Proposition 110.76.1. Let $A$ be a ring. Let $S$ be a multiplicative subset of $A$. Let $\text{Mod}_ A$ denote the category of $A$-modules and $\mathcal{T}$ its Serre subcategory of modules for which any element is annihilated by some element of $S$. Then there is a canonical equivalence $\text{Mod}_ A/\mathcal{T} \rightarrow \text{Mod}_{S^{-1}A}$.
Proof. The functor $\text{Mod}_ A \to \text{Mod}_{S^{-1}A}$ given by $M \mapsto M \otimes _ A S^{-1}A$ is exact (by Algebra, Proposition 10.9.12) and maps modules in $\mathcal{T}$ to zero. Thus, by the universal property given in Homology, Lemma 12.10.6, the functor descends to a functor $\text{Mod}_ A/\mathcal{T} \to \text{Mod}_{S^{-1}A}$.
Conversely, any $A$-module $M$ with $M \otimes _ A S^{-1}A = 0$ is an object of $\mathcal{T}$, since $M \otimes _ A S^{-1}A \cong S^{-1} M$ (Algebra, Lemma 10.12.15). Thus Homology, Lemma 12.10.7 shows that the functor $\text{Mod}_ A/\mathcal{T} \to \text{Mod}_{S^{-1}A}$ is faithful.
Furthermore, this embedding is essentially surjective: a preimage to an $S^{-1}A$-module $N$ is $N_ A$, that is $N$ regarded as an $A$-module, since the canonical map $N_ A \otimes _ A S^{-1}A \to N$ which maps $x \otimes a/s$ to $(a/s) \cdot x$ is an isomorphism of $S^{-1}A$-modules. $\square$
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