On \(x^6+x+a\) in characteristic three
type: Item
instance of: scholarly article
title: On \(x^6+x+a\) in characteristic three
zbMATH Open document ID: 1046.12004
DOI: 10.1023/A:1024711426896
published in: Designs, Codes and Cryptography
publication date: 2003-09-07T00:00:00Z
review text: Let \(F\) be a field of characteristic 3 and let \(a\in F\), \(a\neq 0\). The author shows that the 10 ways to factor \(x^6+ x +a\) into two cubics over the algeraic closure \(\overline F\) are in natural Galois bijection with the 10 roots of \(x^{10}+ ax+ 1\). The author uses this result to (1) prove the two polynomials have the same splitting field; (2) prove that a difference set constructed by Arasu and Player using the polynomial \(x^6+ x+ a\) is isomorphic to a difference set constructed by Dillon using the polynomial \(x^{10}+ x+ a\); (3) obtain a natural realization for the accidental isomorphism between the alternating group \(A_6\) and the special linear group \(\text{PSL}_2(9)\); and (4) characterize how \(x^6+ x+ a\) factors when \(F= \text{GF}(3^m)\) with \(m\) odd. For example, \(x^6+ x+ a\) is irreducible if and only if \(a\) can be written as \(\delta^{-36}+ \delta^4\) with \(\delta\in F^\times\) and \(\text{Tr}(\delta^5)\neq 0\). In particular, the author shows that the monic cubic factors of \(x^6+ x+ a\) over \(\overline F\) are in one-to-one correspondence with the roots of the polynomial \(x^{20}+ ax^2+ 1\) and that this correspondence is Galois-equivariant. If \(\mu\) is a root of \(x^{20}+ ax^2+ 1\), then the corresponding cubic factor of \(x^6+ x+ a\) is \(f_\mu(x)= x^3+ \mu^3 x^2+ (\mu- \mu^6)x- \mu^9+ \mu^4+ 1/\mu\). Also, \(x^6+ x+ a= f_\mu(x)f_{-\mu}(x)\). The author also shows that the roots of \(x^{10}+ ax+ 1\) are in one-to-one correspondence with the set of factorizations of \(x^6+ x+ a\) into two cubics and this correspondence is Galois-equivariant. If \(v\) is a root of \(x^{10}+ ax+ 1\), then \(\pm\sqrt{v}\) are roots of \(x^{20}+ ax^2+ 1\) and the factorization of \(x^6+ x+ a\) which corresponds to \(v\) is \(x^6+ x+ a= f(x)_{\sqrt{v}}f(x)_{-\sqrt{v}}\). The author shows that the polynomials \(x^6+ x+ a\), \(x^6+ x- a\), \(x^{20}+ ax^2+ 1\), \(x^{40}+ ax^4+ 1\), and \(x^{10}+ ax+ 1\) all have the same splitting field over \(F\) an…
reviewed by: John N. Mordeson
Mathematics Subject Classification ID: 12F10
Mathematics Subject Classification ID: 05B10
Mathematics Subject Classification ID: 12E05
Mathematics Subject Classification ID: 12E10
zbMATH DE Number: 1975929
zbMATH Keywords: difference set
zbMATH Keywords: projective polynomial
zbMATH Keywords: accidental isomorphism
zbMATH Keywords: splitting field
zbMATH Keywords: Galois group
author: Antonia Wilson Bluher
MaRDI profile type: Publication
description: scientific article; zbMATH DE number 1975929