Let $M, N$ be $A$-modules, where $A$ is a commutative ring. Let $S$ be a multiplicative subset of $A$. I want to study the relation between $S^{-1}\left(A^{M \times N} \right)$ and $\left(S^{-1}A\right)^{(S^{-1}M) \times (S^{-1}N)}$.
My analysis shows the following. $S^{-1}\left(A^{M \times N} \right)$ is still free on $$ as $S^{-1}A$-module. We can inject $S^{-1}\left(A^{M \times N} \right)$ into $\left(S^{-1}A\right)^{(S^{-1}M) \times (S^{-1}N)}$ by mapping the generator $$ of $S^{-1}\left(A^{M \times N} \right)$ to the generator $<\frac{x}{1},\frac{y}{1}>$ of $\left(S^{-1}A\right)^{(S^{-1}M) \times (S^{-1}N)}$. However, this injection is proper (i.e. not surjective) since there is no way to represent elements of the form $<\frac{x}{s},\frac{y}{s}>$ for $s \in S, s \neq 1$ as a linear combination over $S^{-1}A$ of elements $<\frac{x}{1},\frac{y}{1}>$.
Consequently, $\left(S^{-1}A\right)^{(S^{-1}M) \times (S^{-1}N)}$ is "larger" than $S^{-1}\left(A^{M \times N} \right)$.
Is the above reasoning sound? Thanks.
abstract-algebra
commutative-algebra