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I'm learning about injective resolutions and derived functor sheaf cohomology, and it seems that every source on injective resolutions gives no examples. I feel like just one good example would make the concept so much less abstract and much easier to understand. There's the trivial example of $0 \to I \to I \to 0$, but that doesn't really help. And the world of injective modules is so rich, a good injective resolution would be a cool way to see a few. If I'm not mistaken, the following are injective $\mathbb Z$ modules:

  • $\mathbb Q$
  • $\mathbb Q / \mathbb Z$
  • The submodule of $\mathbb Q / \mathbb Z$ where we only allow the denominators to have certain prime factors
  • $0$
  • Direct sums of other injective $\mathbb Z$ modules

I think that maybe the reason nobody gives examples is that it's harder to come up with injective sheafs of $O$ modules for a sheaf of rings $O$, which seems to be the main case that people care about. I haven't tried to come up with any, so I'm not actually sure if it's harder. A good example in that category would really be amazing for my understanding of sheaf cohomology, but a good example for $\mathbb Z$ modules would help my understanding a lot too. So does anyone have a good example?

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    $\begingroup$ Products of skyscraper sheaves with injective stalks. $\endgroup$
    – Antonius
    Commented Mar 29 at 16:10
  • $\begingroup$ $\operatorname{Hom}_{\mathbb Z}(-, \mathbb Q / \mathbb Z)$ is a faithfully exact functor which sends flats to projectives. In the case of a noetherian scheme it is basically what's written in the comment above, but works over any ring just as well. $\endgroup$
    – Denis T
    Commented Mar 29 at 23:51
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    $\begingroup$ For a finite dimensional Hopf algebra $H$ over a field, projectives are injective. If $M$ is a finite dimensional $H$-module, you can take a $H$-projective resolution $P\to M^\ast$ of the dual $M^\ast=Hom_H(M,H)$. Then, dualizing yields an $H$-injective resolution $M\to P^\ast$. $\endgroup$
    – tj_
    Commented Mar 30 at 15:47

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It is a well-known fact that the category of abelian groups has enough injectives, hence actually any abelian group has an injective resolution. Furthermore, the global dimension of the ring $\mathbb{Z}$ is $1$, so actually these injective resolutions are bounded (and, actually, very short). For instance, we may consider the following injective resolution of the free abelian group $\mathbb{Z}$

$$ 0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0 \to \cdots $$

Where $\mathbb{Z} \to \mathbb{Q}$ is the canonical inclusion and $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ is the canonical projection map induced by the quotient. As you have already pointed out, $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$ are injective abelian groups and it is trivial that the sequence is exact. I don’t know whether this injective resolution is a “good example” but I would argue that is a rather canonical example of an injective resolution in the literature. For a nice application of this resolution, you may look up at Basic Homological Algebra, Osborne 3.4.Example 11 in which uses this resolution to prove that $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z}) \cong \mathbb{R}$ as abelian groups.

I am sorry I lack enough background to provide you with a more geometrical example. Nonetheless, I would like to provide also another elementary example with a rather different nature, which I believe to be very illustrative. We are going to consider now modules over the ring $\mathbb{Z}/(4)$. It is easy to see using Baer’s criterion that the free $\mathbb{Z}/(4)$-module $\mathbb{Z}/(4)$ is also injective. You can now verify that the following complex is, indeed, an injective resolution for the $\mathbb{Z}/(4)$-module $\mathbb{Z}/(2)$:

$$ 0 \to \mathbb{Z}/(2) \xrightarrow{\cdot2} \mathbb{Z}/(4) \xrightarrow{\cdot2} \mathbb{Z}/(4) \xrightarrow{\cdot 2} \cdots $$

Something interesting is that this resolution cannot be “improved” replacing it by a finite injective resolution. A way of proving this fact is noticing that $\mathrm{Ext}^n_{\mathbb{Z}/(4)}(\mathbb{Z}/(2),\mathbb{Z}/(2))$ never vanishes. It is quite remarkable that, even though $\mathbb{Z}/(4)$-modules can be though as abelian groups which have a trivial action with respect to the ideal $(4)$, the homological behaviour of both rings is incredibly different. One can see a similar behaviour with the rings $k[T]/(T^k),k \geq 2$.

I hope my answer has given you some insight. I would be happy to fill in the necessary gaps to help your understanding.

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