# Reference/ elementary proof of a result about projective dimension in group rings

Hello- I've had to use a result that sounds like it should be well-known, but I couldn't find any references, and my proof is rather unsatisfactory. I was hoping someone here could help! The problem is as follows:

Let $(K,O,k)$ be a p-modular system for a finite group $G$, and that $k$ is algebraically closed. Then $O$ is a complete discrete valuation ring with residue field $k$. For an $OG$-module $M$, let $O(M)$ be the projective dimension of $M$ as an $OG$-module. Since $O$ is a (local) PID, it follows that $O(M)$ is either infinite or 1 if $M$ has torsion. My question concerns $O(M)$ for a $kG$ module $M$:

In particular, is $O(M)=1$ if and only if $M$ is projective as a $kG$ module?

To hopefully persuade you that this is not trivial, consider the following. Call an $OG$-module $P$ $\textit{weakly projective}$ if it is a summand of a module induced from the trivial group, and note that a weakly projective $kG$-module is simply projective as a $kG$-module. Then $O(P)=1$, again because $O$ is a PID. It is however, possible to construct a non-weakly projective module $N$ with $O(N)=1$.

I can prove this, but I've had to use some rather heavy machinery- I've not so much used a sledgehammer to crack the nut, I've nuked it. I was hoping someone could offer a more elementary proof, or (better) a reference?

Perhaps a more intuitive way of thinking about it is the following equivalent problem: If $0\rightarrow P\rightarrow P\rightarrow M\rightarrow 0$ is a sequence of $OG$-modules with $P$ projective (and hence free as an $O$-module) and $M$ a $kG$-module, then the sequence is exact only if $M$ is projective (and hence isomorphic to $P\otimes_O k$, so that the first map in the sequence is multiplication by $\pi$, the generator of the maximal ideal in $O$.).

I hope the question is clear, and I would appreciate any insight!

-
I do not understand your equivalent reformulation. What are the morphisms in the sequence P -> P -> M? Or what do we know about this sequence? We must know something (which is not clear to me from what you've written) if there is to be any hope that a property exclusively of M will tell us whether or not the sequence is exact. Also, are all the P's in that paragraph isomorphic? –  Hugh Thomas Dec 15 '12 at 21:36
Sorry, I realise I wasn't particularly clear. My point is that any torsion module M with O(M)=1 would have a projective resolution of that form (The Ps being isomorphic follows from the fact that they would then afford the same character). So my question boils down to whether M being kG-projective is a necessary condition for there to be such an exact sequence. A priori, I do not assume anything else about the sequence. –  user26223 Dec 18 '12 at 13:41
I've also edited that last paragraph slightly- I hope it makes things clearer. –  user26223 Dec 18 '12 at 13:43
Thanks. So the equivalent statement is that a kG-module M admits a resolution of that form iff M is projective as a kG-module. Is that right? Further up in the question, you say that O(M) will be 1 or infinity. I'm confused. I would expect the projective dimension of a projective to be zero, so I would expect the projective dimension of O(M) to be zero or one. Can you clarify this? I should apologize in advance that, even once I understand the question, I may well not be able to help a lot in solving it. –  Hugh Thomas Dec 19 '12 at 5:57
Well, an OG-module can only be projective if it is free as an O-module (since free modules are O-free). Hence any module with torsion (i.e. a kG-module, which is annihilated by $\pi$) must have higher projective dimension. Thanks for the interest! –  user26223 Dec 19 '12 at 11:02