Une introduction à la théorie analytique des nombres

Ilan Vardi

IHES, Bures-sur-Yvette

14 décember 1998

Résumé de Cyril Banderier et Ilan Vardi

Introduction

``Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe.'' (Jacques Hadamard)

Cette citation illustre la puissance que l'analyse peut apporter lorsque l'on est confronté à des questions de théorie des nombres. Le plus ancient et le plus fondamontal de ses thèmes est celui des nombres premiers. La première question qui se pose est : "Y a-t-il une infinité de nombres premiers ?" Il y a de multiples preuves élémentaires :

Euclide : Supposons qu'il n'existe qu'un nombre fini de nombres premiers p₁,..., p_n , alors p₁ p₂...p_n+1 n'est divisible par aucun des p_i, ainsi, ses diviseurs premiers donnent un nouveau nombre premier (Euclide considéra seulement le cas n=3).
Pólya : Les nombres de Fermat F_n=2^2ⁿ+1sont premiers entre eux, ainsi l'ensemble de leurs facteurs premiers doit être infini.
Erdös : Fixons x et considérons les nombres premiers p₁,...,p_n £ x. Puisque chaque entier est le produit d'un carré parfait et d'un nombre sans facteur carré, on peut écrire chaque entier m<=x sous la forme m=p₁^e_¹ ··· p_n^e_ⁿ Q². Il y a 2ⁿ choix pour les e_i et sqrt(x) choix pour Q, il s'ensuit donc que $n \geq\frac{\ln(x)}{2\ln(2)}.$
Euler : On a l'identité formelle suivante

$\begin{displaymath} \sum_n \frac{1}{n^s} =\prod_{p\; {\rm prime}} \frac{1}{1-p^{-s}}\end{displaymath}$ (1)

qui a un sens analytique pour $\Re(s) \gt 1$ . Lorsque $s\to 1$ , le membre gauche de (1) tend vers $\infty$ puisuqe la série harmonique diverge, il y a par conséquent un nombre infini de facteurs à droite.
Cette preuve peut être modifée en remarquant que $\zeta(2) = \pi^2/6$ , où $\zeta(s) = \sum 1/n^s$ . S'il y avait un nombre fini de nombres premiers alors (1) impliquerait que $\pi^2$ serait rationnel, or Legendre a prouvé que ce n'était pas le cas en 1797, voyez également [6]. Diverses autres preuves sont données dans [7].
Une version améliorée de cet argument est due à Mertens: la version finie de (1) donne
$\begin{displaymath} \prod_p \frac{1}{1-p^{-1}}\gt\sum_{n<x} \frac{1}{n} \sim \ln(x)\,,\end{displaymath}$
en prenant les logarithmes, on obtient

$\begin{displaymath} \sum_{p<x} \frac{1}{p}\sim \ln\ln(x)\,,\end{displaymath}$ (2)

et il y a donc une infinité de nombres premiers.

Laquelle de ces preuves est la "meilleure" ? Un point de vue sage serait de choisir celle qui offre la meilleure généralisation. Par exemple, la preuve d'Euclide montre facilement qu'il y a une infinité de nombres premiers de la form 4k+3 (considérez 4 p₁ p₂ ... p_n-3, mais il faut légèrement adapter l'argument pour prouver l'infinitude l'infinitude des nombres premiers de la forme 4k+1 (il faut considérer cette fois 4 (p₁ p₂ ... p_n)²+1). Plus généralement, on aimerait démontrer l'assertion de Dirichlet (ce qu'il a effectivement fait en 1837, dans [3]) ``il existe une infinité de nombres premiers de la forme ak+b, où a et b sont premiers entre eux.'' Il s'avère que la preuve de fait profond utilise une généralisation de la méthode d'Euler, i.e. l´équation (2):

$\begin{displaymath} \displaystyle{ \sum_{p\equiv a\; ({\rm mod}\, q)}\frac{1}{p}... ...rrow \text{ there are an infinite number of primes in $ak+q$}}.\end{displaymath}$

Le théorème de Dirichlet

Soit $\chi$ un caractère multiplicatif modulo q, c'est-à-dire une fonction à valeurs complexes $\chi(n)$ vérifiant $\chi(mn)=\chi(m)\chi(n)$ and $\chi(0)=0$ (ce qui implique que si $\chi(n)\ne 0$ alors c'est une racine de l'unité et a donc norme 1). Un exemple est le symbole de Legendre (ou de Jacobi si q n'est pas premier)

$\begin{displaymath} \chi(n):=\left(\frac{n}{q}\right) =\begin{cases} 0 & \text{... ... q) \text{ for some $x$ }\\ -1 &\text{ otherwise. }\end{cases}\end{displaymath}$

En fait, il y a exactement $\phi(q)$ caractères multiplicatif modulo q, tous donnés par $\chi(n):=\omega^{\ln_\rho(n)}$ où $\rho$ est une racine primitive et $\omega$ est tel que $\omega^{\phi(q)}\equiv 1 \; ({\rm mod}\, q)$ . L'importance des caractères se présent au regard des relations d'orthogonalité:

$\begin{displaymath} \frac{1}{\phi(q)} \sum_\chi \overline{\chi(a)}\chi(n) =\begi... ...equiv a \; ({\rm mod}\, q)$,}\\ 0 &\text{otherwise.}\end{cases}\end{displaymath}$

(3)

qui permettent de distinguer une progression arithmétique particulière. Dans sa preuve, Dirichlet introduisit ce qui est de nos jours appelé L-fonctions de Dirichlet, définies par

$\begin{displaymath} L(s,\chi):=\sum_{n\geq 1} \frac{\chi(n)}{n^s}=\prod_p \frac{1}{1-\chi(p)p^{-s}}.\end{displaymath}$

En prenant les logarithmes, on obtient $\ln L(s,\chi)=\sum_p -\ln(1-\chi(p)p^{-s})$ , ainsi on a

$\begin{displaymath} \frac{1}{\phi(q)}\sum_\chi \overline{\chi(a)} \ln L(s,\chi)=... ...i(q)}\sum_\chi \overline{\chi(a)} \sum_p -\ln(1-\chi(p)p^{-s}) \end{displaymath}$

$\begin{displaymath} = \sum_p\sum_{k\geq 1}\frac{1}{\phi(q)}\sum_\chi \overline{\chi(a)} \frac{\chi(p^k )p^{-sk}}{k}\end{displaymath}$

et une simple application de la relation (3) donne

$\begin{displaymath} \frac{1}{\phi(q)}\sum_\chi \overline{\chi}(a) \ln L(s,\chi)=... ... q)} p^{-s} +O(1) \text{ when $s\rightarrow 1$ from the right.}\end{displaymath}$

Par conséquent, en séparant la somme selon les caractères réels et complexes, on obtient

$\begin{displaymath} \sum_{p \equiv a \; ({\rm mod}\, q)} p^{-1}=\frac{1}{\phi(q)... ...xt{ complex}} \right) \overline{\chi}(a) \ln L(1,\chi) + O(1).\end{displaymath}$

(4)

$\chi_0$ est appelé caractère principal et vaut 1 (tant que n n'est pas congru à 0 mod q) et 0 sinon. La première somme (sur $\chi_0$ ) vaut +l'infini, puisque $L(s,\chi_0)=\zeta(s)\prod_{p\vert q} (1-p^{-s})$. Ce terme infini devrait impliquer qu'il y a une infinité de nombres premiers dans la progression arithmétique. Le seul problème pourrait venir d'une annulation is that one of the other terms could cancel this one by being zero at s = 1 (partial summation shows that $L(1, \chi)$ is finite). One therefore has to show that $L(1, \chi)\ne 0$ .

This is definitely true for complex characters since otherwise, $L(1, \chi) = 0$ would imply that $L(1, \overline\chi) = 0$ and since these terms are different, this would imply that $\prod_{\chi \; ({\rm mod}\, q)} L(1, \chi) = 0$ , which is false as taking logs gives

$\begin{displaymath} \sum_{\chi \; ({\rm mod}\, q)} \ln L(s, \chi) = \sum_{p^n \equiv 1\; ({\rm mod}\, q)} \frac{1}{p^s} \gt 0\,, \quad s \gt 1\,,\end{displaymath}$

and so the value at s=1 must be positive, hence the last sum in relation (4) is bounded.

The real problem is then to bound the middle sum in relation (4), that is to say to show that $L(1,(\cdot , q))\ne 0$ . Dirichlet proved this result by a very ingenious method: He evaluated this number in closed form! This is now known as Dirichlet's class number formula:

$\begin{displaymath} 0\neq L\left(1,(\cdot/q)\right)= \begin{cases} \frac{\pi h }... ...\sqrt q} &\text{when $q \equiv 3\; ({\rm mod}\,4)$} \end{cases}\end{displaymath}$

where h is the class number of ${\bf Q}(\sqrt{(-1)^{(q+1)/2}\, q})$ and $\epsilon$ its fundamental unit and w the number of roots of unity in this field (see the canonical reference [2]). Since each of these quantities counts something, so they are positive, the result now follows:

$\begin{displaymath} \sum_{p\equiv a \; ({\rm mod}\, q)} p^{-1}=+\infty.\end{displaymath}$

Simpler proofs using only complex analysis are also possible. The idea is to use Landau's theorem that a Dirichlet series with positive terms has a pole at its abscissa of convergence and apply it to $\prod_{\chi\; {\rm mod}\, q} L(s, \chi)$ which has just been shown to have positive coefficients.

The Prime Number Theorem

The distribution of primes is quite irregular, so it is easier to study their statistical behaviour. In this direction, let $\pi(x)$ be the number of primes $\leq x$ . Gauss conjectured that $\pi(x)\sim \int_2^x \frac{dt}{\ln t }=:\operatorname{Li}(x).$ This assertion simply says: ``the probability that n is prime is about $1/\ln n$ .'' This result was finally proved by Hadamard and de la Vallée Poussin in 1896. Both of them used fundamental ideas of Riemann who was the first to introduce complex analysis in the study of the distribution of prime numbers.

Using Perron's formula, namely

$\begin{displaymath} \sum_{p^n \le x} \ln(p) = \frac{1}{2i\pi}\int_{c-i\infty}^{c+i\infty}\frac{-\zeta'(s)}{\zeta(s)} \frac{x^sds}{s}\end{displaymath}$

and using residues, Riemann essentially found what is perhaps the most important formula in analytic number theory (the von Mangoldt explicit formula):

$\begin{displaymath} \sum_{p^n \le x} \ln(p) = x-\frac{\zeta'(0)}{\zeta(0)}-\su... ...\Re(\rho)\gt} \frac{x^\rho}{\rho} -\frac{1}{2}\ln(1-x^{-2})\,,\end{displaymath}$

(5)

where sum on the right is over the zeroes of the Riemann $\zeta$ function. These zeroes can be split up into two types: The first are the trivial zeroes at -2, -4, -6, ..., and the zeroes with $0 \le \Re \le 1$ (the right hand side of (5) reflects this dichotomy). This formula has many interesting properties and reflects the following principles of analytic number theory:

1.: Primes should always be counted with weight $\ln(p)$ ;
2.: Primes and prime powers should be counted together;
3.: There are much less prime powers than primes;
4.: The zeroes of the $\zeta$ function are the ``fundamental frequencies'' of the primes, and in this sense are dual to the primes.

Following Chebyshev, one defines $\theta(x) = \sum_{p \le x} \ln(p)$ and $\psi(x) = \sum_{p^n \le x} \ln(p) = \sum_{n \le x} \Lambda(n)$ ,where $\Lambda(n) = \ln(p)$ when n=p^m, and zero otherwise. A fairly straightforward partial summation shows that the prime number theorem is equivalent to $\psi(x) \sim x$ (note that trivially, $\psi(x) = \theta(x) + O(\sqrt{x})$ ), and that more generally,

$\begin{displaymath} \psi(x) = x + R(x) \Longleftrightarrow \pi(x) \sim \operatorname{Li}(x) + O(R(x)/\ln(x))\,.\end{displaymath}$

One can then see from the explicit formula (5) that the prime number theorem would follow if one can bound $\Re \rho < 1$ , since each error term would then be of order < x. The prime number theorem would then be equivalent to showing that $\zeta(1 + it) \ne 0$ for $t\ne 0$ . In fact, this is an equivalence (as was later shown by Wiener) and Hadamard and de la Vallée Poussin were able to prove that $\zeta(1 + it) \ne 0$ using some ingenious trigonometric identities. We will give a proof due to Mertens, in 1898. Set $\rho=1+it$ , then $\zeta(\rho)=0 \Longrightarrow \Re \ln \zeta(\sigma+it)\rightarrow -\infty$ when $\sigma\rightarrow 1$ (we restrict to $\Re(\sigma)\gt 1$ ). But, by the Euler identity, one has $\ln \zeta(s)=\sum_p \sum_{m\geq 1} m^{-1} p^{-m\sigma} \exp(-itm\ln(p))$ and so

$\begin{displaymath} \Re \ln \zeta(s)=\sum_p \sum_{m\geq 1} m^{-1} p^{-m\sigma} \cos(-tm\ln(p)).\end{displaymath}$

Mertens' trick consists in noticing that $2(1+\cos \beta)^2=3 +4 \cos \beta + \cos 2\beta \geq 0$ , thus $3 \ln\zeta(\sigma)+4\Re\ln\zeta(\sigma+it)+\Re\ln\zeta(\sigma+2it)\geq 0$ , hence $\zeta^3(\sigma) \vert\zeta^4(\sigma+it) \zeta(\sigma+2it)\vert \geq 1$ .
But, as $\sigma\rightarrow 1$ , one has $\zeta(\sigma)\sim (\sigma-1)^{-1}$ and $\vert\zeta(\sigma+it)\vert<A (\sigma-1)$ for a some constant A. So one should have $\zeta(\sigma+2it)\rightarrow \infty$ , this contradicts the fact that $\zeta(1+2it)$ is bounded. In conclusion, the $\zeta$ function has no zero with $\Re(\rho)=1$ , the PNT is proved. An elementary (i.e., without complex analysis) proof of the PNT was subsequently found by Erdös and Selberg in 1949 (see [4] and [9]).

Chebyshev's bias

All numerical evidence shows that $\pi(x)<\operatorname{Li}(x)$ and it was long believed that this would be true for all x. Similarly, Chebyshev noted that the number of primes of the form 4k+3 seemed to be more abundant than the primes of the form 4k+1, more precisely, let $\pi_{q,a}(x) = \vert\{p \le x:\ p\equiv a\; ({\rm mod}\, q)\}\vert$ then $\pi_{4, 3}(x) \ge \pi_{4, 1}(x)$ .

In fact, Littlewood proved in 1914 that $\pi(x) - \operatorname{Li}(x)$ changes sign infinitely often and the same is true for $\pi_{4, 3}(x) - \pi_{4, 1}(x)$ . In 1957 Leech showed that $\pi_{4, 1} (x) \gt \pi_{4,3}(x)$ is first true for x = 26861, and that the similar inequality $\pi_{3, 1} (x) \gt \pi_{3,2}(x)$ is first true for x = 608981813029 was shown by Bays and Hudson in 1978. No example of $\pi(x) \gt \operatorname{Li}(x)$ is known. Skewes first gave an upper bound which was later reduced by Sherman-Lehman and then te Riele [10] who gave an upper bound of 10³⁷⁰.

This behaviour can easily be explained using explicit formulas. In the case of $\pi(x)$ , the point is the following: The explicit formula (5) expresses $\psi(x)$ as a sum of powers $x^\rho$ . Assuming the Riemann Hypothesis, one can write this as

$\begin{displaymath} \psi(x) = x - x^{1/2} \left( \sum_{\zeta(1/2+i\gamma) = 0} \frac{x^{i\gamma}}{1/2+i\gamma}\right) +o(x^{1/2})\,.\end{displaymath}$

One can now see the reason for the bias: The function $\psi(x)$ does not count primes but prime powers so what one really wants is the behaviour of $\theta(x)$ which is given by

$\begin{displaymath} \theta(x) = \psi(x) - \theta(\sqrt{x}) +O(x^{1/3})\,,\end{displaymath}$

so that

$\begin{displaymath} \theta(x) = x - x^{1/2} \left(1+ \sum_{\zeta(1/2+i\gamma) = 0} \frac{x^{i\gamma}}{1/2+i\gamma}\right) +o(x^{1/2})\,.\end{displaymath}$

The function

$\begin{displaymath} \sum_{\zeta(1/2+i\gamma) = 0} \frac{e^{i\gamma\ln(x)}}{1/2+i\gamma}\, ,\end{displaymath}$

is a very slowly oscillating trigonometric series which should be zero on average, so the extra term biases $\theta(x)$ to be smaller than x on average. A simple description is that $\operatorname{Li}(x)$ counts the number of prime powers $\le x$ , so the number of primes should be slightly less since the number of prime squares is of the same order as the error term.

There is a similar explanation for the bias in arithmetic progressions. There is an explicit formula

$\begin{displaymath} \sum_{p^n\le x} \chi(n) \ln(p) = - x^{1/2} \left( \sum_{L(1... ...frac{x^{i\gamma_\chi}}{1/2+i\gamma_\chi}\right) + o(x^{1/2})\,,\end{displaymath}$

where the Generalised Riemann Hypothesis has been assumed (there is no x term since $L(1, \chi)$ is no longer a pole if $\chi \ne \chi_0$ ). As before one has

$\begin{displaymath} \psi_{q, a}(x) = \sum_{\substack{p^n \equiv a\;({\rm mod}\... .../2+i\gamma_\chi) = 0} \frac{x^{i\gamma_\chi}}{1/2+i\gamma_\chi}\end{displaymath}$

but one really wants to look at

$\begin{displaymath} \theta_{q,a} (x) = \sum_{\substack{\scriptstyle p \equiv a ... ... O(x^{1/3}) = \psi_{q, a}(x) - c_{q, a} x^{1/2} + O(x^{1/3})\,,\end{displaymath}$

y²= a (mod q). In particular, the same argument shows that there will always be fewer primes in the progression qn + a when a is a residue than when a is a nonresidue. Simply put, the ``balanced'' count is the set of prime powers = a (mod q) when a is quadratic residue since the number of prime squares congruent to a is of the same order as the error term in the analytic formulas.

In 1994, Rubinstein and Sarnak (see [8]) were able to make Chebyshev's bias precise. Assuming GRH (if this is false, then there is no bias) and also the Grand Simplicity Hypothesis (GSH: All the ordinates of zeroes of L-function are linearly independent over $\mathbb{Q}$ ), then

$\begin{displaymath} \frac{1}{\ln(x)} \sum_{\substack{\scriptstyle \pi(n) \gt \op... ..._{4,3}(n) \gt \pi_{4,1}(n)\\ \scriptstyle n \le x}} \to .9959.\end{displaymath}$

References

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Cyril Banderier
3/30/1999