Proposition 109.76.1. 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)