Let $R$ be a commutative and local ring, and $M$ is a right $R$-module and $N$ is a left $R$-module. If $M$ and $N$ are finitely generated, then $M\otimes_{R} N=0$ if and only if $M=0$ or $N=0$.
If $R$ is not local, then this proposition is true? Why?
I want to prove this proposition. But I don't know how to start. Please help me.
Thank you.
Let $R$ be a local ring with maximal ideal $\mathfrak m$. By tensoring with $R / \mathfrak m$ and using associativity, we get $$ (M / \mathfrak m M) \otimes_{R / \mathfrak m} (N / \mathfrak m N) = 0 $$ as vector spaces. It follows that $\dim (M / \mathfrak m M) = 0$ or $\dim (N / \mathfrak m N) = 0$. Now Nakayama's lemma gives the desired result.