Theorem 51.11.6. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Set $Z = V(I) \subset \mathop{\mathrm{Spec}}(A)$. Let $M$ be a finite $A$-module. Set $s = s_{A, I}(M)$ as in (51.11.1.1). Assume that

$A$ is universally catenary,

the formal fibres of the local rings of $A$ are Cohen-Macaulay.

Then $H^ i_ Z(M)$ is finite for $0 \leq i < s$ and $H^ s_ Z(M)$ is not finite.

**Proof.**
By Lemma 51.7.2 we may assume that $A$ is a local ring.

If $A$ is a Noetherian complete local ring, then we can write $A$ as the quotient of a regular complete local ring $B$ by Cohen's structure theorem (Algebra, Theorem 10.160.8). Using Lemma 51.11.2 and Dualizing Complexes, Lemma 47.9.2 we reduce to the case of a regular local ring which is a consequence of Lemma 51.11.5 because a regular local ring is Gorenstein (Dualizing Complexes, Lemma 47.21.3).

Let $A$ be a Noetherian local ring. Let $\mathfrak m$ be the maximal ideal. We may assume $I \subset \mathfrak m$, otherwise the lemma is trivial. Let $A^\wedge $ be the completion of $A$, let $Z^\wedge = V(IA^\wedge )$, and let $M^\wedge = M \otimes _ A A^\wedge $ be the completion of $M$ (Algebra, Lemma 10.97.1). Then $H^ i_ Z(M) \otimes _ A A^\wedge = H^ i_{Z^\wedge }(M^\wedge )$ by Dualizing Complexes, Lemma 47.9.3 and flatness of $A \to A^\wedge $ (Algebra, Lemma 10.97.2). Hence it suffices to show that $H^ i_{Z^\wedge }(M^\wedge )$ is finite for $i < s$ and not finite for $i = s$, see Algebra, Lemma 10.83.2. Since we know the result is true for $A^\wedge $ it suffices to show that $s_{A, I}(M) = s_{A^\wedge , I^\wedge }(M^\wedge )$. This follows from Lemma 51.11.4.
$\square$

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