R is semidualizing.

Example. A dualizing complex is semidualizing.

Semidualizing complexes arise in the study of the homologicalalgebra of ring homomorphisms, e.g., in the compositionquestion for local ring homomorphisms of finite G-dimension.

Much of my recent research has been devoted to the analysisof the set of semidualizing complexes.





The Completion of a Local Ring

R is a local noetherian ring with maximal ideal m. For r , s ∈ R

ord(r) = sup{n ≥ 0 | r ∈ mn} dist(r , s) = 2− ord(r−s)

The function dist(−,−) is a metric on R. The topologicalcompletion of R is denoted R.

R is a noetherian local ring equipped with a canonical ringhomomorphism ϕR : R → R and maximal ideal ϕR(m)R.

Example. If R = k [x1, . . . , xn](x1,...,xn)/(f1, . . . , fm), thenR ∼= k [[x1, . . . , xn]]/(f1, . . . , fm).

If m = (x1, . . . , xn)R, then the induced map on Koszulcomplexes K R → K bR is a homology isomorphism.





Ascent of Semidualizing Complexes

A semidualizing complex over R has the following shape.

C = · · ·∂C

3−→ Rβ2∂C

2−→ Rβ1∂C

1−→ Rβ0 → 0

The completion of C is a semidualizing complex over R.

C = C ⊗R R = · · ·∂C

3−→ Rβ2∂C

2−→ Rβ1∂C

1−→ Rβ0 → 0

Problem. Provide conditions on R such that every semi-dualizing complex over R is isomorphic to one of the form C.

R has the approximation property when, for every finite systemof polynomial equations S = {fi(X1, . . . ,XN) = 0}ti=1, if S has asolution in R, then it has a solution in R.





A Descent Theorem for Semidualizing Complexes

Theorem. (L.W.Christensen-SSW, 2006) If R has theapproximation property, then every semidualizing complex overR is isomorphic to C for some semidualizing complex C over R.

Sketch of Proof. Let B be a semidualizing complex over R.Let x1, . . . , xn ∈ R be a minimal generating set for m.Set K R = K R(x1, . . . , xn) and K bR = K bR(ϕR(x1), . . . , ϕR(xn)).The map ϕR : R → R provides a commutative diagram

C - C ∼= B

R - R

K R?

- KbR? C ⊗bR K

bR?

∼=A?

- A⊗K R KbR ∼= B ⊗bR K

bR?

QEDSean Sather-Wagstaff Homotopy-theory techniques in commutative algebra



Summary

Free resolutions allow us to use linear algebra to studymodules that are not free.The DG algebra structure on the Koszul complex allows usto solve certain problems about commutative rings byleaving the realm of rings.

OutlookQuestion. Is the set of isomorphism classes ofsemidualizing complexes over R a finite set?The proof of a special case suggests that one needs tobuild a deformation theory for DG algebras.

Tak!




The Augmented Koszul Complex May Not Be Exact.

Example. Set R = k [X1,X2]/(X1X2) and let xi = Xi for i = 1,2.The relation x1x2 = 0 in R makes the augmented Koszulcomplex non-exact in degree 1.

0→ R1∂K

2 =h x2−x1

i−−−−−−−→ R2 ∂K

1 =[ x1 x2 ]−−−−−−−→ R1 τ0−→ R/I → 0

The vectors[ x2

0

]and

[0x1

]are in Ker(∂K

1 ) because

[ x1 x2 ][ x2

0

]=[

00

]and [ x1 x2 ]

[0x1

]=[

00

]However,

[ x20

],[

0x1

]6∈ Im(∂K

2 ).

Again, this is good! The complexity of the Koszul complexshould mirror the complexity of the sequence x1, . . . , xn.




The Endomorphism Complex

Let C be a chain complex over R.The endomorphism complex of C is the chain complex End(C):

modules End(C)i =∏j∈Z

HomR(Cj ,Ci+j) 3 (ψj)j∈Z

· · ·∂C

j+1- Cj∂C

j - Cj−1∂C

j−1- · · ·

elements

· · ·∂C

j+i+1- Cj+i

ψj? ∂C

j+i- Cj+i−1

ψj−1? ∂C

j+i−1- · · ·

differentials ∂End(C)i :

∏j∈Z

HomR(Cj ,Ci+j)→∏j∈Z

HomR(Cj ,Ci+j−1)

∂End(C)i ((ψj)j∈Z) = (∂C

i+j ◦ ψj − (−1)iψj−1 ◦ ∂Cj )j∈Z


Homotopy-theory techniques in commutative algebrassatherw/DOCS/CPHColloq2007print… · The...

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