Is a $p$-torsion-free $\mathbb{Z}_{(p)}G$-module with finite projective dimension projective? Let $G$ be a finite group, $\mathbb{Z}_{(p)}$ be the ring of p-local integers (localization of $\mathbb{Z}$ at $p\mathbb{Z}$).
Let $M$ be a $p$-torsion-free (i.e. $pm = 0$ implies $m=0$) $\mathbb{Z}_{(p)}G$-module with finite projective dimension. Does it follow that $M$ must be projective itself?
 A: This is a very old question but it hasn't been answered and it was just edited so let me take a shot at it. 
Tensor products will be taken over $\mathbb Z$ unless stated otherwise. 
First let me note that we will need to assume $M$ is finitely generated (equivalently finitely presented) over $\mathbb Z_{(p)}$ (equivalently $\mathbb Z_{(p)}[G]$), otherwise the result is false. Indeed, $\mathbb Z_{(p)}[G]$ is flat over $\mathbb Z_{(p)}$ so any resolution $0\to P_1\to P_0 \to \mathbb Q\to 0$ over $\mathbb Z_{(p)}$ with $P_1,P_0$ free yields a finite free resolution of $\mathbb Q[G]$ by tensoring with $\mathbb Z_{(p)}[G]$; but of course $\mathbb Q[G]$ is not projective (it has no maps to any free module)
With this in mind, assume $M$ is finitely presented. 
Let $0\to P_n\to ...\to P_0\to M\to 0$ be a finite projective resolution.
Given $N$ a $\mathbb Z_{(p)}[G]$-module, $|G|$ kills $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)$ for $*\geq 1$ (for general reasons  if $*>1$ : $\mathbb Z_{(p)}$ is a PID and $\mathbb Z_{(p)}[G]$-projective implies $\mathbb Z_{(p)}$-projective - for $*=1$, we have to have something like $M$ projective over $\mathbb Z_{(p)}$, which is implied by the hypotheses since $M$ is finitely generated over $\mathbb Z_{(p)}$ and $p$-torsion free, using the structure theorem over a PID)
But also for any prime $q\neq p$, $q$ is an isomorphism of $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)$ (for obvious reasons). So we will now focus on multiplication by $p$ : if it's an iso, then multiplication by $|G|$ will be an iso, and there aren't that many groups with zero as an automorphism. In fact, a lot of the time it won't be an iso, but it will be surjective, which is enough for our purposes. 
For that, we will first study $\mathbb F_p\otimes M$ as as $\mathbb F_p[G]$-module.
Since $M$ is $p$-torsion free, $\mathrm{Tor}^{\mathbb Z_{(p)}}_*(M,\mathbb F_p) = 0$ for $*>0$ so that $0\to P_n\otimes \mathbb F_p\to...\to P_0\otimes\mathbb F_p\to M\otimes \mathbb F_p\to 0$ is still exact. 
Now $-\otimes \mathbb F_p$ is left adjoint to restriction, which is exact, therefore it preserves projectives, so our exact sequence is in fact a projective resolution of $M\otimes \mathbb F_p$ over $\mathbb F_p[G]$. But over $\mathbb F_p[G]$, projectives are injectives (this is a general result, it uses the fact that $\mathbb F_p$ is a field and that $G$ is finite), so that a finite projective resolution implies that $M\otimes \mathbb F_p$ is $\mathbb F_p[G]$-projective. 
Moreover, if we take the same projective resolution, it shows that $\mathrm{Ext}^*(M,N\otimes \mathbb F_p)$ can be computed over $\mathbb F_p[G]$ with the resolution $(P_i)$, that is, it coincides with the computation over $\mathbb Z_{(p)}[G]$ . In particular, $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N\otimes \mathbb F_p) = 0$ for $*>0$. 
Now take $N$ any finitely generated module. Since $\mathbb Z_{(p)}$ is a PID, $N$ is of the form $\mathbb Z_{(p)}^n\oplus \bigoplus_i \mathbb Z/p^{n_i}$ as a $\mathbb Z_{(p)}$-module. Therefore $p^kN$ is $p$-torsion-free for $k$ large enough. 
But note that we have a short exact sequence $0\to pN\to N\to N\otimes \mathbb F_p \to 0$, so our above computations show that $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,pN)\to \mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)$ is surjective for any $*\geq 1$ (in fact it's an iso for $*>1$)
It follows (by iteration) that $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,p^kN)\to \mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)$ is also surjective, for the $k$ large enough that we chose above. 
So if we can show that $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)=0, *>0$ for $N$ $p$-torsion free, it will be enough for finitely generated modules because $p^k N$ is $p$-torsion free. 
But now if $N$ is $p$-torsion free, we have a short exact sequence $0\to N\to N\to N\otimes \mathbb F_p\to 0$, which yields that multiplication by $p$ is surjective on $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)$ by the same analysis as above.
Combining that with what we already said, it follows that multiplication by $|G|$ is surjective, but it's also $0$, so $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)=0, *>0$ for $N$ finitely generated $p$-torsion free, therefore for $N$ finitely generated, via the $p^k N$ trick above. 
We will again use the finite generation of $M$ to go from finitely generated $N$ to arbitrary $N$. 
Indeed, since $M$ is finitely presented (because of course over $\mathbb Z_{(p)}[G]$, finitely generated implies finitely presented), $\hom_G(M,-)$ commutes with filtered colimits. Since those colimits are exact over the category of modules, the same holds of $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,-)$.
Since any module is the filtered colimit of its finitely generated submodules, the $0$-ness result goes from finitely generated modules to arbitrary ones : $\mathrm{Ext}^*_{\mathbb Z_{(p)}[G]}(M,N)=0, *>0$ for any $N$. In other words, $M$ is projective, and we are done.
There might be a simpler proof, but hopefully I at least haven't made a mistake in this one. 
