Irreducible polynomials with several prescribed coefficients

0.21 MB·

by  Junsoo Ha

Irreducible polynomials with several prescribed
6
1
coefficients
0
2

n
Junsoo Ha
a
J

6 January 27, 2016
2

]
T Abstract
N
We study the number of irreducible polynomials over F with some
q
.
h coefficients prescribed. Using the technique developed by Bourgain, we
t
a showthatthereisanirreduciblepolynomialofdegreenwithrcoefficients
m prescribed in any location when r [(1/4 ǫ)n] for any ǫ > 0 and q
≤ −
[ is large; and when r δn for some δ > 0 and for any q. The result
≤
1 improves earlier work of Pollack stating that a similar result holds for
v r [(1 ǫ)√n].
7 ≤ −
6
8
1 Introduction and Statement of Result
6
0
.
1 Theproblemoffindingirreduciblepolynomialswithcertainpropertieshasbeen
0
studied by numerous authors. One of the interesting problems among them is
6
1 the existence of an irreducible polynomial with certain coefficients being pre-
v: scribed.
i The early form of this problem is as follows. Let F be the finite field of q
X q
r elements and n be a given integer, and write a polynomial P = k≤nxkTk.
a
Then we ask if we canfind an irreducible for any givenpairof integers j, n and
P
a F satisfying x = a, except when a = 0 and j = 0. This problem, widely
q j
∈
known as the Hansen-Mullen conjecture, see [1], has been settled by Wan [2]
when n 36 or q > 19; the remaining cases were verified by Ham and Mullen
≥
[3].
One may ask if we can find an irreducible with several preassigned coeffi-
cients. Inotherwords,westudythenumberofirreduciblepolynomialsofdegree
n satisfyingx =a for alli , whenthe index set 0,1,...,n 1 anda
i i
∈I I ⊂{ − }
finite sequence a F for i are given. Unless we assume that the location
i q
∈ ∈I
1
of prescribed coefficients has certain properties,the best knownuniform bound
is due to Pollack [4], who proved that when n is large, there is an irreducible
polynomial with (1 ǫ)√n prescribed coefficients.
⌊ − ⌋
The analogue of the Hansen-Mullen conjecture in number theory is to find
rational primes with prescribed (binary) digits. Recently, Bourgain [5] showed
that for some δ > 0 and for large n, there is a prime of n digits with δn digits
prescribedwithout anyrestrictionontheir position. Thus itis believedthatan
analogous improvement holds for polynomials in finite fields.
In this paper, we show that we can prescribe a positive proportion of coef-
ficients. The result presented here is the combination of several known ideas.
The underlying setup in this type of problems is the circle method over F [T],
q
which can be found in Hayes [6]. A recent application of this method, among
others, can be found in Liu and Wooley [7] on Waring’s problem. The work of
Pollack [4] is also implicitly based on this method.
The main element of this paper is the combination of Pollack’s estimate
and the interpretation of the result of Bourgain [5] in finite fields, though it is
greatly simplified thanks to the Weil bound, i.e., the analogue of the Riemann
Hypothesis for irreducible polynomials.
Our main theorem is as follows.
Theorem1.1. Let beanonemptysubsetof 0,...,n 1 andchoosea F
i q
I { − } ∈
for each i . We write as S the set of monic degree n polynomials with Ti
∈ I
coefficient given by a for each i . Then if ρ:= /n 1/4,
i
∈I |I| ≤
1 = Sqn−|I| 1+O logq ρ1 +1 +O q3n/4 , (1)
  n  q1/ρ−(cid:16)4/(cid:17)(ρ+1)
P isPiXr∈reSducible  (cid:16) (cid:17)
  
 
where the implied constants are absolute, and
1 0 /
∈I
S=1+ 1 0 and a =0 (2)
 q−1 ∈I 0 6
0 0 and a =0.
0
∈I
Then the following corollary is a direct consequence of this theorem.
Corollary 1.2. We have the following.
1. There is δ > 0 so that for any q, n, there is an irreducible polynomial
2
of degree n with [δn] prescribed coefficients, unless the constant term is
prescribed to 0.
2. For any n 8, 0 < ǫ < 1/4 and q q (ǫ) for some large q , there is an
0 0
≥ ≥
irreducible polynomial of degreen with [(1/4 ǫ)n] prescribed coefficients,
−
unless the constant term is prescribed to 0.
3. When n is large and r =o(n), the number of irreducibles of degree n with
r prescribed coefficients is Sqn−r(1+o(1))/n .
Theimpliedconstantscanbeexplicitlycomputedandweconcludeasfollows.
Theorem 1.3. Suppose n 8, q 16, and r n/4 log n 1. Then there
≥ ≥ ≤ − q −
exists monic irreducible polynomial of degree n with r prescribed coefficients,
except when 0 is assigned in the constant term. We conclude the same when
q 5, n 97, and r n/5; or when n 52, r n/10 for arbitrary q.
≥ ≥ ≤ ≥ ≤
1.1 Notation
From now on, let T be an indeterminate and we denote the ring of polynomi-
als over F by F [T]. The polynomials play a parallel role of integers in this
q q
paper, so we keep the polynomials in F [T] as lowercase Latin letters whereas
q
parameters are usually written in capital letters. In particular, we substitute
the variable n in Theorem 1.1 by X. Also, we use m for monic polynomials
and ̟ for monic irreducible polynomials. The variable g usually means the
modulus, and is assumed to be monic.
Following the setup of Hayes [6], we let K be the formal power series
∞
F ((1/T))= a Ti ,whichisthecompletionofF [T]intheusualnorm
q i≪∞ i q
(cid:8)P (cid:9)
m =qdegm
| |
for polynomial m (with convention 0 =0.) We extend this norm to K by
∞
| |
x =qL
| |
where L is the largest index so that x = 0 (here and from now on, whenever
L
6
x K , the subscripted x denotes its Tk-coefficient.)
∞ k
∈
We define T by x K : x <1 , and fix an additive Haar measure,
∞
{ ∈ | | }
3
normalized so that dx=1. Finally we take a nontrivial additive character
T
´
2πi
e(x)=exp p trFq/Fp;(x−1) ,
(cid:18) (cid:19)
where p is the characteristic of F . Then e(x) has similar property as t
q
7→
exp(2πit)innumbertheory. Foronething,wehaveforapolynomiala F [T],
q
∈
1 a=0
e(ax)dx=
ˆT 0 a=0.
 6
We alsoadoptaconvenientnotationfromLiuandWooley[7]thatXˆ =qX and
(Z) = max log Z, 0 . For instance, X = (Xˆ) for X 1, and (m) =
L q L ≥ L | |
degm. This is useful as we can write m =Xˆ in place of degm=X. We also
(cid:0) (cid:1)
| |
use π (X) for the number of monic irreducible polynomials with degree X.
q
Now we define for α T,
∈
(α)= e(̟α)
S
|̟X|=Xˆ
and
(α)= e(mα)
I
S
m∈S
X
where S is as defined in Theorem1.1. Then the number of irreducible polyno-
mials in S is represented by the integral
N = (α) (α)dα. (3)
ˆTS SI
WeuseC todenotepositiveconstants,whichmaydependonmanyparameters
i
but is always absolutely bounded. For instance, we allow C =q/(q 1), which
−
isaconstantdependingonq butisabsolutelyboundedby2;however,wedonot
allowC =2q asitisnotabsolutelybounded. Duetotheirabundantappearance,
we label
qB
C = (4)
(qB) qB 1
−
for positive B for the remainder of this paper, which is absolutely bounded by
1+1/(2B 1) if B is bounded below by some positive constant.
−
4
2 Preliminaries
The following lemmas are counterparts in F [T] for well-known theorems in
q
number theory.
Lemma 2.1 (Rational Approximation). For each α T, there exist unique a,
∈
g F [T] so that g is monic, a < g Xˆ1/2 and
q
∈ | | | |≤
a 1
α < .
− g g Xˆ1/2
(cid:12)(cid:12) (cid:12)(cid:12) | |
(cid:12) (cid:12)
Proof. See Lemma 3 of [4] (cid:12) (cid:12)
We define a Farey arc
a a 1
,Rˆ = α T : α < . (5)
F g ∈ − g Rˆ
(cid:18) (cid:19) (cid:26) (cid:12) (cid:12) (cid:27)
(cid:12) (cid:12)
From Lemma 2.1, we decompose T into Far(cid:12)(cid:12)ey arcs(cid:12)(cid:12)
a
T= , g Xˆ1/2 .
F g | |
|a|<|g[|≤Xˆ1/2 (cid:18) (cid:19)
The Farey arcs in the above decomposition are pairwise disjoint; to prove, if α
lies in two Farey arcs centered at distinct fractions a /g and a /g ,
1 1 2 2
1 a a a a 1
1 2 1 2
>max α , α (6)
Xˆ1/2min(|g1|,|g2|) (cid:18)(cid:12) − g1(cid:12) (cid:12) − g2(cid:12)(cid:19)≥(cid:12)g1 − g2(cid:12)≥ |g1g2|
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
by the ultrametric inequality, w(cid:12)hich con(cid:12)tr(cid:12)adicts g(cid:12) X(cid:12)ˆ1/2. (cid:12)
i
| |≤
We define the set of major arcs by
a
M:= ,Xˆ
F g
|a|<|g[|≤Xˆ1/2 (cid:18) (cid:19)
and the minor arcs by
a a
m:=T M= , g Xˆ1/2 ,Xˆ .
− F g | | −F g
|a|<|g[|≤Xˆ1/2 (cid:18) (cid:19) (cid:18) (cid:19)
In Section 3, we use the fact that (α) is small on minor arcs, and is well
S
approximatedon majorarcs,and that the most contributionof the integral(3)
5
came from the major arcs.
Lemma 2.2 (Prime Number Theorem). We have
Xˆ Xˆ1/2 Xˆ
2 π (X) .
(Xˆ) − (Xˆ) ≤ q ≤ (Xˆ)
L L L
Proof. See Lemma 4 of [4].
Let φ(m) be the number of reduced residue classes mod m, i.e.,
1
φ(m)= m 1 .
| | − ̟
̟Y|m(cid:18) | |(cid:19)
In number theory, Euler totient function ϕ(n) has a lower bound (see Theorem
2.9 of [8])
n 1
ϕ(n) e−γ 1+O
≥ loglogn loglogn
(cid:18) (cid:18) (cid:19)(cid:19)
for n 3. The similar estimate holds for φ(m) as well.
≥
Lemma 2.3. For degm q and T ∤m, we have
≤
m
| | <e.
φ(m)
If degm>q, and T ∤m, we have
m
| | eγ( (m)+1).
φ(m) ≤ LL | |
Proof. The proof below is analogous to Theorem 2.9 of [8]. We write P =
A
̟ for a positive integerA, andwe say m is product ofirreducibles
̟6=T,|̟|≤Aˆ
with smallest possible degrees if P m and mP for some A, and let be
Q A| | A+1 R
the set of polynomials m satisfying
m m
1
| | | | (7)
φ(m) ≥ φ(m )
1
for any m F [T] such that m < m.
1 q 1
∈ | | | |
We claim that contains only the polynomials that are products of irre-
R
ducibles with smallest possible degrees. If m is a polynomial with k distinct
prime factors, we take m to be a polynomial with k prime factors that is the
1
productofirreduciblesofsmallestpossibledegrees. Thenifmisnottheproduct
6
of smallest possible degrees, m > m and
1
| | | |
−1 −1
m 1 1 m
1
| | = 1 < 1 = | |
φ(m) − ̟ − ̟ φ(m )
̟Y|m(cid:18) | |(cid:19) iY≤k(cid:18) | i|(cid:19) 1
where̟ isthechoiceofkpolynomialswithsmallestpossibledegrees. Therefore
i
m / and an element of is necessarily the product of smallest possible
∈ R R
degrees.
We now show that it suffices to prove the lemma for m . For simplicity
∈R
let f be the right hand side of the inequality, i.e., f(m) = eγ( (m)+1) if
LL | |
degm > q and e if degm q. Note that f is an increasing function of m,
≤ | |
since 2eγ > e. Suppose that we proved the lemma for all polynomials in ,
R
and that the lemma is false. Then there is a counterexample and let m be the
0
counterexamplewhosedegreeis smallest. Fromthe assumption,m / . Thus
0
∈R
thereisapolynomialm suchthat m < m and m /φ(m )> m /φ(m ).
1 1 0 1 1 0 0
| | | | | | | |
Then
m m
1 0
| | > | | >f(m ) f(m )
0 1
φ(m ) φ(m ) ≥
1 0
and thus m is also a counterexample for the lemma, which contradicts the
1
choice of m .
0
It remains to prove the lemma for m . Since each polynomial in is
∈ R R
the product of irreducibles with smallest possible degrees, we prove the lemma
in this case.
When m=P ,
1
−1 q−1
m 1 1
= 1 = 1+ <e,
φ(m) − ̟ q 1
deg̟=1,̟6=T(cid:18) | |(cid:19) (cid:18) − (cid:19)
Y
and thus all m < m satisfies (7).
1
| | | |
When m=P with A>1,
A
m 1 πq(r)
| | <e 1+ .
φ(m) qr 1
1<r≤A(cid:18) − (cid:19)
Y
Since rπ (r) qr 1, π (r)log(1+1/(qr 1)) 1/r and thus
q q
≤ − − ≤
m
| | <eP1≤r≤A1/r.
φ(m)
7
From Euler-Maclaurin formula, we have
1 1
logA+γ+
r ≤ 2A
r≤A
X
and that for 0<x<3,
x2
ex 1+x+ .
≤ 2(1 x/3)
−
Combining these two estimate, we have
m 1 1
| | <eP1≤r≤A1/r <A+ + .
φ(m) 2 8(A 1/6)
−
Since
degP =( 1)+ kπ (k) kπ (k)=qA 1,
A q q
− ≥ −
rX≤A Xk|A
we have
1
A log (degP +1) (P )+
≤ q A ≤LL | A| qA 1
−
and thus m /φ(m)<eγ( (m)+1).
| | LL | |
Finally, for mP and P m, we have
A A−1
| |
m P 1 degm−degPA−1
A−1
| | = | | 1+ .
φ(m) φ(P ) qA 1
A−1 (cid:18) − (cid:19)
Therefore we observe that log (m /φ(m)) is linear in degm. To be pre-
q | |
cise, let g(D) be the piecewise linear continuous function defined on D
≥
degP whose breakpoints are D = degP for A 1 and satisfies g(degP ) =
1 A A
≥
log(P /φ(P )). From construction, we have log(m /φ(m)) g(degm).
A A
| | | | ≤
Also, we have g(D) log(log D + 1) on each breakpoint of g and since
≤ q
log(log D +1) is convex, we have g(D) log(log D +1) for any D P .
q ≤ q ≥ 1
Therefore we conclude that
m
| | eg(degm) eγ log degm+1 =eγ( (m)+1)
φ(m) ≤ ≤ q LL | |
(cid:0) (cid:1)
as desired.
8
2.1 Analysis on
I
S
The norm of can be explicitly computed.
I
S
Lemma 2.4. We have
(α) dα=1.
ˆT|SI |
Proof. The set S can be rewritten as
S =m : m =Xˆ, m=TX + a Tj+ x Tj for some x F ,
j j j q
 | | Xj∈I Xj∈/I ∈ 
j<X
so we have 
(α)=e(αTX) e(a Tjα) e(x Tjα)
I j j
S
jY∈I jY∈/IxXj∈Fq
qX−|I|e(αTX) e(a Tjα) α =0 for all j /
= j∈I j −j−1 ∈I (8)

0 Q otherwise.

From the definition, (α) depends only on the first X coefficients of Laurent
I
|S |
series expansion, and thus it is constant on the range α a/TX < 1/Xˆ for
−
each polynomial a F [T]. Therefore
q (cid:12) (cid:12)
∈ (cid:12) (cid:12)
1 a
(α) dα= =q−|I| 1=1
ˆT|SI | Xˆ SI TX
|aX|<Xˆ(cid:12) (cid:16) (cid:17)(cid:12) |aX|<Xˆ
(cid:12)(cid:12) (cid:12)(cid:12) aX−j=0∀j∈/I
which proves the lemma.
We need the following coveringlemma to apply Bourgain’stechnique where
he simply used κ = 2 in the theorem below. We slightly improve the constant
so that we may apply when the density /X is close to 1/4.
|I|
Lemma 2.5 (Covering Lemma). Let x, y be integers with 1 y x and
≤ ≤
I [1,x] be a given set of integers. Then there exists a set of consecutive
⊆
integers J of length y so that
I J
| ∩ | κρ
J ≤
| |
9
where ρ= I /x and κ 2 is given by
| | ≤
2 1<x/y <2
ρ+1
κ(x, y, ρ)= 2u u 1<x/y <u
(u+1)ρ+(u−1) −
1 y x.
|
Proof. If x is multiple of y, the result is trivial because [1,x] can be covered
by nonoverlapping intervals of length y, and by the box principle, at least one
subinterval,sayJ satisfiesthe density I J / J ρ;soweassumeotherwise.
| ∩ | | |≤
We write I = z and u = x/y 2. Now, we cover [1,x] into u intervals of
| | ⌈ ⌉ ≥
length y, say J , , J where the smallest element of each J is [(i 1)x/u].
1 u i
··· −
Then we set κ to satisfy
0
1
κ ρ= max min I J .
0 i
y I 1≤i≤u{| ∩ |}
Theexactformulaforκ isabitcomplicatedasy,z vary,butwecanfindsome
0
upper bound, and any upper bound for κ would work for κ in our lemma.
0
If u=2, as I varies,the minimum density min I J /y gets largestwhen
i i
| ∩ |
the intersection of I and J J is as large as possible. Let ρ = (2y x)/x,
1 2 y
∩ −
whichisthedensityoftheoverlappingintervaloutofthetotallength. Ifρ ρ ,
y
≤
we get the trivial bound
I 2ρ
κ ρ= | | = .
0
y ρ +1
y
If ρ > ρ , the minimum density gets largest when we take I to fill the overlap
y
and equally split the remaining to J J and J J ; then
1 2 2 1
− −
I + J J ρ+ρ
1 2 y
κ ρ= | | | ∩ | .
0
2 ≤ ρ +1
(cid:20) (cid:21) y
Thus in either case, κ=2/(ρ+1) works.
If u 3, the minimum density gets largest when the intersection of I and
≥
(J J ) is as large as possible; that is when I is small, I intersects each
i i+1
∩ | |
J J and the two tails J J , J J equally by I /(u+1), and when
Si∩ i+1 1− 2 u− u−1 | |
I is large,I coversalloverlappingintervals anddistribute remainingso that I
| |
intersects each J by almost equal length. To compute, let ρ =(uy x)/x. If
i y
−
10