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 $R$.


1 Answer 1


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 $L=R/\mathfrak m\otimes_RM$.

  • 1
    $\begingroup$ 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? $\endgroup$
    – mathcourse
    Jul 2, 2018 at 21:26

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