On a factorization result of Ştefănescu–II

Sanjeev Kumar^† 0000-0001-6882-4733

~{}^{\dagger}

Department of Mathematics, SGGS College, Sector-26, Chandigarh-160019, India sanjeev_kumar_19@yahoo.co.in and Jitender Singh^‡,∗ 0000-0003-3706-8239

~{}^{\ddagger}

Department of Mathematics, Guru Nanak Dev University, Amritsar-143005, India jitender.math@gndu.ac.in

Abstract.

Ştefănescu proved an elegant factorization result for polynomials over discrete valuation domains [CASC’2014, Lecture Notes in Computer Science, Ed. by V. Gerdt, W. Koepf, W. Mayr, and E. Vorozhtsov, Springer, Berlin, Vol. 8660, pp. 460–471, 2014.] In this paper, a generalization of Ştefănescu’s result is proved to cover a larger class of polynomials over discrete valuation domains. Such results are useful in devising algorithms for polynomial factorization.

2010 Mathematics Subject Classification:

Primary 30C10; 12E05; 11C08

³³footnotetext: ^,∗Corresponding author: jitender.math@gndu.ac.in; sonumaths@gmail.com²²footnotetext: sanjeev_kumar_19@yahoo.co.in

1. Introduction

Let $(R,v)$ be a discrete valuation domain. Let $f=a_{0}+a_{1}x+\cdots+a_{n}x^{n}\in{R}[x]$ be a nonconstant polynomial. Newton polygon $N_{f}$ of the polynomial $f$ is defined as the lower convex hull of the set $\{(i,v(a_{i}))~{}|~{}a_{i}\neq 0\}$ . The slopes of the underlying Newton polygon are the slopes of some line segments. Note that the slope of the line joining the points $(n,v(a_{n}))$ and $(i,v(a_{i}))$ is $m_{i}(f)=(v(a_{n})-v(a_{i}))/(n-i)$ for each $i=0,1,\ldots,n-1$ . The Newton index $e(f)$ of the polynomial $f$ is defined as

\displaystyle e(f)=\max_{1\leq i\leq n}m_{i}(f).

It follows from the definition of $N_{f}$ and $e(f)$ that for nonconstant polynomials $f,g\in R[x]$ , one has $e(fg)=\max(e(f),e(g))$ . From the application point of view, Newton index has been used in devising algorithms for factoring polynomials [1]. As a generalization of the classical result of Dumas [2], Ştefănescu [3] proved a factorization result for polynomials over a discrete valuation domain using Newton index. Further, using the method of [3], Kumar and Singh [4] extended the result of Ştefănescu to include a wider class of polynomials over discrete valuation domains. In [1], Ştefănescu proved the following elegant factorization results.

Theorem A.

Let $(R,v)$ be a discrete valuation domain. Let $f=a_{0}+a_{1}x+\cdots+a_{n}x^{n}\in{R}[x]$ be a nonconstant polynomial with $a_{0}a_{n}\neq 0$ and $n\geq 2$ . Assume that there exists an index $s\in\{0,1,2,\ldots,n-1\}$ for which each of the following conditions is satisfied.

(a)

$m_{i}(f)<m_{s}(f)$ for all $i\in\{0,1,2,\ldots,n-1\},~{}i\neq s$ ,
(b)

$n(n-s)(m_{s}(f)-m_{0}(f))=1$ ,
(c)

$\gcd(v(a_{s})-v(a_{n}),n-s)=1$ .

Then the polynomial $f$ is either irreducible in $R[x]$ , or $f$ has a factor whose degree is a multiple of $n-s$ .

Theorem B.

(a)

$m_{i}(f)<m_{s}(f)$ for all $i\in\{0,1,2,\ldots,n-1\},~{}i\neq s$ ,
(b)

$u=n(n-s)(m_{s}(f)-m_{0}(f))\geq 2$ ,
(c)

$\gcd(v(a_{s})-v(a_{n}),n-s)=1$ .

Then either $f$ is irreducible in $R[x]$ , or $f$ has a divisor whose degree is a multiple of $n-s$ , or $f$ admits a factorization $f=f_{1}f_{2}$ such that $\alpha_{2}\deg(f_{1})-\alpha_{1}\deg(f_{2})$ is a multiple of $n-s$ for some $\alpha_{1},\alpha_{2}\in\{1,\ldots,u-1\}$ .

This note ameliorates and extends the aforementioned factorization results on the lines of [4]. Our main results are the following:

Theorem 1.

Let $(R,v)$ be a discrete valuation domain and let $f=a_{0}+a_{1}x+\cdots+a_{n}x^{n}\in{R}[x]$ be a nonconstant polynomial with $a_{0}a_{n}\neq 0$ and $n\geq 2.$ Assume that there exists an index $s\in\{0,1,2,\ldots,n-1\}$ such that the following conditions are satisfied:

(a)

$m_{i}(f)<m_{s}(f)$ for all $i\in\{0,1,2,\ldots,n-1\},~{}i\neq s$ ,
(b)

$d=\gcd(v(a_{s})-v(a_{n}),n-s)$ satisfies

$\displaystyle d=\begin{cases}n(n-s)(m_{s}(f)-m_{0}(f)),&~{}\text{if}~{}s\neq 0;\\ 1,&~{}\text{if}~{}s=0.\end{cases}$

Then the polynomial $f$ is either irreducible in $R[x]$ , or $f$ has a factor in $R[x]$ whose degree is zero or a multiple of $(n-s)/d$ .

Theorem 2.

Let $(R,v)$ be a discrete valuation domain and let $f=a_{0}+a_{1}x+\cdots+a_{n}x^{n}\in{R}[x]$ be a nonconstant polynomial with $a_{0}a_{n}\neq 0$ and $n\geq 2$ . Assume that there exists an index $s\in\{1,2,\ldots,n-1\}$ such that each of the following conditions is satisfied.

(a)

$m_{i}(f)<m_{s}(f)$ for all $i\in\{0,1,2,\ldots,n-1\},~{}i\neq s$ ,

(b)

$d=\gcd(v(a_{s})-v(a_{n}),n-s)$ satisfies the following:

\displaystyle d=\begin{cases}\text{a proper divisor of }u,\text{ where }u=n(n-s)(m_{s}(f)-m_{0}(f))\geq 2,&\text{ if }s\neq 0;\\ 1,&\text{ if }s=0.\end{cases}

Then either $f$ is irreducible in $R[x]$ , or $f$ has a divisor whose degree is zero or a multiple of $(n-s)/d$ , or $f$ admits a factorization $f=f_{1}f_{2}$ such that $\alpha_{2}\deg(f_{1})-\alpha_{1}\deg(f_{2})$ is a multiple of $(n-s)/d$ for some $\alpha_{1},\alpha_{2}\in\{1,\ldots,(u/d)-1\}$ with $\alpha_{1}+\alpha_{2}=u/d$ .

We observe that Theorems 1 and 2 reduce to Theorems A and B, respectively for $d=1$ and $s\neq 0$ .

Further, Theorem 1 reduces to the main result of [4] incase $v(a_{n})=0$ .

In view of Theorem 1, if we take $d=2$ and $0<s<n/2$ so that $(n-s)>n/2$ , then either $f$ is irreducible, or $f$ has a factor whose degree is a multiple of $(n-s)/2$ , say $m(n-s)/2$ for some positive integer $m$ . If possible, suppose that $m\geq 4$ , then we have $n>m(n-s)/2\geq mn/4\geq n$ , which is absurd, and so, we must have $m<4$ . So, either $f$ is irreducible, or $f$ has a factor whose degree is equal to one of $(n-s)/2$ and $n-s$ .

Example 1.

For a prime $p$ , let $v=v_{p}$ denotes the $p$ -adic valuation on $\mathbb{Q}$ . For $n\geq 5$ , consider the polynomial

\displaystyle X

\displaystyle=

\displaystyle a_{0}+p^{n-4}a_{1}x+p^{n(n-3)-1}(a_{2}x^{2}+a_{n}x^{n})\in\mathbb{Z}[x],

where $a_{0},a_{1},a_{2},a_{n}\in\{1,2,\ldots,p-1\}$ . Here, we have

$\displaystyle m_{2}(f)$	$\displaystyle=$	$\displaystyle\frac{v_{p}(p^{n(n-3)-1}a_{n})-v_{p}(p^{n(n-3)-1}a_{2})}{n-2}=0,$
$\displaystyle m_{1}(f)$	$\displaystyle=$	$\displaystyle\frac{v_{p}(p^{n(n-3)-1}a_{n})-v_{p}(p^{n-4}a_{1})}{n-1}=n-3,$
$\displaystyle m_{0}(f)$	$\displaystyle=$	$\displaystyle\frac{v_{p}(p^{n(n-3)-1}a_{n})-v_{p}(a_{0})}{n-0}=n-3-\frac{1}{n},$

which shows that $e(X)=m_{1}(f)$ , and so, $s=1$ . Further, we have $n(n-1)(m_{1}(f)-m_{0}(f))=n-1$ , and $\gcd(v_{p}(p^{n-4}a_{1})-v_{p}(p^{n(n-3)-1}a_{n}),n-1)=\gcd((n-1)(n-3),n-1)=n-1$ . By Theorem 1, the polynomial $X$ is either irreducible, or $X$ has a factor whose degree is a multiple of $n-1$ .

Example 2.

For a prime $p$ , let $v_{p}$ be the $p$ -adic valuation on $\mathbb{Q}$ . For a positive integer $d\geq 2$ , we consider the polynomial

\displaystyle Y_{d}

\displaystyle=

\displaystyle p^{d+1}+p^{d-1}x^{d+1}+x^{d(d+1)}\in\mathbb{Z}[x].

Here $n=d(d+1)$ , $a_{0}=p^{d+1}$ , $a_{d+1}=p^{d-1}$ , $a_{n}=1$ , and $a_{j}=0$ for all $j\not\in\{0,d+1,d(d+1)\}$ . So, we have

	$\displaystyle m_{d+1}(f)$	$\displaystyle=$	$\displaystyle\frac{v_{p}(a_{n})-v_{p}(a_{d+1})}{d(d+1)-(d+1)}=-\frac{1}{d+1},$
	$\displaystyle m_{0}(f)$	$\displaystyle=$	$\displaystyle\frac{v_{p}(a_{n})-v_{p}(a_{0})}{d(d+1)-0}=-\frac{1}{d},$

which shows that $e(X)=m_{d+1}(f)$ , and so, $s=d+1$ . Further, $u=(d-1)(d+1)\geq 2$ , since $d\geq 2$ . Furthermore, $\gcd(v_{p}(a_{s})-v_{p}(a_{n}),n-s)=\gcd(d-1,d(d+1)-d-1)=d-1$ , which divides $u$ . Thus, by Theorem 2, the polynomial $Y_{d}$ is irreducible, or has a factor whose degree is a multiple of $d+1$ , or $f$ has a factorization $f=f_{1}f_{2}$ in $\mathbb{Z}[x]$ such that $\alpha_{2}\deg f_{1}-\alpha_{1}\deg f_{2}$ is a multiple of $d+1$ , for some $\alpha_{1},\alpha_{2}\in\{1,2,\ldots,d\}$ with $\alpha_{1}+\alpha_{2}=d+1$ .

2. Proof of Theorems 1 and 2

Proof of Theorem 1.

Our method of proof is similar to that of [4]. If $s=0$ , then the Newton polygon of $f$ is a straight line segment joining the points $(0,v(a_{0}))$ and $(n,v(a_{n}))$ , and so, by the classical result of Dumas [2] on factorization of polynomials via Newton polygons, it follows that $f$ is either irreducible, or $f$ has a factor of degree zero.

Now assume that $s>0$ . Suppose that $f$ is not irreducible in $R[x]$ so that $f$ admits a factorization $f=f_{1}f_{2}$ in $R[x]$ with $\min\{\deg f_{1},f_{2}\}\geq 1$ . For each $i=1,2$ , let $n_{i}=\deg(f_{i})$ so that $n=n_{1}+n_{2}$ , and $f_{i}=\sum_{j=0}^{n_{i}}a_{ij}x^{j}$ . Consequently, we have $a_{0}=a_{10}a_{20}$ and $a_{n}=a_{1n_{1}}a_{2n_{2}}$ so that $v(a_{0})=v(a_{10})+v(a_{20})$ and $v(a_{n})=v(a_{1n_{1}})+v(a_{2n_{2}})$ . If we let

\displaystyle c_{s}=v(a_{n})-v(a_{s}),~{}c_{n}=v(a_{n})-v(a_{0}),~{}c_{i0}=v(a_{in_{i}})-v(a_{i0}),~{}i=1,2,

then it follows that $c_{n}=c_{10}+c_{20}$ . By the hypothesis (a) and the identity $e(f_{1}f_{2})=\max(e(f_{1}),e(f_{2}))$ , we have

\displaystyle\dfrac{c_{s}}{n-s}=\dfrac{v(a_{n})-v(a_{s})}{n-s}=e(f)\geq e(f_{1})\geq m_{0}(f_{1})=\dfrac{c_{10}}{n_{1}}.

We then have $\dfrac{c_{s}}{n-s}-\dfrac{c_{10}}{n_{1}}\geq 0$ , and so $\dfrac{c_{s}}{d}n_{1}-\dfrac{n-s}{d}c_{10}\geq 0$ . Note that from the hypothesis (b), $c_{s}/d$ and $(n-s)/d$ are both integers. Since $e(f)\geq e(f_{2})$ , we must have $c_{s}n_{2}-(n-s)c_{20}\geq 0$ and $\dfrac{c_{s}}{d}n_{2}-\dfrac{(n-s)}{d}c_{20}\geq 0$ . By the hypothesis (b), we have the following:

(1)

\displaystyle 1=\dfrac{c_{s}}{d}n-\dfrac{(n-s)}{d}c_{n}=\Big{(}\dfrac{c_{s}}{d}n_{1}-\dfrac{(n-s)}{d}c_{10}\Big{)}+\Big{(}\dfrac{c_{s}}{d}n_{2}-\dfrac{(n-s)}{d}c_{20}\Big{)},

which shows that one of the nonnegative integers $\dfrac{c_{s}}{d}n_{1}-\dfrac{(n-s)}{d}c_{10}$ and $\dfrac{c_{s}}{d}n_{2}-\dfrac{(n-s)}{d}c_{20}$ is zero.

First assume that $\dfrac{c_{s}}{d}n_{1}-\dfrac{(n-s)}{d}c_{10}=0$ . Then from (1), we have $\dfrac{c_{s}}{d}n_{1}=\dfrac{(n-s)}{d}c_{10}$ . Since $\gcd({c_{s}}/{d},{(n-s)}/{d})=1$ , the integer ${(n-s)}/{d}$ must divide $n_{1}$ . Similarly, if we assume that $\dfrac{c_{s}}{d}n_{2}=\dfrac{(n-s)}{d}c_{20}$ , then it follows from (1) that ${(n-s)}/{d}$ must divide $n_{2}$ . This completes the proof. ∎

Proof of Theorem 2.

Assume that $f=f_{1}f_{2}$ for some nonconstant polynomials $f_{1}$ and $f_{2}$ in $R[x]$ . For $s=0$ , Theorem 2 reduces to Theorem 1. So, we assume that $s>0$ . We use the notation same as in the proof of Theorem 1 so that we arrive at the following:

\displaystyle\dfrac{c_{s}}{d}n-\dfrac{(n-s)}{d}c_{n}=\dfrac{u}{d};~{}\dfrac{c_{s}}{d}n_{i}-\dfrac{(n-s)}{d}c_{i0}\geq 0,~{}i=1,2,

where $n=n_{1}+n_{2},c_{n}=c_{10}+c_{20}$ , and

(2)

\displaystyle\Big{(}\dfrac{c_{s}}{d}n_{1}-\dfrac{(n-s)}{d}c_{10}\Big{)}+\Big{(}\dfrac{c_{s}}{d}n_{2}-\dfrac{(n-s)}{d}c_{20}\Big{)}=\dfrac{u}{d}.

If $\dfrac{c_{s}}{d}n_{i}-\dfrac{(n-s)}{d}c_{i0}=0$ for any $i\in\{1,2\}$ , then as in Theorem 1, we deduce that the degree of a divisor of $f$ must be divisible by ${(n-s)}/{d}$ .

If $\dfrac{c_{s}}{d}n_{1}-\dfrac{(n-s)}{d}c_{10}=1$ , then $\dfrac{c_{s}}{d}n_{2}-\dfrac{(n-s)}{d}c_{20}=\dfrac{u}{d}-1$ , and so, from (2), we have

\displaystyle\dfrac{c_{s}}{d}\bigl{(}n_{2}-\bigl{(}\dfrac{u}{d}-1\bigr{)}n_{1}\bigr{)}=\dfrac{(n-s)}{d}\bigl{(}c_{20}-\bigl{(}\dfrac{u}{d}-1\bigr{)}c_{10}\bigr{)},

which in view of the fact that $c_{s}/d$ and $(n-s)/d$ are coprime, shows that ${(n-s)}/{d}$ divides $n_{2}-(\dfrac{u}{d}-1)n_{1}$ . More generally, if we let $\alpha_{i}=\dfrac{c_{s}}{d}n_{i}-\dfrac{(n-s)}{d}c_{i0}$ , $i=1,2$ with $\alpha_{1}+\alpha_{2}={u}/{d}$ , then using (2) one has the following:

\displaystyle\dfrac{c_{s}}{d}(\alpha_{2}n_{1}-\alpha_{1}n_{2})=\dfrac{(n-s)}{d}(\alpha_{2}c_{10}-\alpha_{1}c_{20}).

This in view of the fact that $\gcd({(n-s)}/{d},{c_{s}}/{d})=1$ shows that ${(n-s)}/{d}$ must divide $\alpha_{2}n_{1}-\alpha_{1}n_{2}$ . This completes the proof. ∎

Acknowledgments.

The present research is supported by Science and Engineering Research Board(SERB), a statutory body of Department of Science and Technology (DST), Govt. of India through the project grant no. MTR/2017/000575 awarded to the second author under MATRICS Scheme.

Disclosure

The authors declare to have no competing interest.

References

[1] D. Ştefănescu, Applications of the Newton index to the construction of irreducible polynomials, in CASC’2014, Lecture Notes in Computer Science, Ed. by V. Gerdt, W. Koepf, W. Mayr, and E. Vorozhtsov, Springer, Berlin, Vol. 8660 (2014), 460–471.
[2] G. Dumas, Sur quelques cas d’irréductibilité des polynomes á coefficients rationnels, J. Math. Pure Appl. 2 (1906), 191–258.
[3] D. Ştefănescu, On the irreducibility of bivariate polynomials, Bull. Math. Soc. Sci. Math. Roumanie 56:104 (2013), 377–384.
[4] S. Kumar and J. Singh, On a factorization result of Ştefănescu, Communications in Algebra 50:11 (2022) 4648–4651.