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

  • $\begingroup$ I don't understand your remark. Where are you assuming or showing that $R$ has characteristic $0$? $\endgroup$ Jun 12, 2016 at 4:48
  • $\begingroup$ @Qiaochu Yuan: Thank you --- I edited the setup to add "of characteristic 0." Some sort of characteristic-zero counter-example was produced in 1977 by C L Chuang and P H Lee. Sadly, I'm not able to get access to this article or the two mentioned in the remark, so I can't study them. $\endgroup$
    – Qayum Khan
    Jun 12, 2016 at 5:16
  • 2
    $\begingroup$ The only hypothesis you need is $R$ finitely generated (no characteristic 0, or japanese, or...). See Bourbaki's Commutative algebra V, §1, no. 9, Theorem 2. $\endgroup$
    – abx
    Jun 12, 2016 at 7:41
  • $\begingroup$ @abx: That is an absolutely perfect answer. It has taken me some time to get the book and verify the statement and the proof. Please shift your comment to an official answer, so that it is credited as answer and the question can be closed. For the record, the second part of that Theorem 2 (with ground ring $K=\mathbb{Z}$) states that the fixed subring $R^G$ is also finitely generated as a ring. $\endgroup$
    – Qayum Khan
    Jun 14, 2016 at 0:37
  • $\begingroup$ What you want is the Artin-Tate Lemma. $\endgroup$ Jun 14, 2016 at 5:57

1 Answer 1


At the request of Qayum Khan, I put my comment into an answer: the only hypothesis needed is $R$ finitely generated (no characteristic $0$, or japanese, or...). See Bourbaki's Commutative algebra V, §1, no. 9, Theorem 2.


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