Commutation of tensor products with inverse limits in a specific case For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow
$$
R^X\otimes_R R^Y\rightarrow R^{X\times Y}
$$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).
What happens if $R$ is a general ring ? ($X,Y$ being infinite).
This question is related to that one
Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding
 A: The key point is to generalize the problem in order to make more effective the use of limits.  Consider more generally for any $R$-module $R$ the natural map
$$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$
given by $m \otimes (r_y) \mapsto (r_y m)$.  In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective. 
If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$.  Hence, it suffices to treat the case when $M$ is finitely generated.  In case $M$ is finitely presented then $f_{M,Y}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy).  So this gives an affirmative answer when $R$ is noetherian.  
Are you interested in non-noetherian $R$?
