Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $R \otimes_R D$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.