## 15.53 Abelian categories of modules

Let $R$ be a ring. The category $\text{Mod}_ R$ of $R$-modules is an abelian category. Here are some examples of subcategories of $\text{Mod}_ R$ which are abelian (we use the terminology introduced in Homology, Definition 12.10.1 as well as Homology, Lemmas 12.10.2 and 12.10.3):

The category of coherent $R$-modules is a weak Serre subcategory of $\text{Mod}_ R$. This follows from Algebra, Lemma 10.90.3.

Let $S \subset R$ be a multiplicative subset. The full subcategory consisting of $R$-modules $M$ such that multiplication by $s \in S$ is an isomorphism on $M$ is a Serre subcategory of $\text{Mod}_ R$. This follows from Algebra, Lemma 10.9.5.

Let $I \subset R$ be a finitely generated ideal. The full subcategory of $I$-power torsion modules is a Serre subcategory of $\text{Mod}_ R$. See Lemma 15.87.5.

In some texts a

*torsion module*is defined as a module $M$ such that for all $x \in M$ there exists a nonzerodivisor $f \in R$ such that $fx = 0$. The full subcategory of torsion modules is a Serre subcategory of $\text{Mod}_ R$.If $R$ is not Noetherian, then the category $\text{Mod}^{fg}_ R$ of finitely generated $R$-modules is

**not**abelian. Namely, if $I \subset R$ is a non-finitely generated ideal, then the map $R \to R/I$ does not have a kernel in $\text{Mod}^{fg}_ R$.If $R$ is Noetherian, then coherent $R$-modules agree with finitely generated (i.e., finite) $R$-modules, see Algebra, Lemmas 10.90.5, 10.90.4, and 10.31.4. Hence $\text{Mod}^{fg}_ R$ is abelian by (1) above, but in fact,in this case the category $\text{Mod}_ R^{fg}$ is a (strong) Serre subcategory of $\text{Mod}_ R$.

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