In Atiyah-Macdonald's Commutative Algebra, on page 39, Proposition 3.5, it says
Let $M$ be an $A$-module. Then the $S^{-1}A$-modules: $S^{-1}M$ and $S^{-1}A \otimes_{A}M$ are isomorphic.
Since $S^{-1}A$ and $M$ are $A$-modules so is $S^{-1}A\otimes_{A}M$.
My question is: how can one tell $S^{-1}A\otimes_{A} M$ is an $S^{-1}A$-module?
Is it true that if $M$ is an $A$-module and $S$ being a multiplicatively closed subset of $A$. Then is $M$ also $S^{-1}A$-module?
Best Answer
Not at all. This is just a particular instance of a general result:
If $A\longrightarrow B$ is a (commutative) $A$-algebra, and $M$ an $A$-module, the $A$-module $B\otimes_AM$ is in a natural way a $B$-module through the scalar multiplication $$\beta((b\otimes m) \overset{\text{def}}{=}(\beta b)\otimes m.$$ This $B$-module is said to be ‘obtained by extension of scalars’ from the $A$-module $M$.