On the monogenicity of power-compositional Shanks polynomials

IF 0.5 Q3 MATHEMATICS
Lenny Jones
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引用次数: 1

Abstract

Let $f(x)\in \mathbb{Z}[x]$ be a monic polynomial of degree $N$ that is irreducible over $\mathbb{Q}$. We say $f(x)$ is \emph{monogenic} if $\Theta=\{1,\theta,\theta^2,\ldots ,\theta^{N-1}\}$ is a basis for the ring of integers $\mathbb{Z}_K$ of $K=\mathbb{Q}(\theta)$, where $f(\theta)=0$. If $\Theta$ is not a basis for $\mathbb{Z}_K$, we say that $f(x)$ is \emph{non-monogenic}.Let $k\ge 1$ be an integer, and let $(U_n)$be the sequence defined by \[U_0=U_1=0,\qquad U_2=1 \qquad \text{and}\qquad U_n=kU_{n-1}+(k+3)U_{n-2}+U_{n-3} \qquad \text{for $n\ge 3$}.\] It is well known that $(U_n)$ is periodic modulo any integer $m\ge 2$, and we let $\pi(m)$ denote the length of this period. We define a \emph{$k$-Shanks prime} to be a prime $p$ such that $\pi(p^2)=\pi(p)$. Let $\mathcal{S}_k(x)=x^{3}-kx^{2}-(k+3)x-1$ and $\mathcal{D}=(k^2+3k+9)/\gcd(3,k)^2$. Suppose that $k\not \equiv 3 \pmod{9}$ and that $\mathcal{D}$ is squarefree. In this article, we prove that $p$ is a $k$-Shanks prime if and only if $\mathcal{S}_k(x^p)$ is non-monogenic, for any prime $p$ such that $\mathcal{S}_k(x)$ is irreducible in $\mathbb{F}_p[x]$. Furthermore, we show that $\mathcal{S}_k(x^p)$ is monogenic for any prime divisor $p$ of $\mathcal{D}$. These results extend previous work of the author on $k$-Wall-Sun-Sun primes.
幂-复合Shanks多项式的单性性
设$f(x)\in \mathbb{Z}[x]$为次为$N$的一元多项式,在$\mathbb{Q}$上不可约。如果$\Theta=\{1,\theta,\theta^2,\ldots ,\theta^{N-1}\}$是$K=\mathbb{Q}(\theta)$的整数环$\mathbb{Z}_K$的基,我们说$f(x)$是\emph{单基因}的,其中$f(\theta)=0$。如果$\Theta$不是$\mathbb{Z}_K$的基础,我们说$f(x)$\emph{是非单基因}的。设$k\ge 1$为整数,$(U_n)$为\[U_0=U_1=0,\qquad U_2=1 \qquad \text{and}\qquad U_n=kU_{n-1}+(k+3)U_{n-2}+U_{n-3} \qquad \text{for $n\ge 3$}.\]定义的序列。众所周知,$(U_n)$对任意整数$m\ge 2$取周期模,我们设$\pi(m)$表示这个周期的长度。我们定义\emph{$k$-Shanks素数}为一个质数$p$,使得$\pi(p^2)=\pi(p)$。让$\mathcal{S}_k(x)=x^{3}-kx^{2}-(k+3)x-1$和$\mathcal{D}=(k^2+3k+9)/\gcd(3,k)^2$。假设$k\not \equiv 3 \pmod{9}$和$\mathcal{D}$是无平方的。在这篇文章中,我们证明$p$是一个$k$ -Shanks素数当且仅当$\mathcal{S}_k(x^p)$是非单基因的,对于任何素数$p$使得$\mathcal{S}_k(x)$在$\mathbb{F}_p[x]$上是不可约的。进一步,我们证明$\mathcal{S}_k(x^p)$对于$\mathcal{D}$的任何素数$p$都是单基因的。这些结果扩展了作者先前关于$k$ -Wall-Sun-Sun素数的工作。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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来源期刊
CiteScore
0.80
自引率
20.00%
发文量
14
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