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?
Best Answer
Congratulations on having noticed this subtle point, rarely discussed in textbooks.
As is often the case, a more general statement is clearer; for your question take $P=S^{-1}M, Q=S^{-1}N$ in the following
General statement Suppose $P,Q$ are $S^{-1}A$-modules. Then there is a canonical $S^{-1}A$- isomorphism $P \otimes _A Q\to P \otimes_ {S^{-1}A} Q$
Preliminary remark An $A$-module $E$ can have at most $one$ $S^{-1}A$-module structure compatible with its $A$-module structure.
Proof of Preliminary remark: we must have $\frac{a}{s} \ast e = (s\bullet)^{-1} (ae)$ (The existence of an $S^{-1}A$-module structure on $E$ forces multiplication by $s$ to be an $A$-linear automorphism $(s\bullet)$ of the $A$-module $E$)
Proof of General statement The preliminary remark shows that the $S^{-1}A$-module structures on $P \otimes_A Q$ coming from $P$ or from $Q$ coincide. Hence there are canonical ${S^{-1}A}$- morphisms
$P \otimes _A Q\to P \otimes_ {S^{-1}A} Q: p\otimes q\mapsto p\otimes q$ and
$P \otimes_ {S^{-1}A} Q \to P \otimes _A Q\ : p\otimes q\mapsto p\otimes q$ which are mutually inverse $S^{-1}A$- isomorphisms ; this proves the General statement.