# Abelian category without enough injectives

What is an example of an abelian category that does not have enough injectives? An example must exist, but I haven't been able to find one. If possible, a brief explanation of why the abelian category lacks enough injectives would be very appreciated.

Consider the category of finitely generated abelian groups, i.e. finitely generated $\mathbb{Z}$-modules. An injective object in this category must be an injective object in the full category of $\mathbb{Z}$-modules, e.g. by Baer's criterion. However, there are no nonzero finitely generated injective $\mathbb{Z}$-modules - see e.g. this answer.

• Thank you for your answer. Do you know if there is a category of sheaves without enough injectives? – user4601931 Sep 9 '14 at 20:33
• @dmdmdmdmdmd: This category is equivalent to coherent $\mathcal{O}_{\text{Spec}(\mathbb{Z})}-$modules – zcn Sep 9 '14 at 20:35
• Right, of course. Thanks again! – user4601931 Sep 9 '14 at 20:36
• But if $X$ is any ringed space, then the category $\mathcal{O}_X$-modules has enough injectives. If $X$ is a scheme, then the subcategory of quasi-coherent modules also has enough injectives. – Martin Brandenburg Sep 9 '14 at 23:48
• Another viewpoint would be that the category of sheaves is a Grothendieck category, which all have enough injectives. – archipelago Sep 10 '14 at 11:01

Take the category of finitely generated $\mathbf{Z}$-modules. Since $\mathbf{Z}$ is Noetherian, it's an Abelian category.

But an injective object $I$ in this category must be a divisible Abelian group. For given $a \in I$, let $\varphi \colon \mathbf{Z} \to I$ be defined by $\varphi(1) = a$. The morphism must be able to be extended to a second copy of $\mathbf{Z}$ in which the first is embedded via the multiplication map by $n$.

On the other hand, no nonzero finitely generated Abelian group can be divisible. This results from the structure theorem for such groups.

• Thank you for taking the time to answer my question. – user4601931 Sep 9 '14 at 20:33