Proposition 107.69.1. 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 $\text{Mod}_ A/\mathcal{T} \rightarrow \text{Mod}_{S^{-1}A}$.

## 107.69 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 15.52. 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 10.11.15). 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 107.69.2. 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 107.69.1 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 107.69.3. 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 107.69.2 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 15.52) 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 107.69.4. 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 107.69.2.
$\square$

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