ELEMENTS OF ANALYTIC NUMBER THEORY - nsc.ruvdovin/lectures/numth_eng.pdf · 2018-11-28 ·...

ELEMENTS OF ANALYTIC NUMBERTHEORY

P. S. Kolesnikov, E. P. Vdovin

Lecture course

Novosibirsk, Russia2013

Contents

Chapter 1. Algebraic and transcendental numbers 4§ 1.1. Field of algebraic numbers. Ring of algebraic integers 4

1. Preliminary information 42. Minimal polynomial 63. Algebraic complex numbers 94. Algebraic integers 11

§ 1.2. Diophantine approximations of algebraic numbers 131. Diophantine approximation of degree ν 132. Dirichlet approximation theorem 153. Liouville theorem on Diophantine approximation of algebraic

numbers 18§ 1.3. Transcendentality of e and π 20

1. Hermite identity 202. Transcendentality of e 233. Symmetrized n-tuples 254. 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 312. Equivalence of the asymptotic behavior of Chebyshev functions

and of the prime-counting function 323. Von Mangoldt function 34

§ 2.2. Riemann function: Elementary properties 351. Riemann function in Re z > 1 352. Distribution of the Dirichlet series of a multiplicative function 363. Convolution product and the Mobius inversion formula 374. Euler identity 385. Logarithmic derivative of the Riemann function 39

2

Contents 3

6. Expression of the integral Chebyshev function via the Riemannfunction 40

§ 2.3. Riemann function: Analytic properties 431. Analytic extension of the Riemann function 432. Zeros of the Riemann function 473. Estimates of the logarithmic derivative 484. 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 562. Characters 583. Characters modulo m 60

§ 3.2. Dirichlet series 601. Convergence of L-series 602. Landau Theorem 663. Proof of the Dirichlet Theorem 68

Chapter 4. p-adic numbers 71§ 4.1. Valuation fields 71

1. Basic properties 712. Valuations over rationals 743. The replenishment of a valuation field 76

§ 4.2. Construction and properties of p-adic fields 791. Ring of p-adic integers and its properties 802. The field of p-adic rationals is the replenishment of rationals in

p-adic metric 813. 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 fromAbstract 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 variablex with coefficients in F . If f(x) = a0 + a1x+ · · ·+ anx

n ∈ F [x], ai ∈ F , ischosen so that an 6= 0, then n is called the degree of f(x), it is denoted bydeg f , while an ∈ F is called the leading coefficient of f(x), and if an = 1then 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 thewell-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 thering of integers Z and in the ring of polynomials F [x]. Indeed, these are par-ticular examples of Euclidean rings, and there are many common featuresand problems that can be solved in similar ways for integers and polynomi-als.

In particular, the greatest common divisor (gcd) d of two polynomialsf, 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 forevery 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 onthe 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] suchthat

f(x)p(x) + g(x)s(x) = d(x).

Exercise 1.1. Let f1, . . . , fn ∈ F [x] be a finite family of polynomialsover a field F . Prove that there exists a unique monic greatest commondivisor of f1, . . . , fn.

Suppose R is a commutative ring with an identity (e.g., R = Z orR = F [x] as above). A subset I ⊆ R is called an ideal of R if a ± b ∈ Ifor every a, b ∈ I, and ax ∈ I for every a ∈ I, x ∈ R. For example, the setof all even integers is an ideal of Z; the set f(x) ∈ F [x] | f(α) = 0 is anideal 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 everyset M ⊆ R there exists minimal ideal of R which contains M , it is denotedby (M). It is easy to note that

(M) =

∑i

xiai | xi ∈ R, ai ∈M.

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 simplya domain) if ab = 0 implies a = 0 or b = 0 for all a, b ∈ R. In particular, Zand F [x] are integral domains. An integral domain R such that every idealof R is principal is called a principal ideal domain.

Exercise 1.2. Prove that Z and F [x] (where F is a field) are principalideal 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. Inorder to construct it we start with the Cartesian product R× (R\0) of R(here each pair (a, b) corresponds to fraction a

b ). Now define an equivalencerelation (a1, b1) ∼ (a2, b2) ⇐⇒ a1b2 = a2b1. LetQ be the set of equivalenceclasses of R × (R \ 0) under this equivalence. Define the addition andmultiplication on representatives by

(a1, b1) + (a2, b2) = (a1b2 + a2b1, b1b2), (a1, b1) · (a2, b2) = (a1a2, b1b2).


We leave for the reader to prove that all operations defined are correct andthat Q is a field under these operations.

Let I be an ideal of R. Then R is split into a disjoint union of congruenceclasses a+I = a+x | x ∈ I, a ∈ R, and the set of all these classes (denotedby 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 Rsuch that I ⊂ J . For example, if R = Z then (n) is maximal if and onlyif n = ±p, where p is a prime natural number; if R = F [x] then (f) ismaximal 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 isa field. Indeed, if a+ I 6= 0 (i.e., a /∈ I) then J = xa+ b | x ∈ R, b ∈ I isan ideal of R such that I ⊂ J . Therefore, J = R, and thus all equations ofthe 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 thefield C of complex numbers.

2. Minimal polynomial. A complex number α ∈ C is algebraic ifthere exists a nonzero polynomial f(x) ∈ Q[x] such that f(α) = 0. Anon-algebraic complex number is said to be transcendental.

An algebraic number α is called an algebraic integer if there exists amonic 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 andonly if it is an integer.

Exercise 1.4. (1) Prove that√

2 and1

2+ ι

√3

2are algebraic in-

tegers.(2) Prove that if α ∈ C is an algebraic number then Reα and Imα

are algebraic numbers. Whether the same statements are true foran algebraic integer α?

(3) Show that the cardinality of the set of all algebraic numbers iscountable.

Since the cardinality of the entire set of complex numbers is uncountable(continuum), transcendental numbers do exist. However, it is not so easyto show an example of such a number accompanied with reasonable proof.


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 positivedegree from h(x) ∈ I(α, x) is a generator of the ideal I(α, x), it is called theminimal 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 monicpolynomial 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, wehave h(x) = hα(x).

For example, if α ∈ Q then hα = x−α. For α =√

2, hα = x2 − 2. The

number α =1

2+ ι

√3

2satisfies the equation α3 + 1 = 0, but its minimal

polynomial is hα(x) = x2 − x+ 1.

Exercise 1.6. Prove that a minimal polynomial does not have multipleroots.

If α is an algebraic number, and β ∈ C is a root of hα(x) then β issaid to be conjugate to α. Therefore, every algebraic number α of degreen has exactly n pairwise different conjugate complex numbers α1, . . . , αn,


including α itself. Moreover,

hα(x) =

n∏i=1

(x− αi) ∈ Q[x].

Given an algebraic number α, the minimal polynomial hα ∈ Q[x] isuniquely defined. If α is an algebraic integer then there also exists a monicpolynomial f ∈ Z[x] of minimal degree such that hα | f . In order to provethat these f and hα coincide, we need the following observation.

Let h(x) ∈ Q[x] be a monic polynomial with rational coefficients,

h(x) =p0

q0+p1

q1x+ · · ·+ pn−1

qn−1xn−1 + xn, gcd(pi, qi) = 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

qh(x) = a0 + a1x+ · · ·+ an−1xn−1 + anx

n ∈ 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 alli = 0, . . . , n − 1, so qi | b, and, finally, q | b = q/d, which is impossible ford > 1.

Polynomials with relatively prime integral coefficients are studied inAbstract Algebra, they are called primitive. The following statement iswell-known.

Exercise 1.7 (Hauss Lemma ). Prove that the product of primitivepolynomials is also a primitive polynomial.

Proposition 1.8. Let α be an algebraic integer. Then the minimalpolynomial hα(x) has integral coefficients.

Proof. Suppose f(x) ∈ Z[x] be a monic polynomial with integral co-efficients 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.Denote q = q(hα), q′ = q(g), where the function q(·) is given by (1.2).

According to the remark above, qhα(x) and q′g(x) are primitive polynomialsin Z[x]. However,

qq′f(x) = (qhα(x))(q′g(x)).


The Hauss Lemma implies qq′f(x) to be primitive, while f(x) has integralcoefficients itself. Therefore, qq′ = 1, i.e., all coefficients of hα and g areintegral.

Thus, there is no need to define separately integral minimal polynomialand degree for algebraic integers. Let us also note that for every algebraicnumber α there exists c ∈ N such that cα is an algebraic integer: It isenough to consider c = q(hα)deg hα .

3. Algebraic complex numbers. Let us denote the set of all alge-braic numbers by A. Recall that the Fundamental Theorem of Algebrastates C to be an algebraically closed field, i.e., every non-constant polyno-mial over C has a root in C. In this section, we will prove that A ⊂ C is theminimal algebraically closed subfield of C, i.e., A is the algebraic closureof 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 divi-sion algorithm, deg r < deg hα. Hence, f(α) = r(α), and the latter is alinear combination over Q of 1, α, . . . , αn−1, where n = deg hα. Therefore,dimQ[α] 6 n. Moreover, 1, α, . . . , αn−1 are linearly independent since n isthe minimal possible degree of a polynomial over Q annihilating α, so

dimQ[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

a0 · 1 + a1α+ · · ·+ anαn = 0,

end at least one of ai is nonzero. Hence, h(x) = a0 +a1x+ · · ·+anxn ∈ 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], thenf ∈ 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.


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 thecase when α 6= 0. Since α−1 ∈ Q[α], there exists f(x) ∈ Q[x] such thatα−1 = 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, sincedimQ[α1, . . . , αn−1] <∞ and dimQ[αn] <∞,

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

Exercise 1.11. Prove that Q[α1, . . . , αn] is a subfield of C providedthat 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 provethe following two statements:(1) If α and β are algebraic numbers then α± β and αβ are also algebraicnumbers;(2) If β 6= 0 is an algebraic number then 1/β is also an algebraic number.

(1) SinceQ[α± β],Q[α · β] ⊆ Q[α, β] ⊆ C,

we have dimQ[α ± β] < ∞, dimQ[α · β] < ∞ by Corollary 1.10. Thus byLemma 1.9 α± β, αβ ∈ A.

(2) By Lemma 1.9, β−1 ∈ Q[β], so Q[β−1] ⊆ Q[β] which is finite-dimensional over Q. Hence, dimQ[β−1] <∞ and thus β−1 ∈ A.

Theorem 1.13. The field A is algebraically closed.

Proof. Suppose α0, . . . , αn ∈ A, n > 1, αn 6= 0, ϕ(x) = α0 + α1x +· · ·+ αnx

n ∈ 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

thatQ[β] ⊆ Q[α0, . . . , αn−1, β].


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

βn = −α0 − α1β − · · · − αn−1βn−1,

we have

dimQ[α0, . . . , αn−1, β] 6 ndimQ[α0, . . . , αn−1] <∞.Hence, Q[β] is a finite-dimensional vector space over Q, and by Lemma 1.9β ∈ 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 polyno-mials in several variables over a ring K. For f ∈ K[x1, . . . , xn] denote bydeg f the maximal sum of the degrees of the variables that appear in aterm of f with a nonzero coefficient. Namely, f may be uniquely written asf = f0 + f1xn + · · ·+ fmx

mn , where fi ∈ K[x1, . . . , xn−1]. Assuming deg fi

are defined by induction, set deg f = maxi=0,...,m

(i+ deg fi).

Theorem 1.14. The set of all algebraic integers is a subring of thefield C.

Proof. Ii is enough to show that if α, β are algebraic integers thenα ± β and αβ are algebraic integers as well. We will prove a more generalfact: Every number of the form

γ =∑k,l

cklαkβl, ckl ∈ Z,

is an algebraic integer.Let

hα(x) = a0 + a1x+ · · ·+ an−1xn−1 + xn,

hβ(x) = b0 + b1x+ · · ·+ bm−1xm−1 + xm.

By Proposition 1.8, ai, bj ∈ Z. Lemma 1.5 implies that hα and hβ areirreducible polynomials over Q, thus they have no multiple roots.

Denote by α1, . . . , αn all complex roots of hα(x), and let β1, . . . , βmstand for all complex roots of hβ(x). To be more precise, set α1 = α,β1 = β. Consider the polynomial

p(x) =

n∏i=1

m∏j=1

x−∑k,l

cklαki β

lj

∈ C[x].


It is clear that p(γ) = 0 and the leading coefficient of p(x) is equal to theidentity. 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,

f(x) =

n∏i=1

m∏j=1

x−∑k,l

cklyki zlj

∈ Z[x, y1, . . . , yn, z1, . . . , zm].

Moreover, every permutation of the variables y1, . . . , yn or z1, . . . , zm doesnot change the polynomial f , i.e., it is symmetric with respect to yi andwith respect to zi. Hence,

f(x, y1, . . . , yn, z1, . . . , zm) =

nm∑a=0

ga(y1, . . . , yn, z1, . . . , zm)xa,

where every polynomial ga is symmetric with respect to yi and with respectto zi. On the other hand, for every a we have

ga(y1, . . . , yn, z1, . . . , zm) =∑

d1,...,dm>1

ga,d1,...,dm(y1, . . . , yn)zd11 . . . zdmm ,

where every polynomial ga,d1,...,dm(y1, . . . , yn) is symmetric (with respect toyi) and has integral coefficients.

Lemma 1.15. Let Ψ(y1, . . . , yn) ∈ Z[y1, . . . , yn] be a symmetric poly-nomial on y1, . . . , yn, deg Ψ = N , and let α1, . . . , αn ∈ C be the rootsof a polynomial h(x) = a0 + a1x + · · · + anx

n ∈ Z[x], an 6= 0. ThenaNn Ψ(α1, . . . , αn) ∈ Z.

Proof. By the Fundamental Theorem of Symmetric Polynomials, thereexists a polynomial G(t1, . . . , tn) ∈ Z[t1, . . . , tn] such that

Ψ(y1, . . . , yn) = G(σ1, . . . , σn), degG 6 N.

where σi(y1, . . . , yn) are the elementary symmetric polynomials on y1, . . . , yn.

The Viet formulae imply σk(α1, . . . , αn) = (−1)kan−kan

, k = 1, . . . , n. Since

degG 6 N , we obtain

aNn Ψ(α1, . . . , αn) = aNn G

(−an−1

an, . . . , (−1)n

a0

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) toobtain ga,d1,...,dm(α1, . . . , αn) ∈ Z (in this case, an = 1 since αi are algebraicintegers). 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 rationalnumbers Q is a dense subset of the set of real numbers R, i. e., for every

α ∈ R and for every ε > 0 there existsp

q∈ Q such that∣∣∣∣α− p

q

∣∣∣∣ < ε. (1.3)

Given a natural number N , the set QN = p/q | p ∈ Z, 0 < q 6 N, has

the following obvious property: |a − b| > 1

N2for all a, b ∈ QN . Hence, for

every α /∈ QN (in particular, for an irrational one), we have

mina∈QN

|a− α| > 0.

Therefore, in order to satisfy (1.3) for small ε, the numberp

qshould have a

an unboundedly large denominator q. This observation raises the followingnatural question: How to measure the accuracy of a rational approximationrelative to the growing denominator values?

1. Diophantine approximation of degree ν. To estimate the ac-curacy 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 thefollowing quantity:

εα,ν(q) = minp∈Z,p/q 6=α

qν∣∣∣∣α− p

q

∣∣∣∣ , ν > 0. (1.4)

It turns out that the study of the behavior of εα,ν(q) as q approaches infinityleads to a necessary condition for α to be algebraic.


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

limq→∞

εα,ν(q) <∞. (1.5)

Let us state a useful criterion that allows to determine whether a givennumber possesses a Diophantine approximation of a given degree.

Lemma 1.16. A real number α possesses a Diophantine approximationof degree ν > 0 if and only if there exists a constant c > 0 such that theinequality ∣∣∣∣α− p

q

∣∣∣∣ < c

qν(1.6)

holds for infinitely many rationalsp

q∈ Q.

Proof. Denote by M = M(α, c, ν) the set of all pairs (p, q) ∈ Z × Nsuch that (1.6) holds.

Suppose α possesses a Diophantine approximation of degree ν > 0. ByDefinition 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 suchq ∈ N there exists p ∈ Z satisfying the inequality (1.6).

Therefore, the set M is infinite, and there is no upper bound for the setq | (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 zerosince 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 infinitelymany rationals p/q. Then the set M defined in the first part of the proof isinfinite.

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 theinequality |a1 − a2| > 1/N2 holds. Hence, any infinite subset S of QN hasinfinite diameter, i.e., for any d > 0 one may find p1/q1 and p2/q2 in S suchthat |p1/q1 − p2/q2| > d.


In particular, the set S = p/q | (p, q) ∈ M ⊆ QN contains p1/q1 andp2/q2 such that |p1/q1 − p2/q2| > 2c. Since (1.6) holds for pi/qi, i = 1, 2,we have ∣∣∣∣α− p1

q1

∣∣∣∣ < c

qν1,

∣∣∣∣α− p2

q2

∣∣∣∣ < c

qν2which implies

2c <

∣∣∣∣p1

q1− p2

q2

∣∣∣∣ < c

(1

qν1+

1

qν2

)< 2c,

a contradiction.Therefore, the set of denominators q | (p, q) ∈ M for some p is in-

finite, so there exist infinitely many q such that εα,ν(q) < c. Hence, thesequence εα,ν(q)q∈N has a finite accumulation point, and (1.5) holds.

Exercise 1.17. Whether a rational number possesses a Diophantineapproximation of degree 1?

2. Dirichlet approximation theorem.

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

q

∣∣∣∣ < 1

qN, q 6 N.

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 numbersξ0, . . . , ξN are set into N intervals, there exists at least one interval whichcontains 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 ∣∣∣∣α− p

q

∣∣∣∣ =1

q|α(k2 − k1)− [k2α] + [k1α]| = 1

q|ξk2 − ξk1 | <

1

Nq.


Corollary 1.19. If α ∈ R \Q then α possesses a Diophantine approx-imation of degree ν = 2.

Proof. Theorem 1.18 implies that for every natural number N thereexist pN ∈ Z and qN ∈ N, qN 6 N , such that∣∣∣∣α− pN

qN

∣∣∣∣ < 1

NqN. (1.7)

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

minp/q∈QM

∣∣∣∣α− p

q

∣∣∣∣ < 1

NqN6

1

N→

N→∞0,

which is impossible.Therefore, (1.7) holds for infinitely many numbers qN . Since qN 6 N ,

we have

q2N

∣∣∣∣α− pNqN

∣∣∣∣ < 1

for infinitely many qN .

Proposition 1.20. Let α ∈ Q. Then for every function ϕ : N → R+

satisfying limq→∞

qϕ(q) = 0 there exist only a finite number of rationalsp

qsuch that ∣∣∣∣α− p

q

∣∣∣∣ < ϕ(q). (1.8)

Proof. Denote by S the set of all rationals p/q, gcd(p, q) = 1, satisfy-ing (1.8). Assume S is infinite. It is easy to see that the distance betweenany two numbers from S does not exceed 2 max

q>1ϕ(q) which is finite. Note

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.

Let α =a

b. Then for all rationals

p

qexcept α itself the following in-

equality holds: ∣∣∣∣ab − p

q

∣∣∣∣ =

∣∣∣∣aq − pbbq

∣∣∣∣ > 1

bq.

Since b is fixed, we have1

bq> ϕ(q),


which is impossible for sufficiently large q. The contradiction obtainedproves that S is finite.

Corollary 1.21. A rational number α does not have a Diophantineapproximation of degree ν > 1.

The result obtained is in some sence paradoxical: All irrational numberspossess Diophantine approximations of degree 2, but neither of rationals hasa Diophantine approximation of degree ν > 1. Therefore, the existence ofa Diophantine approximation of degree ν > 2 is a criterion of irrationality.

Let us apply the criterion above to the base of the natural logarithm

e = 1 +1

1!+

1

2!+ · · ·+ 1

n!+ . . . , (1.9)

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) andreduce to the common denominator:

1 +1

1!+

1

2!+ · · ·+ 1

n!=pnn!.

Then∣∣∣e− pnn!

∣∣∣ =1

(n+ 1)!

(1

n+ 2+

1

(n+ 2)(n+ 3)+ . . .

)<

1

(n+ 1)!

(1

2+

1

22+ . . .

)=

2

(n+ 1)!.

Define a function ϕ : N→ R as follows:

ϕ(1) = 1,

ϕ(q) =2

(n+ 1)!for (n− 1)! < q 6 n!, n > 2.

If q ∈ ((n− 1)!, n!] then

qϕ(q) 6 n!2

(n+ 1)!=

2

n+ 1,

so limq→∞

qϕ(q) = 0. However, ∣∣∣∣e− p

q

∣∣∣∣ < ϕ(q)


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 alge-braic numbers.

Theorem 1.23 (Liouville Theorem). Let α be an algebraic number ofdegree n > 2. Then α has no Diophantine approximation of degree ν > n.

Proof. First, let us find a constant M > 0 such that∣∣∣∣α− p

q

∣∣∣∣ > M

qn(1.10)

for all p ∈ Z and q ∈ N.Set h(x) to be a multiple of the minmal polynomial for α with inte-

ger coefficients, e.g., h(x) = q(hα)hα(x). Suppose α1, . . . , αn ∈ C are thecomplex roots of this polynomial, and assume α1 = α. Then

h(x) = an

n∏k=1

(x− αk) = an(x− α)

n∏k=2

(x− αk),

where an is the leading coefficient of h(x). Denote by M the followingquantity:

M =

(|an|

n∏k=2

(|α|+ |αk|+ 1)

)−1

,

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,∣∣∣∣pq

∣∣∣∣ < 1 + |α|

and thus

|h(p/q)| = |an|∣∣∣∣α− p

q

∣∣∣∣ n∏k=2

∣∣∣∣αk − p

q

∣∣∣∣6

∣∣∣∣α− p

q

∣∣∣∣ |an| n∏k=2

(|αk|+

∣∣∣∣pq∣∣∣∣) <

∣∣∣∣α− p

q

∣∣∣∣ 1

M.


Lemma 1.5 implies h(x) to be irreducible over Q, hence, it has no rationalroots. Therefore,

|h(p/q)| =∣∣∣∣a0 + a1

p

q+ · · ·+ an

pn

qn

∣∣∣∣ > 1

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 > 0such that ∣∣∣∣α− p

q

∣∣∣∣ < c

qν

holds for infinitely many p/q ∈ Q. Then, as it was shown in the proofof Lemma 1.16, the last inequality holds for infinitely many denominatorsq ∈ N. Then for sufficiently large q we have

c

qν<M

qn,

in contradiction to (1.10).

The Liouville Theorem is a powerful tool that allows constructing ex-plicit examples of transcendental numbers. Namely, we obtain the followingsufficient condition of transcendentality.

Corollary 1.24. Let α be a real number. If for everyN ∈ N it possessesa Diophantine approximation of degree ν > N then α is transcendental.

Example 1.2. The following number is transcendental:

α =

∞∑n=1

10−n!.

Proof. Given N ∈ N, consider

pNqN

=N∑n=1

10−n! =pN

10N !.

Note that ∣∣∣∣α− pNqN

∣∣∣∣ =

∞∑n=N+1

10−n! < 2 · 10−(N+1)! =2

qN+1N

.

§ 1.3. Transcendentality of e and π 20

Therefore, for every ν > 0 there exist infinitely many rationalspNqN

, N > ν,

satisfying ∣∣∣∣α− pNqN

∣∣∣∣ < 2

qN+1N

<2

qνN.

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 num-ber to be transcendental. However, this condition is not necessary. Thereexist different methods to prove transcendentality of a series of importantconstants, e.g., e and π. In this section, we are going to study one of thesemethods known as the Hermite Method.

Given an analytic function f(z) on the complex plane, denote by

x∫x0

f(z) dz

the Riemann integral of f(z) along the straight segment starting at x0 ∈ Cand ending at x ∈ C (the Cauchy Integral Theorem implies that this integraldoes 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∫0

f(t)e−αt dt = F (0)− F (x)e−αx, (1.11)

where

F (x) =f(x)

α+f ′(x)

α2+ · · ·+ f (n)(x)

αn+1.

Proof. According to the Fundamental Theorem of Calculus (the Newton—Leibniz Formula),

d

dx

x∫0

f(t)e−αt dt = f(x)e−αx.


On the other hand,

d

dxF (x)e−αx = (F ′(x)− αF (x))e−αx = −f(x)e−αx.

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

H(h(x)) =1

(p− 1)!a(n−1)pn xp−1h(x)p, (1.12)

h(x) ∈ Z[x], n = deg h, p ∈ N. Let us establish the properties of H(h(x)).

Lemma 1.26. Let h(x) = a0 + a1x + · · · + anxn ∈ Z[x], a0, an 6= 0,

n > 1, and let β1, . . . , βn ∈ C be the entire collection of roots of h(x) inwhich 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 followingproperties:

(1) f (j)(0) = 0, 0 6 j 6 p− 2;(2) f (j)(βi) = 0, 0 6 j 6 p− 1, i = 1, . . . , n;

(3) f (p−1)(0) = a(n−1)pn ap0;

(4) f (j)(0) ∈ pZ, j > p;

(5)n∑i=1

f (j)(βi) ∈ pZ, j > p.

Proof. It is easy to see from the construction of f(x) that 0 is its rootof 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 ofa polynomial f(x) is also a root of f ′(x), f ′′(x), . . . , f (k−1)(x). This implies(1) and (2).

To prove (3), let us distribute all brackets in the definition of f(x) andfind the term of lowest degree in x, namely, the term is

1

(p− 1)!a(n−1)pn ap0x

p−1.

Its (p − 1)th derivative is equal to a(n−1)pn ap0. 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


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

(xm)(j) = j!

(m

j

)xm−j ∈ Z[x],

and the jth derivatives of xm, m < j, turn into zero.To prove (4), consider

Φ(x) =(p− 1)!

a(n−1)pn

f(x) = xp−1h(x)p ∈ Z[x].

According to the remark stated above, Φ(j)(x) ∈ j!Z[x] and thus

f(0) ∈ j!

(p− 1)!Z ⊆ jZ

for j > p,Finally, note that

h(x) = an

n∏i=1

(x− βi).

Hence,

f(x) =anpn

(p− 1)!

n∏i=1

(x− βi)pxp−1 =anpn

(p− 1)!Θ(x, β1, . . . , βn),

where

Θ = xp−1n∏i=1

(x− yi)p ∈ Z[x, y1, . . . , yn].

The polynomial

Ψ(y1, . . . , yn) =

n∑i=1

∂jΘ

∂xj(yi, y1, . . . , yn) ∈ Z[y1, . . . , yn]

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

applied to1

j!Ψ(y1, . . . , yn), we obtain

anpnj!

Ψ(β1, . . . , βn) ∈ Z.

Since j > p,anpn

(p− 1)!Ψ(β1, . . . , βn) ∈ pZ,


and it remains to note

n∑i=1

f (j)(βi) =anpn

(p− 1)!Ψ(β1, . . . , βn),

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 inthe obvious way as

f j(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 algebraicnumbers. 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

bnhe(x) = b0 + b1x+ · · ·+ bnxn ∈ 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

f(x) = H(h(x)) =1

(p− 1)!xp−1h(x)p

as in (1.12), where βi = i, i = 1, . . . , n. For every k = 0, 1, . . . , n write theHermite identity from Lemma 1.25:

k∫0

f(t)e−t dt = F (0)− F (k)e−k.


Multiply each of these equations by bkek and add the results:

n∑k=0

bk

k∫0

f(t)ek−t dt =

n∑k=0

bkek(F (0)− F (k)e−k)

= F (0)

n∑k=0

bkek −

n∑k=0

bkF (k) = F (0)bnhe(e)−n∑k=0

bkF (k).

Therefore,

n∑k=0

k∫0

bkf(t)ek−t dt = −n∑k=0

bkF (k). (1.13)

Recall that

F (x) =∑j>0

f (j)(x).

Consider the right-hand side of (1.13). Lemma 1.12 and Remark 1.3 imply

n∑k=0

bkF (k) = b0f(p−1)(0) + b0

∑j>p

f (j)(0) +

n∑k=1

bk∑j>p

f (j)(k).

In the last expression, b0f(p−1)(0) = (−1)npb0(n!)p 6≡ 0 (mod p) by the

choice of p, all other summands are integer multiples of p. Hence, for everysufficiently large prime p the right-hand side of (1.13) is a nonzero integernumber.

Now, let us estimate the absolute value of the left-hand side of (1.13):∣∣∣∣∣∣n∑k=0

bk

k∫0

f(t)ek−t dt

∣∣∣∣∣∣ 6n∑k=0

|bk|k maxt∈[0,k]

|f(t)ek−t|

6 (n+1) maxk=0,...,n

|bk|n maxt∈[0,n]

|f(t)|en 6 C 1

(p− 1)!np−1(n−1)np 6 C

Cp1(p− 1)!

,

where C and C1 do not depend on the choice of p. Since limp→∞

Cp1(p− 1)!

= 0,

∣∣∣∣∣∣n∑k=0

bk

k∫0

f(t)ek−t dt

∣∣∣∣∣∣ < 1


when p is sufficiently large, but the right-hand side of (1.13) is a nonzerointeger and thus its absolute value is greater or equal to 1. The contradictionobtained proves the theorem.

3. Symmetrized n-tuples.

Definition 1.4. An N -tuple (β1, . . . , βN ) ∈ CN is called symmetrizedif

N∏j=1

(x− βj) ∈ Q[x].

It is clear that a symmetrized tuple remains symmetrized after everypermutation 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 symmetrizedtuple.

To prove the transcendence of π we need the following properties ofsymmetrized tuples.

Lemma 1.28. Let (α1, . . . , αn) ∈ Cn be a symmetrized tuple, andlet σ = σk ∈ Z[x1, . . . , xn], k ∈ 1, . . . , n, be an elementary symmetricpolynomial in x1, . . . , xn. Then

σ(eα1 , . . . , eαn

)=

N∑j=1

eβj , N =

(n

k

),

where (β1, . . . , βN ) is a symmetrized tuple.

Proof. Recall that

σ = σk =∑

16i1<···<ik6n

xi1 . . . xik

contains N =(nk

)summands. For every collection of indexes 1 6 i1 < · · · <

ik 6 n, consider the following polynomial:

ηi1,...,ik(t1, . . . , tn) = ti1 + · · ·+ tik ∈ Z[t1, . . . , tn].

Thenσ(eα1 , . . . , eαn

)=

∑16i1<···<ik6n

eηi1,...,ik (α1,...,αn).

It remains to show that complex numbers

ηi1,...,ik(α1, . . . , αn), 1 6 i1 < · · · < ik 6 n,

form a symmetrized tuple.


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 forpolynomials ηi1,...,ik : If (i1, . . . , ik) has number j then ηj = ηi1,...,ik .

If Φ(y1, . . . , yN ) is a symmetric polynomial in y1, . . . , yN with integercoefficients 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 twovariables tl and ti. The set of all polynomials η1, . . . , ηN may be divided intothree groups: The first group contains all those polynomials that either donot depend in both ti and tl, or containt both these variables; The secondgroup consists of all ηjs that depend in ti, but do not depend in tl; The thirdgroup includes all remaining polynomials ηj . When exchanging ti and tl,the polynomials of the first group do not change, the polynomials of thesecond group turn into polynomials of the third group, and, conversely, allpolynomials of the third group move into the second group. Finally, weobtain a permutation of η1, . . . , ηN . Since Φ is symmetric, Ψ is invariantunder the transposition of tl and ti. It is well-known that every permutationcan be obtained by a series of transpositions (all transpositions generate thesymmetric 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 ra-tional 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 nonzerocomplex numbers. If, in addition,

A :=

N∑k=1

eβk 6= 0,

then A /∈ Q.

Proof. Assume A ∈ Q \ 0. Without loss of generality, we can sup-pose A ∈ Z: If A = a/b then one may just repeat the symmetrized N -tupleb times to get a new symmetrized Nb-tuple with integer sum of exponents.


Consider a polynomial

h(x) = bN

N∏k=1

(x− βk) = b0 + b1x+ · · ·+ bNxN ∈ Z[x].

Here b0 = (−1)NbN · β1 · . . . · βN 6= 0.Choose a prime number p ∈ Z such that p > |bN |, p > |b0|, p > |A|,

and construct a polynomial

f(x) =1

(p− 1)!b(N−1)pN xp−1h(x)p

as in Lemma 1.26. The Hermite identity for f(x) has the following form:

βk∫0

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

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

N∑k=1

βk∫0

f(t)eβk−t dt = AF (0)−N∑k=1

F (βk). (1.14)

Consider the right-hand side of equation (1.14):

AF (0)−N∑k=1

F (βk) = Af (p−1)(0) +A∑j>p

f (j)(0)−N∑k=1

∑j>0

f (j)(βk)

.

By Lemma 1.26, the first summand Af (p−1)(0) = Ab(N−1)pN bp0 is an integer

number that cannot be divided by p, but all other summands are integersdivisible 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

from above as CCp1

(p− 1)!in the very same way as it was done in the proof

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 algebraicnumber then eα may not be a negative rational number.


Proof. Let α be a root of a polynomial h(x) = a0 +a1x+ · · ·+anxn ∈

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 α1 = α.

Recall that eα 6= 0 for all α ∈ C. Assume eα = −ab

, a, b ∈ N, where

gcd(a, b) = 1. Consider

0 =

n∏k=1

(eαk +

a

b

).

Distribute the brackets and and multiply by bn to obtain

0 = an +

n−1∑k=1

akbn−kσn−k(eα1 , . . . , eαn). (1.15)

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

σn−k(eα1 , . . . , eαn) =

Nk∑j=1

eβkj .

It may happen that some of βkj are zero. Denote by mk > 0 the numberof zeros among βkj , j = 1, . . . , Nk. Let (γ1, . . . , γN ) stand for the collectionof all nonzero βkj in which every βkj appears akbn−k times. This is asymmetrized N -tuple. Now, we may rewrite (1.15) as

an +

n−1∑k=1

akbn−kmk = −N∑j=1

cjeγj , cj ∈ Z.

The left-hand side is an integer A which is not divisible by b, hence, A 6= 0.Thus,

−N∑j=1

cjeγj = A 6= 0,

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 imaginaryidentity, is also algebraic as a product of two algebraic numbers. The Euleridentity eιπ = −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 com-pass. 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 andcompass must have algebraic length. Hence, a segment of length π1/2 (theside of a desired square) cannot be constructed. A finer statement, whichis a good exercise for a reader, says that if a segment of length a may beconstructed with a straightedge and compass then a is an algebraic numberof degree 2k (a Pythagorean number). In this way, one may prove that anarbitrary angle in the plane may not be divided into three equal parts, anda 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 integercoefficients. Show that α is an algebraic integer.

(2) Find the degree of the algebraic number e2πι/n, where ι is theimaginary identity and n is a natural number.

(3) Show that if α is a nonzero algebraic number then eα is transcen-dental.

(4) Prove the Weierstrass—Lindemann Theorem: If α1, . . . , αn arepairwise different algebraic numbers then eα1 , . . . , eαn are linearlyindependent over the field of algebraic numbers.

CHAPTER 2

Asymptotic law of distribution of primenumbers

Prime integers are “atoms” of the numeric Universe and thus they havebeen attracting attention of researchers for more than two thousand years.Ancient Greek mathematicians obtained a lot of nice results, e.g., Euclid (IIIBC) proved that there exist infinitely many primes, Eratosthenes (II BC)invented an algorithm to find all primes smaller than a given natural N (theSieve of Eratosthenes). These results are now considered as “elementary”.

The aim of this section is to present a precise answer to the follow-ing question: How often prime numbers occur in the series of all naturalnumbers?

This question was raised by Hauss and Legendre at the end of XVIII,and the Hauss conjecture was proved by Jacques Hadamard and CharlesJean de la Vallee-Poussin in 1896. The result is known as the Prime Num-ber Theorem. To explain the statement, recall the notion of asymptoticequivalence.

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. Thesefunctions are said to be asymptotically equivalent if

limx→∞

f(x)

g(x)= 1.

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) isthe 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:

π(x) ∼ x

ln(x).

Exercise 2.1. Prove thatx

lnx∼ li(x), where

li(x) =

x∫2

dt

ln t.

§ 2.1. Chebyshev functions

1. Definition and estimates. Here we will establish important rela-tions between the function π(x) and the following functions defined on allpositive real numbers:

• ψ(x) =∑

(p,m)∈Qx

ln p, where Qx = (p,m) | p ∈ P,m ∈ N, pm 6 x.

The function ψ is called the Chebyshev function;

• ψ(x) =

x∫1

ψ(t)

tdt is known as the integral Chebyshev function.

Note that π(x) may be presented in a similar way as∑p∈P,p6x

1.

Note that (p,m) ∈ Qx if and only if p 6 x and ln(pm) = m ln p 6 lnx.Therefore, the sum ∑

(p,m)∈Qx

ln p = ψ(x)

contains each ln p as many times as the count of all m ∈ N such thatm ln p 6 lnx. If x > 1 (i.e., lnx > 0) there exist [lnx/ ln p] of such ms (here[·] stands for the integral part of a real number). Hence,

ψ(x) =∑

p∈P,p6x

[lnx

ln p

]ln p, x > 1. (2.1)

Proposition 2.2. The following statements hold:

(1) ψ(x) 6 π(x) lnx, ψ(x) 6 π(x) ln2 x for every x > 1;

(2) limx→∞

ψ(x)/xz = limx→∞

ψ(x)/xz = 0 for every z ∈ C such that

Re z > 1.


Proof. (1) If we omit [·] in (2.1) then the value of this sum may justincrease since [x] 6 x, i.e.,

ψ(x) =∑p6x

[lnx/ ln p] ln p 6∑p6x

ln p 6∑p6x

lnx = π(x) lnx

(hereinafter, when we use “p” for summation index, we assume p rangesover prime numbers, as in (2.1)).

For the integral Chebyshev function, note that

ψ(x) 6 ψ(x)

x∫1

dt

t= ψ(x) lnx.

The statement (2) immediately follows from (1): If z = 1+a+ιb, a > 0,then

|ψ(x)/xz| = ψ(x)/x1+a 6 π(x) lnx/x1+a 6 x lnx/x1+a = lnx/xa → 0

as x→∞. For ψ(x), the proof is completely similar.

Exercise 2.3. Prove that limx→∞

ψ(x)/xz = 0 for Re z > 1.

2. Equivalence of the asymptotic behavior of Chebyshev func-tions 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) lnx for x > 1.Hence,

ψ(x)

x6

π(x)

x/ lnx,

and the same inequality holds for upper and lower limits of these functionsas x→∞. Namely,

(A1)⇒ limx→∞

ψ(x)

x6 limx→∞

ψ(x)

x6 1,

(A2)⇒ limx→∞

π(x)

x/ lnx> limx→∞

π(x)

x/ lnx> 1.


On the other hand, choose a parameter 0 < a < 1 and consider

S(x, a) =∑

xa<p6x

ln p, x > 1.

Then ψ(x) > S(x, a), but

S(x, a) >∑

xa<p6x

ln(xa) = ln(xa)∑

xa<p6x

1 = a lnx(π(x)− π(xa)).

Therefore,ψ(x)

x> a

π(x) lnx

x− aπ(xa) lnx

x. (2.2)

Note that the second summand in the right-hand side of (2.2) approacheszero as x→∞ since π(xa) 6 xa.

Assume (A1) holds and B = limx→∞

ψ(x)/x. Then (2.2) implies that

B > a for every positive a < 1, i.e., B > 1. But we have already shownthat (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

ψ(x) =

x∫1

ψ(t)

tdt = x− 1 +

x∫1

r(t)

tdt = x+R(x)− 1,

where R′(x) = r(x)/x. It is enough to check that R = o(x), which is indeedthe 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

I+ε (x) =

x+εx∫x

ψ(t)

tdt = ψ(x+ εx)− ψ(x) = εx+ o(x).

Since ψ is an increasing function, the integral I+ε (x) may be estimated from

below as

I+ε (x) > ψ(x)

x+εx∫x

dt

t= ψ(x) ln(1 + ε).

Similarly, the integral

I−ε (x) =

x∫x−εx

ψ(t)

tdt = ψ(x)− ψ(x− εx) = εx+ o(x)


may be estimated as

I−ε (x) 6

x∫x−εx

dt

t= ψ(x) ln

1

1− ε.

Hence,

ψ(x)

x6

I+ε (x)

x ln(1 + ε)=

ε

ln(1 + ε)+ o(1),

ψ(x)

x>

I−ε (x)

−x ln(1− ε)=

ε

− ln(1− ε)+ o(1).

The upper and lower limits of ψ(x)/x as x→∞ also satisfy these inequal-ities for all ε ∈ (0, 1). It remains evaluate the limit as ε → +0 to obtainψ(x) ∼ x.

Exercise 2.5. Prove the implication (A2)⇒(A1) in Theorem 2.4.

3. Von Mangoldt function. Yet another form of the Chebyshevfunction comes directly from the definition:

ψ(x) =∑

n∈N,n6xΛ(n), (2.3)

where

Λ(n) =

ln p, n = pm, p ∈ P, m ∈ 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

ψ(x) =∑

n∈N,n6xΛ(n) ln(x/n).

Proof. Denote by δ+(y) the step function given by

δ+(y) =

0, y < 0,

1, y > 0.

§ 2.2. Riemann function: Elementary properties 35

Then

ψ(x) =

x∫1

ψ(t)

tdt =

x∫1

1

t

∑n6t

Λ(n) dt

=

x∫1

1

t

∑n6x

Λ(n)δ+(t− n) dt =∑n6x

Λ(n)

x∫1

δ+(t− n)

tdt

=∑n6x

Λ(n)

x∫n

dt

t=∑n6x

Λ(n) ln(x/n),

and the Proposition is proved.

§ 2.2. Riemann function: Elementary properties

1. Riemann function in Re z > 1. We have already seen that theasymptotic 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 func-tions are of the form

f(x) =∑n6x

fn,

where fn are some real coefficients, and the summation is made over naturalnumbers.

One of the most fruitful ideas that lie in the foundation of the ana-lytic number theory is to study a function f(x) as above via the followingcomplex-valued function called Dirichlet series:

ϕ(z) =

∞∑n=1

fnnz.

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

ζ(z) =

∞∑n=1

1

nz, Re z > 1. (2.4)

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 withrespect to z. Hence, (2.4) is uniformly converging on every compact subset


in Re z > 1. By the well-known Weierstrass Theorem, the limit of a sequenceof analytic functions which is uniformly converging on every compact subsetin a given domain is again an analytic function. Therefore, (2.4) defines ananalytic function in the semiplane Re z > 1. Later we will see how to extendζ analytically into the semiplane Re z > 0.

2. Distribution of the Dirichlet series of a multiplicative func-tion. Recall some notions from the elementary number theory. An arbi-trary map f : N → C is called an arithmetic function. An arithmeticfunction is called multiplicative if f(1) = 1 and f(nm) = f(n)f(m) pro-vided that n,m ∈ N are relatively prime. The Fundamental Theorem ofArithmetic implies any multiplicative function f to be uniquely determinedby 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 Mobius function 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∞∑n=1

f(n)n−z is absolutely converging. Then

∞∑n=1

f(n)n−z =∏p∈P

( ∞∑d=0

f(pd)p−dz

).

Proof. It is easy to see that for every p ∈ P the series∞∑d=0

f(pd)p−dz

contains a part of the initial series and thus converges absolutely. Enumerateprime numbers in the increasing order:

P = pn | n ∈ N, p1 = 2, p2 = 3, . . . ,

and consider the partial product

PN =

N∏n=1

( ∞∑d=0

f(pdn)p−dzn

).

Since a product of absolutely converging series is distributive, we may dis-tribute the brackets in the last expression to obtain

PN =

∞∑d1,...,dN=0

f(pd11 ) . . . f(pdNN )p−d1z1 . . . p−dNzN =∑n∈MN

f(n)n−z,


where MN ⊂ N consists of all natural numbers of the form pd11 . . . pdNN ,di > 0. The least natural number that is not in MN is equal to pN+1,hence, ∣∣∣∣∣

∞∑n=1

f(n)n−z − PN

∣∣∣∣∣ 6∞∑

n=pN+1

|f(n)n−z| → 0

as N →∞ (since pN+1 →∞).

3. Convolution product and the Mobius inversion formula.Given two functions f, g : N → C, their convolution product is an arith-metic function defined by

(f g)(n) =∑d|n

f(d)g(n/d), n ∈ N. (2.5)

where the summation index d ranges over the set of all divisors of n.

Exercise 2.8. Prove that the convolution product is associative andcommutative.

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 seriesand the convolution product.

Lemma 2.10. Let f and g be arithmetic functions such that theirDirichlet series

∞∑n=1

f(n)

nz,

∞∑n=1

g(n)

nz

are absolutely converging for some z ∈ C. Then( ∞∑n=1

f(n)

nz

)( ∞∑n=1

g(n)

nz

)=

∞∑n=1

(f g)(n)

nz. (2.6)

Proof. Since the product of absolutely converging series is distribu-tive, we may write( ∞∑

n=1

f(n)

nz

)( ∞∑n=1

g(n)

nz

)=

∞∑n,m=1

f(n)g(m)

nzmz.


Introduce new indexes d = n, N = nm and change the order of summation(it is possible due to absolute convergence of the product series):

∞∑n,m=1

f(n)g(m)

nzmz=

∞∑N=1

1

Nz

∑d|N

f(d)g(N/d) =

∞∑N=1

1

Nz(f g)(N).

Therefore, (2.6) is proved.

Moreover, it is known that the set of all multiplicative functions formsa group with respect to the convolution product, i.e., if f and g are multi-plicative functions then so is f g, and for every multiplicative function fthere exists a multiplicative function f−1 such that f f−1 = f−1 f = e.

Lemma 2.11 (Mobius inversion formula). If f : N→ C is an arithmeticfunction and f I = g then g µ = f .

Proof. Note that I µ = e. Indeed, if the canonical form of n isqm11 . . . qmrr , q1, . . . , qr ∈ P, r > 1, then

(I µ)(n) =∑d|n

µ(d) = µ(1)−∑j

µ(qj) +∑j1<j2

µ(qj1qj2 − . . .

= 1− r +

(r

2

)−(r

3

)+ · · ·+ (−1)r = (1− 1)r = 0

by the Newton binomial formula.Therefore, g µ = (f I) µ = f (I µ) = f e = f .

Exercise 2.12. The Euler function ϕ is defined by

ϕ(n) = |m ∈ N | m 6 n, gcd(m,n) = 1|, n ∈ N.Prove that ϕ I = E, where E(n) = n for all n ∈ N. Deduce the explicitformula for ϕ(n).

4. Euler identity.

Theorem 2.13. If Re z > 1 then

ζ(z) =∏p∈P

(1− p−z)−1 (the Euler identity), (2.7)

1

ζ(z)=

∞∑n=1

µ(n)

nz, (2.8)

where µ(n) is the Mobius function. In particular, ζ has no zeros in thesemiplane Re z > 1.


Proof. Apply Lemma 2.7 for f = I to obtain

ζ(z) =∏p∈P

(1 + p−z + p−2z + . . . ) =∏p∈P

(1− p−z)−1,

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

since |µ(n)| 6 1 and thus the series∞∑n=1

µ(n)n−z is absolutely converging

for Re z > 1. Hence,∞∑n=1

µ(n)n−z =∏p∈P

( ∞∑d=0

µ(pd)p−dz

)=∏p∈P

(1− p−z).

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) = lnn.

Proof. For n = 1 the statement is obvious. If n > 1 then considerthe canonical distribution of n, n = qa11 . . . qarr . By the definitions of theconvolution product and of the von Mangoldt function, we have

(Λ I)(n) = 0 +

r∑j=1

aj∑a=1

Λ(qaj ) · 1 =

r∑j=1

aj ln qj .

On the other hand, lnn = ln(qa11 . . . qarr ) = a1 ln q1 + · · ·+ ar ln qr.

Theorem 2.15. In the semiplane Re z > 1, the following identity holds:

ζ ′(z)

ζ(z)= −

∞∑n=1

Λ(n)

nz. (2.9)

Proof. As we have already noted, (2.4) converges uniformly with re-spect to z in the domain Re z > s for every s > 1. Hence (as we know fromcomplex analysis) the series (2.4) allows term-by-term derivation at everypoint of the semiplane Re z > 1:

ζ ′(z) = −∞∑n=1

lnn

nz.

Multiply the expression obtained by (2.8) for 1/ζ(z):

ζ ′(z)

ζ(z)= −

( ∞∑n=1

lnn

nz

)( ∞∑n=1

µ(n)

nz

).


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

ζ ′(z)

ζ(z)=

∞∑d1,d2=1

µ(d2) ln d1

(d1d2)z

=

∞∑n=1

∑d|n

µ(d) ln(n/d)

1

nz=

∞∑n=1

(ln µ)(n)

nz. (2.10)

By Lemma 2.14, (Λ I)(n) = lnn. Lemma 2.11 implies ln µ = Λ, andit 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 inthe complex plane, and consider La as a path of integration in the upwarddirection, i.e., from a− i∞ to a+ i∞.

Theorem 2.16. For every a > 1 the following identity holds:

ψ(x) =1

2πι

∫La

(−ζ′(z)

ζ(z)

)xz

z2dz for x > 1. (2.11)

Proof. Let Ia(x) stand for the improper integral in the right-handside of (2.11).

It follows from (2.9) that∣∣∣∣−ζ ′(z)ζ(z)

xz

z2

∣∣∣∣ 6 C 1

|z|2,

for z ∈ La (Re z = a > 1), where C is a constant which does not dependon z.

Since the integral ∣∣∣∣∣∣∫La

dz

z2

∣∣∣∣∣∣ 6∞∫−∞

dy

a2 + y2

converges, Ia(x) is absolutely converging. Therefore, it can be adequatelyevaluated via the Cauchy principal value. By (2.9),

Ia(x) = limB→∞

∫LBa

∞∑n=1

Λ(n)

nzxz

z2dz, (2.12)


where LBa = z | Re z = a,−B 6 Im z 6 B. In this expression, theintegrand series is uniformly converging with respect to z ∈ La since∣∣∣∣Λ(n)

nzxz

z2

∣∣∣∣ 6 xa

a2

lnn

na

for every z ∈ La (recall that the series∞∑n=1

lnn/na converges for a > 1).

Hence, the integral in the right-hand side of (2.12) can be evaluated in thetermwise way:

Ia(x) = limB→∞

∞∑n=1

∫LBa

Λ(n)

nzxz

z2dz. (2.13)

On the other hand, z = a+ ιy, and thus partial integrals over the segmentsLBa may be estimated as follows:∣∣∣∣∣∣∣

∫LBa

Λ(n)

nzxz

z2dz

∣∣∣∣∣∣∣ 6∞∫−∞

Λ(n)(xn

)a 1

a2 + y2dy 6 Cxa

Λ(n)

na,

where C is a constant not depending on n and B. Since Λ(n) 6 lnn, theseries in the right-hand side of (2.13) converges uniformly with respect toB, hence, the limit as B →∞ may be evaluated in the termwise way:

Ia(x) =

∞∑n=1

Λ(n)

∫La

(x/n)z1

z2dz. (2.14)

The integrand gn,x(z) = (x/n)z1

z2has the only singular point at z = 0.

The Laurent series for gn,x(z) at z = 0 has the form

(x/n)z1

z2=

(1 +

∞∑k=1

1

k!lnk(x/n)zk

)z−2

= z−2 + ln(x/n)z−1 +1

2!ln2(x/n) + . . . ,

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

radius R =√B2 + a2 centered at the origin and denote by CBa the arc of this

that lies leftward to the line Re z = a. Being combined with an appropriate


Figure 1. Integration path in Case 1

segment L′ = LBa , this arc forms a closed integration path which includesthe origin for sufficiently large B (see Fig. 1). Then∣∣∣∣∣∣∣

∫CBa

(x/n)z1

z2dz

∣∣∣∣∣∣∣ 6∣∣∣∣∣∣

2π−θ∫θ

(x/n)z1

R2e2ιϕιReιϕ dϕ

∣∣∣∣∣∣6 (x/n)a

2π∫0

1

Rdϕ = O(1/R) = O(1/B).

By the Cauchy integral theorem,

1

2πι

∫LBa

(x/n)z1

z2dz +

∫CBa

(x/n)z1

z2dz

= ln(x/n).

The second summand in the left-hand side approaches zero as B →∞, andthe 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:

It consists of the segment LBa (as in the previous case) and of the arc

|z| = R =√B2 + a2 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

path which contains no singularities of the integrand (x/n)z 1z2 . 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

2πι

∫La

(x/n)z1

z2dz =

ln(x/n), n 6 x,

0, n > x.

Plug in these expressions into (2.14) to obtain

1

2πι

∫La

(−ζ′(z)

ζ(z)

)xz

z2dz =

∑n6x

Λ(n) ln(x/n).

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 statea general observation that will be useful later. Suppose f(u, z) is a functiondepending 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


u ∈ [a, b] the function f(u, z) is analytic in z ∈ D. In addition, supposethat 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

F (z) =

b∫a

f(u, z) du

is an analytic function in z ∈ D such that

F ′(z) =

b∫a

∂f(u, z)

∂zdu.

Indeed, the integral over [a, b] is a limit of a sequence of Riemann sums

ΣN =

N∑j=1

f(uj , z)∆u, ∆u = |b− a|/N.

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

ΣN →N→∞

F (z) =

b∫a

f(u, z) du.

The Weierstrass theorem implies F (z) to be an analytic function such thatthe derivative of F (z) is the limit of Σ′N with respect to z. Every Σ′N isequal to a Riemann sum for the function ∂f/∂z, and thus

limN→∞

S′N =

b∫a

∂f(u, z)

∂zdu.

The following important statement says that the Riemann function de-fined by (2.4) in the semiplane Re z > 1 may be analytically extended intoa 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) isanalytic at all points of Re z > 0 except for a simple pole at z = 1, in whichResz=1ζ(z) = 1.


Proof. Denote ρ(u) = 1/2 − u, u > 0, where u = u − [u] is thefractional part of a real number u, [u] is the largest integer not greaterthan u.

Let us fix two natural numbers N < M and consider the followingexpression:

I(N,M, z) :=

M+1/2∫N+1/2

1

uz+ z

ρ(u)

uz+1du =

M+1/2∫N+1/2

1

uz+ z

1/2− u+ [u]

uz+1du

=

M+1/2∫N+1/2

1− zuz

du+1

2

M+1/2∫N+1/2

z

uz+1du+

M+1/2∫N+1/2

z[u]

uz+1du

=1

uz−1

∣∣∣∣M+1/2

N+1/2

− 1

2

1

uz

∣∣∣∣M+1/2

N+1/2

+

N+1∫N+1/2

z[u]

uz+1du

+M−1∑k=N+1

k+1∫k

z[u]

uz+1du+

M+1/2∫M

z[u]

uz+1du

=1

(M + 1/2)z−1− 1

(N + 1/2)z−1− 1/2

(M + 1/2)z+

1/2

(N + 1/2)z

− N

(N + 1)z+

N

(N + 1/2)z+

M−1∑k=N+1

(− k

(k + 1)z+

k

kz

)− M

(M + 1/2)z+M

Mz.

Reduce similar terms to obtain

I(N,M, z) = − N

(N + 1)z+

M−1∑k=N+1

(− k

(k + 1)z+

1

kz−1

)+

1

Mz−1

= −N + 1− 1

(N + 1)z+

M−1∑k=N+1

(−k + 1− 1

(k + 1)z+

1

kz−1

)+

1

Mz−1

= − 1

(N + 1)z−1+

1

(N + 1)z+

M−1∑k=N+1

(1

kz−1− 1

(k + 1)z−1

)

+

M−1∑k=N+1

1

(k + 1)z+

1

Mz−1.


Finally,

I(N,M, z) =

M∑k=N+1

1

kz

Hence, if Re z > 1 then

ζ(z) =

N∑n=1

1

nz+ limM→∞

I(N,M, z).

On the other hand, for Re z > 1

limM→∞

I(N,M, z) =(N + 1/2)−z+1

z − 1+ z

∞∫N+1/2

ρ(u)

uz+1du.

Therefore,

ζ(z) =

N∑n=1

1

nz+

(N + 1/2)−z+1

z − 1+ z

∞∫N+1/2

ρ(u)

uz+1du, Re z > 1, (2.16)

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 to1), and it remains to analyze the third summand.

The integral in the last summand of the right-hand side of (2.16) maybe considered as a series:

∞∫N+1/2

ρ(u)

uz+1du =

N+1∫N+1/2

ρ(u)

uz+1du+

N+2∫N+1

ρ(u)

uz+1du+ . . . .

On every interval of integration ([N + 1/2, N + 1), [N + 1, N + 2), and soon), ρ(u)/uz+1 is a continuous function. Moreover, for every s > 0, the lastseries converges uniformly with respect to z in Re z > s, and the integrandsatisfies the condition (2.15). Hence, the third summand in the right-handside of (2.16) is an analytic function in Re z > 0 and

d

dz

∞∫N+1/2

ρ(u)

uz+1du =

∞∫N+1/2

d

dz

ρ(u)

uz+1du.

Thus, the right-hand side of (2.16) is the desired function ζ(z).

In what follows, we will identify ζ and ζ.


2. Zeros of the Riemann function.

Lemma 2.18. For every s > 1 and for every t ∈ R we have

|ζ3(s)ζ4(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

A =∏p∈P

(|1− p−s|3|1− p−s−ιt|4|1− p−s−2ιt|

)−1.

Let us evaluate the natural log of the both sides of the last expression usingthe well-known relation ln |w| = Re lnw, w ∈ C \ 0. Then

lnA = −Re∑p∈¶

(3 ln(1− p−s) + 4 ln(1− p−s−ιt) + ln(1− p−s−2ιt)

).

Recall that the Taylor series for ln(1 − z) in a neighborhood of z = 0 isgiven by

ln(1− z) = −z − z2

2− z3

3− . . . .

This series is absolutely converging when |z| < 1. Apply this distributionfor z = p−s, z = p−s−ιt, and z = p−s−2ιt:

lnA =∑p∈P

Re

∞∑n=1

(3p−ns

n+ 4

p−n(s+ιt)

n+p−n(s+2ιt)

n

)

=∑p∈P

∞∑n=1

p−ns

n

(3 + 4 Re p−ιnt + Re p−2ιnt

)=∑p∈P

∞∑n=1

p−ns

n(3 + 4 cos(nt ln p) + cos(2nt ln p)) . (2.18)

Note that

3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 > 0,

and thus all summands in the right-hand side of (2.18) are non-negative.Hence, lnA > 0, i.e., A > 1.

Theorem 2.19. Riemann function ζ(z) has no zeros in the line Re z = 1.


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

∞∑n=1

1

ns6 1 +

∞∫1

1

xsdx = 1 +

1

s− 16

2

s− 1

(since a right Riemann sum for a decreasing function is no greater that itsintegral).

Assume ζ(1 + ιt) = 0 for some t 6= 0. Since ζ is analytic at 1 + ιt, itsderivative is bounded in a neighborhood of this point. In particular,∣∣∣∣ζ(s+ ιt)− ζ(1 + ιt)

s− 1

∣∣∣∣ 6 C, 1 < s < 2.

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 , suchthat |ζ(s+ 2ιt)| 6M 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|,

|ζ(s)| 6 2

s− 1,

|ζ(s+ 2ιt)| 6M.

Then

A := |ζ(s)3ζ(s+ ιt)4ζ(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. Thecontradiction obtained proves the theorem.

3. Estimates of the logarithmic derivative. To prove the asymp-totic law of distribution of prime numbers, we need an upper estimate ofthe logarithmic derivative of Riemann function far off the pole z = 1. Letus start with estimates of ζ(z) itself and its derivative. We have alreadyseen 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|,(2) |ζ ′(s+ ιt)| 6 C2 ln2 |t|

for 1 6 s 6 2, |t| > 3.


Proof. (1) The explicit expression for ζ(s + ιt) is given by Theorem2.17:

ζ(s+ ιt) =

N∑k=1

1

ks+ιt+

(N + 1/2)1−s−ιt

s− 1 + ιt+

∞∫N+1/2

(s+ ιt)ρ(u)

us+1+ιtdu. (2.19)

Suppose N = [|t|], and estimate the absolute values of all summands.First,∣∣∣∣∣

N∑k=1

1

ks+ιt

∣∣∣∣∣ 6N∑k=1

1

k6 1 +

N∫1

dx

x= 1 + lnN 6 1 + ln |t|.

Next, ∣∣∣∣ (N + 1/2)1−s−ιt

s− 1 + ιt

∣∣∣∣ 6 1

3,

since s > 1 and |s+ ιt− 1| > 3.Finally,∣∣∣∣∣∣∣∞∫

N+1/2

(s+ ιt)ρ(u)

us+1+ιtdu

∣∣∣∣∣∣∣ 6 (s+ |t|)∞∫

|t|−1

du

us+1=

s+ |t|s(|t| − 1)s

62 + |t||t| − 1

6 C

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). ButB + 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 sincethe improper integral in the third summand is uniformly converging. Thus,

ζ ′(z) = −N∑k=1

ln k

kz+

d

dz

(N + 1/2)1−z

z − 1+

d

dz

∞∫N+1/2

zρ(u)

uz+1du.

As above, let z = s + ιt, N = [|t|]. The second and third summands inthe right-hand side of the last expression are bounded (as it was shown inthe proof of (1)). To estimate the first summand, consider the integral over

[3,∞) of the functionlnx

x, which is decreasing on x > 3:

N∑k=1

ln k

ks6

ln 2

2+

N∑k=3

ln k

k6 C ln2N.


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

Proposition 2.21. There exist constants T0 > 3, C3, C4 > 0 such that

(1) |ζ(s+ ιt)| > C3 ln−31/4 |t| > C3 ln−8 |t|,

(2)

∣∣∣∣ζ ′(s+ ιt)

ζ(s+ ιt)

∣∣∣∣ 6 C4 ln10 |t|

for 1 6 s 6 2, |t| > T0.

Proof. (1) Recall the inequality deduced in the proof of Theorem 2.19:

|ζ(s)3ζ(s+ ιt)4ζ(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,

|ζ(s+ ιt)| > |ζ(s)|−3/4|ζ(s+ 2ιt)|−1/4 > C

(2

s− 1

)−3/4

ln−1/4 |t|

for 1 < s 6 2, |t| > 3, where C is a constant which does not depend on sand t. Let us fix t, |t| > 3, and consider the interval 1 + δ 6 s 6 2, where

δ =2

ln10 |t|In this interval,

|ζ(s+ ιt)| > C ln−31/4 |t|.It remains to estimate from below the quantity |ζ(s + ιt)| in the interval1 6 s 6 1 + δ. Proposition 2.20(2) allows to estimate an increment of ζ(z)in this interval:

|ζ(1 + δ + ιt)− ζ(s+ ιt)| =

∣∣∣∣∣∣1+δ∫s

ζ ′(z) dz

∣∣∣∣∣∣ 6 δ maxIm z=t,s6Re z61+δ

|ζ ′(z)|

6 2 ln−10 |t|C2 ln2 |t| = 2C2 ln−8 |t|.Therefore,

|ζ(s+ ιt)| > |ζ(1 + δ + ιt)| − 2C2 ln−8 |t| > C ln−31/4 |t| − 2C2 ln−8 |t|.Since −31/4 < −8, for any constants C and C2 the second summand inthe right-hand side becomes negligible for sufficiently large |t|. Hence, thereexists T0 > 3 such that

|ζ(s+ ιt)| > (C/2) ln−31/4 |t|,


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 tocomplete the proof of the main statement of this chapter.

Theorem 2.22. The function π(x) is asymptotically equivalent tox/ lnx.

Proof. By Theorem 2.4, it is enough to show that limx→∞

ψ(x)

x= 1.

Theorem 2.16 implies

ψ(x)

x=

1

2πι

∫La

(−ζ′(z)

ζ(z)

)xz−1

z2dz

for every x > 1, a > 1. Here La stands for the line Re z = a, integration ismade from a− ι∞ to a+ ι∞.

Choose real numbers U > T > T0, where T0 is the constant fromProposition 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 , theset of zeros has a condensation point in the same region W . Then thewell-known uniqueness theorem of an analytic function implies ζ to be zeroon 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 quantityη > 0 which is chosen in such a way that the interior of Γ (and Γ itself)does not contain zeros of the Riemann function.


Figure 3. Integration path Γ

Consider the integral

I =1

2πι

∫Γ

(−ζ′(z)

ζ(z)

)xz−1

z2dz

The only singularity of the integrand inside the closed path Γ is at the pointz = 1, and

Resz=1

(−ζ′(z)

ζ(z)

)xz−1

z2= 1. (2.20)

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 con-sider these segments separately.

Segment 1. As we have already mentioned,

limU→∞

I1 =ψ(x)

x.


Segment 2. By Proposition 2.21(2),

|I2| 6 21

2π

a∫1

C4 ln10 Uxa−1

s2 + U2ds

Hence, |I2| = O(ln10 U/U2), i.e.,

limU→∞

I2 = 0.

Segment 3. By Proposition 2.21(2),

|I3| 6 21

2π

U∫T

C4 ln10 t

1 + t2dt 6 C

∞∫T

dt

t3/2= 2C

1

T 1/2

(since ln10 t/t1/2 is a bounded function). Note that the constant C does notdepend on x.

Segment 4. Segments (4) and (5) form a compact set, on which ζ(z) isa nonzero analytic function. Hence, its logarithmic derivative is continuouson these segments. By the Weierstrass theorem on a continuous function,there exists a constant M = M(T, η) such that∣∣∣∣ζ ′(z)ζ(z)

∣∣∣∣ 6M(T, η).

Then

|I4| 6 21

2π

1∫1−η

M(T, η)xs−1

s2 + T 2ds 6

M(T, η)

πT 2

1∫1−η

xs−1 ds 6C

lnx.

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,

|I5| 6 Cx−η,

where C depends on T and η.

Now, evaluate the limit of (2.21) as U →∞:

I = 1 =ψ(x)

x+ 0 + J3 + J4 + J5,


where Jk = limU→∞

Ik. As it was shown above,

|J3| 6 2C1

T 1/2,

|J4| 6C(T, η)

lnx,

|J5| 6 C(T, η)x−η.

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, ∣∣∣∣∣1− ψ(x)

x

∣∣∣∣∣ 6 |J3|+ |J4|+ |J5| < ε

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 lnn.

Proof. By the definition of π, π(pn) = n. Moreover, the Euclid theo-rem implies pn →

n→∞∞. By Theorem 2.22,

n = π(pn) =pn

ln pn(1 +Rn), lim

n→∞Rn = 0.

Then

lnn = lnπ(pn) = ln pn − ln ln pn + ln(1 +Rn).

Multiply these expressions to obtain

n lnn = pn(1 +Rn)ln pn − ln ln pn + ln(1 +Rn)

ln pn.

It is easy to see that

n lnn

pn= (1 +Rn)

(1− ln ln pn

ln pn+

ln(1 +Rn)

ln pn

)→

n→∞1.

§ 2.4. Problems 55

§ 2.4. Problems

(1) Prove that ∑p∈P,p6x

ln p

p= lnx+O(1).

(2) Show that there exists a constant C > 0 such that∑p∈P,p6x

1

p= C + ln(lnx) +O(1/ lnx).

(3) The famous Riemann hypothesis states that all zeros of ζ(z) in thesemiplane Re z > 0 lie in the line Re z = 1

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 numberof primes in an arithmetic progression with coprime difference and the firstmember. 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 algebraicbinary 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 notationis used very often, so the operation is denoted by “+”, the identity elementis denoted by “0”, and the inverse element is denoted by “−a”. By |G| wedenote the cardinality of G.

A one-generated group is called cyclic, i.e. a group G is called cyclicif 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 theset gn | n ∈ Z appears to be a subgroup of G. This subgroup is called thecyclic subgroup generated by g and is denoted by 〈g〉. The cardinality |〈g〉|is called the order of g and is denoted by |g|. Clearly 〈Z,+〉 is an infinitecyclic group and, for every n ∈ N, 〈Zn,+〉 is a finite cyclic group of order n.

Exercise 3.1. Let G be a group and g be an element of G. Then thefollowing hold.

56

§ 3.1. Abelian groups 57

(1) Either |g| = ∞, or |g| is the minimal positive integer k such thatgk = e.

(2) If |g| =∞, then 〈g〉 ' Z; if |g| = n, then 〈g〉 ' Zn1.(3) If gm = e, then |g| divides m.(4) If g1, g2 ∈ G are chosen so that g1·g2 = g2·g1, and 〈g1〉∩〈g2〉 = e,

then |g1 ·g2| = lcm(|g1|, |g2|). In particular, if |g1|, |g2| are coprimethen |g1 · g2| = |g1| · |g2|.

If G1, . . . Gn are groups, then

G1 × . . .×Gn = (g1, . . . , gn) | g1 ∈ G1, . . . , gn ∈ Gnwith coordinate-wise multiplication is called the direct product ofG1, . . . , 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 subgroupsG1, . . . , 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 ofsubgroups 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 existd1, . . . , dn ∈ N such that G ' Zd1 × . . .×Zdn . Moreover such d1 . . . , dn canbe uniquely determined by the following condition: for every i = 1, . . . , n−1,di divides di+1.

Recall that by Z∗n we denote the (multiplicative) group of invertibleelements in Zn.

Exercise 3.4. Prove that |Z∗n| = ϕ(n), where ϕ(n) is the Euler func-tion. If n = pα1

1 · . . . · pαkk is the canonical decomposition of n into the

product of primes, then ϕ(n) = ϕ(pα11 ) · . . . · ϕ(pαkk ) 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).


Hint: Prove that ϕ I = E, where I(n) = 1, E(n) = n for all naturaln. Here ”” stands for the convolution product of arithmetic functionsmentioned 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 G. Define a multiplication on G by

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

Theorem 3.5. G is an abelian group. Moreover, G and G are isomor-phic.

Proof. The multiplication on G is clearly an algebraic operation, andit is associative and commutative (we leave the proof for the reader). Denoteby 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 G. For every χ ∈ G defineχ−1 by χ−1(g) = 1/χ(g). Evidently, χ−1 is the inverse element for χ. Thus

G is an abelian group.In view of the theorem about the structure of finite abelian groups

(Theorem 3.3) we have G = 〈g1〉 × . . . × 〈gr〉 for suitable g1, . . . , gr ∈ G.Denote |gk| by hk, let εk = exp( 2πι

hk). For k = 1, . . . , r define the map

χk : G → C∗ by χk(ga11 . . . garr ) = εakk . It is straightforward that χk is a

character of G for k = 1, . . . , r. It is also clear that χk, χ2k, . . . , χ

hkk = χe are

distinct characters, since their values on gk are distinct.

Consider the map ϕ : G → G defined by ϕ : ga11 . . . garr 7→ χa11 . . . χarr .By the definition, ϕ preserves the multiplication. The map ϕ is clearlyinjective: if e 6= x ∈ G, then x = ga11 . . . garr and there exists ak such thatgakk 6= e; therefore

ϕ(x)(gk) = (χa11 . . . χarr )(gk) = exp

(2akπι

hk

)6= 1.

In order to prove that ϕ is surjective note that ghkk = e implies χ(ghkk ) =

χ(gk)hk = 1 for every χ ∈ G. Hence, χ(gk) is a complex root of unity of orderhk, and thus χ(gk) = εakk for appropriate ak. It is straightforward to checkthat χ(g) = (χa11 . . . χarr )(g) for all g ∈ G. Therefore, χ = χa11 . . . χarr =

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

ϕ(g1) · ϕ(g2)


ϕ(ga11 . . . garr ), 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 dualitybetween discrete and continuous abelian groups. The group of characters

G can be considered as a dual group for G and χ1, . . . , χr is the dual basisfor g1, . . . , gr.

In the proof of the theorem we also derive the following

Corollary 3.7. For every nonidentity g ∈ G there exists χ0 ∈ G suchthat χ0(g) 6= 1.

In the theory of characters for finite groups the following propositionplays an important role.

Proposition 3.8. (Orthogonality relations) The following hold

(1) for every χ ∈ G we have∑g∈G

χ(g) =

|G|, if χ = χe;

0, if χ 6= χe.

(2) for every g ∈ G we have∑χ∈G

χ(g) =

|G|, if g = e;

0, if g 6= e.

Proof. Clearly we need to proof the first identity in case χ 6= χe, andthe second in case g 6= e.

(1) If χ 6= χe, then there exists g0 such that χ(g0) 6= 1. Then

∑g∈G

χ(g) =∑g∈G

χ(gg0) =

∑g∈G

χ(g)

χ(g0),

so∑g∈G χ(g) = 0.

(2) If g 6= e, then Corollary 3.7 implies that the existence χ0 such thatχ0(g) 6= 1. Then

∑χ∈G

χ(g) =∑χ∈G

(χ0χ)(g) =

∑χ∈G

χ(g)

χ0(g),

so∑χ∈G χ(g) = 0.

§ 3.2. Dirichlet series 60

3. Characters modulo m. Consider G = Z∗m. Denote by n the resid-

ual of n modulo m. Then each character χ ∈ G can be extended to anarithmetic function χ : N→ C by

χ(n) =

χ(n), if gcd(n,m) = 1;

0, otherwise.(3.1)

This construction lead us to the notion of a character modulo m. Moreformally, 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 definitionthat each character modulo m is a character of Z∗m extended to all posi-tive integers by using (3.1). So the following proposition is an immediatecorollary to the properties of characters of finite abelian groups obtainedabove.

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 multipli-

cation: (χ1 · χ2)(n) = χ1(n) · χ2(n).(3) If Ωm is a full system of residuals modulo m, then∑

n∈Ωm

χ(n) =

ϕ(m), if χ = χe,

0, otherwise.

(4) For every n ∈ Z we have∑χ∈Gm

χ(n) =

ϕ(m), if n ≡ 1 (mod m),

0, otherwise.

§ 3.2. Dirichlet series

1. Convergence of L-series. Given χ ∈ Gm define an L-series L(z, χ),where z ∈ C, by

L(z, χ) =

∞∑n=1

χ(n)

nz. (3.2)

The series L(z, χ) is called an L-series of character χ.


Theorem 3.10. Let χ be a character modulo m. Then the followingstatements 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 ana-lytic continuation in domain Re z > 0, z 6= 1 with the uniquesimple pole z = 1.

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

L(z, χ) =

∞∑n=1

χ(n)

nz,

set s(x) :=∑n6x χ(n). The main idea of the proof comes from the harmonic

series. In view of Proposition 3.9(3) we have

s(m) =

m∑k=1

χ(k) =∑k∈Z∗

m

χ(k) = 0.

Moreover, for every q ∈ N we have∑mk=1 χ(qm + k) = 0. Therefore, for

each N ∈ N the inequality

|N∑k=1

χ(k)| < m (3.3)

holds.


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

N∑n=1

χ(n)

nz=

N∑n=1

s(n)− s(n− 1)

nz=s(N)

Nz− s(N − 1)

Nz+

s(N − 1)

(N − 1)z− s(N − 2)

(N − 1)z+ . . .+

s(2)

2z− s(1)

2z+s(1)

1=

s(N)

Nz−N−1∑n=1

s(n) ·[

1

(n+ 1)z− 1

nz

]=

s(N)

Nz+

N−1∑n=1

s(n) ·∫ n+1

n

z

xz+1dx =

s(N)

Nz+

N−1∑n=1

z ·∫ n+1

n

s(x)

xz+1dx,

in the last step we use the identity s(x) = s(n) for x ∈ [n, n+ 1). Set

In(z) := z ·∫ n+1

n

s(x)

xz+1dx.

Now we can bound |In(z)|. In view of (3.3) we have

|In(z)| 6 |z| ·∫ n+1

n

m

xs+1dx =

m|z|s·(

1

ns− 1

(n+ 1)s

).

ThereforeN−1∑n=1

|In(z)| < m|z|s

(1− 1

Ns

)<m|z|s

,

so the series L(z, χ) is uniformly converging.(3) The series L(z, χe) is absolutely converging for Re z > 1, so

L(z, χe) =

∞∑n=1

χe(n)

nz=∏p∈P

( ∞∑d=0

χe(pd) · p−dz

)=

∏p∈P

(1− χe(p) · p−z)−1 =∏

p∈P,p-m

(1− p−z)−1,

whence Euler identity (see Theorem 2.13) implies that the following identityholds

L(z, χe) ·∏

p∈P,p|m

(1− p−z)−1 = ζ(z). (3.4)


We can define now L(z, χe) on Re z > 0 by

L(z, χe) = ζ(z) ·

∏p∈P,p|m

(1− p−z)

,

whence item (3) of the theorem.

Lemma 3.11. Set

L(z,Gm) =∏χ∈H

L(z, χ).

Then for Re z > 1 the series L(z,Gm) can be written in the following form

∞∑n=0

annz,

where every an is a nonnegative integer and, moreover, if n = kϕ(m) and(k,m) = 1, then an > 1. Moreover,

L(z,Gm)(k) = (−1)k∞∑n=1

an(lnn)k

nz, k = 1, 2, . . .

Proof. The statement about derivations follows from the fact that theseries

∞∑n=0

annz

is absolutely converging for Re z > 1 (as a product of a finite number ofabsolutely converging series). Thus we need to show that

L(z,Gm) =

∞∑n=0

annz.

Since χ is multiplicative for every χ ∈ Gm, Lemma 2.7 implies that

L(z, χ) =∏p∈P

(1− χ(p)

pz

)−1

.

Therefore,

L(z,Gm) =∏χ∈H

∏p∈P

(1− χ(p)

pz

)−1

=∏p∈P

∏χ∈Gm

(1− χ(p)

pz

)−1

,


where we change the order of multiplication, since Gm is finite and∏p∈P

(1− χ(p)

pz

)−1

is absolutely converging.For every p ∈ P denote by fp the multiplicative order of p in Z∗m, 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 themap

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 homo-morphism. Notice that ψ is surjective. Indeed, we can define a characterχp on a power of a prime rs by

χp(rs) =

exp(2πι/fp)s, if r = p1 if r 6= p and (r,m) = 10 if r divides m,

and extend it on all integers by multiplicativity. Then 〈χp〉 is a subgroupof Gm and ψ(〈χp〉) = 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 ev-ery k = 0, 1, . . . , fp − 1 there exists exactly gp characters χ ∈ Gm withχ(p)− exp(2πιk/fp). Therefore, a polynomial∏

χ∈Gm

(1− χ(p)t)

equals (1− tfp)gp . Thus

L(z,Gm) =∏

p∈P;(p,m)=1

(1− p−fpz)−gp .

Now the Taylor series for (1− z)−g equals

∞∑k=0

(g + k − 1)!

(g − 1)!k!zk,


and the series is absolutely converging for |z| < 1. Since |p−fpz| < 1, we canapply the Taylor series to the expression of L(z,Gm). We obtain

(1− p−fpz)−gp =

∞∑k=0

(gp + k − 1)!

(gp − 1)!k!p−fpkz =

∞∑k=0

up,kpkz

,

where

up,k =

0, if k is not divisible by fp,

(gp+r−1)!(gp−1)!r! , if k = r · fp.

Since for every p ∈ P the series

∞∑k=0

up,kpkz

is absolutely converging, we obtain the following identity for every N :∏p≤N ;(p,m)=1

(1− p−fpz)−gp =

∞∑n=1

annz,

where

an =

0, if (n,m) > 1,

up1,k1 · . . . · upl,kl , if (n,m) = 1 and n = pk11 · . . . · pkll .

Consider 1 < s ∈ R. Then L(s,Gm) is converging. On the other hand, forevery M ,

M∑n=1

anns≤ L(s,Gm),

so the series∞∑n=1

anns

is converging, and so the series

∞∑n=1

annz

is absolutely converging for every z with Re z > 1. It follows that

L(z,Gm) =

∞∑n=1

annz,


where the coefficients an-s are defined above. From the definition we obtainthat an are nonnegative integers. Since fp divides ϕ(m), we obtain that

an 6= 0, if n = kϕ(m) for (k,m) = 1.

2. Landau Theorem. Consider

L(z) := L(z,Gm) = L(z, χe) ·∏

χ∈Gm\χe

L(z, χ).

Theorem 3.10 implies that L(z, χe) is analytic in semiplane Re z > 0 witha unique simple pole z = 1, while each L(z, χ) for χ 6= χe is analytic in thesemiplane Re z > 0, i.e.

∏χ∈Gm\χe L(z, χ) is analytic in the semiplane

Re z > 0. Thus L(z) is analytic in the semiplane Re z > 0 with one possiblesimple pole z = 1. In order to prove that z = 1 is indeed a simple pole forL(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) isanalytic in Re z > 0 and suppose that for Re z > 1 we can write F (z) as aseries

F (z) =

∞∑n=1

annz, (3.5)

where an > 0 for every n. Assume also that in the semiplane Re z > 1 wehave

F (k)(z) = (−1)k∞∑n=1

an(lnn)k

nz,

i.e. series (3.5) can be differentiated term by term in the semiplane Re z > 1.Then the series

∑∞n=1

anns is converging in the interval s ∈ (0, 2).

Proof. We consider the Taylor series for F (s) about s = 2:

F (s) =

∞∑k=0

F (k)(2)

k!(s− 2)k. (3.6)

The Taylor series of an analytic function is converging in the circle of radiusr 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

F (k)(2) =

∞∑n=1

(−1)nan(lnn)k

n2,


hence

F (s) =

∞∑k=0

∞∑n=1

(−1)k

k!· an(lnn)k

n2· (s− 2)k.

The series is converging, moreover, (−1)k and (s−2)k for s ∈ (0, 2) have thesame sign, so all terms in the series are nonnegative. Therefore the seriesis absolutely converging and we can change the order of summation. So weobtain

F (s) =

∞∑n=1

∞∑k=0

(−1)k

k!· an(lnn)k

n2· (s− 2)k =

∞∑n=1

ann2

( ∞∑k=0

(lnn)k(2− s)k

k!

)=

// the inner sum is equal to exp((2− s) lnn) = n2−s// =∞∑n=1

ann2· n2−s =

∞∑n=1

anns,

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 analyticin the semiplane Re z > 0. Moreover,

L(z) =

∞∑n=1

annz,

where an > 0 and further an > 1 for n = kϕ(m) and gcd(k,m) = 1. Everyseries 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 bedifferentiated term by term any number of times. By the Landau theoremthe series

∞∑n=1

anns

is converging for 0 < s < 2. For n = (km+ 1)ϕ(m) we have an > 1, so

N∑n=1

anns> //for s =

1

ϕ(m)// >

[Nm−1]∑k=1

1

km+ 1→

N→∞∞,


a contradiciton.

3. Proof of the Dirichlet Theorem. First we recall the statementof the theorem.

Theorem 3.14. (Dirichlet Theorem) Assume that a,m are naturalcoprime 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

L(z, χ) =

∞∑n=1

χ(n)

nz,

where χ ∈ Gm and Re z > 1. First we prove that

1

L(z, χ)=

∞∑n=1

χ(n) · µ(n)

nz, (3.7)

where µ(n) is the Mobius function. Indeed, the series∑∞n=1

χ(n)·µ(n)nz and∑∞

n=1χ(n)nz are absolutely converging, therefore by Lemma 2.10 we have( ∞∑

n=1

χ(n)

nz

)·

( ∞∑n=1

χ(n) · µ(n)

nz

)=

∞∑n=1

(χ (χ · µ))(n)

nz.

Now χ (χ · µ) = (χ · I) (χ · µ) = χ · (I µ), and the Mobius inversionformulae (Lemma 2.11) implies that I µ = e, so χ · (I µ) = χ · e = e, sowe get (3.7). For Re z > 1 we have

L′(z, χ) = −∞∑n=1

χ(n) lnn

nz.

Thus,

L′(z, χ)

L(z, χ)=

( ∞∑n=1

χ(n) · µ(n)

nz

)·

(−∞∑n=1

χ(n) lnn

nz

)=

−∞∑n=1

((χ · ln) (χ · µ))(n)

nz= −

∞∑n=1

(χ · (ln µ))(n)

nz=

//Lemmas 2.11 and 2.14// = −∞∑n=1

(χ · Λ)(n)

nz,


where Λ(n) is the von Mangoldt function, i.e. we obtain the identity

L′(z, χ)

L(z, χ)= −

∞∑n=1

(χ · Λ)(n)

nz.

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

− L′(z, χ)

L(z, χ)=∑p∈P

∞∑k=1

χ(p)k · ln ppkz

=

//the series is absolutely converging for Re z > 1,

so we can change the order of summation// =∞∑k=1

∑p∈P

χ(p)k · ln ppkz

=∑p∈P

χ(p) · ln ppz

+

∞∑k=2

∑p∈P

χ(p)k · ln ppkz

.

Denote the second summand by R(z, χ). We show that R(z, χ) is absolutelyconverging for Re z > 1

2 + ε for every ε > 0. Indeed, we need to show thatthe series ∑

p∈Pln p

∞∑k=2

(χ(p)

pz

)kis absolutely converging, since in this case we can change the order of sum-

mation and derive that R(z, χ) =∑∞k=2

∑p∈P

χ(p)k·ln ppkz

is absolutely con-

verging as well. Now we have

∑p∈P

ln p

∞∑k=2

(χ(p)

pz

)k=∑p∈P

ln p · χ(p)2

p2z· 1

1− χ(p)pz

.

Since |1− χ(p)pz | > |1−

1p1/2|, we have∣∣∣∣∣ln p · χ(p)2

p2z· 1

1− χ(p)pz

∣∣∣∣∣ 6 ln p

p1+2ε· 1

|1− 1√p |6

4 ln p

p1+2ε,

and thus we obtain that the series R(z, χ) is absolutely converging. Bythe condition of the theorem gcd(a,m) = 1, so we can choose b so that


b · a ≡ 1 (mod m). Then we have∑χ∈Gm

χ(b) ·(−L′(z, χ)

L(z, χ)

)=

∑χ∈Gm

∑p∈P

χ(b) · χ(p) · ln ppz

+∑χ∈Gm

χ(b) ·R(z, χ) =

∑p∈P

ln p

pz·

∑χ∈Gm

χ(b) · χ(p)

+∑χ∈Gm

χ(b) ·R(z, χ) =

//recall that∑χ∈Gm

χ(bp) equals ϕ(m)

if bp ≡ 1 (mod m) and 0 otherwise // =∑p∈P,p≡a (mod m)

ϕ(m) · ln ppz

+∑χ∈Gm

χ(b) ·R(z, χ).

The second summand is analytic for Re z > 12 . 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,∑

χ∈Gm

χ(b) ·(L′(z, χ)

L(z, χ)

)=L′(z, χe)

L(z, χe)+

∑χ∈Gm\χe

χ(b) ·(L′(z, χ)

L(z, χ)

).

By Theorem 3.10 and Corollary 3.13 we obtain that the second summandis bounded for z = 1, while for the first summand we have

L′(z, χe)

L(z, χe)= ln(L(z, χe))

′,

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 Fto the field of real numbers. Then (F, v) is called a valuation field, while vis called a valuation of F , if

(1) For every x ∈ F we have v(x) > 0 and v(x) = 0 if and only ifx = 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 alsohave 1 = v(1) = v(x ·x−1) = v(x) ·v(x−1), so v(x−1) = 1

v(x) . The remaining

statement follows immediately.

If F is the field of rationals, then we can define the following valuations:

(1) v(x) =

0, if x = 0,1, otherwise;

the trivial valuation (it can be defined

over arbitrary field).(2) vα(x) = |x|α for 0 < α 6 1.(3) vp,ρ(x) = ρνp(x), where 0 < ρ < 1, p is a prime, and νp(x) ∈ Z is

given by the identity x = pνp(x) · ab , 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 thedefinition. Can α be greater than 1 in item (2)? Can α be less, than 0 initem (2)?

71

§ 4.1. Valuation fields 72

Clearly, every valuation defines a topology on F . Namely, we can definean 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 clearthat F with the topology is a Hausdorff space. Moreover, the operations“+”, “·” (considered as maps F ×F → F ) and “−”, −1 (considered as mapsF → F ) are continuous. In particular, F is a topological field.

Exercise 4.3. Prove that all operations are continuous maps. Provethat induced topology is a Hausdorff space.

Let (F, v) be a valuation field, a ∈ F . A sequence ann>1 is said toconverge to a (under the valuation v, we use notation an →

(v)a), if

limn→∞

v(an − a) = 0.

The following basic properties of limits hold.

Proposition 4.4. The following identities hold:

limn→∞

(an ± bn) = limn→∞

an ± limn→∞

bn;

limn→∞

(an · bn) = limn→∞

an · limn→∞

bn;

limn→∞

a−1n =

(limn→∞

an

)−1

,

where limn→∞

an and limn→∞

bn are assumed to exist and, in the last identity all

an-s and limn→∞

an are assumed to be not equal to 0.

Exercise 4.5. Prove Proposition 4.4.

Let F be a field and v1, v2 be its valuations. The valuations v1, v2 arecalled equivalent (v1 ∼ v2), if, for every sequence ann>1, we have

ann>1 →(v1)

a ⇐⇒ ann>1 →(v2)

a.

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

Proof. Assume that v1 and v2 are equivalent. Then for every x ∈ Fthe inequality v1(x) < 1 is equivalent to xn →

(v1)0. Since v1 and v2 are

equivalent, it follows that xn →(v2)

0, so v2(x) < 1.


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. Thenthere exists x ∈ F such that v1(x) 6= 1. By Proposition 4.1 we obtain thateither v1(x) > 1, or v1(x−1) > 1. Without loss of generality we may assumethat v1(x) > 1 (and so v2(x) > 1). Choose a sequence ann>1 such thatan →

(v1)a. Therefore, for every m ∈ N, we have xm · an →

(v1)xm · a, i.e.

v1(xman − xma) →n→∞

0. Now denote v2(x) by α (recall that v2(x) > 1).

Then for every ε > 0 there exists M ∈ N such that for every m > M wehave 1

αm < ε.Since v1(xman − xma) →

n→∞0, it follows that there exists N ∈ N such

that for every n > N we have v1(xman − xma) < 1. The condition of thelemma implies that for every n > N the inequality v2(xman − xma) < 1holds. Thus we obtain that for every n > N the inequality v2(x)m · v2(an−a) < 1 holds. Hence, for every n > maxM,N we have v2(an− a) < ε, i.e.an →

(v2)a. The implication an →

(v2)a⇒ an →

(v1)a follows by the symmetry.

Corollary 4.7. Every valuation vα of Q, where 0 < α 6 1, is equivalentto v1(x) = |x|. If p is a fixed prime, then for every 0 < ρ < 1 all valuationsvp,ρ are equivalent, the corresponding equivalence class is denoted by vp. Ifp, q are distinct primes, then vp,ρ and vq,ρ are not equivalent. Moreover, vαand 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-adictopology. The p-adic valuation of a number grows with the power of pin denominator. For example, if an = pn, then an →

n→∞0. Valuation

fields (F1, v1) and (F2, v2) are topologically isomorphic, if there exists anisomorphism of fields ϕ : F1 → F2 such that for every sequence ann>1,an →

(v1)a if and only if ϕ(an) →

(v2)ϕ(a). In particular, if F1 = F2 = F ,

then v1 ∼ v2 if and only if (F, v1) is topologically isomorphic to (F, v2) withisomorphism 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 valuationfield (F2, v2), if there exists an injective homomorphism ϕ : F1 → F2 suchthat the restriction of v2 on ϕ(F1) coincides with v1.


2. Valuations over rationals.

Theorem 4.10. (On valuations over rationals) Let (Q, v) be a valua-tion 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. Assumethat v is not trivial. Then there exists n ∈ N such that v(n) 6= 1. Indeed,since v is nontrivial, there exists p

q ∈ Q such that v(pq ) 6= 1. Hence either

v(p) 6= 1, or v(q) = v( 1q )−1 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 existsn ∈ N such that v(n) 6= 1. Now one of the following two cases holds: eitherthere exists n ∈ N such that v(n) > 1, or for every n ∈ N the inequalityv(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︸︷︷︸n times

) 6 1 + . . .+ 1︸︷︷︸n times

= n,

so there exists α ∈ (0, 1] such that v(n) = nα. By Proposition 4.1 we obtainthat v(nk) = nkα for every k ∈ Z. Assume that m ∈ N. Then there existsk > 0 such that nk 6 m < nk+1. Consider the expansion of m in base n,we obtain:

m = a0 + a1n+ . . .+ aknk,

where 0 6 ai 6 n− 1 for i = 0, 1, . . . , k; and ak 6= 0. Now we have

v(m) 6 v(a0) + v(a1) · nα + . . .+ v(ak) · nkα 6

//for every i, v(ai) 6 ai 6 n− 1// 6 (n− 1)(1 + nα + . . .+ nkα) 6

n− 1

nα − 1(n(k+1)α − 1) <

[n− 1

nα − 1nα]· nkα 6 C ·mα.

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

v(m) 6 mα. (4.2)

Indeed, assume that there exists k ∈ N such that v(k) > kα, i.e. v(k)kα =

D > 1. Then for every s ∈ N we have

v(ks)

ksα= Ds →

s→∞∞.

On the other hand, v(ks)ksα < C, a contradiction. Thus (4.2) holds for every

m ∈ N.


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

v(m) > n(k+1)α − v(m1) > n(k+1)α −mα1 >

nkα(nα − (n− 1)α) = n(k+1)α

(1−

(n− 1

n

)α)> C1 ·mα.

It follows that there exists a constant C1 > 0 such that for every m ∈ N wehave v(m) > C1 ·mα. We state that

v(m) > mα. (4.3)

Indeed, assume that there exists k ∈ N such that v(k) < kα, i.e. v(k)kα =

D1 < 1. So for every s ∈ N we have

v(ks)

ksα= Ds

1 →s→∞

0.

On the over hand, v(ks)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,

so for every pq ∈ Q, the identity v

(pq

)=∣∣∣pq ∣∣∣α holds.

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 ofprimes n = pα1

1 · . . . · pαkk . Then v(n) = v(p1)α1 · . . . · v(pk)αk , so there

exists a prime p with v(p) = ρ < 1. We show first that for every primeq 6= p the identity v(q) = 1 holds. Otherwise there would exist q 6= p withv(q) = µ < 1. Since both ρ and µ are less than 1, there exists k ∈ N suchthat both ρk, µk are less than 1

2 . On the other hand, gcd(pk, qk) = 1, so

there exist a, b ∈ Z such that a · pk + b · qk = 1. Thus we obtain

1 = v(1) = v(a · pk + b · qk) 6 v(a) · v(pk) + v(b) · v(qk) 6

1 · ρk + 1 · µk < 1

2+

1

2= 1,

a contradiction. Now for every mn ∈ Q we have m = pα1 ·m1, n = pα2 · n1,

where gcd(m1 · n1, p) = 1, therefore

v(mn

)= ρα1−α2 · v(m1)

v(n1)= ρα1−α2 = ρνp(

mn )



3. The replenishment of a valuation field. A classical result claimthat every topological space admits a replenishment. Now we constructa replenishment for a valuation field and prove that the replenishment isunique (up to a topological isomorphism).

Let (F, v) be a valuation field. A sequence ann>1, where an ∈ F iscalled fundamental, if for every ε > 0 there exists N such that for everyn,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 . Asequence ann>1 is fundamental in v1 if and only if it is funda-mental in v2.

(2) If ann>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, thenthere exists x ∈ F such that v1(x) = α > 1. Then for every ε > 0 thereexists m ∈ N such that α−m < ε. If the sequence ann>1 is fundamentalin v1, then xm · ann>1 is also fundamental in v1, so there exists N ∈ Nsuch that for every n, k > N we have v1(xman − xmak) < 1. Whencev2(xman − xmak) < 1, and so v2(an − ak) < α−m < 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 sequencein G is converging, i.e. it has the limit in F . If a valuation field is embeddedinto a complete valuation field (F , v) so that F is dense in F then (F , v) iscalled a replenishment of (F, v).

Theorem 4.12. For every valuation field (F, v) there exists a uniqueup to isomorphism replenishment (F , v).

Proof. We construct a replenishment using the standard constructionfor the real numbers. Let

F = ann>1 | ann>1 is a fundamental sequence of (F, v)be a set of all fundamental sequences consisting of elements from F . Definethe addition and multiplication on F by

ann>1 + bnn>1 = an + bnn>1,ann>1 · bnn>1 = an · bnn>1,


Clearly both an + bnn>1 and an · bnn>1 are fundamental sequence. Weprove, for example, that an ·bnn>1 is fundamental provided both ann>1,bnn>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 ann>1, bnn>1 are fundamental sequences, Lemma 4.11(2) impliesthat v(an)n>1, v(bn)n>1 are fundamental sequences as well. Thereforethe sequences v(an)n>1 and v(bn)n>1 are uniformly bounded by anabsolute constant C. Since for every ε there exists N such that for everym,n > N we have v(an − am) < ε

2C and v(bn − bm) < ε2C . So

v(an − am) · v(bn) + v(bn − bm)v(am) <ε

2+ε

2= ε,

hence an · bnn>1 is fundamental.Thus F is a commutative ring with 1. Consider

I0 = ann>1 | an →n→∞

0.

Clearly I0 is an ideal of F . We show that F/I0 is a field. Since F is acommutative ring with 1, we remain to prove that each nonzero element ofF/I0 has inverse. Consider ann>1+I0 6= 0+I0. Since ann>1 ∈ F\I0, itfollows that an 6→

n→∞0, i.e. there exists ε > 0 and N ∈ N such that for every

n > N we have v(an) > ε. Consider the sequence 1ann>N . The sequence

is well-defined, since an 6= 0 for n > N . Moreover, for every n,m > N wehave

v

(1

an− 1

am

)=v(am − an)

v(an · am)6v(am − an)

ε2,

so the sequence 1ann>N is fundamental, i.e. 1

ann>N ∈ F . Define the

sequence bnn>1 by: bn = 1 for n < N and bn = 1an

for n > N . Then

bnn>1 ∈ F and an · bn − 1 = 0 for all n > N . Hence,

(ann>1 + I0) · (bnn>1 + I0) = an · bnn>1 + I0 = 1n>1 + I0,

i.e. bnn>1 + I0 = (ann>1 + I0)−1. Therefore F = F/I0 is a field. The

embedding F → F is defined by a 7→ ann>1 + I0, where an = a for alln ∈ N (we use the notation an>1 below). Define a valuation on F in thefollowing way:

if α = ann>1 + I0, then v(α) = limn→∞

v(an).


Clearly, v does not depend on the representative of a coset ann>1 + I0,since the identity

limn→∞

(v(an) + v(bn)) = limn→∞

v(an) + limn→∞

v(bn)

holds. It is also evident that the restriction of v on F coincides with v. Nowwe check the properties of a valuation.

(1) v(ann>1 + I0) > 0 and v(ann>1 + I0) = 0 if and only ifann>1 +I0 = I0, i.e. an →

n→∞0. Indeed, since for each n we have

v(an) > 0, it follows that limn→∞ v(an) = v(ann>1 + I0) > 0. Ifv(ann>1 + I0) = 0, then limn→∞ v(an) = 0, so ann>1 ∈ I0.

(2)

v(ann>1 + I0 + bnn>1 + I0) =

v(an + bnn>1 + I0) =

limn→∞

v(an + bn) 6

limn→∞

(v(an) + v(bn)) =

limn→∞

v(an) + limn→∞

v(bn) =

v(ann>1 + I0) + v(bnn>1 + I0).

(3)

v(ann>1 + I0) · v(bnn>1 + I0) =

v(an · bnn>1 + I0) =

limn→∞

v(an · bn) =

limn→∞

v(an) · v(bn) =

limn→∞

v(an) · limn→∞

v(bn) =

v(ann>1 + I0) · v(bnn>1 + I0).

So v is a valuation.Now we show that F is dense in F . If α = ann>1 is a fundamental

sequence, then the sequence αnn>1, where αn = ank>1 ∈ F (recall thatank>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 thatαnn>1 is a fundamental sequence in F . Since F is dense in F , for everyn we can choose an ∈ F such that v(αn−ank>1) < 1

n . Let α = ann>1.

§ 4.2. Construction and properties of p-adic fields 79

We claim that α ∈ F (so α+ I0 ∈ F ) and αn →(v)

α. Indeed, for every ε

there exists M such that v(αn − αm) < ε3 for all m,n > M . Further, there

exists N >M such that 1N < ε

3 . So for n,m > N we have

v(an − am) = v(ank>1 − αn + αn − αm + αm − amk>1) 6

v(ank>1 − αn) + v(αn − αm) + v(αm − amk>1) < ε,

so α is a fundamental sequence. Now, by construction, we have

v(αn − α) 6 v(αn − ank>1) + v(ank>1 − α) 6

1

n+ limm→∞

v(an − am).

Since for every ε > 0 there exists N such that 1N < ε

2 and for every n,m >N , v(an − am) < ε

2 , we obtain that for every n > N the inequality v(αn −α) < ε holds. Thus αn →

(v)α. This completes the proof of existence.

Now we prove the uniqueness. Let (F1, v1) and (F2, v2) be two replenish-ments of (F, v). We want to construct a topological isomorphism σ. Defineσ 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 ann>1, whereeach an lies in F . We extend σ to the map σ : F1 → F2 by assuming that forα ∈ F1 such that α = limn→∞ an that σ(α) = limn→∞ σ(an). Clearly thisdefinition is correct, since σ(an)n>1 is a fundamental sequence in F2 andsince for two fundamental sequences ann>1, bnn>1 with limn→∞ an =α = limn→∞ bn the sequence cnn>1, where c2n−1 = an, c2n = bn, is fun-damental and limn→∞ cn = α. It is technical to prove that σ is a bijectionand that σ preserves the operation, and we leave the detailed proof to thereader.

Exercise 4.13. Complete the technical details of the proof of Theorem4.12.

§ 4.2. Construction and properties of p-adic fields

Clearly, if valuations v1 and v2 of a field F are equivalent, then thereplenishments of (F, v1) and (F, v2) are topologically isomorphic. Recallthat we denote the class of equivalent valuations vp,ρ of Q by vp. A replen-

ishment 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


isomorphism. In this section we construct Qp explicitly and derive some itsproperties.

1. Ring of p-adic integers and its properties. Throughout we fixa prime p. Consider

Zp =ann>0 | an ∈ Z, an ≡ an−1 (mod pn) for n > 1

.

Define addition and multiplication on Zp as the addition and the mul-tiplication of sequences, i.e. ann>0 + bnn>0 = an + bnn>0 andann>0 · bnn>0 = an · bnn>0. Clearly Zp is a ring under these op-erations. Consider

Ip =ann>0 ∈ Zp | an ≡ 0 (mod pn+1)

.

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→ ann>0, wherea0 = a1 = . . . = m. Clearly this embedding preserves the addition andmultiplication. Moreover, the unit of Z coincides with the unit of Zp. Now

we choose a canonical representation for every element from Zp. Namely, forevery sequence ann>0 ∈ Zp there exists a unique sequence xnn>0 ∈ Zpsuch that for every n we have 0 6 xn < pn+1 and an ≡ xn (mod pn+1),in particular ann>0 + Ip = xnn>0 + Ip. The sequence xnn>0 with0 6 xn < pn+1 is a canonical representative of the coset ann>0 + Ip.

Theorem 4.14. Let Zp be the ring of p-adic integers. Then the fol-lowing hold.

(1) An element xnn>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 ∈ Z∗psuch 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 xnn>0 6∈ Z∗p.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) = 1for every n. Therefore, for every n we have gcd(xn, p

n+1) = 1. Hence, forevery n > 0 there exists 0 < yn < pn+1 such that xn · yn ≡ 1 (mod pn+1).Since xn ≡ xn−1 (mod pn) and both xn−1 · yn−1 and xn · yn are equivalent


1 modulo pn, it follows that yn ≡ yn−1 (mod pn), i.e. ynn>0 ∈ Zp. Sincexn · yn ≡ 1 (mod pn+1) it follows that there exists zn such that xn · yn =1 + zn · pn+1. Therefore xnn>0 · ynn>0 = 1n>0 + zn · pn+1n>0 andthe second summand in the right hand side lies in Ip.

(2) If α = xnn>0 ∈ Zp \ 0, then there exists a minimal n such thatxn 6= 0. If n = 0 then α = p0ξ and ξ = α. Suppose n > 0, then, bydefinition, xn ≡ xn−1 (mod pn) and in view of our choice xn−1 = 0, so xn isdivisible by pn and xn 6≡ 0 (mod pn+1), in particular, xn

pn 6≡ 0 (mod p) and

gcd(xn/pn, p) = 1. Now xn+1 ≡ xn (mod pn+1), so xn+1/p

n 6≡ 0 (mod p).Arguing in the same way by induction for every k > n we obtain thatxk/p

n 6≡ 0 (mod p). Moreover, since 0 6 xk < pk+1, we have 0 < xk/pn <

pk−n+1. Consider

ξ =

(xnpn,xnpn, . . . ,

xnpn,xn+1

pn,xn+2

pn, . . .

)(first n+ 1 terms are equal).

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 thefirst nonzero term of pn · ξ has number n, while the first nonzero term ofpm · ζ has number m, so n = m. Let ξ = xnn>0 and ζ = ynn>0. Sincepn · ξ = pn · ζ, we obtain that xm = ym for m > n. Moreover

xn−1 = (xn mod pn) = (yn mod pn) = yn−1.

Arguing in the same way we obtain that for all i < n the equality xi = yiholds (indeed, if xi = yi, then xi−1 = (xi mod pi) = (yi mod pi) = yi−1).Thus ξ = ζ.

(3) Assume that α, β ∈ Zp \0 and α ·β = 0. In view of item (2) of thetheorem, α = pn ·ξ, β = pm ·ζ, hence α·β = pn+m(ξ ·ζ) = 0. Multiplying theleft hand side of the identity by ξ−1 ·ζ−1 we obtain pn ·pm = 0. But (n+m)-th term of pn+m equals pn+m 6≡ 0 (mod pn+m+1), a contradiction.

Corollary 4.15. An integer a ∈ Z is invertible in Zp if and only ifgcd(a, p) = 1.

2. The field of p-adic rationals is the replenishment of rationalsin p-adic metric. Corollary 4.15 implies that rational numbers of the formab 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.


In view of Theorem 4.14 all elements of Zp have form pn · ξ, so we needto add the inverses for powers of p, i.e.

Qp = pn · ξ | n ∈ Z, ξ ∈ Z∗p ∪ 0. (4.4)

Given α = pn · ξ ∈ Q∗p set νp(α) = n, and let νp(0) = ∞. For brevity

we use ν instead of νp below. Choose 0 < ρ < 1 and set vp(α) = ρνp(α).Every a

b ∈ Q can be uniquely written as pn a1b1 , where p does not divide

a1 · b1. Then both a1 and b1 belong to Z∗p, and thus a1b1

= a1b−11 = ξ is an

invertible p-adic integer. Therefore, ab corresponds to pnξ ∈ 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 follow-ing 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 thatn > m. Then we have

α+ β = pn · ξ + pm · ζ = pm(pn−m · ξ + ζ),

and pn−m · 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 equalto 0, therefore pn−m · ξ + ζ is invertible in Zp and ν(α+ β) = m.

(3) First we check that vp is a valuation. We have

vp(α+ β) 6 ρmin(ν(α),ν(β)) = max(vp(α), vp(β)) 6 vp(α) + vp(β).vp(α · β) = ρν(α)+ν(β) = 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 weleave the details of the proof for the reader.

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

Theorem 4.18. The valuation field (Qp, vp) is a replenishment of(Q, vp,ρ).


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

Qp. Let αnn>1 be a fundamental sequence in Qp and every αn has the

form pmn ·ξn, where mn ∈ Z and ξn ∈ Z∗p∪0. Since the sequence αnn>1

is fundamental, it follows that vp(αn − αm) →n,m→∞

0. Now vp(αn − αm) =

ρν(αn−αm). If the sequence mnn> is not stabilizing, then for every n thereexists k > n such that mn 6= mk. In view of Theorem 4.16(2) we obtain

ρν(αn−αm) = ρmin(mn,mk) →n,k→∞

0.

Therefore mn →n→∞

∞, i.e. αn →(vp)

0. Assume that the sequence mnn>1

is stabilizing, i.e. there exist N ∈ N,M ∈ Z such that for every n > N wehave mn = M . Then for every n,m > N we have

vp(αn − αm) = vp(pM (ξn − ξm)) = vp(p

M ) · vp(ξn − ξm),

i.e. the sequence ξnn>1 is fundamental. Hence for every m ∈ N thereexists Nm ∈ N such that for all n1, n2 > Nm we have vp(ξn1 − ξn2) < ρm.

So ξn1− ξn2

= pm+1 · ξ, where ξ ∈ Zp. Let ξn = x(n)k k>0, consider

ξ = xkk>0, where xm := x(Nm)m . We claim that ξ ∈ Zp and that ξn →

n→∞ξ.

Show that xm−1 ≡ xm (mod pm) (and so ξ ∈ Zp). Indeed, by definition,

xm−1 = x(Nm−1)m−1 , i.e. for each k > Nm−1 we have v(ξNm−1 − ξk) < ρm−1,

therefore ξNm−1− ξk = pm · ζ for some ζ ∈ Zp. In particular, x

(Nm−1)m−1 −

x(k)m−1 ≡ 0 (mod pm). Choose k = Nm, then, since ξNm ∈ Zp, we have

x(Nm)m−1 ≡ x

(Nm)m (mod pm), whence x

(Nm−1)m−1 ≡ x

(Nm)m (mod pm). Now if k >

Nm, then vp(ξk − ξ) 6 ρm, so ξn →(vp)

ξ and αn →(vp)

pM · ξ.

We remain to show that Q is dense in Qp. Let α = pn · ξ, where n ∈ Zand ξ = (x0, x1, x2, . . .) ∈ Z∗p. Set αm = pn · xm ∈ Q. Then

vp(αm − α) = ρn · vp(0, . . . , 0, xm − xm+1, xm − xm+2, . . .) 6 ρn+m →

m→∞0,


Exercise 4.19. Prove that for every α ∈ Zp there exists a sequence

x(m)m>0, x(m) ∈ Z, such that x(m)−α ∈ pm+1Zp (in particular, x(m) →(vp)

α as m→∞).


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

Theorem 4.20. Assume that F (x1, . . . , xn) ∈ Z[x1, . . . , xn] is a poly-nomial with integer coefficients. The congruence

F (x1, . . . , xn) ≡ 0 (mod pm) (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 ∈Zp. Denote by Im the ideal pm · Zp of Zp. Then for every m there exist

x(m)1 , . . . , x

(m)n ∈ Z such that αi + Im = x

(m)i + Im (this statement follows

from Exercise 4.19). Therefore

F (x(m)1 , . . . , x(m)

n ) + Im = F (α1, . . . , αn) + Im = 0.

Now F (x(m)1 , . . . , x

(m)n ) is an integer, so F (x

(m)1 , . . . , x

(m)n ) ∈ Im ∩ Z =

pm · Z, i.e. x(m)1 , . . . , x

(m)n 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.

Consider the sequence (x(m)1 , . . . , x

(m)n )m>0 of solutions for congruences

(4.5). Notice that we may choose a subsequence (x(mk)1 , . . . , x

(mk)n )k>0

such that x(mk)i is converging to αi ∈ Zp for 1 6 i 6 n. Indeed, con-

sider zero coordinates of (x(m)1 , . . . , x

(m)n )m>0. We obtain the set of tu-

ples of the form (a1, . . . , an), where 0 6 ai < p for each i = 1, . . . , n. Since

(x(m)1 , . . . , x

(m)n )m>0 is infinite, there exists an n-tuple (a1, . . . , an) such

that the zero coordinate of x(m)i equals ai for infinitely many m and they

form a subsequence (x(m0)1 , . . . , x

(m0)n )m0>0 of (x(m)

1 , . . . , x(m)n )m>0. We

repeat the arguments for the first coordinates of (x(m0)1 , . . . , x

(m0)n )m0>0

and derive an infinite subsequence (x(m1)1 , . . . , x

(m1)n )m1>0. We repeat the

arguments for every k > 0. Now we take any element from k-th subsequence

and obtain the subsequence (x(mk)1 , . . . , x

(mk)n )k>0 of (x(m)

1 , . . . , x(m)n )m>0

satisfying the condition: for every i = 1, . . . , n the first k coordinates of

x(mt)i are equal to the coordinates of x

(mk)i for every t > k. Therefore

vp(x(mt)i − x(ms)

i ) < ρt →t,s→∞

0, i.e. for every i the sequence x(mk)i k>0

§ 4.3. Problems 85

is fundamental. So the sequences x(mk)1 k>0, . . . , x(mk)

n k>0 have limits

α1, . . . , αn respectively and, by construction, each αi lies in Zp. Again byconstruction 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 Hausdorffspace. Find a topological field that is not homeomorphic to avaluation field, i.e. there exist topological fields that do not possessa valuation inducing the same topology.

2. Prove that (Zp, vp) is a compact topological space.

3. How many distinct solutions in Z5 has the equation x2 + y2 = 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

an →(v)

a, 72

C, 4

f g, 37

deg, 4

g | f , 4

e(n), 36

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

Qp, 81

R, 4

∼, 30

vα, 71vp,ρ, 71

Z, 4ζ(z), 35

Zp, 80

87

Index

abelian group, 56

algebraic

integer, 6

number, 6

arithmetic function, 36

asymptotically equivalent functions, 30

character modulo m, 60

character of a finite abelian group, 58

Chebyshev function, 31

integral, 31

complete field, 76

conjugate numbers, 7

convolution product, 37

cyclic group, 56

degree

of a Diophantine approximation, 14

of a polynomial, 4

of an algebraic number, 7

Diophantine approximation, 14

Dirichlet approximation theorem, 15

Dirichlet series, 35

Dirichlet theorem, 68

division algorithm, 4

Euler function, 57

Euler identity, 38

factor ring, 6

field of p-adic numbers, 81

field of p-adic numbers (p-adic field),

79

field of fractions, 5

Hermite identity, 20

homomorphism

of groups, 58

ideal, 5

identity function, 36

isomorphism of groups, 57

leading coefficient, 4

Lindemann theorem, 27

Liouville theorem, 18

L-series of character, 60

Mobius function, 36

Mangoldt function, 34

maximal ideal, 6

minimal polynomial, 7

monic polynomial, 4

multiplicative function, 36

p-adic valuation, 71

prime-counting function, 30

principal character, 58

principal ideal, 5

domain, 5

Riemann

hypothesis, 55

zeta-function, 35

ring of p-adic integers, 80

symmetrized tuple, 25

88

Index 89

transcendental number, 6

triangle inequality, 71

trivial valuation, 71

valuation, 71

valuation field, 71

ELEMENTS OF ANALYTIC NUMBER THEORY - nsc.ruvdovin/lectures/numth_eng.pdf · 2018-11-28 ·...

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