Let $f:R\rightarrow S$ be a homomorphism of (commutative) rings with unity and suppose that $G$ is a group acting on $R$ and $S$ in such a way that $f\sigma=\sigma f$ for every $\sigma\in G$. Denote by $R^{G}$ and $S^{G}$ the respective rings of invariants of $G$, this is, $R^{G}=\left\{r\in R:\sigma r=r\textrm{ for every }\sigma\in G\right\}$ and $S^{G}=\left\{s\in R:\sigma s=s\textrm{ for every }\sigma\in G\right\}$. Then it is induced a ring homomorphism $f^{G}:R^{G}\rightarrow S^{G}$. Suppose also that $S$ is a projective $R$-module. Is it true that $S^{G}$ is a projective $R^{G}$-module?


Thank you.



Is projectivity preserved by invariants?
That is not true. Let $G$ be a cyclic group with $2$ elements, $\{e,\sigma\}.$ Let $k$ be a field of characteristic different from $2$. Let $R$ be $k[x,y]$ with $$\sigma(x)=-x,\ \ \sigma(y)=-y.$$ Let $S$ be $R[z]/\langle z^2-1\rangle$ with $$\sigma(x)=-x,\ \ \sigma(y)=-y,\ \ \sigma(z)=-z.$$ Then $R^G$ equals $$k[u,v,w]/\langle uw-v^2\rangle, \ \ u=x^2,\ v=xy,\ w=y^2.$$ Also $S^G$ equals $$k[s,t], \ \ s = xz, \ t=yz.$$ Although $S$ is a rank-$2$, free $R$-module with basis $1$ and $z$, the rank-$2$ $R^G$-module $S^G$ is not free. Indeed, the quotient of $S^G$ by the maximal ideal $\mathfrak{m}=\langle u,v,w\rangle$ of $R^G$ is $k[s,t]/\langle s^2,st,t^2\rangle$. This has rank $3$ as a $k$-vector space, not rank $2$.


