Tensor Product of Two Finitely Generated Modules Over a Local Ring

[caffeinemachine]

Problem. 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.
$\newcommand{\mf}{\mathfrak}$
I am thinking the following: We can define an $R$-bilinear map $M\times N\to M/\mf m M/\otimes_{R/\mf m} N/\mf mN$, where hte latter gets an $R$-module structure by restriction of scalars via the map $R\to R/\mf m$. The map is defined by declaring $(\bar m, \bar n)\mapsto \bar m\otimes \bar n$. This can be checked to be well-defined and $R$-bilinear. 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 surjective and since the domain is $0$, so is the target and we are done.[FONT=MathJax_Math-italic].efined simply a. This map is surjectiv and thus the latter is 0.

[/FONT]Is this correct? Is there a better way to see the same thing? Thank you.

 

[jvicens]

Your approach is correct and a good application of the Nakayama lemma. Another way to see this is by using the fact that $M\otimes_R N$ is a finitely generated $R$-module since $M$ and $N$ are both finitely generated. Since $M\otimes_R N=0$, this means that $M\otimes_R N$ is a finitely generated $R$-module with a basis of 0, which implies that $M=0$ or $N=0$ by definition. This approach is more direct and does not require the use of the Nakayama lemma. However, your approach is more general and can be applied in situations where $M\otimes_R N$ is not necessarily a finitely generated $R$-module.