if $M \otimes_A (A_m/mA_m)=0$ for every maximal ideal $m \subset A$, then $M=0$, $M$ finitely generated

Question

Suppose $M$ is a finitely generated $A$-module. Prove that if $M \otimes_A (A_m/mA_m)=0$ for every maximal ideal $m \subset A$, then $M=0$. Subscrpit $_m$ means localization at $m$.

First consider the exact sequence $$m \rightarrow A \rightarrow A/m \rightarrow 0.$$ Since localization of modules is an exact functor, we have $$mA_m \rightarrow A_m \rightarrow A_m/mA_m \rightarrow 0$$ is exact. Taking tensor product gives us $$M \underset{A}{\otimes}mA_m \rightarrow M \underset{A}{\otimes}A_m \rightarrow M \underset{A}{\otimes}A_m/mA_m \rightarrow0$$ which is exact. Note that $M \underset{A}{\otimes}A_m \simeq M_m$ and by assumption $M \underset{A}{\otimes}A_m/mA_m=0$.

Now using $M$ is finitely generated, how can I show that $M_m=0$?

If I can show this then since $M=0 \Leftrightarrow (M_m=0$ for all maximal ideals $m$ of $A$), I can complete the proof.

Answer 1

Hint: $$M \otimes_A A_{\mathfrak m}/\mathfrak mA_{\mathfrak m}\simeq M_{\mathfrak m}/\mathfrak mM_{\mathfrak m}.$$ Then use Nakayama's lemma.

Answer 2

The last exact sequence with the appropriate identifications becomes $mM_m\rightarrow M_m\rightarrow0$ and so $M_m=mM_m$. Now apply Nakayama's lemma.