# Homotopy-theory techniques in commutative algebra

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1 Homotopy-theory techniques in commutative algebra Department of Mathematical Sciences Kent State University 09 January 2007 Departmental Colloquium Joint with Lars W. Christensen arxiv: math.ac/

2 Outline Free Resolutions 1 Free Resolutions Rings and Modules Examples Resolutions of Modules 2 Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex 3 Semidualizing Complexes Descent of Semidualizing Complexes

3 Set-Up for the Talk Free Resolutions Rings and Modules Examples Resolutions of Modules Throughout this talk, let R be a commutative ring with identity. Examples include: The rings of integers Z and p-adic integers Z p A field k like Q, R, C, or Z/pZ Polynomial rings A[x 1,..., x n ] with coefficients in a commutative ring A with identity Rings of formal power series A[[x 1,..., x n ]] Quotient rings A[x 1,..., x n ]/I and A[[x 1,..., x n ]]/J R-modules are objects that can be acted upon by the ring R. Modules are like vector spaces, but more interesting.

4 Examples of Modules Rings and Modules Examples Resolutions of Modules If k is a field, then M is a k-module if and only if it is a k-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 the quotient 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[x 1,..., x n ]. If H is a Hilbert space and T 1,..., T n are pairwise commuting operators on H, then H is an R-module f ξ = f (T 1,..., T n )ξ.

5 Rings and Modules Examples Resolutions of Modules 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 finite basis e 1,..., e r then F = R r 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 a basis 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 the R-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 should mirror the complexity of the ring.

6 Rings and Modules Examples Resolutions of Modules 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 τ 0 : R r M. Ker(τ 0 ) is a submodule of the free module R r. Since R is a PID, there is an isomorphism Ker(τ 0 ) = R s. Hence, there is an exact sequence 0 R s ι R r τ 0 M 0 meaning that the kernel of each map equals the image of the preceding map: ι is injective, τ 0 is surjective, and Ker(τ 0 ) = Im(ι). The map ι: R s R t is given by a matrix A. Linear algebra!

7 Rings and Modules Examples Resolutions of Modules Resolutions: Use Linear Algebra to Study Modules R is Noetherian and M is a finitely generated R-module. For example, R = A[x 1,..., x n ]/I or R = A[[x 1,..., x n ]]/J where A = k or A = Z or A = Z p There exists a surjection τ 0 : R r 0 M. R is Noetherian, so Ker(τ 0 ) is finitely generated. If Ker(τ 0 ) is free, then stop. If Ker(τ 0 ) is not free, then repeat. There exists a surjection τ 1 : R r 1 Ker(τ 0 ). The composition R r τ 1 1 Ker(τ 0 ) R r 0 is given by a matrix. Repeating the process yields a free resolution of M A3 R r 2 A 2 R r 1 A 1 R r τ 0 0 M 0 which is an exact sequence that may or may not terminate.

8 A Computation of a Resolution Rings and Modules Examples Resolutions of Modules Example. R = A[x, y] and I = (x, y)r and M = R/I. τ 0 : R R/I is the canonical surjection, and Ker(τ 0 ) = I. A surjection τ 1 : R 2 I is given by τ 1 ([ fg ]) = f x + g y. The relation yx xy = 0 implies ([ τ y ]) [ 1 x = yx xy = 0 = y ] x Ker(τ1 ). For all f R we have f [ y ] x Ker(τ1 ), and furthermore Ker(τ 1 ) = R [ y ] = x R. So Ker(τ 1 ) is free! This gives a free resolution 0 R h i y x R 2 [ x y ] R τ 0 R/I 0

9 Example: Resolutions of R/I Rings and Modules Examples Resolutions of Modules R = A[x 1,..., x n ] or R = A[[x 1,..., x n ]] and I = (x 1,..., x n )R. n = 1: 0 R 1 [ x 1 ] R 1 τ 0 R/I 0 n = 2: n = 3: 0 R 1» x3 x 2 h x2 x 1 i R 2 [ x 1 x 2 ] R 1 τ 0 R/I 0 " x2 x 3 0 x 1 0 x 3 0 R 1 x 1 R 3 0 x 1 x 2 R 3 [ x 1 x 2 x 3 ] R 1 τ 0 R/I 0 #

10 Example: Resolutions of R/I Rings and Modules Examples Resolutions of Modules R = A[x 1,..., x n ] or R = A[[x 1,..., x n ]] and I = (x 1,..., x n )R. n = 1: 0 R 1 [ x 1 ] R 1 τ 0 R/I 0 n = 2: n = 3: 0 R 1» x3 x 2 h x2 x 1 i R 2 [ x 1 x 2 ] R 1 τ 0 R/I 0 " x2 x 3 0 x 1 0 x 3 0 R 1 x 1 R 3 0 x 1 x 2 R 3 [ x 1 x 2 x 3 ] R 1 τ 0 R/I 0 Binomial coefficients! #

11 Exterior Powers How the Binomial Coefficients Arise Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex Fix an integer n 1 and consider the free module R n. Let e 1,..., e n R n be the standard basis. For each integer d the dth exterior power of R n is the free R-module d R n = R (n d) with basis {e i1 e i2 e id 1 i 1 < i 2 < < i d n}. d R n d = 4 d = 3 d = 2 d = 1 d = 0 d = 1 n = 1 0 R 1 R 1 0 n = 2 0 R 1 R 2 R 1 0 n = 3 0 R 1 R 3 R 3 R 1 0 basis {e i1 e i2 e i3 } {e i1 e i2 } {e i } {1} Let x 1,..., x n R and set I = (x 1,..., x n )R.

12 The Augmented Koszul Complex Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex Define homomorphisms K d : d R n d 1 R n for d = 1,..., n. 0 R (n n) K n R ( n 1) n K n 1 2 K R (n 1) K 1 R (n 0) τ0 R/I 0 deg n deg n 1 deg 1 deg 0 deg 1 K 1 (e i) = x i R n R 2 K (e i 1 e i2 ) = x i1 e i2 x i2 e i1 R (n 2) R n d d K (e i 1 e i2 e id ) = ( 1) j+1 x ij e i1 e i2 ê ij e id j=1 d 1 K K d = 0 for d = 2,..., n so this is a chain complex. Ker(τ 0 ) = I = Im( 1 K ) If R = A[x 1,..., x n ], this is the free resolution of R/I.

13 The Wedge Product Free Resolutions Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex The Koszul complex K R = K R (x 1,..., x n ) is the chain complex 0 K n+1 R (n n) K n R ( n 1) n K n 1 K 2 R (n 1) K 1 R (n 0) K 0 0 The wedge product provides a product on the Koszul complex. ( d R n ) ( d R n ) d+d R n (v, w) v w =: vw This gives rise to elements of the form (e i1 e id )(e j1 e jd ) = e i1 e id e j1 e jd which are not necessarily basis elements because the subscripts may not be in strictly ascending order. Use the following two relations: e i1 e ij e ij+1 e id+d = e i1 e ij+1 e ij e id+d e i1 e ij e ij e id+d = 0

14 Differential Graded Algebras The Homotopy-Theoretic Methods Basic properties. For v d R n and w d R n wv = ( 1) dd vw v 2 = 0 when d is odd Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex Leibniz Rule: K d+d (vw) = K d (v)w ( 1)d v K d (w) This gives the Koszul complex the structure of a differential graded 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 ι K R 0 R (n n) R (n 1) = R (n 0) 0

15 Differential Graded Algebras (cont.) Compatibility with Ring Homomorphisms Construction of the Koszul Complex Differential Graded Algebra Structure on the Koszul Complex A ring homomorphism ϕ: R S induces a DG algebra homomorphism K R (x 1,..., x n ) K S (ϕ(x 1 ),..., ϕ(x n )) K R 0 R (n n) R (n 1) R (n 0) 0 K ϕ K S ϕ (n n) ϕ (n 1) ϕ (n 0) 0 S (n n) S (n 1) S (n 0) 0 The maps ι and K ϕ make the following diagram commute. R ϕ S ι K R K ϕ ι K S

16 The Homothety Homomorphism Semidualizing Complexes Descent of Semidualizing Complexes Let C be a chain complex over R. The ith homology module of C is the R-module H i (C) = Ker( C i )/ Im( C i+1 ). The endomorphism complex of C is denoted End(C). For each r R there is a commutative diagram C j+1 C j C j C j 1 C j 1 r r j+1 C C j C j 1 C j C j 1 ( ) End(C) 0. The homothety C r C is in Ker End(C) 0 The map R H 0 (End(C)) given by r (C r C) is an R-module homomorphism.

17 Semidualizing Complexes Semidualizing Complexes Descent of Semidualizing Complexes A chain complex C over R is semidualizing if each C i is a free R-module of finite rank, C i = 0 for each i < 0, H i (C) = 0 for i 0, R = H 0 (End(C)), and H i (End(C)) = 0 for each i 0. Example. R is semidualizing. Example. A dualizing complex is semidualizing. Semidualizing complexes arise in the study of the homological algebra of ring homomorphisms, e.g., in the composition question for local ring homomorphisms of finite G-dimension. Much of my recent research has been devoted to the analysis of the set of semidualizing complexes.

18 The Completion of a Local Ring Semidualizing Complexes Descent of Semidualizing Complexes R is a local noetherian ring with maximal ideal m. For r, s R ord(r) = sup{n 0 r m n } dist(r, s) = 2 ord(r s) The function dist(, ) is a metric on R. The topological completion of R is denoted R. R is a noetherian local ring equipped with a canonical ring homomorphism ϕ R : R R and maximal ideal ϕ R (m) R. Example. If R = k[x 1,..., x n ] (x1,...,x n)/(f 1,..., f m ), then R = k[[x 1,..., x n ]]/(f 1,..., f m ). If m = (x 1,..., x n )R, then the induced map on Koszul complexes K R K br is a homology isomorphism.

19 Semidualizing Complexes Descent of Semidualizing Complexes Ascent of Semidualizing Complexes A semidualizing complex over R has the following shape. C = C 3 R β 2 C 2 R β 1 C 1 R β 0 0 The completion of C is a semidualizing complex over R. Ĉ = C R R C = 3 R β 2 C 2 R β 1 C 1 R β 0 0 Problem. Provide conditions on R such that every semidualizing complex over R is isomorphic to one of the form Ĉ. R has the approximation property when, for every finite system of polynomial equations S = {f i (X 1,..., X N ) = 0} t i=1, if S has a solution in R, then it has a solution in R.

20 Semidualizing Complexes Descent of Semidualizing Complexes A Descent Theorem for Semidualizing Complexes Theorem. (L.W.Christensen-SSW, 2006) If R has the approximation property, then every semidualizing complex over R is isomorphic to Ĉ for some semidualizing complex C over R. Sketch of Proof. Let B be a semidualizing complex over R. Let x 1,..., x n R be a minimal generating set for m. Set K R = K R (x 1,..., x n ) and K br = K br (ϕ R (x 1 ),..., ϕ R (x n )). The map ϕ R : R R provides a commutative diagram C Ĉ = B R K R A R R K b R Ĉ br K b = R A K R K b = B R br K b QED

21 Free Resolutions Tak! Free resolutions allow us to use linear algebra to study modules that are not free. The DG algebra structure on the Koszul complex allows us to solve certain problems about commutative rings by leaving the realm of rings. Outlook Question. Is the set of isomorphism classes of semidualizing complexes over R a finite set? The proof of a special case suggests that one needs to build a deformation theory for DG algebras.

22 The Augmented Koszul Complex May Not Be Exact. Example. Set R = k[x 1, X 2 ]/(X 1 X 2 ) and let x i = X i for i = 1, 2. The relation x 1 x 2 = 0 in R makes the augmented Koszul complex non-exact in degree 1. h i 0 R 1 K 2 = x2 x 1 R 2 K 1 =[ x 1 x 2 ] R 1 τ 0 R/I 0 The vectors [ x 2 ] [ 0 and 0x1 ] are in Ker( K 1 ) because [ x 1 x 2 ] [ x 2 ] [ 0 = 00 ] and [ x 1 x 2 ] [ ] [ 0 x1 = 00 ] However, [ x 2 0 ], [ 0x1 ] Im( K 2 ). Again, this is good! The complexity of the Koszul complex should mirror the complexity of the sequence x 1,..., x n.

23 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 Hom R (C j, C i+j ) (ψ j ) j Z C j+1 C j C j C j 1 C j 1 elements differentials C j+i+1 C j+i ψ j ψ j 1 C j+i C j+i 1 C j+i 1 End(C) i : Hom R (C j, C i+j ) Hom R (C j, C i+j 1 ) j Z j Z End(C) i ((ψ j ) j Z ) = ( i+j C ψ j ( 1) i ψ j 1 j C ) j Z

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