# Is the category of torsion abelian groups a Grothendieck category?

I thought that I had shown that the category of $$\mathcal{A}$$ of all torsion abelian groups is a Grothendieck category:

1. All coproducts exist, they are just the coproducts of abelian groups;
2. The colimits are also just the colimits in the category of abelian groups, since the cokernel of a morphism of torsion groups is also a torsion group;
3. There is a generator.

The third point is the least obvious. Let $$G=\bigoplus_{n\in \mathbb{N}}\mathbb{Z}/n\mathbb{Z}$$ and $$A$$ be an arbitrary torsion abelian group. The morphism $$\bigoplus_{0\neq a\in A}\mathbb{Z}\to A$$, which takes the 1 of the summand corresponding to $$a$$ to $$a,$$ is an epimorphism in the category of abelian groups. But since $$A$$ is torsion, this filters through the map $$\bigoplus_{0\neq a\in A}\mathbb{Z}/\text{ord}(a)\mathbb{Z}\to A$$, which then naturally extends to a morphism $$\bigoplus_{0\neq a\in A}G\to A.$$

However, one of the comments at Complete but not cocomplete category says:

For example the category of torsion abelian groups is not grothendieck (there is no progenerator).

Although I do agree that there is no progenerator, the axioms of a Grothendieck category seemingly do not require its existence, only the existence of a not necessarily projective generator.

So in the end, is the category of torsion abelian groups actually Grothendieck?

• You have not shown the "Grothendieck" condition AB5: that direct unions preserve finite intersections, or equivalently that filtered colimits are exact. Jul 7, 2022 at 19:27
• @VladimirSotirov I thought that's what I did in 2: the colimits in this category are just the colimits in the category of all abelian groups, and those are exact. Jul 7, 2022 at 21:25

Given a (not necessarily commutative) ring $$A$$, a torsion class is a full subcategory of (left) $$A$$-modules closed under quotients, coproducts, and extensions. In particular it is closed under colimits. A hereditary torsion class is one that is also closed under subobjects. In particular, the fact the filtered colimits of $$A$$-modules are exact implies filtered colimits of such torsion modules are exact. Finally, any hereditary torsion class is generated by the cyclic $$A$$-modules that are in the class (see chapter VI, Proposition 3.6. in Bo Stenstrom's book Rings of Quotients - An Introduction to Methods of Ring Theory). Thus hereditary torsion classes should be Grothendieck categories.