Skip to main content
MathOverflow

Return to Question

replaced http://mathoverflow.net/ with https://mathoverflow.net/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

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 #241618Question #241618

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

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

@Qiaochu Yuan: added "of characteristic 0" to the setup. 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 in the remark.
Source Link
Qayum Khan
Qayum Khan
  • 687
  • 5
  • 13

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

SETUP: Let $R$ be a finitely generated, noetherian, integral domain. 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

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

Source Link
Qayum Khan
Qayum Khan
  • 687
  • 5
  • 13

Module-finiteness over the fixed subring

SETUP: Let $R$ be a finitely generated, noetherian, integral domain. 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