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If $G$ is a profinite group, then the category $Mod(G)$ of discrete $G$-modules has sufficiently many injectives (Neukirch, Schmidt, Wingberg: Cohomology of Number Fields, 2.6.5).

Since the cited book says nothing on projectives within this category, I guess $Mod(G)$ doesn't have sufficiently many projectives. Can you give me an example of a discrete $G$-module $M$ such that there is no epimorphism $P \to M$ with $P$ projective ? If there are little projectives, is it even possible to classify them ?

Added: Note that a discrete $G$-module $M$ is an abelian group (with the discrete topology) with a continuous $G$-action $G \times M \to M$ (N-S-W, 1.1.5). For example $\mathbb{Z}$ with trivial $G$-action is a discrete $G$-module.

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Discrete modules over a profinite group $G$ are the same thing as comodules over its group coalgebra $\mathbb Z(G)$. There aren't supposed to be many (or, generally speaking, any) projectives in the abelian category of comodules. One way to see this is to notice that infinite products of discrete $G$-modules aren't exact functors. Infinite products are always exact in an abelian category with enough projectives.

There are enough projectives in the abelian category of $\mathbb Z(G)$-contramodules, though. This is another kind of module category associated with a coalgebra or coring (and, in particular, with a profinite group). So the familiar algebra/discrete group situation of a module category with enough projectives and injectives splits in two halves when one passes to coalgebras/profinite groups---with two different kinds of module categories, one having enough injectives and another enough projectives.

Reference: Eilenberg, Moore. Foundations of relative homological algebra. Memoirs AMS vol.55, 1965.

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  • $\begingroup$ Why is a discrete $G$-module the same as a comodule over the coalgebra $\mathbb{Z}G$ ? $\endgroup$
    – KBuck
    Commented Mar 13, 2013 at 21:58
  • $\begingroup$ Having a discrete $G$-module $M$ means that to every element of $M$ you have assigned a locally constant function on $G$ with values in $M$. So does having a $\mathbb Z(G)$-comodule $M$. (One still has to check that the associativity and unity equations are the same in these two points of view, of course.) $\endgroup$
    – Leonid Positselski
    Commented Mar 14, 2013 at 14:24
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Here is a link a proof that the category $\text{Mod}(G)$ does not have enough projectives, when $G$ is an infinite profinite group. I think the proof is fairly elementary. There might be a few details missing though - sorry about that.

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  • $\begingroup$ The link is not working. $\endgroup$
    – Dietrich Burde
    Commented Aug 30, 2016 at 22:39

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