Let $R$ be a commutative ring, $M,M',N,N'$ be $R$-modules. Under what hypotheses on $M,M',N,N'$ the canonical map $$ {\rm Hom}_R(M,N)\otimes_R{\rm Hom}_R(M',N')\to {\rm Hom}_R(M\otimes_RM',N\otimes_RN')$$ is an isomorphism?

The ring $R$ should be arbitrary (at most noetherian if it simplifies a lot). For instance, I know that it holds in any of the following cases:

(i) $M$ and $M'$ are projective of finite type

(ii) $M$ and $N$ are projective of finite type

(iii) $M$ and $M'$ are finitely presented, $N'$ and ${\rm Hom}_R(M,N)$ flat. I am looking for (reasonable) different or weaker hypotheses. Any natural answer generalizing at once (i) and (iii) would be (probably) preferred.

A second question, related to (iii), is: when ${\rm Hom}_R(M,N)$ is flat? I know that a sufficient condition is $M$ and $N$ injective (at least if $R$ is noetherian).

(There are at least two questions discussing this, but there is no answer to my question: The relationship of $\hom(M\otimes_RN,M'\otimes_RN')$ and $\hom_R(M,M')\otimes\hom_R(N,N')$. and Is there a relation between $End(M)$ and $M$ under tensor products? )


1 Answer 1


As asserted in Bourbaki, Algebra, ch. II Linear Algebra, §4, n°4, prop.4, you also have the case $M', N'$ are finitely generated projective.

For your subsidiary question there's the obvious case $M$ finitely generated projective and $N$ flat. You can remove ‘finitely generated’ if $R$ is a coherent ring.

  • 1
    $\begingroup$ Thank you. I had not listed this case since it is equivalent to (ii). For the second question I was looking for different answers (I'm sorry for not pointing it in my question). $\endgroup$
    – A.G
    Jul 5, 2017 at 20:47

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