Ideals and Class Groups of Number Fields

Home , Algebraic number field, Class number problem, Fractional ideal, Free abelian group, Ideal class group, Ideal theory

Ideals and class groups of number fields

A thesis submitted To Kent State University in partial Fulfillment of the requirements for the Degree of Master of Science

Minjiao Yang

August, 2018

Thesis written by Minjiao Yang B.S., Kent State University, 2015 M.S., Kent State University, 2018

Approved by Gang Yu , Advisor Andrew Tonge , Chair, Department of Mathematics Science James L. Blank , Dean, College of Arts and Science

TABLE OF CONTENTS…………………………………………………………...….…...iii

ACKNOWLEDGEMENTS…………………………………………………………....…...iv

CHAPTER

I. Introduction…………………………………………………………………...... 1

II. Algebraic Numbers and Integers……………………………………………...... 3

III. Rings of Integers……………………………………………………………...... 9

Some basic properties…………………………………………………………...9

Factorization of algebraic integers and the unit group………………….……....13

Quadratic integers……………………………………………………………….17

IV. Ideals……..…………………………………………………………………...... 21

A review of ideals of commutative……………………………………………...21

Ideal theory of integer ring 풪푘…………………………………………………..22

V. Ideal class group and class number…………………………………………...... 28

Finiteness of 퐶푙푘………………………………………………………………....29

The Minkowski bound…………………………………………………………...32

Further remarks…………………………………………………………………..34

BIBLIOGRAPHY…………………………………………………………….…………...... 36

iii

ACKNOWLEDGEMENTS

I want to thank my advisor Dr. Gang Yu who has been very supportive, patient and encouraging throughout this tremendous and enchanting experience. Also, I want to thank my thesis committee members Dr. Ulrike Vorhauer and Dr. Stephen Gagola who help me correct mistakes

I made in my thesis and provided many helpful advices.

Chapter 1. Introduction Algebraic number theory is a branch of number theory which leads the way in the world of mathematics. It uses the techniques of abstract algebra to study the integers, rational numbers, and their generalizations. Concepts and results in algebraic number theory are very important in learning mathematics.

Algebraic number theory started with Diophantine, Fermat, Gauss and Dirichlet. The beginnings of algebraic number theory can be traced to Diophantine equations, was conjectured by the 3rd-century famous mathematician, Diophantus, who studied them and developed methods for the solution of some of Diophantine equations. Diophantine equations have been studied for thousands of years and only a portion of his major work Arithmetica survived. Fermat’s last theorem, named after 17th century mathematician, Pierre de Fermat, was first discovered in the margin in his copy of an edition of Diophantus, and included the statement that the margin was too small to include the proof. No successful proof was published until 1995 and the unsolved problem stimulated the development of algebraic number theory in the 19th century. One of the founding works of algebraic number theory, Disquisitiones Arithmeticae was written by Carl

Friedrich Gauss in 1798, in which Gauss brings together the work of number theory from his predecessors with his own important new result into a systematic framework, filled in gaps and extended the subject in numerous ways.

Algebraic number theory reached its first peak in the work of Kummer about cyclotomic

1 fields. However, Kummer was not at all interested in a general theory of algebraic numbers. He restricted his investigations to algebraic numbers connected with the nth roots of unity and the nth roots of such numbers. On the other hand, Dirichlet considered the group of units of the ring generated over ℤ by an arbitrary integral algebraic number and determined the structure of this group. It was Dedekind who understood the basic notion of the theory of algebraic number field, which was absent in the investigation of his predecessor. Dedekind formulated the theory by means of ideals, an approach which is now generally accepted. During this same period, the basic notions and theorems were done in equivalent ways by Kronecker with adjunction of variables and Zolotarev with the notion of exponential valuation theory.

In the 19th century, David Hilbert unified the field of algebraic number theory and resolved a significant number-theory problem. In later time, Emil Artin established his Artin reciprocity law, which is a general theorem in number theory that forms a central part of global class field theory.

When writing this thesis, I was motivated by the desire to learn more about the principles of number theory. It is a compilation of various topics in number theory which were encountered during my course of study, particularly in the area of number theory and I aim to analyze the ideal class group in this thesis.

Chapter 2. Algebraic Numbers and integers

In this chapter, we will briefly introduce algebraic numbers and algebraic integers.

Definition 2.1. An algebraic number is any complex number that is a root of a non-zero polynomial equation with integer (or, equivalently, rational) coefficient, i.e., 훼 is an algebraic number if and only if it satisfies 푓(훼) = 0, where

푛 푛−1 푛−2 푓(푥) = 푎푛푥 + 푎푛−1푥 + 푎푛−2푥 + ⋯ + 푎1푥 + 푎0 ∈ ℤ[푥] ----- (2.1)

with 푎푛 > 0. The minimal such 푛 is called the degree of 훼. An algebraic integer is a complex number that is a root of some monic polynomial equation with integer coefficients, i.e., 훼 is an algebraic integer if and only if it satisfies 푓(훼) = 0 with

푚 푚−1 푚−2 푓(푥) = 푥 + 푏푚−1푥 + 푏푚−2푥 + ⋯ + 푏1푥 + 푏0 ∈ ℤ[푥] for some positive integer 푚. Clearly, algebraic integers are included in algebraic numbers.

Remark 2.2. If 푓(훼) = 0 for some 푓(푥) ∈ ℤ[푥] , then from the unique factorization of ℤ[푥], there is a unique irreducible factor 푔(푥) ∈ ℤ[푥] (with positive leading coefficient) of 푓(푥) such that 푔(훼) = 0. This 푔(푥) is called the minimal polynomial of 훼, and the degree of 푔(푥) is the degree of 훼. Apparently, the minimal polynomial of an algebraic integer is a monic irreducible polynomial with integer coefficients.

We think of algebraic integers and algebraic numbers as generalizations of ordinary

3 integers and rational numbers. All (ordinary or “rational”) integers belong to algebraic integers, and all rational numbers belong to algebraic numbers. Some other examples of algebraic integers

2휋 1+푖 1 4+√7 2휋 are 푖, √2, 4 + √7 and 2 cos . The algebraic numbers include , √ , and sin . The 9 2 5 2 7 numbers e and 휋 are not algebraic−they are transcendental, meaning they are not solutions of any non-constant polynomial equation with rational coefficients.

Next we prove several simple properties of algebraic numbers. The first result is a relatively trivial fact about ordinary rational numbers.

Proposition 2.3. A rational number which is an algebraic integer is an integer.

푎 Proof. Write 푥 = in lowest terms, assume that 푥 is not an integer. Then 푏 is an integer greater 푏 than 1. We prove the result by contradiction.

Since 푥 is an algebraic integer, then there exists 푐0, 푐1, … 푐푛−1 ∈ ℤ such that

푛 푛−1 푛−2 푓(푥) = 푥 + 푐푛−1푥 + 푐푛−2푥 + ⋯ + 푐1푥 + 푐0 = 0.

푎 We plug in for all x and then we have 푏

푎 푎 푛 푎 푛−1 푎 푛−2 푎 푓 ( ) = ( ) + 푐 ( ) + 푐 ( ) + ⋯ + 푐 ( ) + 푐 = 0. 푏 푏 푛−1 푏 푛−2 푏 1 푏 0

Multiply 푏푛 for both sides and we get

푛 푛−1 푛−2 2 푛−1 푛 푎 + 푐푛−1푎 푏 + 푐푛−2푎 푏 + ⋯ + 푐1푎푏 + 푐0푏 = 0.

Which implies that 푎푛 ≡ 0 (푚표푑 푏). Since 푥 is not an integer, 푏 ≠ 1 and thus there is a prime number 푝 such that 푝|푏. So we have 푝|푎푛, and hence 푝|푎. Now we have 푝 dividing both a

4 and b, which contradicts the assumption that a/b is in lowest terms. Therefore, the rational number 푥 is an integer. ∎

Proposition 2.4. For any algebraic number 푢, there is an algebraic integer 푣 such that 푢푣 is an algebraic integer.

Proof. It suffices to show that there is a rational integer 푣, such that 푣푢 is an algebraic integer.

Suppose

푛 푛−1 푛−2 푓(푥) = 푎푛푥 + 푎푛−1푥 + 푎푛−2푥 + ⋯ + 푎1푥 + 푎0 ∈ ℤ[푥]

is the minimal polynomial of 훼 (so 푎푛 is a positive integer). Let

푛 푛−1 푛−2 2 푛−3 푛−2 푛−1 푔(푥) = 푥 + 푎푛−1푥 + 푎푛−2푎푛푥 + 푎푛−3푎푛푥 … + 푎1푎푛 푥 + 푎0푎푛 , then we see that 푔(푥) ∈ ℤ[푥], and

푛−1 푔(푎푛훼) = 푎푛 푓(훼) = 0.

Note that 푔(푥) is monic, we thus have proved the theorem with 푣 = 푎푛. ∎

Next, we show that algebraic numbers (integers) are closed under addition and multiplication.

Theorem 2.5. The sum and product of two algebraic integers are algebraic integers.

Proof. Say α and 훽 are two algebraic integers, we need to prove 훼 + 훽 and α훽 are also algebraic integers. Let ℤ[α, 훽] ≔ { 푓(α, 훽)| 푓(x, 푦) 휖 ℤ[푥, 푦]}. It suffices to show γ is an algebraic integer for all γ ϵ ℤ[α, β].

Since α, 훽 are algebraic integers, then there exist 푎0, 푎1, … 푎푛−1 ∈ ℤ and

푏0, 푏1, … 푏푚−1 ∈ ℤ such that

푛 푛−1 푛−2 α + 푎푛−1α + 푎푛−2α + ⋯ + 푎1α + 푎0 = 0. and

푚 푚−1 푚−2 훽 + 푏푚−1훽 + 푏푚−2훽 + ⋯ + 푏1훽 + 푏0 = 0.

Thus

푛 푛−1 푛−2 α = −푎푛−1α − 푎푛−2α − ⋯ − 푎1α − 푎0, and

푚 푛−1 푛−2 훽 = −푏푚−1훽 − 푏푚−2훽 − ⋯ − 푏1훽 − 푏0.

Hence

푖 푗 ℤ[α, 훽] = { ∑푖<푛,푗<푚 푐푖푗훼 훽 | 푐푖푗 휖 ℤ}.

Suppose γ ϵ ℤ[α, β]. Note that for any non-negative integers 푘, 푙, we have

k 푙 k 푙 푘푙 푖 푗 γα 훽 휖 ℤ[훼, 훽]. Thus γα 훽 = ∑푖<푛,푗<푚 푐푖푗 α 훽 , i.e.,

k 푙 푘푙 푖 푗 γα 훽 − ∑푖<푛,푗<푚 푐푖푗 α 훽 = 0 ------(2.2)

푘푙 푖 푗 0 0 1 0 푛−1 0 0 1 푛−1 푚−1 for some integers 푐푖푗 . Ordering α 훽 by α 훽 , α 훽 , … , α 훽 , α 훽 , … , α 훽 , then from

(2.2) , with 푘, 푙 respectively running from 0 to 푛 − 1 and 푚 − 1, we get

00 00 0 0 γ − 푐00 −푐10 … α 훽 [ 10 10 ] [ 1 0] = 0. −푐00 γ − 푐10 … α 훽 ⋮ ⋮ ⋱ ⋮

Then the matrix is singular, and thus it has determinant 0. On the other hand, the determinant of the matrix is a degree 푚푛 monic polynomial of 훾 with integer coefficients, hence

γ is an algebraic integer. ∎

푬풙풂풎풑풍풆 ퟐ. ퟔ. Let α = 1 + √2 and 훽 = 1 + √3, They are both algebraic integers, and we have

α2 − 2α − 1 = 0, and 훽2 − 2훽 − 2 = 0.

Let γ = αβ = 1 + √2 + √3 + √6. Rewrite this as γ − 1 − √6 = √2 + √3 and square both sides to obtain

(γ − 1)2 − 2√6(γ − 1) + 6 = 5 + 2√6,

Which simplifies to γ2 − 2γ + 2 = 2√6γ. Squaring both sides, we obtain γ4 − 4γ3 −

16γ2 − 8γ + 4 = 0.

From Theorem 2.5 and the proof of Proposition 2.4, or directly with a proof nearly identical to that for Theorem 2.5, we have

Theorem 2.7. The sum and product of two algebraic numbers are algebraic numbers.

We end this chapter by showing that, for a given algebraic number 훼, ℚ[α] is a field.

Theorem 2.8. Given any algebraic number 훼, ℚ[α] is a field.

Proof: Without loss of generality, we suppose that 훼 is an algebraic number of degree 푛 ≥ 2, with minimal polynomial 푓(푥). From the proof of Theorem 2.5, we clearly have

n−1 ℚ[α] = {c0 + c1α + ⋯ + cn−1α |c0, c1, … , cn−1 ∈ ℚ} ----- (2.3)

To show that this is a field, it suffices to show that, for any Α, Β ∈ ℚ[α], we have Α ±

Β, ΑΒ and 1/Β (if Β ≠ 0) are in ℚ[α].

Suppose

n−1 n−1 Α = h(훼) = a0 + a1α + ⋯ + an−1α , Β = g(α) = b0 + b1α + ⋯ + bn−1α

7 for some rational numbers a0, a1, … , an−1, b0, b1, … , bn−1. It is clear that Α ± Β is also in the form of (2.3), and thus is in ℚ[α]. From the Euclidean algorithm for polynomials over ℚ, there are 푞(푥), 푟(푥) ∈ ℚ[x], with deg(푟) ≤ 푛 − 1 such that

푔(푥)ℎ(푥) = 푞(푥)푓(푥) + 푟(푥).

Thus

ΑΒ = 푔(훼)ℎ(훼) = 푞(훼)푓(훼) + 푟(훼) = 푟(훼) ∈ ℚ[α].

Now suppose Β ≠ 0, then 푔(푥) and 푓(푥) must be coprime, thus there are 푠(푥), 푡(푥) ∈

ℚ[x] , with deg(푡) ≤ 푛 − 1, such that

푠(푥)푓(푥) + 푡(푥)푔(푥) = 1.

We hence have

1 = 푠(훼)푓(훼) + 푡(훼)푔(훼) = 푡(훼)푔(훼) = 푡(훼)Β.

1 Thus ∈ ℚ[α], ∎ 퐵

Remark 2.9. From Theorem 2.8 (and its proof), we see that ℚ[α] = ℚ(α), the field generated by

α over ℚ. From now on, we will use the notation ℚ(α) for this field. It can be shown that, for a finite number of algebraic numbers 훼1, … , 훼푘, the field obtained by completing the set ℚ ∪

{훼1, … , 훼푘} via +, −, ×, / is also in the form ℚ(α) for a single algebraic number 훼. We shall not prove this, but will assume this fact when it is needed. For an algebraic number 훼, ℚ(α) is called an algebraic number field, or simply a number field.

Chapter 3. Rings of Integers

In this chapter, we study the algebraic integers in a number field ℚ(α) where 훼 is a given algebraic number which, without loss of generality, is supposed to have degree 푛 ≥ 2.

Let 훼1, … , 훼푛 be the conjugates of 훼. The fields ℚ(훼푗) are very similar to ℚ(훼), each being generated by an element (over ℚ) with the same minimum polynomial. In fact, they are all isomorphic. Note that if 푔1, 푔2 ∈ ℚ[푋] and 푔1(훼) = 푔2(훼), then for any if 푗 ≤ 푛, we have

푔1(훼푗) = 푔2(훼푗) which follows from the fact that 훼 and 훼푗 have the same minimum polynomial.

Therefore, it is fairly easy to verify that the map 휎푗 ∶ ℚ(훼) → ℚ(훼푗), defined by 휎푗(푔(훼)) =

푔(훼푗) for 푔 ∈ ℚ[푋], is an isomorphism. We often let 훼1 = 훼, and then 휎1 is the identity map on

ℚ(α).

3.1. Some basic properties

Let 훼1, … , 훼푛 be the conjugates of a given degree 푛 algebraic number 훼, and 휎1, … , 휎푛 be the isomorphisms defined above.

Definition 3.1.1. Let 훽 ∈ ℚ(α). Then 푛표푟푚 of 훽, denoted by 푁(훽), is defined by

푛

푁(훽) = ∏ 휎푗(훽) 푗=1

9 and 푇(훽), the 푡푟푎푐푒 of 훽, is defined by

푛

푇(훽) = ∑ 휎푗(훽). 푗=1

Proposition 3.1.2. We have the following properties for the trace and norm of algebraic numbers

(1). 푁(훽훾) = 푁(훽)푁(훾) for all 훽, 훾 ∈ ℚ(α),

(2). 푁(푐훽) = 푐푛푁(훽) for all 푐 ∈ ℚ and 훽 ∈ ℚ(α),

(3). 푇(훽 + 훾) = 푇(훽) + 푇(훾) for all 훽, 훾 ∈ ℚ(α), and

(4). 푇(푐훽) = 푐푇(훽) for all 푐 ∈ ℚ and 훽 ∈ ℚ(α).

Proof: These all easily follow from the definitions and the facts that the 휎푗 preserve summation and multiplication. ∎

1 Clearly, 푁(0) = 0 and 푁(1) = 1. If 훽 ≠ 0 then 1 = 푁(1) = 푁(훽)푁( ) so that 푁(훽) ≠ 훽

0. Next, we state without proof a very important fact about the trace and norm of an algebraic number.

Proposition 3.1.3. Suppose 훼 is an algebraic number. For 훽 ∈ ℚ(훼), we have 푁(훽) ∈ ℚ and

푇(훽) ∈ ℚ. Moreover, if 훽 is an algebraic integer, then 푁(훽) ∈ ℤ and 푇(훽) ∈ ℤ.

Remark 3.1.4. In fact, if 훼 is a degree 푛 algebraic integer, with minimal polynomial

푛 푛−1 푛−2 푓(푥) = 푥 + 푎푛−1푥 + 푎푛−2푥 + ⋯ + 푎1푥 + 푎0, then we have

−푎푛−1 = 훼1 + ⋯ + 훼푛 = 푇(훼) ∈ ℤ,

And

푛 (−1) 푎0 = 훼1훼2 ⋯ 훼푛 = 푁(훼) ∈ ℤ.

Before we bring out the topic of the ring of integers of a number field, let us recall a little about the theory of commutative rings.

Definition 3.1.5. A 푐표푚푚푢푛푡푎푡푖푣푒 푟푖푛푔 푅 is a set with two operations, addition and multiplication, such that:

(1) 푅 is an abelian group under addition,

(2) 푎푏 = 푏푎 for all 푎, 푏 ∈ 푅,

(3) 푎(푏푐) = (푎푏)푐 for any 푎, 푏, 푐 ∈ 푅,

(4) There is an element 1 ∈ 푅 with 1 ≠ 0 and with 1 ⋅ 푎 = 푎 ⋅ 1 = 푎 for any 푎 ∈ 푅,

(5) 푎(푏 + 푐) = 푎푏 + 푎푐 for any 푎, 푏, 푐 ∈ 푅.

For a given a number field 퐾, consider the collection of all algebraic integers contained in

퐾. From Theorem 2.5, we can check that all conditions for a commutative ring is satisfied by the set under the ordinary summation and multiplication of complex numbers. We thus have the following

Definition 3.1.6. Given a number field 퐾, the 푟푖푛푔 표푓 푖푛푡푒푔푒푟푠 풪푘 of 퐾 is the commutative ring consisting of the algebraic integers in 퐾, under the ordinary summation and multiplication.

In the special case that 퐾 = ℚ, we have 풪푘 = ℤ. While many arithmetic/algebraic properties of ℤ are well known to us, we will see that general rings of integers have many important properties in common with ℤ.

Proposition 3.1.7. The ring of integers 풪푘 is a finitely-generated ℤ-module. Indeed, it is a free

ℤ-module, and thus has an integral basis ω1, … , 휔푛 ∈ 풪푘of the ℚ-vector space 퐾 such that each element 푥 in 풪푘 can be uniquely represented as

푛 푥 = ∑푖=1 푎푖휔푖 ,

with 푎푖 ∈ ℤ. The rank 푛 of 풪푘 as a free Z-module is equal to the degree of 퐾 over ℚ, in where

[퐾 ∶ ℚ] = 푛.

Proposition 3.1.8. The ring of integers in a number field is a 퐷푒푑푒푘푖푛푑 푑표푚푎푖푛.

Suppose ω1, … , 휔푛 ∈ 풪푘 are a basis for 풪푘 as a ℤ-module. Consider the embedding 훔: K →

n C defined by 흈(푎) = (휎1(푎), 휎2(푎), … , 휎푛(푎) ), where 휎1, 휎2, … , 휎푛 are the distinct embeddings of 퐾 into 퐶. Let 퐴 be the 푛 × 푛 matrix consisting of rows 흈(휔1), … , 흈(휔푛), then

det(퐴퐴푇) = det( ∑ 휎 (휔 )휎 (휔 )) = det ( ∑ 휎 (휔 휔 )) = det(푇푟(휔 휔 ) ) 푘 푖 푘 푗 푘 푖 푗 푖 푗 1≤푖,푗≤푛 푘=1,…,푛 푘=1,…,푛 which shows that det(퐴퐴푇) can be defined in terms of the trace, and we also conclude that det(퐴퐴푇) is a rational integer. Moreover, because changing the integral basis is equivalent to multiplying to 퐴 (either to its left or right) an integer matrix 푈 of determinant ±1, thus det(퐴퐴푇) is independent of the choice of the integral basis.

Definition 3.1.9. The discriminant of a number field 퐾, denoted by 푑퐾 or Δ퐾, is the integer given by the det(퐴퐴푇).

Δ퐾 is one of the most important invariants of a number field 퐾. In our Chapter 5 examples, when we try to find the class group/number of a (quadratic) fields, the size of Δ퐾 is a key factor that more or less determines the complexity of the computation.

3.2. Factorization of algebraic integers and the unit group

In this section, we will discuss divisibility and factorization of algebraic integers. For rational integers, we have the Fundamental Theorem of Arithmetic -- each positive integer can be uniquely expressed as the product of prime numbers. A rational prime number 푝 is a positive integer satisfying any one of the following two conditions.

(1) If 푝 = 푎푏 with 푎, 푏 ∈ ℤ then 푎 = ±1 or 푏 = ±1, ----- (3.2.1)

(2) If 푝 | 푐푑 with 푐, 푑 ∈ ℤ then 푝 | 푐 or 푝 | 푑. ------(3.2.2)

These two conditions are equivalent, but their analogs in the ring of integers 풪푘 are not necessarily equivalent.

Definition 3.2.3. For 훽, 훾 ∈ 풪푘 with 훽 ≠ 0, we say that 훽 divides 훾, or 훾 is divisible by 훽 (with

훾 1 notation 훽 | 훾), if ∈ 풪 . β ∈ 풪 is a 푢푛푖푡 in 풪 if 훽 | 1 or, equivalently, ∈ 풪 . 훽 푘 푘 푘 β 푘

Lemma 3.2.4. Let 훽, 훾 ∈ 풪푘. If 훽 | 훾, then 푁(훽) | 푁(γ) as integers.

Proof: This is obvious from the definition of divisibility and Proposition 3.1.3 and 3.1.2(1). ∎

It is fairly easy to show that the set of units of 풪푘, denoted by 푈(풪푘), forms an Abelian group under multiplication. We call 푈(풪푘) the unit group of 풪푘, or the unit group of 퐾.

Lemma 3.2.5. Suppose that 퐾 is a number field. Then 훽 ∈ 푈(풪푘) if and only if 푁(훽) = ±1.

Proof: If 훽 ∈ 푈(풪푘), then there is a 훾 ∈ 풪푘 such that 훽훾 = 1. The lemma then follows directly from Proposition 3.1.3, and the fact that 푁(1) = 1. ∎

Definition 3.2.6. An element 훽 ∈ 풪푘 is 푖푟푟푒푑푢푐푖푏푙푒 if

(1) 훽 ≠ 0,

(2) 훽 is not a unit,

(3) If 훽 = 훾훿 with 훾, 훿 ∈ 풪푘 then either 훾 or 훿 is a unit.

In ℤ, the irreducible elements have the form ±푝, where 푝 is a prime number. From Lemma

3.2.5, we obtain that if 푁(훽) is a prime number then 훽 is irreducible. The converse, however, is not true in general. For example, for 퐾 = ℚ(푖) we have 풪푘 = ℤ[푖] (we will see this later in section 3.3). One can check that 3 is irreducible in 풪푘, but 푁(3) = 9 is not prime.

Similar to the Fundamental Theorem of Arithmetic, every element in 풪푘 can be factored into a product of irreducibles.

Theorem 3.2.7. Let 퐾 be a number field. Suppose 훽 ∈ 풪푘 with 훽 ≠ 0 and 훽 ∉ 푈(풪푘). Then there exist irreducible elements 훾1, … , 훾푘 ∈ 풪푘 such that 훽 = 훾1 ⋅⋅⋅ 훾푘.

Proof: We prove by induction on |푁(훽)|. Since 훽 ≠ 0 and 훽 ∉ 푈(풪푘), |푁(훽)| ≥ 2. Now we have two cases, if 훽 is irreducible then take 푘 = 1 and 훾1 = 훽. If 훽 is reducible then 훽 = 훽1훽2 where 훽1, 훽2 ∈ 풪푘 and 훽1, 훽2 ∉ 푈(풪푘). Then |푁(훽1)|, |푁(훽2)| > 1 and |푁(훽)| =

|푁(훽1)||푁(훽2)| implies |푁(훽1)| < |푁(훽)| and |푁(훽2)| < |푁(훽)|. By induction hypothesis, 훽1 and 훽2 are products of irreducible elements. Combine these factorizations we see that 훽 is also a product of irreducible elements. ∎

A natural question about the factorization (into a product of irreducible integers) is whether it is unique. Before we go any further, let us define the uniqueness here. It is obvious that if 훽 is irreducible and 휉 ∈ 푈(풪푘) then 휉훽 is irreducible as well. We can adjust factorizations

14 by multiplying factors by units. For example, let 훽 = 훾1훾2훾3 where 훾1, 훾2, 훾3 are irreducible, and ∈ 푈(풪푘) . Then

−1 −1 훽 = 훾1훾2훾3 휉휉 = (휉훾1)훾2(휉 훾3) ------(3.2.3) gives two formally different factorizations of 훽into irreducibles. These two factorizations are essentially the same if we consider factorizations modulo the unit group. We call two algebraic integers associated (or one is an associate of the other) if they differ by a unit factor. In the factorization of algebraic integers, we treat two associated integers as the same, and thus the two factorizations in (3.2.5) are regarded as the same. So, when it comes to the uniqueness of factorization, two factorizations are different only if one is not a re-ordering of associates of the integers in the other factorization.

Example 3.2.8. Let 퐾 = ℚ(√−5). 2, 3, 1 + √−5, and 1 − √−5 are irreducible integers in 풪퐾, and they are not associated. Observing that 6 = 2 · 3 = (1 + √−5)(1 − √−5), 풪퐾 does not have unique factorization into irreducibles.

Example 3.2.9. Let 퐾 = ℚ(√−6), from section 3.3 we will see that 풪푘 = ℤ[√−6]. Now 6 =

2 × 3 = √−6(−√−6). These two factorizations of 6 into irreducible elements are inequivalent.

Remark 3.2.10. Examples 3.2.8 and 3.2.9 show that, in number fields, while every integer can be expressed as a product of irreducible integers, such factorization is not necessarily unique.

Thus, if we pursue an analog of the Fundamental Theorem of Arithmetic in number fields, irreducible integers in number fields, as an extension of rational prime numbers following the definition (3.2.3), do not serve the same role as rational prime numbers in ℤ. Next, let us consider the algebraic integers satisfying a condition similar to (3.2.4). We call such integers primes in the number field.

Definition 3.2.11. Let 훽 ∈ 풪푘, 훽 is 푝푟푖푚푒 if

(1) 훽 ≠ 0,

(2) 훽 is not a unit,

(3) If 훽|훾훿 with 훾, 훿 ∈ 풪푘 then 훽|훾 or 훽|훿.

Primes are always irreducible.

Lemma 3.2.12. Let 퐾 be number field. If 훽 is a prime element of 풪푘 then 훽 is irreducible in 풪푘.

Therefore 훽 is irreducible. ∎

Clearly, each 풪푘 is an integral domain. While we see from Examples 3.2.8 and 3.2.9 that the unique factorization (into a product of irreducibles) does not hold in 풪푘 in general, if some condition is satisfied, however, 풪푘 becomes a UFD. (Recall that a UFD, namely a unique factorization domain, is a domain in which every non-zero element can be uniquely expressed as a product of irreducible elements.)

Theorem 3.2.13. Let 퐾 be number field and suppose that every irreducible element of 풪푘 is prime. Then 풪푘 has unique factorization.

Proof: We prove by contradiction. Let

훽 = 훾1훾2 ⋅⋅⋅ 훾푠 = 훿1훿2 ⋅⋅⋅ 훿푘,

where 훾푗 and 훿푟 are irreducible elements in 풪푘 for 푗 = 1, … , 푠 and 푘 = 1, … , 푘.

16 be two factorizations of 훽. We need to show these are equivalent by induction on |푁(훽)|. Since

훾1 is prime and 훾1 | 훿1 ⋅⋅⋅ 훿푘 then 훾1 | 훿푟 for some 푟. By shuffling the 훿푟 we may assume that

훾1 | 훿1 and since 훿1 is irreducible, we obtain 훿1 = 휇훾1 where is a unit. Thus

훽/훾1 = 훾2 ⋅⋅⋅ 훾푠 = 휇훿2 ⋅⋅⋅ 훿푘

is a factorization into irreducibles and |푁(훽/훾1)| < |푁(훽)|. By the inductive hypothesis, these factorizations of 훽/훾1 are equivalent, and hence the given factorizations of 훽 are equivalent.∎

We end this section by stating Dirichlet’s unit theorem. We mentioned before that 푈(풪푘), the unit group of a number field 퐾, is a finitely generated Abelian group. In fact, the following theorem of Dirichlet gives a precise description to the structure of 푈(풪푘).

Theorem 3.2.14. (Dirichlet) For any number field 퐾, the unit group 푈(풪푘) is the product of a finite cyclic group of roots of unity with a free Abelian group of rank 푟 + 푠 − 1, where 푟 is the number of real embeddings of 퐾, and 푠 is the number of complex conjugate pairs of embeddings.

Remark 3.2.15. Suppose 훼 is algebraic of degree n. Among the conjugates 훼1, … , 훼푛 of 훼, suppose there are 푟 real roots and 2푠 imaginary roots. Then 푛 = 푟 + 2푠, and Dirichlet’s unit theorem says that, for 퐾 = ℚ(훼), we have

푟+푠−1 푈(풪푘) ≅ W ⨁ ℤ , where 푊 is a finite cyclic group.

3.3. Quadratic integers

In this section we particularly focus on the fields 퐾 = ℚ(α) that have degree 2 over ℚ.

Such a field 퐾is called a quadratic field. In this case, 훼 can be chosen as √푑 for a unique square- free integer (either positive or negative) 푑. Namely, every quadratic field is in the form

ℚ(√푑) = {푎 + 푏√푑: 푎, 푏 ∈ ℚ}.

Depending on the sign of 푑, there are two types of quadratic fields. If d > 0, then all element of ℚ(√푑) are real numbers, and we call ℚ(√푑) a real quadratic field; If d < 0, then

ℚ(√푑) contains non-real numbers (imaginary numbers)−it is an imaginary quadratic field.

For the integer ring of a quadratic field 퐾 = ℚ(√d), rather than 풪푘, we would like to use a notation 푅푑 (to indicate the associated square-free number 푑).

Formally, for a number = 푐 + 푓√푑 ∈ ℚ(√푑) , we call 훽∗ = 푐 − 푓√푑 the conjugate of 훽.

Theorem 3.3.1. Suppose 푑 is a square-free integer (possibly with a negative sign). If 푑 ≢

1 (푚표푑 4), then

Rd = {푎 + 푏√푑 ∶ 푎, 푏 ∈ ℤ}.

And If 푑 ≡ 1 (푚표푑 4) then

1+ 푑 R = {푎 + 푏 ( √ ) ∶ 푎, 푏 ∈ ℤ}. d 2

Proof: A direct calculation shows that both sets consist of algebraic integers. We need to show, conversely, that every element in Rd has the indicated form.

∗ ∗ 2 2 Let α = 푥 + 푦√푑 ∈ 푅푑, so 훼 + 훼 = 2푥 ∈ ℤ 푎푛푑 훼훼 = 푥 − 푑푦 ∈ ℤ. The first condition means 푥 is half an ordinary integer, so either 푥 ∈ ℤ or 푥 is half an odd number.

The importance of 푑 being squarefree is that if 푑푟2 ∈ ℤ for some 푟 ∈ ℚ then 푟 ∈ ℤ. Indeed, if 푟 has a prime factor 푝 in its (reduced form) denominator then dr2 ∉ ℤ since 푝2 is not canceled by 푑 in 푑푟2. So 푑푟2 ∈ ℤ implies 푟 has no prime in its denominator, i.e., 푟 ∈ ℤ.

푎 If 푥 is half an odd number, namely 푥 = with 푎 odd, then from the fact that 2

푎2 푁(훼) = 푥2 − 푑푦2 = − 푑푦2 ∈ ℤ, 4 we have

푎2 − 푑(2푦)2ϵ 4ℤ

Thus 푑(2푦)2 is an integer congruent to 1 modulo 4. Since 푑 is square-free, 2푦 ϵ ℤ and, moreover, 푦 is half an odd number and 푑 ≡ 1 (mod 4) (because all odd squares are 1 mod 4).

Returning to α,

푎 푏 푎 − 푏 1 + √푑 α = 푥 + 푦√푑 = + √푑 = + b ( ) , 2 2 2 2 which has the necessary form because 푎 − 푏 is even.

We are done when 푑 ≡ 1 (mod 4). When 푑 ≢ 1 (mod 4) we can’t have 푥 be half an odd number, so 푥 ϵ ℤ. From 푁(훼) = x2 − 푑푦2ϵ ℤ, it follows that 푑푦2ϵ ℤ, so from 푑 being square- free we get 푦 ϵ ℤ. Therefore α has the necessary form, with 푎 = 푥 and 푏 = 푦. ∎

To unify the notation, set

√푑, 푖푓 푑 ≢ 1 (푚표푑 4), ω = {1+ 푑 √ , 푖푓 푑 ≡ 1 (푚표푑 4). 2

Then, 푅푑 = {푎 + 푏ω ∶ 푎, 푏 ϵ ℤ} = ℤ[ω].

A quick consequence of this is the following

Theorem 3.3.2. Suppose 푑 is a square-free integer (either positive or negative). Then for the quadratic field 퐾 = ℚ(√푑), we have Δ퐾 = 푑 if 푑 ≡ 1 푚표푑 4 ; and Δ퐾 = 4푑 if 푑 ≡ 2, 3 푚표푑 4.

Let us turn to the unit group next. By 푈푑, we denote the unit group of ℚ(√푑). Dirichlet’s unit theorem shows that, if 푑 < 0, then 푈푑 has rank 0 and thus is finite; if 푑 > 0, then 푈푑 has rank 1. In fact, we have some much more precise results in this case.

Theorem 3.3.3. Suppose −푑 > 0 is a square-free integer. Then we have

{±1, ±√−1 }, 푖푓 푑 = −1, 2 Ud = {{±1, ±훼, ±훼 }, 푖푓 푑 = −3, {±1}, 푖푓 푑 ≠ −1, −3, where 훼 = (−1 + √−3 )/2.

Theorem 3.3.4. Suppose 푑 > 1 is a square-free integer. Then

r Ud = {±ζ : r ∈ ℤ},

where ζ is the smallest possible unit of Ud satisfying ζ > 1, called the fundamental unit of Ud.

Chapter 4. Ideals

We have seen from examples in Chapter 3 that the factorization of an algebraic integer into a product of irreducibles is not necessarily unique. In this chapter, I will introduce ideals in number fields (rings), and study some of their properties. In particular, we will see that the

Fundamental Theorem of Arithmetic finds a good analog in the factorizations of ideals in number rings.

4.1. A review of ideals of commutative rings

First, we recall that an 푖푑푒푎푙 of a commutative ring 푅 is a subset 퐼 of 푅 such that

(1) 퐼 is a subgroup of 푅,

(2) if 푎, 푏 ∈ 퐼, then 푎 + 푏 ∈ 퐼,

(3) if 푎 ∈ 퐼 and 푥 ∈ 푅 then 푥푎 ∈ 퐼.

Norm of an ideal is the cardinality of the quotient ring 푅/퐼, i.e. 푁(퐼) = [푅 ∶ 퐼]. And

푁(퐼퐽) = 푁(퐼)푁(퐽).

A principal ideal of a commutative ring 푅 is an ideal generated by a single element

〈푎〉 = {푥푎 ∶ 푥 ∈ 푅} for some 푎 ∈ 푅.

By trivial ideals we mean the zero ideal 〈0〉 = {0} and 푅 itself. All other ideals of 푅 are regarded nontrivial.

Ideals of commutative rings satisfy some basic (yet important) rules, such as

(1). If 퐼, 퐽 are ideals of 푅, then 퐼 + 퐽 = {푎 + 푏 ∶ 푎 ∈ 퐼, 푏 ∈ 퐽} is also an ideal of 푅.

(2). If 퐼, 퐽 be ideals of 푅, then 퐼 ∩ 퐽 is also an ideal of 푅.

Moreover, if we define the product of two ideals as

IJ= {푎1푏1 + 푎2푏2 + ⋯ + 푎푛푏푛 ∶ 푎1, … , 푎푛 ∈ 퐼, 푏1, … , 푏푛 ∈ 퐽}, then 퐼퐽 is the smallest ideal containing all 푎푏.

The sum and product satisfy some formal properties:

(1) 퐼 + 퐼 = 퐼, where 퐼 is an ideal of 푅,

(2) 퐼 + 퐽 = 퐽 + 퐼, where 퐼 푎푛푑 퐽 are ideals of 푅,

(3) 퐼1 + (퐼2 + 퐼3) = (퐼1 + 퐼2) + 퐼3 , where 퐼1, 퐼2, 퐼3 are ideals of 푅,

(4) 퐼퐽 ⊆ 퐼 ∩ 퐽, where 퐼 푎푛푑 퐽 are ideals of 푅,

(5) 퐼퐽 = 퐽퐼, where 퐼 푎푛푑 퐽 are ideals of 푅,

(6) 퐼1(퐼2퐼3) = (퐼1퐼2)퐼3 , where 퐼1, 퐼2, 퐼3 are ideals of 푅,

(7) 퐼1(퐽1 + 퐽2) = 퐼퐽1 + 퐼퐽2 , where 퐼, 퐽1, 퐽2 are ideals of 푅.

4.2. Ideal theory of integer ring 풪퐾

In this section, we will prove the fundamental theorem of ideal theory in number fields: every non-zero proper ideal in the integer ring of a number field can be uniquely factored into a product of prime ideals.

Definition 4.2.1. For ideals 푎 and 푏 in a commutative ring, write 푎|푏 if 푏 = 푎푐 for an ideal 푐.

Thus, divisibility and primality in 풪퐾 can be expressed in terms of ideals. For two integers 훽, 훾,

훽 | 훾 if and only if 훾 ∈ 〈훽〉 which occurs if and only if 〈훽〉 | 〈훾〉. Similarly, 〈훽〉 = 〈훾〉 if and only if 훾/훽 is a unit in 풪퐾.

Definition 4.2.2. We say that an ideal P of 풪퐾 is 푝푟푖푚푒 if

(1) 푃 ≠ 〈0〉,

(2) 푃 ≠ 〈1〉,

(3) If 훾, 훿 ∈ 풪퐾 and 훾훿 ∈ 푃, then 훾 ∈ 푃 or 훿 ∈ 푃.

Before we prove the unique factorization of ideals in a general number ring, we remark that, if 풪퐾 is a PID, then it is fairly easy to show that every irreducible element of 풪퐾 is prime, and then each nontrivial ideal of 풪퐾 is a product of prime ideals.

To prove for a general number field the fundamental theorem of ideals. We need some preparations.

Lemma 4.2.3. Let 퐾 be a number field and 푃 be a prime ideal of 풪퐾. If 퐼 and 퐽 are ideals of 풪퐾 with 퐼퐽 ⊆ 푃, then 퐼 ⊆ 푃 or 퐽 ⊆ 푃. In general, if 퐼1퐼2 ⋅⋅⋅ 퐼푚 ⊆ 푃, then 퐼푘 ⊆ 푃 for some 푘.

Proof: Suppose 퐼퐽 ⊆ 푃 but 퐼 ⊈ 푃 and 퐽 ⊈ 푃. There exist 훽 ∈ 퐼, 훾 ∈ 퐽 with 훽 ∉ 푃 and 훾 ∉ 푃, then 훽훾 ∈ 퐼퐽. But 훽훾 ∉ 푃 contradict 퐼퐽 ⊆ 푃. The case of an 푚-term product 퐼1퐼2 ⋅⋅⋅ 퐼푚 now follows by induction. ∎

Definition 4.2.4. An ideal 퐼 of 풪퐾 is said to be an 푚푎푥푖푚푎푙 ideal if, for all ideals 퐽 of 풪퐾 such that I⊆ 퐽 ⊆ 풪퐾, either 퐽 = 퐼 or 퐽 = 풪퐾.

Lemma 4.2.5. Let 퐾 be a number field. An ideal 퐼 of 풪퐾 is prime if and only if it is maximal.

Proof: First suppose that 퐼 is maximal. Let 훽, 훾 ∈ 풪퐾 with 훽훾 ∈ 퐼 and 훽 ∉ 퐼. To show that 퐼 is prime it suffices to show that 훾 ∈ 퐼. Let 퐽 = 퐼 + 〈훽〉. Then 퐽 is an ideal and 퐼 ⊆ 퐽, but 퐼 ≠ 퐽 since

훽 ∈ 퐽. By maximality of 퐼, 퐽 = 풪퐾. Hence 1 ∈ 퐽 so 1 = 휇 + 훿훽 where 휇 ∈ 퐼 and 훿 ∈ 풪퐾. Then

1 ≡ 훿훽 (푚표푑 퐼). Consequently, 훾 = 1훾 ≡ 훿훽훾 ≡ 0 (푚표푑 퐼), as 훽훾 ∈ 퐼. We conclude that 훾 ∈ 퐼 and that 퐼 is prime.

Conversely, suppose that 퐼 is prime, 퐽 is an ideal of 풪퐾 with 퐼 ⊆ 퐽 and 퐼 ≠ 퐽. We need to show that 퐽 = 풪퐾, or equivalently, that 1 ∈ 퐽. Let 훽 ∈ 퐽 and 훽 ∉ 퐼. Then 퐽 ⊇ 퐼 + 〈훽〉. The ideal 퐼 has finite index, 푚, in 풪퐾. Let 훾1, 훾2, … , 훾푚 be coset representative for 퐼 in 풪퐾. That is each element of 풪퐾 is congruent modulo 퐼 to exactly one 훾푗. In particular, 훾푗 ≡ γk (푚표푑 퐼) if and only if 푗 = 푘. If 훽훾푗 ≡ 훽훾푘 (푚표푑 퐼) then 훽(훾푗 − 훾푘) ∈ 퐼 and as 퐼 is prime and 훽 ∉ 퐼 then 훾푗 −

훾푘 ∈ 퐼 and so 푗 = 푘. The numbers 훽훾1, 훽훾2, … , 훽훾푚 lie in distinct cosets of 퐼, and so they represent all cosets. In particular 1 ≡ 훽훾푗 (푚표푑 퐼) for some 푗, and so 1 = 휇 + 훾푗훽 for some 휇 ∈

퐼. Thus 1 ∈ 퐼 + 〈훽〉 and 퐼 is maximal.∎

Lemma 4.2.6. Let 퐾 be a number field and 퐼 be a nontrivial ideal of 풪퐾. Then there is a prime ideal 푃 of 풪퐾 such that 퐼 ⊆ 푃.

Proof: Consider the nontrivial ideals 퐽 of 풪퐾 with 퐼 ⊆ 퐽. There is at least one since 퐽 = 퐼 suffices. Take one, 푃, with least possible norm. Then 푃 is maximal, for if 푃 ⊆ 퐽1 with 퐽1 ≠ 푃 an ideal of 풪퐾, then 푁(퐽1) < 푁(푃) and so 퐽1 = 풪퐾. ∎

Lemma 4.2.7. Let 퐾 be a number field and 퐼 be a nontrivial ideal of 풪퐾. Then 퐼 ⊇ 푃1푃2 ⋅⋅⋅ 푃푚 where 푃푗 for 푗 = 1, 2, … are prime ideals of 풪퐾.

Proof: Let’s use induction on 푁(퐼). If 퐼 is prime, then we can take 푚 = 1 and 푃1 = 퐼. If 퐼 is not prime, then there exist 훽, 훾 ∈ 풪퐾 with 훽 ∉ 퐼, 훾 ∉ 퐼 but 훽훾 ∈ 퐼. Let 퐽1 = 〈훽〉 + 퐼 and 퐽2 = 〈훾〉 +

퐼. Then 퐼 ⊆ 퐽1 and 퐼 ⊆ 퐽2, but 퐼 ≠ 퐽1 and 퐼 ≠ 퐽2. Hence 푁(퐽1) < 푁(퐼) and 푁(퐽2) < 푁(퐼). But

2 퐽1퐽2 = 〈훽훾〉 + 훽퐼 + 훾퐼 + 퐼 ⊆ 퐼 as 훽훾 ∈ 퐼. By inductive hypothesis, 퐽1 ⊇ 푃1푃2 ⋅⋅⋅ 푃푟 and 퐽2 ⊇

푄1푄2 ⋅⋅⋅ 푄푠 where the 푃푗 and 푄푘 are prime. Therefore, 퐼 ⊇ 퐽1퐽2 ⊇ 푃1푃2 ⋅⋅⋅ 푃푟푄1푄2 ⋅⋅⋅ 푄푠.∎

To prove that a non-zero proper ideal in 풪퐾 has a unique factorization into a product of prime ideals, we need to show how to invert a prime ideal. The “inverse” of a prime ideal will be an 풪퐾-module in 퐾 (but not in 풪퐾). To this end, we need to introduce the 풪퐾-modules which are essentially ideals with denominators.

Definition 4.2.8. A 푓푟푎푐푡푖표푛푎푙 푖푑푒푎푙 of 퐾 is a set of the form 훽퐼 where 훽 is a nonzero element of 퐾 and 퐼 is a nonzero ideal of 풪퐾.

Lemma 4.2.9. Let 퐾 be a number field, then 퐼 is a fractional ideal of 퐾 if and only if

(1) 퐼 is a nonzero subgroup of K closed under addition,

(2) If 훽 ∈ 퐼 and 훾 ∈ 풪퐾, then 훾훽 ∈ 퐼,

(3) There exists a nonzero 휇 ∈ 퐾 such that 훽/휇 ∈ 풪퐾 for each 훽 ∈ 퐼.

Proof: If 퐼 = 휇퐽 is a fractional ideal of 퐾, with 휇 ∈ 퐾 and J an ideal of 풪퐾, then the three properties follow with the same value of 휇. Conversely, suppose the three properties hold. Then

−1 퐽 = 휇 퐼 = {훽/휇: 훽 ∈ 퐼} is a nonzero ideal of 풪퐾 and so 퐼 = 휇퐽 is a fractional ideal. ∎

Theorem 4.2.10. Every fractional ideal is invertible. That is, for a given fractional ideal 퐼, there is a fractional ideal 퐽 with 퐼퐽 = 〈1〉.

Lemma 4.2.11. Let 퐾 be a number field and 퐼 be a nonzero ideal of 풪퐾. If 훾퐼 ⊆ 퐼 for some 훾 ∈

퐾, then 훾 ∈ 풪퐾.

Proof: We know each nonzero ideal of 풪퐾 is a free abelian group of rank n under the operation of addition. 퐼 ⊆ 풪퐾 is a free abelian group, let 훽1, … , 훽푛 be an integral basis of 퐼. Then 훾훽푗 ∈ 퐼

푛 for all 푗, so 훾훽푗 = ∑푘=1 푎푗푘훽푘 where the 푎푗푘 ∈ ℤ. Hence 훾퐯 = 퐴퐯 where 퐯 is the column vector with entries the 훽푗 and 퐴 is the matrix with entries the 푎푗푘. Thus 훾 is an eigenvalue of 퐴 which is a matrix with integer entries. Thus 훾 is an algebraic integer and so 훾 ∈ 퐾⋂퐁 = 풪퐾, where 퐁 is basis of 퐴. ∎

Lemma 4.2.12. Let 퐾 be a number field and 푃 be a prime ideal of 풪퐾. Then there is a fractional ideal 퐽 of 퐾 with 푃퐽 = 〈1〉.

∗ ∗ ∗ Proof: Let 푃 = {훽 ∈ 퐾 ∶ 훽푃 ⊆ 풪퐾}. Then 푃 is a fractional ideal of 퐾, 풪퐾 ⊆ 푃 and 푃 ⊆

∗ ∗ ∗ 푃푃 ⊆ 풪퐾. By the maximality of the prime ideal 푃, either 푃푃 = 푃 or 푃푃 = 풪퐾. We show that the latter is true, so to obtain a contradiction, suppose that 푃푃∗ = 푃.

∗ ∗ Then if 훾 ∈ 푃 we have 훾푃 ⊆ 푃 and so 훾 ∈ 풪퐾 by Lemma 4.2.11 hence 푃 ⊆ 풪퐾 and we

∗ ∗ conclude that 푃 = 풪퐾. To obtain the desired contradiction, it suffices to find an element in 푃 but not in 풪퐾.

Let 훽 be a nonzero element of 푃. Then by Lemma 4.2.7 〈훽〉 contains a product 푃1푃2 ⋅⋅⋅

푃푟 of prime ideals. Choose such a product with fewest possible factors. Then 푃 ⊇ 〈훽〉 ⊇ 푃1푃2 ⋅⋅⋅

푃푟 and so 푃 ⊇ 푃푗 for some 푗 by Lemma 4.2.3. We assume, without loss of generality, that 푃 ⊇

푃1. Then by maximality of the prime ideal 푃1, 푃 = 푃1. Thus 〈훽〉 ⊇ 푃퐼 where 퐼 = 푃2 ⋅⋅⋅ 푃푟. As 푟

27 was chosen to be minimal then 〈훽〉 ⊉ 퐼. Thus, there exists 훾 ∈ 퐼 but 훾 ∉ 〈훽〉. Hence 훿 = 훾/훽 ∉

−1 ∗ 풪퐾. But 훾푃 ⊆ 푃퐼 ⊆ 〈훽〉 and so 훿푃 = 훽 훾푃 ⊆ 풪퐾. So 훿 ∈ 푃 , but as 훾 ∉ 〈훽〉 then 훿 ∉ 풪퐾.

∗ ∗ The assumption that 푃 = 풪퐾 has led to a contradiction. We cannot then have 푃푃 = 푃

∗ and we conclude that 푃푃 = 풪퐾 = 〈1〉. ∎

Theorem 4.2.13. (The Fundamental Theorem of Ideals in Number Fields) Let 퐾 be a number field and 퐼 be a nonzero ideal of 풪퐾. Then 퐼 = 푃1푃2 ⋅⋅⋅ 푃푚, where 푃푗 for 푗 = 1, 2 … are prime ideals and, if the order of the prime ideals is unimportant, this factorization is unique

Proof: We use induction on 푁(퐼). When 퐼 is prime, we are done. Assume not, by Lemma 4.2.6

−1 퐼 ⊆ 푃 for some prime ideal 푃. Let 푃 be the inverse of 푃 as a fractional ideal. Since 푃 ⊆ 풪퐾

−1 −1 −1 −1 then 풪퐾 = 푃푃 ⊆ 풪퐾푃 . Let 퐽 = 퐼푃 , since 퐼 ⊆ 푃 then 퐽 ⊆ 푃푃 = 풪퐾 and 퐽 is an ideal of

풪퐾. Also 퐼 = 푃퐽, thus 퐽 is a proper ideal of 풪퐾.

−1 −1 −1 We know 푃 ⊇ 풪퐾 but 푃 ⊈ 풪퐾 for otherwise 푃 would be 풪퐾 which is not an

−1 −1 inverse of 푃. Hence 퐽 = 퐼푃 ⊇ 퐼풪퐾 = 퐼. If we had 퐽 = 퐼 then 훾퐼 ⊆ 퐼 for all 훾 ∈ 푃 and so

−1 −1 푃 ⊆ 풪퐾 by Lemma 4.2.11, this contradicts 푃 ⊈ 풪퐾. Therefore, 퐼 ⊆ 퐽 but 퐼 ≠ 퐽 and so

푁(퐽) < 푁(퐼). By the inductive hypothesis 퐽 is a product of primes, and hence so is 퐼 = 푃퐽. ∎

The unique factorization theorem has some quick consequences

Corollary 4.2.14. Let 퐾 be a number field and 퐼1, 퐼2 be nonzero ideals of 풪퐾. Then 퐼1 ⊇ 퐼2 if and only if there is an ideal 퐽 of 풪퐾 with 퐼2 = 퐼1 퐽.

Proof: If 퐼2 = 퐼1퐽 for an ideal 퐽 then 퐼1 | 퐼2, we obtain 퐼1 ⊇ 퐼2. Conversely, suppose 퐼1 ⊇ 퐼2, then

−1 −1 −1 −1 풪퐾 = 퐼1퐼1 ⊇ 퐼2퐼1 . Say 퐽 = 퐼2퐼1 is an ideal of 풪퐾 and 퐼1퐽 = 퐼1퐼2퐼 = 퐼2. ∎

Theorem 4.2.15. Let 퐾 be a number field and 푚 be a positive integer. There are only finitely many ideals 퐼 in 풪퐾 with 푁(퐼) = 푚.

푒1 푒2 푒푔 Proof: Let 〈푚〉 = 푝1 푝2 ⋅⋅⋅ 푝푔 be a product of primes. 〈푚〉 ⊆ 퐼 implies 퐼 | 〈푚〉, then we obtain

푓1 푓2 푓푔 퐼 = 푝1 푝2 ⋅⋅⋅ 푝푔 where 0 ≤ 푓푖 ≤ 푔푖. So there are only finitely many possibilities for 퐼. ∎

Chapter 5. Ideal class group and class number

In this chapter, we will introduce ideal class group and some of its important properties.

For a number field 퐾, we denote by 퐼푘 the abelian group formed by the set of fractional ideals of

퐾 under multiplication; by 푃푘 we define the subgroup of 퐼푘 consisting of the principal fractional ideals.

Definition 5.0.1. The 푐푙푎푠푠 푔푟표푢푝 of 퐾, denoted by 퐶푙퐾, is the quotient group 퐼푘/푃푘; the elements of 퐶푙퐾, namely the cosets of 푃푘 in 퐼푘, are called 푖푑푒푎푙 푐푙푎푠푠푒푠; the number of ideal classes (i.e., the cardinality of 퐶푙퐾), denoted by ℎ퐾, is called the class number of 퐾.

An ideal class is an equivalence class of fractional ideals under the equivalence relation

퐼 ∼ 퐽 if 퐼 = 훼퐽 for some nonzero element 훼 in 퐾. The set of principal fractional ideals forms an ideal class, the 푝푟푖푛푐푖푝푎푙 푖푑푒푎푙 푐푙푎푠푠. We denote the ideal class containing the fractional ideal

퐼 by [퐼]. We can see 퐶푙퐾 as the set of such symbols [퐼] follow the rules:

(1) [퐼] = [퐽] whenever 퐽 = 훼퐼, and the operation [퐼][퐽] = [퐼퐽] is well defined,

(2) the identity element of 퐶푙퐾 is [〈1〉] = [풪퐾],

(3) if 퐼 = 훼퐽 for some ideal 퐽, then [퐼] = [퐽] and so each ideal class contains ideals.

Example 5.0.2. ℤ, ℤ[ω], and ℤ[√−1] have trivial ideal class groups, where ω is a primitive cubic root of 1.

5.2. Finiteness of 퐶푙퐾

Proposition 5.1.1. Let 퐾 be a number field. There is a positive number 푚 ∈ 퐾 such that, in each nonzero ideal 퐼 of 풪퐾, there is a nonzero element 훽 with |푁(훽)| ≤ 푚푁(퐼).

Proof: Let 훼1, 훼2, … , 훼푛 form an integral basis of 풪퐾. Let 푞 = 푁(퐼) and let 푟 be the integer part

1 1 of 푞푛, i.e., 푟 ≤ 푞푛 < 푟 + 1. Consider set

푆 = {푏1훼1 + 푏2훼2 + ⋯ + 푏푛훼푛 ∶ 푏푗 ∈ ℤ, 0 ≤ 푏푗 ≤ 푟}.

We can see that S has (푟 + 1)푛 elements and so |푆| > 푞. The elements of S cannot lie in distinct cosets of 퐼 in 풪퐾. Hence there are 훽1, 훽2 ∈ 푆 with 훽1 ≠ 훽2 but 훽1 ≡ 훽2(푚표푑 퐼). Set 훽 =

푛 훽1 − 훽2. Then 훽 ≠ 0 but 훽 ∈ 퐼. Also 훽 = ∑푗=1 푐푗훼푗 where 푐푗 ∈ ℤ and |푐푗| ≤ 푟.

푛 푛 Now 푁(훽) = ∏푘=1 휎푘( 훽) and 휎푘(훽) = ∑푗=1 푐푗휎푘(훼푗). Thus

푛 푛 |휎푘(훽)| ≤ ∑푗=1 |푐푗||휎푘(훼푗)| ≤ 푟 ∑푗=1 |휎푘(훼푗)|.

Multiplying all together, we obtain

푛 푛 푛 |푁(훽)| ≤ 푟 ∏ ∑ |휎푘(훼푗)| ≤ 푚푞 = 푚푁(퐼) 푘=1 푗=1

푛 푛 With 푚 = ∏푘=1 ∑푗=1|휎푘(훼푗)| . ∎

Proposition 5.1.2. If m is as in the above proposition, then every ideal class in 퐶푙퐾 contains an ideal of norm ≤ 푚.

Proof: Let 퐼 be an ideal of 풪퐾 and consider its ideal class [퐼]. By Proposition 5.1.1. there exists 훽 ∈ 퐼, 훽 ≠ 0 with |푁(훽)| ≤ 푚푁(퐼). Since 퐼 ⊇ 〈훽〉, by Lemma 5.2.12 there is an ideal

퐽 ⊆ 풪퐾 with 〈훽〉 = 퐼퐽. Then as [퐼][퐽] = [〈훽〉] = [풪퐾] it follows that the class [퐽] is the inverse of

푁(〈훽〉) |푁(훽)| [퐼], [퐽] = [퐼]−1. Now 푁(퐽) = = ≤ 푚. Therefore, the ideal class [퐼]−1 has an ideal of 푁(퐼) 푁(퐼) norm at most m. ∎

Theorem 5.1.3. Let 퐾 be a number field, then 퐶푙퐾 is a finite abelian group.

Proof: By Propositions 5.1.1 and 5.1.2, there is a positive number 푚 with each class of 퐾 containing an ideal of 풪퐾 with norm at most 푚. By Lemma 5.2.1, there are only finitely many ideals 퐼 of 풪퐾 of each given norm, and so there are only finitely many ideals of norm at most 푚.

Hence 퐾 has only finitely many ideal classes, that is 퐶푙퐾 is a finite abelian group. ∎

Remark 5.1.4. The above results (and their proofs) suggest a method to determine 퐶푙퐾. Given 퐾, we calculate the bound 푚 (defined as in the proof of Proposition 5.1.1), and then find all integral ideals of norm at most 푚. Since every ideal class has a representative among these ideals, we can find 퐶푙퐾 by determining how many classes these ideals belong to.

Remark 5.1.5. In the quadratic case that 퐾 = ℚ(√푑), where |푑| is a square-free integer, the bound 푚 is equal to

2 푚 = (|휎1(1)| + |휎1(√푑)|)(|휎2(1)|+|휎2(√푑)|) = (1 + √|푑|) ----- (5.1.1) if 푑 ≡ 2, 3 푚표푑 4, and

(3+√|푑|)2 푚 = (|휎 (1)| + |휎 ((1 + √푑)/2)|)(|휎 (1)|+|휎 ((1 + √푑)/2)|) = ----- (5.1.2) 1 1 2 2 4 if 푑 ≡ 1 푚표푑 4.

Example 5.1.6. Let 퐾 = ℚ(√−6). Then 풪퐾 has integral basis 1, √−6. Using the proof of

Proposition 5.1.3 we can take

2 푚 = (|휎1(1)| + |휎1(√−6)|)(|휎2(1)|+|휎2(√−6)|) = (1 + √6) ≈ 11.9.

We shall find all prime ideals of norm at most 11. We find that 2 and 3 split, 5 and 7 and

2 2 11 split. In detail 〈2〉 = 푃2 where 푃2 = 〈2, √−6 〉, 〈3〉 = 푃3 where 푃3 = 〈3, √−6 〉, 〈5〉 = 푃5푄5 where 푃5 = 〈5, √−6 + 2〉 and 푄5 = 〈5, √−6 − 2〉, 〈7〉 = 푃7푄7 where 푃7 = 〈7, √−6 + 1 〉 and

푄7 = 〈7, √−6 − 1 〉, and 〈11〉 = 푃11푄11 where 푃11 = 〈11, √−6 + 4〉 and 푄11 = 〈11, √−6 − 4 〉.

Thus 푃2, 푃3, 푃5, 푄5, 푃7, 푄7, 푃11 푎푛푑 푄11 are prime ideals of norm at most 11, and each ideal class of 퐾 is equal to one of [푃2], [푃3] … [푄11] as a product of some of those, e.g. 푃2푃3 if the norm is ≤

11.

Now we look at the factorization of principal ideals of small norm to find the relations

2 2 between these ideal classes. We’ve already know that [푃2] = [〈2〉] = [〈1〉], [푃3] = [〈3〉] =

[〈1〉], [푃5][푄5] = [〈5〉] = [〈1〉], [푃7][푄7] = [〈7〉] 푎푛푑 [푃11][푄11] = [〈11〉] = [〈1〉]. Consider

〈√−6〉, It has norm 6 = 2 ∙ 3, that is the product of prime ideals of norms 2 and 3. Hence √−6 =

2 푃2푃3 and [푃2][푃3] = [〈1〉]. As [푃2] = [〈1〉] then [푃2] = [푃3]. Next 〈1 + √−6〉 has norm 7, so this is a prime ideal of norm 7. As 1 + √−6 ∈ 푃7 then 푃7 = 〈1 + √−6〉 is principal, and so

−1 [푃7] = [〈1〉] and [푄7] = [〈1〉][푃7] = [〈1〉]. Next 〈2 + √−6〉 has norm 10 and so is the product of prime ideals of norms 2 and 5. Hence it is either 푃2푃5 or 푃2푄5. But 2 + √−6 ∈ 푃5 and so

−1 −1 〈2 + √−6〉 ⊆ 푃5. Hence 〈2 + √−6〉 = 푃2푃5 and [푃5] = [푃2] = [푃2]. Also [푄5] = [푃5] =

−1 [푃2] = [푃2]. Then 〈4 + √−6〉 has norm 22 and so is the product of prime ideals of norms 2 and 11. It is either 푃2푃11 or 푃2푄11. But 4 + √−6 ∈ 푃11 and so 〈4 + √−6〉 ⊆ 푃11. Hence

−1 −1 −1 〈4 + √−6〉 = 푃2푃11 and so [푃11] = [푃2] = [푃2]. Also [푄11] = [푃11] = [푃2] = [푃2]. In conclusion, we have [푃2] = [푃3] = [푃5] = [푄5] = [푃11] = [푄11] and [푃7] = [푄7] = [〈1〉]. We

2 also have [푃2] = [〈1〉]. Therefore, 퐶푙푘 = {[〈1〉], [푃2]}. We have seen that 푃2 is non-principal so that [푃2] ≠ [〈1〉]. So the class group has two element and ℎ푘 = 2.

5.2. The Minkowski bound

In the previous section, we have seen that there is a positive number 푚 with each class of

퐾 containing an ideal of 풪퐾 with norm at most 푚. This yields a method for us to find 퐶푙퐾. This whole process, however, is often very complicated. In this section, we introduce a bound that improves the 푚 used in section 5.1 a little bit. In applications, this new bound largely reduces the complexity of computation.

Let number field 퐾 = ℚ(훼), where 훼 has degree 푛. Consider the conjugates 훼1, 훼2, … , 훼푛 of 훼. Denote 푠 as the number of 훼푗 which are real. Since the nonreal conjugates come in pairs, there are 2푡 nonreal conjugates of 훼 and 푠 + 2푡 = 푛. The 푀푖푛푘표푤푠푘푖 푏표푢푛푑 of K is

4 푡 푛! 푀 = ( ) √|∆ | , 퐾 휋 푛푛 푘

where ∆푘 is the discriminant of 퐾.

1 In particular, when 퐾 is real quadratic then 푀 = √|∆ | , and when 퐾 is imaginary 퐾 2 푘

2 quadratic, we have 푀 = √|∆ |. Comparing these with the 푚 given by (5.1.1) and (5.1.2), one 퐾 휋 푘 sees that 푀퐾 is significantly smaller in general.

Theorem 5.2.1. Each ideal class of 퐾 contains an ideal 퐼 of 풪퐾 with 푁(퐼) ≤ 푀퐾.

Example 5.2.2. Let 퐾 = ℚ(√82). We will show the class group is cyclic of order 4.

Note that 푛 = 2, 푡 = 0, ∆퐾= 4 ⋅ 82, we have the Minkowski bound approximately 9.055. Now

34 we look at the primes lying over 2, 3, 5, and 7. The following table describes how (푝) factors from the way 푇2 − 82 factor modulo 푝.

푝 푇2 − 82 mod 푝 (푝)

2 푇2 픭2 2 ′ 3 (푇 − 1)(푇 + 1) 픭3픭3

5 Irreducible Prime

7 Irreducible Prime

2 Thus, the class group of ℚ(√82) is generated by [픭2] and [픭3], with 픭2 =< 2 > ~ < 1 >

′ −1 and 픭3 ~ 픭3 .

2 Since 푁퐾/ℚ(10 + √82) = 18 = 2 ⋅ 3 , and 10 + √82 is not divisible by 3, (10 + √82) is

2 −2 divisible by just one of 픭3 and 픭3′. Let 픭3 be that prime, so (10 + √82) = 픭2픭3, Thus 픭2~픭3 , so the class group of 퐾 is generated by [픭3] and we have the formulas

2 2 [픭2] = 1, [픭3] = [픭2].

Hence [픭3] has order dividing 4.

We will show 픭2 is non-principal, so [픭3] has order 4, and thus 퐾 has a class group 〈[픭3]〉 ≅

2 2 ℤ/4ℤ. If 픭2 = (푎 + 푏√82), then 푎 − 82푏 = ±2, so 2 or −2 is congruent to a square module

41. This is no contradiction, since 2 ≡ 172 푚표푑 41. We need a different idea.

2 2 The idea is to use the known fact that 픭2 is principal. If 픭2 = (푎 + 푏√82), then (2) = 픭2 =

2 2 ((푎 + 푏√82) ), so 2 = (푎 + 푏√82) 푢, where 푢 is a unit.

Taking norms here 푁(푢) must be positive, so 푁(푢) = 1. The unit group of ℤ[√82] is

ℤ ±(9 + √82) , with 9 + √82 having norm −1. Therefore, the positive units of norm 1 are the

2 integral powers of (9 + √82) , which are all squares. A unit square can be absorbed into the

2 2 (푎 + 푏√82) terms, so we have to be able to solve 2 = (푎 + 푏√82) in integers 푎 and 푏. This implies √2 lies in ℤ[√82]. Thus, 픭2 is not principal.

Example 5.2.3. Let 퐾 = ℚ(√−14). We will show the class group is cyclic of order 4. 푛 =

2, 푡 = 1, ∆푘= 4 ⋅ −14 = −56, so the Minkowski bound is approximately 4.764. So the class group is generated by primes dividing (2) and (3) . The following table describes how (2) and

2 (3) factor in 풪퐾 from the way 푇 − 82 factor modulo 2 and modulo 3.

푝 푇2 + 14 mod 푝 (푝)

2 2 2 푇 픭2

′ 3 (푇 − 1)(푇 + 1) 픭3픭3

2 −1 ′ −1 Since 픭2 ~ 1 and 픭2 ~ 픭2 . Since 픭3픭3~ 1, 픭3 ~ 픭3 . Therefore, the class group of 퐾 is

2 2 generated by [픭2] and [픭3]. Both 픭2 and 픭3 are non-principal, since the equations 푎 + 14푏 = 2

2 2 and 푎 + 14푏 = 3 have no integral solutions. To find relations between 픭2 and 픭3, we use

2 푁퐾/ℚ(2 + √−14) = 18 = 2 ⋅ 3 . And the ideal (10 + √82 ) is divisible by just one of 픭3 and

픭3′. Since 10 + √82 is not a multiple of 3. We may let 픭3 be the prime of norm 3 dividing

2 2 −1 (2 + √−14). Then 픭2픭3 ~ 1, so 픭3 ~ 픭2 ~ 픭2 and the class group of 퐾 is generated by [픭3].

2 Since 픭2 is non-principal and 픭2 ~ 1, [픭3] has order 4. Hence, the class group of 퐾 is cyclic of order 4.

5.3. Further remarks

Other than that the class number of a given number field is finite, more about the values of class numbers is known to us.

Take the quadratic case as an example. The classical class number problem of Gauss asks for a classification of all quadratic fields with a given class number 푘. In particular, Gauss conjectured

(1). ℎ푑 → ∞ as −푑 → ∞ (thus ℎ푑 can be arbitrarily large);

(2). For each positive integer 푘, there are finitely many imaginary quadratic fields with

class number equal to 푘. In particular, there are exactly 9 imaginary quadratic fields with

class number one, namely: ℚ(√−1), ℚ(√−2), ℚ(√−3), ℚ(√−7), ℚ(√−11),

ℚ(√−19), ℚ(√−43), ℚ(√−67), and ℚ(√−163);

(3). There are infinitely many real quadratic fields with class number one.

These conjectures have stimulated the development of number theory. We do not want to review the complete history here, but it is worth mentioning that conjecture (1) was proved by

Heilbronn in 1934; The class number 1 problem of imaginary quadratic fields was solved from the work of Baker (1966), Stark (1967), and Heegner (1952); The class number 2 problem was solved from the work of Baker (1971), and Stark (1971); In 2004, Mark Watkins found the complete list of imaginary quadratic fields with class number up to 100.

In the real quadratic case, we have a completely different story. Lack of understanding of the fundamental units, people know very little about how the class numbers are distributed. In view of this, conjecture (3) is still big open.

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