I've been doing some tensoring, but am having a hard time understanding the following isomorphism.

Suppose $A$ is a commutative $R$-algebra, and for any $R$-module $M$, denote by $M_A=A\otimes_R M$. Then for any $R$-module and $A$-module $N$, $M_A\otimes_A N\cong M\otimes_R N$.

I tried to explain this to myself by defining some $\varphi_a\colon M\times N\to A\otimes_R M\otimes_A N$ defined by $m\otimes n\mapsto a\otimes m\otimes n$ to get some module homomorphism $M\otimes_R N\to A\otimes_R M\otimes_A N$. I was thinking then I could use some sort of projection $g\colon A\otimes_R M\otimes_A N\to M\otimes_R N$ given by $a\otimes m\otimes n\mapsto m\otimes n$ to find an inverse. However, I'm feeling shaky since the maps seem to depend on a certain $a\in A$, so maybe what I'm saying is nonsense.

What is the proper explanation for this isomorphism?


The real trick is to do algebra with modules, rather than in modules.

$$ M_A \otimes_A N \cong (A \otimes_R M) \otimes_A N \cong (M \otimes_R A) \otimes_A N \cong M \otimes_R (A \otimes_A N) \cong M \otimes_R N $$

That said, to aid with working through your ideas, keep in mind that, usually, when rewriting the product of $a$, $m$, and $n$, it still has to have $a$, $m$, and $n$ in it. The only place we can stick an $a$ without resorting to $\otimes$ is $$ a \otimes m \otimes n \mapsto m \otimes (an) $$

Edit: as Steve notes in the comment, I took heavy advantage of commutivity here, so that I can view things as bimodules reorder my tensors with abandon. When you aren't in such a nice situation, things become messier.

  • $\begingroup$ Perhaps it would be good to note that "associativity" holds here because $A$ is an $R$-$A$ bimodule. $\endgroup$ – user641 Mar 9 '12 at 22:34
  • $\begingroup$ The only reordering that happened was when you turned $A \otimes_R M$ into $M \otimes_R A$. If you define $M_A$ as $M \otimes_R A$ then everything still works even when $R$ and $A$ are not commutative. $\endgroup$ – Zhen Lin Mar 9 '12 at 23:03
  • $\begingroup$ Many thanks Hurkyl, that was very clean! $\endgroup$ – Jacqueline Pauwels Mar 16 '12 at 22:03

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