Orders in the Euclidean court

Abstract Following Johnson, Queen, and Sevilla, we call an order $$\mathcal {O}$$ O in a number field K if, for all $$\alpha ,\beta \in \mathcal {O}$$ α , β ∈ O with $$\beta \ne 0$$ β ≠ 0 for which $$\alpha \mathcal {O}+\beta \mathcal {O}$$ α O + β O is a principal ideal of $$\mathcal {O}$$ O , there are $$\gamma , \rho \in \mathcal {O}$$ γ , ρ ∈ O with $$\alpha =\beta \gamma + \rho $$ α = β γ + ρ and $$|N_{K/\mathbb {Q}}(\rho )| < |N_{K/\mathbb {Q}}(\beta )|$$ | N K / Q ( ρ ) | < | N K / Q ( β ) | . We prove that only finitely many quadratic orders are generalized Euclidean. Additionally, assuming GRH, we show that for each fixed $$d > 1,$$ d > 1 , there are only finitely many integers m for which $$T^d-m$$ T d - m is irreducible over $$\mathbb {Q}$$ Q and $$\mathbb {Z}[m^{1/d}] = \mathbb {Z}[T]/(T^d-m)$$ Z [ m 1 / d ] = Z [ <

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Publication Details

Journal
Archiv der Mathematik
Published
2026-09-12
DOI
https://doi.org/10.1007/s00013-026-02288-1
Primary Topic
Algebraic Geometry and Number Theory
Type
article
Field-Weighted Citation Impact
0.00
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article

Orders in the Euclidean court

Enrique Treviño, Kevin J. McGown, Paul Pollack, Carlo Francisco Adajar
Archiv der Mathematik
Algebraic Geometry and Number Theory
article

Orders in the Euclidean court

Enrique Treviño, Kevin J. McGown, Paul Pollack, Carlo Francisco Adajar
article en

Abstract

Abstract Following Johnson, Queen, and Sevilla, we call an order $$\mathcal {O}$$ O in a number field K if, for all $$\alpha ,\beta \in \mathcal {O}$$ α , β ∈ O with $$\beta \ne 0$$ β ≠ 0 for which $$\alpha \mathcal {O}+\beta \mathcal {O}$$ α O + β O is a principal ideal of $$\mathcal {O}$$ O , there are $$\gamma , \rho \in \mathcal {O}$$ γ , ρ ∈ O with $$\alpha =\beta \gamma + \rho $$ α = β γ + ρ and $$|N_{K/\mathbb {Q}}(\rho )| < |N_{K/\mathbb {Q}}(\beta )|$$ | N K / Q ( ρ ) | < | N K / Q ( β ) | . We prove that only finitely many quadratic orders are generalized Euclidean. Additionally, assuming GRH, we show that for each fixed $$d > 1,$$ d > 1 , there are only finitely many integers m for which $$T^d-m$$ T d - m is irreducible over $$\mathbb {Q}$$ Q and $$\mathbb {Z}[m^{1/d}] = \mathbb {Z}[T]/(T^d-m)$$ Z [ m 1 / d ] = Z [ <

Archiv der Mathematik
Lake Forest College (US), University of Georgia (US), California State University, Chico (US)
Openalex Percentile: Top 6%
Algebraic Geometry and Number Theory
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