Let $\mathcal A$ be any Grothendieck abelian category and $0 \neq M \in \cal A$ an object. It is true that $M$ admits a simple subquotient?

It is certainly true for $\mathcal A=R-Mod$ since $M$ contains a cyclic module and any (left)ideal is contained in a maximal.

Motivation: In the category of modules over a ring $R$ the following are equivalent for an object $M$.

1) $M$ is a (maybe infinite) direct sum of simple modules.

2) Every short exact equence $$0 \rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$$ splits.

I want to generalize this statement to any abelian category and the above seems crucial for $2) \Rightarrow 1)$.

Edit: Hanno Becker informed me, that there are abelian categories without any irreducible objects, see Jeremy Rickards answer to this question. I changed my question accordingly.


Even with the extra condition that $\mathcal{A}$ is a Grothendieck category, it may still have no simple objects. I think the following is the easiest example I know.

Let $R$ be a (necessarily non-noetherian) commutative local ring with non-zero maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=\mathfrak{m}$. Let $\mathcal{C}$ be the category of $R$-modules and let $\mathcal{D}$ be the full subcategory of modules annihilated by $\mathfrak{m}$; i.e., of semisimple modules.

Then $\mathcal{D}$ is a full abelian subcategory of $\mathcal{C}$ closed under coproducts. An extension in $\mathcal{C}$ of two objects of $\mathcal{D}$ is a module for $R/\mathfrak{m}^2=R/\mathfrak{m}$, and so $\mathcal{D}$ is closed under extensions. So $\mathcal{D}$ is a localizing subcategory of $\mathcal{C}$, which implies that the quotient category $\mathcal{A}=\mathcal{C}/\mathcal{D}$ is a Grothendieck category.

Suppose $M$ is an $R$-module. $M$ has a maximal semisimple quotient $M'=M/M\mathfrak{m}$, and in turn $M\mathfrak{m}$ has a maximal semisimple submodule $M''=\operatorname{soc}(M\mathfrak{m})$.

Let $N=M\mathfrak{m}/M''$. Then $M\cong M\mathfrak{m}\cong N$ in $\mathcal{A}$. Since we know that semisimple modules are closed under extensions, $N$ can have no non-zero semisimple quotients or submodules without contradicting the maximality of the quotient $M/M\mathfrak{m}$ or the submodule $\operatorname{soc}(M\mathfrak{m})$.

Suppose $N'$ is a proper non-zero $R$-submodule of $N$. Since neither $N'$ nor $N/N'$ can be in $\mathcal{D}$, $N$ is not simple in $\mathcal{A}$.

So $\mathcal{A}$ is a Grothendieck category with no simple objects.


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