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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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Okay, I'm a dummy and can't read. Deleting my comments. –  Thomas Andrews Oct 31 '11 at 17:12
    
I have edited my question completely since the first version was totally errorneous. –  Martin Brandenburg Nov 1 '11 at 10:54
1  
Dear Martin: Are you assuming that $A$ is commutative? –  Pierre-Yves Gaillard Nov 1 '11 at 11:45
    
Dear Martin: Did you look at Bourbaki, Algèbre, Section II.5.3? –  Pierre-Yves Gaillard Nov 1 '11 at 12:06
    
1) All rings are commutative here. Hence also the tag 'commutative algebra'. 2) I've looked at the section in Bourbaki. It says that $\alpha_{M,N}$ is an isomorphism when $B$ is a finitely generated projective $A$-module, even without assuming that $M$ is finitely presented. Well this assertions is trivial and is only related to my question, which basically asks for a converse. If $B$ is finitely generated projective over $A$, then in particular it is flat. So in this case (1) => (2) is OK. –  Martin Brandenburg Nov 1 '11 at 12:38
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