Lemma 10.62.4. Let $R$ be a Noetherian ring. Let $I \subset R$ be an ideal. Let $M$ be a finite $R$-module. Then $I^ nM = 0$ for some $n \geq 0$ if and only if $\text{Supp}(M) \subset V(I)$.

**Proof.**
It is clear that $I^ nM = 0$ for some $n \geq 0$ implies $\text{Supp}(M) \subset V(I)$. Suppose that $\text{Supp}(M) \subset V(I)$. Choose a filtration $0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$ as in Lemma 10.62.1. Each of the primes $\mathfrak p_ i$ is contained in $V(I)$ by Lemma 10.62.2. Hence $I \subset \mathfrak p_ i$ and $I$ annihilates $M_ i/M_{i - 1}$. Hence $I^ n$ annihilates $M$.
$\square$

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