This problem is from Vakil's Foundations of Algebraic Geometry (Exercise 1.3.L.):
Problem If $S$ is a multiplicative subset of a ring $A$ and $M$ is an $A$-module, describe a natural isomorphism $$(S^{-1}A)\otimes_A M \simeq S^{-1}M$$ (as $S^{-1}A$-modules and as $A$-modules).
My progress:
- I suspect the isomorphism we are looking for to be $\frac{a}{s} \otimes m \mapsto \frac{am}{s}$. This map exists since the map $(S^{-1}A)\times M$, $(\frac{a}{s}, m) \mapsto \frac{am}{s}$ is bilinear, so the existence follows from the universal property of tensor products.
- It can be seen easily that the predicted map is surjective. However, I find it not easy to see that it is injective.
- That is why I tried to find an inverse instead and think that $$ S^{-1}M \to (S^{-1}A)\otimes_A M, \quad \frac{m}{s} \mapsto \frac{1}{s} \otimes m$$ is the map we are looking for. Now I have trouble to show that this map is well-defined, i.e. for $\frac{m}{s} = \frac{m'}{s'}$ (equivalently, there is a $t \in S$ such that $t(s'm - sm') =0$) we must have $\frac{1}{s'} \otimes m' = \frac{1}{s} \otimes m$. I cannot see how this can be shown. My hope is it to use the equivalence relation defined on $S^{-1}A$, but without success.
Could you please help me with this exercise? That would be nice!