So this may be a trivial question but I am new to the idea of group rings. Suppose we have a ring $R$ and a group $G$, I was wondering if the trivial $RG$-module $R$ is projective? In which case, how should I go about proving this? Thanks for any help in this.
Sam
The free module $RG$ is certainly projective, and there is a surjective $RG$-module homomorphism $\varphi:RG\to R$ where $$\varphi\left(\sum_{g\in G}a_gg\right)=\sum_{g\in G}a_g,$$ so $R$ is projective if and only if $\varphi$ splits. By considering possible splitting maps, it's easy to see that this is the case if and only if $G$ is finite and $|G|$ is a unit in $R$:
For suppose $\theta:R\to RG$ is a splitting map, so that $\varphi\theta=\operatorname{id}_R$. For $\theta$ to be an $RG$-module homomorphism, $\theta(1)g=\theta(1)$ for all $g\in G$. If $G$ is infinite then there are no non-zero elements of $RG$ with this property, and if $G$ is finite then the only possibility is $\theta(1)=b\sum_{g\in G}g$ for some $b\in R$.
In the second case, $1=\varphi\theta(1)=b|G|$, and so $|G|$ is a unit. Conversely, if $|G|$ is a unit, then we can take $b=|G|^{-1}$ to get a splitting map.