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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