If $N$ is a finitely generated $R-$module then... Show that $\left(\prod_\alpha M_\alpha\right)\otimes_R N\cong \prod_\alpha\left(M_\alpha\otimes_R N\right)$
I'd also like a counterexample where this isn't true for arbitrary $N$. I don't even really know how to start with this, this has come up in algebraic topology course and it's been quite a while since I done too much abstract algebra or tensors.
Any help is appreciated!
 A: Depending on how you define the tensor product this might be helpful.
Let $n_1,\ldots,n_\ell$ generate $N$.
It should be clear that an arbitrary element of $(\Pi_\alpha M_\alpha)\otimes_R N$ can be written as
$\sum_{i=1}^n(\prod_\alpha m_{\alpha,i})\otimes n_i,$
where $m_{\alpha,i}$ are some arbitrary element in $M_\alpha$. Similarly it should be clear that an arbitrary element in $\Pi_\alpha (M_\alpha\otimes_R N)$ can be written as
$\sum_{i=1}^n\Pi_\alpha (m_{\alpha,i}\otimes n_i).$
Note that the way to do this in general are far from unique, to reach these expressions simply split up arbitrary expressions and collect the terms involving a particular $n_i$.
At this point it should be clear what your isomorphism should be (send $\sum_{i=1}^n(\prod_\alpha m_{\alpha,i})\otimes n_i$ to $\sum_{i=1}^n\Pi_\alpha (m_{\alpha,i}\otimes n_i)$). What remains is to show that this map is well defined (this might be a bit technical) and that it is in fact an isomorphism (this should be easy). While I don't immediately see a super easy counterexample it should be clear that this "obvious" construction will not work when $N$ is not finitely generated.
