**SETUP:** Let $R$ be a finitely generated, noetherian, integral domain of characteristic $0$. Let $G$ be a finite group that acts on $R$ by ring automorphisms.

**QUESTION:** *Is $R$ a finitely generated module over the fixed subring $R^G$?*

**REMARK:** Since $R$ has characteristic $0$, K R Nagarajan's 1968 counter-example does not apply. Since the order of $G$ is not a unit in $R$, G M Bergman's 1971 positive result does not apply, either. My secondary source for this information is page 13 of an e-presentation of Dobbs--Shapiro.

**COROLLARY 1:** *$R^G$ is noetherian,* since $R$ is, by the Eakin–Nagata theorem.

**COROLLARY 2:** *$R$ is integral over $R^G$,* by the so-called Determinantal Trick. Since $R$ is a domain, then this is true independently, by Exercise 13.2 in Eisenbud's book on commutative algebra. In particular, $\dim(R^G)=\dim(R)$ by the Cohen–Seidenberg theorems [Eisenbud: 4.15, 4.18].

**MOTIVATION:** I am a topologist, and the class of examples of the Question that occur for me are the group rings $R=\mathbb{Z}[A]$, where $A$ is a finite-rank free-abelian group, and where $G$ acts on $A$ by group automorphisms. *Is the Question true at least in this setting?* (Note that $R$ is not a hereditary ring.)

**SIMILAR POST:** Question #241618

Commutative algebraV, §1, no. 9, Theorem 2. $\endgroup$ – abx Jun 12 '16 at 7:41