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Math 711: Lecture of October 25, 2006
Proposition. Let M be an R-module. Let
(a) If M has a finite filtration with factors Nj , 1 ≤ j ≤ s, and x is a nonzerodivisor on every Nj , then M/xM has a filtration with s factors Nj /xNj , and M/xnM has a filtration with ns factors: there are n copies of every Nj /xNj , 1 ≤ j ≤ s. (b) If x 1 ,... , xd is a regular sequence on M and n 1 ,... , nd are nonnegative integers, then M/(xn 1 1 ,... , xn d d)M has a filtration by n 1 · · · nd copies of M/(x 1 ,... , xd)M.
Proof. (a) By induction on the number of factors, this comes down to the case where there are two factors. That is, one has 0 → N 1 → M → N 2 → 0. This has an isomorphic subcomplex 0 → xN 1 → xM → xN 2 → 0, and the desired statement now follows from the exactness of the quotient complex. It follows as well that M/xnM has a filtration by the modules Nj /xnNj , and each of these has a filtration with n factors, xkNj /xk+1Nj ∼= Nj /xNj , 0 ≤ k ≤ n − 1.
For part (b) we use induction on d. The case d = 1 has already been handled in part (a). For the inductive step, we know that M/(xn 1 1 ,... , x nd− 1 d− 1 )M^ has a filtration by^ n^1 · · ·^ nd−^1 copies of M/(x 1 ,... , xd− 1 )M , and x = xd is a nonzerodivisor on each of these. The result now follows from the last statement in part (a), with n = nd.
We next observe:
Lemma. Let R be a Noetherian ring and let M be a finitely generated R-module of di- mension d > 0.
(a) M contains a maximum submodule N such that dim (N ) < d, and M/N has pure dimension d, i.e., for every P ∈ Ass (M/N ), dim (R/P ) = d. (b) Let W be a multiplicative system of R consisting of nonzerodivisors and suppose that M and M ′^ are R-modules such that W −^1 M ∼= W −^1 M ′. Then there exist exact sequences 0 → M ′^ → M → C 1 → 0 and 0 → M → M ′^ → C 2 → 0 such that each of C 1 and C 2 is killed by a single element of W.
(c) Let (R, m, K) be a complete local ring of dimension d, and let M be a finitely generated faithful R-module of pure dimension d. Let x 1 ,... , xd be a system of parameters for R. If R contains a field there is a coefficient field K ⊆ R for R, and M is a torsion- free module over A = K[[x 1 ,... , xd]], so that for some integer ρ > 0 , M and Aρ become isomorphic when we localize at W = A − {{ 0 }.
In mixed characteristic, there exists Cohen-Macaulay ring A ⊆ R containing x 1 ,... , xd as a system of parameters such that A has the form B/(f ) where B is regular and f 6 = 0. Moreover, if W is the multiplicative system of nonzerodivisors in A then W consists of nonzero divisors of on M and W −^1 M is a finite direct sum of modules of 1
the form W −^1 B/gj B where each gj is a divisor of f. In particular, M ′^ =
j B/gj^ B^ is a Cohen-Macaulay module over A of pure dimension d such that W −^1 M and W −^1 M ′ are isomorphic as A-modules.
Proof. To prove (a), first note that since M has ACC on submodules, it has a maximal submodule N of dimension less than d: it may be 0. If N ′^ is another submodule of M of dimension < d, then d > dim (N ⊕ N ′) ≥ dim (N + N ′), and so N + N ′^ ⊆ M contradicts the maximality of N. Thus, N contains every submodule of M of dimension < d. If M/N had any nonzero submodule of dimension less than d, its inverse image in M would be strictly larger than N and of dimension less than d as well.
(b) Since M ⊆ W −^1 M ∼= W −^1 M ′, we have an injection M ↪→ W −^1 M ′. Let u 1 ,... , uh be generators of M. Suppose that ui maps to vi/wi, 1 ≤ i ≤ h, where vi ∈ M ′^ and wi ∈ W. Let w = w 1 · · · wh. Then M ∼= wM ↪→
i Rvi^ ⊆^ M^
′. The map W − (^1) M → W −M ′ (^) that
this induces is still an isomorphsim, since w is a unit in W −^1 R. It follows that the cokernel C 1 of the map M → M ′^ that we constructed is such that W −^1 C 1 = 0. Since C 1 is finitely generated, there is a single element of W that kills C 1. An entirely similar argument yields 0 → M ′^ → M → C 2 such that C 2 is killed by an element of W.
(c) Let u 1 ,... , uh generate M. Then the map R → M ⊕h^ sending r 7 → (ru 1 ,... , ruh) is injective. It follows that Ass (R) ⊆ Ass (M ⊕h) = Ass (M ), so that R is also of pure dimension d. Choose a field or discrete valuation ring V that maps onto a coefficient ring for R (so that the residue class field of V maps isomorphically to the residue class field of R), and let X 1 ,... , Xd be formal indeterminates over V. Then the map V → R extends uniquely to a continuous map B = V [[X 1 ,... , Xd]] → R such that Xi 7 → xi, 1 ≤ i ≤ d. Let MB be the maximal ideal of B. Since the map B → R induces an isomorphism of residue class fields, and since R/mB R has finite length over B (the xi generate an m- primary ideal of R), R is module-finite over the image A of B in R. Moreover, we must have dim (A) = dim (R).
In the equal characteristic case, where V = K is a field, we must have B ∼= A = K[[x 1 ,... , xd]]. Moreover, M must be torision-free over A, since a nonzero torsion sub- module would have dimension smaller than d. Hence M and Aρ, where ρ is the torsion-free rank of M over A, become isomorphic when we localize at A − { 0 }.
We suppose henceforth that we are in the mixed characteristic case. We know that the ring A has pure dimension d. It follows that A = B/J, where J is an ideal all of whose associated primes in B have height one. Since B is regular, it is a UFD. Height one primes are principal, and any ideal primary to a height one prime has the form gk, where g generates the prime and k is a nonnegative integer. It follows that A = B/f B, where f = f 1 k 1 · · · f (^) hk kis the factorization of f into prime elements. Let W be the multiplicative system consisting of the complement of the union of the fj B. The associated primes of A are the Pj = fj A, and these are also the associated primes of M. Then W −^1 A is an Artin ring and is the product of the local rings AP j : each of these may be thought of as
obtained by killing f (^) jk j in the DVR obtained by localizing B at the prime fj B. M , as a B-module, is then a product of modules over the various APj , each of which is a direct sum
two elements of odd degree in reverse order reverses the sign on the product). An R-linear map d of Λ into itself that lowers degrees of homogeneous elements by one and satisfies
(#) d(uv) = (du)v + (−1)deg(u)u dv
when u is a form is called an R-derivation of Λ.
Then
(G) is an N -graded skew-commutative R-algebra, and it is easy to verify that the differential is an R-derivation. By the R-bilinearity of both sides in u and v, it suffices to verify (#) when u = uj 1 ∧ · · · ∧ ujh and v = uk 1 ∧ · · · ∧ uki with j 1 < · · · < jh and k 1 < · · · < ki. It is easy to see that this reduces to the assertion (∗∗) that the formula (∗) above is correct even when the sequence j 1 ,... , ji of integers in { 1 , 2 ,... , n} is allowed to contain repetitions and is not necessarily in ascending order: one then applies (∗∗) to j 1 ,... , jh, k 1 ,... , ki. To prove (∗∗), note that if we switch two consecutive terms in the sequence j 1 ,... , ji every term on both sides of (∗) changes sign. If the j 1 ,... , ji are mutually distinct this reduces the proof to the case where the elements are in the correct order, which we know from the definition of the differential. If the elements are not all distinct, we may reduce to the case where jt = jt+1 for some t. But then uj 1 ∧ · · · ∧ uji = 0, while all but two terms in the sum on the right contain ujt ∧ ujt+1 = 0, and the remaining two terms have opposite sign.
Once we know that d is a derivation, we obtain by a straightforward induction on k that if v 1 ,... , vk are forms of degrees a 1 ,... , ak, then
(∗ ∗ ∗) d(v 1 ∧ · · · ∧ vi) =
t=i
(−1)a^1 +···+at−^1 vj 1 ∧ · · · ∧ vjt− 1 ∧ dvjt ∧ vjt+1 ∧ · · · ∧ vji.
Note that the formula (∗) is a special case in which all the given forms have degree 1.
It follows that the differential on the Koszul complex is uniquely determined by what it does in degree 1, that is, by the map G → R, where G is the free R-module K 1 (x; R), together with the fact that it is a derivation on
(G). Any map G → R extends uniquely to a derivation: we can choose a free basis u 1 ,... , un for G, take the xi to be the values of the map on the ui, and then the differential on K•(x 1 ,... , xn; R) gives the extension we want. Uniqueness follows because the derivation property forces (∗ ∗ ∗) to hold, and hence forces (∗) to hold, thereby determining the values of the derivation on an R-free basis.
Thus, instead of thinking of the Koszul complex K(x 1 ,... , xn; R) as arising from a sequence of elements x 1 ,... , xn of R, we may think of it as arising from an R-linear map of a free module θ : G → R (we might have written d 1 for θ), and we write K•(θ; R) for the corresponding Koszul complex. The sequence of elements is hidden, but can be recovered by choosing a free basis for G, say u 1 ,... , un, and taking xi = θ(ui), 1 ≤ i ≤ n. The exterior algebra point of view makes it clear that the Koszul complex does not depend on the choice of the sequence of elements: only on the map of the free module G → R. Different choices of basis produce Koszul complexes that look different from the “sequence
of elements” point of view, but are obviously isomorphic. In particular, up to isomprphism, permuting the elements does not change the complex.
We write K•(x 1 ,... , xd; M ), where M is an R-module, for K•(x 1 ,... , xn; R) ⊗ M. The homology of this complex is denoted H•(x 1 ,... , xn; M ). Let x = x 1 ,... , xn and Let I = (x)R.
We have the following comments:
(1) The complex is finite: if M is not zero, it has length n. The i the term is the direct sum of
n i
copies of M. Both the complex and its homology are killed by AnnRM.
(2) The map from degree 1 to degree 0 is the map M n^ → M sending
(u 1 ,... , un) 7 → x 1 u 1 + · · · xnun.
The image of the map is IM , and so H 0 (x 1 ,... , xd; M ) ∼= M/IM. (3) The map from degree n to degree n − 1 is the map M → M n^ that sends
u 7 → (x 1 u 1 , −x 2 u 2 , · · · , ±xnun),
and so Hn(x 1 ,... , xn; M ) ∼= AnnM I.
(4) Given a short exact sequence of modules 0 → M ′^ → M → M ′′^ → 0 we may tensor with the free complex K•(x 1 ,... , xn; R) to obtain a short exact sequence of complexes
K•(x 1 ,... , xn; M ′) → K•(x 1 ,... , xn; M ) → K•(x 1 ,... , xn; M ′′) → 0.
The snake lemma then yields a long exact sequence of Koszul homology:
· · · → Hi(x; M ′) → Hi(x; M ) → Hi(x; M ′′) → Hi− 1 (x; M ′) → · · ·
→ H 1 (x; M ′) → H 1 (x; M ) → H 1 (x; M ′′) → M ′/IM ′^ → M/IM → M ′′/IM ′′^ → 0
(5) I kills Hi(x; M ) for every i and every R-module M. It suffices to see that xd kills the homology: the argument for xi is similar. Let z ∈ Ki(x; M ), and consider z ∧ un ∈ Ki+1(x; M ). Then (∗) d(z ∧ un) = dz ∧ un + (−1)ixnz.
Hence, if z is a cycle, d(z ∧ un) = (−1)ixnz, which shows that xnz is a boundary. (6) Let x−^ denote x 1 ,... , xn− 1. Let G−^ ⊆ G be the free module on the free basis u 1 ,... , un− 1. Then K•(x−; M ) may be identified with ∧• (G−) ⊗R M ⊆
(G) ⊗ M = K•(x; M ). This subcomplex is spanned by all terms that involve only u 1 ,... , un− 1. The quotient complex my be identified with K•(x−; M ) as well: one lets uj 1 ∧ · · · ∧ uji− 1 ∧ un ⊗ w
