Difference between revisions

campbell-galois

(page created)

Revision 241 · 2026-09-30 05:09 UTC
atlas import import · claude-opus-5-5
“Written for the atlas by Claude Opus 5.5 at the direction of Danielle Fong (content/wiki/Statement__campbell-galois.md).”

Status and infobox

status: — → proved
tier: — → V3
kind: — → theorem
year: — → 1973
statement: — → Let $F:\mathbb C^n\to\mathbb C^n$ be polynomial with $\det JF\in\mathbb C^{\times}$. If the field extension $\mathbb C(F)\subset\mathbb C(x)$ is Galois, then $F$ is a polynomial automorphism.
short: — → If ℂ(x)/ℂ(F) is Galois, a Keller map F is an automorphism

Relations

+ special case of jc-n: Campbell's theorem says that if this extension is Galois, the map is invertible, so the Jacobian conjecture holds for every Keller map whose function field extension is Galois.

Source +17 −0

+---

+title: Campbell's theorem: Galois Keller maps are automorphisms

+short: If ℂ(x)/ℂ(F) is Galois, a Keller map F is an automorphism

+kind: theorem

+status: proved

+tier: V3

+year: 1973

+authors: L. A. Campbell (over ℂ, 1973); M. Razar (1979) and D. Wright (1981) over any field of characteristic 0

+statement: Let $F:\mathbb C^n\to\mathbb C^n$ be polynomial with $\det JF\in\mathbb C^{\times}$. If the field extension $\mathbb C(F)\subset\mathbb C(x)$ is Galois, then $F$ is a polynomial automorphism.

+---

+For a Keller map, the degree of $\mathbb C(x)/\mathbb C(F)$ is the number of preimages of a general point. Campbell's theorem says that if this extension is Galois, the map is invertible, so [[special-case-of::Statement:jc-n|the Jacobian conjecture holds for every Keller map whose function field extension is Galois]].<ref>L. A. Campbell, A condition for a polynomial map to be invertible, Math. Ann. 205 (1973) 243–248, [doi:10.1007/BF01349234](https://doi.org/10.1007/BF01349234). M. Razar, Polynomial maps with constant Jacobian, Israel J. Math. 32 (1979) 97–106. D. Wright, On the Jacobian conjecture, Illinois J. Math. 25 (1981) 423–440.</ref>

+Every extension of degree 2 in characteristic zero is Galois. So no counterexample, in any dimension, has geometric degree 2, and the least possible degree is 3. Alpöge's map attains it, and its cubic extension is not Galois: its Galois closure has group $S_3$. The research for this page deduced that from the theorem and the fiber count; it is not quoted from a source.

+In the plane, more is known. Degree 3 was excluded by Orevkov (1986) and degree 4 by Domrina, and Żołądek proved the conjecture for every topological degree up to 5. A preprint claims every prime degree is excluded (see [[Statement:prime-degree-plane|prime degree in the plane]]).

+<references />

Revision 241 as rendered

For a Keller map, the degree of $\mathbb C(x)/\mathbb C(F)$ is the number of preimages of a general point. Campbell's theorem says that if this extension is Galois, the map is invertible, so the Jacobian conjecture holds for every Keller map whose function field extension is Galois.^[1]

Every extension of degree 2 in characteristic zero is Galois. So no counterexample, in any dimension, has geometric degree 2, and the least possible degree is 3. Alpöge's map attains it, and its cubic extension is not Galois: its Galois closure has group $S_3$ . The research for this page deduced that from the theorem and the fiber count; it is not quoted from a source.

In the plane, more is known. Degree 3 was excluded by Orevkov (1986) and degree 4 by Domrina, and Żołądek proved the conjecture for every topological degree up to 5. A preprint claims every prime degree is excluded (see prime degree in the plane).

↑ L. A. Campbell, A condition for a polynomial map to be invertible, Math. Ann. 205 (1973) 243–248, doi:10.1007/BF01349234. M. Razar, Polynomial maps with constant Jacobian, Israel J. Math. 32 (1979) 97–106. D. Wright, On the Jacobian conjecture, Illinois J. Math. 25 (1981) 423–440.