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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.

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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 $2$.

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  • $\begingroup$ Thank you for the answer, but there is something that I am not understanding. The inclusion given in the example induces a ring homomorphism $R^{G}\rightarrow S^{G}$ which is the restriction of the aforementioned inclusion. This is also an integral extension and $S^{G}$ is a finitely generated $R^{G}$-algebra. Namely, $S^{G}=R^{G}[xz,yz]$. Does it not imply that $S^{G}$ is a free $R^{G}$-module? $\endgroup$ – Don Rogelio Aug 15 '17 at 18:08
  • $\begingroup$ @DonRogelio: "Does it not imply that $S^G$ is a free $R^G$-module?" The tensor product of $S^G$ with the fraction field of $R^G$ is a $2$-dimensional vector space generated by the images of $1$ and either $xz$ or $yz = (v/u)\cdot xz$. If $S^G$ were a free $R^G$-module, it would be free of rank $2$. However, the tensor product of $S^G$ with the $R^G$-algebra $R^G/\mathfrak{m}$ is a $3$-dimensional vector space over the field $R^G/\mathfrak{m}$. Thus $S^G$ is not a free $R^G$-module. $\endgroup$ – Jason Starr Aug 16 '17 at 19:57
  • $\begingroup$ I see now that it is not free. Why is it not projective? $\endgroup$ – Don Rogelio Aug 19 '17 at 21:01
  • $\begingroup$ "I see now that it is not free. Why is it not projective?" Every finitely generated projective module is locally free. So if it were projective, there would be a nonzero divisor $f$ in $R^G\setminus \mathfrak{m}$ such that $S^G[f^{-1}]$ is free as a module over $R^G[f^{-1}]$. This leads to precisely the same contradiction with regards to the fraction field of $R^G[f^{-1}]$ and the residue field $R^G[f^{-1}] / \mathfrak{m} R^G[f^{-1}]$. $\endgroup$ – Jason Starr Aug 20 '17 at 9:56

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