Examples proving why the tensor product does not distribute over direct products. I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't. 
What are some simple counterexamples to why this property isn't true? I know that there is a natural homomorphism 
$$
\left(\prod M_i\right)\otimes N\to \prod (M_i\otimes N)
$$
given by $(\prod m_i)\otimes n\mapsto \prod (m_i\otimes n)$ when $M$ and $N$ are modules over some commutative ring $R$. Are there standard examples where this homomorphism is not injective/surjective and hence not an isomorphism?
 A: Let $X$ and $Y$ be indeterminates. 
For an example where your map is not surjective, take $M_i:=R$ for all $i\in\mathbb N$, and $N:=R[Y]$. 
Then you get the natural map 
$$
R[[X]][Y]\to R[Y][[X]],
$$
and 
$$
\sum_{i\in\mathbb N}\ X^i\ Y^i
$$
is not in the image.
EDIT. Same example with different notation: Put 
$$
A:=\left(\prod M_i\right)\otimes N,\quad B:=\prod\ (M_i\otimes N),
$$
and, for all $i,j\in\mathbb N$, 
$$
M_i=R_i=R_j=R_{ij}=R.
$$
Set also $N:=\bigoplus R_j$. Then we have canonical isomorphisms
$$
A=\bigoplus_j\ \prod_i\ R_{ij},\quad B=\prod_i\ \bigoplus_j\ R_{ij}. 
$$
We also have the inclusions
$$
A\subset B\subset\prod_{i,j}\ R_{ij},
$$
and your map becomes the first inclusion. 
Note that the Kronecker symbol $(\delta_{ij})$ is in $B$ but not in $A$.
A: We consider $\mathbb{Z}$-modules (i.e., abelian groups). 
Since $\mathbb{Q}$ is divisible, if $A$ is a torsion abelian group, then $A\otimes\mathbb{Q}$ is trivial.
Let $G$ be the direct product of cyclic group of order $p^n$, with $p$ a prime, and $n$ increasing; that is:
$$G = \prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}.$$
Then 
$$\prod_{n=1}^{\infty}\left(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}\right) = 0.$$
But $G\otimes\mathbb{Q}$ is not trivial: if we let $x$ be the element that corresponds to the class of $1$ in every coordinate, then $x$ has infinite order. Therefore,
$$\langle x\rangle \otimes\mathbb{Q}\cong \mathbb{Z}\otimes\mathbb{Q} \cong\mathbb{Q};$$
but tensoring with $\mathbb{Q}$ over $\mathbb{Z}$ is exact; therefore, the embedding $\langle x\rangle \hookrightarrow G$ induces an embedding $\langle x\rangle\otimes \mathbb{Q}\hookrightarrow G\otimes \mathbb{Q}$. Therefore, $G\otimes\mathbb{Q}\neq 0$. Thus, we have
$$\left(\prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}\right)\otimes \mathbb{Q}\not\cong \prod_{n=1}^{\infty}(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}).$$
