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CONSTRUCTION OF IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS VIA POLYNOMIAL COMPOSITION

by

FUNDA ¨ OZDEM˙IR

Submitted to the Graduate School of Engineering and Natural Sciences in partial fulfillment of

the requirements for the degree of Master of Science

Sabancı University

Spring 2012

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CONSTRUCTION OF IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS VIA POLYNOMIAL COMPOSITION

APPROVED BY

Prof. Dr. Henning Stichtenoth ...

(Thesis Supervisor)

Prof. Dr. Alev Topuzoˇ glu ...

Assoc. Prof. Dr. Cem G¨ uneri ...

Assist. Prof. Dr. Alp Bassa ...

Assoc. Prof. Dr. Berrin Yanıkoˇ glu ...

DATE OF APPROVAL: May 30, 2012

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Funda ¨ c Ozdemir 2012

All Rights Reserved

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CONSTRUCTION OF IRREDUCIBLE POLYNOMIALS OVER FINITE FIELDS VIA POLYNOMIAL COMPOSITION

Funda ¨ Ozdemir

Mathematics, Master Thesis, 2012

Thesis Supervisor: Prof. Dr. Henning Stichtenoth

Keywords: Finite fields, irreducible polynomials, polynomial composition methods, linearized polynomials, primitive polynomials, composed product.

Abstract

The construction of irreducible polynomials over finite fields is currently a strong subject of interest with important applications including coding theory and cryptog- raphy. One of the most popular methods of construction of irreducible polynomials is the method of composition of polynomials where irreducible polynomials of relatively higher degrees are generated from irreducible polynomials of relatively lower degrees.

In this thesis, we give some polynomial composition methods and several applications

of them.

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SONLU C˙IS˙IMLER ¨ UZER˙INDE POL˙INOM B˙ILES ¸ ˙IM˙I METODU ˙ILE

˙IND˙IRGENEMEZ POL˙INOM ˙INS¸ASI

Funda ¨ Ozdemir

Matematik, Y¨ uksek Lisans Tezi, 2012 Tez Danı¸smanı: Prof. Dr. Henning Stichtenoth

Anahtar Kelimeler: Sonlu cisimler, indirgenemez polinomlar, polinom bile¸simi y¨ ontemleri, doˇ grusalla¸stırılmı¸s polinomlar, ilkel polinomlar, bile¸ske ¸carpım.

Ozet ¨

Sonlu cisimler ¨ uzerinde indirgenemez polinomların in¸sası, kodlama teorisi ve krip-

tografideki ¨ onemli uygulamaları da dahil olmak ¨ uzere son zamanlarda g¨ u¸cl¨ u bir ilgi

odaˇ gı olu¸sturmaktadır. ˙Indirgenemez polinomların in¸sasında en pop¨ uler y¨ ontemlerden

biri olan polinom bile¸simi metodunda, d¨ u¸s¨ uk dereceli indirgenemez polinomlardan

y¨ uksek dereceli indirgenemez polinomlar elde edilir. Bu tezde, bir takım polinom

bile¸simi y¨ ontemleri ile bunların uygulamalarına yer verilmi¸stir.

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to my parents and my husband

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Acknowledgements

First of all, I would like to express my appreciation to my supervisor Prof. Dr.

Henning Stichtenoth for his patience, understanding, and guidance throughout my thesis. It has been an honor to work with him.

I also thank Prof. Dr. Alev Topuzoˇ glu who has supported me during my thesis with her knowledge and guidance.

Special thanks also to all my graduate friends in the Mathematics Program for their helps and friendships.

Finally, the most special thanks goes to my family who have motivated and sup- ported me unconditionally throughout my whole life, and my husband for his endless love and support.

vii

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Table of Contents

Abstract iv

Ozet ¨ v

Acknowledgements vii

Introduction ix

1 First Composition Method 1

2 Irreducibility of Polynomials of the Form g(x)

n

P (f (x)/g(x)) 7

3 Recursive Constructions 12

4 Composed Product of Polynomials 16

Bibliography 20

viii

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Introduction

Let F

q

be the finite field of order q = p

s

and of characteristic p, where p is a prime and s is a positive integer, F

q

be its multiplicative group which is cyclic. A generator of the cyclic group F

q

is called a primitive element of F

q

and its minimal polynomial over F

p

is called a primitive polynomial.

Throughout this thesis, we assume, unless otherwise specified, that the considered polynomials are monic, i.e. with leading coefficient 1. Let f (x) be an irreducible polynomial of degree n over F

q

and let β be a root of f (x). The field F

q

(β) = F

qn

is a degree n extension of F

q

and can be viewed as a vector space of dimension n over F

q

. Moreover, the conjugates of β with respect to F

q

, namely β, β

q

, . . . , β

qn−1

, are all the roots of f (x).

The subject of irreducible polynomials over finite fields along with several construc- tion methods has been of considerable interest in recent years. Such polynomials, which have both theoretical and practical importance, are used to perform arithmetic in finite fields and are found in many applications, including coding theory and cryptography.

One of the most popular methods of construction is the method of composition of poly- nomials where irreducible polynomials of relatively higher degree are produced from given irreducible polynomials of relatively lower degrees. There is a detailed literature on the problem of irreducibility of polynomial composition by several authors including Cohen, Kyuregyan-Kyureghyan, Varshamov who have approached this problem from different aspects. In this thesis, we intend to give a survey of works about polynomial composition methods.

• In Chapter 1, we present the approach of Kyuregyan-Kyureghyan [5] to the con- struction of irreducible polynomials over F

q

. Theorem 1.4 is used to obtain explicit families of irreducible polynomials of degrees n(q

n

− 1) and n(q

n

+ 1) over F

q

, where n is a natural number. At the end of this chapter, the result of Cohen [3] which is one of the most applicable results in this area is proved using Theorem 1.4.

• In Chapter 2, by using the result of Cohen [3] in the previous chapter and some auxiliary results, the irreducibility of compositions of irreducible polynomials in the form P (f /g) := (g(x))

n

P (f (x)/g(x)) is studied for some specified relatively prime polynomials f and g, and any degree n polynomial P .

• In Chapter 3, we present how to construct recursively irreducible polynomials, using the irreducibility criteria developed in Chapter 2.

• In the final chapter, we introduce first the notion of composed product by Braw- ley and Carlitz [2] and state an important theorem, again due to Brawley and Carlitz [2], which says how to construct irreducible polynomials of degree mn from irreducible polynomials of degrees m and n with gcd(m, n) = 1 through the use of composed product. Moreover, we restate a result of Varshamov in [10] and a result in [5] more directly, and we prove them by using a consequence of the theorem of Brawley and Carlitz.

ix

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1

First Composition Method

We say that the degree of an element α over F

q

is equal to k and write deg

q

(α) = k if F

q

(α) = F

qk

or equivalently α ∈ F

qk

and α / ∈ F

qν

for any proper divisor ν of k.

Similarly, we say that the degree of a subset A = {α

1

, α

2

, . . . , α

r

} ⊂ F

qk

over F

q

is equal to k and write deg

q

1

, α

2

, . . . , α

r

) = k, if for any proper divisor ν of k there exists at least one element α

u

∈ A such that α

u

∈ F /

qν

.

We begin with the following well known results which can be found in [6].

Proposition 1.1 ( [6], Theorem 3.46). Let f (x) be a monic irreducible polynomial of degree n over F

q

and let k ∈ N. Then f (x) factors into d irreducible polynomials in F

qk

[x] of the same degree n/d, where d = gcd(n, k).

Proposition 1.2 ( [6], Corollary 3.47). An irreducible polynomial over F

q

of degree n remains irreducible over F

qk

if and only if k and n are relatively prime.

Given 0 ≤ ν ≤ k − 1 and g(x) = P

m

i=0

b

i

x

i

∈ F

qk

[x], we use the notation g

(ν)

(x) =

m

X

i=0

b

qiν

x

i

,

where g(x) = g

(0)

(x).

Lemma 1.3. Let f (x) be a monic irreducible polynomial of degree dk over F

q

. Then there is a monic irreducible divisor g(x) of degree k of f (x) in F

qd

[x]. Moreover, every irreducible factor of f (x) in F

qd

[x] is given by g

(ν)

(x) for some 0 ≤ ν ≤ d − 1. In particular, the factorization of f (x) in F

qd

[x] is

f (x) =

d−1

Y

ν=0

g

(ν)

(x)

Proof. By Proposition 1.1, f (x) factors into d monic irreducible polynomials in F

qd

[x]

of the same degree k. Let α ∈ F

qdk

be a root of f (x). Then all the roots of f (x) are the conjugates of α with respect to F

q

, namely α, α

q

, α

q2

, . . . , α

qdk−1

. Let g(x) be a monic irreducible divisor of f (x) of degree k in F

qd

[x] assuming α as a root. Then all the roots of g(x) are the conjugates of α with respect to F

qd

, which are α, α

qd

, α

q2d

, . . . , α

q(k−1)d

.

1

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Hence we can write the factorization of g(x) and g

(ν)

(x), for 0 ≤ ν ≤ d − 1, over F

qdk

as

g(x) = (x − α)(x − α

qd

)(x − α

q2d

) · · · (x − α

qdk−d

) g

(1)

(x) = (x − α

q

)(x − α

qd+1

)(x − α

q2d+1

) · · · (x − α

qdk−d+1

) g

(2)

(x) = (x − α

q2

)(x − α

qd+2

)(x − α

q2d+2

) · · · (x − α

qdk−d+2

)

.. .

g

(d−1)

(x) = (x − α

qd−1

)(x − α

q2d−1

)(x − α

q3d−1

) · · · (x − α

qdk−1

) Both polynomials f (x) and Q

d−1

ν=0

g

(ν)

(x) of the same degree dk have the same dk distinct roots in F

qdk

. Therefore they are equal.

The converse of Lemma 1.3 does not hold in general: Given an irreducible poly- nomial of degree k over F

qd

, the product Q

d−1

ν=0

g

(ν)

(x) is a polynomial over F

q

, but it is not necessarily irreducible over F

q

. To ensure the converse statement, g(x) must be described precisely as stated in the following theorem.

Theorem 1.4 ( [5], Lemma 1). A monic polynomial f (x) ∈ F

q

[x] of degree n = dk is irreducible over F

q

if and only if there is a monic irreducible polynomial g(x) = P

k

i=0

g

i

x

i

over F

qd

of degree k such that F

q

(g

0

, . . . , g

k

) = F

qd

and f (x) = Q

d−1

ν=0

g

(ν)

(x) in F

qd

[x].

Proof. Suppose f (x) is irreducible over F

q

. Then by Lemma 1.3 there is an irreducible polynomial g(x) = P

k

i=0

g

i

x

i

of degree k over F

qd

such that f (x) =

d−1

Y

ν=0

g

(ν)

(x) (1.1)

over F

qd

. Next we show that the set of coefficients of g(x) generates F

qd

. Suppose, on the contrary, that F

q

(g

0

, . . . , g

k

) = F

qs

, for some proper divisor s of d with d = rs.

Then, because of F

qs

[x] ⊂ F

qd

[x], the polynomial g(x) is also irreducible over F

qs

and by Lemma 1.3

f (x) =

s−1

Y

ω=0

h

(ω)

(x) (1.2)

over F

qs

and h

(ω)

(x) = P

rk

j=0

h

qjω

x

j

, 0 ≤ ω ≤ s − 1, are distinct irreducible polynomials of degree rk over F

qs

. Then, by combining the equations (1.1) and (1.2), we get

f (x) =

s−1

Y

ω=0

h

(ω)

(x) =

d−1

Y

ν=0

g

(ν)

(x)

in F

qs

[x], which contradicts to the uniqueness of the decomposition into irreducible factors in F

qs

[x].

2

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For the proof of the converse, let g(x) = P

k

i=0

g

i

x

i

be an irreducible polynomial of degree k over F

qd

with F

q

(g

0

, . . . , g

k

) = F

qd

and let α ∈ F

qdk

be a zero of g(x). Further, let f (x) be the minimal polynomial of α over F

q

of degree n. We want to prove that n = dk. Let l = gcd(n, k). Then f has exactly l irreducible factors in F

qd

[x], by Proposition 1.1. Since g divides f over F

qd

, we get

f (x) =

l−1

Y

ν=0

g

(ν)

(x).

However, f factors into l irreducible polynomials also over F

ql

⊆ F

qd

. The condition F

q

(g

0

, . . . , g

k

) = F

qd

forces l = d which means d divides n. Hence we have shown that F

qd

is a subfield of F

qn

= F

q

(α), implying that F

q

(α) = F

qd

(α), i.e. F

qn

= F

qdk

and consequently n = dk.

Now we obtain explicit families of irreducible polynomials of degree n(q

n

− 1) from a given primitive polynomial of degree n over F

q

, using Theorem 1.4 and the following theorem.

Theorem 1.5 ( [1] Chapter 5, Theorem 24 (Dickson’s theorem)). Let q = p

s

, m be a divisor of s and p

m

6= 2. Suppose β, θ ∈ F

q

and θ is a primitive element of F

q

. Then the polynomial

f (x) = x

pm

− θx + β

is the product of a linear polynomial and an irreducible polynomial of degree p

m

− 1 over F

q

.

Theorem 1.6 ( [5], Theorem 7). Let q

n

> 2, β, γ ∈ F

q

, β 6= −γ and f (x) be a primitive polynomial of degree n over F

q

. Set h(x) = f ((β + γ)x + 1) and h

(x) = x

n

h(

x1

). Then the polynomial

F (x) = (x − γ)

n

f



(x − γ)

−1

(x

qn

+ β)

 

h

(x − γ) 

−1

is an irreducible polynomial of degree n(q

n

− 1) over F

q

.

Proof. Let α be a root of f (x). Then f (x) =

n−1

Y

ν=0



x − α

qν



(1.3) holds in F

qn

[x]. Substituting (x − γ)

−1

(x

qn

+ β) for x in (1.3), and multiplying both sides of the equation by (x − γ)

n

, we get

(x − γ)

n

f



(x − γ)

−1

(x

qn

+ β)



=

n−1

Y

ν=0



x

qn

− α

qν

x + β + γα

qν



=

n−1

Y

ν=0



x

qn

− αx + β + γα



(ν)

3

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Since q

n

> 2 and α

qν

is a primitive element in F

qn

, by Theorem 1.5 each of the polynomials g

(ν)

(x) := (x

qn

− α

qν

x + β + γα

qν

) is product of a linear polynomial and an irreducible polynomial of degree q

n

− 1 over F

qn

. Also if θ is a root of g(x) in F

qn

, then θ

qν

∈ F

qn

is a root of g

(ν)

(x), where θ

qν

= (β + γα

qν

)(α

qν

− 1)

−1

. Thus the linear factor of g

(ν)

is x − θ

qν

and the irreducible factor of g

(ν)

is

Q

(ν)

(x) = x

qn

− α

qν

x + β + γα

qν

x − θ

qν

= x

qn

− θ

qn+ν

− α

qν

(x − θ

qν

) x − θ

qν

over F

qn

. Note that the constant term of Q

(ν)

(x) is 1 − α

qν

, and in particular the degree of the set of its coefficients is n over F

q

. Therefore, by Theorem 1.4 the polynomial Q

n−1

ν=0

Q

(ν)

(x) is irreducible over F

q

. To complete the proof observe that (β +γ)

−1

(α−1) is a root of h(x) = f ((β +γ)x+1) and so θ = (β +γ)(α −1)

−1

+γ is a root of (h

(x−γ).

Then in F

qn

[x] it holds

n−1

Y

ν=0



x − θ

qν



= h

(x − γ)

which yields

F (x) =

(x − γ)

n

f 

(x − γ)

−1

(x

qn

+ β)  Q

n−1

ν=0



x − θ

qν

 =

n−1

Y

ν=0

Q

(ν)

(x)

Finally, the irreducibility of F (x) over F

q

follows from Theorem 1.4.

Further we use the following result by Sidelnikov [9] that enables explicit construc- tions of irreducible polynomials of degree n(q

n

+ 1) .

Theorem 1.7. Let ω ∈ F

q

and x

0

∈ F

q2

\ F

q

such that x

q+10

= 1. Then the polynomial f (x) = x

q+1

− ωx

q

− (x

0

+ x

q0

− ω)x + 1 ∈ F

q

[x]

is irreducible if and only if

ω−x

q 0

ω−x0

is a generating element of the multiplicative subgroup S := {y ∈ F

q2

|y

q+1

= 1} of F

q2

.

Theorem 1.8 ( [5], Theorem 9). Let f (x) be an irreducible polynomial of degree 2n over F

q

of order e(q

n

+ 1). Further let ψ(x) ∈ F

q

[x] be the minimal polynomial of β

qn

+ β + 1, where β = α

e

for a root α ∈ F

q2n

of f (x). Then the polynomial

x

qn+1

+ x

qn

− (β

qn

+ β + 1)x + 1

is irreducible over F

qn

. Moreover, ψ(x) and F (x) = x

n

ψ(

xqn+1+xx qn+1

) are irreducible polynomials over F

q

of degrees n and n(q

n

+ 1), respectively.

4

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Proof. Since ord(f (x)) = e(q

n

+ 1) and f (x) is irreducible, we have that α

e(qn+1)

= β

qn+1

= 1. Thus ord

q

(β) = q

n

+ 1 which does not divide q

k

− 1 for k ≤ n but q

2n

− 1.

Hence deg

q

(β) = 2n. Because β ∈ F

q2n

, (β

qn

+ β + 1)

qn

= β

qn

+ β + 1 which means λ := β

qn

+ β + 1 ∈ F

qn

. Next we show that deg

q

λ = n. Indeed, suppose that λ ∈ F

qd

for some divisor d of n. We have

βλ = β

qn+1

+ β

2

+ β = 1 + β

2

+ β,

and consequently, β

2

+ (1 − λ)β + 1 = 0. Therefore β is a root of the quadratic polynomial x

2

− (1 − λ)x + 1 over F

qd

, implying that [F

q2n

: F

qd

] ≤ 2 and thus d = n.

Since ψ(x) is the minimal polynomial of λ, deg(ψ(x)) = n.

Next we show that the conditions of Theorem 1.7 are fulfilled also. Indeed, since β ∈ F

q2n

\ F

qn

such that β

qn+1

= 1, choose x

0

= β and ω = −1. It remains to note that

ω−xqn0

ω−x0

=

−1−β−1−βqn

= β

qn

generates S. Therefore, by Theorem 1.7, x

qn+1

+ x

qn

− (β

qn

+ β + 1)x + 1 is irreducible over F

qn

.

To complete the proof, we show that F (x) is irreducible of degree n(q

n

+ 1) over F

q

. Since ψ(x) is the minimal polynomial of β

qn

+ β + 1 over F

q

,

ψ(x) =

n−1

Y

ν=0

(x − (β

qn

+ β + 1)

qν

). (1.4)

Substituting

xqn+1+xx qn+1

for x in (1.4), and multiplying both sides of the equation by x

n

, we obtain

F (x) = x

n

ψ( x

qn+1

+ x

qn

+ 1

x ) =

n−1

Y

ν=0

(x

qn+1

+ x

qn

− (β

qn

+ β + 1)

qν

x + 1)

=

n−1

Y

ν=0

(x

qn+1

+ x

qn

− (β

qn

+ β + 1)x + 1)

(ν)

.

By Theorem 1.4, F (x) is irreducible over F

q

since x

qn+1

+ x

qn

− (β

qn

+ β + 1)x + 1 is irreducible over F

qn

and deg

q

qn

+ β + 1) = n.

The following result by S. Cohen [3] was employed by several authors to give iter- ative constructions of irreducible polynomials over finite fields and Theorem 1.4 yields a proof for this result.

Theorem 1.9 ( [3], Lemma 1). Let f (x), g(x) ∈ F

q

[x] be relatively prime polynomials and let P (x) ∈ F

q

[x] be an irreducible polynomial of degree n. Then the composition

F (x) = g(x)

n

P (f (x)/g(x))

is irreducible over F

q

if and only if f (x) − λg(x) is irreducible over F

qn

for some root λ ∈ F

qn

of P (x).

5

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Proof. Let λ ∈ F

qn

be a root of P (x). Since all the roots of P (x) are the conjugates of λ, the polynomial P (x) is the product Q

n−1

ν=0

(x − λ

qν

) and thus F (x) = g(x)

n

P (f (x)/g(x)) =

n−1

Y

ν=0



f (x) − λ

qν

g(x) 

=

n−1

Y

ν=0



f (x) − λg(x) 

(ν)

is irreducible over F

q

if and only if f (x) − λg(x) is irreducible over F

qn

, by Theorem 1.4.

6

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2

Irreducibility of Polynomials of the Form g(x)

n

P (f (x)/g(x))

Let f (x), g(x) ∈ F

q

[x] and let P (x) = P

n

i=0

c

i

x

i

∈ F

q

[x] of degree n. Then the following composition

P (f /g) := g(x)

n

P (f (x)/g(x)) =

n

X

i=0

c

i

f (x)

i

g(x)

n−i

is again a polynomial in F

q

[x]. Theorem 1.9 establishes the conditions under which the composition poynomial P (f /g) is irreducible over F

q

[x].

Definition 2.1. For α ∈ F

qn

the trace of α, denoted by T r

Fqn/Fq

(α), is defined by T r

Fqn/Fq

(α) = α + α

q

+ · · · + α

qn−2

+ α

qn−1

.

For convenience, we denote T r

Fqn/Fq

= T r

qn/q

.

Definition 2.2. A trinomial is a polynomial with three nonzero terms, one of them being the constant term.

Definition 2.3. A polynomial of the form l(x) =

n

X

i=0

a

i

x

qi

with coefficients in F

q

is called a linearized polynomial over F

q

.

Definition 2.4. A polynomial of the form l(x) − b ∈ F

q

[x], where l(x) is a linearized polynomial over F

q

and b ∈ F

q

, is called an affine polynomial over F

q

.

Proposition 2.5 ( [7], Lemma 3.4). Suppose that the linearized polynomial l(x) has no nonzero root in F

q

. Then for any b ∈ F

q

, the affine polynomial l(x) − b has a linear factor x − A, A ∈ F

q

.

Proposition 2.6 ( [7], Theorem 3.5). With the notation of Proposition ??, the trino- mial x

p

− x − α is irreducible in F

q

[x] if and only if T r

q/p

(α) 6= 0.

Proposition 2.7 ( [7], Corollary 3.6). For a, b ∈ F

q

, the trinomial x

p

− ax − b is irreducible over F

q

if and only if a = A

p−1

for some A ∈ F

q

and T r

q/p

(b/A

p

) 6= 0.

7

(18)

Now we consider some special cases of P (f /g):

(a) f (x) = x

2

+ 1 and g(x) = x. Then P (f /g) = x

n

P (x + x

−1

). We distinguish the cases: q even and q odd.

Recall that if h(x) is a polynomial of degree k then its reciprocal is the polynomial h

(x) = x

k

h(1/x), and if h(x) = h

(x) then h(x) is said to be self-reciprocal.

Theorem 2.8. Let q = 2

m

and let P (x) = P

n

i=0

c

i

x

i

∈ F

q

[x] be irreducible over F

q

of degree n and with c

0

6= 0. Then x

n

P (x + x

−1

) is a self-reciprocal polynomial of degree 2n over F

q

, and

(i) x

n

P (x + x

−1

) is irreducible over F

q

if and only if T r

q/2

(c

1

/c

0

) 6= 0.

(ii) x

n

P

(x + x

−1

) is irreducible over F

q

if and only if T r

q/2

(c

n−1

/c

n

) 6= 0.

Proof. Let R(x) = x

n

P (x + x

−1

). Clearly, R(x) is of degree 2n and x

2n

R(1/x) = x

2n

x

−n

P (x + x

−1

) = R(x)

Thus R(x) is self-reciprocal.

Now we prove (i); the proof of (ii) is similar. By Theorem 1.9, R(x) is irreducible over F

q

if and only if x

2

+ 1 − αx is irreducible over F

qn

for some root α ∈ F

qn

of P (x).

By Proposition 2.7, the last condition is equivalent to T r

qn/2

−2

) 6= 0. Since T r

qn/2

−2

) = (T r

qn/2

−1

))

2

= (T r

q/2

(T r

qn/2

−1

)))

2

= (T r

q/2

(−c

1

/c

0

))

2

= (T r

q/2

(c

1

/c

0

))

2

, it is also equivalent to (T r

q/2

(c

1

/c

0

)) 6= 0.

Part (i) of Theorem 2.8 was obtained by Meyn ( [8], Theorem 6), and part (ii) is stated as Theorem 3.10(ii) in [7].

Theorem 2.9 ( [8], Theorem 8). Let q be a power of an odd prime and P (x) be an irreducible polynomial of degree n over F

q

. Then x

n

P (x + x

−1

) is irreducible over F

q

if and only if P (2)P (−2) / ∈ F

∗2q

.

Proof. By Theorem 1.9, x

n

P (x + x

−1

) is irreducible over F

q

if and only if x

2

− αx + 1 is irreducible over F

qn

, where α is a root of P (x). This is equivalent to the condition α

2

− 4 / ∈ F

∗2qn

, which is true if and only if

−1 = (α

2

− 4)

(qn−1)/2

= {[(2 − α)(−2 − α)]

(qn−1)/(q−1)

}

(q−1)/2

= {

n−1

Y

i=0

[(2 − α)(−2 − α)]

qi

}

(q−1)/2

= {

n−1

Y

i=0

(2 − α

qi

)(−2 − α

qi

)}

(q−1)/2

= {P (2)P (−2)}

(q−1)/2

8

(19)

that is, P (2)P (−2) / ∈ F

∗2q

.

Corollary 2.10 ( [7], Corollary 3.12). Let q be an odd prime power and P (x) be an irreducible polynomial of degree n over F

q

. Then 2

n

x

n

P ((x + x

−1

)/2) is irreducible over F

q

if and only if P (1)P (−1) / ∈ F

∗2q

.

Proof. Let P

0

(x) = 2

n

P (x/2) and apply Theorem 2.9 to P

0

(x).

(b) f (x) = x

p

− x − b and g(x) = 1. Then P (f /g) = P (x

p

− x − b).

Theorem 2.11 ( [7], Theorem 3.13). Let P (x) = x

n

+ a

n−1

x

n−1

+ · · · + a

0

be an irreducible polynomial over F

q

of characteristic p and let b ∈ F

q

. Then the polynomial P (f /g) = P (x

p

− x − b) is irreducible over F

q

if and only if T r

q/p

(nb − a

n−1

) 6= 0.

Proof. Let α be a root of P (x) in F

qn

. By Theorem 1.9, P (x

p

− x − b) is irreducible over F

q

if and only if x

p

− x − b − α is irreducible over F

qn

. By Proposition 2.6 this is equivalent to the condition

T r

qn/p

(b + α) = T r

q/p

(T r

qn/q

(b + α))

= T r

q/p

(nb − a

n−1

) 6= 0.

(c) f (x) = l(x) is a linearized polynomial and g(x) = 1. The irreducibility of these types of polynomials was established by Agou in a series of papers in 1977, 1978, 1980. First we consider the simple case l(x) = x

p

− ax, where a ∈ F

q

. Then P (f /g) = P (x

p

− ax).

Theorem 2.12 ( [7], Theorem 3.14). Let P (x) = x

n

+ c

n−1

x

n−1

+ · · · + c

0

be an irreducible polynomial over F

q

of characteristic p and let α be a root of P (x). Then for any a ∈ F

q

, P (x

p

− ax) is irreducible over F

q

if and only if

a

n1(q−1)/(p−1)

= 1 and T r

qn/p

(α/A

p

) 6= 0,

where n

1

= gcd(n, p − 1) and A ∈ F

qn

such that A

p−1

= a. In particular, if A ∈ F

q

then P (x

p

− A

p−1

x) is irreducible over F

q

if and only if T r

q/p

(c

n−1

/A

p

) 6= 0.

Proof. By Theorem 1.9, P (x

p

− ax) is irreducible over F

q

if and only if x

p

− ax − α is irreducible over F

qn

. By Proposition 2.7, this is equivalent to a = A

p−1

for some A ∈ F

qn

and T r

qn/p

(α/A

p

) 6= 0. Clearly, a = A

p−1

for some A ∈ F

qn

if and only if

a

(qn−1)/(p−1)

= 1 (2.1)

Since a ∈ F

q

, a

q−1

= 1. Thus (2.1) holds if and only if a

h

= 1, where h = gcd( q

n

− 1

p − 1 , q − 1) = q − 1

p − 1 gcd( q

n

− 1

q − 1 , p − 1)

9

(20)

But (q

n

−1)/(q−1) = q

n−1

+q

n−2

+· · ·+1 ≡ n (mod p−1). Hence h = n

1

(q−1)/(p−1).

Moreover, if A ∈ F

q

then a

n1(q−1)/(p−1)

= A

n1(q−1)

= 1 holds automatically and T r

qn/p

(α/A

p

) = T r

q/p

(T r

qn/q

(α/A

p

))

= T r

q/p

(T r

qn/q

(α)/A

p

)

= −T r

q/p

(c

n−1

/A

p

) Therefore, the last assertion also holds.

Now we turn to the general case, i.e. l(x) is any linearized polynomial. To determine when P (l(x)) is irreducible for any linearized polynomial l(x), we need some preliminary results in [7].

Lemma 2.13. Given a linearized polynomial l(x) over F

q

, there exists another lin- earized polynomial g(x) over F

q

and an element r in F

q

such that

l(x) = g(x

p

− x) + rx.

Proof. Let l(x) = a

ν

x

pν

+ a

ν−1

x

pν−1

+ · · · + a

0

x. We use induction on ν to prove the lemma. The case ν = 0 is trivial. Suppose ν ≥ 1 and put

l(x) = l(x) − a

ν

(x

p

− x)

pν−1

= (a

ν−1

+ a

ν

)x

pν−1

+ a

ν−2

x

pν−2

+ · · · ,

another linearized polynomial but of degree (at most) p

ν−1

. By induction, there is a linearized polynomial g(x) such that l(x) = g(x

p

− x) + rx. Then l(x) = g(x

p

− x) + a

ν

(x

p

− x)

pν−1

+ rx. Put g(x

p

− x) = g(x

p

− x) + a

ν

(x

p

− x)

pν−1

where g is the required linearized polynomial for the conclusion.

Lemma 2.14. Suppose the linearized polynomial l(x) over F

q

has a non-zero root A in F

q

. Then there exists a linearized polynomial g(x) such that l(x) = g(x

p

− A

p−1

x).

Proof. l(Ax) is a linearized polynomial over F

q

with 1 as a root. By Lemma 2.13, there exists another linearized polynomial ˜ g(x) and r ∈ F

q

such that l(Ax) = ˜ g(x

p

− x) + rx.

In fact, r = 0 because the substitution x = 1 yields 0 = ˜ g(0) + r = r. Thus l(Ax) =

˜

g(x

p

− x), which yields that l(x) = ˜ g(

xp−AApp−1x

) = g(x

p

− A

p−1

x) for some linearized polynomial g(x) = ˜ g(

Axp

).

Lemma 2.15. Suppose l(x) is a linearized polynomial over F

q

of degree p

ν

with ν ≥ 2.

Then for any b in F

q

, l(x) − b is irreducible over F

q

if and only if (i) p = ν = 2, and (ii) l(x) has the form

l(x) = x(x + A)(x

2

+ Ax + B) (2.2)

where A, B ∈ F

q

such that the quadratics x

2

+ Ax + B and x

2

+ Bx + b are both irreducible over F

q

.

10

(21)

Proof. By Proposition 2.5 we may assume that l(x) has a nonzero root A in F

q

. Using Lemma 2.14, we write l(x) = g(x

p

− A

p−1

x) and put g(x) = g(x) − b for some linearized polynomial g(x) over F

q

. Then l(x) − b = g(x

p

− A

p−1

x). Next, we apply the last assertion of Theorem 2.12 with P (x) = g(x) = x

n

+ b

n−1

x

n−1

+ · · · + b

1

x − b and n = deg(g(x)) = p

ν−1

. Since g is an affine polynomial, the coefficient b

n−1

of x

n−1

in g is zero unless p

ν−1

−1 = p

ν−2

which occurs only if p = ν = 2. Hence, T r

q/p

(b

n−1

/A

p

) = 0 and l(x) − b is reducible except when p = ν = 2. Now suppose that p = ν = 2, and g(x) = x

2

+ Bx, where B ∈ F

q

. Hence g(x) = x

2

+ Bx + b and

l(x) = g(x

2

− Ax) = x(x + A)(x

2

+ Ax + B)

By Theorem 2.12 again, l(x) − b = g(x

2

− Ax) is irreducible over F

q

if and only if g(x) = x

2

+ Bx + b is irreducible over F

q

and T r

q/p

(B/A

2

) 6= 0. The latter condition, by Proposition 2.7, is equivalent to x

2

+ Ax + B being irreducible over F

q

. This completes the proof.

Theorem 2.16 ( [7], Theorem 3.18). Let P (x) = x

n

+ P

n−1

i=0

c

i

x

i

be a monic irreducible polynomial of degree n over F

q

, and let l(x) be a monic linearized polynomial over F

q

of degree p

ν

with ν ≥ 2. Then P (l(x)) is irreducible over F

q

if and only if (i) p = ν = 2, (ii) n is odd, and (iii) l(x) has the form (2.2) where A, B ∈ F

q

and both x

2

+ Ax + B and x

2

+ Bx + c

n−1

are irreducible over F

q

.

Proof. By Theorem 1.9, P (l(x)) is irreducible over F

q

if and only if l(x)−α is irreducible over F

qn

, for some α ∈ F

qn

such that P (α) = 0. Applying Lemma 2.15 to l(x) − α, we conclude that P (l(x)) is irreducible over F

q

if and only if p = ν = 2, and l(x) has the form (2.2) where A, B ∈ F

qn

with both x

2

+ Ax + B and x

2

+ Bx + α irreducible over F

qn

.

Assume now that p = ν = 2. Then deg(l(x)) = 4 and deg(l(x)/x) = 3. If l(x)/x is irreducible over F

q

or a product of three linear factors over F

q

, then it remains so over F

qn

. So for l(x)/x to have a quadratic irreducible factor over F

qn

, it must be a product of a linear factor and a quadratic irreducible factor over F

q

, and, by Proposition 1.2, n must be odd so that the quadratic remains irreducible over F

qn

. Now assume further that l(x) is of the form (2.2) where A, B ∈ F

qn

with both x

2

+ Ax + B and x

2

+ Bx + α irreducible over F

qn

. Then A, B ∈ F

q

, x

2

+ Ax + B is irreducible over F

q

, and n is odd.

Finally, by Proposition 2.7, x

2

+ Bx + α is irreducible over F

qn

if and only if T r

qn/p

(α/B

2

) 6= 0. But

T r

qn/p

(α/B

2

) = T r

q/p

(T r

qn/q

(α/B

2

))

= T r

q/p

(T r

qn/q

(α)/B

2

)

= −T r

q/p

(c

n−1

/B

2

).

By Proposition 2.7 again, T r

q/p

(c

n−1

/B

2

) 6= 0 if and only if x

2

+Bx+c

n−1

is irreducible over F

q

. This completes the proof.

11

(22)

3

Recursive Constructions

Based on the irreducibility criteria developed in the previous chapter, we study how to recursively construct irreducible polynomials of arbitrarily large degrees.

First we introduce the following recursive construction of Varshamov [10].

Theorem 3.1. Let p be a prime and let f (x) = x

n

+ P

n−1

i=0

c

i

x

i

be irreducible over F

p

. Suppose that there exists an element a ∈ F

p

such that (na + c

n−1

)f

0

(a) 6= 0. Further let g(x) = x

p

− x + a and define f

k

(x) for k = 0, 1, 2, . . . recursively by

f

0

(x) = f (g(x)),

f

k

(x) = f

k−1

(g(x)) for k ≥ 1,

where f

k−1

(x) is the reciprocal polynomial of f

k−1

(x). Then for all k ≥ 0, f

k

(x) is irreducible over F

p

of degree np

k+1

.

Proof. For any k ≥ 0, let degf

k

(x) = n

k

and

f

k

(x) =

nk

X

i=0

b

ki

x

i

.

Denote by (P

k

) the family of claims:

• b

k1

= f

k0

(a) 6= 0,

• both f

k

(x) and f

k0

(x) are constant on F

p

,

• f

k

(x) is irreducible over F

p

,

• n

k

= np

k+1

.

We prove (P

k

) by induction on k.

When k = 0, we have

f

00

(x) = f

0

(g(x))g

0

(x)

12

(23)

Then

b

01

= (f

00

(x))|

x=0

= (f

0

(g(x))g

0

(x))|

x=0

= −f

0

(a) (since g(0) = a, g

0

(0) = −1) and

f

00

(a) = (f

0

(g(x))g

0

(x))|

x=a

= −f

0

(a) (since g(a) = a, g

0

(a) = −1)

Thus b

01

= f

00

(a) = −f

0

(a) 6= 0, by assumption. Clearly g(x) is constant on F

p

and g

0

(x) = −1, hence both f

0

(x) = f (g(x)) and f

00

(x) are constant on F

p

. Since degf

0

(x) = np, n

0

= np. From Theorem 2.11, f

0

(x) = f (g(x)) is irreducible over F

p

if and only if T r

p/p

(na + c

n−1

) = na + c

n−1

6= 0. By assumption na + c

n−1

6= 0, so f

0

(x) is irreducible over F

p

.

Now assume that (P

k

) is true for k ≥ 0. We prove that (P

k+1

) is also true. Since f (x) and f

(x) have the same degree and by induction hypothesis n

k

= np

k+1

, f

k+1

(x) = f

k

(g(x)) is of degree n

k+1

= np

k+2

. The constant term b

k0

6= 0 since f

k

(x) is irreducible, and also b

k1

6= 0 by induction hypothesis. Thus b

−1k0

f

k

(x) is monic and the coefficient of x

nk−1

is b

−1k0

b

k1

6= 0. Then

T r

p/p

(n

k

a + b

−1k0

b

k1

) = T r

p/p

(np

k+1

a + b

−1k0

b

k1

) = b

−1k0

b

k1

6= 0,

It follows from Theorem 2.11 that f

k+1

(x) = f

k

(g(x)) is irreducible over F

p

. By definition

f

k+1

(x) = f

k

(g(x)) =

nk

X

i=0

b

ki

g(x)

nk−i

Thus

f

k+10

(x) =

nk−1

X

i=0

b

ki

(n

k

− i)g(x)

nk−i−1

g

0

(x)

=

nk−1

X

i=0

b

ki

ig(x)

nk−i−1

(since g

0

(x) = −1)

Because g(x) is constant on F

p

, so are f

k+1

(x) and f

k+10

(x). Moreover, b

k+1,1

= (f

k+10

(x))|

x=0

= (f

k0

(g(x))g

0

(x))|

x=0

= −f

k0

(a)

= f

k0

(a

−1

)a

nk−2

= f

k0

(a)a

nk−2

,

13

(24)

which is nonzero by the induction hypothesis. Similarly,

f

k+10

(a) = (f

k+10

(x))|

x=a

= (f

k0

(g(x))g

0

(x))|

x=a

= −f

k0

(a) which is again non-zero as above. This completes the proof of (P

k+1

).

By induction (P

k

) holds for all k ≥ 0. In particular, for all k ≥ 0, f

k

(x) is irreducible over F

p

of degree np

k+1

.

The next construction is over F

q

, for q even, and is based on Theorem 2.8.

Theorem 3.2 ( [11], Theorem 10.26). Let q = 2

m

and let f (x) = P

n

i=0

c

i

x

i

be ir- reducible over F

q

of degree n with c

0

c

n

6= 0. Suppose that T r

q/2

(c

1

/c

0

) 6= 0 and T r

q/2

(c

n−1

/c

n

) 6= 0. For all k ≥ 0, define polynomials recursively:

f

0

(x) = f (x),

f

k

(x) = x

n2k−1

f

k−1

(x + x

−1

) for k ≥ 1.

Then f

k

(x) is a self-reciprocal irreducible polynomial of degree n2

k

over F

q

for all k ≥ 1.

Proof. It is easily seen by Theorem 2.8 and by induction on k that f

k

(x) is of degree n2

k

for every k ≥ 0 and f

k

(x) is a self-reciprocal polynomial for every k ≥ 1. We apply induction on k to prove that f

k

(x) is irreducible for every k ≥ 1. Since T r

q/2

(c

1

/c

0

) 6= 0 by assumption, f

1

(x) = x

n

f

0

(x + x

−1

) is irreducible by Theorem 2.8. Let k ≥ 1 and assume that f

k

(x) is irreducible. Let n

k

= n2

k

and f

k

(x) = P

nk

i=0

c

ki

x

i

, k ≥ 0. We have

f

k

(x) = x

nk−1

f

k−1

(x + x

−1

)

= x

nk−1

nk−1

X

i=0

c

k−1,i

(x + x

−1

)

i

= x

nk−1

nk−1

X

i=0

c

k−1,i

((1 + x

2

)/x)

i

=

nk−1

X

i=0

c

k−1,i

(1 + x

2

)

i

x

nk−1−i

=

nk

X

i=0

c

ki

x

i

.

Thus

c

k0

= c

k−1,nk−1

and c

k1

= c

k−1,nk−1−1

(3.1) By Theorem 2.8, f

k+1

(x) = x

n2k

f

k

(x + x

−1

) is irreducible over F

q

T r

q/2

(c

k1

/c

k0

) 6= 0 (3.2)

14

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