Elements of Analytic Number Theory

ELEMENTS OF ANALYTIC NUMBER THEORY

P. S. Kolesnikov, E. P. Vdovin

Lecture course

Novosibirsk, Russia 2013 Contents

Chapter 1. Algebraic and transcendental numbers4 § 1.1. Field of algebraic numbers. Ring of algebraic integers4 1. Preliminary information4 2. Minimal polynomial6 3. Algebraic complex numbers9 4. Algebraic integers 11 § 1.2. Diophantine approximations of algebraic numbers 13 1. Diophantine approximation of degree ν 13 2. Dirichlet approximation theorem 15 3. Liouville theorem on Diophantine approximation of algebraic numbers 18 § 1.3. Transcendentality of e and π 20 1. Hermite identity 20 2. Transcendentality of e 23 3. Symmetrized n-tuples 25 4. Transcendentality of π 26 § 1.4. Problems 29

Chapter 2. Asymptotic law of distribution of prime numbers 30 § 2.1. Chebyshev functions 31 1. Deﬁnition and estimates 31 2. Equivalence of the asymptotic behavior of Chebyshev functions and of the prime-counting function 32 3. Von Mangoldt function 34 § 2.2. Riemann function: Elementary properties 35 1. Riemann function in Re z > 1 35 2. Distribution of the Dirichlet series of a multiplicative function 36 3. Convolution product and the M¨obiusinversion formula 37 4. Euler identity 38 5. Logarithmic derivative of the Riemann function 39

2 Contents 3

6. Expression of the integral Chebyshev function via the Riemann function 40 § 2.3. Riemann function: Analytic properties 43 1. Analytic extension of the Riemann function 43 2. Zeros of the Riemann function 47 3. Estimates of the logarithmic derivative 48 4. Proof of the Prime Number Theorem 51 § 2.4. Problems 55

Chapter 3. Dirichlet Theorem 56 § 3.1. Finite abelian groups and groups of characters 56 1. Finite abelian groups 56 2. Characters 58 3. Characters modulo m 60 § 3.2. Dirichlet series 60 1. Convergence of L-series 60 2. Landau Theorem 66 3. Proof of the Dirichlet Theorem 68 Chapter 4. p-adic numbers 71 § 4.1. Valuation fields 71 1. Basic properties 71 2. Valuations over rationals 74 3. The replenishment of a valuation field 76 § 4.2. Construction and properties of p-adic fields 79 1. Ring of p-adic integers and its properties 80 2. The field of p-adic rationals is the replenishment of rationals in p-adic metric 81 3. Applications 84 § 4.3. Problems 85 Bibliography 86

Glossary 87 Index 88 CHAPTER 1

Algebraic and transcendental numbers

§ 1.1. Field of algebraic numbers. Ring of algebraic integers 1. Preliminary information. Let us recall some basic notions from Abstract Algebra. Throughout we use the following notations: P is the set of all prime numbers; N is the set of positive integers (the set of natural numbers); Z is the set of all integers; Q is the set of all rational numbers; R is the set of all real numbers; C is the set of all complex numbers, C = R + ιR, ι2 = −1. Given a field F , symbol F [x] denotes the ring of polynomials in variable n x with coefficients in F . If f(x) = a0 + a1x + ··· + anx ∈ F [x], ai ∈ F , is chosen so that an 6= 0, then n is called the degree of f(x), it is denoted by deg f, while an ∈ F is called the leading coefficient of f(x), and if an = 1 then f(x) is called monic. If f(x) = 0 (all coefficients are equal to zero) then the degree of f(x) is said to be −∞. If f(x), g(x) ∈ F [x], g(x) 6= 0, then there exist unique q(x), r(x) ∈ F [x] such that f(x) = g(x)q(x) + r(x), deg r < deg g. (1.1)

These polynomials (quotient q(x) and remainder r(x)) can be found by the well-known division algorithm. If r(x) = 0 then we write g | f (g divides f). One may easily note the similarity between division algorithms in the ring of integers Z and in the ring of polynomials F [x]. Indeed, these are particular examples of Euclidean rings, and there are many common features and problems that can be solved in similar ways for integers and polynomials. In particular, the greatest common divisor (gcd) d of two polynomials f, g ∈ F [x] is deﬁned as a monic common divisor which is divided by every

4 § 1.1. Field of algebraic numbers 5 other common divisor, i.e., d = gcd(f, g) if and only if d | f, d | g, and for every h ∈ F [x] with h | f and h | g it follows that h divides d. To ﬁnd gcd of f and g, one may use the Euclidean algorithm based on the following observation: If f and g are related by (1.1) then gcd(f, g) = gcd(g, r). Moreover, if d = gcd(f, g) then there exist p(x), s(x) ∈ F [x] such that f(x)p(x) + g(x)s(x) = d(x).

Exercise 1.1. Let f1, . . . , fn ∈ F [x] be a finite family of polynomials over a field F . Prove that there exists a unique monic greatest common divisor of f1, . . . , fn. Suppose R is a commutative ring with an identity (e.g., R = Z or R = F [x] as above). A subset I ⊆ R is called an ideal of R if a ± b ∈ I for every a, b ∈ I, and ax ∈ I for every a ∈ I, x ∈ R. For example, the set of all even integers is an ideal of Z; the set {f(x) ∈ F [x] | f(α) = 0} is an ideal of F [x], where α is an element of some extension field of F . Since an intersection of any family of ideals is again an ideal, for every set M ⊆ R there exists minimal ideal of R which contains M, it is denoted by (M). It is easy to note that X (M) = xiai | xi ∈ R, ai ∈ M . i An ideal I of R is said to be principal if there exists a ∈ R such that I = (a), where (a) stands for ({a}). Recall that a commutative ring R is called an integral domain (or simply a domain) if ab = 0 implies a = 0 or b = 0 for all a, b ∈ R. In particular, Z and F [x] are integral domains. An integral domain R such that every ideal of R is principal is called a principal ideal domain. Exercise 1.2. Prove that Z and F [x] (where F is a field) are principal ideal domains. In particular, if f(x), g(x) ∈ F [x] then ({f, g}) = (gcd(f, g)). If R is a domain, then we can consider the field of fractions of R. In order to construct it we start with the Cartesian product R × (R \{0}) of R a (here each pair (a, b) corresponds to fraction b ). Now define an equivalence relation (a1, b1) ∼ (a2, b2) ⇐⇒ a1b2 = a2b1. Let Q be the set of equivalence classes of R × (R \{0}) under this equivalence. Define the addition and multiplication on representatives by

(a1, b1) + (a2, b2) = (a1b2 + a2b1, b1b2), (a1, b1) · (a2, b2) = (a1a2, b1b2). § 1.1. Field of algebraic numbers 6

We leave for the reader to prove that all operations defined are correct and that Q is a field under these operations. Let I be an ideal of R. Then R is split into a disjoint union of congruence classes a+I = {a+x | x ∈ I}, a ∈ R, and the set of all these classes (denoted by R/I) is a ring with respect to natural operations (a + I) + (b + I) = (a + b) + I, (a + I)(b + I) = ab + I. The ring R/I obtained is called a factor ring of R over I. For example, Z/(n) = Zn, the ring of remainders modulo n. A proper ideal I of R is maximal if there are no proper ideals J of R such that I ⊂ J. For example, if R = Z then (n) is maximal if and only if n = ±p, where p is a prime natural number; if R = F [x] then (f) is maximal if and only if the polynomial f is irreducible over F . Note that if I is a maximal ideal of a commutative ring R then R/I is a field. Indeed, if a + I 6= 0 (i.e., a∈ / I) then J = {xa + b | x ∈ R, b ∈ I} is an ideal of R such that I ⊂ J. Therefore, J = R, and thus all equations of the form (a + I)X = c + I, c ∈ R, have solutions in R/I. Exercise 1.3. Prove that R[x]/(x2 + x + 1) is a field isomorphic to the field C of complex numbers. 2. Minimal polynomial. A complex number α ∈ C is algebraic if there exists a nonzero polynomial f(x) ∈ Q[x] such that f(α) = 0. A non-algebraic complex number is said to be transcendental. An algebraic number α is called an algebraic integer if there exists a monic polynomial f(x) ∈ Z[x] such that f(α) = 0. Every rational number α ∈ Q ⊂ C is obviously an algebraic one. More- over, as we will see later, a rational number is an algebraic integer if and only if it is an integer. √ √ 1 3 Exercise 1.4. (1) Prove that 2 and + ι are algebraic in- 2 2 tegers. (2) Prove that if α ∈ C is an algebraic number then Re α and Im α are algebraic numbers. Whether the same statements are true for an algebraic integer α? (3) Show that the cardinality of the set of all algebraic numbers is countable. Since the cardinality of the entire set of complex numbers is uncountable (continuum), transcendental numbers do exist. However, it is not so easy to show an example of such a number accompanied with reasonable proof. § 1.1. Field of algebraic numbers 7

Let α be an algebraic number. Denote by x a formal variable, and let

I(α, x) = {f(x) ∈ Q[x] | f(α) = 0}. It is clear that I(α, x) is an ideal of the ring Q[x]. Since Q[x] is a principal ideal domain, the ideal I(α, x) is generated by a single polynomial. Namely, the monic polynomial of minimal positive degree from h(x) ∈ I(α, x) is a generator of the ideal I(α, x), it is called the minimal polynomial for α. Given an algebraic number α, denote its minimal polynomial by hα(x) and say deg hα to be the degree of α. Lemma 1.5. Let α be an algebraic number, and let h(x) be a monic polynomial from Q[x]. Then the following conditions are equivalent:

(1) h(x) = hα(x); (2) h(α) = 0, and h divides every f ∈ I(α, x); (3) h(α) = 0, and h(x) is irreducible over Q. Proof. (1) ⇒ (2) It is obvious by deﬁnition. (2) ⇒ (3) Assume h(x) is reducible over Q, i.e., it can be decomposed into nonscalar factors as follows:

h(x) = h1(x)h2(x), where hi(x) ∈ Q[x], deg hi > 1.

Then h(α) = h1(α)h2(α) = 0, so for either of i = 1, 2 we have hi(α) = 0. Therefore, the corresponding polynomial hi(x) belongs to I(α, x), hence, hi(x) is a multiple of h(x), which is impossible due to deg hi < deg h. (3) ⇒ (1) Let hα(x) be the minimal polynomial for α. Then h(x) ∈ I(α, x) = (hα(x)), so hα | h. Since h(x) is irreducible and deg hα > 0, we have h(x) = hα(x). √ 2 For example, if √α ∈ Q then hα = x − α. For α = 2, hα = x − 2. The 1 3 number α = + ι satisﬁes the equation α3 + 1 = 0, but its minimal 2 2 2 polynomial is hα(x) = x − x + 1. Exercise 1.6. Prove that a minimal polynomial does not have multiple roots.

If α is an algebraic number, and β ∈ C is a root of hα(x) then β is said to be conjugate to α. Therefore, every algebraic number α of degree n has exactly n pairwise diﬀerent conjugate complex numbers α1, . . . , αn, § 1.1. Field of algebraic numbers 8 including α itself. Moreover, n Y hα(x) = (x − αi) ∈ Q[x]. i=1

Given an algebraic number α, the minimal polynomial hα ∈ Q[x] is uniquely defined. If α is an algebraic integer then there also exists a monic polynomial f ∈ Z[x] of minimal degree such that hα | f. In order to prove that these f and hα coincide, we need the following observation. Let h(x) ∈ Q[x] be a monic polynomial with rational coefficients, p0 p1 pn−1 n−1 n h(x) = + x + ··· + x + x , gcd(pi, qi) = 1. q0 q1 qn−1 Denote by q(h) the least common multiple of the coefficients’ denominators:

q(h) = lcm (q0, . . . , qn−1). (1.2) Then for q = q(h) we have n−1 n qh(x) = a0 + a1x + ··· + an−1x + anx ∈ Z[x], where the gcd of all coefficients is equal to the identity: (a0, . . . , an) = 1. Indeed, assume a0, . . . , an have a common divisor d > 1. Then d | q = an, and for b = q/d ∈ Z we have bh(x) ∈ Z[x]. Therefore, qi | bpi for all i = 0, . . . , n − 1, so qi | b, and, finally, q | b = q/d, which is impossible for d > 1. Polynomials with relatively prime integral coefficients are studied in Abstract Algebra, they are called primitive. The following statement is well-known. Exercise 1.7 (Hauss Lemma ). Prove that the product of primitive polynomials is also a primitive polynomial. Proposition 1.8. Let α be an algebraic integer. Then the minimal polynomial hα(x) has integral coefficients. Proof. Suppose f(x) ∈ Z[x] be a monic polynomial with integral coefficients such that f(α) = 0. Then Lemma 1.5 implies hα | f, i.e.,

f(x) = hα(x)g(x), g(x) ∈ Q[x].

Since both f and hα are monic polynomials, so is g. 0 Denote q = q(hα), q = q(g), where the function q(·) is given by (1.2). 0 According to the remark above, qhα(x) and q g(x) are primitive polynomials in Z[x]. However, 0 0 qq f(x) = (qhα(x))(q g(x)). § 1.1. Field of algebraic numbers 9

The Hauss Lemma implies qq0f(x) to be primitive, while f(x) has integral 0 coefficients itself. Therefore, qq = 1, i.e., all coefficients of hα and g are integral. Thus, there is no need to define separately integral minimal polynomial and degree for algebraic integers. Let us also note that for every algebraic number α there exists c ∈ N such that cα is an algebraic integer: It is deg hα enough to consider c = q(hα) .

3. Algebraic complex numbers. Let us denote the set of all algebraic numbers by A. Recall that the Fundamental Theorem of Algebra states C to be an algebraically closed field, i.e., every non-constant polynomial over C has a root in C. In this section, we will prove that A ⊂ C is the minimal algebraically closed subfield of C, i.e., A is the algebraic closure of Q. Lemma 1.9. The following statements are equivalent for α ∈ C: (1) α ∈ A; (2) Q[α] := {f(α) | f(x) ∈ Q[x]} is a finite-dimensional vector space over Q; (3) Q[α] is a subfield of C.

Proof. (1) ⇒ (2). Note that f(x) = q(x)hα(x) + r(x) by the division algorithm, deg r < deg hα. Hence, f(α) = r(α), and the latter is a n−1 linear combination over Q of 1, α, . . . , α , where n = deg hα. Therefore, n−1 dim Q[α] 6 n. Moreover, 1, α, . . . , α are linearly independent since n is the minimal possible degree of a polynomial over Q annihilating α, so

dim Q[α] = deg hα. (2) ⇒ (1). It is enough to note that 1, α, α2,... are linearly dependent, so there exist a0, a1, . . . , an ∈ Q such that n a0 · 1 + a1α + ··· + anα = 0, n end at least one of ai is nonzero. Hence, h(x) = a0 +a1x+···+anx ∈ Q[x] is a nonzero polynomial annihilating α. (1) ⇒ (3) Obviously, Q[α] is a subring of C. Let 0 6= f(α) ∈ Q[α], then f ∈ Q[x] \ I(α, x). Therefore, hα does not divide f. Since hα is irreducible, we have gcd(f, hα) = 1. Hence, there exist polynomials u(x), v(x) ∈ Q[x] such that

u(x)f(x) + v(x)hα(x) = 1. § 1.1. Field of algebraic numbers 10

Then for x = α we obtain u(α)f(α) = 1, i.e., f(α)−1 = u(α) ∈ Q[α], i.e., Q[α] is a ﬁeld. (3) ⇒ (1) Suppose Q[α] is a subﬁeld of C. It is enough to consider the case when α 6= 0. Since α−1 ∈ Q[α], there exists f(x) ∈ Q[x] such that −1 α = f(α), i.e., h(α) = 0 for h(x) = xf(x) − 1, deg h > 1.

Corollary 1.10. If α1, . . . , αn ∈ A then Q[α1, . . . , αn] := {f(α1, . . . , αn) | f(x1, . . . , xn) ∈ Q[x1, . . . , xn]} is a ﬁnite-dimensional vector space over Q. Proof. For n = 1, it follows from Lemma 1.9. By induction, since dim Q[α1, . . . , αn−1] < ∞ and dim Q[αn] < ∞,

dim Q[α1, . . . , αn] 6 dim Q[α1, . . . , αn−1] · dim Q[αn] < ∞ (all dimensions are over Q).

Exercise 1.11. Prove that Q[α1, . . . , αn] is a subfield of C provided that α1, . . . , αn ∈ A. Theorem 1.12. The set of all algebraic numbers is a subfield of C. Proof. Since 1, 0 ∈ C are obviously algebraic, it is enough to prove the following two statements: (1) If α and β are algebraic numbers then α ± β and αβ are also algebraic numbers; (2) If β 6= 0 is an algebraic number then 1/β is also an algebraic number. (1) Since Q[α ± β], Q[α · β] ⊆ Q[α, β] ⊆ C, we have dim Q[α ± β] < ∞, dim Q[α · β] < ∞ by Corollary 1.10. Thus by Lemma 1.9 α ± β, αβ ∈ A. (2) By Lemma 1.9, β−1 ∈ Q[β], so Q[β−1] ⊆ Q[β] which is finite- −1 −1 dimensional over Q. Hence, dim Q[β ] < ∞ and thus β ∈ A. Theorem 1.13. The field A is algebraically closed.

Proof. Suppose α0, . . . , αn ∈ A, n > 1, αn 6= 0, ϕ(x) = α0 + α1x + n ··· + αnx ∈ A[x]. It is enough to show that ϕ(x) has a root in A. Without loss of generality, assume αn = 1. Indeed, by Theorem 1.12 we may divide ϕ(x) by αn, and the result is still in A[x]. By the Fundamental Theorem of Algebra, ϕ(x) has a root β ∈ C. Note that Q[β] ⊆ Q[α0, . . . , αn−1, β]. § 1.1. Field of algebraic numbers 11

Since βn can be expressed as a linear combination of βk, k = 0, . . . , n − 1, with coeﬃcients depending on αi, i = 0, . . . , n − 1, as n n−1 β = −α0 − α1β − · · · − αn−1β , we have

dim Q[α0, . . . , αn−1, β] 6 n dim Q[α0, . . . , αn−1] < ∞. Hence, Q[β] is a ﬁnite-dimensional vector space over Q, and by Lemma 1.9 β ∈ A Theorems 1.12 and 1.13 imply that A is the algebraic closure of Q, which is often denoted by Q.

4. Algebraic integers. Suppose K[x1, . . . , xn] is the ring of polynomials in several variables over a ring K. For f ∈ K[x1, . . . , xn] denote by deg f the maximal sum of the degrees of the variables that appear in a term of f with a nonzero coefficient. Namely, f may be uniquely written as m f = f0 + f1xn + ··· + fmxn , where fi ∈ K[x1, . . . , xn−1]. Assuming deg fi are defined by induction, set deg f = max (i + deg fi). i=0,...,m Theorem 1.14. The set of all algebraic integers is a subring of the field C. Proof. Ii is enough to show that if α, β are algebraic integers then α ± β and αβ are algebraic integers as well. We will prove a more general fact: Every number of the form X k l γ = cklα β , ckl ∈ Z, k,l is an algebraic integer. Let n−1 n hα(x) = a0 + a1x + ··· + an−1x + x , m−1 m hβ(x) = b0 + b1x + ··· + bm−1x + x . By Proposition 1.8, ai, bj ∈ Z. Lemma 1.5 implies that hα and hβ are irreducible polynomials over Q, thus they have no multiple roots. Denote by α1, . . . , αn all complex roots of hα(x), and let β1, . . . , βm stand for all complex roots of hβ(x). To be more precise, set α1 = α, β1 = β. Consider the polynomial n m   Y Y X k l p(x) = x − cklαi βj ∈ C[x]. i=1 j=1 k,l § 1.1. Field of algebraic numbers 12

It is clear that p(γ) = 0 and the leading coeﬃcient of p(x) is equal to the identity. It remains to show that p(x) ∈ Z[x]. It follows from the deﬁnition of p(x) that

p(x) = f(x, α1, . . . , αn, β1, . . . , βm), where f ∈ Z[x, y1, . . . , yn, z1, . . . , zm]. Namely,

n m   Y Y X k l f(x) = x − cklyi zj ∈ Z[x, y1, . . . , yn, z1, . . . , zm]. i=1 j=1 k,l

Moreover, every permutation of the variables y1, . . . , yn or z1, . . . , zm does not change the polynomial f, i.e., it is symmetric with respect to yi and with respect to zi. Hence, nm X a f(x, y1, . . . , yn, z1, . . . , zm) = ga(y1, . . . , yn, z1, . . . , zm)x , a=0 where every polynomial ga is symmetric with respect to yi and with respect to zi. On the other hand, for every a we have

X d1 dm ga(y1, . . . , yn, z1, . . . , zm) = ga,d1,...,dm (y1, . . . , yn)z1 . . . zm , d1,...,dm>1 where every polynomial ga,d1,...,dm (y1, . . . , yn) is symmetric (with respect to yi) and has integral coeﬃcients.

Lemma 1.15. Let Ψ(y1, . . . , yn) ∈ Z[y1, . . . , yn] be a symmetric polynomial on y1, . . . , yn, deg Ψ = N, and let α1, . . . , αn ∈ C be the roots n of a polynomial h(x) = a0 + a1x + ··· + anx ∈ Z[x], an 6= 0. Then N an Ψ(α1, . . . , αn) ∈ Z. Proof. By the Fundamental Theorem of Symmetric Polynomials, there exists a polynomial G(t1, . . . , tn) ∈ Z[t1, . . . , tn] such that

Ψ(y1, . . . , yn) = G(σ1, . . . , σn), deg G 6 N. where σi(y1, . . . , yn) are the elementary symmetric polynomials on y1, . . . , yn. k an−k The Viet formulae imply σk(α1, . . . , αn) = (−1) , k = 1, . . . , n. Since an deg G 6 N, we obtain N N an−1 n a0 an Ψ(α1, . . . , αn) = an G − ,..., (−1) . an an The latter is an integer number. § 1.2. Diophantine approximations 13

Now we can use Lemma 1.15 for polynomials ga,d1,...,dm (y1, . . . , yn) to obtain ga,d1,...,dm (α1, . . . , αn) ∈ Z (in this case, an = 1 since αi are algebraic integers). Hence,

ga(α1, . . . , αn, z1, . . . , zm) ∈ Z[z1, . . . , zm] are symmetric polynomials on z1, . . . , zm, and by Lemma 1.15 we have

ga(α1, . . . , αn, β1, . . . , βm) ∈ Z. Therefore, p(x) ∈ Z[x].

§ 1.2. Diophantine approximations of algebraic numbers It is well-known from the course of Analysis that the set of rational numbers is a dense subset of the set of real numbers , i. e., for every Q p R α ∈ and for every ε > 0 there exists ∈ such that R q Q

p α − < ε. (1.3) q

Given a natural number N, the set QN = {p/q | p ∈ Z, 0 < q 6 N}, has 1 the following obvious property: |a − b| for all a, b ∈ . Hence, for > N 2 QN every α∈ / QN (in particular, for an irrational one), we have min |a − α| > 0. a∈QN p Therefore, in order to satisfy (1.3) for small ε, the number should have a q an unboundedly large denominator q. This observation raises the following natural question: How to measure the accuracy of a rational approximation relative to the growing denominator values?

1. Diophantine approximation of degree ν. To estimate the accuracy of an approximation by rationals, we will compare the diﬀerence |α − p/q| with a decreasing function q−ν , ν > 0. Namely, let us consider the following quantity: ν p εα,ν (q) = min q α − , ν > 0. (1.4) p∈Z,p/q6=α q

It turns out that the study of the behavior of εα,ν (q) as q approaches inﬁnity leads to a necessary condition for α to be algebraic. § 1.2. Diophantine approximations 14

Definition 1.1. A real number α ∈ R possesses a Diophantine approximation of degree ν > 0 if

lim εα,ν (q) < ∞. (1.5) q→∞ Let us state a useful criterion that allows to determine whether a given number possesses a Diophantine approximation of a given degree. Lemma 1.16. A real number α possesses a Diophantine approximation of degree ν > 0 if and only if there exists a constant c > 0 such that the inequality

p c α − < (1.6) q qν p holds for inﬁnitely many rationals ∈ . q Q Proof. Denote by M = M(α, c, ν) the set of all pairs (p, q) ∈ Z × N such that (1.6) holds. Suppose α possesses a Diophantine approximation of degree ν > 0. By Deﬁnition 1.1, there exists c > 0 such that

εα,ν (q) < c for infinitely many q ∈ N. Let us fix this c and note that for every such q ∈ N there exists p ∈ Z satisfying the inequality (1.6). Therefore, the set M is infinite, and there is no upper bound for the set {q | (p, q) ∈ M for some p}. It remains to show that the set {p/q | (p, q) ∈ M} is also infinite. Indeed, if it were finite then the left-hand side of (1.6) has a positive lower bound, but the right-hand side of (1.6) approaches zero since ν > 0 and q may be chosen to be as large as we need. Conversely, suppose there exists c such that (1.6) holds for infinitely many rationals p/q. Then the set M defined in the first part of the proof is infinite. Assume the set {q | (p, q) ∈ M for some p} has an upper bound N, i.e.,

{p/q | (p, q) ∈ M} ⊆ QN , where QN = {p/q ∈ Q | p ∈ Z, 0 < q 6 N}. As we have already mentioned above, for every distinct a1, a2 ∈ A the 2 inequality |a1 − a2| > 1/N holds. Hence, any infinite subset S of QN has infinite diameter, i.e., for any d > 0 one may find p1/q1 and p2/q2 in S such that |p1/q1 − p2/q2| > d. § 1.2. Diophantine approximations 15

In particular, the set S = {p/q | (p, q) ∈ M} ⊆ QN contains p1/q1 and p2/q2 such that |p1/q1 − p2/q2| > 2c. Since (1.6) holds for pi/qi, i = 1, 2, we have

p1 c p2 c α − < ν , α − < ν q1 q1 q2 q2 which implies p1 p2 1 1 2c < − < c ν + ν < 2c, q1 q2 q1 q2 a contradiction. Therefore, the set of denominators {q | (p, q) ∈ M for some p} is infinite, so there exist infinitely many q such that εα,ν (q) < c. Hence, the sequence {εα,ν (q)}q∈N has a finite accumulation point, and (1.5) holds. Exercise 1.17. Whether a rational number possesses a Diophantine approximation of degree 1? 2. Dirichlet approximation theorem.

Theorem 1.18 (Dirichlet Approximation Theorem). For every α ∈ R and for every N ∈ N there exist p ∈ Z and q ∈ N such that

p 1 α − < , q N. q qN 6 Proof. Consider the fractional parts of the numbers kα, k = 0,...,N:

ξk = {kα} = kα − [kα] ∈ [0, 1). Divide the interval [0, 1) into N intervals of length 1/N as follows: [k/N, (k + 1)/N), k = 0,...,N − 1. According to the combinatorial Dirichlet’s Principle, when N + 1 numbers ξ0, . . . , ξN are set into N intervals, there exists at least one interval which contains at least two of these numbers, i.e.,

|ξk1 − ξk2 | < 1/N for some k1, k2, 0 6 k1 < k2 6 N. Let p = [k2α] − [k1α], q = k2 − k1. Then

p 1 1 1 α − = |α(k2 − k1) − [k2α] + [k1α]| = |ξk2 − ξk1 | < . q q q Nq § 1.2. Diophantine approximations 16

Corollary 1.19. If α ∈ R \ Q then α possesses a Diophantine approximation of degree ν = 2. Proof. Theorem 1.18 implies that for every natural number N there exist pN ∈ Z and qN ∈ N, qN 6 N, such that

pN 1 α − < . (1.7) qN NqN

Let us show that the sequence {qN }N>1 is not bounded. Assume the converse, i.e., suppose there exists a constant M such that qN 6 M for all N. Then (1.7) implies

p 1 1 min α − < 6 → 0, p/q∈QM q NqN N N→∞ which is impossible. Therefore, (1.7) holds for inﬁnitely many numbers qN . Since qN 6 N, we have

2 pN qN α − < 1 qN for inﬁnitely many qN .

Proposition 1.20. Let α ∈ Q. Then for every function ϕ : N → R+ p satisfying lim qϕ(q) = 0 there exist only a ﬁnite number of rationals q→∞ q such that

p α − < ϕ(q). (1.8) q Proof. Denote by S the set of all rationals p/q, gcd(p, q) = 1, satisfying (1.8). Assume S is infinite. It is easy to see that the distance between any two numbers from S does not exceed 2 max ϕ(q) which is finite. Note q>1 that the set of denominators of all fractions from S has no upper bound: Otherwise, if S ⊆ QN , where QN introduced in the proof of Lemma 1.16, then S has an infinite diameter. a p Let α = . Then for all rationals except α itself the following in- b q equality holds:

a p aq − pb 1 − = . b q bq > bq Since b is fixed, we have 1 > ϕ(q), bq § 1.2. Diophantine approximations 17 which is impossible for sufficiently large q. The contradiction obtained proves that S is finite. Corollary 1.21. A rational number α does not have a Diophantine approximation of degree ν > 1. The result obtained is in some sence paradoxical: All irrational numbers possess Diophantine approximations of degree 2, but neither of rationals has a Diophantine approximation of degree ν > 1. Therefore, the existence of a Diophantine approximation of degree ν > 2 is a criterion of irrationality. Let us apply the criterion above to the base of the natural logarithm 1 1 1 e = 1 + + + ··· + + ..., (1.9) 1! 2! n! also called the Euler’s number. Proposition 1.22. The Euler’s number is irrational. Proof. Consider the sum of the first n + 1 summands in (1.9) and reduce to the common denominator: 1 1 1 p 1 + + + ··· + = n . 1! 2! n! n! Then p 1 1 1 e − n = + + ... n! (n + 1)! n + 2 (n + 2)(n + 3) 1 1 1 2 < + + ... = . (n + 1)! 2 22 (n + 1)! Define a function ϕ : N → R as follows: ϕ(1) = 1, 2 ϕ(q) = for (n − 1)! < q n!, n 2. (n + 1)! 6 > If q ∈ ((n − 1)!, n!] then 2 2 qϕ(q) n! = , 6 (n + 1)! n + 1 so lim qϕ(q) = 0. However, q→∞

p e − < ϕ(q) q § 1.2. Diophantine approximations 18 for inﬁnitely many p/q = pn/n!, n > 2. Hence, e is irrational by Proposi- tion 1.20. 3. Liouville theorem on Diophantine approximation of algebraic numbers. Theorem 1.23 (Liouville Theorem). Let α be an algebraic number of degree n > 2. Then α has no Diophantine approximation of degree ν > n. Proof. First, let us ﬁnd a constant M > 0 such that

p M α − > (1.10) q qn for all p ∈ Z and q ∈ N. Set h(x) to be a multiple of the minmal polynomial for α with integer coeﬃcients, e.g., h(x) = q(hα)hα(x). Suppose α1, . . . , αn ∈ C are the complex roots of this polynomial, and assume α1 = α. Then n n Y Y h(x) = an (x − αk) = an(x − α) (x − αk), k=1 k=2 where an is the leading coeﬃcient of h(x). Denote by M the following quantity: n !−1 Y M = |an| (|α| + |αk| + 1) , k=2 and let us show that (1.10) holds. If p and q meet the inequality |α − p/q| > 1 then (1.10) is valid since M < 1 (|α| > 0, n > 2). Assume p and q satisfy the condition |α − p/q| < 1. In this case,

p < 1 + |α| q and thus n p Y p |h(p/q)| = |an| α − αk − q q k=2 n p Y p p 1 α − |an| |αk| + < α − . 6 q q q M k=2 § 1.2. Diophantine approximations 19

Lemma 1.5 implies h(x) to be irreducible over Q, hence, it has no rational roots. Therefore, n p p 1 |h(p/q)| = a0 + a1 + ··· + an , q qn > qn and (1.10) follows. Finally, apply (1.10) to show that α has no Diophantine approximation of degree ν > n. Assume the converse: Let there exist ν > n and c > 0 such that

p c α − < q qν holds for infinitely many p/q ∈ Q. Then, as it was shown in the proof of Lemma 1.16, the last inequality holds for infinitely many denominators q ∈ N. Then for sufficiently large q we have c M < , qν qn in contradiction to (1.10).

The Liouville Theorem is a powerful tool that allows constructing explicit examples of transcendental numbers. Namely, we obtain the following suﬃcient condition of transcendentality.

Corollary 1.24. Let α be a real number. If for every N ∈ N it possesses a Diophantine approximation of degree ν > N then α is transcendental. Example 1.2. The following number is transcendental: ∞ X α = 10−n!. n=1 Proof. Given N ∈ N, consider N pN X pN = 10−n! = . q 10N! N n=1 Note that ∞ pN X −n! −(N+1)! 2 α − = 10 < 2 · 10 = N+1 . qN n=N+1 qN § 1.3. Transcendentality of e and π 20

pN Therefore, for every ν > 0 there exist inﬁnitely many rationals , N > ν, qN satisfying

pN 2 2 α − < N+1 < ν . qN qN qN Hence, α possesses a Diophantine approximation of any degree. Theo- rem 1.23 implies α is not algebraic.

§ 1.3. Transcendentality of e and π The Liouville Theorem provides a suﬃcient condition for a real number to be transcendental. However, this condition is not necessary. There exist diﬀerent methods to prove transcendentality of a series of important constants, e.g., e and π. In this section, we are going to study one of these methods known as the Hermite Method. Given an analytic function f(z) on the complex plane, denote by x Z f(z) dz

x0 the Riemann integral of f(z) along the straight segment starting at x0 ∈ C and ending at x ∈ C (the Cauchy Integral Theorem implies that this integral does not depend on the choice of a path with the same endpoints x0 and x, we choose the straight segment for convenience). 1. Hermite identity.

Lemma 1.25 (Hermite Identity). Let α ∈ C, α 6= 0, and let f(x) ∈ C[x], deg f > 1. Then for every x ∈ C we have x Z f(t)e−αt dt = F (0) − F (x)e−αx, (1.11)

0 where f(x) f 0(x) f (n)(x) F (x) = + + ··· + . α α2 αn+1 Proof. According to the Fundamental Theorem of Calculus (the Newton— Leibniz Formula), x d Z f(t)e−αt dt = f(x)e−αx. dx 0 § 1.3. Transcendentality of e and π 21

On the other hand, d F (x)e−αx = (F 0(x) − αF (x))e−αx = −f(x)e−αx. dx Hence, the derivatives with respect to x of the both sides of (1.11) coincide. It remains to compare the values at x = 0 to obtain the desired equality. The main idea of the Hermite’s method is to apply the Hermite identity (1.11) to a polynomial of the form 1 H(h(x)) = a(n−1)pxp−1h(x)p, (1.12) (p − 1)! n h(x) ∈ Z[x], n = deg h, p ∈ N. Let us establish the properties of H(h(x)). n Lemma 1.26. Let h(x) = a0 + a1x + ··· + anx ∈ Z[x], a0, an 6= 0, n > 1, and let β1, . . . , βn ∈ C be the entire collection of roots of h(x) in which every root of multiplicity k appears k times. Then for every p ∈ N, p > 2, the polynomial f(x) = H(h(x)) defined by (1.12) has the following properties: (j) (1) f (0) = 0, 0 6 j 6 p − 2; (j) (2) f (βi) = 0, 0 6 j 6 p − 1, i = 1, . . . , n; (p−1) (n−1)p p (3) f (0) = an a0; (j) (4) f (0) ∈ pZ, j > p; n P (j) (5) f (βi) ∈ pZ, j > p. i=1 Proof. It is easy to see from the construction of f(x) that 0 is its root of multiplicity p − 1 and every βi is a root of f(x) of multiplicity at least p. As we know from the Abstract Algebra course, a root of multiplicity k of a polynomial f(x) is also a root of f 0(x), f 00(x), . . . , f (k−1)(x). This implies (1) and (2). To prove (3), let us distribute all brackets in the definition of f(x) and find the term of lowest degree in x, namely, the term is 1 a(n−1)papxp−1. (p − 1)! n 0

(n−1)p p Its (p − 1)th derivative is equal to an a0. For all other terms in f(x), their (p − 1)th derivatives contain x and thus turn into zero at x = 0. Before we proceed with the proof of the remaining statements, note the following general fact. Given a polynomial Φ(x) ∈ Z[x], its jth derivative § 1.3. Transcendentality of e and π 22

Φ(j)(x), j ∈ N, belongs to j!Z[x], i.e., all nonzero coeﬃcients of Φ(j)(x) contain the factor j!. Indeed, for all m > j we have m (xm)(j) = j! xm−j ∈ [x], j Z and the jth derivatives of xm, m < j, turn into zero. To prove (4), consider (p − 1)! Φ(x) = f(x) = xp−1h(x)p ∈ [x]. (n−1)p Z an According to the remark stated above, Φ(j)(x) ∈ j!Z[x] and thus j! f(0) ∈ ⊆ j (p − 1)!Z Z for j > p, Finally, note that n Y h(x) = an (x − βi). i=1 Hence, n anp Y anp f(x) = n (x − β )pxp−1 = n Θ(x, β , . . . , β ), (p − 1)! i (p − 1)! 1 n i=1 where n p−1 Y p Θ = x (x − yi) ∈ Z[x, y1, . . . , yn]. i=1 The polynomial n X ∂jΘ Ψ(y , . . . , y ) = (y , y , . . . , y ) ∈ [y , . . . , y ] 1 n ∂xj i 1 n Z 1 n i=1 is symmetric with respect to y1, . . . , yn, deg Ψ = deg Φ−j = np+p−1−j < np for j > p, and all coeﬃcients of Ψ are divisible by j!. By Lemma 1.15 1 applied to Ψ(y , . . . , y ), we obtain j! 1 n anp n Ψ(β , . . . , β ) ∈ . j! 1 n Z Since j > p, anp n Ψ(β , . . . , β ) ∈ p , (p − 1)! 1 n Z § 1.3. Transcendentality of e and π 23 and it remains to note n X anp f (j)(β ) = n Ψ(β , . . . , β ), i (p − 1)! 1 n i=1 which proves (5).

Remark 1.3. Upon the conditions of Lemma 1.26, assume that βi ∈ Z, i = 1, . . . , n. Then the statement (5) of Lemma 1.26 may be enhanced in the obvious way as

j f (βi) ∈ pZ, i = 1, . . . , n, j > p. 2. Transcendentality of e.

Theorem 1.27. If α ∈ Q \{0} then eα is a transcendental number.

Proof. Suppose β = ea/b is an algebraic number for some a/b ∈ Q, a 6= 0. Then e is a root of the equation xa−βb = 0 with algebraic coeﬃcients. By Theorem 1.13, all roots of such equation (in particular, e) are algebraic numbers. Hence, it is enough to show that e itself is transcendental. Assume e is algebraic, and let he(x) ∈ Q[x] be its minimal polynomial. Then there exists bn ∈ Z such that n bnhe(x) = b0 + b1x + ··· + bnx ∈ Z[x].

It is clear that b0 6= 0. Choose a prime number p ∈ Z such that p > n and p > |b0| (it is possible to make such a choice since the set of primes is inﬁnite). Consider the polynomial

h(x) = (x − 1)(x − 2) ... (x − n) and construct 1 f(x) = H(h(x)) = xp−1h(x)p (p − 1)! as in (1.12), where βi = i, i = 1, . . . , n. For every k = 0, 1, . . . , n write the Hermite identity from Lemma 1.25:

k Z f(t)e−t dt = F (0) − F (k)e−k.

0 § 1.3. Transcendentality of e and π 24

k Multiply each of these equations by bke and add the results:

k n Z n X k−t X k −k bk f(t)e dt = bke (F (0) − F (k)e ) k=0 0 k=0 n n n X k X X = F (0) bke − bkF (k) = F (0)bnhe(e) − bkF (k). k=0 k=0 k=0 Therefore, k n Z n X k−t X bkf(t)e dt = − bkF (k). (1.13) k=0 0 k=0 Recall that X F (x) = f (j)(x). j>0 Consider the right-hand side of (1.13). Lemma 1.12 and Remark 1.3 imply

n n X (p−1) X (j) X X (j) bkF (k) = b0f (0) + b0 f (0) + bk f (k). k=0 j>p k=1 j>p (p−1) np p In the last expression, b0f (0) = (−1) b0(n!) 6≡ 0 (mod p) by the choice of p, all other summands are integer multiples of p. Hence, for every suﬃciently large prime p the right-hand side of (1.13) is a nonzero integer number. Now, let us estimate the absolute value of the left-hand side of (1.13):

k n Z n X k−t X k−t bk f(t)e dt 6 |bk|k max {|f(t)e |} t∈[0,k] k=0 0 k=0 p n 1 p−1 np C1 6 (n+1) max |bk|n max {|f(t)|}e 6 C n (n−1) 6 C , k=0,...,n t∈[0,n] (p − 1)! (p − 1)! p C1 where C and C1 do not depend on the choice of p. Since lim = 0, p→∞ (p − 1)!

k n Z X k−t bk f(t)e dt < 1

k=0 0 § 1.3. Transcendentality of e and π 25 when p is suﬃciently large, but the right-hand side of (1.13) is a nonzero integer and thus its absolute value is greater or equal to 1. The contradiction obtained proves the theorem. 3. Symmetrized n-tuples. N Definition 1.4. An N-tuple (β1, . . . , βN ) ∈ C is called symmetrized if N Y (x − βj) ∈ Q[x]. j=1 It is clear that a symmetrized tuple remains symmetrized after every permutation of its components. If we add (or remove) a rational number to (or from) a symmetrized tuple then the tuple obtained is symmetrized. Also, the concatenation of two or more symmetrized tuples is again a symmetrized tuple. To prove the transcendence of π we need the following properties of symmetrized tuples. n Lemma 1.28. Let (α1, . . . , αn) ∈ C be a symmetrized tuple, and let σ = σk ∈ Z[x1, . . . , xn], k ∈ {1, . . . , n}, be an elementary symmetric polynomial in x1, . . . , xn. Then N X n σeα1 , . . . , eαn = eβj ,N = , k j=1 where (β1, . . . , βN ) is a symmetrized tuple. Proof. Recall that X σ = σk = xi1 . . . xik 16i1<···

ηi1,...,ik (t1, . . . , tn) = ti1 + ··· + tik ∈ Z[t1, . . . , tn]. Then α1 αn X ηi ,...,i (α1,...,αn) σ e , . . . , e = e 1 k .

16i1<···

ηi1,...,ik (α1, . . . , αn), 1 6 i1 < ··· < ik 6 n, form a symmetrized tuple. § 1.3. Transcendentality of e and π 26

Let us enumerate all k-tuples (i1, . . . , ik), 1 6 i1 < ··· < ik 6 n, by integer numbers j = 1,...,N. We will use the same enumeration for polynomials ηi1,...,ik : If (i1, . . . , ik) has number j then ηj = ηi1,...,ik . If Φ(y1, . . . , yN ) is a symmetric polynomial in y1, . . . , yN with integer coeﬃcients then

Ψ = Φ(η1(t1, . . . , tn), . . . , ηN (t1, . . . , tn)) ∈ Z[t1, . . . , tn] is a symmetric polynomial in t1, . . . , tn. Indeed, let us transpose in Ψ two variables tl and ti. The set of all polynomials η1, . . . , ηN may be divided into three groups: The ﬁrst group contains all those polynomials that either do not depend in both ti and tl, or containt both these variables; The second group consists of all ηjs that depend in ti, but do not depend in tl; The third group includes all remaining polynomials ηj. When exchanging ti and tl, the polynomials of the ﬁrst group do not change, the polynomials of the second group turn into polynomials of the third group, and, conversely, all polynomials of the third group move into the second group. Finally, we obtain a permutation of η1, . . . , ηN . Since Φ is symmetric, Ψ is invariant under the transposition of tl and ti. It is well-known that every permutation can be obtained by a series of transpositions (all transpositions generate the symmetric group Sn). Hence, Ψ is also symmetric. Therefore, for every symmetric Φ ∈ Z[y1, . . . , yN ] we have

Φ(β1, . . . , βN ) = Ψ(α1, . . . , αn) ∈ Q.

In particular, all elementary symmetric polynomials in y1, . . . , yN take rational values at β1, . . . , βN , i.e., (β1, . . . , βN ) is a symmetrized N-tuple.

4. Transcendentality of π.

Lemma 1.29. Let (β1, . . . , βN ) be a symmetrized N-tuple of nonzero complex numbers. If, in addition,

N X A := eβk 6= 0, k=1 then A/∈ Q.

Proof. Assume A ∈ Q \{0}. Without loss of generality, we can suppose A ∈ Z: If A = a/b then one may just repeat the symmetrized N-tuple b times to get a new symmetrized Nb-tuple with integer sum of exponents. § 1.3. Transcendentality of e and π 27

Consider a polynomial

N Y N h(x) = bN (x − βk) = b0 + b1x + ··· + bN x ∈ Z[x]. k=1

N Here b0 = (−1) bN · β1 · ... · βN 6= 0. Choose a prime number p ∈ Z such that p > |bN |, p > |b0|, p > |A|, and construct a polynomial 1 f(x) = b(N−1)pxp−1h(x)p (p − 1)! N as in Lemma 1.26. The Hermite identity for f(x) has the following form:

β Z k −t −βk f(t)e dt = F (0) − F (βk)e , k = 1,...,N. 0

Let us multiply each of these expressions by eβk for the corresponding k, and compute the sum of all values obtained for k = 1,...,N. Then

β N Z k N X βk−t X f(t)e dt = AF (0) − F (βk). (1.14) k=1 0 k=1 Consider the right-hand side of equation (1.14):

N N   X (p−1) X (j) X X (j) AF (0) − F (βk) = Af (0) + A f (0) −  f (βk) . k=1 j>p k=1 j>0

(p−1) (N−1)p p By Lemma 1.26, the ﬁrst summand Af (0) = AbN b0 is an integer number that cannot be divided by p, but all other summands are integers divisible by p. Hence, the right-hand side of (1.14) is a nonzero integer. The absolute value of the left-hand side of (1.14) may be estimated Cp from above as C 1 in the very same way as it was done in the proof (p − 1)! of Theorem 1.27. Therefore, if p is suﬃciently large then (1.14) does not hold.

Theorem 1.30 (Lindemann Theorem). If α is a nonzero algebraic number then eα may not be a negative rational number. § 1.3. Transcendentality of e and π 28

n Proof. Let α be a root of a polynomial h(x) = a0 +a1x+···+anx ∈ Z[x] which is irreducible over Q, and let (α1, . . . , αn) be all roots of h(x), i.e., all numbers conjugate to α, where α = α. 1 a Recall that eα 6= 0 for all α ∈ . Assume eα = − , a, b ∈ , where C b N gcd(a, b) = 1. Consider n Y a 0 = eαk + . b k=1 Distribute the brackets and and multiply by bn to obtain n−1 n X k n−k α1 αn 0 = a + a b σn−k(e , . . . , e ). (1.15) k=1

By Lemma 1.28, for each k = 1, . . . , n − 1 there exist a number Nk and a symmetrized Nk-tuple (βk1, . . . , βkNk ) such that

Nk α1 αn X βkj σn−k(e , . . . , e ) = e . j=1

It may happen that some of βkj are zero. Denote by mk > 0 the number of zeros among βkj, j = 1,...,Nk. Let (γ1, . . . , γN ) stand for the collection k n−k of all nonzero βkj in which every βkj appears a b times. This is a symmetrized N-tuple. Now, we may rewrite (1.15) as n−1 N n X k n−k X γj a + a b mk = − cje , cj ∈ Z. k=1 j=1 The left-hand side is an integer A which is not divisible by b, hence, A 6= 0. Thus, N X γj − cje = A 6= 0, j=1 which is a contradiction to Lemma 1.29. Corollary 1.31. The number π is transcendental. Proof. If π were algebraic then α = ιπ, where ι is the imaginary identity, is also algebraic as a product of two algebraic numbers. The Euler ιπ identity e = −1 ∈ Q implies a contradiction to Theorem 1.30. As an immediate corollary, we conclude that the famous ancient prob- lem of squaring the circle may not be solved with a straightedge and compass. Indeed, given a segment of unit length in the plane (the radius of § 1.4. Problems 29 a given circle), every segment one may construct with a straightedge and compass must have algebraic length. Hence, a segment of length π1/2 (the side of a desired square) cannot be constructed. A ﬁner statement, which is a good exercise for a reader, says that if a segment of length a may be constructed with a straightedge and compass then a is an algebraic number of degree 2k (a Pythagorean number). In this way, one may prove that an arbitrary angle in the plane may not be divided into three equal parts, and a right heptagon may not be constructed with a straightedge and compass (for more details, see Exercise 2 below).

§ 1.4. Problems (1) Suppose α is a root of a monic polynomial with algebraic integer coefficients. Show that α is an algebraic integer. (2) Find the degree of the algebraic number e2πι/n, where ι is the imaginary identity and n is a natural number. (3) Show that if α is a nonzero algebraic number then eα is transcendental. (4) Prove the Weierstrass—Lindemann Theorem: If α1, . . . , αn are pairwise different algebraic numbers then eα1 , . . . , eαn are linearly independent over the field of algebraic numbers. CHAPTER 2

Asymptotic law of distribution of prime numbers

Prime integers are “atoms” of the numeric Universe and thus they have been attracting attention of researchers for more than two thousand years. Ancient Greek mathematicians obtained a lot of nice results, e.g., Euclid (III BC) proved that there exist infinitely many primes, Eratosthenes (II BC) invented an algorithm to find all primes smaller than a given natural N (the Sieve of Eratosthenes). These results are now considered as “elementary”. The aim of this section is to present a precise answer to the following question: How often prime numbers occur in the series of all natural numbers? This question was raised by Hauss and Legendre at the end of XVIII, and the Hauss conjecture was proved by Jacques Hadamard and Charles Jean de la Vallée-Poussin in 1896. The result is known as the Prime Num- ber Theorem. To explain the statement, recall the notion of asymptotic equivalence. Suppose f(x) and g(x) are two real-valued functions defined on a ray [a, ∞) ⊂ R such that f(x), g(x) 6= 0 for all sufficiently large x. These functions are said to be asymptotically equivalent if

f(x) lim = 1. x→∞ g(x)

In this case, we write f(x) ∼ g(x).

Recall that P ⊂ N stands for the set of all primes, P = {2, 3, 5, 7, 11,... }. The main object of study in this chapter is the prime-counting function π(x), x > 0, describing the distribution of prime numbers. Namely, π(x) is the cardinality of the set of all primes p ∈ P such that p 6 x, x ∈ R, x > 0.

30 § 2.1. Chebyshev functions 31

The main purpose of this section is to prove the following equivalence: x π(x) ∼ . ln(x) x Exercise 2.1. Prove that ∼ li(x), where ln x x Z dt li(x) = . ln t 2 § 2.1. Chebyshev functions 1. Deﬁnition and estimates. Here we will establish important relations between the function π(x) and the following functions deﬁned on all positive real numbers: X m • ψ(x) = ln p, where Qx = {(p, m) | p ∈ P, m ∈ N, p 6 x}. (p,m)∈Qx The function ψ is called the Chebyshev function; x Z ψ(t) • ψ˜(x) = dt is known as the integral Chebyshev function. t 1 Note that π(x) may be presented in a similar way as X 1. p∈P,p6x m Note that (p, m) ∈ Qx if and only if p 6 x and ln(p ) = m ln p 6 ln x. Therefore, the sum X ln p = ψ(x)

(p,m)∈Qx contains each ln p as many times as the count of all m ∈ N such that m ln p 6 ln x. If x > 1 (i.e., ln x > 0) there exist [ln x/ ln p] of such ms (here [·] stands for the integral part of a real number). Hence, X ln x ψ(x) = ln p, x > 1. (2.1) ln p p∈P,p6x Proposition 2.2. The following statements hold: ˜ 2 (1) ψ(x) 6 π(x) ln x, ψ(x) 6 π(x) ln x for every x > 1; (2) lim ψ(x)/xz = lim ψ˜(x)/xz = 0 for every z ∈ such that x→∞ x→∞ C Re z > 1. § 2.1. Chebyshev functions 32

Proof. (1) If we omit [·] in (2.1) then the value of this sum may just increase since [x] 6 x, i.e., X X X ψ(x) = [ln x/ ln p] ln p 6 ln p 6 ln x = π(x) ln x p6x p6x p6x (hereinafter, when we use “p” for summation index, we assume p ranges over prime numbers, as in (2.1)). For the integral Chebyshev function, note that x Z dt ψ˜(x) ψ(x) = ψ(x) ln x. 6 t 1 The statement (2) immediately follows from (1): If z = 1+a+ιb, a > 0, then

z 1+a 1+a 1+a a |ψ(x)/x | = ψ(x)/x 6 π(x) ln x/x 6 x ln x/x = ln x/x → 0 ˜ as x → ∞. For ψ(x), the proof is completely similar.

Exercise 2.3. Prove that lim ψ˜(x)/xz = 0 for Re z > 1. x→∞ 2. Equivalence of the asymptotic behavior of Chebyshev functions and of the prime-counting function. Theorem 2.4. The following statements are equivalent: (A1) π(x) ∼ x/lnx; (A2) ψ(x) ∼ x; (A3) ψ˜(x) ∼ x.

Proof. (A1)⇔(A2) By Proposition 2.2, ψ(x) 6 π(x) ln x for x > 1. Hence, ψ(x) π(x) , x 6 x/ ln x and the same inequality holds for upper and lower limits of these functions as x → ∞. Namely, ψ(x) ψ(x) (A1) ⇒ lim 6 lim 6 1, x→∞ x x→∞ x π(x) π(x) (A2) ⇒ lim > lim > 1. x→∞ x/ ln x x→∞ x/ ln x § 2.1. Chebyshev functions 33

On the other hand, choose a parameter 0 < a < 1 and consider X S(x, a) = ln p, x > 1. a x S(x, a), but X a a X a S(x, a) > ln(x ) = ln(x ) 1 = a ln x(π(x) − π(x )). a a x x x Note that the second summand in the right-hand side of (2.2) approaches a a zero as x → ∞ since π(x ) 6 x . Assume (A1) holds and B = lim ψ(x)/x. Then (2.2) implies that x→∞ B > a for every positive a < 1, i.e., B > 1. But we have already shown that (A1) implies B 6 1. Thus, B = 1 and (A2) holds. In a completely similar way, (A2) implies (A1). (A2)⇒(A3) Suppose ψ(x) = x + r(x), where r = o(x). Then x x Z ψ(t) Z r(t) ψ˜(x) = dt = x − 1 + dt = x + R(x) − 1, t t 1 1 where R0(x) = r(x)/x. It is enough to check that R = o(x), which is indeed the case by the L’Hopital’s Rule. (A3)⇒(A2) Let ψ˜(x) = x + R(x), R = o(x). Fix a parameter ε, 0 < ε < 1, and consider x+εx Z ψ(t) I+(x) = dt = ψ˜(x + εx) − ψ˜(x) = εx + o(x). ε t x + Since ψ is an increasing function, the integral Iε (x) may be estimated from below as x+εx Z dt I+(x) ψ(x) = ψ(x) ln(1 + ε). ε > t x Similarly, the integral x Z ψ(t) I−(x) = dt = ψ˜(x) − ψ˜(x − εx) = εx + o(x) ε t x−εx § 2.1. Chebyshev functions 34 may be estimated as

x Z dt 1 I−(x) = ψ(x) ln . ε 6 t 1 − ε x−εx Hence, ψ(x) I+(x) ε ε = + o(1), x 6 x ln(1 + ε) ln(1 + ε) ψ(x) I−(x) ε ε = + o(1). x > −x ln(1 − ε) − ln(1 − ε) The upper and lower limits of ψ(x)/x as x → ∞ also satisfy these inequalities for all ε ∈ (0, 1). It remains evaluate the limit as ε → +0 to obtain ψ(x) ∼ x.

Exercise 2.5. Prove the implication (A2)⇒(A1) in Theorem 2.4. 3. Von Mangoldt function. Yet another form of the Chebyshev function comes directly from the deﬁnition: X ψ(x) = Λ(n), (2.3) n∈N,n6x where ( ln p, n = pm, p ∈ , m ∈ , Λ(n) = P N 0, otherwise is called the von Mangoldt function. It is also possible to express the integral Chebyshev function ψ˜(x) via Λ(n).

Proposition 2.6. For every x > 1 we have X ψ˜(x) = Λ(n) ln(x/n). n∈N,n6x Proof. Denote by δ+(y) the step function given by ( 0, y < 0, δ+(y) = 1, y > 0. § 2.2. Riemann function: Elementary properties 35

Then x x Z ψ(t) Z 1 X ψ˜(x) = dt = Λ(n) dt t t 1 1 n6t x x Z 1 X X Z δ+(t − n) = Λ(n)δ+(t − n) dt = Λ(n) dt t t 1 n6x n6x 1 x X Z dt X = Λ(n) = Λ(n) ln(x/n), t n6x n n6x and the Proposition is proved. § 2.2. Riemann function: Elementary properties 1. Riemann function in Re z > 1. We have already seen that the asymptotic behaviors of the functions π(x) and ψ(x) are equivalent. The latter is asymptotically equivalent to the integral Chebyshev function ψ˜(x). It follows from the relation (2.3) and from Proposition 2.6 that these functions are of the form X f(x) = fn, n6x where fn are some real coeﬃcients, and the summation is made over natural numbers. One of the most fruitful ideas that lie in the foundation of the analytic number theory is to study a function f(x) as above via the following complex-valued function called Dirichlet series: ∞ X fn ϕ(z) = . nz n=1 The new variable z takes its values in an appropriate domain in C. If f = ψ then fn = Λ(n), and the corresponding function ϕ turns to be closely related with Riemann zeta-function given by ∞ X 1 ζ(z) = , Re z > 1. (2.4) nz n=1 It is easy to see that the series (2.4) is absolutely converging for Re z > 1. In the semiplane Re z > s, s > 1, the series (2.4) converges uniformly with respect to z. Hence, (2.4) is uniformly converging on every compact subset § 2.2. Riemann function: Elementary properties 36 in Re z > 1. By the well-known Weierstrass Theorem, the limit of a sequence of analytic functions which is uniformly converging on every compact subset in a given domain is again an analytic function. Therefore, (2.4) deﬁnes an analytic function in the semiplane Re z > 1. Later we will see how to extend ζ analytically into the semiplane Re z > 0.

2. Distribution of the Dirichlet series of a multiplicative function. Recall some notions from the elementary number theory. An arbitrary map f : N → C is called an arithmetic function. An arithmetic function is called multiplicative if f(1) = 1 and f(nm) = f(n)f(m) provided that n, m ∈ N are relatively prime. The Fundamental Theorem of Arithmetic implies any multiplicative function f to be uniquely determined by its values f(pm), p ∈ P, m ∈ N. Examples of multiplicative functions are given by: • the function I(n), I(pm) = 1; • the identity function e(n), e(pm) = 0 (while e(1) = 1); • the M¨obiusfunction µ(n), µ(p) = −1, µ(pm) = 0 for m > 1.

Lemma 2.7. Let f be a multiplicative function and let z ∈ C. Suppose ∞ the series P f(n)n−z is absolutely converging. Then n=1 ∞ ∞ ! X Y X f(n)n−z = f(pd)p−dz . n=1 p∈P d=0 ∞ P d −dz Proof. It is easy to see that for every p ∈ P the series f(p )p d=0 contains a part of the initial series and thus converges absolutely. Enumerate prime numbers in the increasing order:

P = {pn | n ∈ N}, p1 = 2, p2 = 3,..., and consider the partial product N ∞ ! Y X d −dz PN = f(pn)pn . n=1 d=0 Since a product of absolutely converging series is distributive, we may distribute the brackets in the last expression to obtain ∞ X d1 dN −d1z −dN z X −z PN = f(p1 ) . . . f(pN )p1 . . . pN = f(n)n , d1,...,dN =0 n∈MN § 2.2. Riemann function: Elementary properties 37

d1 dN where MN ⊂ N consists of all natural numbers of the form p1 . . . pN , di > 0. The least natural number that is not in MN is equal to pN+1, hence,

∞ ∞ X −z X −z f(n)n − PN 6 |f(n)n | → 0 n=1 n=pN+1 as N → ∞ (since pN+1 → ∞).

3. Convolution product and the M¨obiusinversion formula. Given two functions f, g : N → C, their convolution product is an arithmetic function deﬁned by X (f ◦ g)(n) = f(d)g(n/d), n ∈ N. (2.5) d|n where the summation index d ranges over the set of all divisors of n.

Exercise 2.8. Prove that the convolution product is associative and commutative.

Exercise 2.9. For every arithmetic function f, show f ◦ e = e ◦ f = f. (This is the reason why e is called the identity function.)

The following statement shows a nice relation between Dirichlet series and the convolution product.

Lemma 2.10. Let f and g be arithmetic functions such that their Dirichlet series ∞ ∞ X f(n) X g(n) , nz nz n=1 n=1 are absolutely converging for some z ∈ C. Then ∞ ! ∞ ! ∞ X f(n) X g(n) X (f ◦ g)(n) = . (2.6) nz nz nz n=1 n=1 n=1 Proof. Since the product of absolutely converging series is distributive, we may write

∞ ! ∞ ! ∞ X f(n) X g(n) X f(n)g(m) = . nz nz nzmz n=1 n=1 n,m=1 § 2.2. Riemann function: Elementary properties 38

Introduce new indexes d = n, N = nm and change the order of summation (it is possible due to absolute convergence of the product series): ∞ ∞ ∞ X f(n)g(m) X 1 X X 1 = f(d)g(N/d) = (f ◦ g)(N). nzmz N z N z n,m=1 N=1 d|N N=1 Therefore, (2.6) is proved. Moreover, it is known that the set of all multiplicative functions forms a group with respect to the convolution product, i.e., if f and g are multiplicative functions then so is f ◦ g, and for every multiplicative function f there exists a multiplicative function f −1 such that f ◦ f −1 = f −1 ◦ f = e.

Lemma 2.11 (M¨obiusinversion formula). If f : N → C is an arithmetic function and f ◦ I = g then g ◦ µ = f. Proof. Note that I ◦ µ = e. Indeed, if the canonical form of n is m1 mr q1 . . . qr , q1, . . . , qr ∈ P, r > 1, then X X X (I ◦ µ)(n) = µ(d) = µ(1) − µ(qj) + µ(qj1 qj2 − ...

d|n j j1 1 then Y ζ(z) = (1 − p−z)−1 (the Euler identity), (2.7) p∈P ∞ 1 X µ(n) = , (2.8) ζ(z) nz n=1 where µ(n) is the M¨obiusfunction. In particular, ζ has no zeros in the semiplane Re z > 1. § 2.2. Riemann function: Elementary properties 39

Proof. Apply Lemma 2.7 for f = I to obtain Y Y ζ(z) = (1 + p−z + p−2z + ... ) = (1 − p−z)−1, p∈P p∈P when Re z > 1. This proves (2.7). To prove (2.8), apply Lemma 2.7 to f = µ. It is possible to do so ∞ P −z since |µ(n)| 6 1 and thus the series µ(n)n is absolutely converging n=1 for Re z > 1. Hence, ∞ ∞ ! X Y X Y µ(n)n−z = µ(pd)p−dz = (1 − p−z). n=1 p∈P d=0 p∈P Thus (2.7) implies (2.8). 5. Logarithmic derivative of the Riemann function. Lemma 2.14. For all n ∈ N we have (Λ ◦ I)(n) = ln n. Proof. For n = 1 the statement is obvious. If n > 1 then consider a1 ar the canonical distribution of n, n = q1 . . . qr . By the deﬁnitions of the convolution product and of the von Mangoldt function, we have

r aj r X X a X (Λ ◦ I)(n) = 0 + Λ(qj ) · 1 = aj ln qj. j=1 a=1 j=1

a1 ar On the other hand, ln n = ln(q1 . . . qr ) = a1 ln q1 + ··· + ar ln qr. Theorem 2.15. In the semiplane Re z > 1, the following identity holds: ∞ ζ0(z) X Λ(n) = − . (2.9) ζ(z) nz n=1 Proof. As we have already noted, (2.4) converges uniformly with respect to z in the domain Re z > s for every s > 1. Hence (as we know from complex analysis) the series (2.4) allows term-by-term derivation at every point of the semiplane Re z > 1: ∞ X ln n ζ0(z) = − . nz n=1 Multiply the expression obtained by (2.8) for 1/ζ(z): ∞ ! ∞ ! ζ0(z) X ln n X µ(n) = − . ζ(z) nz nz n=1 n=1 § 2.2. Riemann function: Elementary properties 40

Since both series in the right-hand side are absolutely converging, we may distribute the brackets and collect similar terms with nz to obtain

0 ∞ ζ (z) X µ(d2) ln d1 = z ζ(z) (d1d2) d1,d2=1 ∞   ∞ X X 1 X (ln ◦µ)(n) = µ(d) ln(n/d) = . (2.10)   nz nz n=1 d|n n=1 By Lemma 2.14, (Λ ◦ I)(n) = ln n. Lemma 2.11 implies ln ◦µ = Λ, and it remains to apply (2.10) to complete the proof. 6. Expression of the integral Chebyshev function via the Rie- mann function. Denote by La (a ∈ R) the vertical line {z | Re z = a} in the complex plane, and consider La as a path of integration in the upward direction, i.e., from a − i∞ to a + i∞. Theorem 2.16. For every a > 1 the following identity holds: 1 Z ζ0(z) xz ψ˜(x) = − dz for x 1. (2.11) 2πι ζ(z) z2 > La

Proof. Let Ia(x) stand for the improper integral in the right-hand side of (2.11). It follows from (2.9) that 0 z ζ (z) x 1 − C , ζ(z) z2 6 |z|2 for z ∈ La (Re z = a > 1), where C is a constant which does not depend on z. Since the integral ∞ Z Z dz dy

z2 6 a2 + y2 La −∞ converges, Ia(x) is absolutely converging. Therefore, it can be adequately evaluated via the Cauchy principal value. By (2.9), ∞ Z X Λ(n) xz Ia(x) = lim dz, (2.12) B→∞ nz z2 B n=1 La § 2.2. Riemann function: Elementary properties 41

B where La = {z | Re z = a, −B 6 Im z 6 B}. In this expression, the integrand series is uniformly converging with respect to z ∈ La since z a Λ(n) x x ln n

nz z2 6 a2 na ∞ P a for every z ∈ La (recall that the series ln n/n converges for a > 1). n=1 Hence, the integral in the right-hand side of (2.12) can be evaluated in the termwise way: ∞ X Z Λ(n) xz Ia(x) = lim dz. (2.13) B→∞ nz z2 n=1 B La On the other hand, z = a + ιy, and thus partial integrals over the segments B La may be estimated as follows:

∞ Z Λ(n) xz Z x a 1 Λ(n) dz Λ(n) dy Cxa , z 2 6 2 2 6 a n z n a + y n B −∞ La where C is a constant not depending on n and B. Since Λ(n) 6 ln n, the series in the right-hand side of (2.13) converges uniformly with respect to B, hence, the limit as B → ∞ may be evaluated in the termwise way: ∞ X Z 1 I (x) = Λ(n) (x/n)z dz. (2.14) a z2 n=1 La 1 The integrand g (z) = (x/n)z has the only singular point at z = 0. n,x z2 The Laurent series for gn,x(z) at z = 0 has the form

∞ ! 1 X 1 (x/n)z = 1 + lnk(x/n)zk z−2 z2 k! k=1 1 = z−2 + ln(x/n)z−1 + ln2(x/n) + ..., 2! and thus the residue at z = 0 (the coeﬃcient at z−1) is equal to ln(x/n). Let us now evaluate the integrals in the right-hand side of (2.14) by means of the Cauchy integral theorem. Case 1: x n, x/n 1. Consider the circle in the complex plane of √ > > 2 2 B radius R = B + a centered at the origin and denote by Ca the arc of this that lies leftward to the line Re z = a. Being combined with an appropriate § 2.2. Riemann function: Elementary properties 42

Figure 1. Integration path in Case 1

0 B segment L = La , this arc forms a closed integration path which includes the origin for suﬃciently large B (see Fig.1). Then

2π−θ Z 1 Z 1 (x/n)z dz (x/n)z ιReιϕ dϕ 2 6 2 2ιϕ z R e B Ca θ 2π Z 1 (x/n)a dϕ = O(1/R) = O(1/B). 6 R 0 By the Cauchy integral theorem,   1 Z 1 Z 1  (x/n)z dz + (x/n)z dz = ln(x/n). 2πι  z2 z2  B B La Ca The second summand in the left-hand side approaches zero as B → ∞, and the limit of the ﬁrst summand is the desired integral over La. Case 2: x < n, x/n < 1. Consider the integration path shown in Fig.2: B It consists√ of the segment La (as in the previous case) and of the arc 2 2 |z| = R = B + a which is located to the right of La. This is a closed § 2.3. Riemann function: Analytic properties 43

Figure 2. Integration path in Case 2

z 1 path which contains no singularities of the integrand (x/n) z2 . By the same reasons as those used in Case 1, the integral in the right-hand side of (2.14) is equal to zero for x < n. Summarizing Case 1 and Case 2, conclude that ( 1 Z 1 ln(x/n), n x, (x/n)z dz = 6 2πι z2 0, n > x. La Plug in these expressions into (2.14) to obtain 1 Z ζ0(z) xz X − dz = Λ(n) ln(x/n). 2πι ζ(z) z2 n x La 6 ˜ By Proposition 2.6, the last expression is equal to ψ(x).

§ 2.3. Riemann function: Analytic properties 1. Analytic extension of the Riemann function. Let us ﬁrst state a general observation that will be useful later. Suppose f(u, z) is a function depending on a real variable u ∈ [a, b] and on a complex variable z ∈ D, where [a, b] is an interval and D is a domain in C. Assume that for every § 2.3. Riemann function: Analytic properties 44 u ∈ [a, b] the function f(u, z) is analytic in z ∈ D. In addition, suppose that for every ε > 0 there exists δ > 0 such that |∆u| < δ, u + ∆u ∈ [a, b] ⇒ |f(u + ∆u, z) − f(u, z)| < ε (2.15) for all z ∈ D, u ∈ [a, b]. Then

b Z F (z) = f(u, z) du

a is an analytic function in z ∈ D such that b Z ∂f(u, z) F 0(z) = du. ∂z a Indeed, the integral over [a, b] is a limit of a sequence of Riemann sums

N X ΣN = f(uj, z)∆u, ∆u = |b − a|/N. j=1

Each of ΣN is an analytic function in z ∈ D. Condition (2.15) guarantees uniform convergence (with respect to z ∈ D)

b Z ΣN → F (z) = f(u, z) du. N→∞ a The Weierstrass theorem implies F (z) to be an analytic function such that 0 0 the derivative of F (z) is the limit of ΣN with respect to z. Every ΣN is equal to a Riemann sum for the function ∂f/∂z, and thus

b Z 0 ∂f(u, z) lim SN = du. N→∞ ∂z a The following important statement says that the Riemann function de- ﬁned by (2.4) in the semiplane Re z > 1 may be analytically extended into a wider region in which the series (2.4) is diverging. Theorem 2.17. There exists a function ζ˜(z) deﬁned in the semiplane Re z > 0, z 6= 1, such that ζ˜(z) = ζ(z) for Re z > 1. Moreover, ζ˜(z) is analytic at all points of Re z > 0 except for a simple pole at z = 1, in which ˜ Resz=1ζ(z) = 1. § 2.3. Riemann function: Analytic properties 45

Proof. Denote ρ(u) = 1/2 − {u}, u > 0, where {u} = u − [u] is the fractional part of a real number u,[u] is the largest integer not greater than u. Let us ﬁx two natural numbers N < M and consider the following expression:

M+1/2 M+1/2 Z 1 ρ(u) Z 1 1/2 − u + [u] I(N, M, z) := + z du = + z du uz uz+1 uz uz+1 N+1/2 N+1/2 M+1/2 M+1/2 M+1/2 Z 1 − z 1 Z z Z z[u] = du + du + du uz 2 uz+1 uz+1 N+1/2 N+1/2 N+1/2 N+1 M+1/2 M+1/2 Z 1 1 1 z[u] = z−1 − z + z+1 du u N+1/2 2 u N+1/2 u N+1/2 M+1/2 M−1 k+1 X Z z[u] Z z[u] + du + du uz+1 uz+1 k=N+1 k M 1 1 1/2 1/2 = − − + (M + 1/2)z−1 (N + 1/2)z−1 (M + 1/2)z (N + 1/2)z M−1 N N X k k M M − + + − + − + . (N + 1)z (N + 1/2)z (k + 1)z kz (M + 1/2)z M z k=N+1 Reduce similar terms to obtain

M−1 N X k 1 1 I(N, M, z) = − + − + + (N + 1)z (k + 1)z kz−1 M z−1 k=N+1 M−1 N + 1 − 1 X k + 1 − 1 1 1 = − + − + + (N + 1)z (k + 1)z kz−1 M z−1 k=N+1 M−1 1 1 X 1 1 = − + + − (N + 1)z−1 (N + 1)z kz−1 (k + 1)z−1 k=N+1 M−1 X 1 1 + + . (k + 1)z M z−1 k=N+1 § 2.3. Riemann function: Analytic properties 46

Finally, M X 1 I(N, M, z) = kz k=N+1 Hence, if Re z > 1 then N X 1 ζ(z) = + lim I(N, M, z). nz M→∞ n=1 On the other hand, for Re z > 1 ∞ (N + 1/2)−z+1 Z ρ(u) lim I(N, M, z) = + z du. M→∞ z − 1 uz+1 N+1/2 Therefore,

N ∞ X 1 (N + 1/2)−z+1 Z ρ(u) ζ(z) = + + z du, Re z > 1, (2.16) nz z − 1 uz+1 n=1 N+1/2 for every natural N. In (2.16), the ﬁrst summand is an analytic function in the entire C, the second one has a unique simple pole z = 1 (in which the residue is equal to 1), and it remains to analyze the third summand. The integral in the last summand of the right-hand side of (2.16) may be considered as a series: ∞ N+1 N+2 Z ρ(u) Z ρ(u) Z ρ(u) du = du + du + .... uz+1 uz+1 uz+1 N+1/2 N+1/2 N+1 On every interval of integration ([N + 1/2,N + 1), [N + 1,N + 2), and so on), ρ(u)/uz+1 is a continuous function. Moreover, for every s > 0, the last series converges uniformly with respect to z in Re z > s, and the integrand satisﬁes the condition (2.15). Hence, the third summand in the right-hand side of (2.16) is an analytic function in Re z > 0 and ∞ ∞ d Z ρ(u) Z d ρ(u) du = du. dz uz+1 dz uz+1 N+1/2 N+1/2 ˜ Thus, the right-hand side of (2.16) is the desired function ζ(z). In what follows, we will identify ζ and ζ˜. § 2.3. Riemann function: Analytic properties 47

2. Zeros of the Riemann function.

Lemma 2.18. For every s > 1 and for every t ∈ R we have 3 4 |ζ (s)ζ (s + ιt)ζ(s + 2ιt)| > 1. (2.17) Proof. Denote A = |ζ3(s)ζ4(s+ιt)ζ(s+2ιt)|. The Euler identity (2.7) implies Y −1 A = |1 − p−s|3|1 − p−s−ιt|4|1 − p−s−2ιt| . p∈P Let us evaluate the natural log of the both sides of the last expression using the well-known relation ln |w| = Re ln w, w ∈ C \{0}. Then X ln A = − Re 3 ln(1 − p−s) + 4 ln(1 − p−s−ιt) + ln(1 − p−s−2ιt) . p∈¶

Recall that the Taylor series for ln(1 − z) in a neighborhood of z = 0 is given by z2 z3 ln(1 − z) = −z − − − .... 2 3 This series is absolutely converging when |z| < 1. Apply this distribution for z = p−s, z = p−s−ιt, and z = p−s−2ιt:

∞ X X p−ns p−n(s+ιt) p−n(s+2ιt) ln A = Re 3 + 4 + n n n p∈P n=1 ∞ X X p−ns = 3 + 4 Re p−ιnt + Re p−2ιnt n p∈P n=1 ∞ X X p−ns = (3 + 4 cos(nt ln p) + cos(2nt ln p)) . (2.18) n p∈P n=1 Note that 2 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ) > 0, and thus all summands in the right-hand side of (2.18) are non-negative. Hence, ln A > 0, i.e., A > 1.

Theorem 2.19. Riemann function ζ(z) has no zeros in the line Re z = 1. § 2.3. Riemann function: Analytic properties 48

Proof. Let us estimate ζ(s) in a neighborhood of the pole z = 1, namely, in the interval 1 < s < 2. It is easy to see that

∞ ∞ X 1 Z 1 1 2 1 + dx = 1 + ns 6 xs s − 1 6 s − 1 n=1 1 (since a right Riemann sum for a decreasing function is no greater that its integral). Assume ζ(1 + ιt) = 0 for some t 6= 0. Since ζ is analytic at 1 + ιt, its derivative is bounded in a neighborhood of this point. In particular,

ζ(s + ιt) − ζ(1 + ιt) C, 1 < s < 2. s − 1 6 Hence, |ζ(s + ιt)| 6 C|s − 1|. Moreover, since ζ(z) is analytic, it is in particular continuous on the interval z = s + 2ιt, 1 6 s 6 2, and thus there exists a constant M, such that |ζ(s + 2ιt)| 6 M for 1 < s < 2. Therefore, we have obtained the following estimates of the factors in (2.17) for 1 < s < 2: |ζ(s + ιt)| 6 C|s − 1|, 2 |ζ(s)| , 6 s − 1 |ζ(s + 2ιt)| 6 M. Then 3 4 A := |ζ(s) ζ(s + ιt) ζ(s + 2ιt)| 6 C1|s − 1| → 0 as s → 1 + 0, where C1 is a constant, but Lemma 2.18 implies A > 1 for all s > 1. The contradiction obtained proves the theorem. 3. Estimates of the logarithmic derivative. To prove the asymptotic law of distribution of prime numbers, we need an upper estimate of the logarithmic derivative of Riemann function far oﬀ the pole z = 1. Let us start with estimates of ζ(z) itself and its derivative. We have already seen that |ζ(s)| 6 2/(s − 1) for 1 < s < 2.

Proposition 2.20. There exist constants C1,C2 > 0 such that (1) |ζ(s + ιt)| 6 C1 ln |t|, 0 2 (2) |ζ (s + ιt)| 6 C2 ln |t| for 1 6 s 6 2, |t| > 3. § 2.3. Riemann function: Analytic properties 49

Proof. (1) The explicit expression for ζ(s + ιt) is given by Theorem 2.17: N ∞ X 1 (N + 1/2)1−s−ιt Z (s + ιt)ρ(u) ζ(s + ιt) = + + du. (2.19) ks+ιt s − 1 + ιt us+1+ιt k=1 N+1/2 Suppose N = [|t|], and estimate the absolute values of all summands. First,

N N N X 1 X 1 Z dx 1 + = 1 + ln N 1 + ln |t|. s+ιt 6 6 6 k k x k=1 k=1 1 Next, 1−s−ιt (N + 1/2) 1 , s − 1 + ιt 6 3 since s > 1 and |s + ιt − 1| > 3. Finally,

∞ ∞ Z (s + ιt)ρ(u) Z du s + |t| 2 + |t| du (s + |t|) = C s+1+ιt 6 s+1 s 6 6 u u s(|t| − 1) |t| − 1 N+1/2 |t|−1 since N + 1/2 > |t| − 1, |s + ιt| 6 s + |t|, and s 6 2. Therefore, the absolute value of the right-hand side of (2.19) does not exceed B + ln |t| for some constant B (B does not depend on s and t). But B + ln |t| < (B + 1) ln |t| since ln |t| > 1 for |t| > 3. (2) The derivative of (2.19) may be evaluated in the termwise way since the improper integral in the third summand is uniformly converging. Thus,

N ∞ X ln k d (N + 1/2)1−z d Z zρ(u) ζ0(z) = − + + du. kz dz z − 1 dz uz+1 k=1 N+1/2 As above, let z = s + ιt, N = [|t|]. The second and third summands in the right-hand side of the last expression are bounded (as it was shown in the proof of (1)). To estimate the ﬁrst summand, consider the integral over ln x [3, ∞) of the function , which is decreasing on x 3: x > N N X ln k ln 2 X ln k + C ln2 N. ks 6 2 k 6 k=1 k=3 § 2.3. Riemann function: Analytic properties 50

Hence, there exists a constant C2 > 0 such that (2) holds.

Proposition 2.21. There exist constants T0 > 3, C3,C4 > 0 such that −31/4 −8 (1) |ζ(s + ιt)| > C3 ln |t| > C3 ln |t|, 0 ζ (s + ιt) 10 (2) C4 ln |t| ζ(s + ιt) 6 for 1 6 s 6 2, |t| > T0. Proof. (1) Recall the inequality deduced in the proof of Theorem 2.19: 3 4 |ζ(s) ζ(s + ιt) ζ(s + 2ιt)| > 1. It holds for s > 1 and for all real t, in particular, for |t| > 3. Moreover, ζ(s) 6 2/(s − 1) when 1 < s 6 2. By Proposition 2.20(1), |ζ(s + 2ιt)| 6 C1 ln(2|t|) 6 2C1 ln |t| for s > 1, |t| > 3. Hence, 2 −3/4 |ζ(s + ιt)| |ζ(s)|−3/4|ζ(s + 2ιt)|−1/4 C ln−1/4 |t| > > s − 1 for 1 < s 6 2, |t| > 3, where C is a constant which does not depend on s and t. Let us ﬁx t, |t| > 3, and consider the interval 1 + δ 6 s 6 2, where 2 δ = ln10 |t| In this interval, −31/4 |ζ(s + ιt)| > C ln |t|. It remains to estimate from below the quantity |ζ(s + ιt)| in the interval 1 6 s 6 1 + δ. Proposition 2.20(2) allows to estimate an increment of ζ(z) in this interval: 1+δ Z 0 0 |ζ(1 + δ + ιt) − ζ(s + ιt)| = ζ (z) dz 6 δ max |ζ (z)| Im z=t,s6Re z61+δ s −10 2 −8 6 2 ln |t|C2 ln |t| = 2C2 ln |t|. Therefore, −8 −31/4 −8 |ζ(s + ιt)| > |ζ(1 + δ + ιt)| − 2C2 ln |t| > C ln |t| − 2C2 ln |t|.

Since −31/4 < −8, for any constants C and C2 the second summand in the right-hand side becomes negligible for suﬃciently large |t|. Hence, there exists T0 > 3 such that −31/4 |ζ(s + ιt)| > (C/2) ln |t|, § 2.3. Riemann function: Analytic properties 51 for |t| > T0. (2) This estimate immediately follows from (1) and Proposition 2.20(2).

4. Proof of the Prime Number Theorem. Now we are ready to complete the proof of the main statement of this chapter.

Theorem 2.22. The function π(x) is asymptotically equivalent to x/ ln x.

ψ˜(x) Proof. By Theorem 2.4, it is enough to show that lim = 1. x→∞ x Theorem 2.16 implies

ψ˜(x) 1 Z ζ0(z) xz−1 = − dz x 2πι ζ(z) z2 La for every x > 1, a > 1. Here La stands for the line Re z = a, integration is made from a − ι∞ to a + ι∞. Choose real numbers U > T > T0, where T0 is the constant from Proposition 2.21. Since z = 1 is a simple pole, there exists a neighborhood Uδ (a circle |z − 1| < δ) in which |ζ(z)| > 0. Note that the region

W = {z ∈ C | 1/2 6 Re z 6 1, | Im z| 6 T, |z − 1| > δ} ⊂ C may contain only a ﬁnite number of zeros of the Riemann function. Other- wise, if there are inﬁnitely many zeros in a closed bounded region W , the set of zeros has a condensation point in the same region W . Then the well-known uniqueness theorem of an analytic function implies ζ to be zero on W , which is not the case. Therefore, for every T > 0 there exists η > 0 such that

S(T, η) = {z ∈ C | 1 − η 6 Re z 6 1, | Im z| 6 T } contains no zeros of the Riemann function. Consider the integration path Γ shown at Fig.3. It depends on three parameters a, U, and T , where 1 < a < 2, U > T > T0, and on the quantity η > 0 which is chosen in such a way that the interior of Γ (and Γ itself) does not contain zeros of the Riemann function. § 2.3. Riemann function: Analytic properties 52

Figure 3. Integration path Γ

Consider the integral 1 Z ζ0(z) xz−1 I = − dz 2πι ζ(z) z2 Γ The only singularity of the integrand inside the closed path Γ is at the point z = 1, and ζ0(z) xz−1 Res − = 1. (2.20) z=1 ζ(z) z2 By the Cauchy integral theorem, I = 1 for all a, U, T . Let us now present the integral I as a sum of ﬁve integrals over straight segments (1)–(5) of the path Γ (see Fig.3):

I = I1 + I2 + I3 + I4 + I5 (2.21)

(for k = 2, 3, 4, Ik contains integrals over both segments (k)). Let us consider these segments separately. Segment 1. As we have already mentioned, ψ˜(x) lim I1 = . U→∞ x § 2.3. Riemann function: Analytic properties 53

Segment 2. By Proposition 2.21(2), a 1 Z xa−1 |I | 2 C ln10 U ds 2 6 2π 4 s2 + U 2 1

10 2 Hence, |I2| = O(ln U/U ), i.e.,

lim I2 = 0. U→∞ Segment 3. By Proposition 2.21(2),

U ∞ 1 Z C ln10 t Z dt 1 |I | 2 4 dt C = 2C 3 6 2π 1 + t2 6 t3/2 T 1/2 T T (since ln10 t/t1/2 is a bounded function). Note that the constant C does not depend on x. Segment 4. Segments (4) and (5) form a compact set, on which ζ(z) is a nonzero analytic function. Hence, its logarithmic derivative is continuous on these segments. By the Weierstrass theorem on a continuous function, there exists a constant M = M(T, η) such that 0 ζ (z) M(T, η). ζ(z) 6 Then 1 1 1 Z xs−1 M(T, η) Z C¯ |I | 2 M(T, η) ds xs−1 ds . 4 6 2π s2 + T 2 6 πT 2 6 ln x 1−η 1−η The constant C¯ in this expression depends on T and η, since it is propor- tional to M(T, η). Segment 5. By the same reasons as for the previous segment, ˜ −η |I5| 6 Cx , where C˜ depends on T and η. Now, evaluate the limit of (2.21) as U → ∞:

ψ˜(x) I = 1 = + 0 + J + J + J , x 3 4 5 § 2.3. Riemann function: Analytic properties 54 where Jk = lim Ik. As it was shown above, U→∞ 1 |J | 2C , 3 6 T 1/2 C¯(T, η) |J | , 4 6 ln x ˜ −η |J5| 6 C(T, η)x .

Hence, for every small ε > 0 one may choose T such that |J3| 6 ε/2. For a given T , the quantities C¯ and C˜ are ﬁxed, so

|J4| + |J5| 6 ε/2 for a suﬃciently large x. Finally,

ψ˜(x)

1 − 6 |J3| + |J4| + |J5| < ε x for a suﬃciently large x.

Corollary 2.23 (Asymptotic formula for nth prime). Let pn, n = 1, 2, 3,... , stand for the nth prime number (p1 = 2, p2 = 3 and so on). Then

pn ∼ n ln n.

Proof. By the deﬁnition of π, π(pn) = n. Moreover, the Euclid theorem implies pn → ∞. By Theorem 2.22, n→∞ pn n = π(pn) = (1 + Rn), lim Rn = 0. ln pn n→∞ Then

ln n = ln π(pn) = ln pn − ln ln pn + ln(1 + Rn). Multiply these expressions to obtain

ln pn − ln ln pn + ln(1 + Rn) n ln n = pn(1 + Rn) . ln pn It is easy to see that n ln n ln ln pn ln(1 + Rn) = (1 + Rn) 1 − + → 1. pn ln pn ln pn n→∞ § 2.4. Problems 55

§ 2.4. Problems (1) Prove that X ln p = ln x + O(1). p p∈P,p6x (2) Show that there exists a constant C > 0 such that X 1 = C + ln(ln x) + O(1/ ln x). p p∈P,p6x (3) The famous Riemann hypothesis states that all zeros of ζ(z) in the 1 semiplane Re z > 0 lie in the line Re z = 2 . Assuming Riemann hypothesis is true, prove ψ(x) = x + O(xε+0.5), π(x) = ln(x) + O(xε+0.5) for every ε > 0. CHAPTER 3

Dirichlet Theorem

In this chapter we prove the famous Dirichlet theorem on the number of primes in an arithmetic progression with coprime difference and the first member. We start with the structure and properties of finite abelian groups.

§ 3.1. Finite abelian groups and groups of characters 1. Finite abelian groups. Recall that the set G with an algebraic binary operation “·” is called a group (we often omit “·” and write g1g2 instead of g1 · g2), if the following axioms are satisfied: (a) for every a, b, c ∈ G the identity (a · b) · c = a · (b · c) holds; (b) there exists e ∈ G such that for every a ∈ G we have a·e = e·a = a, such element e is called the identity element or the neutral element; (c) for every a ∈ G there exists a−1 ∈ G such that a·a−1 = a−1 ·a = e, such a−1 is called the inverse element. If an additional axiom of commutativity holds: (d) for every a, b ∈ G we have a · b = b · a, then the group is called abelian. If G is abelian, then the additive notation is used very often, so the operation is denoted by “+”, the identity element is denoted by “0”, and the inverse element is denoted by “−a”. By |G| we denote the cardinality of G. A one-generated group is called cyclic, i.e. a group G is called cyclic if there exists a ∈ G such that G = {an | n ∈ Z} (in additive notation, G = {na | n ∈ Z}. Notice that for every group G and for every g ∈ G the set {gn | n ∈ Z} appears to be a subgroup of G. This subgroup is called the cyclic subgroup generated by g and is denoted by hgi. The cardinality |hgi| is called the order of g and is denoted by |g|. Clearly hZ, +i is an infinite cyclic group and, for every n ∈ N, hZn, +i is a finite cyclic group of order n. Exercise 3.1. Let G be a group and g be an element of G. Then the following hold.

56 § 3.1. Abelian groups 57

(1) Either |g| = ∞, or |g| is the minimal positive integer k such that gk = e. 1 (2) If |g| = ∞, then hgi ' Z; if |g| = n, then hgi ' Zn . (3) If gm = e, then |g| divides m. (4) If g1, g2 ∈ G are chosen so that g1·g2 = g2·g1, and hg1i∩hg2i = {e}, then |g1 ·g2| = lcm(|g1|, |g2|). In particular, if |g1|, |g2| are coprime then |g1 · g2| = |g1| · |g2|.

If G1,...Gn are groups, then

G1 × ... × Gn = {(g1, . . . , gn) | g1 ∈ G1, . . . , gn ∈ Gn} with coordinate-wise multiplication is called the direct product of G1,...,Gn. If G is abelian and G1,...,Gn are subgroups of G, then the set G1 ·...·Gn = {g1 · ... · gn | g1 ∈ G1, . . . , gn ∈ Gn} forms a subgroup of G. Exercise 3.2. Let G be an abelian group. Assume that subgroups G1,...,Gn of G satisfy to the following

(1) G = G1 · ... · Gn. (2) for every i we have Gi ∩ (G1 · ... · Gi−1 · Gi+1 · ... · Gn) = {e}.

Then G ' G1 × ... × Gn. In such case G is also called a direct product of subgroups G1,...,Gn. Hint: Prove that conditions (1) and (2) are equivalent to the statement “for every g ∈ G there exist unique g1 ∈ G1, . . . , gn ∈ Gn such that g = g1 · ... · gn”. The proof of the following theorem can be found in many algebra text- books and we do not provide the proof here. Theorem 3.3. Let G be a ﬁnite abelian group. Then there exist d1, . . . , dn ∈ N such that G ' Zd1 × ... × Zdn . Moreover such d1 . . . , dn can be uniquely determined by the following condition: for every i = 1, . . . , n−1, di divides di+1. ∗ Recall that by Zn we denote the (multiplicative) group of invertible elements in Zn. ∗ Exercise 3.4. Prove that |Zn| = ϕ(n), where ϕ(n) is the Euler func- α1 αk tion. If n = p1 · ... · pk is the canonical decomposition of n into the α1 αk product of primes, then ϕ(n) = ϕ(p1 ) · ... · ϕ(pk ) and for every prime p and positive integer k, ϕ(pk) = pk − pk−1.

1Recall that groups G, H are isomorphic if there exists a bijection ϕ : G → H preserving operation, i.e. ϕ(g1 · g2) = ϕ(g1) · ϕ(g2). § 3.1. Abelian groups 58

Hint: Prove that ϕ ◦ I = E, where I(n) = 1, E(n) = n for all natural n. Here ”◦” stands for the convolution product of arithmetic functions mentioned in Chapter 2.

2. Characters. Let G be a ﬁnite abelian group. A homomorphism2 χ : G → C∗ is called a character of G. The set of all characters of G is denoted by Gb. Deﬁne a multiplication on Gb by

for every χ1, χ2 ∈ Gb and g ∈ G we have (χ1 · χ2)(g) = χ1(g) · χ2(g).

Theorem 3.5. Gb is an abelian group. Moreover, G and Gb are isomorphic.

Proof. The multiplication on Gb is clearly an algebraic operation, and it is associative and commutative (we leave the proof for the reader). Denote by χe the principal character, it is defined by χe(g) = 1 for all g ∈ G. It is immediate that χe is the identity element of Gb. For every χ ∈ Gb define χ−1 by χ−1(g) = 1/χ(g). Evidently, χ−1 is the inverse element for χ. Thus Gb is an abelian group. In view of the theorem about the structure of finite abelian groups (Theorem 3.3) we have G = hg1i × ... × hgri for suitable g1, . . . , gr ∈ G. 2πι Denote |gk| by hk, let εk = exp( ). For k = 1, . . . , r define the map hk ∗ a1 ar ak χk : G → C by χk(g1 . . . gr ) = εk . It is straightforward that χk is a 2 hk character of G for k = 1, . . . , r. It is also clear that χk, χk, . . . , χk = χe are distinct characters, since their values on gk are distinct. a1 ar a1 ar Consider the map ϕ : G → Gb defined by ϕ : g1 . . . gr 7→ χ1 . . . χr . By the definition, ϕ preserves the multiplication. The map ϕ is clearly a1 ar injective: if e 6= x ∈ G, then x = g1 . . . gr and there exists ak such that ak gk 6= e; therefore 2a πι a1 ar k ϕ(x)(gk) = (χ1 . . . χr )(gk) = exp 6= 1. hk

hk hk In order to prove that ϕ is surjective note that gk = e implies χ(gk ) = h χ(gk) k = 1 for every χ ∈ Gb. Hence, χ(gk) is a complex root of unity of order ak hk, and thus χ(gk) = εk for appropriate ak. It is straightforward to check a1 ar a1 ar that χ(g) = (χ1 . . . χr )(g) for all g ∈ G. Therefore, χ = χ1 . . . χr =

2Homomorphism of groups is a map, preserving the operation. Namely, the map ϕ : G → H is called a homomorphism, if for every g1, g2 ∈ G we have ϕ(g1 · g2) = ϕ(g1) · ϕ(g2) § 3.1. Abelian groups 59

a1 ar ϕ(g1 . . . gr ), and ϕ is surjective. This completes the proof of the theorem. Exercise 3.6. Check all technical statements in the proof. Remark 3.1. Theorem 3.5 is a particular case of Pontryagin duality between discrete and continuous abelian groups. The group of characters Gb can be considered as a dual group for G and χ1, . . . , χr is the dual basis for g1, . . . , gr. In the proof of the theorem we also derive the following

Corollary 3.7. For every nonidentity g ∈ G there exists χ0 ∈ Gb such that χ0(g) 6= 1. In the theory of characters for ﬁnite groups the following proposition plays an important role. Proposition 3.8. (Orthogonality relations) The following hold (1) for every χ ∈ Gb we have X |G|, if χ = χ ; χ(g) = e 0, if χ 6= χe. g∈G (2) for every g ∈ G we have X |G|, if g = e; χ(g) = 0, if g 6= e. χ∈Gb

Proof. Clearly we need to proof the ﬁrst identity in case χ 6= χe, and the second in case g 6= e. (1) If χ 6= χe, then there exists g0 such that χ(g0) 6= 1. Then   X X X χ(g) = χ(gg0) =  χ(g) χ(g0), g∈G g∈G g∈G P so g∈G χ(g) = 0. (2) If g 6= e, then Corollary 3.7 implies that the existence χ0 such that χ0(g) 6= 1. Then   X X X χ(g) = (χ0χ)(g) =  χ(g) χ0(g), χ∈Gb χ∈Gb χ∈Gb so P χ(g) = 0. χ∈Gb § 3.2. Dirichlet series 60

∗ 3. Characters modulo m. Consider G = Zm. Denote byn ¯ the resid- ual of n modulo m. Then each character χ ∈ Gb can be extended to an arithmetic function χ : N → C by χ(¯n), if gcd(n, m) = 1; χ(n) = (3.1) 0, otherwise. This construction lead us to the notion of a character modulo m. More formally, an arithmetic function χ : N → C is called a character modulo m, if the following conditions are satisfied: (1) χ(n) 6= 0 if gcd(n, m) = 1; (2) χ(n) = 0 if gcd(n, m) > 1; (3) χ(n1) = χ(n2) if n1 ≡ n2 (mod m); (4) χ(n1n2) = χ(n1)χ(n2) for all n1n2 ∈ N. Define the set of all characters modulo m by Gm. It follows by definition ∗ that each character modulo m is a character of Zm extended to all positive integers by using (3.1). So the following proposition is an immediate corollary to the properties of characters of finite abelian groups obtained above. Proposition 3.9. The following hold (1) There exist exactly ϕ(m) distinct characters modulo m. (2) All characters modulo m forms a group Gm under usual multiplication: (χ1 · χ2)(n) = χ1(n) · χ2(n). (3) If Ωm is a full system of residuals modulo m, then X ϕ(m), if χ = χ , χ(n) = e 0, otherwise. n∈Ωm (4) For every n ∈ Z we have X ϕ(m), if n ≡ 1 (mod m), χ(n) = 0, otherwise. χ∈Gm

§ 3.2. Dirichlet series

1. Convergence of L-series. Given χ ∈ Gm deﬁne an L-series L(z, χ), where z ∈ C, by ∞ X χ(n) L(z, χ) = . (3.2) nz n=1 The series L(z, χ) is called an L-series of character χ. § 3.2. Dirichlet series 61

Theorem 3.10. Let χ be a character modulo m. Then the following statements hold.

(1) The series L(z, χ) is absolutely converging for Re z > 1. (2) If χ 6= χe, then L(z, χ) is uniformly converging in every compact domain of the semiplane Re z > 0. (3) If χ = χe, then L(z, χ) is uniformly converging in semiplane Re z > s for every s > 1. Moreover, L(z, χ) possesses an analytic continuation in domain Re z > 0, z 6= 1 with the unique simple pole z = 1.

Proof. (1) is evident, since |χ(n)| 6 1. (2) Consider the series

∞ X χ(n) L(z, χ) = , nz n=1 set s(x) := P χ(n). The main idea of the proof comes from the harmonic n6x series. In view of Proposition 3.9(3) we have

m X X s(m) = χ(k) = χ(k¯) = 0. k=1 ¯ ∗ k∈Zm

Pm Moreover, for every q ∈ N we have k=1 χ(qm + k) = 0. Therefore, for each N ∈ N the inequality

N X | χ(k)| < m (3.3) k=1 holds. § 3.2. Dirichlet series 62

Now consider χ(n) = s(n) − s(n − 1). We obtain

N N X χ(n) X s(n) − s(n − 1) s(N) s(N − 1) = = − + nz nz N z N z n=1 n=1 s(N − 1) s(N − 2) s(2) s(1) s(1) − + ... + − + = (N − 1)z (N − 1)z 2z 2z 1 N−1 s(N) X 1 1 − s(n) · − = N z (n + 1)z nz n=1 N−1 s(N) X Z n+1 z + s(n) · dx = N z xz+1 n=1 n N−1 s(N) X Z n+1 s(x) + z · dx, N z xz+1 n=1 n in the last step we use the identity s(x) = s(n) for x ∈ [n, n + 1). Set Z n+1 s(x) In(z) := z · z+1 dx. n x

Now we can bound |In(z)|. In view of (3.3) we have Z n+1 m m|z| 1 1 |In(z)| 6 |z| · s+1 dx = · s − s . n x s n (n + 1) Therefore N−1 X m|z| 1 m|z| |I (z)| < 1 − < , n s N s s n=1 so the series L(z, χ) is uniformly converging. (3) The series L(z, χe) is absolutely converging for Re z > 1, so

∞ ∞ ! X χe(n) Y X L(z, χ ) = = χ (pd) · p−dz = e nz e n=1 p∈P d=0 Y −z −1 Y −z −1 (1 − χe(p) · p ) = (1 − p ) , p∈P p∈P,p-m whence Euler identity (see Theorem 2.13) implies that the following identity holds Y −z −1 L(z, χe) · (1 − p ) = ζ(z). (3.4) p∈P,p|m § 3.2. Dirichlet series 63

We can deﬁne now L(z, χe) on Re z > 0 by   Y −z L(z, χe) = ζ(z) ·  (1 − p ) , p∈P,p|m whence item (3) of the theorem. Lemma 3.11. Set Y L(z, Gm) = L(z, χ). χ∈H

Then for Re z > 1 the series L(z, Gm) can be written in the following form ∞ X an , nz n=0 ϕ(m) where every an is a nonnegative integer and, moreover, if n = k and (k, m) = 1, then an > 1. Moreover, ∞ k X an(ln n) L(z, G )(k) = (−1)k , k = 1, 2,... m nz n=1 Proof. The statement about derivations follows from the fact that the series ∞ X an nz n=0 is absolutely converging for Re z > 1 (as a product of a ﬁnite number of absolutely converging series). Thus we need to show that ∞ X an L(z, G ) = . m nz n=0

Since χ is multiplicative for every χ ∈ Gm, Lemma 2.7 implies that −1 Y χ(p) L(z, χ) = 1 − . pz p∈P Therefore, −1 −1 Y Y χ(p) Y Y χ(p) L(z, G ) = 1 − = 1 − , m pz pz χ∈H p∈P p∈P χ∈Gm § 3.2. Dirichlet series 64 where we change the order of multiplication, since Gm is ﬁnite and −1 Y χ(p) 1 − pz p∈P is absolutely converging. ∗ For every p ∈ P denote by fp the multiplicative order of p in Zm, i.e. the minimal positive k with pk ≡ q (mod m). Then

1 = χ(1) = χ(pfp ) = χ(p)fp , so χ(p) = exp(2πιk/fp) for some k = 0, 1, . . . , fp = 1. It follows that the map ∗ ψ : Gm → C acting by ψ : χ 7→ χ(p) maps Gm into {exp(2πιk/fp) | k = 0, 1, . . . , fp − 1}. Clearly ψ is a homomorphism. Notice that ψ is surjective. Indeed, we can deﬁne a character s χp on a power of a prime r by  s exp(2πι/fp) , if r = p s  χp(r ) = 1 if r 6= p and (r, m) = 1  0 if r divides m, and extend it on all integers by multiplicativity. Then hχpi is a subgroup of Gm and ψ(hχpi) = {exp(2πιk/fp) | k = 0, 1, . . . , fp − 1}. So the ker- nel of ψ has order |Gm|/fp = ϕ(m)/fp =: gp. It follows that for every k = 0, 1, . . . , fp − 1 there exists exactly gp characters χ ∈ Gm with χ(p) − exp(2πιk/fp). Therefore, a polynomial Y (1 − χ(p)t)

χ∈Gm equals (1 − tfp )gp . Thus

Y −fpz −gp L(z, Gm) = (1 − p ) . p∈P;(p,m)=1 Now the Taylor series for (1 − z)−g equals ∞ X (g + k − 1)! zk, (g − 1)!k! k=0 § 3.2. Dirichlet series 65 and the series is absolutely converging for |z| < 1. Since |p−fpz| < 1, we can apply the Taylor series to the expression of L(z, Gm). We obtain ∞ (g + k − 1)! ∞ u −fpz −gp X p −fpkz X p,k (1 − p ) = p = kz , (gp − 1)!k! p k=0 k=0 where ( 0, if k is not divisible by fp, up,k = (gp+r−1)! , if k = r · fp. (gp−1)!r! Since for every p ∈ P the series ∞ X up,k pkz k=0 is absolutely converging, we obtain the following identity for every N: ∞ Y X an (1 − p−fpz)−gp = , nz p≤N;(p,m)=1 n=1 where 0, if (n, m) > 1, an = k1 kl up1,k1 · ... · upl,kl , if (n, m) = 1 and n = p1 · ... · pl .

Consider 1 < s ∈ R. Then L(s, Gm) is converging. On the other hand, for every M, M X an ≤ L(s, G ), ns m n=1 so the series ∞ X an ns n=1 is converging, and so the series ∞ X an nz n=1 is absolutely converging for every z with Re z > 1. It follows that ∞ X an L(z, G ) = , m nz n=1 § 3.2. Dirichlet series 66 where the coefficients an-s are defined above. From the definition we obtain that an are nonnegative integers. Since fp divides ϕ(m), we obtain that ϕ(m) an 6= 0, if n = k for (k, m) = 1. 2. Landau Theorem. Consider Y L(z) := L(z, Gm) = L(z, χe) · L(z, χ).

χ∈Gm\{χe}

Theorem 3.10 implies that L(z, χe) is analytic in semiplane Re z > 0 with a unique simple pole z = 1, while each L(z, χ) for χ 6= χe is analytic in the semiplane Re z > 0, i.e. Q L(z, χ) is analytic in the semiplane χ∈Gm\{χe} Re z > 0. Thus L(z) is analytic in the semiplane Re z > 0 with one possible simple pole z = 1. In order to prove that z = 1 is indeed a simple pole for L(z) we need to prove that for every χ ∈ Gm \{χe} we have L(1, χ) 6= 0. The next theorem helps us to prove the desired statement. Theorem 3.12. (Landau Theorem) Assume that a function F (z) is analytic in Re z > 0 and suppose that for Re z > 1 we can write F (z) as a series ∞ X an F (z) = , (3.5) nz n=1 where an > 0 for every n. Assume also that in the semiplane Re z > 1 we have ∞ k X an(ln n) F (k)(z) = (−1)k , nz n=1 i.e. series (3.5) can be diﬀerentiated term by term in the semiplane Re z > 1. P∞ an Then the series n=1 ns is converging in the interval s ∈ (0, 2). Proof. We consider the Taylor series for F (s) about s = 2: ∞ X F (k)(2) F (s) = (s − 2)k. (3.6) k! k=0 The Taylor series of an analytic function is converging in the circle of radius r centered at s0, where r is the distance to the nearest singular point, i.e. series (3.6) is converging for every s ∈ (0, 2). Now

∞ k X an(ln n) F (k)(2) = (−1)n , n2 n=1 § 3.2. Dirichlet series 67 hence ∞ ∞ k k X X (−1) an(ln n) F (s) = · · (s − 2)k. k! n2 k=0 n=1 The series is converging, moreover, (−1)k and (s−2)k for s ∈ (0, 2) have the same sign, so all terms in the series are nonnegative. Therefore the series is absolutely converging and we can change the order of summation. So we obtain ∞ ∞ k k X X (−1) an(ln n) F (s) = · · (s − 2)k = k! n2 n=1 k=0 ∞ ∞ k k ! X an X (ln n) (2 − s) = n2 k! n=1 k=0 // the inner sum is equal to exp((2 − s) ln n) = n2−s// = ∞ ∞ X an X an · n2−s = , n2 ns n=1 n=1 and the theorem follows.

Now we can prove that for every χ ∈ Gm \{χe} we have L(1, χ) 6= 0.

Corollary 3.13. If χ ∈ Gm \{χe} then L(1, χ) 6= 0.

Proof. Assume that L(1, χ) = 0. Then L(z) = L(z, Gm) is analytic in the semiplane Re z > 0. Moreover, ∞ X an L(z) = , nz n=1 ϕ(m) where an > 0 and further an > 1 for n = k and gcd(k, m) = 1. Every series L(z, χ) is absolutely converging in the semiplane Re z > 1, so L(z) is absolutely converging for Re z > 1. Therefore the series L(z) can be differentiated term by term any number of times. By the Landau theorem the series ∞ X an ns n=1 ϕ(m) is converging for 0 < s < 2. For n = (km + 1) we have an > 1, so N N [ m −1] X an 1 X 1 > //for s = // > → ∞, ns ϕ(m) km + 1 N→∞ n=1 k=1 § 3.2. Dirichlet series 68 a contradiciton. 3. Proof of the Dirichlet Theorem. First we recall the statement of the theorem. Theorem 3.14. (Dirichlet Theorem) Assume that a, m are natural coprime numbers. Then there exist infinitely many primes of the form a + km, or, equiv- alently, there exist infinitely many primes p such that p ≡ a (mod m). Proof. Consider the series ∞ X χ(n) L(z, χ) = , nz n=1 where χ ∈ Gm and Re z > 1. First we prove that ∞ 1 X χ(n) · µ(n) = , (3.7) L(z, χ) nz n=1 P∞ χ(n)·µ(n) where µ(n) is the Möbiusfunction. Indeed, the series n=1 nz and P∞ χ(n) n=1 nz are absolutely converging, therefore by Lemma 2.10 we have ∞ ! ∞ ! ∞ X χ(n) X χ(n) · µ(n) X (χ ◦ (χ · µ))(n) · = . nz nz nz n=1 n=1 n=1 Now χ ◦ (χ · µ) = (χ · I) ◦ (χ · µ) = χ · (I ◦ µ), and the Möbiusinversion formulae (Lemma 2.11) implies that I ◦ µ = e, so χ · (I ◦ µ) = χ · e = e, so we get (3.7). For Re z > 1 we have ∞ X χ(n) ln n L0(z, χ) = − . nz n=1 Thus,

∞ ! ∞ ! L0(z, χ) X χ(n) · µ(n) X χ(n) ln n = · − = L(z, χ) nz nz n=1 n=1 ∞ ∞ X ((χ · ln) ◦ (χ · µ))(n) X (χ · (ln ◦µ))(n) − = − = nz nz n=1 n=1 ∞ X (χ · Λ)(n) //Lemmas 2.11 and 2.14// = − , nz n=1 § 3.2. Dirichlet series 69 where Λ(n) is the von Mangoldt function, i.e. we obtain the identity

∞ L0(z, χ) X (χ · Λ)(n) = − . L(z, χ) nz n=1

By deﬁnition we have χ(pk) = χ(p)k and χ(p) 6= 0 if and only if gcd(p, m) = 1. Therefore

∞ L0(z, χ) X X χ(p)k · ln p − = = L(z, χ) pkz p∈P k=1 //the series is absolutely converging for Re z > 1, so we can change the order of summation// = ∞ ∞ X X χ(p)k · ln p X χ(p) · ln p X X χ(p)k · ln p = + . pkz pz pkz k=1 p∈P p∈P k=2 p∈P

Denote the second summand by R(z, χ). We show that R(z, χ) is absolutely 1 converging for Re z > 2 + ε for every ε > 0. Indeed, we need to show that the series ∞ k X X χ(p) ln p pz p∈P k=2 is absolutely converging, since in this case we can change the order of sum- k mation and derive that R(z, χ) = P∞ P χ(p) ·ln p is absolutely con- k=2 p∈P pkz verging as well. Now we have

∞ k X X χ(p) X χ(p)2 1 ln p = ln p · · . pz p2z 1 − χ(p) p∈P k=2 p∈P pz

χ(p) 1 Since |1 − pz | > |1 − p1/2 |, we have

χ(p)2 1 ln p 1 4 ln p ln p · · 6 · 6 , p2z χ(p) p1+2ε |1 − √1 | p1+2ε 1 − pz p and thus we obtain that the series R(z, χ) is absolutely converging. By the condition of the theorem gcd(a, m) = 1, so we can choose b so that § 3.2. Dirichlet series 70 b · a ≡ 1 (mod m). Then we have X L0(z, χ) χ(b) · − = L(z, χ) χ∈Gm X X χ(b) · χ(p) · ln p X + χ(b) · R(z, χ) = pz χ∈Gm p∈P χ∈Gm   X ln p X X · χ(b) · χ(p) + χ(b) · R(z, χ) = pz   p∈P χ∈Gm χ∈Gm X //recall that χ(bp) equals ϕ(m)

χ∈Gm if bp ≡ 1 (mod m) and 0 otherwise // = X ϕ(m) · ln p X + χ(b) · R(z, χ). pz p∈P,p≡a (mod m) χ∈Gm 1 The second summand is analytic for Re z > 2 . If the theorem is false, then the first summand is finite and therefore it has a precise value for z = 1. On the other hand, 0 0 0 X L (z, χ) L (z, χe) X L (z, χ) χ(b) · = + χ(b) · . L(z, χ) L(z, χe) L(z, χ) χ∈Gm χ∈Gm\{χe} By Theorem 3.10 and Corollary 3.13 we obtain that the second summand is bounded for z = 1, while for the first summand we have 0 L (z, χe) 0 = ln(L(z, χe)) , L(z, χe) 0 and ln(L(z, χe)) has a pole at z = 1 since L(z, χe) has a pole at z = 1. CHAPTER 4

p-adic numbers

§ 4.1. Valuation ﬁelds

1. Basic properties. Let F be a field and v : F → R a map from F to the field of real numbers. Then (F, v) is called a valuation field, while v is called a valuation of F , if (1) For every x ∈ F we have v(x) > 0 and v(x) = 0 if and only if x = 0. (2) v(x + y) 6 v(x) + v(y) (triangle inequality). (3) v(x · y) = v(x) · v(y). We collect evident properties of a valuation in the next proposition. Proposition 4.1. Let (F, v) be a valuation field. Then v(1) = 1, v(−1) = 1, and, for every x ∈ F ∗ and every k ∈ Z we have v(xk) = (v(x))k. Proof. Since v(x) = v(x · 1) = v(x) · v(1) we obtain that v(1) = 1. Now v(−1)2 = v((−1)2) = 1 and v(−1) > 0, whence v(−1) = 1. We also −1 −1 −1 1 have 1 = v(1) = v(x·x ) = v(x)·v(x ), so v(x ) = v(x) . The remaining statement follows immediately. If F is the field of rationals, then we can define the following valuations: 0, if x = 0, (1) v(x) = the trivial valuation (it can be defined 1, otherwise; over arbitrary field). α (2) vα(x) = |x| for 0 < α 6 1. νp(x) (3) vp,ρ(x) = ρ , where 0 < ρ < 1, p is a prime, and νp(x) ∈ Z is νp(x) a given by the identity x = p · b , where p does not divide a · b, and vp,ρ(0) := 0; the p-adic valuation. Exercise 4.2. Check, that all defined above valuations satisfy to the definition. Can α be greater than 1 in item (2)? Can α be less, than 0 in item (2)?

71 § 4.1. Valuation ﬁelds 72

Clearly, every valuation deﬁnes a topology on F . Namely, we can deﬁne an open sphere of radius ε ∈ R>0 centered at a ∈ F by

Bε(a) = {x ∈ F | v(a − x) < ε}, (4.1) and consider the family of all such spheres as a basis of a topology. It is clear that F with the topology is a Hausdorff space. Moreover, the operations “+”, “·” (considered as maps F ×F → F ) and “−”, −1 (considered as maps F → F ) are continuous. In particular, F is a topological field. Exercise 4.3. Prove that all operations are continuous maps. Prove that induced topology is a Hausdorff space.

Let (F, v) be a valuation ﬁeld, a ∈ F . A sequence {an}n>1 is said to converge to a (under the valuation v, we use notation an → a), if (v)

lim v(an − a) = 0. n→∞ The following basic properties of limits hold. Proposition 4.4. The following identities hold:

lim (an ± bn) = lim an ± lim bn; n→∞ n→∞ n→∞ lim (an · bn) = lim an · lim bn; n→∞ n→∞ n→∞ −1 −1 lim an = lim an , n→∞ n→∞ where lim an and lim bn are assumed to exist and, in the last identity all n→∞ n→∞ an-s and lim an are assumed to be not equal to 0. n→∞ Exercise 4.5. Prove Proposition 4.4.

Let F be a ﬁeld and v1, v2 be its valuations. The valuations v1, v2 are called equivalent (v1 ∼ v2), if, for every sequence {an}n>1, we have

{an}n>1 → a ⇐⇒ {an}n>1 → a. (v1) (v2)

Lemma 4.6. Valuations v1, v2 of F are equivalent if and only if for every x ∈ F we have v1(x) < 1 ⇔ v2(x) < 1.

Proof. Assume that v1 and v2 are equivalent. Then for every x ∈ F n the inequality v1(x) < 1 is equivalent to x → 0. Since v1 and v2 are (v1) n equivalent, it follows that x → 0, so v2(x) < 1. (v2) § 4.1. Valuation ﬁelds 73

Now we prove the converse statement. If v1 is trivial, then v2, clearly, is also trivial and the claim is evident. Assume that v1 is nontrivial. Then there exists x ∈ F such that v1(x) 6= 1. By Proposition 4.1 we obtain that −1 either v1(x) > 1, or v1(x ) > 1. Without loss of generality we may assume that v1(x) > 1 (and so v2(x) > 1). Choose a sequence {an}n>1 such that m m an → a. Therefore, for every m ∈ N, we have x · an → x · a, i.e. (v1) (v1) m m v1(x an − x a) → 0. Now denote v2(x) by α (recall that v2(x) > 1). n→∞ Then for every ε > 0 there exists M ∈ N such that for every m > M we 1 have αm < ε. m m Since v1(x an − x a) → 0, it follows that there exists N ∈ such n→∞ N m m that for every n > N we have v1(x an − x a) < 1. The condition of the m m lemma implies that for every n > N the inequality v2(x an − x a) < 1 m holds. Thus we obtain that for every n > N the inequality v2(x) · v2(an − a) < 1 holds. Hence, for every n > max{M,N} we have v2(an − a) < ε, i.e. an → a. The implication an → a ⇒ an → a follows by the symmetry. (v2) (v2) (v1)

Corollary 4.7. Every valuation vα of Q, where 0 < α 6 1, is equivalent to v1(x) = |x|. If p is a ﬁxed prime, then for every 0 < ρ < 1 all valuations vp,ρ are equivalent, the corresponding equivalence class is denoted by vp. If p, q are distinct primes, then vp,ρ and vq,ρ are not equivalent. Moreover, vα and vp,ρ are not equivalent. Exercise 4.8. Prove Corollary 4.7

The topology induced on Q by a p-adic valuation is called a p-adic topology. The p-adic valuation of a number grows with the power of p n in denominator. For example, if an = p , then an → 0. Valuation n→∞ ﬁelds (F1, v1) and (F2, v2) are topologically isomorphic, if there exists an isomorphism of ﬁelds ϕ : F1 → F2 such that for every sequence {an}n>1, an → a if and only if ϕ(an) → ϕ(a). In particular, if F1 = F2 = F , (v1) (v2) then v1 ∼ v2 if and only if (F, v1) is topologically isomorphic to (F, v2) with isomorphism Id, where Id is the identical map.

Exercise 4.9. Prove that (Q, vp) is not topologically isomorphic to (Q, vq) for distinct primes p, q.

We say that a valuation field (F1, v1) can be embedded into a valuation field (F2, v2), if there exists an injective homomorphism ϕ : F1 → F2 such that the restriction of v2 on ϕ(F1) coincides with v1. § 4.1. Valuation fields 74

2. Valuations over rationals.

Theorem 4.10. (On valuations over rationals) Let (Q, v) be a valuation ﬁeld. Then either v is the trivial valuation, or v = vα, or v = vp,ρ. Proof. If v is the trivial valuation, we have nothing to prove. Assume that v is not trivial. Then there exists n ∈ N such that v(n) 6= 1. Indeed, p p since v is nontrivial, there exists q ∈ Q such that v( q ) 6= 1. Hence either 1 −1 v(p) 6= 1, or v(q) = v( q ) 6= 1. Now p, q ∈ Z, hence for some p ∈ Z, we obtain v(p) 6= 1. By Proposition 4.1 we have v(p) = v(−p), so there exists n ∈ N such that v(n) 6= 1. Now one of the following two cases holds: either there exists n ∈ N such that v(n) > 1, or for every n ∈ N the inequality v(n) 6 1 holds. Consider these cases separately. Case 1. There exists n such that v(n) > 1. We have v(n) = v(1 + ... + 1) 6 1 + ... + 1 = n, | {z } | {z } n times n times so there exists α ∈ (0, 1] such that v(n) = nα. By Proposition 4.1 we obtain that v(nk) = nkα for every k ∈ Z. Assume that m ∈ N. Then there exists k k+1 k > 0 such that n 6 m < n . Consider the expansion of m in base n, we obtain: k m = a0 + a1n + ... + akn , where 0 6 ai 6 n − 1 for i = 0, 1, . . . , k; and ak 6= 0. Now we have α kα v(m) 6 v(a0) + v(a1) · n + ... + v(ak) · n 6 α kα //for every i, v(ai) 6 ai 6 n − 1// 6 (n − 1)(1 + n + ... + n ) 6 n − 1 n − 1 (n(k+1)α − 1) < nα · nkα C · mα. nα − 1 nα − 1 6 Thus there exists a constant C such that for every m ∈ N we have v(m) < C · mα. We state that for every m we have α v(m) 6 m . (4.2) α v(k) Indeed, assume that there exists k ∈ N such that v(k) > k , i.e. kα = D > 1. Then for every s ∈ N we have v(ks) = Ds → ∞. ksα s→∞ v(ks) On the other hand, ksα < C, a contradiction. Thus (4.2) holds for every m ∈ N. § 4.1. Valuation ﬁelds 75

k+1 k k+1 Now n = m + m1, where 0 < m1 6 n (n − 1). Therefore v(n ) 6 v(m) + v(m1), so

(k+1)α (k+1)α α v(m) > n − v(m1) > n − m1 > n − 1α nkα(nα − (n − 1)α) = n(k+1)α 1 − > C · mα. n 1

It follows that there exists a constant C1 > 0 such that for every m ∈ N we α have v(m) > C1 · m . We state that α v(m) > m . (4.3) α v(k) Indeed, assume that there exists k ∈ N such that v(k) < k , i.e. kα = D1 < 1. So for every s ∈ N we have s v(k ) s = D1 → 0. ksα s→∞ v(ks) On the over hand, ksα > C1, a contradiction. Combining inequalities (4.2) and (4.3) we obtain that for every m ∈ N the identity v(m) = mα holds. By Proposition 4.1 it follows that v(−m) = v(m) and v(m−1) = (v(m))−1, α so for every p ∈ , the identity v p = p holds. q Q q q Case 2. For every n ∈ N we have v(n) 6 1. Choose n so that v(n) 6= 1 (hence v(n) < 1). Consider the decomposition of n into the product of α1 αk α1 αk primes n = p1 · ... · pk . Then v(n) = v(p1) · ... · v(pk) , so there exists a prime p with v(p) = ρ < 1. We show ﬁrst that for every prime q 6= p the identity v(q) = 1 holds. Otherwise there would exist q 6= p with v(q) = µ < 1. Since both ρ and µ are less than 1, there exists k ∈ N such k k 1 k k that both ρ , µ are less than 2 . On the other hand, gcd(p , q ) = 1, so there exist a, b ∈ Z such that a · pk + b · qk = 1. Thus we obtain

k k k k 1 = v(1) = v(a · p + b · q ) 6 v(a) · v(p ) + v(b) · v(q ) 6 1 1 1 · ρk + 1 · µk < + = 1, 2 2

m α1 α2 a contradiction. Now for every n ∈ Q we have m = p · m1, n = p · n1, where gcd(m1 · n1, p) = 1, therefore

m α −α v(m1) α −α ν m v = ρ 1 2 · = ρ 1 2 = ρ p( n ) n v(n1) and the theorem follows. § 4.1. Valuation ﬁelds 76

3. The replenishment of a valuation ﬁeld. A classical result claim that every topological space admits a replenishment. Now we construct a replenishment for a valuation ﬁeld and prove that the replenishment is unique (up to a topological isomorphism).

Let (F, v) be a valuation ﬁeld. A sequence {an}n>1, where an ∈ F is called fundamental, if for every ε > 0 there exists N such that for every n, m > N we have v(an − am) < ε. Lemma 4.11. The following hold:

(1) Assume that v1 and v2 are equivalent valuations of a ﬁeld F .A

sequence {an}n>1 is fundamental in v1 if and only if it is fundamental in v2.

(2) If {an}n>1 is a fundamental sequence in (F, v), then {v(an)}n>1 is a fundamental sequence in R under valuation v(x) = |x|.

Proof. We prove item (1) first. If v1 is the trivial valuation, then v2 is also trivial and we have nothing to prove. If v1 is nontrivial, then there exists x ∈ F such that v1(x) = α > 1. Then for every ε > 0 there −m exists m ∈ N such that α < ε. If the sequence {an}n>1 is fundamental m in v1, then {x · an}n>1 is also fundamental in v1, so there exists N ∈ N m m such that for every n, k > N we have v1(x an − x ak) < 1. Whence m m −m v2(x an − x ak) < 1, and so v2(an − ak) < α < ε. Now we turn to (2). Since v(an) = v(an−am+am) 6 v(am)+v(an−am), and, by symmetry, v(am) 6 v(an) + v(an − am), it follows that |v(an) − v(am)| 6 v(an − am). A valuation field (F, v) is called complete if every fundamental sequence in G is converging, i.e. it has the limit in F . If a valuation field is embedded into a complete valuation field (F, v¯) so that F is dense in F then (F, v¯) is called a replenishment of (F, v). Theorem 4.12. For every valuation field (F, v) there exists a unique up to isomorphism replenishment (F, v¯). Proof. We construct a replenishment using the standard construction for the real numbers. Let

F = {{an}n>1 | {an}n>1 is a fundamental sequence of (F, v)} be a set of all fundamental sequences consisting of elements from F . Deﬁne the addition and multiplication on F by

{an}n>1 + {bn}n>1 = {an + bn}n>1, {an}n>1 ·{bn}n>1 = {an · bn}n>1, § 4.1. Valuation ﬁelds 77

Clearly both {an + bn}n>1 and {an · bn}n>1 are fundamental sequence. We prove, for example, that {an ·bn}n>1 is fundamental provided both {an}n>1, {bn}n>1 are fundamental. We have

v(an · bn − am · bm) = v(an · bn − am · bn + am · bn − am · bm) 6 v(an · bn − am · bn) + v(am · bn − am · bm) =

v(an − am) · v(bn) + v(bn − bm)v(am).

Since {an}n>1, {bn}n>1 are fundamental sequences, Lemma 4.11(2) implies that {v(an)}n>1, {v(bn)}n>1 are fundamental sequences as well. Therefore the sequences {v(an)}n>1 and {v(bn)}n>1 are uniformly bounded by an absolute constant C. Since for every ε there exists N such that for every ε ε m, n > N we have v(an − am) < 2C and v(bn − bm) < 2C . So ε ε v(a − a ) · v(b ) + v(b − b )v(a ) < + = ε, n m n n m m 2 2 hence {an · bn}n>1 is fundamental. Thus F is a commutative ring with 1. Consider

I0 = {{an}n 1 | an → 0}. > n→∞

Clearly I0 is an ideal of F. We show that F/I0 is a ﬁeld. Since F is a commutative ring with 1, we remain to prove that each nonzero element of

F/I0 has inverse. Consider {an}n>1 +I0 6= 0+I0. Since {an}n>1 ∈ F \I0, it follows that an 6→ 0, i.e. there exists ε > 0 and N ∈ N such that for every n→∞ 1 n N we have v(an) > ε. Consider the sequence { }n N . The sequence > an > is well-deﬁned, since an 6= 0 for n > N. Moreover, for every n, m > N we have 1 1 v(am − an) v(am − an) v − = 6 2 , an am v(an · am) ε 1 1 so the sequence { }n N is fundamental, i.e. { }n N ∈ F. Deﬁne the an > an > 1 sequence {bn}n 1 by: bn = 1 for n < N and bn = for n N. Then > an > {bn}n>1 ∈ F and an · bn − 1 = 0 for all n > N. Hence,

({an}n>1 + I0) · ({bn}n>1 + I0) = {an · bn}n>1 + I0 = {1}n>1 + I0, −1 i.e. {bn}n>1 + I0 = ({an}n>1 + I0) . Therefore F = F/I0 is a field. The embedding F → F is defined by a 7→ {an}n>1 + I0, where an = a for all n ∈ N (we use the notation {a}n>1 below). Define a valuation on F in the following way:

if α = {an}n 1 + I0, thenv ¯(α) = lim v(an). > n→∞ § 4.1. Valuation ﬁelds 78

Clearly,v ¯ does not depend on the representative of a coset {an}n>1 + I0, since the identity

lim (v(an) + v(bn)) = lim v(an) + lim v(bn) n→∞ n→∞ n→∞ holds. It is also evident that the restriction ofv ¯ on F coincides with v. Now we check the properties of a valuation.

(1)¯v({an}n>1 + I0) > 0 andv ¯({an}n>1 + I0) = 0 if and only if {an}n 1 +I0 = I0, i.e. an → 0. Indeed, since for each n we have > n→∞ v(an) > 0, it follows that limn→∞ v(an) =v ¯({an}n>1 + I0) > 0. If v¯({an}n>1 + I0) = 0, then limn→∞ v(an) = 0, so {an}n>1 ∈ I0. (2)

v¯({an}n>1 + I0 + {bn}n>1 + I0) =

v¯({an + bn}n>1 + I0) = lim v(an + bn) n→∞ 6

lim (v(an) + v(bn)) = n→∞

lim v(an) + lim v(bn) = n→∞ n→∞

v¯({an}n>1 + I0) +v ¯({bn}n>1 + I0). (3)

v¯({an}n>1 + I0) · v¯({bn}n>1 + I0) =

v¯({an · bn}n>1 + I0) = lim v(an · bn) = n→∞

lim v(an) · v(bn) = n→∞

lim v(an) · lim v(bn) = n→∞ n→∞

v¯({an}n>1 + I0) · v¯({bn}n>1 + I0). Sov ¯ is a valuation.

Now we show that F is dense in F . If α = {an}n>1 is a fundamental sequence, then the sequence {αn}n>1, where αn = {an}k>1 ∈ F (recall that {an}k>1 is a sequence such that all its members are equal to an), is clearly converging to α. We remain to show that F is complete. Assume that

{αn}n>1 is a fundamental sequence in F . Since F is dense in F , for every 1 n we can choose an ∈ F such thatv ¯(αn − {an}k>1) < n . Let α = {an}n>1. § 4.2. Construction and properties of p-adic ﬁelds 79

We claim that α ∈ F (so α + I0 ∈ F ) and αn → α. Indeed, for every ε (¯v) ε there exists M such thatv ¯(αn − αm) < 3 for all m, n > M. Further, there 1 ε exists N > M such that N < 3 . So for n, m > N we have

v(an − am) =v ¯({an}k>1 − αn + αn − αm + αm − {am}k>1) 6

v¯({an}k>1 − αn) +v ¯(αn − αm) +v ¯(αm − {am}k>1) < ε, so α is a fundamental sequence. Now, by construction, we have

v¯(αn − α) 6 v¯(αn − {an}k>1) +v ¯({an}k>1 − α) 6 1 + lim v(an − am). n m→∞ 1 ε Since for every ε > 0 there exists N such that N < 2 and for every n, m > ε N, v(an − am) < 2 , we obtain that for every n > N the inequalityv ¯(αn − α) < ε holds. Thus αn → α. This completes the proof of existence. (¯v) Now we prove the uniqueness. Let (F1, v1) and (F2, v2) be two replenishments of (F, v). We want to construct a topological isomorphism σ. Define σ in F as the identity map. By definition, F is dense in both F1 and F2, so every element of F1 is a limit of a fundamental sequence {an}n>1, where each an lies in F . We extend σ to the map σ : F1 → F2 by assuming that for α ∈ F1 such that α = limn→∞ an that σ(α) = limn→∞ σ(an). Clearly this definition is correct, since {σ(an)}n>1 is a fundamental sequence in F2 and since for two fundamental sequences {an}n>1, {bn}n>1 with limn→∞ an = α = limn→∞ bn the sequence {cn}n>1, where c2n−1 = an, c2n = bn, is fundamental and limn→∞ cn = α. It is technical to prove that σ is a bijection and that σ preserves the operation, and we leave the detailed proof to the reader. Exercise 4.13. Complete the technical details of the proof of Theorem 4.12.

§ 4.2. Construction and properties of p-adic ﬁelds

Clearly, if valuations v1 and v2 of a field F are equivalent, then the replenishments of (F, v1) and (F, v2) are topologically isomorphic. Recall that we denote the class of equivalent valuations vp,ρ of Q by vp. A replenishment of (Q, vp) is denoted by Qp and is called a field of p-adic numbers or a p-adic field. Theorem 4.12 implies that Qp exists and is unique up to § 4.2. Construction and properties of p-adic fields 80 isomorphism. In this section we construct Qp explicitly and derive some its properties. 1. Ring of p-adic integers and its properties. Throughout we fix a prime p. Consider n n o Zp = {an} | an ∈ , an ≡ an−1 (mod p ) for n 1 . n>0 Z >

Deﬁne addition and multiplication on Zp as the addition and the multiplication of sequences, i.e. {an}n>0 + {bn}n>0 = {an + bn}n>0 and {an}n>0 ·{bn}n>0 = {an · bn}n>0. Clearly Zp is a ring under these operations. Consider n n+1 o Ip = {an} ∈ Zp | an ≡ 0 (mod p ) . n>0

It is also evident that Ip is an ideal of Zp. Deﬁne Zp := Zp/Ip, a ring of p-adic integers. The embedding Z → Zp is deﬁned by m 7→ {an}n>0, where a0 = a1 = ... = m. Clearly this embedding preserves the addition and multiplication. Moreover, the unit of Z coincides with the unit of Zp. Now we choose a canonical representation for every element from Zp. Namely, for every sequence {an}n>0 ∈ Zp there exists a unique sequence {xn}n>0 ∈ Zp n+1 n+1 such that for every n we have 0 6 xn < p and an ≡ xn (mod p ), in particular {an}n>0 + Ip = {xn}n>0 + Ip. The sequence {xn}n>0 with n+1 0 6 xn < p is a canonical representative of the coset {an}n>0 + Ip.

Theorem 4.14. Let Zp be the ring of p-adic integers. Then the following hold. (1) An element {xn}n 0 ∈ Zp is invertible in Zp if and only if x0 6= 0. > ∗ (2) For every 0 6= α ∈ Zp there exist the unique n ∈ N and ξ ∈ Zp such that α = pn · ξ. (3) Zp is an integral domain. Proof. Recall that if m ∈ Z, then we identify m with its image (m, m, . . .) ∈ Zp. In particular, 1 = (1, 1,...). (1) Necessity. If x0 = 0, then for every n ∈ Z we have x0 · n = 0, ∗ therefore {xn}n>0 6∈ Zp. Sufficiency. Assume that x0 6= 0. Since 0 < x0 < p, it follows that gcd(x0, p) = 1. Now, by definition, x1 ≡ x0 (mod p), therefore gcd(x1, p) = 1. Repeating the argument, by induction, we obtain that gcd(xn, p) = 1 n+1 for every n. Therefore, for every n we have gcd(xn, p ) = 1. Hence, for n+1 n+1 every n > 0 there exists 0 < yn < p such that xn · yn ≡ 1 (mod p ). n Since xn ≡ xn−1 (mod p ) and both xn−1 · yn−1 and xn · yn are equivalent § 4.2. Construction and properties of p-adic fields 81

n n 1 modulo p , it follows that yn ≡ yn−1 (mod p ), i.e. {yn}n>0 ∈ Zp. Since n+1 xn · yn ≡ 1 (mod p ) it follows that there exists zn such that xn · yn = n+1 n+1 1 + zn · p . Therefore {xn}n>0 ·{yn}n>0 = {1}n>0 + {zn · p }n>0 and the second summand in the right hand side lies in Ip.

(2) If α = {xn}n>0 ∈ Zp \{0}, then there exists a minimal n such that 0 xn 6= 0. If n = 0 then α = p ξ and ξ = α. Suppose n > 0, then, by n deﬁnition, xn ≡ xn−1 (mod p ) and in view of our choice xn−1 = 0, so xn is n n+1 xn divisible by p and xn 6≡ 0 (mod p ), in particular, pn 6≡ 0 (mod p) and n n+1 n gcd(xn/p , p) = 1. Now xn+1 ≡ xn (mod p ), so xn+1/p 6≡ 0 (mod p). Arguing in the same way by induction for every k > n we obtain that n k+1 n xk/p 6≡ 0 (mod p). Moreover, since 0 6 xk < p , we have 0 < xk/p < pk−n+1. Consider x x x x x ξ = n , n ,..., n , n+1 , n+2 ,... (ﬁrst n + 1 terms are equal). pn pn pn pn pn

In view of item (1) of the theorem, ξ is invertible in Zp and, by construction, α = pn · ξ. Now we show the uniqueness of n and ξ. If α = pn · ξ = pm · ζ, then the ﬁrst nonzero term of pn · ξ has number n, while the ﬁrst nonzero term of m p · ζ has number m, so n = m. Let ξ = {xn}n>0 and ζ = {yn}n>0. Since n n p · ξ = p · ζ, we obtain that xm = ym for m > n. Moreover n n xn−1 = (xn mod p ) = (yn mod p ) = yn−1.

Arguing in the same way we obtain that for all i < n the equality xi = yi i i holds (indeed, if xi = yi, then xi−1 = (xi mod p ) = (yi mod p ) = yi−1). Thus ξ = ζ. (3) Assume that α, β ∈ Zp \{0} and α·β = 0. In view of item (2) of the theorem, α = pn ·ξ, β = pm ·ζ, hence α·β = pn+m(ξ·ζ) = 0. Multiplying the left hand side of the identity by ξ−1 ·ζ−1 we obtain pn ·pm = 0. But (n+m)- n+m n+m n+m+1 th term of p equals p 6≡ 0 (mod p ), a contradiction.

Corollary 4.15. An integer a ∈ Z is invertible in Zp if and only if gcd(a, p) = 1. 2. The field of p-adic rationals is the replenishment of rationals in p-adic metric. Corollary 4.15 implies that rational numbers of the form a b with gcd(b, p) = 1 are naturally embedded into Zp. Since Zp is an integral domain, it possesses a field of fractions, denote it by Qp. We introduce a valuation vp on Qp and show that Qp is a replenishment of Q under a p-adic valuation, i.e. we show that Qp is a field of p-adic numbers. § 4.2. Construction and properties of p-adic fields 82

n In view of Theorem 4.14 all elements of Zp have form p · ξ, so we need to add the inverses for powers of p, i.e. n ∗ Qp = {p · ξ | n ∈ Z, ξ ∈ Zp} ∪ {0}. (4.4) n ∗ Given α = p · ξ ∈ Qp set νp(α) = n, and let νp(0) = ∞. For brevity νp(α) we use ν instead of νp below. Choose 0 < ρ < 1 and set vp(α) = ρ . Every a ∈ can be uniquely written as pn a1 , where p does not divide b Q b1 ∗ a1 −1 a1 · b1. Then both a1 and b1 belong to , and thus = a1b = ξ is an Zp b1 1 a n invertible p-adic integer. Therefore, b corresponds to p ξ ∈ Qp and this correspondence gives the canonical embedding of Q into Qp.

Theorem 4.16. In the above notations for every α, β ∈ Qp the following hold. (1) ν(α · β) = ν(α) · ν(β). (2) ν(α+β) > min(ν(α), ν(β)), moreover for ν(α) 6= ν(β) the equality holds. (3) vp is a valuation of Qp and the restriction of vp on Q coincides with the valuation vp,ρ of Q. Proof. (1), (2). If α = pn · ξ, β = pm · ζ, then α · β = pn+m · (ξ · ζ), therefore ν(α · β) = ν(α) + ν(β). Without loss of generality assume that n > m. Then we have α + β = pn · ξ + pm · ζ = pm(pn−m · ξ + ζ), n−m and p · ξ + ζ lies in Zp, so ν(α + β) > m = min(ν(α), ν(β)). If n > m, then the ﬁrst term of pn−m · ξ + ζ equals the ﬁrst term of ζ, so is not equal n−m to 0, therefore p · ξ + ζ is invertible in Zp and ν(α + β) = m. (3) First we check that vp is a valuation. We have min(ν(α),ν(β)) vp(α + β) 6 ρ = max(vp(α), vp(β)) 6 vp(α) + vp(β). ν(α)+ν(β) vp(α · β) = ρ = vp(α) · vp(β). vp(α) > 0 and vp(α) = 0 if and only if α = 0. So vp is a valuation. The restriction of vp on Q coincides with vp,ρ and we leave the details of the proof for the reader.

Exercise 4.17. Prove that the restriction of vp on Q coincides with vp,ρ.

Theorem 4.18. The valuation ﬁeld (Qp, vp) is a replenishment of (Q, vp,ρ). § 4.2. Construction and properties of p-adic ﬁelds 83

Proof. We need to prove that Qp is complete and that Q is dense in

Qp. Let {αn}n>1 be a fundamental sequence in Qp and every αn has the ∗ mn form p ·ξn, where mn ∈ Z and ξn ∈ Zp ∪{0}. Since the sequence {αn}n>1 is fundamental, it follows that vp(αn − αm) → 0. Now vp(αn − αm) = n,m→∞ ν(αn−αm) ρ . If the sequence {mn}n> is not stabilizing, then for every n there exists k > n such that mn 6= mk. In view of Theorem 4.16(2) we obtain

ρν(αn−αm) = ρmin(mn,mk) → 0. n,k→∞

Therefore mn → ∞, i.e. αn → 0. Assume that the sequence {mn}n>1 n→∞ (vp) is stabilizing, i.e. there exist N ∈ N,M ∈ Z such that for every n > N we have mn = M. Then for every n, m > N we have

M M vp(αn − αm) = vp(p (ξn − ξm)) = vp(p ) · vp(ξn − ξm), i.e. the sequence {ξn}n>1 is fundamental. Hence for every m ∈ N there m exists Nm ∈ N such that for all n1, n2 > Nm we have vp(ξn1 − ξn2 ) < ρ . m+1 (n) So ξn1 − ξn2 = p · ξ, where ξ ∈ Zp. Let ξn = {xk }k>0, consider (Nm) ξ = {xk}k 0, where xm := xm . We claim that ξ ∈ p and that ξn → ξ. > Z n→∞ m Show that xm−1 ≡ xm (mod p ) (and so ξ ∈ Zp). Indeed, by deﬁnition, (Nm−1) m−1 xm−1 = xm−1 , i.e. for each k > Nm−1 we have v(ξNm−1 − ξk) < ρ , m (Nm−1) therefore ξNm−1 − ξk = p · ζ for some ζ ∈ Zp. In particular, xm−1 − (k) m xm−1 ≡ 0 (mod p ). Choose k = Nm, then, since ξNm ∈ Zp, we have (Nm) (Nm) m (Nm−1) (Nm) m xm−1 ≡ xm (mod p ), whence xm−1 ≡ xm (mod p ). Now if k > m M Nm, then vp(ξk − ξ) 6 ρ , so ξn → ξ and αn → p · ξ. (vp) (vp) n We remain to show that Q is dense in Qp. Let α = p · ξ, where n ∈ Z ∗ n and ξ = (x0, x1, x2,...) ∈ Zp. Set αm = p · xm ∈ Q. Then

n n+m vp(αm − α) = ρ · vp(0,..., 0, xm − xm+1, xm − xm+2,...) ρ → 0, 6 m→∞ and the theorem follows.

Exercise 4.19. Prove that for every α ∈ Zp there exists a sequence (m) (m) (m) m+1 (m) {x }m>0, x ∈ Z, such that x − α ∈ p Zp (in particular, x → (vp) α as m → ∞). § 4.2. Construction and properties of p-adic ﬁelds 84

3. Applications. The construction of p-adic numbers shows that they are closely related to residuals modulo powers of p. This connection becomes clear due to the following theorem.

Theorem 4.20. Assume that F (x1, . . . , xn) ∈ Z[x1, . . . , xn] is a polynomial with integer coeﬃcients. The congruence m F (x1, . . . , xn) ≡ 0 (mod p ) (4.5) has a solution in Z for every m > 0 if and only if

F (x1, . . . , xn) = 0 (4.6) has a solution in Zp.

Proof. Assume that the equation (4.6) has a solution α1, . . . , αn ∈ m Zp. Denote by Im the ideal p · Zp of Zp. Then for every m there exist (m) (m) (m) x1 , . . . , xn ∈ Z such that αi + Im = xi + Im (this statement follows from Exercise 4.19). Therefore (m) (m) F (x1 , . . . , xn ) + Im = F (α1, . . . , αn) + Im = 0. (m) (m) (m) (m) Now F (x1 , . . . , xn ) is an integer, so F (x1 , . . . , xn ) ∈ Im ∩ Z = m (m) (m) p · Z, i.e. x1 , . . . , xn is a solution for the congruence F (x1, . . . , xn) ≡ 0 (mod pm). Now assume that the congruences (4.5) have solution for each m > 0. (m) (m) Consider the sequence {(x1 , . . . , xn )}m>0 of solutions for congruences (mk) (mk) (4.5). Notice that we may choose a subsequence {(x1 , . . . , xn )}k>0 (mk) such that xi is converging to αi ∈ Zp for 1 6 i 6 n. Indeed, con- (m) (m) sider zero coordinates of {(x1 , . . . , xn )}m>0. We obtain the set of tuples of the form (a1, . . . , an), where 0 6 ai < p for each i = 1, . . . , n. Since (m) (m) {(x1 , . . . , xn )}m>0 is infinite, there exists an n-tuple (a1, . . . , an) such (m) that the zero coordinate of xi equals ai for infinitely many m and they (m0) (m0) (m) (m) form a subsequence {(x1 , . . . , xn )}m0>0 of {(x1 , . . . , xn )}m>0. We (m0) (m0) repeat the arguments for the first coordinates of {(x1 , . . . , xn )}m0>0 (m1) (m1) and derive an infinite subsequence {(x1 , . . . , xn )}m1>0. We repeat the arguments for every k > 0. Now we take any element from k-th subsequence (mk) (mk) (m) (m) and obtain the subsequence {(x1 , . . . , xn )}k>0 of {(x1 , . . . , xn )}m>0 satisfying the condition: for every i = 1, . . . , n the first k coordinates of (mt) (mk) xi are equal to the coordinates of xi for every t > k. Therefore (mt) (ms) t (mk) vp(x − x ) < ρ → 0, i.e. for every i the sequence {x }k 0 i i t,s→∞ i > § 4.3. Problems 85

(mk) (mk) is fundamental. So the sequences {x1 }k>0,..., {xn }k>0 have limits α1, . . . , αn respectively and, by construction, each αi lies in Zp. Again by construction F (α1, . . . , αn) lies in every ideal Im, so F (α1, . . . , αn) = 0. § 4.3. Problems 1. Prove that a valuation defines on a field the structure of Hausdorff space. Find a topological field that is not homeomorphic to a valuation field, i.e. there exist topological fields that do not possess a valuation inducing the same topology. 2. Prove that (Zp, vp) is a compact topological space. 2 2 3. How many distinct solutions in Z5 has the equation x + y = 0? Bibliography

[1] Apostol T. V. Introduction to Analytic Number Theory. Springer, New York, 1976. [2] Bateman P. D., Diamond H. G. Analytic Number Theory – An Introductionary Course. World Scientiﬁc Publ., Singapore, 2002. [3] Bicadze A. V. Foundations of the Theory of Analytic Functions of a Complex Vari- able [Russian]. M.: Nauka, 1969. [4] Borevich Z. I., Shafarevich I. R. Number Theory [Russian]. M.: Nauka, 1985. [5] Buhshtab A. A. Number Theory [Russian]. M.: Prosveschenie, 1966. [6] Chandrasekharan K. Introduction to Analytic Number Theory. Springer, Berlin- Heidelberg, 2012. [7] Galochkin A. I., Nesterenko Yu. F., Shidlovskii A. B. Introduction to Number Theory [Russian]. M.: MSU, 1984. [8] Gelfond A. O. Transcendental and Algebraic Numbers [Russian]. M.: Gostehizdat, 1952. [9] Ingam A. V. The Distribution of Prime Numbers [Russian translation]. M.: Li- brokom, 2009. [10] Karatsuba A. L. Foundations of Analytic Number Theory [Russian]. M.: URSS, 1983. [11] Kostrikin A. I. Introduction to Algebra (Part 3) [Russian]. M.: Fizmatlit, 2004. [12] Vinogradov I. M. Elements of Number Theory [Russian]. M.: Nauka, 1972.

86 Glossary

A, 9 Qp, 81 an → a, 72 (v) R, 4 C, 4 ∼, 30 f ◦ g, 37 vα, 71 deg, 4 vp,ρ, 71 g | f, 4 Z, 4 e(n), 36 ζ(z), 35 Zp, 80 gcd, 4 hα(x), 7

I(n), 36 ι, 4

Λ(n), 34 li, 31 L(z, χ), 60 L(z, Gm), 63

µ(n), 36

N, 4 |g|, 56

P, 4 π(x), 30 ψ(x), 31 ψ˜(x), 31

Q, 4

87 Index

abelian group, 56 field of fractions,5 algebraic integer,6 Hermite identity, 20 number,6 homomorphism arithmetic function, 36 of groups, 58 asymptotically equivalent functions, 30 ideal,5 character modulo m, 60 identity function, 36 character of a finite abelian group, 58 isomorphism of groups, 57 Chebyshev function, 31 leading coefficient,4 integral, 31 Lindemann theorem, 27 complete field, 76 Liouville theorem, 18 conjugate numbers,7 L-series of character, 60 convolution product, 37 cyclic group, 56 Möbiusfunction, 36 Mangoldt function, 34 degree maximal ideal,6 of a Diophantine approximation, 14 minimal polynomial,7 of a polynomial,4 monic polynomial,4 of an algebraic number,7 multiplicative function, 36 Diophantine approximation, 14 Dirichlet approximation theorem, 15 p-adic valuation, 71 Dirichlet series, 35 prime-counting function, 30 Dirichlet theorem, 68 principal character, 58 division algorithm,4 principal ideal,5 domain,5 Euler function, 57 Euler identity, 38 Riemann hypothesis, 55 factor ring,6 zeta-function, 35 field of p-adic numbers, 81 ring of p-adic integers, 80 field of p-adic numbers (p-adic field), 79 symmetrized tuple, 25

88 Index 89 transcendental number,6 triangle inequality, 71 trivial valuation, 71 valuation, 71 valuation ﬁeld, 71