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