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?

  • 9
    $\begingroup$ More likely is that this condition is equivalent to $R$ being coherent, but I am not sure if that's the case. $\endgroup$
    – Wojowu
    Jan 23, 2021 at 15:10

1 Answer 1


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.

  • 2
    $\begingroup$ 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. $\endgroup$ Jan 24, 2021 at 9:27
  • 1
    $\begingroup$ @JeremyRickard of course, thanks! (For those of you who like myself are confused about this, take a look at this example on math SE.) $\endgroup$ Jan 24, 2021 at 18:10

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