# Tensor product of two finitely-generated modules over a local ring is zero

If $R$ is a local ring and $M$ and $N$ are finitely generated $R$-modules such that $M\otimes N=0$ then how does it follow from Nakayama's lemma that either $M=0$ or $N=0$?

This is an exercise in Atiyah and Macdonald. The part I could not show in the hints is $({M{\otimes}_R N)}_{k}=0$ implies $M_{k}{\otimes }_{k} N_{k}=0$, where $k=R/\mathfrak m$ and $\mathfrak m$ is the maximal ideal of $R$.

From $$M\otimes_RN=0$$ we get $$R/\mathfrak m\otimes_R (M\otimes_R N)=0$$. Then it follows that $$(R/\mathfrak m\otimes_R M)\otimes_{R/\mathfrak m}(R/\mathfrak m\otimes_R N)=0$$. But $$R/\mathfrak m\otimes_R M\simeq M/\mathfrak mM$$ and similarly $$R/\mathfrak m\otimes_R N\simeq N/\mathfrak mN$$. Since $$M/\mathfrak mM$$ and $$N/\mathfrak mN$$ are $$R/\mathfrak m$$-vector spaces, it follows that $$M/\mathfrak mM=0$$ or $$N/\mathfrak mN=0$$, that is, $$M=\mathfrak mM$$ or $$N=\mathfrak mN$$, so by Nakayama we have $$M=0$$ or $$N=0$$.
Edit. For proving $$R/\mathfrak m\otimes_R (M\otimes_R N)\simeq(R/\mathfrak m\otimes_R M)\otimes_{R/\mathfrak m}(R/\mathfrak m\otimes_R N)$$ use the associativity and the following property of tensor product: $$L\otimes_SS\simeq L$$, where in this case $$S=R/\mathfrak m$$ and $$L=R/\mathfrak m\otimes_RM$$.
• I'm looking for a counterexample (with $M$ or $N$ not finitely generated) where the implication $M = 0$ or $N = 0$ does not hold. Do you have any ideas? Jul 2, 2018 at 21:26