On the Galois group of Generalised Laguerre polynomials II

TL;DR

This paper investigates the algebraic structure, specifically the Galois group, of certain generalized Laguerre polynomials with parameters related to negative integers, contributing to the understanding of their symmetries and algebraic properties.

Contribution

It provides new insights into the Galois groups of generalized Laguerre polynomials with specific parameter choices, extending previous algebraic studies of these polynomials.

Findings

Determined the Galois group structure for specific parameter cases.

Extended algebraic understanding of Laguerre polynomials.

Connected polynomial properties to algebraic symmetry concepts.

Abstract

For real number $\alpha,$ Generalised Laguerre Polynomials (GLP) is a family of polynomials defined by \begin{align*} L_n^{(\alpha)}(x)=(-1)^n\displaystyle\sum_{j=0}^{n}\binom{n+\alpha}{n-j}\frac{(-x)^j}{j!}. \end{align*}These orthogonal polynomials are extensively studied in Numerical Analysis and Mathematical Physics. In 1926, Schur initiated the study of algebraic properties of these polynomials. We consider the Galois group of Generalised Laguerre Polynomials $ L_n^{(\frac{1}{2}+u)}(x^2)$ when $u$ is a negative integer.

Equations

$$L_n^{(\alpha )}(x)=(-1{)}^n\sum _{j=0}^n(\genfrac{}{}{0ex}{}{n+\alpha }{n-j})\frac{(-x{)}^j}{j!}.$$

$$x{y}^{{}^{\prime \prime }}+(\alpha +1-x)y^{{}^{\prime }}+ny=0,y=L_n^{(\alpha )}(x)$$

$$L_n^{(\alpha )}(x)-L_n^{(\alpha -1)}(x)=L_{n-1}^{(\alpha )}(x).$$

$${\mathcal{L}}_n^{(u)}(x)=\sum _{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(1+2(u+n))(1+2(u+n-1))\cdots (1+2(u+j+1))x^j$$

$${\Delta }_n^{(\alpha )}=\prod _{j=1}^n{j}^j(\alpha +j{)}^{j-1}$$

$$\alpha =u+\frac{1}{2}$$

$$H_{2n}(x)=(-1{)}^n{2}^{2n}n!L_n^{(-\frac{1}{2})}(x^2)\text{and}{H}_{2n+1}(x)=(-1{)}^n{2}^{2n+1}n!x{L}_n^{(\frac{1}{2})}(x^2).$$

$$x<p<\frac{6}{5}x\text{for}x\ge 25.$$

$$v=-u.$$

$$p\ge \max \left\{\frac{2}{3}n,2v-1\right\}.$$

$$c_j=\prod _{i=j+1}^n(1+2(u+i))=\prod _{i=j+1}^n(1-2(v-i)),$$

$$3p>2(n-v)+3.$$

$$p<n\le 2(n-v)+1.$$

$${\Delta }_n^{(u)}:=\prod _{j=2}^n{j}^j(\frac{2u+1+2j}{2}{)}^{j-1}.$$

$${\Delta }_n^{(u)}=1\cdot 3\cdot 5\cdots n_o\cdot (2u+1+4)(2u+1+8)\cdots (2u+1+2n_e)\times 2^{\frac{-n_e}{2}}\times \square$$

$$D_v=1\cdot 3\cdot 5\cdots n_o(-2v+1+4)(-2v+1+8)\cdots (-2v+1+2n_e).$$

$$D_v=1\cdot 3\cdots n_o(-2v+1+4)(-2v+1+8)\cdots (r-4)\cdot r\cdot (r+4)\cdots (-2v+1+2n_e).$$

Full text

[TABLE]

These orthogonal polynomials are extensively studied in Numerical Analysis and Mathematical Physics. In 1926, Schur initiated the study of algebraic properties of these polynomials. We consider the Galois group of Generalised Laguerre Polynomials $L_{n}^{(\frac{1}{2}+u)}(x^{2})$ when $u$ is a negative integer.

2000 Mathematics Subject Classification: Primary 11A41, 11B25, 11N05, 11N13, 11C08, 11Z05.

Keywords: Irreducibility, Laguerre Polynomials, Primes, Newton Polygons.

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Laishram, Nair and Shorey

1. Introduction

For real number $\alpha$ and integer $n\geq 1,$ the Generalised Laguerre Polynomials (GLP) is a family of polynomials defined by

[TABLE]

These orthogonal polynomials have been extensively studied in various branches of analysis and mathematical physics where they play an important role. They are used in Gaussian quadrature to numerically compute integrals of the form $\displaystyle\int_{0}^{\infty}f(x)e^{-x}dx$. They satisfy second order linear differential equation

[TABLE]

and the difference equation

[TABLE]

Schur [12], [13] was the first to establish interesting and important algebraic properties of these polynomials. We define

[TABLE]

and observe that $\mathcal{L}_{n}^{(u)}(2x)=2^{n}n!L_{n}^{(\alpha)}(-x)$ and thus the irreducibility of $\mathcal{L}_{n}^{(u)}(x)$ implies irreducibility of ${L}_{n}^{(\alpha)}(x)$ and their assosciated Galois groups are same. Schur gave a formula for the discriminant $\Delta_{n}^{(\alpha)}$ of $\mathcal{L}_{n}^{(\alpha)}(x)$ by

[TABLE]

and calculated their assosciated Galois groups. Schur [12], [13] showed that for every positive integer $n$, the polynomial $L_{n}^{(0)}(x)$ is irreducible and has associated Galois group the symmetric group $S_{n}$. He showed that $L_{n}^{(1)}(x)$ is irreducible for all positive integers $n$ and has associated Galois group the alternating group $A_{n}$ if $n>1$ and $n$ is odd, and $S_{n}$ otherwise. Further, he showed that the polynomials $L_{n}^{(-1-n)}(x)$ have associated Galois group $A_{n}$ if $n\equiv 0\ ({\rm mod}\ 4)$ and $S_{n}$ otherwise. Gow [5] showed that if $n$ is an even integer $>2$, then the Galois group associated with $L_{n}^{(n)}(x)$ is $A_{n}$ provided that the polynomial $L_{n}^{(n)}(x)$ is irreducible. Filaseta, Kidd and Trifonov [4] proved that for every integer $n>2$ with $n\equiv 2\ ({\rm mod}\ 4),$ the polynomial $L_{n}^{(n)}(x)$ is irreducible. For $n=2,$ $L_{2}^{(2)}(x)$ is reducible but its Galois group is $A_{2}=\{e\}.$ These results settled the inverse Galois problem for $A_{n}$ explicitly that for every positive integer $n>1,$ there exists an explicit Laguerre polynomial of degree $n$ whose Galois group is the alternating group $A_{n}.$

From now onwards, we always assume that

[TABLE]

where $u$ is an integer . We recall that Hermite polynomials $H_{2n}(x)$ and $H_{2n+1}(x)$ are given by

[TABLE]

Schur [12], [13], proved that $L_{n}^{(-\frac{1}{2})}(x^{2})$ and $L_{n}^{(\frac{1}{2})}(x^{2})$ are irreducible and these imply the irreducibility of $H_{2n}(x)$ and $H_{2n+1}(x)/x.$ We observe that $u\in\{-1,0\}$ in these results and Laishram [8] proved that the Galois group is $S_{n}$ when $u\in\{-1,0\}.$ Laishram, Nair and Shorey [9],[10] showed that $L_{n}^{(\alpha)}(x)$ with $\alpha=u+\frac{1}{2}$ and $1\leq u\leq 45$ are irreducible except when $(u,n)=(10,3)$ and have associated Galois group $S_{n}$ other than in the case of $(u,n)=(10,3)$ where the Galois group is $\mathbb{Z}_{2}$. Further Nair and Shorey [11] proved that $L_{n}^{(\alpha)}(x)$ with $-18\leq u\leq-2$ are irreducible. In this paper, we compute the Galois groups of $L_{n}^{(\alpha)}(x)$ with $-18\leq u\leq-2$.

Theorem 1**.**

$(n,u)=$* Let $\alpha=u+\frac{1}{2}$ and $-18\leq u\leq-2$. Then the Galois group of $L^{(\alpha)}_{n}(x)$ is $S_{n}$ except when $(n,u)\in\{(8,-5),(8,-6),(9,-5),(9,-6)(16,-9),(16,-10),\\ (24,-13),(24,-14),(25,-13),(25,-14),(32,-17),(32,-18)\}$ in which cases the Galois group is $A_{n}$.*

2. Preliminaries

We will use a result of Hajir [6] which gives a criterion for an irreducible polynomial to have large Galois group using Newton polygons. We restate the result which is [6, Lemma 3.1]. For an integer $x,$ let $\nu(x)=\nu_{p}(x)$ be the highest power of $p$ dividing $x$ and we write $\nu(0)=\infty.$

Lemma 2.1**.**

Let f(x)=\sum^{m}_{j=0}\binom{m}{j}c_{j}x^{j}\in\mbox{\mathbb{Q}}[X] be an irreducible polynomial of degree $m$. Let $p$ be a prime with $\frac{m}{2}<p<m-2$ such that

• $(i)$

$\nu_{p}(c_{j})\geq 0$* for $j=0,1,\ldots,m,$*

• $(ii)$

$\nu_{p}(c_{0})=1$,

• $(iii)$

$\nu_{p}(c_{j})\geq 1$* for $0\leq j\leq m-p$,*

• $(iv)$

$\nu_{p}(c_{p})=0$.

Then the Galois group of $f$ contains $A_{m}.$ Further Galois group is $A_{m}$ if discriminant of f\in\mbox{\mathbb{Q}}^{*2} and $S_{m}$ otherwise.

Next, we state a result which we deduce from a result due to Harborth and Kemnitz and state it as a lemma so that we can use it easily.

Lemma 2.2**.**

There exists a prime $p$ satisfying $\frac{2}{3}n\leq p<n-2\ \text{for}\ n\geq 14.$

Proof.

In [7], Harborth and Kemnitz proved that there exists a prime $p$ satisfying

[TABLE]

We take $x=\frac{3}{4}n.$ Then for $n\geq 34$, we conclude that there exists a prime $p$ in $(\frac{3}{4}n,\frac{9}{10}n)\subset[\frac{2}{3}n,n-2).$ For $14\leq n\leq 33,$ we check that $[\frac{2}{3}n,n-2)$ contains a prime. ∎

Lemma 2.3**.**

Let $r\in\{1,3\}.$ The interval $(x,1.048x]$ contain primes congruent to $r$ modulo $4$ when $x\geq 887.$

This follows from Cullinan and Hajir [1, Theorem 1] with $k=4.$ ∎

Lemma 2.4**.**

For $-18\leq u\leq-2,$ the polynomials $L_{n}^{(\frac{1}{2}+u)}(x)$ are irreducible.

This follows from Nair and Shorey [11, Theorem 1].

3. Galois group of $L_{n}^{(\alpha)}(x)$

The results in this section are more general than required for the proof of Theorem 1. For a fixed $u,$ if $n\leq 13$ we can compute Galois group directly using MAGMA. Thus we assume $n\geq 14$ and we always write

[TABLE]

Now we state our main lemma.

Lemma 3.1**.**

Assume $\mathcal{L}_{n}^{(u)}(x)$ is irreducible and $n\geq\max\{14,2v-1\}$. If there exists a prime $\in(2v-3,n-2)$, then the Galois group of $\mathcal{L}_{n}^{(u)}(x)$ contains $A_{n}.$

Proof.

By Lemma 2.2, there exists a prime $\in[\frac{2}{3}n,n-2)$. So we choose $p\in(2v-3,n-2)$ such that

[TABLE]

We write

[TABLE]

$p=2v+l$ and $l\geq-1.$ We observe that $p<1+2(p-v+1)$ since $p>2v-3.$ Further, $4v+\frac{3}{2}(l-1)=\frac{3}{2}(2v+l)+v-\frac{3}{2}=\frac{3}{2}p+v-\frac{3}{2}>n$ since $p\geq\frac{2}{3}n$ and $v\geq 2.$ Thus, it follows that

[TABLE]

Therefore, we conclude that $p\nmid c_{p}.$

Next we show that $\nu_{p}(c_{j})\geq\nu_{p}(c_{0})=1$ for $1\leq j\leq n-p.$ We have $c_{0}=\pm 1\cdot 3\cdot(2v-3)\cdot 1\cdot 3\cdot(1+2(n-v))$ and $c_{n-p}=(1+2(n+1-p-v))\cdots(1+2(n-v))$. Since $n\geq 2v-1$, we get

[TABLE]

This together with $p>2v-3$ and (3), it follows that $\nu_{p}(c_{0})=1.$ Now it suffices to show that $p|c_{n-p}.$ If $n+1<p+v$, then $1+2(n-v-p+1)<1$ which together with (4) imply $p|c_{n-p}.$ So, we assume that $n+1\geq p+v$ and in this case $c_{n-p}$ is the product of $p$ consecutive odd numbers and hence divisible by $p.$ Therefore we conclude that the Galois group of $\mathcal{L}_{n}^{u}(x)$ contains $A_{n}$ by Lemma 2.1. ∎

We recall that the discriminant $\Delta_{n}^{(u)}$ of $\mathcal{L}_{n}^{(u)}(x)$ is given by

[TABLE]

Lemma 3.2**.**

Let $2n\geq\max\{28,2v-5\}$ and $n\notin\{17,18,26,27\}$. Then, $\Delta^{(u)}_{n}$ is not a square for all pairs of $(n,u)$ satisfying above conditions.

Proof.

Given a positive integer $n,$ let $n_{o}$ and $n_{e}$ denote the largest odd and even number less than or equal $n$ respectively. Then

[TABLE]

where $\square$ denotes a term in $\mathbb{Q}^{*2}.$ If $n\equiv 2,3\ (\text{mod}\ 4),$ then $\frac{n_{e}}{2}$ is odd and hence $\Delta^{(u)}_{n}$ is not a square. Hence $n\equiv 0,1\ (\text{mod}\ 4).$ We consider

[TABLE]

It suffices to show that $D_{v}$ is not a square. Let $r\in\{1,3\}$ be such that $-2v+1\equiv r\ (\text{mod}\ 4).$ Then

[TABLE]

Let $p_{0}$ be the largest prime with $\frac{2}{3}n\leq p<n$ and $p\equiv 3r\ (\text{mod}\ 4)$. We take $x=\frac{2}{3}n$ and $n>1330$ in Lemma 2.3 to conclude that the interval $[\frac{2}{3}n,n-2)$ contain both primes congruent to $1$ and $3$ modulo $4$. We check that for $14\leq n\leq 1330,$ the interval $[\frac{2}{3}n,n-2)$ contain both primes congruent to $1$ and $3$ modulo $4$ except for $n\in\{17,18,26,27\}$. Let $n\notin\{17,18,26,27\}$. Then $p_{0}||1\cdot 3\cdot 5\cdots n_{o}$. Further $3p_{0}>2n>2n_{e}-2v+1$ and $2v-5\leq 2n<3p_{0}$. Since $p\equiv 3r\ (\text{mod}\ 4)$ and $-2v+1\equiv r\ (\text{mod}\ 4)$, the above conditions imply that $p_{0}||D_{v}$ and hence $D_{v}$ is not a square. Therefore $\Delta^{(u)}_{n}$ is not a square. ∎

The following result is obtained by direct computation using MAGMA.

Lemma 3.3**.**

Let $2\leq n\leq 40$ and $-18\leq u\leq-2$. Then the Galois group of $\mathcal{L}^{(\alpha)}_{n}(x)$ is $S_{n}$ except when $(n,u)\in\{(8,-5),(8,-6),(9,-5),(9,-6)(16,-9),(16,-10),(24,-13),\\ (24,-14),(25,-13),(25,-14),(32,-17),(32,-18)\}$ in which cases the Galois group is $A_{n}$.

We now prove Theorem 1.

Proof of Theorem 1:

Let $-18\leq u\leq-2$. Then $\mathcal{L}_{n}^{(u)}(x)$ are irreducible by Lemma 2.4. Further, by Lemma 3.3, we can always assume that $n\geq 41.$ Since $v\leq 18,$ we have $n\geq 2v-1.$ If $n\geq 50,$ then $2v-3<\frac{2}{3}n$ and hence using Lemma 2.2, we conclude that there exists a prime in $(2v-3,n-2).$ If $41\leq n\leq 49,$ we directly check that such a prime exists as $2v-3\leq 33.$ Hence, by Lemmas 3.1 and 3.2, the Galois group of $\mathcal{L}_{n}^{(u)}(x)$ is $S_{n}.$ ∎

Acknowledgements

The first named author acknowledges the support of the MATRICS Grant of SERB, DST, India.