Example of a Grothendieck category which is not Gabriel?

Question

Following Gabriel, for $\mathcal C$ a Grothendieck category, set $\mathcal T(\mathcal C)$ to be the localizing subcategory generated by the objects of finite length, and $\mathcal C' = \mathcal C / \mathcal T(\mathcal C)$ to be the localization thereat. This construction may be iterated: set $\mathcal C^{(0)} = \mathcal C$, $\mathcal C^{(\alpha+1)} = \mathcal C^{(\alpha)\prime}$ and at limit ordinals take intersections. By abstract nonsense this chain eventually stabilizes at some ordinal $\delta(\mathcal C)$, and $\mathcal C$ is said to be a Gabriel category if $\mathcal C^{(\delta(\mathcal C))} = 0$. The ordinal $\delta(\mathcal C)$ is called the Gabriel dimension of $\mathcal C$ (this term is usually reserved for the case when $\mathcal C$ is a Gabriel category, but let's agree to apply it even when $\mathcal C^{(\delta(\mathcal C))} \neq 0$).

Question: What is an example of a Grothendieck category which is not Gabriel?

• Equivalently, by looking at the $\mathcal C^{(\delta(\mathcal C))}$, I’m looking for a nonzero Grothendieck category with no simple objects.

• Gabriel shows that Gabriel categories are closed under localization at a localizing subcategory. So by the Gabriel-Popescu theorem, if there is a non-Gabriel category, then there is a non-Gabriel module category.

• I have a proof that every locally finitely generated Grothendieck category (in particular, every module category and hence every Grothendieck category) is Gabriel but I don’t trust it. The proof proceeds by observing that finite length objects are always finitely generated and finitely generated Grothendieck categories are closed under localization at finitely generated objects and every nonzero finitely generated object in a Grothendieck category admits a simple quotient.

Answer 1

If $B$ is a complete boolean algebra without atoms (seen as a locale), then the categories $Sh(B,Ab)$ or $Sh(B,R\text{-Mod})$ are Grothendieck abelian categories that have no simple objects, in fact no indecomposable objects.

If I understand correctly what you are saying, It follows that for any non-discrete regular topological space or locale $X$, the category $Sh(X,R\text{-Mod})$ is a Grothendieck abelian category that is not Gabriel, indeed if you localize it as the double negation topology (and kill off the eventual isolated point of X by localizing away from them) you'll get the example above, and as you said that the localization of a Gabriel category is again a Gabriel category, this implies that $Sh(X,R\text{-Mod})$ is not a Gabriel category. A more concrete special case of this is that if $B$ is an atomeless boolean algebra (ordinary this time, not a complete one), and you see $B$ as a Boolean ring, then $B$-Mod is not a Gabriel category, as $B$-Mod $\simeq Sh(Spec(B),\mathbb{Z}/2\mathbb{Z})$

Here is the proof of the claim above:

Essentially the point is that you can always decompose object in this category as follows: Assume $F$ is any object in $Sh(B,Ab)$ then for any element $U \in B$ we can construct the restriction $F|_U$ defined by $F|_U = i_* i^* U$ where $i:U \to X$ is the inclusion. or more explicitly $F|_U (V) = F(U \cap V)$, and we can always write that $F = F|_U \oplus F|_{\neg U}$.

To make an actual proof out of this, assume $F$ is simple (or just indecomposable) it follows from the abofe that for every $U \in B$ we have either $F|_U = 0 $ or $F|_{\neg U} =0$.

From this we can deduce that if $F$ is an indecomposable object, then if we define $P = \{V \in B| F|_V \neq 0 \}$ we can check that $P$ is a point of $B$

Indeed, $1 \in P$ because $F$ is non-trivial, if $\bigcup U_i \in P$ then $F$ has to be non trivial on at least one of the $U_i$. and if $F$ is non-trivial on $V$ and on $W$, then $F|_{\neg V} = 0$, so $F$ has to be concentrated on $W \cap V$. hence $W \cap V \in P$.

This concludes the proof as points of complete boolean algebra corresponds to atoms, and we assume there was no atoms in $B$.

Answer 2

I posted the following example of a Grothendieck category with no simple objects in answer to a question on math.stackexchange about seven and a half years ago. I seem to have said at the time that it was the easiest example I knew, and I was probably telling the truth.

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.