4
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.