Lemma 51.10.2. Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be a finite $A$-module, let $\mathfrak p \subset A$ be a prime ideal, and let $s \geq 0$ be an integer. Assume
$A$ has a dualizing complex,
$\mathfrak p \not\in V(I)$, and
for all primes $\mathfrak p' \subset \mathfrak p$ and $\mathfrak q \in V(I)$ with $\mathfrak p' \subset \mathfrak q$ we have
\[ \text{depth}_{A_{\mathfrak p'}}(M_{\mathfrak p'}) + \dim ((A/\mathfrak p')_\mathfrak q) > s \]
Then there exists an $f \in A$, $f \not\in \mathfrak p$ which annihilates $H^ i_{V(I)}(M)$ for $i \leq s$.
Proof.
Consider the sets
\[ T = V(I) \quad \text{and}\quad T' = \bigcup \nolimits _{f \in A, f \not\in \mathfrak p} V(f) \]
These are subsets of $\mathop{\mathrm{Spec}}(A)$ stable under specialization. Observe that $T \subset T'$ and $\mathfrak p \not\in T'$. Assumption (3) says that hypothesis (2) of Proposition 51.10.1 holds. Hence we can find $J \subset A$ with $V(J) \subset T'$ such that $J H^ i_{V(I)}(M) = 0$ for $i \leq s$. Choose $f \in A$, $f \not\in \mathfrak p$ with $V(J) \subset V(f)$. A power of $f$ annihilates $H^ i_{V(I)}(M)$ for $i \leq s$.
$\square$
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