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The Stacks project

Lemma 43.16.2. Let A be a Noetherian local ring. Let I = (f_1, \ldots , f_ r) be an ideal generated by a regular sequence. Let M be a finite A-module. Assume that \dim (\text{Supp}(M/IM)) = 0. Then

e_ I(M, r) = \sum (-1)^ i\text{length}_ A(\text{Tor}_ i^ A(A/I, M))

Here e_ I(M, r) is as in Remark 43.15.6.

Proof. Since f_1, \ldots , f_ r is a regular sequence the Koszul complex K_\bullet (f_1, \ldots , f_ r) is a resolution of A/I over A, see More on Algebra, Lemma 15.30.7. Thus the right hand side is equal to

\sum (-1)^ i\text{length}_ A H_ i(K_\bullet (f_1, \ldots , f_ r) \otimes _ A M)

Now the result follows immediately from Theorem 43.15.5 if I is an ideal of definition. In general, we replace A by \overline{A} = A/\text{Ann}(M) and f_1, \ldots , f_ r by \overline{f}_1, \ldots , \overline{f}_ r which is allowed because

K_\bullet (f_1, \ldots , f_ r) \otimes _ A M = K_\bullet (\overline{f}_1, \ldots , \overline{f}_ r) \otimes _{\overline{A}} M

Since e_ I(M, r) = e_{\overline{I}}(M, r) where \overline{I} = (\overline{f}_1, \ldots , \overline{f}_ r) \subset \overline{A} is an ideal of definition the result follows from Theorem 43.15.5 in this case as well. \square


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