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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.

Eric Wofsey
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user4601931
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  • http://en.wikipedia.org/wiki/Injective_object – Martin Brandenburg Sep 09 '14 at 20:25
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    Pick any category without enough projectives, and consider its dual category :-) See http://mathoverflow.net/questions/5378/when-are-there-enough-projective-sheaves-on-a-space-x for examples. – Mariano Suárez-Álvarez Sep 09 '14 at 20:29
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    A category with no nonzero projective objects, but which has arbitrary products and coproducts is the category of abelian $p$-groups ($p$ a prime). Its opposite category has no non zero injective object. – egreg Sep 09 '14 at 20:30

2 Answers2

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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.

Community
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zcn
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  • Thank you for your answer. Do you know if there is a category of sheaves without enough injectives? – user4601931 Sep 09 '14 at 20:33
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    @dmdmdmdmdmd: This category is equivalent to coherent $\mathcal{O}_{\text{Spec}(\mathbb{Z})}-$modules – zcn Sep 09 '14 at 20:35
  • Right, of course. Thanks again! – user4601931 Sep 09 '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 09 '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
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    It is unclear to me how you use the implication "An injective object in this category must be an injective object in the full category of Z-modules, e.g. by Baer's criterion." in this argument. Could you please explain what this is used for? – gen May 29 '19 at 13:57
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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.

Dave
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