以解析組合的方法討論有限體下多項式的性質

國立交通大學

應用數學系

碩 士 論 文

以解析組合

以解析組合

以解析組合

以解析組合的方法討論

的方法討論

的方法討論有限體下

的方法討論

有限體下

有限體下

有限體下多項

多項

多項

多項式的性質

式的性質

式的性質

式的性質

Random Polynomials over Finite Fields

via Analytic Combinatorics

研

研

研

研

究

究

究

究

生

生

生:

生

:

: 江紀葳

:

江紀葳

江紀葳

江紀葳

指導教授

指導教授

指導教授

指導教授:

:

:

: 符麥克

符麥克

符麥克

符麥克

教授

教授

教授

教授

中 華 民 國 一 ○ 一 年 十 一 月

以解析組合的方法討論有限體下多項式的性質

Random Polynomials over Finite Fields

via Analytic Combinatorics

研 究 生: 江紀葳

Student: Chi-Wei Chiang

指導教授: 符麥克 Advisor: Michael Fuchs

國 立 交 通 大 學

應用數學系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics National Chiao Tung University

in Partial Fulfillment of the Requirements for the Degree of

Master in

Applied Mathematics November 2012

R

ANDOM

P

OLYNOMIALS OVER

F

INITE

F

IELDS

VIA

A

NALYTIC

C

OMBINATORICS

Chi-Wei Chiang

Department of Applied Mathematics,

National Chiao Tung University

This thesis was supervised by Dr. Michael Fuchs

November 12, 2012

Preface

Properties of irreducible factors of random polynomials over finite fields (similar to properties of irreducible factors of random integers) have been intensively studied in the mathematical literature. Such properties have applications in computer science, cryptography, coding theory, etc.

The purpose of this thesis is three-fold. First, we want to give a survey of re-sults which have been obtained in the literature on properties of irreducible factors of random polynomials over finite fields. Secondly, we want to demonstrate the useful-ness of analytic combinatorics to prove such results. Finally, we want to discuss some applications of these properties to polynomial factorization over finite fields. We will provide detailed proofs of all the results (some of the proofs have been only sketched in the literature).

We give a brief outline of the thesis. In Chapter 1, we will give a short outlook and summarize the results we are going to prove. In Chapter 2, we will recall some tools from analytic combinatorics with detailed proofs. In Chapter 3, we will use the results from Chapter 2 to prove our results on random polynomials. Finally, in Chapter 4, we will discuss applications of the results from Chapter 3.

中文摘要

中文摘要

中文摘要

中文摘要

數學領域經常研究有限體下隨機多項式質因式的性質（如整係數質因式的性質）。 而這些性質常常應用在資訊工程、密碼學、編碼理論...等方面上。 這篇論文有三個研究動機：第一個動機是想整理參考文獻中有關有限體下 隨機多項式質因式分解的性質。第二個動機是想展現如何用解析組合來證明這 些質因式分解性質的結果。第三個動機是想將質因式分解性質的結果應用在質 因式分解的演算法中。這些性質大多在文獻上只有粗淺的說明，而這篇論文將 提供這些性質的詳細的證明及結果。 而這篇論文主要的概述：第一章，我們將給個簡短的概況並整理後面即將 證明的結果。第二章，我們會給解析組合中常用的定理及詳細的證明。第三章， 我們會用第二章中的結果來證明有限體下隨機多項式質因式的性質。最後，第 四章，我們將討論第三章結果的應用。

誌

誌

誌

誌謝

謝

謝

謝辭

辭

辭

辭

首先，我要感謝我的指導教授符麥克老師。在這段時間，不論在課業上或生活 中，都從老師身上學到好多好多東西。 而接下來，感謝我的好同學、好夥伴們豐富了這兩年的生活。講到這裡， 就開始懷念以前一起出去玩、一起吃飯、打球、串門子、玩桌遊的日子。都是 因為有你們，害我撐到現在…哈！開玩笑的啦～謝謝你們讓我能有動力撐到現在。 而現在，遊學的遊學、工作的工作、實習的實習、當兵的當兵，約大家有空的 時間好好聚聚似乎很不容易。相信這兩年這些甜美的回憶將會永遠保存在大家 的心中！ 最後，感謝受我身體髮膚（還有零用錢、生活費、電腦），同時也持續支 持我的父母，讓我能很順利在求學的過程中學習，不必擔心生活方面的問題。 很開心能完成碩士學位，這在我人生中是一個很重要的里程碑。希望大家之後 能一起在各自的未來打拼！

Contents

1 Introduction 1

2 Some Tools from Analytic Combinatorics 5

2.1 Singularity Analysis . . . 6

2.2 Darboux’s and Hybrid Method . . . 20

2.3 Useful Functions . . . 24

3 Properties of Random Polynomials over Finite Fields 28 3.1 The Number of Irreducible Factors . . . 28

3.2 Squarefree Polynomials and K-free Polynomials . . . 33

3.3 The Degree of the Irreducible Factors . . . 37

3.4 Other Restrictions on the Degree of Irreducible Factors . . . 51 4 Application to Polynomial Factorization 59

Chapter 1

Introduction

Factorizing polynomials over the finite field Fq, where q is a prime power, may make a lot of effort in many fields such as cryptography [7, 15, 17], coding theory [4], number theory [6] and polynomial factorization over the integers [8, 13, 14, 19]. Consequently, algorithms for factorizing polynomials have been studied by many authors, e.g., see Bach and Shoup [2], Berlekamp [3], and R´onyai[18].

In the analysis of these algorithms properties of random polynomials over a fi-nite field have played an important role. The main purpose of this thesis is to survey these properties and to show how to use generating functions and analytic combina-torics to prove these results.

Generating functions are very helpful in this context, for instance take the prob-lem of counting monic polynomials over a finite field Fq as an example. Let P (z) denote the corresponding generating function, i.e., the n-th coefficient of P (z) is the number of monic polynomials over Fq of degree n. Then, monic polynomials can be counted by the irreducible factors in their unique prime factorization. This yields

P (z) =Y k≥1 1 + zk+ z2k+ z3k+ · · ·
Ik =Y k≥1  1 1 − zk Ik ,

where Ik is the number of monic irreducible polynomials of degree k. Of course, monic polynomials can also be counted directly which gives

P (z) = 1 + qz + q2z2+ q3z3 + · · · = 1 1 − qz.

Now, this simple example lends itself to many generalizations. For instance, consider the question of counting monic polynomials excluding irreducible factors of degree k1, k2, k3, ... , krin their prime factorization. Similar as above, we get

∞ Y k = 1 k 6= k1, · · · , kr 1 + zk+ z2k+ z3k+ · · ·
Ik = 1 1 − qz · r Y j=1 1 − zkj
Ikj_.

Observe that the right hand side is a meromorphic function with a simple pole at z = 1/q. Moreover, for z → 1/q, 1 1 − qz · r Y j=1 1 − zkj
Ikj _∼ 1 1 − qz · r Y j=1 1 − q−kj
Ikj_.

Using a method called singularity analysis, which will be introduced in Section 2, the latter asymptotic relation remains true on the coefficient level. This yields that the number of monic polynomials excluding irreducible factors of degree k1, k2, k3, ... , kr in their prime factorization is asymptotically equal to

qn r Y j=1

(1 − q−kj₎Ij_.

(Alternatively, this result can be obtained by the Inclusion-Exclusion principle.) Sometimes, however, generating functions in one variable are not enough. For instance, suppose we want to count the number of monic polynomials with a fixed number of irreducible factors in their prime factorization. Then, we need bivariate generating functions. More precisely, consider a second variable u which counts the

number of the irreducible factors in the prime factorization of a polynomial. Then, we have for the bivariate generating function.

P (z, u) =Y k≥1 1 + uzk+ u2z2k+ u3z3k+ · · ·
Ik =Y k≥1  1 1 − uzk Ik . By taking partial derivative with respect to u and letting u = 1, we obtain as coefficient of zn the cumulative number of irreducible factors in the prime factorization of all polynomials of degree n. We will see later that again an asymptotic expansion can be easily derived by singularity analysis. From this, one obtains then the expected value of the number of irreducible factors in the prime factorization of a random polynomial. Moreover, by taking higher derivatives, higher moments can be derived as well.

We conclude by giving a sketch of the thesis. First, in Chapter 2, we will in-troduce singularity analysis and some other analytic methods we need for deriving our results, such as Darboux’s method and the Hybrid method. Then, in Chapter 3, there will be four different topics. The first topic is mainly concerned with the number of irreducible factors of random polynomials. For instance, we will derive the probability of a polynomial of degree n being irreducible. The second topic will be about k-free polynomials. There are two cases in this section, the first case is when k = 2 and the second case will be the more general case. The third topic is discussing the degree of the irreducible factors of the polynomial, such as the maximal degree of the irreducible factors not greater than m, the maximal degree of the irreducible factors equals m and the maximal degree Dn[1] of the irreducible factors equals m1 and the second largest degree D[2]n of the irreducible factors equals m2. The fourth topic is about the degree of the irreducible factors being distinct and related questions. Finally, in Chapter 4, we are going to apply the results from Chaper 3 to polynomial factorization.

The results, we are going to present in Section 3 are summarized in the follow-ing table. (Xn is the number of irreducible factors in a random polynomial of degree n and ρ denotes the Dickman function.)

Properties Results Section Prob(Xn = 1) = In/q n 1 n+ O(q −n/2₎ 3.1 E (Xn) log n + O(1)

Var(Xn) log n + O(1)

Section

Prob(x ∈ squarefree) (for n = 0, 1) 1

3.2

Prob(x ∈ squarefree) (for n ≥ 2) 1 − 1/q E(degree of remaining part) Pk≥1

kIk

q2k_−qk

Prob(x ∈ k-free) (for n = 0, 1) 1 Prob(x ∈ k-free) (for n ≥ 2) 1 − 1/qk−1 E(degree of remaining part) Pj≥1

jIj qkj_−q(k−1)j Section Prob(m-smooth) ρ _mn
+ O(log n_m ) 3.3 Prob(D[1]n = m) _m1ρ _mn − 1
+ O log n_m2
Prob(D[1]n = m, D[2]n ≤ m/2) _m1ρ 2n_m − 2
+ O log n_m2
Prob(D[1]n = m1, D [2] n = m2 < m1) _m11_m2ρ  n m2 − m1 m2 − 1  + O_mlog n 1m2₂  Section Prob(D[1]n > Dn[2] > · · · ) Q_k≥1(1 + I_qkk)(1 − 1 qk) Ik 3.4 Prob(D[1]n > Dn[2] > · · · ) Q_k≥1(1 + I_q2k2k)(1 − 1 q2k)I2k

Chapter 2

Some Tools from Analytic

Combinatorics

We first recall Landau’s O notation.

Definition 2.1. Let f (n) and g(n) be two complex-valued functions. 1. If there exist constantsc, n0 ∈ N such that

|f (n)| ≤ c · |g(n)| , ∀n ≥ n0,

then we say thatf (n) is a big-O of g(n), which is denoted by f (n) = O(g(n)). 2. If for all constants > 0, there exists a n0 = n0() ∈ N such that

|f (n)| ≤  · |g(n)| , ∀n ≥ n0,

then we say thatf (n) is a small-o of g(n), which is denoted by f (n) = o(g(n)). 3. Iff (n)/g(n) → 1 as n → ∞, then we say that f (n) is asymptotic to g(n), which

As already mentioned in Chapter 1, in this thesis, generating functions will play an important role. We will frequently need their n-th coefficients. Therefore, we will recall the following standard notation from combinatorics.

Definition 2.2. Given a generating function f (z) , fn = [zn]f (z) denotes the coeffi-cient ofzninf (z).

2.1

Singularity Analysis

In this section, we treat generating functions as analytic functions. In the sequel, we will often encounter generating functions with a singularity at ζ and (local) behavior

f (z) =  1 − z ζ −α log 1 1 − z_ζ !β , where α and β are arbitrary complex numbers.

For convenience, we use a transformation to let the singularity z = ζ be on the unit circle |z| = 1. More precisely, note that by the scaling rule, we have

g(z) ≡ f (zζ) =  1 −zζ ζ −α log 1 1 −zζ_ζ !β = (1 − z)−α  log 1 1 − z β .

So, g(z) has a singularity at z = 1, and in this way, we can get fn = [zn]f (z) as follows,

fn= [zn]f (z) = ζ−n[zn]f (zζ) = ζ−n[zn]g(z) = ζ−ngn,

where gn = [zn]g(z). Since all generating functions can be brought in this form, we only need to discuss generating functions with a singularity at z = 1 and (local) behavior (1 − z)−α  log 1 1 − z β ,

where α and β are arbitrary complex numbers.

First, we consider the special case (1 − z)−r, where r ∈ Z≥1. Then, (1 − z)−r = ∞ X n=0 n + r − 1 n  zn. Consequently, the coefficients are given by

[zn] (1 − z)−r =n + r − 1 n  = (n + r − 1) (n + r − 2) · · · (n + 1) (r − 1)! = n r−1 (r − 1)!  1 + O 1 n  .

Now, for (1 − z)−α, where α ∈ C, we expect a similar result: [zn] (1 − z)−α =n + α − 1 α − 1  = cnα−1  1 + O 1 n  , (2.1) where c is a constant which will turn out to be related to the Gamma function Γ(α) which is defined as follows

Γ(α) := Z ∞

0

e−ttα−1dt

for <(α) > 0. Moreover, we recall the following integral representation 1 Γ(α) = 1 2πi I C (−t)−αe−tdt, (2.2) where the contour C comes from ∞ + i, goes around 0 in counterclockwise direction, and then goes back to ∞ − i (see [12, p. 745]).

The formula (2.1) is made precise in the following theorem.

Theorem 2.3. Given a function f (z) = (1 − z)−αwith_{α ∈ C \ Z}≤0, we have for the n-th coefficient fn= [zn]f (z) ∼ nα−1 Γ(α) 1 + X k≥1 ek nk ! , whereekis a polynomial inα of degree 2k.

Proof. We first prove the case where k = 1. By Cauchy’s coefficient formula, we have fn = 1 2πi Z C (1 − z)−α dz zn+1 , where we choose the contour C as follows:

C

R R

1 n

0 1 0 1 1

Figure 2.1: Our contour C is the curve on the left and we separate it into two parts, the center one and the right one.

Next, we break C into two parts eC and H, where e

C =z

z = Reiθ, R > 1, arcsin(1/nR) ≤ θ ≤ 2π − arcsin(1/nR) and H =  z     z = 1 +e iθ n , π/2 ≤ θ ≤ 3π/2  ∪ ( z      1 ≤ <(z) ≤ r R2₋ 1 n2, I(z) = ± 1 n ) .

Then, we consider the integral over eC and H separately. e

C : Since |z| = R, z−n_{is bounded by R}−n_{. Thus, the integral over eC is exponentially} small. More precisely,

    1 2πi Z e C (1 − z)−α dz zn+1     ≤ cR · 1 Rn+1 = O  1 Rn  , where c = sup|z|=R|(1 − z)−α|.

H : We are going to break H into 3 parts:          H+_{(n) =}n_z z = ω +_ni, 1 ≤ ω ≤ q R2₋ 1 n2 o ; H−_{(n) =}n_z z = ω −_ni, 1 ≤ ω ≤ q R2₋ 1 n2 o ; Ho_{(n) =}n_z z = 1 + e_niθ, θ ∈ [π₂,3π₂ ]o.

By the change of variable z = 1 + _nt, the integral over H becomes 1 2πi Z H (1 − z)−α dz zn+1 = 1 2πi Z ¯ H  −t n −α 1/n dt (1 + t/n)n+1 = n α−1 2πi Z ¯ H (−t)−α  1 + t n −n−1 dt with ¯H = ¯Ho_{∪ ¯H}+_{∪ ¯H}−_{, where} ¯ H+₌n_t t = ω + i, 0 ≤ ω ≤ √n2_R2_{− 1 − n}o_; ¯ H−=ntt = ω − i, 0 ≤ ω ≤ √n2_R2_{− 1 − n}o_; ¯ Ho =  tt = eiθ, θ ∈ [π 2, 3π 2 ]  .

Let t = ω + i = √ω2_{+ 1 e}iθ _{where θ = arg(t). The absolute value of |(−t)}−α_| is  (−t)−α=     √ ω2_{+ 1 e}i(θ+π)−<(α)−i=(α)     =    √ ω2_{+ 1}−<(α)−i=(α)   e−i(θ+π)(<(α)+i=(α)) =√ω2_{+ 1}−<(α)_e=(α)θ+=(α)π_{. (2.3)}

We will break ¯H+_{(and ¯H}−_{) into two parts according to whether <(t) ≤ log}2 n or not. Then, the integral of the part with <(t) > log2n is at most

    Z ∞+i log2_n+i (−t)−α  1 + t n −n−1 dt     = Z ∞ log2_n  (−ω − i)−α     1 + ω + i n     −n−1 dω

= O Z ∞ log2n √ ω2_{+ 1}−<(α) 1 + ω n    −n−1 dω  = O Z ∞ log2_n √ ω2_{+ 1}−<(α)_e−ω−ωn dω  = O Z ∞ log2_n ω−<(α)e−ωdω  . Now, if <(α) ≥ 0, it is easy to see that both ω−<(α) and e−ω will make the integral exponentially small; on the other hand, if <(α) < 0, by integration by parts Z ∞ log2_n ω−<(α)e−ωdω = −ω−<(α)e−ω     ∞ log2n − <(α) Z ∞ log2_n ω−<(α)−1e−ωdω = · · · = O(log2n)−<(α)e− log2n= O (log n)−2<(α)n− log n
, (2.4) which is exponentially small, too. The same estimate holds for the corresponding part of ¯H−_.

Next, for t with <(t) ≤ log2n, we use the following asymptotic expansion for 1 + _nt
−n−1:  1 + t n −n−1 = e−(n+1) log(1+t/n) = exp −(n + 1) " − − t n
1 − −t n
2 2 − −t n
3 3 − · · · #! = exp  −t + t 2 2n − t3 3n2 + · · ·  exp  −t n + t2 2n2 − t3 3n3 + · · ·  = e−t· exp t 2_{− 2t} 2n − 2t3_{− 3t}2 6n2 + 3t4_{− 4t}3 12n3 − · · ·  = e−t  1 + t 2_{− 2t} 2n + t4_{− 4t}3_{+ 4t}2 8n2 − 2t3_{− 3t}2 6n2 + · · ·  = e−t  1 + t 2_{− 2t} 2n + 3t4− 20t3_{+ 24t}2 24n2 + · · ·  . (2.5)

For k = 1, we only need (1 + t/n)−n−1= e−t1 + Olog_n4n. After plugging this in, we can add the part with <(t) > log2n since this part is again

expo-nentially small. Overall, we can let the contour become eH = eHo_{∪ eH}+_{∪ eH}− , where e H+_=t t = ω + i, ω ≥ 0 ; e H−=tt = ω − i, ω ≥ 0 ; e Ho ₌  tt = eiθ, θ ∈ [π 2, 3π 2 ]  . Then, Z e H (−t)−αe−tdt  1 + O log 4 n n  = 1 Γ(α)  1 + O log 4 n n  , where the last line follows from (2.2).

Finally, by putting every thing together, we obtain fn = nα−1 Γ(α)  1 + O log 4 n n  .

This concludes the proof of k = 1. For the general case, one only needs to use more terms in (2.5). This then gives

fn ∼ nα−1 Γ(α) 1 + X k≥1 ek nk ! .

Theorem 2.4. Given a function f (z) = (1 − z)−α 1_zlog_1−z1
β with _{α ∈ C \ Z}≤0, β ∈ C, we have for the n-th coefficient

fn = [zn]f (z) ∼ nα−1 Γ(α)(log n) β 1 +X k≥1 dk logkn ! , wheredkis a polynomial inα of degree 2k.

Proof. Again by Cauchy’s coefficient formula, fn = 1 2πi Z C (1 − z)−α 1 z log 1 1 − z β dz zn+1,

where the contour C = eC ∪ H is as in Theorem 2.3: e

C =z

z = Reiθ, R > 1, arcsin(1/nR) ≤ θ ≤ 2π − arcsin(1/nR) , and H =  z     z = 1 + e iθ n , π/2 ≤ θ ≤ 3π/2  ∪ ( z      1 ≤ <(z) ≤ r R2₋ 1 n2, I(z) = ± 1 n ) .

As in the proof of Theorem 2.3, we consider the integral over eC and H separately. e

C : Since |z| = R, z−n_{is bounded by R}−n_{. Thus, the integral over eC is exponentially} small     1 2πi Z e C (1 − z)−α 1 zlog 1 1 − z β dz zn+1     ≤ cR · 1 Rn+1 = O  1 Rn  , where c = sup|z|=R   (1 − z) −α 1 zlog 1 1−z
β . H : We are going to break H into 3 parts:

         H+_{(n) =}n_z z = ω +_ni, 1 ≤ ω ≤ q R2₋ 1 n2 o ; H−_{(n) =}n_z z = ω −_ni, 1 ≤ ω ≤ q R2₋ 1 n2 o ; Ho_{(n) =}n_z z = 1 + e_niθ, θ ∈ [π₂,3π₂ ] o . By the change of variable z = 1 + _nt,

1 2πi Z H (1 − z)−α 1 z log 1 1 − z β dz zn+1 (2.6) = 1 2πi Z H  −t n −α 1 1 + t/nlog 1 −t/n β 1/n dt (1 + t/n)n+1 = n α−1 2πi (log n) β Z ¯ H (−t)−α  1 −log(−t) log n β 1 + t n −n−1−β dt

with ¯H = ¯Ho_{∪ ¯H}+_{∪ ¯H}− , where ¯ H+₌n_t t = ω + i, 0 ≤ ω ≤ √n2_R2_{− 1 − n}o_; ¯ H−=ntt = ω − i, 0 ≤ ω ≤ √n2_R2_{− 1 − n}o_; ¯ Ho =  tt = eiθ, θ ∈ [π 2, 3π 2 ]  .

By (2.3), we have |(−t)−α| = √ω2_{+ 1}−<(α)_e=(α)θ_{. As before, we break ¯H}+ (and ¯H−

) into two parts according to whether <(t) ≤ log2n or not. The part with <(t) > log2n is at most

    Z ∞+i log2n+i (−t)−α  1 −log(−t) log n β 1 + t n −n−1−β dt     = O Z ∞ log2n | − ω − i|−<(α)     1 − log(−ω − i) log n     <(β)    1 + ω + i n     −n−1−<(β) dω ! = O Z ∞ log2n (ω2+ 1)−<(α)2      1 − log( √ ω2_{+ 1) + iθ} log n      <(β) ·     1 + ω + i n     −n−1−<(β) dω ! = O Z ∞ log2n |ω2_{+ 1|}−<(α) 2   1 − log(√ω2_{+ 1)} log n !2 +  θ log n 2   <(β) 2 ·     1 + ω + i n     −n−1−<(β) dω ! = O Z ∞ log2_n ωc   1 + ω n    −n−1 dω  , where c is a suitable constant.

Since this integral is exponentially small, we only need to consider the part where <(t) ≤ log2_n.

For this part, we use the following asymptotic expansion for 1 + _nt
−n−1−β:  1 + t n −n−1−β = e−(n+1+β) log(1+t/n) = exp −(n + 1 + β) " − − t n
1 − −t n
2 2 − −t n
3 3 − · · · #! = exp  −t + t 2 2n − t3 3n2 + · · · − t(1 + β) n + t2(1 + β) 2n2 − t3(1 + β) 3n3 + · · ·  = e−t  1 + O log 4_n n  .

Moreover, we have (again for <(t) ≤ log2n)  1 − log(−t) n β ∼X k≥0 β k  (−1)k log(−t) n k .

Now, as in the proof of Theorem 2.3, plugging this in and adding the tail <(t) > log2n which is exponentially small, shows that the integral (2.6) is asymptotic to nα−1 2πi (log n) β Z e H (−t)−α X k≥0 β k   −log(−t) log n k! e−tdt =X k≥0 nα−1 2πi (log n) β−k (−1)kβ k  Z e H (−t)−αlogk(−t)e−tdt =X k≥0 nα−1 2πi (log n) β−kβ k  dk dαk Z e H (−t)−αe−tdt  =X k≥0 dk dαk nα−1 Γ(α)(log n) β−kβ k  with eH = eHo_{∪ eH}+_{∪ eH}− , where e H+ =tt = ω + i, ω ≥ 0 ; e H− =t t = ω − i, ω ≥ 0 ;

e Ho =  tt = eiθ, θ ∈ [π 2, 3π 2 ]  . This proves the claimed result.

Definition 2.5. A domain is a ∆-domain at 1 if it can be written as ∆(R, φ) = {z | |z| < R, z 6= 1, | arg(z − 1)| > φ},

whereR > 1 and 0 < φ < π/2. Moreover, if a function is analytic in some ∆-domain, the function is called a∆-analytic function.

Theorem 2.6. Let α , β ∈ R and f (z) be a function that is analytic in ∆ := ∆ (R, φ). If f (z) = O (1 − z)−α  log 1 1 − z β! , wherez ∈ ∆ and approaching 1, then

fn= [zn]f (z) = O  nα−1(log n)β  . Similarly, if f (z) = o (1 − z)−α  log 1 1 − z β! , wherez ∈ ∆-domain and approaching 1, then

fn= [zn]f (z) = o(nα−1(log n)β) . Proof. By Cauchy’s coefficient formula, we have

fn = 1 2πi Z C f (z) dz zn+1,

and C is a closed contour in the unit disc. Since f (z) is not analytic at z = 1, we change the contour C into a union of following 4 parts:

             γ1 = {z 

|z − 1| = 1_n, | arg (z − 1) | ≥ θ } (inner circle) γ2 = {z



|z − 1| ≥ _n1, |z| ≤ r, arg (z − 1) = θ } (top line segment) γ3 = {z | |z| = r, | arg (z − 1) | ≥ θ } (outer circle) γ4 = {z



|z − 1| ≥ _n1, |z| ≤ r, arg (z − 1) = −θ } (bottom line segment), where 1 < r < R, and φ < θ < π₂, so that our contour C lies entirely inside our ∆-domain. We let fn[1] , fn[2] , fn[3] , fn[4] be the integral along γ1, γ2, γ3, γ4, i.e.,

f_n[i] = 1 2πi Z γi f (z) dz zn+1. Then, we have fn= 1 2πi Z C f (z) dz zn+1 = 1 2πi f [1] n + f [2] n + f [3] n + f [4] n  . So now, we will discuss the integrals separately.

1. Inner circle (γ1):

The line integral of fn[1] will be at most the length of γ1 times the maximum of |f (z)|,

f_n[1] ≤ |γ1| · max|f (z)| 

z ∈ γ1 .

Since f (z) is O(1 − z)−α log_1−z1
β, there is a constant c > 0 such that |f (z)| ≤ c · |1 − z|−α

log_1−z1  β

. Hence, for the contour γ1where |z − 1| = 1/n and | arg (z − 1) | ≥ θ,  f_n[1]≤ |γ₁| · max|f (z)|  z ∈ γ1 ≤ 2π1 n · c  1 − z −α |log 1 − z|β = O 1 n  · nα log |1 − z| + iφ   β = O 1 n  · nα _log2_{|1 − z| + φ}2
β/2 = O nα−1
· log2_{n + φ}2
β/2 ,

where φ ∈ [0, 2π). Next, observe for large n 

 

log2n + φ2
β/2 ≤ log2n + log2n
β/2 = O logβn
, if β ≥ 0, log2n + φ2
β/2

≤ log2n
β/2 = logβn = O logβn
, if β < 0. Consequently, fn[1] = O nα−1logβn
.

2. Rectilinear parts (γ2,γ4):

Again, there is a constant c > 0 such that |f (z)| ≤ c · |1 − z|−αlog_1−z1  β

. Then, by the change of variable z = 1 + _nteiθ, our integral is

 f_n[2]=      1 2πi Z r0 t=1 f  1 + t ne iθ  eiθ ndt 1 + _nteiθ
n+1      ≤ c 2π Z r0 1       t n −α log _−t1 neiθ β    ·     1 + t ne iθ     −n−1    eiθ n     dt = c 2πn α−1 Z r0 1 t−α     log −te iθ n     β ·     1 + t ne iθ     −n−1 dt = c 2πn α−1 Z r0 1

t−αlog(−t) + log eiθ − log n  β ·     1 + t ne iθ     −n−1 dt = c 2πn α−1_logβ n Z r0 1 t−α     1 − log(−t) + iθ log n     β ·     1 + t ne iθ     −n−1 dt = c 2πn α−1_logβ_n Z r0 1 t−α     1 − log t + i(θ + π) log n     β ·     1 + t ne iθ     −n−1 dt = c 2πn α−1_logβ_n Z r0 1 t−α "  1 − log t log n 2 + θ + π log n 2#β/2 ·     1 + t ne iθ     −n−1 dt, where r0 is a constant satisfying z = 1 + r0eiθ with |z| = r. But since as in the proof of Theorem 2.3, the integral over t ∈ log2n, ∞ is exponentially small,

so we can change the range to 1 ≤ t ≤ log2n. Then, obviously "  1 − log t log n 2 + θ + π log n 2#β/2 = O(1). Consequently, f_n[2] = O nα−1(log n)β Z log2n 1 t−α·     1 + t ne iθ     −n−1 dt ! . Since1 + _nteiθ≥ 1 + < _nteiθ
= 1 + _nt cos θ, we have

Z log2n 1 t−α     1 + t ne iθ     −n−1 dt ≤ Z ∞ 1 t−α  1 + t cos θ n −n−1 dt ≤ Z ∞ 1 t−αe−(n+1) log(1+t cos θn )dt ≤ Z ∞ 1 t−αe−(n+1)t cos θn dt ≤ Z ∞ 1 t−αe−t cos θdt,

where this integral is finite since 0 < θ < π₂. This yields,

f_n[2] = O nα−1(log n)β
. 3. Outer circle (γ3):

Since |z| = r, z−n is bounded by r−n. Thus, the integral fn[3] is exponentially small.

Since fn[1], fn[2], fn[3], fn[4]are all O  nα−1_{(log n)}β , so that gives fn = O  nα−1_{(log n)}β . The proof of the small-o part is similarly.

Definition 2.7. A log-power function at 1 is a finite sum of the form σ(z) = r X k=1 ck  log  1 1 − z  (1 − z)αk_,

whereα1 < · · · < αrand eachck(z) is a polynomial. A log-power function at a finite set of pointsZ = {ζ1, · · · , ζm}, is a finite sum

Σ(z) = m X j=1 σj  z ζj  , where eachσj is a log-power function at 1.

Theorem 2.8. If a function f (z) is analytic on a ζ ·∆ domain and there exist log-power functionsσ and τ such that

f (z) = σ(z/ζ) + O (τ (z/ζ)) asz → ζ in ζ · ∆, thenfn= [zn]f (z) will be asymptotic to

fn= ζ−nσn+ O ζ−nτn
, whereσn = [zn]σ(z) and τn = [zn]τ (z).

Proof. Let g(z) = f (ζz). Then, g(z) is ∆-analytic, with a singularity at 1 and g(z) = σ(z) + O (τ (z)) as z → 1. Now, since τ (z) is a log-power function, so

gn= [zn]g(z) = [zn]σ(z) + [zn]O (τ (z)) = [zn]σ(z) + O ([zn]τ (z))

= σn+ O(τn). Finally, since fn = ζ−ngn, we get

fn = ζ−ngn = ζ−n(σn+ O(τn)) = ζ−nσn+ O ζ−nτn
.

According to the latter result, we know how to find fnwhen f (z) has one singu-larity at ζ. Similarly, fn can be found if f (z) has finitely many singularities (see [12, p. 398] for a proof).

Theorem 2.9. Let f (z) be analytic in |z| < ρ with a finite number of singularities ζ1, · · · , ζk on the circle|z| = ρ. Suppose there exists a ∆-domain such that f (z) is analytic in the domain

D = k \ i=1

(ζi· ∆) ,

Moreover, we havek log-power functions σ1, · · · , σk, and a log-power functionτ (z) = (1 − z)−α(log_1−z1 )β _{such that}

f (z) = σi(z/ζi) + O (τ (z/ζi)) as z → ζi in D. Then, the coefficientsfn = [zn]f (z) satisfies

fn= k X i=1 ζ_i−n(σi)n+ O ρ −n nα−1(log n)β
, where(σi)n= [zn]σi.

The last result is quite powerful and has many applications (for some of them see Chapter 3). However, it can be only applied if f (z) is analytic in a domain which is larger than |z| < ρ. Sometimes, however, we only know that a function is analytic on |z| < ρ and analytic extension is either hard to prove or not possible (some examples for this will be given in Section 3.4). Then, singularity analysis cannot be applied and we need other methods. We are going to introduce two such methods in the next section.

2.2

Darboux’s and Hybrid Method

First, we introduce a method called Darboux’s method. In contrast to singularity anal-ysis, this method does not need that the function is analytically continuable beyond the disc of convergence. However, we will need some smoothness on the disc.

Definition 2.10. Let h(z) be a function which is analytic in |z| < 1 and s ∈ N ∪ {0}. If h(k)_{(z) is defined for |z| < 1 and has a continuous extension on |z| ≤ 1 for all integers} from 0 to s, then we callh(z) Cs_{-smooth on the unit disc.}

Remark 2.11. (Riemann-Lebesgue Lemma) If f (z) is L1 _{integrable and supported} on (0, ∞), then

Z ∞ 0

f (z)e−tzdz → 0, as |z| → ∞ within the half-plane =(z) ≥ 0.

Theorem 2.12. (Darboux’s Method) Assume that h(z) is Cs-smooth. Then, hn= [zn]h(z) = o

 1 ns

 . Proof. By Cauchy’s coefficient formula, we have

hn= 1 2πi Z C h(z) dz zn+1, where C is the unit circle. Now, let z = eiθ, so that

hn= 1 2πi Z 2π 0 h eiθ
ie iθ_dθ (eiθ₎n+1 = 1 2π Z 2π 0 h eiθ
e−niθdθ.

When s = 0, we getR₀2πh eiθ
_e−niθ_{dθ → 0 as n → ∞ by the Riemann-Lebesgue} Lemma. When s > 0, we use integration by parts s times and obtain

hn= 1 2π Z 2π 0 h eiθ
e−niθdθ = 1 2π Z 2π 0 1 nih

0 _eiθ
_ieiθ_e−niθ_dθ = 1 2π 1 n Z 2π 0 h0 eiθ
e−i(n−1)θdθ = 1 2π 1 n(n − 1) Z 2π 0 h00 eiθ
e−(n−2)iθdθ = · · · = 1 2π 1 n · · · (n − s + 1) Z 2π 0 h(s) eiθ
e−(n−s)iθdθ.

Then, again by the Riemann-Lebesgue Lemma, Z 2π

0

h(s) eiθ
e−(n−s)iθ_{dθ → 0.} Consequently, hn= o(n−s) as claimed.

Definition 2.13. A function f (z) which is analytic in the open unit disc is said to be of global ordera ≤ 0 if there exists a constant c such that |f (z)| ≤ c (1 − |z|)a for allz satisfying|z| < 1. In other words, for all |z| < 1, we have

f (z) = O ((1 − |z|)a) .

Theorem 2.14. (Hybrid Method) Letf (z) be a function that has a finite number of singularitiesZ = {ζ1, ... , ζm} with |z| = 1 and let U (z), V (z) be analytic functions on |z| < 1 satisfying f = U · V . Assume that V (z) is Cs_{-smooth on the unit disc.} Moreover, assume thatU (z) is of global order a ≤ 0 and that there exists a log-power function eU at Z such that U = eU + R with R a Ct-smooth function on the unit disc. Finally, supposet ≥ s+a₂ ≥ 0. Then, we have

fn= [zn]f (z) = [zn] eU (z) · V (z) + o(n−

s+a

2 ),

whereV is a polynomial.

Proof. _{First, fix a constant c ∈ N with c ≤ s. Next, let V = V + S, where V is a} polynomial of degree c that satisfies

∂i ∂ziV (z)     z=ζj = ∂ i ∂ziV (z)     z=ζj ,

where 0 ≤ i < c and 1 ≤ j ≤ m. Then, since U = eU + R, we have f = U · V =U + Re



· V = eU · V + R · V = eU · V + eU · S + R · V.

e U · V :

Since eU is a log-power function, and V is a polynomial, we can calculate the coefficient of eU · V by singularity analysis.

e U · S :

Since S = V − V and derivatives of order from 0 to c − 1 are the same at ζifor V and V , we have that derivatives of S of order from 0 to c − 1 disappears at ζi. Consequently, we can factorize S(z) into

S(z) = κ(z) m Y j=1 (z − ζj) c ,

where κ(z) is a Cs−c-smooth function. Then,

e U · S = U ·e m Y j=1 (z − ζj)c ! · κ(z).

Since U has a global order a, so eU is O ((z − ζj) a

) near ζj. Thus, eU · Qm

j=1(z − ζj) c

is at least a C(a+c)_{-smooth. Since a C}α_{-smooth function times a C}β_-smooth func-tion is at least a Cmin(α,β)-smooth function, so eU · S is a Cmin{a+c,s−c}-smooth function. From this and Theorem 2.12, we get

[zn] eU · S = o 1 nu  , where u = min {a + c, s − c}. R · V :

Since R is Ct-smooth and V is Cs-smooth, R · V is Cmin{s,t}-smooth. Set v = min{s, t}. Then, again by Theorem 2.12,

[zn] R · V = o 1 nv

 .

Summing up the three part yields

fn = [zn] eU · V + [zn] eU · S + [zn] R · V = [zn] eU · V + o(n−u) + o(n−v) = [zn] eU · V + o(n− min{v,u}) = [zn] eU · V + o(n− min{a+c,s−c,t}).

We want the minimum to be as large as possible, so we choose a + c = s − c which gives c = s−a₂ . Then, we have

fn= [zn] eP · Q + o  n−(a+s−a2 )  = [zn] eP · Q + on−s+a2  .

Note that [zn] eU · V can be obtained with singularity analysis (as already men-tioned in the proof). Hence, the hybrid method combines Darboux’s method with singularity analysis. We will use this method in Section 3.4.

2.3

Useful Functions

Here, we collect some useful functions. The first is the following one. Definition 2.15. The exponential integral E(a) is defined as

E (a) = Z ∞ a e−s s ds, where0 ≤ | arg(a)| < π, a 6= 0.

Remark 2.16. One important property of the exponential integral is that e−E(z) is bounded for all z with <(z) ≥ 0 (see [1] for a proof).

Next, we use the exponential integral to find a representation of the remainder of the logarithmic series.

Lemma 2.17. The remainders of the logarithm series rm(z) =

X k>m

zk k ,

Proof. First note rm(z) = Z z 0 r_m0 (t) dt = Z z 0 X k>m tk−1dt.

Then, by the plugging in z = e−hand using the change of variable t = e−u, we have rm e−h
= Z e−h 0 X k>m tk−1dt = Z h ∞ X k>m e−u(k−1) ! −e−u
du = Z ∞ h X k>m e−kudu = Z ∞ h e−u(m+1) 1 − e−u du = Z ∞ h e−mu eu_{− 1} du = Z ∞ h  e−mu u + e−mu eu_{− 1}− e−mu u  du = Z ∞ h  me −mu mu + e −mu  1 eu _{− 1}− 1 u   du = Z ∞ mh  me −s s + e −s  1 es/m_{− 1} − 1 s/m   1 mds = Z ∞ mh  e−s s + e−s m  1 es/m_{− 1} − 1 s/m   ds = Z ∞ mh e−s s ds + 1 m Z ∞ mh e−s  1 es/m_{− 1} − 1 s/m  ds = E (mh) + 1 m Z ∞ mh e−sφs m  ds, where φ(z) = _ez1₋₁ − 1

z. Now, if the function φ(z) is bounded, then 1 m R∞ mhe −s_φ s m
ds will be O(1/m). In order to show that φ(z) is bounded, note that

φ(z) → 0 as z → ∞, and φ(z) → −1 as z → −∞. Moreover, when z is approaching 0, we have

lim z→0 1 ez_{− 1} − 1 z = limz→0 z − ez+ 1 z (ez_{− 1)} = lim_z→0 1 − ez (ez _{− 1) + ze}z = lim z→0 −ez ez _{+ (e}z_{+ ze}z₎ = lim_z→0 −ez 2ez _{+ ze}z = lim z→0 −1 2 + z = − 1 2.

So, φ(z) is bounded and our claim is proved.

In Section 3, we will need another function which is very similar to φ. Lemma 2.18. Let ψ(z) be defined as

ψ(z) = 1 1 − e−z −

1 z. Thenψ(z) is also bounded.

Proof. First, observe

ψ(z) → 1 as z → ∞, and ψ(z) → 0 as z → −∞. Next, when z is approaching 0, we have

lim z→0 1 1 − e−z − 1 z = limz→0 z − 1 + e−z z (1 − e−z₎ = lim_z→0 1 − e−z (1 − e−z_{) + ze}−z = lim z→0 e−z e−z_{+ (e}−z_{− ze}−z₎ = lim_z→0 e−z 2e−z_{− ze}−z = lim z→0 1 2 − z = 1 2. Thus, ψ(z) is bounded.

Another function which will be needed later is the Dickman function.

Definition 2.19. The Dickman function ρ (u) is the unique continuous solution of the difference-differential equation    ρ (u) = 1 0 ≤ u ≤ 1, uρ0(u) = −ρ (u − 1) u > 1.

Lemma 2.20. The Laplace transform of the Dickman functionρ (s) satisfies s_b ρ (s) =_b e−E(s). Consequently, we have

ρ (u) = 1 2πi Z 1+i∞ 1−i∞ e−E(v) v e uv_dv.

As a final class of functions, we will need polylogarithms (see [16] and [12, p. 408]).

Definition 2.21. The polylogarithm Lim(z), with m ∈ N, is defined as Lim(z) =

X n≥1

zn nm. Remark 2.22. Note that Lim(z) is Cm−2-smooth.

Lemma 2.23. Lim(z) is analytically continuable to C \ [1, ∞). Moreover, the singu-larity expansion atz = 1 is given by

Lim(z) = (−1)m (m − 1)!τ m−1 (log τ − Hm−1) + X j≥0, j6=m−1 (−1)j j! ζ(m − j)τ j ,

whereτ , the harmonic numbers Hm and the Riemann zeta functionζ(s) are defined as τ = − log z = X l≥1 (1 − z)l l , Hm = m X k=1 1 k and ζ(s) = X k≥1 1 ks.

Chapter 3

Properties of Random Polynomials

over Finite Fields

In this chapter, we will show the properties of random polynomials over finite fields from the introduction. We will do this in four sections. More precisely, in Section 3.1, we will discuss the number of irreducible factors of a polynomial, in Section 3.2, we will look at squarefree and k-free polynomials, in Section 3.3, we will discuss the maximal degree of the irreducible factors, and finally in Section 3.4, we will discuss the probability that a random polynomial has all irreducible factors of distinct degrees. Throughout the section all polynomials will be considered to be monic.

3.1

The Number of Irreducible Factors

Definition 3.1. Let Inbe the number of irreducible polynomials of degreen and denote byI(z) its generating function. Moreover, let Xnbe the number of irreducible factors in a random polynomial of degreen (counted with multiplicities).

Theorem 3.2. The number of irreducible polynomials of degree n is In =

qn

n + O q n/2
_.

Proof. As already explained in the introduction, from the uniqueness of the prime factorization, we obtain P (z) =Y k≥1  1 1 − zk Ik = exp X k≥1 Iklog 1 1 − zk ! = exp X k≥1 X j≥1 Ik zkj j ! = exp  I(z) + 1 2I(z 2_{) +} 1 3I(z 3_{) + · · ·}  = 1 1 − qz.

Taking logarithms on both sides of the equality in the second line gives log 1 1 − qz = X k≥1 I(zk₎ k . Next, the right hand side can be written as

X k≥1 I(zk₎ k = X k≥1 X l≥1 1 kIlz kl₌X n≥1 X k|n 1 kIn/kz n_.

Consequently, by M¨obius inversion formula I(z) = log 1 1 − qz + X j≥2 µ(j) j log 1 1 − qzj.

Note that the second term is an analytic function on |z| < q−1/2. Hence, applying singularity analysis gives

Ik = qk

k + O(q k/2_). Example 3.3. The distribution of Xn.

1. The expected value E(Xn):

with the exponent of u counting the numbers of irreducible factors. Then, P (z, u) =Y

k≥1

1 (1 − uzk₎Ik.

Next, differentiating with respect to u and letting u = 1 yields ∂ ∂uP (z, u)     u=1 =Y k≥1 1 (1 − zk₎Ik · X k≥1 Ikzk 1 − zk = P (z) · X k≥1 Ikzk 1 − zk = 1 1 − qz · X k≥1 zk 1 − zk  qk k + O q k/2
 = 1 1 − qz · X k≥1 qk k zk 1 − zk + X k≥1 zk 1 − zkO(q k/2₎ ! = 1 1 − qz · X k≥1 qk k zk 1 − zk + X k≥1 O q k 2zk 1 − zk !! = 1 1 − qz · X j≥1 X k≥1 qk k z jk₊X k≥1 O q k 2zk 1 − zk !! = 1 1 − qz · log 1 1 − qz + X j≥2 log 1 1 − qzj + X k≥1 O q k 2zk 1 − zk !! (3.1) = 1 1 − qz log 1 1 − qz + 1 1 − qz · X j≥2 log 1 1 − qzj + S(z) ! , where S(z) = P k≥1O  qk2zk 1−zk 

. Note that the two terms in the bracket are analytic on |z| < q−1/2. Hence, by singularity analysis, we have

E(Xn) = 1 qn[z n_] ∂ ∂uP (z, u)     u=1 = log n + γ + c + O(n−1), where γ is Euler’s constant and c = P

j≥2log(1 − q1−j)

−1_{+ S(1/q).} 2. The variance of Xn:

P (z, u) =Y k≥1

1 (1 − uzk₎Ik.

Now, we take the second derivative with respect to u and again set u = 1 (the additional factor u after taking the first derivative is needed because we want to compute the second moment).

∂ ∂u  u · ∂ ∂uP (z, u)      u=1 = ∂ ∂u u · Y k≥1 (1 − uzk)−Ik _·X k≥1 Ikzk 1 − uzk !    u=1 =Y k≥1 (1 − uzk)−IkX k≥1 Ikzk 1 − uzk     u=1 + uY k≥1 (1 − uzk)−Ik X k≥1 Ikzk 1 − uzk !2     u=1 + u ·Y k≥1 (1 − uzk)−Ik_·X k≥1 Ikz2k (1 − uzk₎2     u=1 =Y k≥1 (1 − zk)−Ik_·   X k≥1 Ikzk 1 − zk + X k≥1 Ikzk 1 − zk !2 +X k≥1 Ikz2k (1 − zk₎2   = 1 1 − qz ·   X k≥1 Ikzk 1 − zk + X k≥1 Ikzk 1 − zk !2 +X k≥1 Ikz2k (1 − zk₎2  .

Note that the last term in the bracket is analytic on |z| < q−1/2. As for the first two terms, we use what we already obtained in the analysis of the mean,

X k≥1 Ikzk 1 − zk = log 1 1 − qz + X j≥2 log 1 1 − qzj + S(z), Plugging this in and applying singularity analysis gives

E(Xn2) = 1 qn[z n_] 1 1 − qz ·  log 1 1 − qz + S(z)  + 1 qn[z n_] 1 1 − qz · log 2 1 1 − qz + 1 qn[z n ] 1 1 − qz · 2 log 1 1 − qz " X j≥2 log 1 1 − qzj + S(z) #

+ 1 qn[z n_] 1 1 − qz · X j≥2 log 1 1 − qzj + S(z) !2 + 1 qn[z n_] 1 1 − qz · X k≥1 Ikz2k (1 − zk₎2

= log n + O(1) + log2n + 2γ log n + O(1) + 2c log n + O(1) = log2n + (2γ + 2c + 1) log n + O(1).

Consequently, Var(Xn) = E(Xn2) − E(Xn)2

= log2n + (2γ + 2c + 1) log n + O(1) − log2n + 2(γ + c) log n + O(1)
= log n + O(1).

3. The probability of Xn = 1, which is the same as the probability that a random polynomial of degree n is irreducible, is given by

P rob (Xn = 1) = 1 qn · In= 1 qn  qn n + O q n/2
 = 1 n + O q −n/2
_.

4. The probability of Xn = 2, which is the probability of a random polynomial of degree n to be a product of two irreducible factors, is obtained from the generat-ing function I[2](z) = I(z) · I(z) 2! + I(z2₎ 2 = 1 2 log 1 1 − qz + X j≥2 µ(j) j log 1 1 − qzj !2 +1 2 log 1 1 − qz2 + X j≥2 µ(j) j log 1 1 − qz2j ! .

Since the latter term is analytic on |z| < q−1/2, singularity analysis yields P rob (Xn= 2) = 1 qn · [z n_]I[2]_{(z) =} log n n + O  1 n  .

3.2

Squarefree Polynomials and K-free Polynomials

Definition 3.4. A polynomial is called squarefree if each of its irreducible factors ap-pears only once.

Remark 3.5. If a polynomial f (x) is not squarefree, we can sort the irreducible factors of the polynomial into a squarefree part g (x) and a remaining part h (x). The square-free part g (x) gathers all of the irreducible factors only once, and the remaining part h (x) takes the rest.

Remark 3.6. The remaining part h = f /g is not necessarily not squarefree, e.g., if f (x) = (x + 1)2(x + 2)(x2+ x + 1)3, then g(x) = (x + 1)(x + 2)(x2+ x + 1) and consequently h(x) = (x + 1)(x2_{+ x + 1)}2_.

Example 3.7. Here, we want to find the probability of a random polynomial being squarefree and the expected value of the degree of the remaining part.

1. The probability of a random polynomial being squarefree:

The generating functions of squarefree polynomials (denoted by Q(z)) and all polynomials are Q(z) = ∞ Y k=1 (1 + zk)Ik_, P (z) = ∞ Y k=1 (1 − zk)−Ik ₌ 1 1 − qz.

Now, a relation between squarefree polynomials and all polynomials is as fol-lows: every polynomial can be factorized into a square part and a remaining part, which is squarefree, e.g.,

f (x) = x(x + 1)(x + 3)7(x2+ 1)2(x2+ 5)6

=(x + 3)3(x2+ 1)1(x2+ 5)32

| {z }

· x(x + 1)(x + 3)

| {z }

. square part remaining part This yields, P (z) = P (z2) · Q(z). Q(z) = P (z) P (z2₎ = 1 − q(z2₎ 1 − qz = (1 − qz 2_{) · (1 + qz + q}2_z2_{+ q}3_z3_{+ · · · )} =X k≥0 qkzk−X k≥2 qk−1zk.

Hence, the coefficient Qnof znin Q(z) is easily obtained as

Qn=    qn_{, if n = 0, 1;} qn− qn−1_{, if n ≥ 2.}

So, the probability of a random polynomial with degree n being squarefree is 

 

1, when n = 0, 1; 1 −1_q, when n ≥ 2. 2. The expected value of the degree of the remaining part:

We again use a bivariate generating function P (z, u), where the second variable counts the degree of the remaining part. Consequently,

P (z, u) =Y k≥1 1 + zk+ ukz2k+ u2kz3k+ · · ·
Ik =Y k≥1  1 + z k 1 − uk_zk Ik .

∂ ∂uP (z, u)     u=1 =Y k≥1  1 + z k 1 − zk Ik ·X k≥1 Ik  1 + _1−zzkk Ik−1 −zk
1 − zk
−2 −kzk
 1 + _1−zzkk Ik =Y k≥1  1 1 − zk Ik ·X k≥1 kIk z2k 1 − zk
−2 (1 − zk₎−1 = P (z) ·X k≥1 kIk· z2k 1 − zk = 1 1 − qz · X k≥1 kIk· z2k 1 − zk.

Note that the second term is analytic on |z| < q−1/2. Consequently, by singular-ity analysis [zn] ∂ ∂uP (z, u)   u=1 ∼ q nX k≥1 kIk q−2k 1 − q−k = q nX k≥1 kIk q2k _{− q}k.

Hence, the expected value of the degree of the remaining part is asymptotically equal to

X k≥1

kIk q2k_{− q}k.

A natural extension of squarefree polynomials are k-free polynomials. We will consider them next.

Definition 3.8. A polynomial is called k-free if the multiplicity of each irreducible factor is less thank.

Remark 3.9. If a polynomial f (x) is not k-free, we can sort the irreducible factors of the polynomial into a k-free part g (x) and a remaining part h (x). The k-free part g (x) gathers all of the irreducible factors at most k − 1 times and the remaining part h (x) takes the rest.

Example 3.10. Let us find the probability of a random polynomial being k-free and the expected value of the degree of the remaining part. The analysis is similar to the one from Example 3.7.

1. The probability of a random polynomial being k-free: The generating function of k-free polynomials Q[k](z) is

Q[k](z) = Y j≥1

1 + zj + z2j + z3j + · · · + z(k−1)j
Ik.

Then, we again can find a relation between Q[k](z) and P (z). More precisely, since every polynomial can be composed into a k-free polynomial times a poly-nomial of power k, so P (z) = Q[k](z) · P (zk). Q[k](z) = P (z) P (zk₎ = 1 − qzk 1 − qz = (1 − qz k_{) · (1 + qz + q}2_z2 _{+ q}3_z3_{+ · · · )} =X j≥0 qjzj −X j≥0 qj+1zk+j.

Since a polynomial with degree n < k must be k-free, so the probability of a polynomial of degree n < k being k-free is 1. Next, we consider on polynomial of degree n ≥ k. From the above, we get for the number of k-free polynomials (denoted by Q[k]n ):

Q[k]_n = qn− qn−k+1.

So, the probability that a random polynomial of degree n ≥ k is k-free is 1−_qk−11 .

2. The expected value of the degree of the remaining part:

Similar as in Example 3.7, we find the bivariate generating function with u count-ing the degree of the remaincount-ing part.

Then, P (z, u) =Y j≥1 1 + zj+ z2j + · · · + z(k−1)j+ ujzkj+ u2jz(k+1)j+ · · ·
Ij =Y j≥1  1 − z(k−1)j 1 − zj + z(k−1)j 1 − uj_zj Ij .

Next, we differentiate with respect to u and let u = 1. This yields ∂ ∂uP (z, u)     u=1 =Y j≥1  1 1 − zj Ij ·X j≥1 Ij _1−z1j
Ij−1 −z(k−1)j
_{(1 − z}j₎−2_(−jzj₎ 1 1−zj
Ij =Y j≥1  1 1 − zj Ij ·X j≥1 jIjzkj(1 − zj) −2 (1 − zj₎−1 ! = P (z) ·X j≥1  jIj· zkj 1 − zj  = 1 1 − qz · X j≥1  jIj· zkj 1 − zj  .

Applying singularity analysis yields [zn] ∂ ∂uP (z, u)   u=1∼ q nX j≥1 jIj q−kj 1 − q−j = q nX j≥1 jIj qkj_{− q}(k−1)j. Hence, the expected value of the remaining part is asymptotic to

X j≥1

jIj qkj _{− q}(k−1)j.

3.3

The Degree of the Irreducible Factors

Definition 3.11. A polynomial is called m-smooth polynomial if there is no irreducible factor whose degree is greater thanm.

Example 3.12. First, we discuss the number of m-smooth polynomial of degree n. Therefore, let Sm(z) be the generating function of m-smooth polynomials. Then, for |z| < 1 Sm(z) = m Y k=1 1 − zk
−Ik = P (z) · Y k>m 1 − zk
Ik = 1 1 − qz · exp X k>m Iklog 1 − zk
! = 1 1 − qz · exp X k>m Ik·  −zk− z 2k 2 − z3k 3 − · · · ! = 1 1 − qz · exp − X k>m X j≥1 Ik· zkj j ! = 1 1 − qz · exp − X j≥1 1 j X k>m Ikzkj ! = 1 1 − qz · exp − X j≥1 rm[j](z) j ! ,

where rm[j](z) = P_k>mIkzkj. Next, we need suitable estimates for r [j] m. First, we estimate r[1]m(z) for |z| < 1: r_m[1] z q  = X k>m Ik  z q k =X k>m zk k + O q −m/2
_.

Moreover, for rm[j](z) with j ≥ 2, we have r[j]_m  z q  =X k>m O  qkz kj qkj  =X k>m O  zkj qk(j−1)  = O  1 qm(j−1)  ,

for |z| < 1. Overall, we have found that the sum of the error terms of r[j]m(z) with j ≥ 2 are bounded by the error term of r[1]m(z) which is O q−m/2
.

So, the number of m-smooth polynomial of degree n (denoted by Nq(n, m)) is Nq(n, m) = 1 2πi Z C Sm(z) dz zn+1 = qn 2πi Z e C Sm  z q  dz zn+1 = q n 2πi Z e C 1 1 − q(z/q) · exp −r [1] m  z q  −r [2] m(z/q) 2 − rm[3](z/q) 3 − · · · ! dz zn+1 = q n 2πi Z e C 1 1 − z · exp − X k>m zk k + O q −m/2
! dz zn+1 = q n 2πi Z e C 1 1 − z · exp −rm(z) + O q −m/2

dz zn+1,

where the contour eC is z = eiθ−1/n _{with −π ≤ θ ≤ π. Next, we use the change of} variable z = e−h/n. Note that, by Lemma 2.17, we have

rm(z) = rm e−h/n
= E (µh) + O (1/m) ,

where µ = m/n. Moreover, note that since O q−m/2
is exponentially small, so it will be eliminated by O (1/m) in rm(z). Consequently,

Nq(n, m) = qn 2πi Z e C 1 1 − z · exp −rm(z) + O q −m/2

dz zn+1 = q n 2πi Z 1−inπ 1+inπ 1 1 − e−h/n · exp −rm e −h/n
_{+ O q}−m/2

− 1 ne −h/n_dh e−h(1+1/n) = q n 2πi Z 1+inπ 1−inπ exp (−E (µh) + O (1/m)) n (1 − e−h/n₎ e h_dh = q n 2πi Z 1+inπ 1−inπ e−E(µh) e O(1/m) n (1 − e−h/n₎e h_{dh. (3.2)}

Now, set ψ(z) = _1−e1−z − 1_z which is analytic in |z| < 2π. We can rewrite parts of the

integrand of (3.2) into an expression of ψ. 1 n (1 − e−h/n₎ = 1 n  1 h/n + ψ  h n  = 1 h + 1 nψ  h n 

eO(1/m) n (1 − e−h/n₎ =  1 h + 1 nψ  h n  ·  1 + O 1 m  = 1 h+ 1 nψ  h n  + O  1 hm  . (3.3)

Next, substitute (3.3) into (3.2) and separate the integral into three parts Nq(n, m) = qn 2πi Z 1+inπ 1−inπ e−E(µh) e O(1/m) n (1 − e−h/n₎e h_dh = q n 2πi Z 1+inπ 1−inπ e−E(µh) 1 h + 1 nψ  h n  + O  1 hm  ehdh = q n 2πi Z 1+inπ 1−inπ e−E(µh) 1 nψ  h n  ehdh + q n 2πi Z 1+inπ 1−inπ e−E(µh) h e h_dh + q n 2πi Z 1+inπ 1−inπ e−E(µh)O  1 hm  ehdh . Now, we will discuss the 3 integrals separately:

1. The main term: 1 2πi Z 1+inπ 1−inπ e−E(µh) h e h_dh = 1 2πi Z 1+i∞ 1−i∞ e−E(µh) h e h_{dh −} 1 2πi Z L e−E(µh) h e h_dh,

where L is the union of the two semi-vertical lines (1 + inπ, 1 + i∞) and (1 − i∞, 1 − inπ). Then, by partial integration, we obtain for the L part the bound O _n1
. Since the proofs of the positive and negative semi-vertical line are the same, we only consider the positive part.

Z 1+i∞ 1+inπ e−E(µh) h e h_dh = e −E(µh) h e h     1+i∞ 1+inπ − Z 1+i∞ 1+inπ  e−E(µh) h e−µh h − e−E(µh) h2  ehdh = e −E(µh) h e h     1+i∞ 1+inπ + Z 1+i∞ 1+inπ e−E(µh) h2 1 − e −µh
_eh_dh.

In order to go on, note that since e−E(z) is bounded for <(z) > 0, there exists a constant c > 0 such that |e−E(z)| ≤ c. So the absolute value of the first term is

     e−E(µh) h e h     1+i∞ 1+inπ      ≤ c  lim x→∞ |e1+ix_| |1 + ix| + |e1+inπ_| |1 + inπ|  = ce  lim x→∞ 1 √ x2_{+ 1} + 1 √ n2_π2_{+ 1}  = O 1 n  . Moreover, the absolute value of the second term is bounded by

    Z 1+i∞ 1+inπ e−E(µh) h2 1 − e −µh
_eh_dh     ≤ Z ∞ nπ     e−E(µ+iµx) (1 + ix)2 1 − e −µ−iµx
_e1+ix     dx ≤ ce Z ∞ nπ     1 − e−µ−iµx (1 + ix)2     dx ≤ ce Z ∞ nπ 1 |1 + ix|2dx + ce Z ∞ nπ |e−µ−iµx_| |1 + ix|2dx = ce Z ∞ nπ 1 1 + x2dx + ce 1−µ Z ∞ nπ 1 1 + x2dx = ce + ce1−µ
Z ∞ nπ 1 1 + x2dx ≤ ce + ce1−µ
Z ∞ nπ 1 x2dx = ce + ce1−µ nπ = O  1 n  . So, the main term becomes

1 2πi Z 1+i∞ 1−i∞ e−E(µh) h e h_{dh + O} 1 n  = 1 2πi Z 1+i∞ 1−i∞ e−E(µh) µh (e µh_)µ1_{µdh + O} 1 n  = 1 2πi Z 1+i∞ 1−i∞ e−E(µh) µh (e µh_)µ1_{d(µh) + O} 1 n  . Then, by Lemma 2.20, ρ (u) = _2πi1 R_1−i∞1+i∞e−E(v)_v euv_{dv. So, our main term is}

1 2πi Z 1+i∞ 1−i∞ e−E(µh) µh e µh
1_µ d (µh) + O 1 n  = ρ 1 µ  + O 1 n  = ρn m  + O 1 n  .

2. The part containing _n1ψ h_n
: Integration by part gives

1 2πi Z 1+inπ 1−inπ e−E(µh) 1 nψ  h n  ehdh = 1 2nπi Z 1+inπ 1−inπ e−E(µh)ψ h n  ehdh = 1 2nπi e −E(µh) ψ h n  eh     1+inπ 1−inπ − Z 1+inπ 1−inπ  e−E(µh)ψ h n 0 ehdh ! = 1 2nπi  e−E(µh)ψ h n  eh     1+inπ 1−inπ − Z 1+inπ 1−inπ  e−µh h e −E(µh) ψ h n  + e−E(µh) 1 nψ 0 h n  ehdh  = 1 2nπi  e−E(µh)ψ h n  eh     1+inπ 1−inπ − Z 1+inπ 1−inπ e−E(µh) e −µh h ψ  h n  + 1 nψ 0 h n  ehdh  .

We will bound this expression in three steps. In every step we will use that e−E(z) and ψ(z) are bounded:

(a)      1 2nπie −E(µh)_ψ h n
e h     1+inπ 1−inπ      : Here, we have 1 2nπ      e−E(µh)ψ h n  eh     1+inπ 1−inπ      = 1 2nπ     e−E(µ(1+inπ))ψ 1 + inπ n  e1+inπ − e−E(µ(1−inπ))ψ 1 − inπ n  e1−inπ     ≤ 1 2nπ     e−E(µ(1+inπ))ψ 1 + inπ n  e1+inπ     +     e−E(µ(1−inπ))ψ 1 − inπ n  e1−inπ      ≤ c 2nπ  e1+inπ+  e1−inπ
= ce 2nπ  einπ+  e−inπ
= O 1 n  .

So,    1 2nπie −E(µh)_ψ h n
e h  1+inπ 1−inπ   = O 1 n
. (b)    1 2nπi R1+inπ 1−inπ e −E(µh)_ψ h n
_e(1−µ)h h dh   : Here, we have 1 2nπ     Z 1+inπ 1−inπ e−E(µh)ψ h n  e(1−µ)h h dh     ≤ 1 2nπ Z nπ −nπ     e−E(µ+iµx)ψ 1 + ix n  e(1−µ)(1+ix) 1 + ix     dx ≤ c 2nπ Z nπ −nπ     e(1−µ)(1+ix) 1 + ix     dx = ce 1−µ 2nπ Z nπ −nπ 1 √ 1 + x2dx = ce 1−µ 2nπ Z nπ 0 2 √ 1 + x2dx = ce 1−µ nπ Z 1 0 1 √ 1 + x2dx + Z nπ 1 1 √ 1 + x2dx  = ce 1−µ nπ log(x + √ 1 + x2₎     1 0 + Z nπ 1 1 √ 1 + x2dx ! = ce 1−µ nπ  log(1 +√2) + Z nπ 1 1 √ 1 + x2dx  = O log n n  . Thus,    1 2nπi R1+inπ 1−inπ e −E(µh)_ψ h n
_e(1−µ)h h dh   = O log n n
. (c)    1 2n2_πi R1+inπ 1−inπ e −E(µh)_ψ0 h n
e h_dh : Note that ψ0(z) is also bounded. Hence,

1 2n2_π     Z 1+inπ 1−inπ e−E(µh)ψ0 h n  ehdh     ≤ 1 2n2_π Z nπ −nπ     e−E(µ+iµx)ψ0 1 + ix n  e1+ix     dx ≤ c 2n2_π Z nπ −nπ  e1+ixdx = ce 2n2_π Z nπ −nπ  eixdx = O  1 n  . This gives    1 2n2_πi R1+inπ 1−inπ e −E(µh)_ψ0 h n
e h_dh = O 1 n
.

Overall, we obtain _2πi1 R_1−inπ1+inπe−E(µh) 1_nψ h_n
ehdh = O log n_n
. 3. The part containing O _hm1
:

Since e−E(µh)is bounded, we easily get the error term:     1 2πi Z 1+inπ 1−inπ e−E(µh)O  1 hm  ehdh     ≤ c Z nπ −nπ     O  1 (1 + ix)m  e1+ix     dx = ce Z nπ −nπ O  1 √ 1 + x2_m  dx = O log n m  . Hence, by combining the above estimates, we have

Nq(n, m) = qn·  ρ n m  + O log n m   .

Next, we are going to discuss the degree of the largest irreducible factor of a random polynomial of degree n (which we denote by Dn[1]).

Example 3.13. Here, we consider the probability that a random polynomial has the degree of the largest irreducible factor Dn[1] = m.

As before, we first find the generating function Lm(z) of polynomials with the degree of the largest irreducible factor = m. By the previous example, we have the generating function Sm(z) of m-smooth polynomials. Then, Lm(z) is given by

Lm(z) = Sm(z) − Sm−1(z) = Sm(z)  1 − (1 − zm)Im = Sm(z) · X k≥1 Im k  (−1)k+1zkm ! = Sm(z) · Imzm+ X k≥2 I_m k  (−1)k+1zkm ! . Thus, P r D_n[1] = m
= 1 qn[z n_]L m(z) = [zn]Lm  z q  = 1 2πi Z C Lm  z q  dz zn+1 = 1 2πi Z C Sm  z q  Im zm qm + X k≥2 Im k  (−1)k+1 z q km! dz zn+1.

Since Im = qm/m + O qm/2
= O(qm/m), the probability becomes P r D_n[1] = m
= 1 2πi Z C Sm  z q   Im zm qm + O  1 m2  dz zn+1 = 1 2πi Z C Sm  z q   zm m + O q −m/2
_{+ O}  1 m2  dz zn+1 = 1 2πi Z C Sm  z q  zm m dz zn+1  1 + O 1 m  ,

where the contour C is z = e−1/n+iθ with −π ≤ θ ≤ π. By the change of variable z = e−h/n, we have P r D[1]_n = m
= 1 2πi Z C Sm  z q  zm m dz zn+1  1 + O 1 m  = 1 2πi Z 1−inπ 1+inπ Sm  e−h/n q  e−µh m −1 ne −h/n_dh e−h−h/n  1 + O 1 m  = 1 2πi Z 1+inπ 1−inπ Sm  e−h/n q  e(1−µ)h nm dh  1 + O 1 m  , where µ = m/n. Now, as in Example 3.12,

P r D[1]_n = m
= 1 2mπi Z 1+inπ 1−inπ e−E(µh) e O(1/m) n (1 − e−h/n₎e (1−µ)h_dh  1 + O 1 m  = 1 2mπi Z 1+inπ 1−inπ e−E(µh) 1 h + 1 nψ  h n  + O  1 hm  e(1−µ)hdh. As before, we break the integral into three parts and discuss the three parts separately:

1. The main term:

By Example 3.12, we know that the contour of the first part [1 − inπ, 1 + inπ] can be replaced by (1 − i∞, 1 + i∞). Then, by Lemma 2.20,

1 2mπi Z 1+i∞ 1−i∞ e−E(µh)1 he (1−µ)h_{dh =} 1 2mπi Z 1+i∞ 1−i∞ e−E(µh) 1 µhe µh(_µ1−1_{)d (µh)} = 1 m ρ  1 µ− 1  .

2. The term containing _n1ψ h_n
:

This term is similar as in Example 3.12. Integration by part gives 1 2mπi Z 1+inπ 1−inπ e−E(µh) 1 nψ  h n  e(1−µ)hdh = 1 2nmπi Z 1+inπ 1−inπ e−E(µh)ψ h n  e(1−µ)hdh = 1 2nm (1 − µ) πi  e−E(µh)ψ h n  e(1−µ)h     1+inπ 1−inπ − Z 1+inπ 1−inπ e−E(µh) e −µh h ψ  h n  + 1 nψ 0 h n  e(1−µ)hdh  . (a)    1 2nmπie −E(µh)_ψ h n
e (1−µ)h  1+inπ 1−inπ   :

Since e−E(z)and ψ(z) are bounded, we have 1 2nmπ      e−E(µh)ψ h n  e(1−µ)h     1+inπ 1−inπ      = 1 2nmπ     e−E(µ(1+inπ))ψ 1 + inπ n  e(1−µ)(1+inπ) − e−E(µ(1−inπ))ψ 1 − inπ n  e(1−µ)(1−inπ)     ≤ 1 2nmπ     e−E(µ(1+inπ))ψ 1 + inπ n  e(1−µ)(1+inπ)     +     e−E(µ(1−inπ))ψ 1 − inπ n  e(1−µ)(1−inπ)      ≤ c 2nmπ  e(1−µ)(1+inπ)+  e(1−µ)(1−inπ)
= ce 1−µ nmπ = O  1 nm  . So,    1 2nmπie −E(µh)_ψ h n
e (1−µ)h  1+inπ 1−inπ   = O 1 nm
. (b)    1 2nmπi R1+inπ 1−inπ e −E(µh)_ψ h n
e(1−2µ)h h dh   : Since e−E(z)and ψ(z) are bounded, we have

1 2nmπ     Z 1+inπ 1−inπ e−E(µh)ψ h n  e(1−2µ)h h dh     ≤ 1 2nmπ Z nπ −nπ     e−E(µ+iµx)ψ 1 + ix n  e(1−2µ)(1+ix) 1 + ix     dx ≤ c 2nmπ Z nπ −nπ     e(1−2µ)(1+ix) 1 + ix     dx = ce 1−2µ 2nmπ Z nπ −nπ 1 √ 1 + x2dx = ce 1−2µ 2nmπ Z nπ 0 2 √ 1 + x2dx = ce 1−2µ nmπ Z 1 0 1 √ 1 + x2dx + Z nπ 1 1 √ 1 + x2dx  = ce 1−2µ nmπ log(x + √ 1 + x2₎     1 0 + Z nπ 1 1 √ 1 + x2dx ! = ce 1−2µ nmπ  log(1 +√2) + Z nπ 1 1 √ 1 + x2dx  = O log n nm  . This gives    1 2nmπi R1+inπ 1−inπ e −E(µh)_ψ h n
_e(1−2µ)h h dh   = O log n nm
. (c)    1 2n2_mπi R1+inπ 1−inπ e −E(µh)_ψ0 h n
e (1−µ)h_dh :

We know that e−E(z) and ψ0(z) are bounded. Consequently, 1 2n2_mπ     Z 1+inπ 1−inπ e−E(µh)ψ0 h n  e(1−µ)hdh     ≤ 1 2n2_mπ Z nπ −nπ     e−E(µ+iµx)ψ0 1 + ix n  e(1−µ)(1+ix)     dx ≤ c 2n2_mπ Z nπ −nπ  e(1−µ)(1+ix)dx = ce 1−µ 2n2_mπ Z nπ −nπ  eix(1−µ)dx = O  1 nm  . This gives    1 2n2_πi R1+inπ 1−inπ e −E(µh)_ψ0 h n
e h_dh = O 1 nm
. Overall, summing up these three parts gives an error term O log n_nm
.

3. The part containing O _hm1
:

Since e−E(µh)and e(1−µ)h _{are bounded, we obtain the estimate}     1 2mπi Z 1+inπ 1−inπ e−E(µh)O  1 hm  e(1−µ)hdh     ≤ c 2mπ Z nπ −nπ     O  1 m(1 + ix)  e(1−µ)(1+ix)     dx = ce 1−µ 2mπ Z nπ −nπ O  1 m√1 + x2  dx = O log n m2  .

Overall, we get for the probability of D[1]n = m: P r D_n[1] = m
= 1 m ρ  1 µ− 1  + O log n m2  = 1 m ρ n m − 1  + O log n m2  . Example 3.14. Here, we discuss the probability that a random polynomial has D[1]n = m and Dn[2] ≤ m/2, where D[2]n denotes the degree of the second largest irreducible factor.

The generating function eLm(z) of polynomials with D[1]n = m and D[2]n ≤ m/2 is given by e Lm(z) = Sbm/2c(z) · Imzm 1 − zm = Sbm/2c(z) · zm 1 − zm ·  qm m + O q m/2
 = Sbm/2c(z) · qm_zm m(1 − zm₎ · 1 + O(mq −m/2_)
.

Then, the probability of D[1]n = m and D[2]n ≤ m/2 is P r D_n[1] = m, D[2]_n ≤ m/2
= 1 qn[z n ]eLm(z) = [zn]eLm  z q  = 1 2πi Z C e Lm  z q  dz zn+1 = 1 2πi Z C Sbm/2c  z q  zm m (1 − zm_/qm₎ dz zn+1(1 + O(mq −m/2 ))

with the contour C equal to z = e−1/n+iθ with −π ≤ θ ≤ π. Then, by the change of variable z = e−h/n, we have

P r D[1]_n = m, D[2]_n ≤ m/2
= 1 2πi Z C Sbm/2c  z q  zm m (1 − zm_/qm₎ dz zn+1(1 + O(mq −m/2 )) = 1 2πi Z 1−inπ 1+inπ Sbm/2c  e−h/n q  e−µh m (1 − e−µh_/qm₎ −1 ne −h/n_dh e−h−h/n (1 + O(mq −m/2 )) = 1 2πi Z 1+inπ 1−inπ Sbm/2c  e−h/n q  e(1−µ)h m (1 − e−µh_/qm₎ 1 ndh(1 + O(mq −m/2 )) = 1 2mπi Z 1+inπ 1−inπ e−E(µh/2) e O(1/m) n (1 − e−h/n₎e (1−µ)h  1 + e −µh qm + e−2µh q2m + · · ·  dh. Here, the error term O(mq−m/2) is eliminated by the error term O(1/m). Moreover, we can ignoreP

k≥1 e

−µh_/qm
k

since it is exponentially small. So, the probability is P r D[1]_n = m, D[2]_n ≤ m/2
= 1 2mπi Z 1+inπ 1−inπ e−E(µh/2) e O(1/m) n (1 − e−h/n₎e (1−µ)h dh. Next, by (3.3) of Example 3.12, P r D_n[1] = m, D_n[2] ≤ m/2
= 1 2mπi Z 1+inπ 1−inπ e−E(µh/2) 1 h + 1 nψ  h n  + O  1 hm  e(1−µ)hdh.

As before, we break the integral into three parts. For the second and third part, we again obtain O log n_m2
.

For the first part, similar as in Example 3.12 we can replace [1 − inπ, 1 + inπ] by (1 − i∞, 1 + i∞). Then, by the change of variable v = µh/2 and Lemma 2.20,

1 2mπi Z 1+i∞ 1−i∞ e−E(µh/2)1 he (1−µ)h_{dh =} 1 2mπi Z 1+i∞ 1−i∞ e−E(v) 1 2v/µe (1−µ)2v/µ2 µdv = 1 2mπi Z 1+i∞ 1−i∞ e−E(v) v e 2v_(µ1−1_{)dv =} 1 m ρ  2 µ − 2  .

Overall, P r D[1]_n = m, D[2]_n ≤ m/2
= 1 m ρ  2 µ − 2  + O log n m2  = 1 m ρ  2n m − 2  + O log n m2  .

Example 3.15. Finally, we discuss the probability that a random polynomials has Dn[1] = m1and D

[2]

n = m2with m2 < m1.

The generating function Lm1,m2(z) of random polynomials with D

[1] n = m1and Dn[2] = m2is Lm1,m2(z) = Lm2(z) · Im1z m1 1 − zm1 = Lm2(z) · zm1 1 − zm1 ·  qm1 m1 + O(qm1/2₎  = Lm2(z) · qm1_zm1 m1(1 − zm1) · 1 + O(m1q−m1/2)
. Then, the probability of D[1]n = m1and D

[2] n = m2is P r D_n[1] = m1, D[2]n = m2
= 1 qn[z n ]Lm1,m2(z) = [z n ]Lm1,m2  z q  = 1 2πi Z C Lm1,m2  z q  dz zn+1 = 1 2πi Z C Sm2  z q  zm2 m2 zm1 m1(1 − zm1/qm1) dz zn+1  1 + O  1 m2 

with the contour C equal to z = e−1/n+iθ with −π ≤ θ ≤ π. Then, by the change of variable z = e−h/n, we have P r D_n[1] = m1, D[2]n = m2
= 1 2πi Z C Sm2  z q  zm1+m2 m1m2(1 − zm1/qm1) dz zn+1  1 + O  1 m2  = 1 2πi Z 1−inπ 1+inπ Sm2  e−h/n q  e−(µ1+µ2)h m1m2(1 − e−µ1h/qm1) −1 ne −h/n_dh e−h−h/n  1 + O  1 m2  = 1 2πi Z 1+inπ 1−inπ e−E(µ2h) e O(1/m2) n (1 − e−h/n₎ · e(1−µ1−µ2)h m1m2(1 − e−µ1h/qm1) dh,

where µ1 = m1/n and µ2 = m2/n. Note that we can replace 1 − e−µ1h/qm1
−1

by 1 since the remainder is exponentially small. Then, as in Example 3.12, the probability is P r D_n[1] = m1, Dn[2] = m2
= 1 2m1m2πi Z 1+inπ 1−inπ e−E(µ2h) 1 h+ 1 nψ  h n  + O  1 hm2  e(1−µ1−µ2)h_dh.

We again break this integral into three parts, where the second and third part satisfy O_mlog n

1m2

 .

For the first part we replace the contour by (1 − i∞, 1 + i∞). Then, by Lemma 2.20 and change of variable v = µ2h, the main term becomes

1 2m1m2πi Z 1+inπ 1−inπ e−E(µ2h)1 he (1−µ1−µ2)h_dh = 1 2m1m2πi Z 1+i∞ 1−i∞ e−E(v) 1 v/µ2 e(1−µ1−µ2)v/µ2 1 µ2 dv = 1 2m1m2πi Z 1+i∞ 1−i∞ e−E(v) v e v  1−µ1−µ2 µ2  dv = 1 m1m2 ρ 1 − µ1− µ2 µ2  . Overall, we obtain for the probability that D[1]n = m1 and D

[2] n = m2: P r D[1]_n = m1, Dn[2] = m2
= 1 m1m2 ρ 1 µ2 −µ1 µ2 − 1  + O log n m1m22  = 1 m1m2 ρ n m2 − m1 m2 − 1  + O log n m1m22  .

3.4

Other Restrictions on the Degree of Irreducible

Fac-tors

Example 3.16. Here, we will discuss the probability that a random polynomial has irreducible factors of distinct degrees.