Projectives in the category of discrete G-modules  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. 
 A: 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.
A: 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. 
