On power integral bases for certain pure sextic fields

Lhoussain El Fadil

doi:10.5269/bspm.42373

Abstract

In their paper [1], Shahzad Ahmad et al. given a characterization on any pure sextic number field Q(m1/6) with square-free integers m satisfying m 6 ±1 (mod 9) to have a power integral bases or do not. In this paper, for these results, we give a new easier proof than that given in [1]. We further investigate the cases m 1 (mod 4) independently to the satisfaction of m2 1 (mod 9), m 1 (mod 9), and the number fields defined by x2r3t−m, where r, t are two non-negative integers, and m is a square free integer are investigated. The proposed proofs are based on Dedekind’s criterion and on prime ideal factorization.

Highlights

If m ≡ 1(mod 8), 2ZK = p1p2p3p4 is the factorization into product of prime ideals of ZK, with f1 = f2 = 1 and f3 = f4 = 2 being the respective residue degrees
If m ≡ 5(mod 8), 2ZK = p1p2p3 is the factorization into product of prime ideals of ZK, with f1 = f2 = f3 = 2 being the respective residue degrees
In order to show Lemma 3.1, we recall some fundamental notions on Newton polygon techniques

Summary

El Fadil

3. Let f (x) = x2r3t − m ∈ Z[x], where r and t are natural integers. If m ≡ 2 or 3(mod 4) and m ≡ ∓1(mod 9), Z[α] is the ring of integers of K. If m ≡ 1(mod 4) and m ≡ ∓1(mod 9), K is monogenic. If m ≡ 1(mod 4) or m ≡ 1(mod 9), number K is not monogenic. 1. In [1], it was shown that if m ≡ 1(mod 4) satisfying m ≡ ∓1(mod 9), ZK is not monogenic. In Theorem 2.4, we show that if m ≡ 1(mod 4), ZK is not monogenic independently to the satisfaction of the condition m ≡ ∓1(mod 9). The investigation given in [1] does not cover the case m ≡ ∓1(mod 9)

Proofs

Hence for every θ ∈ ZK , 2 divides the index (ZK : Z[θ]) and ZK can not have a power integral basis