# When is the category of finitely presented modules abelian?

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?

• More likely is that this condition is equivalent to $R$ being coherent, but I am not sure if that's the case. Jan 23, 2021 at 15:10

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; and 1 $$\Rightarrow$$ 2 is well known (in the commutative case, see Tags 05CW and 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.

• A minor quibble: an additive functor that preserves monomorphisms is not necessarily left exact. However, a similar proof, applying $\operatorname{Hom}(R,-)$ to an exact sequence $0\to L\to M\to N$, rather than just to a monomorphism, shows that the inclusion is left exact. Jan 24, 2021 at 9:27
• @JeremyRickard of course, thanks! (For those of you who like myself are confused about this, take a look at this example on math SE.) Jan 24, 2021 at 18:10