Let $R$ be an associative ring with identity and $\mathrm{mod}R$ be the category of finitely presented $R$-modules. I would like to know when the category $\mathrm{mod}R$ is abelian.


I know that if $R$ is noetherian, then $\mathrm{mod}R$ is an abelian category. Maybe, it could be the case that if $\mathrm{mod}R$ is abelian, then $R$ must be noetherian. Is it true?



When is the category of finitely presented modules abelian?
Wojowu's idea is right:


**Lemma.** *Let $R$ be a ring, let $\mathbf{Mod}\_R$ be the category of (left) $R$-modules, and let $\mathbf{Mod}\_R^{\text{fp}}$ be the subcategory of finitely presented modules. Then the following are equivalent:*


1. *$R$ is left coherent, i.e. every finitely generated left ideal is finitely presented;*
2. *$\mathbf{Mod}\_R^{\text{fp}}$ is a weak Serre subcategory of $\mathbf{Mod}\_R$;*
3. *$\mathbf{Mod}\_R^{\text{fp}}$ is abelian and the inclusion $\mathbf{Mod}\_R^{\text{fp}} \hookrightarrow \mathbf{Mod}\_R$ is exact;*
4. *$\mathbf{Mod}\_R^{\text{fp}}$ is abelian.*


*Proof.* Implication 3 $\Rightarrow 4$ is obvious; implication 2 $\Rightarrow$ 3 is Tag [0754](https://stacks.math.columbia.edu/tag/0754); and 1 $\Rightarrow$ 2 is well known (in the commutative case, see Tags [05CW](https://stacks.math.columbia.edu/tag/05CW) and [05CX](https://stacks.math.columbia.edu/tag/05CX); the proofs generalise without difficulty to the general case).


For 4 $\Rightarrow$ 1, we note that the inclusion $\mathbf{Mod}\_R^{\text{fp}} \hookrightarrow \mathbf{Mod}\_R$ is left exact, as it is isomorphic to $\operatorname{Hom}\_R(R,-)$. Now if $I \subseteq R$ is a finitely generated left ideal, then $M = R/I$ is finitely presented, i.e. $M \in \mathbf{Mod}\_R^{\text{fp}}$. The morphism $R \to M$ is an epimorphism in $\mathbf{Mod}\_R$, so in particular in $\mathbf{Mod}\_R^{\text{fp}}$, so we get some short exact sequence
$$0 \to K \to R \to M \to 0$$
in $\mathbf{Mod}\_R^{\text{fp}}$. By left exactness, we conclude that $K = I$ is the ideal we started with, which therefore has to be finitely presented. Thus $R$ is left coherent as $I \subseteq R$ was arbitrary. $\square$


(I feel there might be some general reason why $\mathbf{Mod}\_R^{\text{fp}} \to \mathbf{Mod}\_R$ satisfies nice properties, which could shorten the above proof.)


**Example.** Easy examples of coherent rings $R$ that are not Noetherian are valuation rings. Every finitely generated ideal of $R$ is principal, but $R$ is Noetherian if and only if the valuation is discrete.


