Lemma 52.9.1. Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module and let $\mathfrak p \subset A$ be a prime. Let $s$ and $d$ be integers. Assume

$A$ has a dualizing complex,

$\text{cd}(A, I) \leq d$, and

$\text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) > d + s$.

Then there exists an $f \in A \setminus \mathfrak p$ which annihilates $H^ i(R\Gamma _\mathfrak m(M)^\wedge )$ for $i \leq s$ where ${}^\wedge $ indicates $I$-adic completion.

**Proof.**
According to Local Cohomology, Lemma 51.9.4 the function

\[ \mathfrak p' \longmapsto \text{depth}_{A_{\mathfrak p'}}(M_{\mathfrak p'}) + \dim (A/\mathfrak p') \]

is lower semi-continuous on $\mathop{\mathrm{Spec}}(A)$. Thus the value of this function on $\mathfrak p' \subset \mathfrak p$ is $> s + d$. Thus our lemma is a special case of Lemma 52.8.1 provided that $\mathfrak p \not= \mathfrak m$. If $\mathfrak p = \mathfrak m$, then we have $H^ i_\mathfrak m(M) = 0$ for $i \leq s + d$ by the relationship between depth and local cohomology (Dualizing Complexes, Lemma 47.11.1). Thus the argument given in the proof of Lemma 52.8.1 shows that $H^ i(R\Gamma _\mathfrak m(M)^\wedge ) = 0$ for $i \leq s$ in this (degenerate) case.
$\square$

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