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?
Extension of scalars
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abstract-algebra
commutative-algebra
ring-theory
modules
tensor-products
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2All 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. – 2012-11-24
1 Answers
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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 $R \otimes_R D$?