Looking for criterion for $\mathbb{Z}G$-modules to be projective Given a finite group $G$ and a (finitely generated) $\mathbb{Z}G$-module $M$, assume that for each prime $p$ dividing the order $|G|$ of $G$ the $\mathbb{Z}_pG$-module $M^{\mathbb{Z}_p} = M\otimes\mathbb{Z}_p$ is projective.

How can I prove that $M$ is projective?

 A: As Geoff mentioned, there is a lot of material on this type of question in "Methods of Representation Theory" by Curtis and Reiner, though I think Volume 1 is more relevant here.
If we specialise Corollary (25.16) to the case you are interested in, then we get the following:
Let $G$ be a finite group of order $n$. Let $M$ be a finitely generated left $\mathbb{Z}[G]$-module. Suppose that $M$ is free as a $\mathbb{Z}$-module. Then $M$ is $\mathbb{Z}[G]$-projective if and only if $M \otimes_{\mathbb{Z}}\mathbb{Z}_{p}$ is $\mathbb{Z}_{p}[G]$-projective for each prime $p$ dividing $n$.
"Maximal Orders" by Reiner will probably also be a useful reference. 
A: You need to assume projectivity for all primes $p$, otherwise stupid counterexamples are easily found with $G$ the trivial group.
Assuming this, then for any finitely generated $\mathbb Z G$-module $N$, the Ext modules
$Ext^i_{\mathbb Z G}(M,N)$ are finitely generated $\mathbb Z$-modules, whose localisations satisfy
$$Ext^i_{\mathbb Z G}(M,N)\otimes_{\mathbb Z} \mathbb Z_p =
Ext^i_{\mathbb Z_p G}(M \otimes_{\mathbb Z} \mathbb Z_p,N\otimes_{\mathbb Z} \mathbb Z_p)=0.
$$
for $i>0$. To see this, pick a resolution of $M$ by free and finitely generated $\mathbb Z G$-modules and use the flatness of $\mathbb Z_p$ over $\mathbb Z$.
Therefore these Ext modules vanish and $M$ is projective over $\mathbb Z G$.
