I am dealing with the following question:
Let $M$ be an $A$-module and $I$ an ideal of $A$. Suppose that $M_m=0\;\forall m\in \operatorname{Specm}(A)$ s.t. $m\supset I$. Then $M=IM$.
(Here, $M_m$ is the localization by the ideal $m$, i.e., $S^{-1}M$, $S^{-1}=A\setminus m.$)
I've found an answer in this post: $M_m=0$ for every maximal ideal $m$ containing $I$ if and only if $M=IM$, where $M$ is finite.
It's reproduced below:
Consider the $R$-module $\overline{M}=M/IM$ and let $m\subseteq R$ be a maximal ideal of $R$. If $I\not\subseteq m$, then $I_m=R_m$ and $\overline{M}_m=0$. If $I\subseteq m$ then by assumption $M_m=0$ and $\overline{M}_m=0$. Hence $\overline{M}=0$. i.e. $M=IM$.
However, I could not understand the most important part to me:
If $I\not\subseteq m$, then $I_m=R_m$ and $\overline{M}_m=0$.
Please, could someone fill with more details?
Many thanks!
If $I\nsubseteq \mathfrak m$, at least one element of $I$ is not in $\mathfrak m$, hence is invertible in $R_{\mathfrak m}$ and thus $I_{\mathfrak m}$ contains the unit element of $R_{\mathfrak m}$.
If $\overline M$ denotes $M/IM$, its localisation at $\mathfrak m$ is $$(M/IM)_{\mathfrak m}\simeq M_{\mathfrak m}/(IM)_{\mathfrak m}=M_{\mathfrak m}/I_{\mathfrak m}M_{\mathfrak m}=M_{\mathfrak m}/M_{\mathfrak m}=\{0\}.$$