Lemma 16.10.1 (Ogoma). Let $A$ be a Noetherian ring and let $M$ be a finite $A$-module. Let $S \subset A$ be a multiplicative set. If $\pi \in A$ and $\mathop{\mathrm{Ker}}(\pi : S^{-1}M \to S^{-1}M) = \mathop{\mathrm{Ker}}(\pi ^2 : S^{-1}M \to S^{-1}M)$ then there exists an $s \in S$ such that for any $n > 0$ we have $\mathop{\mathrm{Ker}}(s^ n\pi : M \to M) = \mathop{\mathrm{Ker}}((s^ n\pi )^2 : M \to M)$.

Proof. Let $K = \mathop{\mathrm{Ker}}(\pi : M \to M)$ and $K' = \{ m \in M \mid \pi ^2 m = 0\text{ in }S^{-1}M\}$ and $Q = K'/K$. Note that $S^{-1}Q = 0$ by assumption. Since $A$ is Noetherian we see that $Q$ is a finite $A$-module. Hence we can find an $s \in S$ such that $s$ annihilates $Q$. Then $s$ works. $\square$

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