Dirichlet’s Theorem and Algebraic Number Fields

Pedro Sousa Vieira

February 16, 2012

Abstract

In this paper we look at two different fields of Modern Number Theory:Analytic Number Theory and Algebraic Number Theory. Making use ofanalytic methods we demonstrate a classic problem relating prime numberswith arithmetic progressions: Dirichlet’s Theorem. Using algebraic methodswe investigate the structure and properties of algebraic number fields, thebasis of Algebraic Number Theory.

Contents

1 Introduction 21.1 Context and motivation . . . . . . . . . . . . . . . . . . . . . 21.2 Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Dirichlet’s Theorem 32.1 Characters of Finite Abelian Groups . . . . . . . . . . . . . . 32.2 Zeta and L-functions . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Zeta function . . . . . . . . . . . . . . . . . . . . . . . 52.2.2 L-functions . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 Dirichlet’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 7

3 Algebraic Number Fields 103.1 Algebra Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Algebraic Number Fields and the Ring of Algebraic Integers . 11

4 Examples of Algebraic Number Fields 154.1 Quadratic Number Fields . . . . . . . . . . . . . . . . . . . . 154.2 Cyclotomic Fields . . . . . . . . . . . . . . . . . . . . . . . . . 17

Bibliography 19

1

Chapter 1

Introduction

1.1 Context and motivation

The aim of this paper is to introduce two important branches of ModernNumber Theory: Analytic Number Theory and Algebraic NumberTheory. As the names suggest, the first one deals with number theoreticproblems from an analytic perspective (making use of integrals, series, etc)while the second one studies algebraic structures intimately related withNumber Theory such as rings of algebraic integers.

This report is a summary of the work I developed for the course Projectoem Matematica at Instituto Superior Tecnico as a 3rd year undergraduatestudent of Licenciatura em Matematica Aplicada e Computacao under thesupervision of Professor Gustavo Granja, to whom I am very grateful for allthe collaboration, help and support.

1.2 Contents

The paper is divided into three separate chapters. Chapter 2 is dedicatedto proving Dirichlet’s Theorem involving primes in arithmetic progressionsfollowing the classic proof of Dirichlet himself. The proof relies essentiallyon analytic methods, especially on convergence and divergence of Dirichlet’sL-functions. In Chapter 3 we introduce basic concepts of Algebraic NumberTheory with special focus on rings of algebraic integers. We’ll discuss someproperties of these rings and prove that they are Dedekind rings. Finally, inChapter 4 we illustrate two simple but important classes of algebraic numberfields: quadratic fields and cyclotomic fields.

2

Chapter 2

Dirichlet’s Theorem

In this chapter we illustrate a proof of Dirichlet’s Theorem, an outstand-ing and classic problem in Number Theory relating prime numbers witharithmetic progressions.

Theorem 1 (Dirichlet).Given a and m relatively prime positive integers there exist infinitely

many primes p such that p ≡ a (mod m).

As a matter of fact we will prove something a little stronger: primenumbers are ”equally” distributed amongst the invertible classes modulom. Even though this result is somewhat natural, its demonstration is nowalk in the park. The method followed will be the one of Dirichlet himselfwhich uses the properties of the Zeta and L-functions as well as modularcharacters. For a more detailed demonstration the reader should consult[JPS].

2.1 Characters of Finite Abelian Groups

In this section we will talk about characters of finite abelian groups, a simpletool needed for the demonstration of Dirichlet’s Theorem 1. In what follows,let G be a finite abelian group written multiplicatively with identity e.

Definition 1. A character of G is a homomorphism of G into the multi-plicative group C∗. The character χ1 such that χ1(g) = 1 for all g ∈ G iscalled the main character.

The characters of G form a group Hom(G,C∗) known has the dual of Gwhich we’ll denote by G.

Notice that if |G| = n then it follows that (χ(g))n = χ(gn) = χ(e) = 1for all g ∈ G and any character χ of G (remember that gn = e by Lagrange’sTheorem), i.e., the image of a character of G is a subset of the set of nthroots of unity.

3

1 + 5Z 2 + 5Z 3 + 5Z 4 + 5Zχ1 1 1 1 1

χ2 1 i −i −1

χ3 1 −1 −1 1

χ4 1 −i i −1

Example 1. Considering G = (Z/5Z)∗ we have the following 4 characters:

There are two very important properties of characters which are statedin the following propositions. The second one will be particularly useful inthe proof of Dirichlet’s Theorem 1.

Proposition 1. The dual group G is isomorphic to G.

This fact can be easily proven by decomposing G into a product of cyclicsubgroups (using the classification of finite abelian groups). For a cyclicgroup with generator g one can easily check that φ : G → G which assignsχ to χ(g) is an isomorphism.

Proposition 2 (Ortogonality relations). Let |G| = n and χ ∈ G. Then:

∑x∈G

χ(x) =

{n if χ = χ1

0 if χ 6= χ1

Proof. The result is trivial if χ = χ1. If χ 6= χ1 let y ∈ G be such thatχ(y) 6= 1. In this case:∑

x∈Gχ(x) =

∑x∈G

χ(xy) = χ(y)∑x∈G

χ(x)

Since χ(y) 6= 1, it follows that∑

x∈G χ(x) = 0 as intended.

From Proposition 2 above applied to G we get the following corollary.

Corollary 1. Let |G| = n and x ∈ G. Then:

∑χ∈G

χ(x) =

{n if x = e

0 if x 6= e

Given m ≥ 1 we denote by G(m) the multiplicative group of invertibleelements mod m, that is (Z/mZ)∗. In the proof of Dirichlet’s Theorem 1we will consider a particular class of characters known as Modular Char-acters.

Definition 2 (Modular Characters). A modular character is a character ofG(m) for some m ≥ 1. An element χ ∈ G(m) is called a character modulom.

4

Notice that we can view a character modulo m, χ, as a totally multi-plicative function with values in C∗ defined on the set of integers prime tom. It will be convenient for the proof of Dirichlet’s Theorem 1 to extendsuch a function to all of Z by setting χ(a) = 0 if a is not prime to m. Noticethat this function is still totally multiplicative.

Example 2. m = 5. Since G(5) = Z/5Z the characters modulo 5 are theones stated in Example 1. Notice that they extend to functions in Z in thefollowing way:

ψj(x) =

{χj(x+ 5Z) if 5 - x0 if 5 | x

for j = 1, 2, 3, 4.

2.2 Zeta and L-functions

In this section we will talk about the Zeta and L functions which are veryimportant not only for the proof of the Dirichlet Theorem 1 but also in manybranches of Mathematics. For example, the famous Riemann Hypothesisconcerns the locus of zeros of the Zeta function.

In the following we will ignore most questions regarding convergence ofseries or infinite products as these may be tedious. For more rigorous proofsthe reader should consult [JPS] or [MRM].

2.2.1 Zeta function

Definition 3 (Zeta function). For s ∈ C such that <(s) > 1 we define:

ζ(s) =∞∑n=1

1

ns

Absolute convergence of the series for <(s) > 1 follows from comparisonwith the integral of 1

x<(s). An interesting property of the Zeta function is that

it factors as an infinite product due to the unique factorization of integersinto prime numbers.

Proposition 3. For s ∈ C such that <(s) > 1:

ζ(s) =∏

p prime

1

1− 1ps

Proof. Omitting the convergence details we have, due to the unique factor-ization of integers into prime numbers:

ζ(s) =

∞∑n=1

1

ns=

∏p prime

∞∑k=0

p−sk =∏

p prime

1

1− 1ps

5

We can still give a better characterization of the Zeta function.

Proposition 4. The Zeta function is non-zero in the half plane <(s) > 1and verifies for <(s) > 0:

ζ(s) =1

s− 1+ φ(s)

where φ(s) is holomorphic for <(s) > 0.

A proof of the preceding proposition can be found in [JPS]. As a corollarywe obtain:

Corollary 2. As s→ 1, one has∑

p prime

1ps ∼ log

(1s−1

).

Proof. From proposition 3 we have:

log ζ(s) = log

∏p prime

(1− p−s)−1 = −

∑p prime

log(1− p−s

)=

=∑

p prime

∑k≥1

1

kpks=

∑p prime

1

ps+ ψ(s)

where ψ(s) =∑

p prime

∑k≥2

1kpks

. Notice that ψ(s) is dominated by the

series:∑p prime

∑k≥2

1

pks=

∑p prime

1

ps(ps − 1)≤

∑p prime

1

p(p− 1)≤∞∑n=2

1

n(n− 1)= 1

Since ψ(s) is majored as s→ 1, by proposition 4, log ζ(s) ∼ log 1s−1 and the

conclusion follows.

Observe that we can conclude from corollary 2 that the set of primenumbers is infinite. The argument used to prove Dirichlet’s Theorem 1 is

similar. It uses the fact that∑

p≡a (mod m)

1ps ∼

1φ(m) log

(1s−1

)as s→ 1 where

φ is Euler’s totient function.

6

2.2.2 L-functions

In the proof of the Dirichlet Theorem 1 we make use of the Dirichlet’s L-functions. These functions make use of the modular characters discussed insection 2.1.

Let m ≥ 1 and let χ be a character modulo m. We define the corre-sponding L-function by the series:

L(s, χ) =∞∑n=1

χ(n)

ns

Observe that, in this sum, the only integers n that give a non-zero con-tribution are the ones which are prime to m. Note also that for m = 1 theonly character modulo m is the main character χ1 and for this characterL(s, χ1) = ζ(s). For m > 1 we can also give a characterization of L(s, χ1)in terms of ζ(s).

Proposition 5. For m ≥ 1 one has:

L(s, χ1) = F (s)ζ(s)

where F (s) =∏p|m

(1− 1

ps

). From proposition 4, since F (s) is entire, L(s, χ1)

is holomorphic in <(s) > 0 and has a simple pole at s = 1.

When we consider characters modulo m other than χ1, the correspondingL-functions are somewhat better behaved.

Proposition 6. For χ 6= χ1, L(s, χ) converges in the half plane <(s) > 0.In the half plane <(s) > 1, L(s, χ) converges absolutely and one has:

L(s, χ) =∏

p prime

1

1− χ(p)ps

Notice that, in particular, L(1, χ) is finite if χ 6= χ1. In fact, for χ 6= χ1,L(1, χ) in non-zero. This is one of the key points of Dirichlet’s proof. Aproof of this result can be found in [JPS].

2.3 Dirichlet’s Theorem

In this section we prove Dirichlet’s Theorem 1 using the tools developed inthe previous sections. Let P be the set of prime numbers. We know fromCorollary 2 that as s→ 1 ∑

p∈P

1

ps∼ log

1

s− 1

7

It is then natural that given A ⊆ P we define its density in P as

lims→1

∑p∈A

1ps

log 1s−1

whenever the limit exists. Notice that P has density 1 and that if A is finitethen its density is zero.

The idea of Dirichlet’s proof consists in showing that, for m ≥ 1 and aprime with m, the set of prime numbers such that p ≡ a (mod m), hereindenoted Pa,m, has density 1

φ(m) where φ is Euler’s totient function. Thus,Pa,m can’t be finite as it would have density zero. To prove this, we willfirst need three Lemmas.

Let χ be a character modulo m. We set

fχ(s) =∑p - m

χ(p)

ps

this expression making sense for s > 1.

Lemma 1. As s→ 1 we have fχ1 ∼ log 1s−1 .

This follows from Corollary 2 because fχ1 differs from∑

p prime

1ps by a

finite number of terms.

Lemma 2. If χ 6= χ1 then fχ remains bounded when s→ 1.

This follows from Proposition 6 using a similar argument as the one usedin the proof of Corollary 2 and the fact that L(1, χ) 6= 0. A proof can befound in [JPS] and in [MRM].

Now let ga(s) =∑

p∈Pa,m

1ps . We wish to study the behavior of ga(s) as

s→ 1.

Lemma 3. One has:

ga(s) =1

φ(m)

∑χ∈G(m)

χ(a)−1fχ(s)

Proof. After replacing fχ by its definition, the right hand side becomes:

1

φ(m)

∑p - m

1

ps

∑χ∈G(m)

χ(a−1)χ(p)

Since χ(a−1)χ(p) = χ(a−1p) we have by Corollary 1:

∑χ∈G(m)

χ(a−1p) =

{φ(m) if p ≡ a (mod m)

0 if p 6≡ a (mod m)

8

Since gcd(a,m) = 1 we find that the above expression is equal to ga(s) asintended.

We are now ready to show that Pa,m has density 1φ(m) . From Lemmas 1

and 2 we know that fχ ∼ log 1s−1 for χ = χ1 and that fχ remains bounded

for χ 6= χ1. Hence, by Lemma 3 it is clear that ga(s) ∼ 1φ(m) log 1

s−1 , which

means that Pa,m has density 1φ(m) .

We can interpret this result as meaning that the prime numbers aresomewhat equally distributed between the different classes modulo m whichare prime to m.

9

Chapter 3

Algebraic Number Fields

In this chapter we will study algebraic number fields. In the first section we’llintroduce some algebra tools needed for the rest of the section, as well as forchapter 4. These include important concepts as the norm, the trace and thediscriminant. In the second section we’ll be interested in the structure ofthe ring of algebraic integers of a number field, following more or less whatis covered on chapter 12 of [IR].

3.1 Algebra Tools

Let K and L be fields such that L/K is a finite algebraic extension of fields.The dimension of L/K will be denoted by n. To each element α ∈ L we canassign two elements of K with good properties: the norm NL/K(α) and thetrace tL/K(α).

Definition 4. Let α1, ..., αn be a basis for L/K and let aij ∈ K be such that

ααi =n∑j=1

aijαj. We define the norm of α by NL/K(α) := det(aij) and the

trace of α by tL/K(α) := a11 + ...+ ann.

It is an easy exercise to check that this definition does not depend onthe choice of basis α1, ..., αn. In the following we will use N and t to denotethe norm and the trace whenever the extension L/K is clear. The goodproperties of the norm and the trace we mentioned above are the ones weexpect from the usual norm (determinant) and trace of matrices.

Proposition 7. Let α, β ∈ L and a ∈ K. Then the following holds:

1. N(αβ) = N(α)N(β);

2. t(α+ β) = t(α) + t(β);

3. N(aβ) = anN(β);

10

4. t(aα) = at(α);

5. If α 6= 0 then N(α)−1 = N(α−1).

In the case when L is a separable extension over K (that is, the minimalpolynomial over K of any element in L has distinct roots) we can give analternative definition of norm and trace. In fact, let σ1, ..., σn be the distinctisomorphisms of L into a fixed algebraic closure of K which leave K fixed.For α ∈ L we denote the jth conjugate of α, σj(α), by α(j) and considerα(1) = α.

Proposition 8. For any α ∈ L one has:

1. t(α) = α(1) + ...+ α(n);

2. N(α) = α(1)...α(n).

Given an n-tuple α1, ..., αn of elements of L there is an important elementof L to which we can associate.

Definition 5. Let α1, ..., αn be elements of L. We define the discriminantof α1, ..., αn, ∆(α1, ..., αn), as det(t(αiαj)).

In a separable extension of fields the discriminant of an n-tuple is closelyrelated with it forming a basis of L over K.

Proposition 9. If L/K is separable then α1, ..., αn is a basis for L over Kif and only if ∆(α1, ..., αn) 6= 0.

As before, when L/K is separable we can give an alternative definitionfor the discriminant of an n-tuple in terms of its conjugates.

Proposition 10. For α1, ..., αn ∈ L and L/K separable one has

∆(α1, ..., αn) = det(α(j)i )2

3.2 Algebraic Number Fields and the Ring of Al-gebraic Integers

In this section we talk about algebraic number fields and their ring of al-gebraic integers. We’ll study the structure of the ring of algebraic integerswith the goal of showing that it is a Dedekind ring while making use ofimportant concepts in Algebraic Number Theory such as the class number.First, we must introduce some definitions.

Definition 6 (Algebraic Number Field). A subfield F of the complex num-bers is called an algebraic number field if F is a finite dimensional vectorspace over Q.

11

Definition 7 (Ring of Algebraic Integers). Let F be an algebraic numberfield. An element α ∈ F is said to be an algebraic integer if α is the rootof some monic polynomial in Z[x]. The subset of F consisting of algebraicintegers forms a ring D, called the ring of algebraic integers in F .

Example 3. F = Q is an algebraic number field because [Q : Q] = 1. Inthis case we have D = Z.

In fact, if pq ∈ D where p and q are coprime integers (q 6= 0) then there

is some polynomial r(x) = xm + am−1xm−1 + ...+ a1x+ a0 ∈ Z[x] such that

r(pq ) = 0. Multiplying both sides of r(pq ) = 0 by qm we get pm+am−1pm−1q+

...+a1pqm−1 +a0q

m = 0 implying that q | pm. Since p and q where assumedto be coprime we obtain that p

q ∈ Z and so D ⊆ Z. On the other hand, it isobvious that Z ⊆ D.

In a ring of algebraic integers D the norm and the trace defined in theprevious section are very simple, meaning that if α ∈ D then N(α) andt(α) are integers. To see this, notice that if α satisfies a monic polynomialin Z[x] then so do its conjugates, meaning that they are also in D. UsingProposition 8 we see that N(α) and t(α) are in D and using Definition 4we know that they are also in Q. Hence, from Example 3 we conclude thatN(α) and t(α) are integers.

Example 4. F = Q[√−5] is an algebraic number field because [Q[

√−5] :

Q] = 2. In this case, as we shall see in the next chapter, we have D =Z + Z

√−5. For a, b ∈ Z, if α = a + b

√−5 then N(α) = a2 + 5b2 and

t(α) = 2a.

We can see from Example 4 that D is not necessarily a unique fac-torization domain (UFD) because D = Z + Z

√−5 and we have 3 × 3 =

(2 +√−5)(2 −

√−5) and 3, 2 +

√−5 and 2 −

√−5 are irreducible in D.

However, D has a property which is almost as good as being a UFD: it isa Dedekind ring, i.e., every nonzero ideal of D can be written uniquely as aproduct of prime ideals.

We will briefly illustrate how to show this using the important notion inAlgebraic Number Theory of class number.

Definition 8. Two ideals A,B ⊂ D are said to be equivalent, A ∼ B, ifthere exist nonzero α, β ∈ D such that (α)A = (β)B. This is an equivalencerelation on the set of ideals of D. The equivalence classes are called idealclasses. The number of classes, hF , is called the class number of F .

Notice that hF = 1 if and only if D is a principal ideal domain (PID).Thus we can interpret the class number as measuring, in some sense, howfar D is from being a PID. We will show that hF is always finite using twoLemmas, the second of which is due to A. Hurwitz.

Lemma 4. For any nonzero ideal A, D/A is finite.

12

Lemma 5 (A. Hurwitz). There exists a positive integer M depending onlyon F with the following property. Given α, β ∈ D, β 6= 0 there is an integert, 1 ≤ t ≤M , and an element ω ∈ D such that |N(tα− ωβ)| < |N(β)|.

Proofs of these Lemmas can be found in [IR].

Theorem 2. The class number hF is finite.

Proof. Let A be an ideal in D. For α ∈ A, α 6= 0, we know that |N(α)|is a positive integer. Choosing β ∈ A nonzero such that |N(β)| is minimal,we know from Lemma 5 that there are t, 1 ≤ t ≤ M , and ω ∈ D such that|N(tα − ωβ)| < |N(β)|. Since tα − ωβ ∈ A we must have tα − ωβ = 0.It then follows that M !A ⊂ (β) or equivalently (β−1)M !A ⊂ D. Let B =(β−1)M !A. Then B is an ideal and M !A = (β)B. Since β ∈ A, one musthave M !β ∈ (β)B and so M ! ∈ B. Now, by Lemma 4, we know that M !can be contained in at most finitely many ideals (because D/(M !) is finite).Hence, we have shown that A ∼ B where B is one of at most finitely manyideals. Thus hF is finite, as we wanted to show.

Using that the class number is finite it is easy to demonstrate the fol-lowing proposition.

Proposition 11.

1. For any ideal A ⊂ D there is an integer k, 1 ≤ k ≤ hF such that Ak

is a principal ideal.

2. If A and B are ideals such that A ⊂ B then there exists and ideal Csuch that A = BC;

3. Every ideal can be written as a product of prime ideals.

Proof. We will only demonstrate 3 using 2. Let A be a proper ideal. Weknow from Lemma 4 that D/A is finite and so it is contained in some maxi-mal ideal P1. By 2) we know that there is some ideal B1 such that A = P1B1.If B1 6= D we can use similar reasoning to obtain a maximal ideal P2 and anideal B2 such that A = P1P2B2. If B2, B3, ... 6= D we can continue with thisprocess indefinitely. However, since A ⊂ B1 ⊂ B2 ⊂ ... is a proper ascendingchain of ideals we get contradiction with Lemma 4 because D/A if finite andso there must be a t such that Bt = D. Thus A = P1P2...Pt.

It should be stressed that one can show using Proposition 11 (1) thatthe ideal classes can be made into a group. The group structure on the idealclasses is a much studied problem in Modern Algebraic Number Theory.

We are almost done proving that D is a Dedekind ring. We’ve alreadydemonstrated that an ideal of D factors into a product of prime ideals. Itjust remains to show that this factorization is unique. For that, we need todefine the order of a prime ideal in another ideal, an analogous of the orderof a prime number in an integer.

13

Definition 9. Let P be a prime ideal and A an ideal. Then ordPA is definedto be the unique nonnegative integer t such that A ⊂ P t and A 6⊂ P t+1.

It is worthwhile mentioning that this definition makes sense, that is, thatsuch a t always exists and that it is unique. Also, it verifies three ratherexpected properties.

Proposition 12.

1. ordPP = 1;

2. If P ′ 6= P is prime then ordPP′ = 0;

3. ordPAB = ordPA+ ordPB.

Making use of Proposition 12 we can now demonstrate that D is aDedekind ring.

Theorem 3. Let A ⊂ D be an ideal. Then A =∏P

P a(P ) where the product

runs over all prime ideals of D and the a(P ) are nonnegative integers all butfinitely many of which are zero. The integers a(P ) are uniquely determinedby a(P ) = ordPA.

Proof. The product representation follows from 3 of Proposition 11. If P0

is a prime ideal, then to check that a(P0) = ordP0A we just need to applyordP0 to both sides of the product. Using Proposition 12 we obtain:

ordP0A =∑P

a(P )ordP0P = a(P0)

14

Chapter 4

Examples of AlgebraicNumber Fields

In this section we will briefly describe two very simple classes of algebraicnumber fields: quadratic number fields and cyclotomic fields. In quadraticnumber fields the concepts of norm, trace and discriminant are very simpleand will be computed in this section. Using them, we will be able to charac-terize the factorization into prime ideals of the ideals (p) where p is a primenumber. For cyclotomic fields, even though the notions of norm, trace anddiscriminant are not as simple as for quadratic fields, we will still be able tostudy the structure of the ideals (p). In the end, we will see that these twoclasses of algebraic number fields are in some sense related: any quadraticnumber field is contained in some cyclotomic field.

For more information on these algebraic number fields, the interestedreader should consult chapter 13 of [IR].

4.1 Quadratic Number Fields

A quadratic number field F is a number field such that [F : Q] = 2.Using the results from the previous section we will study the structure ofthe ring of integers D of a quadratic field F .

Since [F : Q] = 2 we know that F = Q[α] where α satisfies a quadraticequation in Q[x]. Using the formula for the roots of a quadratic polynomialone can see that in fact F = Q[

√d] where d is a square-free integer. It is

now obvious that, given α = a+ b√d ∈ F we have α(1) = α = a+ b

√d and

α(2) = a− b√d. Also, the norm and the trace of α are, by Proposition 8 of

the previous chapter, N(α) = a2 − db2 and t(α) = 2a.

Example 5. F = Q[√−1] = Q[i] is a quadratic number field. Given α ∈

Q[i], the norm of α is the square of the usual norm in the complex numbers.Its trace is t(α) = 2<(α).

15

In the previous chapter we saw that if α ∈ D then t(α) and N(α) areintegers. For quadratic number field the reciprocal is also true. If t(α) andN(α) are integers then (x−α(1))(x−α(2)) = x2− t(α)x+N(α) ∈ Z[x] andso α ∈ D. Thus α ∈ D if and only if t(α) ∈ Z and N(α) ∈ Z. Using thisresult, it is a simple exercise to describe D explicitly.

Proposition 13.If d ≡ 2, 3 (mod 4) then D = Z + Z

√d.

If d ≡ 1 (mod 4) then D = Z + Z(−1+

√d

2

).

With this result we can compute an important value of a quadraticnumber field known as the discriminant of D and denoted by δF . Given anintegral basis α1, ..., αn for D, that is, α1, ..., αn is a basis for F over Q andD = Zα1 + ...+Zαn, we define the discriminant of D as δF = ∆(α1, ..., αn).This definition is standard for any number field, not just quadratic numberfields. Since for any integral basis α1, ..., αn we have αi ∈ D, we know fromDefinition 5 that δF is an integer. Furthermore, from Proposition 9 it followsthat δF 6= 0. In particular, for quadratic number fields one has:

Proposition 14.If d ≡ 2, 3 (mod 4) then δF = 4d.If d ≡ 1 (mod 4) then δF = d.

Proof. Using Proposition 13 and Proposition 10 one has:

• If d ≡ 2, 3 (mod 4) then δF = ∆(1,√d) = det2

(1√d

1 −√d

)= 4d.

• If d ≡ 1 (mod 4) then δF = ∆(1, −1+√d

2 ) = det2

(1 −1+

√d

2

1 −1−√d

2

)= d.

We can now describe the factorization into prime ideals of some idealsin D in terms of δF . Let p be a prime in Z, P a prime ideal in D containingp and P (2) = {α(2)|α ∈ P} (which is easily seen to be a prime ideal in Dcontaining (p)).

Proposition 15. Suppose p is odd. Then:

1. If p - δF and x2 ≡ d (mod p) is solvable in Z then (p) = PP (2) andP 6= P (2).

2. If p - δF and x2 ≡ d (mod p) is not solvable in Z then (p) = P .

3. If p | δF then (p) = P 2.

Proposition 16. Suppose p = 2. Then:

16

1. If 2 - δF and d ≡ 1 (mod 8) then (2) = PP (2) and P 6= P (2).

2. If 2 - δF and d ≡ 5 (mod 8) then (2) = P .

3. If 2 | δF then (2) = P 2.

The proofs of these propositions can be found in [IR].

4.2 Cyclotomic Fields

Let m be a positive integer and let ζm = e2πim . Note that ζm satisfies the

polynomial equation xm − 1 = 0, as do all of its powers. Thus, one hasxm − 1 = (x − 1)(x − ζm)...(x − ζm−1m ). The field F = Q(ζm) is called thecyclotomic field of mth roots of unity and it is an algebraic numberfield.

Example 6. F = Q[v] where v = e2πi5 . Notice that x5 − 1 = (x − 1)(x4 +

x3 + x2 + x + 1) where the last factor is an irreducible polynomial over Z.Thus, [F : Q] = 4.

Example 7. F = Q[w] where w = e2πi6 . Notice that x6 − 1 = (x − 1)(x +

1)(x2−x+ 1)(x2 +x+ 1) where x2−x+ 1 = (x−w)(x−w5). Since w 6∈ Q,we have [F : Q] = 2. In fact F = Q[

√−3].

Examples 6 and 7 show that the degree of a cyclotomic extension doesnot necessarily increase with m. However, this degree is closely related withm and with the mth cyclotomic polynomial.

Definition 10. Let Φm(x) =∏

gcd(a,m)=1

(x − ζam) where 1 ≤ a ≤ m. This

polynomial is called the mth cyclotomic polynomial.

Notice that Φm(x) has as roots precisely the roots of unity with orderm and that its degree is φ(m). In fact φ(m) is also the degree of Q[ζm] overQ, as a result of the following Theorem.

Theorem 4. The mth cyclotomic polynomial Φm(x) is in Z[x] and is irre-ducible in Z[x].

From the above Theorem we can see that Φm(x) is the minimal polyno-mial of ζm in Z[x] and so [Q[ζm] : Q] = φ(m).

What can be said about the ring of integers D in Q[ζm]? We know thatZ[ζm] ⊆ D because Z ⊆ D and ζm ∈ D but is D = Z[ζm]? The answer isyes but the proof is not simple for general m. A proof when m is prime canbe found in [IR].

Like we did for quadratic number fields in the previous section, we nowwish to study more closely how certain ideals factor in D.

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Theorem 5. Let p be a prime such that p - m. Let f be the smallest positiveinteger such that pf ≡ 1 (mod m). Then in D ⊂ Q(ζm) we have

(p) = P1...Pg,

where each Pi is such that D/Pi has pf elements and g = φ(m)f .

Theorem 5 is a very pleasant result on the decomposition of primes whichdo not divide m. For primes which do divide m we consider only the specialcase when m = l is prime.

Proposition 17. Let l be a prime in Z. Then, in Q[ζl], one has (l) = Ll−1

where L = (1− ζl).

It is interesting to notice that the decomposition of (p) varies drasticallyfrom when p - m to when p | m. In the first case, (p) only decomposesinto unramified factors. In the second case, when m is prime, (p) ramifiescompletely.

To end this chapter we illustrate how quadratic number fields are closelyrelated with cyclotomic fields. First, let’s prove that if p is an odd primethen either

√p ⊂ Q[ζp] or

√−p ⊂ Q[ζp]. From the polynomial identity

1 + x+ ...+ xp−1 =xp − 1

x− 1=

p−1∏j=1

(x− ζjp)

substituting x = 1 we obtain

p =

p−1∏j=1

(1− ζjp)

Grouping together the terms corresponding to j and p − j in the followingway

(1− ζjp)(1− ζp−jp ) = (1− ζjp)(1− ζ−jp ) = −ζ−jp (1− ζjp)2

one gets

p = (−1)p−12 ζbp

p−12∏j=1

(1− ζjp)2, where b = −(1 + 2 + ...+p− 1

2)

Since p is odd, there is c ∈ Z such that 2c ≡ 1 (mod p). Thus, ζb = (ζbc)2.

It follows that (−1)p−12 p is a square in Q[ζp] and hence either

√p or

√−p

are in Q[ζp]. It is now straightforward to prove the next proposition:

Proposition 18. Let p be a prime number. Then Q[√p] ⊂ Q[ζ4p].

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Proof. For p = 2 the result is trivial. For p > 2 we just have to check that√p ∈ Q[ζ4p]. From what was said above we know that either

√p or√−p are

in Q[ζp] ⊂ Q[ζ4p]. Since i ∈ Q[i] ⊂ Q[ζ4p] we conclude that√p ∈ Q[ζ4p].

As a matter of fact, Proposition 18 can be more or less extended to allother quadratic fields.

Proposition 19. Any quadratic number field is contained in a cyclotomicfield.

Proof. Let’s consider the quadratic field Q[√d] where d = ±p1p2...pk where

the pj are prime numbers. Using Proposition 18 we get:

Q[√d] ⊂ Q[

√p1, ...,

√pk] ⊂ Q[ζ4p1 , ..., ζ4pk ] ⊂ Q[ζ4p1p2...pk ]

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Bibliography

[IR] K. Ireland and M. Rosen, A Classical Introduction to Modern Num-ber Theory, New York, Springer-Verlag, 1990

[JPS] J. P. Serre, A Course in Arithmetic, New York, Springer-Verlag, 1973

[MRM] M. Ram Murty, Problems in Analytic Number Theory, 2nd Edition,New York, Springer-Verlag, 2008

[TA] Tom M. Apostol, Introduction to Analytic Number Theory, NewYork, Springer-Verlag, 1976

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