This is part of Exercise 1.3.D in Vakil's Foundations of Algebraic Geometry:
Let $A$ be a ring and $S$ be a multiplicative subset of $A$.
To show: An $S^{-1}A$-module is the same thing as an $A$-module $M$ for which $s \times \_: M \to M$ is an isomorphism for all $s \in S$.
My approach:
- If we have an $S^{-1}A$-module $M$, we obtain an $A$-module $M$ by defining the scalar multiplication by $$a \times_{A} m := \frac{a}{1} \times_{S^{-1}A} m.$$ Then $s \times_A \_$ is isomorphic because $$M \to M, \quad m \mapsto \frac{1}{s} \times_{S^{-1}A} m$$ is its inverse.
- Conversely, if we have an $A$-module $N$ such that $s \times_A \_: N \to N$ is an isomorphism for all $s \in S$, then $N$ is an $S^{-1}A$-module by defining the scalar multiplication by $$ \frac{a}{s} \times_{S^{-1}A} n := a \times_A (s \times_A \_)^{-1}(n) .$$
I got confused by my construction and could not verify after certain attempts that these two maps are isomorphisms.
Could you please help me with this exercise? Thank you!