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On the discriminant and index of a certain class of polynomials

Rupam Barman, Anuj Narode, Vinay Wagh

Abstract

Let $f(x) = (x^{2}+1)^{n} - a x^{n} \in \mathbb{Z}[x]$ and assume $f(x)$ is irreducible. Let $θ$ be a root of $f(x)$, set $K= \mathbb{Q}(θ)$, and denote by $\mathbb{Z}_{K}$ the ring of integers of $K$. The index of $f$, denoted $\operatorname{ind}(f)$, is the index of $\mathbb{Z}[θ]$ in $\mathbb{Z}_{K}$. A polynomial $f(x)$ is said to be monogenic if $\operatorname{ind}(f) = 1$. In this article, we explicitly compute the discriminant of the polynomial $f(x)$, and then derive necessary and sufficient conditions on the parameters $a$ and $n$ for $f(x)$ to be monogenic. Furthermore, we provide a complete description of the primes that divide $\operatorname{ind}(f)$.

On the discriminant and index of a certain class of polynomials

Abstract

Let and assume is irreducible. Let be a root of , set , and denote by the ring of integers of . The index of , denoted , is the index of in . A polynomial is said to be monogenic if . In this article, we explicitly compute the discriminant of the polynomial , and then derive necessary and sufficient conditions on the parameters and for to be monogenic. Furthermore, we provide a complete description of the primes that divide .
Paper Structure (3 sections, 5 theorems, 29 equations, 1 table)

This paper contains 3 sections, 5 theorems, 29 equations, 1 table.

Key Result

Theorem 1.1

Let $f(x) = (x^2 + 1)^n - a x^n \in \mathbb{Z}[x]$, where $n \ge 2$. If $f(x)$ is irreducible, then

Theorems & Definitions (10)

  • Theorem 1.1
  • Theorem 1.2
  • Corollary 1.3
  • Theorem 2.1
  • Proposition 2.2
  • proof : Proof of Theorem \ref{['thm-1.1']}
  • proof : Proof of Theorem \ref{['Thm-1.2']}
  • Example 3.1
  • Remark 3.2
  • Remark 3.3