Let $R$ be a Noetherian ring and $M$ be an $R$-module.
We know that if $M$ is Noetherian then $M_m$ is a Noetherian $R_m$-module for every $m\in\operatorname{Max}(R) $.
Now, I want to know if the converse is true? under what circumstances?
I saw this:
If $|\operatorname{Min}\operatorname{Supp} (M/N) |<\infty$ for every finitely generated submodule $N$ of $M$ then the converse is true.
But I couldn't prove it.
Best Answer
For a simple counterexample, let $R$ be any Noetherian ring with infinitely many maximal ideals (e.g., $\mathbb{Z}$) and let $M=\bigoplus_{m\in\operatorname{max}(R)} R/m$. Then $M$ is not finitely generated, but $M_m=R/m$ is for any maximal ideal $m$ since all the other summands of $M$ are killed by the localization.
Assuming that by "$\min \operatorname{Supp} (M/N) <\infty$" you mean that the support of $M/N$ has finitely many minimal elements, then yes, that condition for all finitely generated $N\subseteq M$ does suffice. Equivalently, this condition means that $\operatorname{Supp} (M/N)$ is closed, since a specialization-closed subset of $\operatorname{Spec} R$ with finitely many minimal elements is closed (since it must be the union of the closures of its finitely many minimal elements). Given this condition, since $R$ is Noetherian, there exists a finitely generated $N$ such that $\operatorname{Supp} (M/N)$ is minimal.
If $M/N=0$, then $M=N$ is finitely generated and we're done. Otherwise, let $m$ be a maximal ideal in the support of $M/N$. By minimality of $\operatorname{Supp} (M/N)$, for any finitely generated $K\subseteq M$ containing $N$, $m$ is still in $\operatorname{Supp} (M/K)$. So, we can recursively choose an infinite sequence $x_0,x_1,\dots$ such that $x_n$ is nonzero in $(M/K_n)_m$ for all $n$ where $K_n$ is the submodule generated by $N$ and $x_0,\dots,x_{n-1}$. But this means the $(K_n)_m$ give a strictly increasing chain of submodules of $M_m$, contradicting the assumption that $M_m$ is Noetherian.