Let $A \rightarrow B$ be a homomorphism of commutative rings. Let $M, N$ be $A$-modules. We denote $M\otimes_A B$ by $M_B$. We regard $M_B$ as a $B$-module. Then $(M\otimes_A N)_B \cong M_B\otimes_B N_B$?
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5This has been asked when the map is a localization here and the answer is the same. – Mariano Suárez-Álvarez Jan 05 '13 at 17:54
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@YACP I didn't say I wouldn't accept it. It's usually not a good idea to accept an answer too soon. – Makoto Kato Jan 07 '13 at 06:37
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There is an isomorphism of $A$-modules
$$M_B \otimes_B N_B \cong (M \otimes_A B) \otimes_B (B \otimes_A N) \cong M \otimes_A (B \otimes_B B) \otimes_A N \cong M \otimes_A B \otimes_A N \cong (M \otimes_A N)_B.$$
By its explicit description on elements (or some diagram) it is easily seen to be $B$-linear.
A more detailed proof can be found in many treatments of algebra, for example N. Bourbaki, Elements of Mathematics, Algebra I, Chapter II, §5, Prop 3.
Martin Brandenburg
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