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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?

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    $\begingroup$ Is $\mathbb{Z}_p$ the $p$-adics or $\mathbb{Z}/p$? If the latter you can't prove this: take $G = C_2$, $M = (\mathbb{Z}/2)G$ which is not projective for $\mathbb{Z}G$ but is for $\mathbb{F}_2 G$. $\endgroup$
    – M T
    Commented Mar 14, 2012 at 8:25
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    $\begingroup$ This is a local to global problem. It is not necessarly the case that if all localizations are projective, then the module itself is projective. You might find it useful to look up "Swan modules". $\endgroup$
    – Geoff Robinson
    Commented Mar 14, 2012 at 8:26
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    $\begingroup$ @mt: $\mathbb{Z}_p$ are the $p$-adic numbers. The finite field I'd call $\mathbb{F}_p$. $\endgroup$
    – j.p.
    Commented Mar 14, 2012 at 8:46
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    $\begingroup$ Youm might try "Methods of Representation Theory" by Curtis and Reiner, esp part II. It's pretty comprehensive. $\endgroup$
    – Geoff Robinson
    Commented Mar 14, 2012 at 9:04
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    $\begingroup$ @Geoff: If I remember correctly, a $RG$-module that is projective as $R$module is projective over $RG$ iff it's projective over $R/pR[G]$ for all $p$ dividing $|G|$. $\endgroup$
    – Ralph
    Commented Mar 14, 2012 at 19:41

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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.

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  • $\begingroup$ This is the relevant reference, though as you imply CR state it in more generality. One naive question is what happens here concerning the projective property when the prime $p$ doesn't divide the group order? In that case the ordinary and $p$-modular representation theories agree over the relevant fields (with modules being automatically projective), but I'm less clear about what is happening over the intermediate ring. $\endgroup$
    – Jim Humphreys
    Commented Mar 14, 2012 at 22:53
  • $\begingroup$ @Jim. If p does not divide the order of the group, a f.g. $\mathbb Z_p G$ module is projective iff the underlying $\mathbb Z_p$-module is. That the condition is necessary is clear. That it is sufficient follows eg from the spectral sequence $ H^i(G,Ext^j_{Z}(M,N)) => Ext^{i+j}_{ZG}(M,N)$ since $G$ has no cohomology on $\mathbb Z_p G$-modules (invariants are the image of the usual idempotent). By the way, the assumption that $M$ is a free $\ZM$-module in the CR reference above corresponds to the "extended" assumption on localisations at all primes in the first sentence of my answer $\endgroup$
    – Jef
    Commented Mar 15, 2012 at 8:06
  • $\begingroup$ A slightly different way of looking at Jef's comment is as follows. If $p$ does not divide the order of $G$, then $Λ:=\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order and hence is a hereditary ring. This implies that every submodule of a free (left) $\Lambda$-module is projective. Of course, if M is a $\Lambda$-module, then M is a submodule of a free Λ module if and only if M is free as a $\mathbb{Z}_p$ module. $\endgroup$
    – Henri Johnston
    Commented Mar 15, 2012 at 10:54
  • $\begingroup$ @Jef: This can also be seen by elementary means: Let $P$ be $\mathbb{Z}_p$-projective and let $f: M \to P$ be an epi. of $\mathbb{Z}_pG$-modules. If $g: P \to M$ a $\mathbb{Z}_p$-linear spliting of $f$ then $1/|G|\;\operatorname{tr}(g): P \to M$ is a $\mathbb{Z}_pG$-linear splitting of $f$ (tr denotes the usual transfer). This works over any Ring in which $|G|$ is invertible. $\endgroup$
    – Ralph
    Commented Mar 15, 2012 at 23:40
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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 $\mathbb Z G$.

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