Lemma 51.7.2. Let $A$ be a Noetherian ring, $I \subset A$ an ideal, $M$ a finite $A$-module, and $n \geq 0$ an integer. Let $Z = V(I)$. The following are equivalent

the modules $H^ i_ Z(M)$ are finite for $i \leq n$, and

for all $\mathfrak p \in \mathop{\mathrm{Spec}}(A)$ the modules $H^ i_ Z(M)_\mathfrak p$, $i \leq n$ are finite $A_\mathfrak p$-modules.

**Proof.**
The implication (1) $\Rightarrow $ (2) is immediate. We prove the converse by induction on $n$. The case $n = 0$ is clear because both (1) and (2) are always true in that case.

Assume $n > 0$ and that (2) is true. Let $N = H^0_ Z(M)$ and $M' = M/N$. By Dualizing Complexes, Lemma 47.11.6 we may replace $M$ by $M'$. Thus we may assume that $H^0_ Z(M) = 0$. This means that $\text{depth}_ I(M) > 0$ (Dualizing Complexes, Lemma 47.11.1). Pick $f \in I$ a nonzerodivisor on $M$ and consider the short exact sequence

\[ 0 \to M \to M \to M/fM \to 0 \]

which produces a long exact sequence

\[ 0 \to H^0_ Z(M/fM) \to H^1_ Z(M) \to H^1_ Z(M) \to H^1_ Z(M/fM) \to H^2_ Z(M) \to \ldots \]

and similarly after localization. Thus assumption (2) implies that the modules $H^ i_ Z(M/fM)_\mathfrak p$ are finite for $i < n$. Hence by induction assumption $H^ i_ Z(M/fM)$ are finite for $i < n$.

Let $\mathfrak p$ be a prime of $A$ which is associated to $H^ i_ Z(M)$ for some $i \leq n$. Say $\mathfrak p$ is the annihilator of the element $x \in H^ i_ Z(M)$. Then $\mathfrak p \in Z$, hence $f \in \mathfrak p$. Thus $fx = 0$ and hence $x$ comes from an element of $H^{i - 1}_ Z(M/fM)$ by the boundary map $\delta $ in the long exact sequence above. It follows that $\mathfrak p$ is an associated prime of the finite module $\mathop{\mathrm{Im}}(\delta )$. We conclude that $\text{Ass}(H^ i_ Z(M))$ is finite for $i \leq n$, see Algebra, Lemma 10.62.5.

Recall that

\[ H^ i_ Z(M) \subset \prod \nolimits _{\mathfrak p \in \text{Ass}(H^ i_ Z(M))} H^ i_ Z(M)_\mathfrak p \]

by Algebra, Lemma 10.62.19. Since by assumption the modules on the right hand side are finite and $I$-power torsion, we can find integers $e_{\mathfrak p, i} \geq 0$, $i \leq n$, $\mathfrak p \in \text{Ass}(H^ i_ Z(M))$ such that $I^{e_{\mathfrak p, i}}$ annihilates $H^ i_ Z(M)_\mathfrak p$. We conclude that $I^ e$ with $e = \max \{ e_{\mathfrak p, i}\} $ annihilates $H^ i_ Z(M)$ for $i \leq n$. By Lemma 51.7.1 we see that $H^ i_ Z(M)$ is finite for $i \leq n$.
$\square$

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