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Let $R$ be a local ring (commutative with identity) ans $M$ and $N$ be finitely generated $R$-modules.

If $M\otimes_R N=0$, then $M=0$ or $N=0$.

The problem clearly seems to be an application of the Nakayama lemma. If we can show that $M=\mathfrak mM$ or $N=\mathfrak mN$, where $\mathfrak m$ is the unique maximal ideal of $M$, then by Nakayama we would have $M=0$ or $N=0$, for the Jacobson radical of $M$ is nothing but $\mathfrak m$ itself.

I am unable to figure out what to do.


marked as duplicate by Pete L. Clark, Eric Wofsey, Claude Leibovici, Community Sep 15 '15 at 10:50

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    $\begingroup$ Well, what happens if you tensor from $R$ to $R/\mathfrak{m}$? $\endgroup$ – Pete L. Clark Sep 15 '15 at 5:34
  • $\begingroup$ @PeteL.Clark Using your suggestion two times gives $M/\mathfrak mM\otimes_R N/\mathfrak mN=0$. Perhaps now I should extend the scalars to get $M/\mathfrak mM\otimes_{R/\mathfrak m}N/\mathfrak mN=0$ somehow (not sure yet about the extension of scalar technique). But if we can do this, then we have a tensor product of two vector spaces (free modules) over $R/\mathfrak m$ vanishes. Thus one of these vector spaces must be 0 and we are done. $\endgroup$ – caffeinemachine Sep 15 '15 at 5:39
  • $\begingroup$ Yes, you are on the right track. $\endgroup$ – Pete L. Clark Sep 15 '15 at 5:48
  • $\begingroup$ @PeteL.Clark $\newcommand{\mf}{\mathfrak}$ I do not see how the extension of scalars on $M/\mf mM\otimes_R N/\mf mN$ can work. I now am thinking something else. We can define an $R$-linear map $M\times N\to M/\mf mM\otimes_{R/\mf m}N/\mf mN$, where the latter gets an $R$-module structre via restriction of scalars using $R\to R/\mf m$. The map is defined simply as $(\bar m, \bar n)\mapsto \bar m\otimes \bar n$. This is a well defined $R$-linear map and thus we get an $R$-linear map $M/\mf mM\otimes_R N/\mf mN\to M/\mf mM\otimes_{R/\mf m}N/\mf mN$. This map is surjectiv and thus the latter is 0. $\endgroup$ – caffeinemachine Sep 15 '15 at 6:18
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    $\begingroup$ I don't see a difference between your two approaches. Recall that an $R/\mathfrak{m}$-module is precisely an $R$-module which is annihilated by $\mathfrak{m}$. So taking the tensor product of two $R/\mathfrak{m}$-modules over $R$ is the same as doing it over $R/\mathfrak{m}$. You were the one who said "extension of scalars"...but yes, that's exactly what you're doing (either way). You wrote "I am unable to figure out what to do" so I gave you a hint. You can take it from here (in fact you are done but just need to understand why). $\endgroup$ – Pete L. Clark Sep 15 '15 at 6:46

$$M\otimes_RN=0\Rightarrow M\otimes_RN\otimes_RR/m=0\Rightarrow M\otimes_RN/mN=0\Rightarrow (M\otimes_RR/m)\otimes_{R/m}N/mN=0\Rightarrow M/mM\otimes_{R/m}N/mN=0$$


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