Identity of tensor products over an algebra I hope this question hasn't been asked before, but I wasn't sure what to search so as to check. Suppose $R$ is a commutative ring with 1 and that $A$, $B$ are unital $R$-algebras. Suppose further that $M$, $M'$ are $A$-modules and $N$, $N'$ are $B$-modules. Is the following isomorphism of $A \otimes_R B$-modules correct, and if so, how to prove it?
$$(M \otimes_R N) \otimes_{A \otimes_R B} (M' \otimes_R N') \cong (M \otimes_A M') \otimes_R (N \otimes_B N').$$
Here we view $A \otimes_R B$ as an $R$-algebra in the typical way, while $M$, $N$, $M'$, $N'$ are $R$-modules via the embedding $r \mapsto r \cdot 1$, and $M \otimes_R N$ is an $A \otimes_R B$-module via the factor-wise action, etc. 
The obvious thing to try is the map $(m \otimes n) \otimes (m' \otimes n') \mapsto (m \otimes m') \otimes (n \otimes n')$, but showing it's well-defined seems an intractable nightmare. Thanks for any advice.
 A: For $M,M',N,N'$ free there is an obvious, well defined isomorphism. In fact, if $M=A^{(I)}$, $M'=A^{(I')}$, $N=B^{(J)}$, $N'=B^{(J')}$, they are both naturally isomorphic to $A\otimes_R B^{(I\times I'\times J\times J')}$.
In the general case, take free resolutions of $M,M',N,N'$, they induce free resolutions of both $(M \otimes_R N) \otimes_{A \otimes_R B} (M' \otimes_R N')$ and $(M \otimes_A M') \otimes_R (N \otimes_B N')$ as $A\otimes_R B$ modules, and we have an obvious isomorphism between the resolutions. Thus, also our modules are isomorphic.
A: I'm going to assume that $M,N$ are right modules and $M',N'$ are left modules.  $(M\otimes_R N) \otimes_{A\otimes_R B} (M'\otimes_R N')$ is governed $R$-multilinear maps $f:(M\times N)\times (M'\times N')\to Z$ for $R$-modules $Z$ that are balanced $A\otimes_R B$-linear maps with respect to the factors $M\times M'$ and $N\times N'$. $(M\otimes_A M')\otimes_R (N\otimes_B N')$ is governed by $R$-multilinear maps $g:(M\times M')\times (N\times N')\to Z$ for $R$-modules $Z$ that are balanced $A$- (respectivey $B$-) linear maps with respect to the factors $M\times M'$ and $N\times N'$. One can construct a compatible bijection between maps like $f$ and maps like $g$.
