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
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
Funda ¨ c Ozdemir 2012
All Rights Reserved
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.
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.
to my parents and my husband
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
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)
nP (f (x)/g(x)) 7
3 Recursive Constructions 12
4 Composed Product of Polynomials 16
Bibliography 20
viii
Introduction
Let F
qbe the finite field of order q = p
sand of characteristic p, where p is a prime and s is a positive integer, F
∗qbe its multiplicative group which is cyclic. A generator of the cyclic group F
∗qis called a primitive element of F
qand its minimal polynomial over F
pis 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
qand let β be a root of f (x). The field F
q(β) = F
qnis a degree n extension of F
qand 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))
nP (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
1
First Composition Method
We say that the degree of an element α over F
qis equal to k and write deg
q(α) = k if F
q(α) = F
qkor equivalently α ∈ F
qkand α / ∈ F
qνfor any proper divisor ν of k.
Similarly, we say that the degree of a subset A = {α
1, α
2, . . . , α
r} ⊂ F
qkover F
qis equal to k and write deg
q(α
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
qand 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
qof degree n remains irreducible over F
qkif and only if k and n are relatively prime.
Given 0 ≤ ν ≤ k − 1 and g(x) = P
mi=0
b
ix
i∈ F
qk[x], we use the notation g
(ν)(x) =
m
X
i=0
b
qiνx
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
qdkbe 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
Hence we can write the factorization of g(x) and g
(ν)(x), for 0 ≤ ν ≤ d − 1, over F
qdkas
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−1ν=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−1ν=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
qif and only if there is a monic irreducible polynomial g(x) = P
ki=0
g
ix
iover F
qdof degree k such that F
q(g
0, . . . , g
k) = F
qdand f (x) = Q
d−1ν=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
ki=0
g
ix
iof degree k over F
qdsuch 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
qsand by Lemma 1.3
f (x) =
s−1
Y
ω=0
h
(ω)(x) (1.2)
over F
qsand h
(ω)(x) = P
rkj=0
h
qjωx
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
For the proof of the converse, let g(x) = P
ki=0
g
ix
ibe an irreducible polynomial of degree k over F
qdwith F
q(g
0, . . . , g
k) = F
qdand let α ∈ F
qdkbe a zero of g(x). Further, let f (x) be the minimal polynomial of α over F
qof 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
qdforces l = d which means d divides n. Hence we have shown that F
qdis a subfield of F
qn= F
q(α), implying that F
q(α) = F
qd(α), i.e. F
qn= F
qdkand 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
m6= 2. Suppose β, θ ∈ F
qand 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
nh(
x1). Then the polynomial
F (x) = (x − γ)
nf
(x − γ)
−1(x
qn+ β)
h
∗(x − γ)
−1is 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)
nf
(x − γ)
−1(x
qn+ β)
=
n−1
Y
ν=0
x
qn− α
qνx + β + γα
qν=
n−1
Y
ν=0
x
qn− αx + β + γα
(ν)3
Since q
n> 2 and α
qνis a primitive element in F
qn, by Theorem 1.5 each of the polynomials g
(ν)(x) := (x
qn− α
qνx + 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
qnis 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
qνand the irreducible factor of g
(ν)is
Q
(ν)(x) = x
qn− α
qνx + β + γα
qνx − θ
qν= x
qn− θ
qn+ν− α
qν(x − θ
qν) x − θ
qνover 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−1ν=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)
nf
(x − γ)
−1(x
qn+ β) Q
n−1ν=0
x − θ
qν=
n−1
Y
ν=0
Q
(ν)(x)
Finally, the irreducibility of F (x) over F
qfollows 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
qand x
0∈ F
q2\ F
qsuch 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−xq 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
qof order e(q
n+ 1). Further let ψ(x) ∈ F
q[x] be the minimal polynomial of β
qn+ β + 1, where β = α
efor a root α ∈ F
q2nof 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
nψ(
xqn+1+xx qn+1) are irreducible polynomials over F
qof degrees n and n(q
n+ 1), respectively.
4
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
qλ = n. Indeed, suppose that λ ∈ F
qdfor 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
qnsuch 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
qngenerates 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+1for x in (1.4), and multiplying both sides of the equation by x
n, we obtain
F (x) = x
nψ( x
qn+1+ x
qn+ 1
x ) =
n−1
Y
ν=0
(x
qn+1+ x
qn− (β
qn+ β + 1)
qνx + 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
qsince x
qn+1+ x
qn− (β
qn+ β + 1)x + 1 is irreducible over F
qnand deg
q(β
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)
nP (f (x)/g(x))
is irreducible over F
qif and only if f (x) − λg(x) is irreducible over F
qnfor some root λ ∈ F
qnof P (x).
5
Proof. Let λ ∈ F
qnbe 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−1ν=0
(x − λ
qν) and thus F (x) = g(x)
nP (f (x)/g(x)) =
n−1
Y
ν=0
f (x) − λ
qνg(x)
=
n−1
Y
ν=0
f (x) − λg(x)
(ν)is irreducible over F
qif and only if f (x) − λg(x) is irreducible over F
qn, by Theorem 1.4.
6
2
Irreducibility of Polynomials of the Form g(x)
nP (f (x)/g(x))
Let f (x), g(x) ∈ F
q[x] and let P (x) = P
ni=0
c
ix
i∈ F
q[x] of degree n. Then the following composition
P (f /g) := g(x)
nP (f (x)/g(x)) =
n
X
i=0
c
if (x)
ig(x)
n−iis 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
qnthe 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
ix
qiwith coefficients in F
qis 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
qand 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
qif and only if a = A
p−1for some A ∈ F
∗qand T r
q/p(b/A
p) 6= 0.
7
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
nP (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
kh(1/x), and if h(x) = h
∗(x) then h(x) is said to be self-reciprocal.
Theorem 2.8. Let q = 2
mand let P (x) = P
ni=0
c
ix
i∈ F
q[x] be irreducible over F
qof degree n and with c
06= 0. Then x
nP (x + x
−1) is a self-reciprocal polynomial of degree 2n over F
q, and
(i) x
nP (x + x
−1) is irreducible over F
qif and only if T r
q/2(c
1/c
0) 6= 0.
(ii) x
nP
∗(x + x
−1) is irreducible over F
qif and only if T r
q/2(c
n−1/c
n) 6= 0.
Proof. Let R(x) = x
nP (x + x
−1). Clearly, R(x) is of degree 2n and x
2nR(1/x) = x
2nx
−nP (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
qif and only if x
2+ 1 − αx is irreducible over F
qnfor some root α ∈ F
qnof P (x).
By Proposition 2.7, the last condition is equivalent to T r
qn/2(α
−2) 6= 0. Since T r
qn/2(α
−2) = (T r
qn/2(α
−1))
2= (T r
q/2(T r
qn/2(α
−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
nP (x + x
−1) is irreducible over F
qif and only if P (2)P (−2) / ∈ F
∗2q.
Proof. By Theorem 1.9, x
nP (x + x
−1) is irreducible over F
qif 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)/28
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
nx
nP ((x + x
−1)/2) is irreducible over F
qif and only if P (1)P (−1) / ∈ F
∗2q.
Proof. Let P
0(x) = 2
nP (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−1x
n−1+ · · · + a
0be an irreducible polynomial over F
qof characteristic p and let b ∈ F
q. Then the polynomial P (f /g) = P (x
p− x − b) is irreducible over F
qif 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
qif 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−1x
n−1+ · · · + c
0be an irreducible polynomial over F
qof 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
qif 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
∗qnsuch that A
p−1= a. In particular, if A ∈ F
∗qthen P (x
p− A
p−1x) is irreducible over F
qif 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
qif and only if x
p− ax − α is irreducible over F
qn. By Proposition 2.7, this is equivalent to a = A
p−1for some A ∈ F
∗qnand T r
qn/p(α/A
p) 6= 0. Clearly, a = A
p−1for some A ∈ F
∗qnif 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
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
∗qthen 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
qand an element r in F
qsuch that
l(x) = g(x
p− x) + rx.
Proof. Let l(x) = a
νx
pν+ a
ν−1x
pν−1+ · · · + a
0x. 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−2x
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−1where g is the required linearized polynomial for the conclusion.
Lemma 2.14. Suppose the linearized polynomial l(x) over F
qhas 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−1x).
Proof. l(Ax) is a linearized polynomial over F
qwith 1 as a root. By Lemma 2.13, there exists another linearized polynomial ˜ g(x) and r ∈ F
qsuch 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−1x) for some linearized polynomial g(x) = ˜ g(
Axp).
Lemma 2.15. Suppose l(x) is a linearized polynomial over F
qof degree p
νwith ν ≥ 2.
Then for any b in F
q, l(x) − b is irreducible over F
qif 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
qsuch that the quadratics x
2+ Ax + B and x
2+ Bx + b are both irreducible over F
q.
10
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−1x) 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−1x). Next, we apply the last assertion of Theorem 2.12 with P (x) = g(x) = x
n+ b
n−1x
n−1+ · · · + b
1x − b and n = deg(g(x)) = p
ν−1. Since g is an affine polynomial, the coefficient b
n−1of x
n−1in g is zero unless p
ν−1−1 = p
ν−2which 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
qif and only if g(x) = x
2+ Bx + b is irreducible over F
qand 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−1i=0
c
ix
ibe a monic irreducible polynomial of degree n over F
q, and let l(x) be a monic linearized polynomial over F
qof degree p
νwith ν ≥ 2. Then P (l(x)) is irreducible over F
qif and only if (i) p = ν = 2, (ii) n is odd, and (iii) l(x) has the form (2.2) where A, B ∈ F
qand both x
2+ Ax + B and x
2+ Bx + c
n−1are irreducible over F
q.
Proof. By Theorem 1.9, P (l(x)) is irreducible over F
qif and only if l(x)−α is irreducible over F
qn, for some α ∈ F
qnsuch that P (α) = 0. Applying Lemma 2.15 to l(x) − α, we conclude that P (l(x)) is irreducible over F
qif and only if p = ν = 2, and l(x) has the form (2.2) where A, B ∈ F
qnwith 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
qor 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
qnwith 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
qnif 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−1is irreducible over F
q. This completes the proof.
11
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−1i=0
c
ix
ibe irreducible over F
p. Suppose that there exists an element a ∈ F
∗psuch 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
pof degree np
k+1.
Proof. For any k ≥ 0, let degf
k(x) = n
kand
f
k(x) =
nk
X
i=0
b
kix
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
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
pand 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
pif and only if T r
p/p(na + c
n−1) = na + c
n−16= 0. By assumption na + c
n−16= 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
k06= 0 since f
k(x) is irreducible, and also b
k16= 0 by induction hypothesis. Thus b
−1k0f
k∗(x) is monic and the coefficient of x
nk−1is b
−1k0b
k16= 0. Then
T r
p/p(n
ka + b
−1k0b
k1) = T r
p/p(np
k+1a + b
−1k0b
k1) = b
−1k0b
k16= 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
kig(x)
nk−iThus
f
k+10(x) =
nk−1
X
i=0
b
ki(n
k− i)g(x)
nk−i−1g
0(x)
=
nk−1
X
i=0
b
kiig(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
k∗0(g(x))g
0(x))|
x=0= −f
k∗0(a)
= f
k0(a
−1)a
nk−2= f
k0(a)a
nk−2,
13
which is nonzero by the induction hypothesis. Similarly,
f
k+10(a) = (f
k+10(x))|
x=a= (f
k∗0(g(x))g
0(x))|
x=a= −f
k∗0(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
pof 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
mand let f (x) = P
ni=0
c
ix
ibe ir- reducible over F
qof degree n with c
0c
n6= 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−1f
k−1(x + x
−1) for k ≥ 1.
Then f
k(x) is a self-reciprocal irreducible polynomial of degree n2
kover F
qfor 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
kfor 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
nf
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
kand f
k(x) = P
nki=0
c
kix
i, k ≥ 0. We have
f
k(x) = x
nk−1f
k−1(x + x
−1)
= x
nk−1nk−1
X
i=0
c
k−1,i(x + x
−1)
i= x
nk−1nk−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)
ix
nk−1−i=
nk
X
i=0