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Let $A \to B$ be a ring homomorphism and $M,N$ be two $A$-modules. Consider the natural map

$\alpha_{M,N} : \mathrm{Hom}_A(M,N) \otimes_A B \to \mathrm{Hom}_B(M \otimes_A B,N \otimes_A B)$

Consider the statements:

(1) $\alpha_{M,N}$ is an isomorphism for all $M,N$ whenever $M$ is finitely presented.

(2) $A \to B$ is flat.

Then it is easy to see that (2) => (1). Namely, $\alpha_{-,N}$ is a natural transformation between finitely cocontinuous functors and $\alpha_{A,N}$ is obviously an isomorphism, thus $\alpha_{M,N}$ is an isomorphism whenever $M$ is finitely presented.

Question. Does also (1) => (2) hold? If this is false for trivial reasons, can we strenghten (1) a little bit so that it becomes true? For example, one might try:

(1') For all $N$ the locus where $\alpha_{-,N}$ is an isomorphism is closed under finite colimits, or equivalently, cokernels.

Then we still have (2) => (1') and one might ask (1') => (2). An affirmative answer to this question would give a proof.

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    Dear Martin: Thanks for your answer. I found it by chance. It's better to use the `@` sign to notify users.2011-11-01

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