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

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.

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

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.

• Thanks both of you and Nakayama!
– Andy
Mar 10, 2019 at 18:41
• Btw doesn't Nakayama's lemma only work for $M=IM$ where $I$ is contained in every maximal ideal (i.e intersection of every maximal ideal)? $m$ is just a random maximal ideal and is not necessarily contained every maximal ideal.
– Andy
Mar 10, 2019 at 20:17
• @Andrew $A_m$ is a local ring. Mar 10, 2019 at 20:25
• Thanks! I got it now.
– Andy
Mar 10, 2019 at 20:29