Proposition 110.77.1. 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 $\text{Mod}_ A/\mathcal{T} \rightarrow \text{Mod}_{S^{-1}A}$.

## 110.77 The category of modules modulo torsion modules

The category of torsion groups is a Serre subcategory (Homology, Definition 12.10.1) of the category of all abelian groups. More generally, for any ring $A$, the category of torsion $A$-modules is a Serre subcategory of the category of all $A$-modules, see More on Algebra, Section 15.53. If $A$ is a domain, then the quotient category (Homology, Lemma 12.10.6) is equivalent to the category of vector spaces over the fraction field. This follows from the following more general proposition.

**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. $\square$

Proposition 110.77.2. Let $A$ be a ring. Let $Q(A)$ denote its total quotient ring (as in Algebra, Example 10.9.8). Let $\text{Mod}_ A$ denote the category of $A$-modules and $\mathcal{T}$ its Serre subcategory of torsion modules. Let $\text{Mod}_{Q(A)}$ denote the category of $Q(A)$-modules. Then there is a canonical equivalence $\text{Mod}_ A/\mathcal{T} \rightarrow \text{Mod}_{Q(A)}$.

**Proof.**
Follows immediately from applying Proposition 110.77.1 to the multiplicative subset $S = \{ f \in A \mid f \text{ is not a zerodivisor in }A\} $, since a module is a torsion module if and only if all of its elements are each annihilated by some element of $S$.
$\square$

Proposition 110.77.3. Let $A$ be a Noetherian integral domain. Let $K$ denote its field of fractions. Let $\text{Mod}_ A^{fg}$ denote the category of finitely generated $A$-modules and $\mathcal{T}^{fg}$ its Serre subcategory of finitely generated torsion modules. Then $\text{Mod}_ A^{fg}/\mathcal{T}^{fg}$ is canonically equivalent to the category of finite dimensional $K$-vector spaces.

**Proof.**
The equivalence given in Proposition 110.77.2 restricts along the embedding $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_ A/\mathcal{T}$ to an equivalence $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Vect}_ K^{fd}$. The Noetherian assumption guarantees that $\text{Mod}_ A^{fg}$ is an abelian category (see More on Algebra, Section 15.53) and that the canonical functor $\text{Mod}_ A^{fg}/\mathcal{T}^{fg} \to \text{Mod}_ A/\mathcal{T}$ is full (else torsion submodules of finitely generated modules might not be objects of $\mathcal{T}^{fg}$).
$\square$

Proposition 110.77.4. The quotient of the category of abelian groups modulo its Serre subcategory of torsion groups is the category of $\mathbf{Q}$-vector spaces.

**Proof.**
The claim follows directly from Proposition 110.77.2.
$\square$

## 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)