I am just reading Ravi-Vakil's note's on algebraic geometry. There is a simple fact which I don't understand. I suppose I am missing something very simple. In one of the exercises, which I solved, he has the following comment: Can someone maybe shed some light on this statement more ?
Furthermore, an $S^{-1}A$-module is the same thing as n $A-$module for which $s \times . : M \rightarrow M$ is an A-module isomorphism for all $s \in S$.
Suppose $M$ is an $S^{-1}A$-module. Then $M$ inherits the structure of an $A$-module via the canonical localization morphism $A \to S^{-1}A$ (i.e., by setting $a \cdot m = (a/1) \cdot m$). For any $s \in A$, multiplication by $s$ is an $A$-module endomorphism of $M$; denote this by $\rho_{s}$. Furthermore, for any $s \in S$, $\rho_{s}$ has a set-theoretic inverse given by $\rho_{1/s} \colon M \to M, m \mapsto (1/s) \cdot m$ (check this! It boils down to how we defined the $A$-module structure on $M$). It is straightforward to verify that $\rho_{1/s}$ is $A$-linear, so $\rho_{s}$ is an $A$-module automorphism for all $s \in S$.
Conversely, suppose $M$ is an $A$-module such that $\rho_{s}$ is an automorphism of $A$-modules for all $s \in S$. Define an $S^{-1}A$-module structure on $M$ by $(a/s) \cdot m = a \cdot \rho_{s}^{-1}(m)$. I leave it to you to check that this is well-defined.