Let $A$ be a commutative ring with $1$ and let $M,N$ be $A$-modules.
Since there is a map $f: A \rightarrow S^{-1}A$, defined by $a \mapsto \frac{a}{1}$ then given any $S^{-1}A$-module we can view it as a $A$ module via restriction of scalars right?
Now $S^{-1}M$ and $S^{-1}N$ are $S^{-1}A$-modules.
My question is if the following isomorphism holds?
$S^{-1}M \otimes_{A} S^{-1}N \cong S^{-1}M \otimes_{S^{-1}A} S^{-1}N$
Is the above valid because any $S^{-1}A$ module is an $A$-module or why? (or perhaps it is false), can you please help?
Following Daniel's hint:
$S^{-1}(M \otimes_{A} N) \cong S^{-1}A \otimes_{A} (M \otimes_{A} N) \cong (S^{-1}A \otimes_{S^{-1}A}) (S^{-1}A \otimes_{A} (M \otimes _{A} N) )$
After this I end up with $(S^{-1}A \otimes _{S^{-1}A} N) \otimes_{A} S^{-1}M$ which is isomorphic to $S^{-1}N \otimes_{A} S^{-1}M$. Where's the error?