# Extension of scalars

Let $A\rightarrow B$ be a ring homomorphism, $M$ and $N$ - modules over $A$. How to prove, that $$(M \otimes_A N) \otimes _A B = (M \otimes_A B) \otimes_B (N\otimes_A B)$$ as $B$-modules in the most simplest way. Of course, it's possible to define 2 maps and prove, that they are correct and are inverse to each other. Is there more understandable proof? For example, using some universal property?

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All the standard tensor product isomorphisms (and there are lots, associativity, transitivity, variants on these) are proved by using the universal property to define some canonical maps in both directions. One then verifies that the composites in both direction are the identity by checking the effect on a simple tensor and concluding by additivity. So it's not totally clear to me what you're looking for. – Keenan Kidwell Nov 24 '12 at 19:33

Hint: $(M \otimes_A B) \otimes_B (N \otimes_A B) = M \otimes_A (B \otimes_B (N\otimes_A B))$. By extension of scalars, $N \otimes_A B$ is a $B$-module. Now if $R$ is any ring and $D$ is an $R$-module, what is $R \otimes_R D$?