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.


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 $2$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.