Tensor Products of Modules (restriction and extension of scalars) Let $\phi:A \rightarrow B$ be a ring homomorphism, $M$ be an $A$-module, and $N$ a $B$-module.
Prove that
$$N \otimes_B (B \otimes_A M) \cong N \otimes_A M$$ as either $A$ or $B$-modules.
We know that $B \otimes_AM$ is a $B$-module and $N$ is an $A$-module via extension and restriction of scalars. We have that $N \cong N \otimes_BB$. 
Is it legal to do this: 
$$N \otimes_B (B \otimes_AM) \cong (N \otimes_BB) \otimes_AM \cong N \otimes_AM$$ 
 A: Yep; and that's exactly the approach I'd use.
But if you're nervous, that argument also suggests how to write down an actual function that gives the isomorphism:


*

*The forward direction is $n \otimes b \otimes m \mapsto nb \otimes m $

*The backward direction is $n \otimes m \mapsto n \otimes 1 \otimes m $


and then you simply need to show that both functions are well-defined and are inverses.

To show these definitions are well-defined, we use the fact that I've expressed their values on pure tensors. Recall:

Let $U$ be a right $R$-module and $V$ be a left $R$-module. There is a one-to-one correspondence between
  
  
*
  
*Functions $f : U \times V \to W$ such that $f(u,v)$ is $R$-linear in both variables
  
*Linear maps $g : U \otimes_R V \to W$ 
  
  
  And these are related by $f(u,v) = g(u \otimes v)$.
Furthermore, if $U,W$ are left $S$-module, then if $f(u,v)$ is $S$-linear in $u$ if and only if $g$ is $S$-linear.

This extends inductively to repeated tensors; e.g.

Let $U$ be a right $R$-module, $V$ be a left $R$ module and a right $S$-module, and $W$ be a left $S$-module. There is a one-to-one correspondence
  
  
*
  
*Functions $f : U \times (V \times W) \to X$ such that $f(u,v,w)$ is $R$-linear in $u$ and $v$, and $S$-linear in $v$ and $w$
  
*Linear maps $g : U \otimes_R (V \otimes_S W) \to X$ 
  
  
  And these are related by $f(u,v,w) = g(u \otimes v \otimes w)$.

You can derive this by applying the two-factor version to $U$ and $V \otimes W$, and then again to $V$ and $W$.
