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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.

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2 Answers 2

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

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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.

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  • $\begingroup$ Thanks both of you and Nakayama! $\endgroup$
    – Andy
    Mar 10, 2019 at 18:41
  • $\begingroup$ 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. $\endgroup$
    – Andy
    Mar 10, 2019 at 20:17
  • $\begingroup$ @Andrew $A_m$ is a local ring. $\endgroup$
    – Rafael
    Mar 10, 2019 at 20:25
  • $\begingroup$ Thanks! I got it now. $\endgroup$
    – Andy
    Mar 10, 2019 at 20:29

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