I've been trying to prove the following statement:
Let $R$ be a ring, $I$ a finitely generated ideal of $R$, and $M$ an $R$-module. Suppose that $M_\mathfrak{p}=0$ for every $\mathfrak{p}\in\text{Spec}(R)\setminus V(I)$. Then for any element $x\in M$, there exists some $n\geq 1$ such that $I^nx=0$.
I've attempted to follow the proof of Lemma 15.87.6 of The Stacks Project, but I'm confused at one particular step.
They claim that, for any $x\in M$, if $\frac{x}{1}=0$ in $M_\mathfrak{p}$ for each prime $\mathfrak{p}\nsupseteq I$, then $\frac{x}{1}=0$ in $M_a$ for all $a\in I$. However I can't see why this is the case.
Any help would be much appreciated.
Update:
In the case where $R$ is Noetherian, we can apply their Lemma 10.62.4 since, for any $x\in M$, we have
$$\text{Supp}(Rx)\subseteq\text{Supp}(M)\subseteq V(I)$$
However they don't impose this condition in the statement of 15.87.6, and the arguments seem quite different.
Best Answer
Let $a\in I$ and $S=\{1,a,a^2,\dots\}$. Let us assume that $\mathrm{Ann}(x)\cap S=\emptyset$. Then there exists a prime ideal $\mathfrak p$ such that $\mathfrak p\supseteq\mathrm{Ann}(x)$ and $\mathfrak p\cap S=\emptyset$. If $\mathfrak p\not\supseteq I$, since $\frac x1=\frac01$ in $M_{\mathfrak p}$ there exists an element $s\notin\mathfrak p$ such that $sx=0$, that is, $s\in\mathrm{Ann}(x)$, a contradiction. Therefore $\mathfrak p\supseteq I$, so $a\in\mathfrak p$, which is a contradiction with $\mathfrak p\cap S=\emptyset$.