About group ring and projective module

Question

Let $R$ be commutative ring and $G$ be finite group. Now $A=R[G]$ is the group ring of $G$ with coefficients in $R$. Give an example of right $A$-module which is finitely generated and projective as $R$-module, but not projective as $A$-module.
Is the other direction possible, that is, projective as $A$-module, but not as $R$-module?

Answer 1

Consider the case $R=\Bbb F_2$, $G=\Bbb Z/2\Bbb Z$. Since $R$ is a field, every $R$-module is projective, so the task reduces to finding a right $R[G]$-module that is not projective. Consider the module $M$ given by $\Bbb F_2$ with a trivial $G$-action. Then we have an exact sequence $0 \to M \to R \to M \to 0$ of right $R[G]$-modules. (Let $g$ be a generator of $G$ and consider the subspace in $R[G]$ spanned by $1+g$. $1+g$ is invariant, thus the span is isomorphic to $M$. Now check that $G$ acts trivially on the quotient)
This exact sequence doesn't split, as $G$ doesn't act trivially on $R$, hence $M$ is not projective.

Every projective $A$-module is projective as an $R$-module. To see this, note that $R[G]$ is free as an $R$-module, so every free $R[G]$-module is a free $R$-module. It follows that every direct summand of a free $R[G]$-module is a direct summand of a free $R$-module, so projective $R[G]$-modules are projective as $R$-modules.