Localization at annihilators of an ideal

Question

I was reading this post and on line +10-11 of the proof of lemma 27.25.1, it seems to claim the following:

Let $A$ be a ring, $I \subseteq A$ an ideal, and $M$ an $A$-module. Let $M_I:=\{ x \in M\,: \, Ix=0 \}$. Let $D(f)$ be a basic open set contained in $\operatorname{Spec}(A) \setminus V(I)$. Then $$ (M_I)_{f} = M_{f},$$ where the subscript denotes localization.

Is this true? If so, how does one show this?

Answer 1

In the case where $A$ is an integral domain and $0\neq I\subseteq A$, then taking $M$ to be some free module over $A$ we would have $M_I=0$ and so $(M_I)_f=0\neq M_f$.