Abelian category without enough injectives

Question

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.

Answer 1

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.

Answer 2

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.