Is every $S^{-1}A$-module the localization of an $A$-module?

Question

Let $A$ be a commutative ring, $S\subseteq A$ closed under multiplication, and consider $S^{-1}A$ the localization of $A$ in $S$. Let $M$ be an $S^{-1}A$-module. Does it necessarily exist an $A$-module $M'$ such that $M = S^{-1}M'$?

Im trying to prove that if $M$ is projective, then its localization at $S$ is an $S^{-1}A$ projective module. In this question I'm not looking for the solution to this problem, but an answer to what I'm asking. I simply add this information for contextualization!

Thanks in advance!

Answer 1

An $S^{-1}A$-module $M$ is also an $A$-module (via the ring homomorphism $A\to S^{-1}A$). Then $S^{-1}M=M$, or strictly speaking the localisation of the $A$-module $M$ is naturally isomorphic to the $S^{-1}A$-module $M$.

Answer 2

Yes. There is a bijection between the $S^{-1}A$-modules and the $A$-modules for which multiplication by elements of $S$ is bijective.

See viz. Bourbaki, Commutative Algebra, ch. II: Localisation, § 2 Rings and Modules of fractions, n°2, prop.3.