# Flat algebras and tensor product

All rings are commutative. Suppose $B$ is a flat $A$-algebra, and that $M$ and $N$ are flat $B$-modules.

Is there a way to compare the two $A$-modules $M \otimes_A N$ and $M \otimes_B N$?

Thanks

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Well, the natural map $M \times N \rightarrow M \otimes_B N$ is $B$-bilinear, hence also $A$-bilinear, so there is an induced $A$-module map $M \otimes_A N \rightarrow M \otimes_B N$. Moreover the map is surjective. Are you looking for more than that as an answer? – Pete L. Clark Aug 2 '11 at 11:57
I am wondering if there are any reasonable conditions on the map $A\to B$ which will make this map an isomorphism. – the L Aug 2 '11 at 12:00
Suitable conditions on $M$ and $N$ will also be good. – the L Aug 2 '11 at 12:02
Hmm. Notice that if $B/A$ is a proper field extension and $M,N$ are nonzero finite-dimensional vector spaces, then the map will not be an isomorphism. In some sense, these modules and rings are as nice as possible. So I think you need to be a bit more specific as to what you're looking for. – Pete L. Clark Aug 2 '11 at 12:06
Moderator Note Let's keep comments civil, constructive and on-topic, or I'll need to have some private chats with folks. – Tim Post Jul 1 '14 at 13:55

If $B$ is a localization of $A$ then the natural map $M\otimes_A N \to M\otimes_B N$ will be an isomorphism (without any assumption on $M$ and $N$).
Well, one way to think about your question is to take $M = N = B$. (You can't get much flatter $B$-modules than this!) Then you are asking that the natural map $B\otimes_A B \to B$ be an isomorphism, which is to say that the diagonal map $$\mathrm{Spec} \, B \hookrightarrow \mathrm{Spec \, B}\otimes_{\mathrm{Spec} \, A} \mathrm{Spec} \, B$$ be an isomorphism. This is asking that $A \to B$ be an epimorphism.
It is not so easy to find flat epimorphisms that are not localizations (see the answers here, especially this one), and so in practice I think that you should consider "$B$ is a localization of $A$" to be the most reasonable answer to the question of when this map is an isomorphism.