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The Hansen-Mullen Conjecture
Giorgos Kapetanakis
Sabanci University
October 11, 2016
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 1 / 45
1 Introduction
2 Wan’s proofPreliminariesWeighted sumsCompletion of the proof
3 The HMC for self-reciprocal irreducible polynomialsMotivationPreliminariesWeighted sumCompletion of the proofThe restriction k ≤ n/2Further research
4 Other extensions of the HMC
5 Different approaches
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 2 / 45
Introduction
Let q be a prime power. We denote by Fq the finite field of qelements.
A generator of the multiplicative group F∗q is called primitive.
A polynomial of degree n of Fq[X] is called primitive if one (henceall) of its roots is a primitive element of Fqn .
Primitive polynomials are irreducible, but not vice versa.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 3 / 45
Introduction
The properties of irreducible polynomials over Fq have proved to be ofgreat theoretical and practical interest. Such properties, that have beeninvestigated, include
primitivity,
normality,
having certain coefficients fixed to given values and
combinations of those properties with other interesting ones.
One such result is the following.
Theorem (Cohen (1990))
Let n ≥ 2 and t ∈ Fq with t 6= 0 if n = 2 or if n = 3 and q = 4. Thenthere exists a primitive polynomial of degree n over Fq with trace t.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 4 / 45
Introduction
Based on the above and on computational evidence, Hansen and Mullen(1992) stated two important conjectures.
Conjecture (Hansen-Mullen (1992))
Let a ∈ Fq, let n ≥ 2 and 0 ≤ j < n. Then there exists an irreduciblepolynomial f(X) = Xn +
∑n−1k=0 akX
k ∈ Fq[X] with aj = a except when
j = a = 0 or
q = 2m, n = 2, j = 1, and a = 0.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 5 / 45
Introduction
Conjecture (Hansen-Mullen (1992))
Let a ∈ Fq, let n ≥ 2 and 0 ≤ j < n. Then there exists a primitivepolynomial f(X) = Xn +
∑n−1k=0 akX
k ∈ Fq[X] with aj = a except when
j = 0 and a 6= (−1)nα, where α ∈ Fq is a primitive element;
n = 2, j = 1, and a = 0;
q = 4, n = 3, j = 2, and a = 0;
q = 4, n = 3, j = 1, and a = 0;
q = 2, n = 4, j = 2, and a = 1.
The exceptions are genuine.
Both conjectures are established.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 6 / 45
Introduction
Cohen (1990) proved the HMC (Hansen-Mullen Conjecture) forj = n− 1.
In order to further support their claim, Hansen and Mullen (1992)proved their conjecture for j = 0.
Wan (1997) proved the HMC for irreducible polynomials if n ≥ 36 orq > 19 with the help of character sums.
Soon, Ham and Mullen (1998) completed the proof, by providingexamples for the remaining cases with the help of computers.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 7 / 45
Introduction
Cohen (1990) proved the HMC for j = n− 1.
In order to further support their claim, Hansen and Mullen (1992)proved their conjecture for j = 0.
The work of Cohen (2000) and Cohen and Huczynska (2003) provethe HMC for j = 1.
Han (1996) and Cohen and Mills (2003) proved the HMC forj = n− 2 and n ≥ 5.
The work of Fan and Han (2004), Mills (2004) and Cohen and King(2004), established the HMC for j = n− 3 and n ≥ 7.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 8 / 45
Introduction
Fan and Han (2003) used p-adic methods to asymptotically prove theHMC for primitive polynomials for fixed n and q large enough, exceptwhen n/2 ≤ j ≤ n/2 + 1.
Fan and Han (2004) proved the HMC for primitive polynomials foreven q and odd n ≥ 7, by using character sums, p-adic methods,sieving techniques and computer searches.
Cohen (2006) proved the HMC for primitive polynomials for n ≥ 9, byusing character sums, p-adic methods, sieving techniques andcomputer searches.
Two years later, Cohen and Presern (2008) completed the proof ofthe Hansen-Mullen conjecture for primitive polynomials.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 9 / 45
Introduction
The main idea behind the techniques of most authors dates back to thework of Carlitz (1952) and still remains popular. Roughly, this method is:
1 Express the characteristic or a characteristic-like function for apolynomial with the desired properties with help of characters.
2 With the the help of character sum estimates or calculations, thisleads to a lower bound of the number of polynomials with the desiredproperties.
3 This implies sufficient asymptotic conditions for the existence of thedesired polynomial.
4 These conditions could be relaxed (i.e. with sieving) and/or give riseto effective results.
5 If desirable, one can we deal with the remaining cases with acase-by-case approach.
Nonetheless, there are also other approaches of the HMC.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 10 / 45
Wan’s proof
Now, we will present Wan’s proof for the HMC for irreducible polynomials.
D. Wan.Generators and irreducible polynomials over finite fields.Math. Comp., 66(219):1195–1212, 1997.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 11 / 45
Wan’s proof Preliminaries
Definition
Let G be a finite abelian group. A character of G is a grouphomomorphism G→ C∗, where C∗ stands for the multiplicative group ofC. The characters of G form a group under multiplication, which isisomorphic to G. This group is called the dual of G and denoted by G.Furthermore, the character χ0 : G→ C∗, where χ0(g) = 1 for all g ∈ G, iscalled the trivial character of G. Finally, by χ we denote the inverse of χ.
A character or exponential sum is a sum that involves characters.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 12 / 45
Wan’s proof Preliminaries
The simplest, albeit very important, form of character sum is the following.
Lemma (Orthogonality relations)
Let χ be a non-trivial character of a group G and g a non-trivial elementof G. Then ∑
x∈Gχ(x) = 0 and
∑χ∈G
χ(g) = 0.
The orthogonality relations are true for arbitrary group G.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 13 / 45
Wan’s proof Preliminaries
Definition
Given some F ∈ Fq[X], a Dirichlet character modulo F is a functionχ : Fq[X]→ C∗, such that
1 χ(G+ FH) = χ(G),
2 χ(GH) = χ(G)χ(H) and
3 χ(G) 6= 0 ⇐⇒ (G,F ) = 1,
for every G,H ∈ Fq[X].
Dirichlet characters were originally defined over Z,
Dirichlet characters modulo F are essentially the characters of(Fq[X]/F Fq[X])∗, extended to zero.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 14 / 45
Wan’s proof Preliminaries
For χ a Dirichlet character modulo M ∈ Fq[X] and n ∈ N set
cn(χ) :=∑d|n
n
d
∑P monic irreducible
deg(P )=n/d
χ(P )d.
In particular the logarithmic derivative of the Dirichlet L-function of χ(multiplied by u) is
∑∞n=0 cn(χ)un. It follows from the Riemann
Hypothesis from function fields, proven by Weil (1948), that:
Theorem (Weil)
Let χ be a Dirichlet character modulo M . Then
1 If χ 6= χ0 then|cn(χ)| ≤ (deg(M)− 1)q
n2 .
2 If χ 6= χ0 and χ(F∗q) = 1, then
|1 + cn(χ)| ≤ (deg(M)− 2)qn2 .
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 15 / 45
Wan’s proof Preliminaries
For H ∈ Fq[X], we define the von Mangoldt function as
Λ(H) =
deg(P ), if H is a power of the irreducible P ,
1, if H = 1,
0, otherwise.
It follows directly from the definition of Λ, that
cn(χ) =∑
H monicdeg(H)=n
Λ(H)χ(H).
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 16 / 45
Wan’s proof Preliminaries
By using Weil’s theorem and this characterization for cn(χ), one can seethat the following character sum estimates hold.
Corollary
Let χ be a Dirichlet character modulo M . Then
1 If χ 6= χ0 then∣∣∣∣∣∣∣∣∑
P monic irreducibledeg(P )=d
χ(P )
∣∣∣∣∣∣∣∣ ≤1
d(deg(M) + 1)qd/2.
2 If χ 6= χ0 and χ(F∗q) = 1, then∣∣∣∣∣∣∣∣1
d+
∑P monic irreducible
deg(P )=d
χ(P )
∣∣∣∣∣∣∣∣ ≤1
ddeg(M)qd/2.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 17 / 45
Wan’s proof Preliminaries
Similarly, we can estimate character sums that run through irreduciblepolynomials with constant term 1 (not necessarily monic).
Corollary
Let χ be a non-trivial Dirichlet character modulo M , such that χ(F∗q) = 1,then ∣∣∣∣∣∣∣∣
∑P irreducible with constant term 1
deg(P )=d
χ(P )
∣∣∣∣∣∣∣∣ ≤1
ddeg(M)qd/2.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 18 / 45
Wan’s proof Weighted sums
From now on, fix n, 1 ≤ j < n and a ∈ Fq and define the followingweighted sum
wa(n, j) :=∑
H monic of degree j
Λ(aH)∑
P≡aH (mod Xj+1)P monic irreducible of degree n
1.
If wa(n, j) 6= 0, then there exists a monic irreducible of degree n with itsj-th coefficient prescribed to a.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 19 / 45
Wan’s proof Weighted sums
From the orthogonality relations, we have that
wa(n, j) =∑H
Λ(aH)∑P
1
qj(q − 1)
∑χ
χ(P )χ(aH)
=1
qj(q − 1)
∑χ
(∑P
χ(P )
)(∑H
Λ(aH)χ(aH)
),
where we note that for a 6= 0,∣∣∣∣∣∑H
Λ(aH)χ(aH)
∣∣∣∣∣ =
∣∣∣∣∣∑H
Λ(H)χ(H)
∣∣∣∣∣ ,while the case a = 0 requires special, but similar, treatment.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 20 / 45
Wan’s proof Weighted sums
We then separate the term of χ0 and apply the previous results and weeventually get.
Proposition
We have ∣∣∣∣wa(n, j)− πnq − 1
∣∣∣∣ < 1
n(j2 + 2j)q(n+j)/2.
The above, combined with known lower bounds for πn, implies.
Corollary
Let a ∈ Fq, n ≥ 3 and 1 ≤ j < n. If
qn−j ≥ (q − 1)2(j + 1)4,
then there exists an irreducible P ∈ Fq[X], with deg(P ) = n and Pj = a,where Pj is the coefficient of Xj .
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 21 / 45
Wan’s proof Weighted sums
The latter is effective for j ≤ n/2. For larger values of j, we consider thereciprocal of a monic irreducible, i.e. an irreducible with constant termequal to 1. Let Gj−1 be the set of polynomials of Fq[X] of degree j − 1with constant term equal to 1 and fix a ∈ Fq, n and 1 ≤ j < n and define
Wa(n, j) :=∑
H∈Gj−1
Λ(H)∑
P irreducible of degree nand constant term equal to 1P≡H+aXj (mod Xj+1)
1.
If Wa(n, j) 6= 0, then there exists an irreducible P ∈ Fq[X] of degree nwith constant term equal to 1 and Pj = a. This means that its reciprocal,i.e. PR := XnP (1/X), is a monic irreducible of degree n with PRn−j = a.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 22 / 45
Wan’s proof Weighted sums
With a similar treatment as with wa(n, j), we show that
Proposition
If j > 1, then ∣∣∣∣Wa(n, j)−πnq
∣∣∣∣ < 1
nj2q(n+j−1)/2.
Again, by using known bounds for πn, the above implies the following.
Corollary
Let a ∈ Fq and 1 ≤ j < n. If
qn−j−1 ≥ (j + 1)4,
then there exists an irreducible P ∈ Fq[X] with constant term equal to 1and its j-th coefficient fixed to a.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 23 / 45
Wan’s proof Completion of the proof
The latter is effective for j ≤ n/2, but since this implies the existence of amonic irreducible polynomial of degree n with the coefficient of Xn−j
prescribed, it is actually effective for j ≥ n/2.By putting everything together, we end up with the following
Corollary
Let a ∈ Fq, n ≥ 3 and 1 ≤ j < n. If either
qn−j ≥ (q − 1)2(j + 1)4 or qj−1 ≥ (n− j + 1)4,
then there exists a monic irreducible P ∈ Fq[X] of degree n with thecoefficient of Xj equal to a.
A few computations verify that at least one of the requirements of theabove is true for any value of j if q > 19 or m ≥ 36, thus.
Theorem (Wan)
If q > 19 or m ≥ 36, then the HMC for irreducible polynomials is true.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 24 / 45
The HMC for self-reciprocal irreducible polynomials
Next, we will present the work of Garefalakis and K. on the HMC forself-reciprocal irreducible polynomials.
T. Garefalakis and G. Kapetanakis.On the Hansen-Mullen conjecture for self-reciprocal irreduciblepolynomials.Finite Fields Appl., 18(4):832–841, 2012.
T. Garefalakis and G. Kapetanakis.A note on the Hansen-Mullen conjecture for self-reciprocal irreduciblepolynomials.Finite Fields Appl., 35(C):61–63, 2015.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 25 / 45
The HMC for self-reciprocal irreducible polynomials Motivation
Given a polynomial Q ∈ Fq[X], its reciprocal QR is defined as
QR(X) = Xdeg(Q)Q(1/X).
One class of polynomials that has been intensively investigated is that ofself-reciprocal irreducible polynomials, that is, irreducible polynomials thatsatisfy QR(X) = Q(X). Besides their theoretical interest, self-reciprocalirreducible polynomials have been useful in applications, and in particularin the construction of error-correcting codes.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 26 / 45
The HMC for self-reciprocal irreducible polynomials Motivation
It is natural to expect that self-reciprocal monic irreduciblepolynomials over finite fields, with some coefficient fixed, exist. Here,we restrict ourselves to the case where q is odd and prove that thereexists a self-reciprocal irreducible monic polynomial over Fq, of degree2n with its k-th coefficient prescribed, provided that
qn−k−1
2 ≥ 16
5k(k + 5) +
1
2.
With this result in mind, we show that we can prescribe the k-thcoefficient of a self-reciprocal irreducible polynomial of degree 2n,provided that k ≤ bn/2c, with a small number of genuine exceptions.
Our proof is based on an estimate of a weighted sum, similar to theone that Wan considers, character sums and Carlitz’s (1967)characterization of self-reciprocal irreducible monic polynomials.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 27 / 45
The HMC for self-reciprocal irreducible polynomials Preliminaries
Carlitz (1967) characterized self-reciprocal irreducible monicpolynomials over Fq (srimp): Q is a srimp iff
Q(X) = XnP (X +X−1)
for some monic irreducible P of degree n, such that ψ(P ) = −1,where ψ(P ) = (P |X2 − 4), the Jacobi symbol of P modulo X2 − 4.
We compute
Q(X) =
n∑i=0
i∑j=0
(i
j
)PiX
n−i+2j ,
which imples
Qk =∑
0≤j≤kk−j∈2Z
(n− jk−j2
)Pn−j .
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 28 / 45
The HMC for self-reciprocal irreducible polynomials Preliminaries
In the last expression, Qk is written in terms of the large degreecoefficients of P . In order to express in terms of the low degreecoefficients of a polynomial, we define.
P = XnP (4X−1)
and prove that Pi = 4n−iPn−i and ψ(P ) = −εψ(P ), where ε = ±1,depending on q and n.
It follows that
Qk =∑
0≤j≤kk−j∈2Z
(n− jk−j2
)4jPj =
k∑j=0
δjPj ,
where
δj =
{(n−jk−j2
)4j , if k − j ≡ 0 (mod 2),
0, if k − j ≡ 1 (mod 2).
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 29 / 45
The HMC for self-reciprocal irreducible polynomials Preliminaries
We define τn,k : Gk → Fq, H 7→∑k
j=0 δjHj , whereGk := {H ∈ Fq[X] : deg(H) ≤ k and H0 = 1}. We have proved:
Proposition
Let n ≥ 2, 1 ≤ k ≤ n, and a ∈ Fq. Suppose there exists an irreducibleP ∈ Fq[X] with P0 = 1, such that ψ(P ) = ε and P ≡ H (mod Xk+1) forsome H ∈ Gk with τn,k(H) = a. Then there exists a srimp Q, of degree2n, with Qk = a.
Next, we need to correlate the inverse image of τn,k with Gk−1. In thisdirection, we prove.
Proposition
Let a ∈ Fq, n ≥ 2 and 1 ≤ k ≤ n. Let F =∑k
i=0 FiXi ∈ Fq[X], with
F0 = 1 and Fi = δk−iδ−1k , 1 ≤ i ≤ k − 1, and Fk = δ−1k (δ0 − a). Then
the map τ−1n,k(a)→ Gk−1 : H 7→ HF (mod Xk+1) is a bijection.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 30 / 45
The HMC for self-reciprocal irreducible polynomials Weighted sum
Let n ≥ 2, 1 ≤ k ≤ n and a ∈ Fq. Inspired by Wan’s work (1997) weintroduce the following weighted sum.
wa(n, k) =∑
H∈τ−1n,k(a)
Λ(FH)∑
ψ(P )=ε, P≡H (mod Xk+1)
1.
It is clear that if wa(n, k) > 0, then there exists some P such that P ≡ H(mod Xk+1) for some H ∈ Gk, with τn,k(H) = a and ψ(P ) = ε. Thenthere exists a srimp Q, of degree 2n with Qk = a.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 31 / 45
The HMC for self-reciprocal irreducible polynomials Weighted sum
Let U be the subgroup of (Fq[X]/Xk+1Fq[X])∗ that contains classes ofpolynomials with constant term equal to 1. The set Gk−1 is a set ofrepresentatives of U . Further, the group of characters of U consistsexactly of those characters of (Fq[X]/Xk+1Fq[X])∗ that are trivial on Fq,that is, U = {χ ∈ (Fq[X]/Xk+1Fq[X])∗ : χ(F∗q) = 1}. Using theseobservations and the orthogonality relations, we get that
wa(n, k) =1
qk
∑χ∈U
∑P∈Jnψ(P )=ε
χ(P )∑
H∈τ−1n,k(a)
Λ(FH)χ(H)
=1
qk
∑χ∈U
∑P∈Jnψ(P )=ε
χ(P )χ(G)∑
H∈Gk−1
Λ(H)χ(H),
where G is the inverse of F modulo Xk+1 and Jn stands for the set ofirreducible polynomials of degree n and constant term equal to 1.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 32 / 45
The HMC for self-reciprocal irreducible polynomials Weighted sum
We separate the term that corresponds to χ0 and we get∣∣∣∣∣∣wa(n, k)− πq(n, ε)
qk
∑H∈Gk−1
Λ(H)
∣∣∣∣∣∣ ≤1
qk
∑χ 6=χo
∣∣∣∣∣∣∑
P∈Jn, ψ(P )=ε
χ(P )
∣∣∣∣∣∣∣∣∣∣∣∣∑
H∈Gk−1
Λ(H)χ(H)
∣∣∣∣∣∣ ,where πq(n, ε) = |{P irreducible of degree n : ψ(P ) = ε}|.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 33 / 45
The HMC for self-reciprocal irreducible polynomials Completion of the proof
We will also use the following result.
Theorem (Garefalakis (2011))
Let ψ(P ) = (P |X2 − 4) be the Jacobi symbol of P modulo X2 − 4 and χbe a non-trivial Dirichlet character modulo Xk+1, where k ≥ 1. Thefollowing bounds hold:
1 For every n ∈ N, n ≥ 2,∣∣∣∣∣∣∑
P monic irreducible, ψ(P )=−1
χ(P )
∣∣∣∣∣∣ ≤ k+5n q
n2 .
2 For every n ∈ N, n ≥ 2, n odd,∣∣∣∣∣∣∑
P monic irreducible, ψ(P )=1
χ(P )
∣∣∣∣∣∣ ≤ k+5n q
n2 .
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 34 / 45
The HMC for self-reciprocal irreducible polynomials Completion of the proof
The latter, along with other results we have already seen yield:
∑H∈Gk−1
Λ(H) =qk − 1
q − 1,
∣∣∣∣∣∣∣∣∑
deg(H)=nH0=1
Λ(H)χ(H)
∣∣∣∣∣∣∣∣ ≤ 1 + kqn2 , for n ≥ 1.
∣∣∣∣∣∣∣∣∑
P irreducibleP0=1, ψ(P )=ε
χ(P )
∣∣∣∣∣∣∣∣ ≤k+5n q
n2 , for n ≥ 2,
where χ is a non-trivial Dirichlet character modulo Xk+1, such thatχ(F∗q) = 1 and either ε = −1, or ε = 1 and n is odd.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 35 / 45
The HMC for self-reciprocal irreducible polynomials Completion of the proof
All of the above combined give:
Theorem
Let n ≥ 2, 1 ≤ k ≤ n, and a ∈ Fq. There exists a srimp Q ∈ Fq[X], ofdegree 2n with Qk = a if the following bound holds.
πq(n,−1) ≥ k(k + 5)
n(√q + 1)q
n+k2 .
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 36 / 45
The HMC for self-reciprocal irreducible polynomials Completion of the proof
Further, Carlitz (1967) computed πq(n, ε) and those computations imply∣∣∣∣πq(n,−1)− qn
2n
∣∣∣∣ ≤ 1
2n
q
q − 1q
n3 ,
hence:
Theorem
Let n ≥ 2, 1 ≤ k ≤ n, and a ∈ Fq. There exists a srimp Q ∈ Fq[X], ofdegree 2n with Qk = a if the following bound holds.
qn−k−1
2 ≥ 16
5k(k + 5) +
1
2.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 37 / 45
The HMC for self-reciprocal irreducible polynomials The restriction k ≤ n/2
Our final step is to content ourselves for k ≤ n/2 and solve the resultingproblem. Using the theory developed earlier, we conclude that there existsa srimp over Fq of degree 2n with its k-th coefficient prescribed, if
πq(n,−1) >bn/2c(bn/2c+ 5)
n(√q + 1)(qbn/2c/2 − 1)qn/2.
This bound is always true for n ≥ 27. For n < 27 this bound is satisfiedfor the pairs (q, n) described below
n 3 4 5 6 7 8 9 10q ≥ 149 ≥ 839 ≥ 37 ≥ 59 ≥ 17 ≥ 23 ≥ 11 ≥ 13
n 11 12 13 14 15 16 17 18q ≥ 9 ≥ 9 ≥ 7 ≥ 7 ≥ 5 ≥ 7 ≥ 5 ≥ 5
n 19 20 21 22 23 24 25 26q ≥ 5 ≥ 5 ≥ 5 ≥ 5 ≥ 5 ≥ 5 ≥ 3 ≥ 5
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 38 / 45
The HMC for self-reciprocal irreducible polynomials The restriction k ≤ n/2
Summing up, we have theoretically proved that:
Corollary
If n ≥ 3 an integer and q a power of an odd prime, then there exists asrimp of degree 2n such that any of its bn/2c low degree coefficients isprescribed, if either n ≥ 27 or q ≥ 839.
For the remaining cases, computers searches have been employed. Thecomputer results, combined with the above imply the following.
Theorem
Let n ≥ 3 an integer and q a power of an odd prime. If k ≤ n/2 anda ∈ Fq, then there exists a srimp of degree 2n such that any of its k-thcoefficient is prescribed to a, unless
1 q = 3, n = 3, a = 0 and k = 1 or
2 q = 3, n = 4, a = 0 and k = 2.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 39 / 45
The HMC for self-reciprocal irreducible polynomials Further research
1 Can we extend this to k ≥ n/2?
2 What about even q?
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 40 / 45
Other extensions of the HMC
Now, we present some of the extensions of the HMC for irreduciblepolynomials that various authors have considered.
Many authors have considered prescribing the norn and the trace ofan irreducible polynomial (Wan (1997), Car (1999), Moisio (2008)and Omidi Koma, Panario and Wang (2010)).
Cohen (2005) proved that there exists an irreducible polynomial ofdegree n with its first k and last m coefficients prescribed whenever2 ≤ k +m ≤ n/3.
Garefalakis (2008) considered consecutive zero coefficients and heproved that for any c with 0 < c < 1 and any positive integer n suchthat (1− 3c)n ≥ 2 + 8 logq n, there exists an irreducible polynomialof degree n over Fq with any bcnc consecutive coefficients (otherthan the first or last) equal to 0.
Panario and Tzanakis (2012), Pollack (2013) and Ha (2016)considered prescribing multiple coefficients of an irreduciblepolynomial.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 41 / 45
Other extensions of the HMC
Similarly, much effort has been given in extending the HMC for primitivepolynomials.
Many authors have considerd prescribing the norm and trace of aprimitive polynomial (Cohen and Hachenberger (1999, 2000), Cohen(2000, 2012), Cohen and Huczynska (2003), Fan and Wang (2009)).
Chou and Cohen (2001) fixed the coefficient of X and the trace to 0,given that n ≥ 5.
There are several results about prescribing the first two (Han (1996,2009), Bao (1997), Cohen and Mills (2003), Cohen and King (2004)and Shuqin and Wenbao (2004)) and the first three (Cohen and King(2004), Fan and Han (2004) and Mills (2004)) coefficients of aprimitive polynomial.
Cohen and King (2004) and Fan and Han (2004) provided asymptoticresults about the existence of primitives with their first b(n− 1)/2ccoefficients prescribed.
Cohen (2004) proved that the first m, where m ≤ n/3, coefficients ofa primitive polynomial can be prescribed.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 42 / 45
Other extensions of the HMC
Another setting that has been investigated is prescribing coefficients ofprimitive and normal polynomials, where normal polynomials are thosewhose roots constitute a normal basis, that is a basis of the form{x, xq, . . . , xqn−1}.
Fan and Wang (2009) proved that there exists a primitive normalpolynomial with a prescribed coefficient when n ≥ 15.
Several efforts have been made in fixing the norm and trace of aprimitive normal polynomial (Cohen (2000), Cohen and Hachenberger(2000) and Cohen and Huczynska (2003)).
Fan and Wang (2009) proved the existence of a primitive normalpolynomial with its first two coefficients prescribed if n ≥ 5.
Asymptotic results about prescribing the last bn/2c (Fan and Han(2007)) and the first bn/2c (Fan (2009)) have also been established.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 43 / 45
Different approaches
Finally, it is worth mentioning that recently several different approacheshave emerged.
Ha [Finite Fields Appl. 40:10–25, 2016] introduced number-theoreticmethods of Bourgain (2015) that enabled him to prescribe rcoefficients, given that r ≤ (1/4− ε)n, for any ε > 0. This workimproved previous work of Pollack (2013), where r ≤ (1/4− ε)
√n.
Tuxanidy and Wang (A new proof of the Hansen-Mullen irreducibilityconjecture, arXiv:1604.0423 [math.NT], unpublished. Irreduciblepolynomials with prescribed sums of coefficients, unpublished) havealso introduced a new promising method to the area.
Giorgos Kapetanakis (Sabanci) The Hansen-Mullen Conjecture October 11, 2016 44 / 45
Thank You!
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