Let $R$ be a local Noetherian ring which contains the field $\mathbb{Q}$ of rational numbers, let $G$ be a finite group acting on $R$, and let $R^G \subseteq R$ be the fixed points for the action of $G$. How do I see that $R^G$ is also a local Noetherian ring, which is Cohen-Macaulay if $R$ is Cohen-Macaulay?
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$\begingroup$ I don't understand the question. Are you not assuming that $R$ is local? In that case, take $G$ to be trivial... $\endgroup$– Qiaochu YuanCommented Jun 7, 2016 at 5:58
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$\begingroup$ @QiaochuYuan ah sorry my bad, you're right. Fixed. $\endgroup$– user64623Commented Jun 7, 2016 at 6:02
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1$\begingroup$ The map $R^G \subseteq R$ is finite and so it follows that $R$ local implies that $R^G$ local. The rest is then the content of the Hochster-Roberts Theorem. $\endgroup$– David Benjamin LimCommented Jun 7, 2016 at 6:05
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In general, the ring of invariants of a finite group acting on a Noetherian ring need not be Noetherian. Counterexamples were given by Nagata. But if $|G|$ is inverible in $R$, as in your case, then $R^G$ is indeed Noetherian. See for example "Nagata, Masayoshi: Some questions on rational actions of groups". Moreover, under these circumstances if $R$ is CM then also $R^G$ is CM (Hochster-Eagon Theorem, see Prop. 12 of "Hochster-Eagon: Cohen-Macaulay Rings, Invariant Theory, and the Generic Perfection of Determinantal Loci".
answered Jun 7, 2016 at 6:42