# When is the augmentation ideal projective as RG-module?

Let $$G$$ be a finite group and let $$R$$ be a commutative ring.

I'd like to ask, if there is a theorem of the following kind:

The augmentation ideal $$I_G$$ is projective as RG-module, if and only if ... ?

This should happen only in rare cases, but I was wondering, if there exists an if-and-only-if criterion.

Thank you very much for the help.

• I'm not sure about general rings, but this happens never for $R=\mathbb{Z}$, or for $R=\mathbb{Z}_p$ and $p$ a prime dividing the order of $G$. This is due to the fact that characters of projective $\mathbb{Z}_p[G]$-modules vanish on elements of order divisible by $p$. Jun 16, 2021 at 20:14
• The general result for arbitrary groups is mentioned in mathoverflow.net/questions/297043/… Jun 16, 2021 at 23:17
• Namely G has projective augmentation ideal iff it is the fundamental group of a graph of finite groups with order invertible in R. Equivalenty G acts on a tree and all the vertex and edge stabilizers are finite groups with order invertible in R Jun 17, 2021 at 0:50
• I mentioned the characterization via cohomological dimension in this post but it is nice to see the more concrete consequence in the solutions below. Jun 22, 2021 at 14:53

Okay, this happens precisely in the obvious case, namely if all primes dividing the order $$|G|$$ are invertible in $$R$$.
To see this, note that $$\operatorname{Ext}^*_{R[G]}(R,R)$$ is group cohomology of $$G$$ with coefficients in $$R$$. If $$I_G$$ were projective, $$I_G\to R[G]$$ would be a projective resolution of $$R$$ and thus the cohomology would be $$1$$-dimensional. However, the cohomology of $$G$$ with coefficients in $$\mathbb{Z}$$ has $$p$$-torsion in arbitrarily high degrees for all primes dividing $$|G|$$ (this is a well-known fact about cohomology of finite groups). So by the universal coefficient theorem, the cohomology with $$R$$-coefficients is unbounded, except if $$p$$ is invertible in $$R$$ for each $$p$$ dividing $$|G|$$.