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Part II — Number Fields

Based on lectures by I. GrojnowskiNotes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Part IB Groups, Rings and Modules is essential and Part II Galois Theory is desirable

Definition of algebraic number fields, their integers and units. Norms, bases anddiscriminants. [3]

Ideals, principal and prime ideals, unique factorisation. Norms of ideals. [3]

Minkowski’s theorem on convex bodies. Statement of Dirichlet’s unit theorem. Deter-mination of units in quadratic fields. [2]

Ideal classes, finiteness of the class group. Calculation of class numbers using statementof the Minkowski bound. [3]

Dedekind’s theorem on the factorisation of primes. Application to quadratic fields. [2]

Discussion of the cyclotomic field and the Fermat equation or some other topic chosen

by the lecturer. [3]

1

Contents II Number Fields

Contents

0 Introduction 3

1 Number fields 4

2 Norm, trace, discriminant, numbers 10

3 Multiplicative structure of ideals 17

4 Norms of ideals 27

5 Structure of prime ideals 32

6 Minkowski bound and finiteness of class group 37

7 Dirichlet’s unit theorem 48

8 L-functions, Dirichlet series* 56

Index 68

2

0 Introduction II Number Fields

0 Introduction

Technically, IID Galois Theory is not a prerequisite of this course. However,many results we have are analogous to what we did in Galois Theory, and wewill not refrain from pointing out the correspondence. If you have not learntGalois Theory, then you can ignore them.

3

1 Number fields II Number Fields

1 Number fields

The focus of this course is, unsurprisingly, number fields. Before we define whatnumber fields are, we look at some motivating examples. Suppose we wanted tofind all numbers of the form x2 + y2, where x, y ∈ Z. For example, if a, b canboth be written in this form, does it follow that ab can?

In IB Groups, Rings and Modules, we did the clever thing of working withZ[i]. The integers of the form x2 + y2 are exactly the norms of integers in Z[i],where the norm of x+ iy is

N(x+ iy) = |x+ iy|2 = x2 + y2.

Then the previous result is obvious — if a = N(z) and b = N(w), then ab =N(zw). So ab is of the form x2 + y2.

Similarly, in the IB Groups, Rings and Modules example sheet, we foundall solutions to the equation x2 + 2 = y3 by working in Z[

√−2]. This is a very

general technique — working with these rings, and corresponding fields Q(√−d)

can tell us a lot about arithmetic we care about.In this chapter, we will begin by writing down some basic definitions and

proving elementary properties about number fields.

Definition (Field extension). A field extension is an inclusion of fields K ⊆ L.We sometimes write this as L/K.

Definition (Degree of field extension). Let K ⊆ L be fields. Then L is a vectorspace over K, and the degree of the field extension is

[L : K] = dimK(L).

Definition (Finite extension). A finite field extension is a field extension withfinite degree.

Definition (Number field). A number field is a finite field extension over Q.

A field is the most boring kind of ring — the only ideals are the trivial oneand the whole field itself. Thus, if we want to do something interesting withnumber fields algebraically, we need to come up with something more interesting.

In the case of Q itself, one interesting thing to talk about is the integers Z.It turns out the right generalization to number fields is algebraic integers.

Definition (Algebraic integer). Let L be a number field. An algebraic integeris an α ∈ L such that there is some monic f ∈ Z[x] with f(α) = 0. We write OLfor the set of algebraic integers in L.

Example. It is a fact that if L = Q(i), then OL = Z[i]. We will prove this inthe next chapter after we have the necessary tools.

These are in fact the main objects of study in this course. Since we say thisis a generalization of Z ⊆ Q, the following had better be true:

Lemma. OQ = Z, i.e. α ∈ Q is an algebraic integer if and only if α ∈ Z.

4


Proof. If α ∈ Z, then x− α ∈ Z[x] is a monic polynomial. So α ∈ OQ.On the other hand, let α ∈ Q. Then there is some coprime r, s ∈ Z such that

α = rs . If it is an algebraic integer, then there is some

f(x) = xn + an−1xn−1 + · · ·+ a0

with ai ∈ Z such that f(α) = 0. Substituting in and multiplying by sn, we get

rn + an−1rn−1s+ · · ·+ a0s

n︸︷︷︸divisible by s

= 0,

So s | rn. But if s 6= 1, there is a prime p such that p | s, and hence p | rn. Thusp | r. So p is a common factor of s and r. This is a contradiction. So s = 1, andα is an integer.

How else is this a generalization of Z? We know Z is a ring. So perhaps OLalso is.

Theorem. OL is a ring, i.e. if α, β ∈ OL, then so is α± β and αβ.

Note that in general OL is not a field. For example, Z = OQ is not a field.The proof of this theorem is not as straightforward as the previous one.

Recall we have proved a similar theorem in IID Galois Theory before with“algebraic integer” replaced with “algebraic number”, namely that if L/K is afield extension with α, β ∈ L algebraic over K, then so is αβ and α± β, as wellas 1

α if α 6= 0.To prove this, we notice that α ∈ K is algebraic if and only if K[α] is a finite

extension — if α is algebraic, with

f(α) = anαn + · · ·+ a0 = 0, an 6= 0

then K[α] has degree at most n, since αn (and similarly α−1) can be written asa linear combination of 1, α, · · · , αn−1, and thus these generate K[α]. On theother hand, if K[α] is finite, say of degree k, then 1, α, · · · , αk are independent,hence some linear combination of them vanishes, and this gives a polynomialfor which α is a root. Moreover, by the same proof, if K ′ is any finite extensionover K, then any element in K ′ is algebraic.

Thus, to prove the result, notice that if K[α] is generated by 1, α, · · · , αnand K[β] is generated by 1, β, · · · , βm, then K[α, β] is generated by {αiβj} for1 ≤ i ≤ n, 1 ≤ j ≤ m. Hence K[α, β] is a finite extension, hence αβ, α ± β ∈K[α, β] are algebraic.

We would like to prove this theorem in an analogous way. We will considerOL as a ring extension of Z. We will formulate the general notion of “being analgebraic integer” in general ring extensions:

Definition (Integrality). Let R ⊆ S be rings. We say α ∈ S is integral over Rif there is some monic polynomial f ∈ R[x] such that f(α) = 0.

We say S is integral over R if all α ∈ S are integral over R.

Definition (Finitely-generated). We say S is finitely-generated over R if thereexists elements α1, · · · , αn ∈ S such that the function Rn → S defined by(r1, · · · , rn) 7→

∑riαi is surjective, i.e. every element of S can be written as a R-

linear combination of elements α1, · · · , αn. In other words, S is finitely-generatedas an R-module.

5


This is a refinement of the idea of being algebraic. We allow the use of ringsand restrict to monic polynomials. In Galois theory, we showed that finitenessand algebraicity “are the same thing”. We will generalize this to integrality ofrings.

Example. In a number field Z ⊆ Q ⊆ L, α ∈ L is an algebraic integer if andonly if α is integral over Z by definition, and OL is integral over Z.

Notation. If α1, · · · , αr ∈ S, we write R[α1, · · · , αr] for the subring of Sgenerated by R,α1, · · · , αr. In other words, it is the image of the homomorphismfrom the polynomial ring R[x1, · · · , xn]→ S given by xi 7→ αi.

Proposition.

(i) Let R ⊆ S be rings. If S = R[s] and s is integral over R, then S isfinitely-generated over R.

(ii) If S = R[s1, · · · , sn] with si integral over R, then S is finitely-generatedover R.

This is the easy direction in identifying integrality with finitely-generated.

Proof.

(i) We know S is spanned by 1, s, s2, · · · over R. However, since s is integral,there exists a0, · · · , an ∈ R such that

sn = a0 + a1s+ · · ·+ an−1sn−1.

So the R-submodule generated by 1, s, · · · , sn−1 is stable under multiplica-tion by s. So it contains sn, sn+1, sn+2, · · · . So it is S.

(ii) Let Si = R[s1, · · · , si]. So Si = Si−1[si]. Since si is integral over R, it isintegral over Si−1. By the previous part, Si is finitely-generated over Si−1.To finish, it suffices to show that being finitely-generated is transitive.More precisely, if A ⊆ B ⊆ C are rings, B is finitely generated over A andC is finitely generated over B, then C is finitely generated over A. Thisis not hard to see, since if x1, · · · , xn generate B over A, and y1, · · · , ymgenerate C over B, then C is generated by {xiyj}1≤i≤n,1≤j≤m over A.

The other direction is harder.

Theorem. If S is finitely-generated over R, then S is integral over R.

The idea of the proof is as follows: if s ∈ S, we need to find a monicpolynomial which it satisfies. In Galois theory, we have fields and vector spaces,and the proof is easy. We can just consider 1, s, s2, · · · , and linear dependencekicks in and gives us a relation. But even if this worked in our case, there is noway we can make this polynomial monic.

Instead, consider the multiplication-by-s map: ms : S → S by γ 7→ sγ. If Swere a finite-dimensional vector space over R, then Cayley-Hamilton tells us ms,and thus s, satisfies its characteristic polynomial, which is monic. Even thoughS is not a finite-dimensional vector space, the proof of Cayley-Hamilton willwork.

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Proof. Let α1, · · · , αn generate S as an R-module. wlog take α1 = 1 ∈ S. Forany s ∈ S, write

sαi =∑

bijαj

for some bij ∈ R. We write B = (bij). This is the “matrix of multiplication byS”. By construction, we have

(sI −B)

α1

...an

= 0. (∗)

Now recall for any matrix X, we have adj(X)X = (detX)I, where the i, jthentry of adj(X) is given by the determinant of the matrix obtained by removingthe ith row and jth column of X.

We now multiply (∗) by adj(sI −B). So we get

det(sI −B)

α1

...αn

= 0

In particular, det(sI−B)α1 = 0. Since we picked α1 = 1, we get det(sI−B) = 0.Hence if f(x) = det(xI −B), then f(x) ∈ R[x], and f(s) = 0.

Hence we obtain the following:

Corollary. Let L ⊇ Q be a number field. Then OL is a ring.

Proof. If α, β ∈ OL, then Z[α, β] is finitely-generated by the proposition. Butthen Z[α, β] is integral over Z, by the previous theorem. So α ± β, αβ ∈Z[α, β].

Note that it is not necessarily true that if S ⊇ R is an integral extension,then S is finitely-generated over R. For example, if S is the set of all algebraicintegers in C, and R = Z, then by definition S is an integral extension of Z, butS is not finitely generated over Z.

Thus the following corollary isn’t as trivial as the case with “integral” replacedby “finitely generated”:

Corollary. If A ⊆ B ⊆ C be ring extensions such that B over A and C over Bare integral extensions. Then C is integral over A.

The idea of the proof is that while the extensions might not be finitely gener-ated, only finitely many things are needed to produce the relevant polynomialswitnessing integrality.

Proof. If c ∈ C, let

f(x) =

N∑i=0

bixi ∈ B[x]

be a monic polynomial such that f(c) = 0. Let B0 = A[b0, · · · , bN ] and letC0 = B0[c]. Then B0/A is finitely generated as b0, · · · , bN are integral over A.Also, C0 is finitely-generated over B0, since c is integral over B0. Hence C0 isfinitely-generated over A. So c is integral over A. Since c was arbitrary, we knowC is integral over A.

7


Now how do we recognize algebraic integers? If we want to show somethingis an algebraic integer, we just have to exhibit a monic polynomial that vanisheson the number. However, if we want to show that something is not an algebraicinteger, we have to make sure no monic polynomial kills the number. How canwe do so?

It turns out to check if something is an algebraic integer, we don’t have tocheck all monic polynomials. We just have to check one. Recall that if K ⊆ Lis a field extensions with α ∈ L, then the minimal polynomial is the monicpolynomial pα(x) ∈ K[x] of minimal degree such that pα(α) = 0.

Note that we can always make the polynomial monic. It’s just that thecoefficients need not lie in Z.

Recall that we had the following lemma about minimal polynomials:

Lemma. If f ∈ K[x] with f(α) = 0, then pα | f .

Proof. Write f = pαh+ r, with r ∈ K[x] and deg(r) < deg(pα). Then we have

0 = f(α) = p(α)h(α) + r(α) = r(α).

So if r 6= 0, this contradicts the minimality of deg pα.

In particular, this lemma implies pα is unique. One nice application of thisresult is the following:

Proposition. Let L be a number field. Then α ∈ OL if and only if the minimalpolynomial pα(x) ∈ Q[x] for the field extension Q ⊆ L is in fact in Z[x].

This is a nice proposition. This gives us an necessary and sufficient conditionfor whether something is algebraic.

Proof. (⇐) is trivial, since this is just the definition of an algebraic integer.(⇒) Let α ∈ OL and pα ∈ Q[x] be the minimal polynomial of α, and

h(x) ∈ Z[x] be a monic polynomial which α satisfies. The idea is to use h toshow that the coefficients of pα are algebraic, thus in fact integers.

Now there exists a bigger field M ⊇ L such that

pα(x) = (x− α1) · · · (x− αr)

factors in M [x]. But by our lemma, pα | h. So h(αi) = 0 for all αi. So αi ∈ OMis an algebraic integer. But OM is a ring, i.e. sums and products of the αi’s arestill algebraic integers. So the coefficients of pα are algebraic integers (in OM ).But they are also in Q. Thus the coefficients must be integers.

Alternatively, we can deduce this proposition from the previous lemma plusGauss’ lemma.

Another relation between Z and Q is that Q is the fraction field of Z. Thisis true for general number fields

Lemma. We have

FracOL =

{α

β: α, β ∈ OL, β 6= 0

}= L.

In fact, for any α ∈ L, there is some n ∈ Z such that nα ∈ OL.

8


Proof. If α ∈ L, let g(x) ∈ Q[x] be its monic minimal polynomial. Then thereexists n ∈ Z non-zero such that ng(x) ∈ Z[x] (pick n to be the least commonmultiple of the denominators of the coefficients of g(x)). Now the magic is toput

h(x) = ndeg(g)g(xn

).

Then this is a monic polynomial with integral coefficients — in effect, we havejust multiplied the coefficient of xi by ndeg(g)−i! Then h(nα) = 0. So nα isintegral.

9

2 Norm, trace, discriminant, numbers II Number Fields

2 Norm, trace, discriminant, numbers

Recall that in our motivating example of Z[i], one important tool was the normof an algebraic integer x + iy, given by N(x + iy) = x2 + y2. This can begeneralized to arbitrary number fields, and will prove itself to be a very usefulnotion to consider. Apart from the norm, we will also consider a number knownas the trace, which is also useful. We will also study numbers associated withthe number field itself, rather than particular elements of the field, and it turnsout they tell us a lot about how the field behaves.

Norm and trace

Recall the following definition from IID Galois Theory:

Definition (Norm and trace). Let L/K be a field extension, and α ∈ L. Wewrite mα : L→ L for the map ` 7→ α`. Viewing this as a linear map of L vectorspaces, we define the norm of α to be

NL/K(α) = detmα,

and the trace to betrL/K(α) = trmα.

The following property is immediate:

Proposition. For a field extension L/K and a, b ∈ L, we have N(ab) =N(a)N(b) and tr(a+ b) = tr(a) + tr(b).

We can alternatively define the norm and trace as follows:

Proposition. Let pα ∈ K[x] be the minimal polynomial of α. Then thecharacteristic polynomial of mα is

det(xI −mα) = p[L:K(α)]α

Hence if pα(x) splits in some field L′ ⊇ K(α), say

pα(x) = (x− α1) · · · (x− αr),

thenNK(α)/K(α) =

∏αi, trK(α)/K(α) =

∑αi,

and hence

NL/K(α) =(∏

αi

)[L:K(α)]

, trL/K = [L : K(α)](∑

αi

).

This was proved in the IID Galois Theory course, and we will just use itwithout proving.

Corollary. Let L ⊇ Q be a number field. Then the following are equivalent:

(i) α ∈ OL.

(ii) The minimal polynomial pα is in Z[x]

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(iii) The characteristic polynomial of mα is in Z[x].

This in particular implies NL/Q(α) ∈ Z and trL/Q(α) ∈ Z.

Proof. The equivalence between the first two was already proven. For theequivalence between (ii) and (iii), if mα ∈ Z[x], then α ∈ OL since it vanisheson a monic polynomial in Z[x]. On the other hand, if pα ∈ Z[x], then so is thecharacteristic polynomial, since it is just pNα .

The final implication comes from the fact that the norm and trace are justcoefficients of the characteristic polynomial.

It would be nice if the last implication is an if and only if. This is in generalnot true, but it occurs, obviously, when the characteristic polynomial is quadratic,since the norm and trace would be the only coefficients.

Example. Let L = K(√d) = K[z]/(z2− d), where d is not a square in K. As a

vector space over K, we can take 1,√d as our basis. So every α can be written

asα = x+ y

√d.

Hence the matrix of multiplication by α is

mα =

(x dyy x

).

So the trace and norm are given by

trL/K(x+ y√d) = 2x = (x+ y

√d) + (x− y

√d)

NL/K(x+ y√d) = x2 − dy2 = (x+ y

√d)(x− y

√d)

We can also obtain this by consider the roots of the minimal polynomial ofα = x+ y

√d, namely (α− x)2 − y2d = 0, which has roots x± y

√d.

In particular, if L = Q(√d), with d < 0, then the norm of an element is just

the norm of it as a complex number.Now that we have computed the general trace and norm, we can use the

proposition to find out what the algebraic integers are. It turns out the result is(slightly) unexpected:

Lemma. Let L = Q(√d), where d ∈ Z is not 0, 1 and is square-free. Then

OL =

{Z[√d] d ≡ 2 or 3 (mod 4)

Z[

12 (1 +

√d)]

d ≡ 1 (mod 4)

Proof. We know x+ y√λ ∈ OL if and only if 2x, x2 − dy2 ∈ Z by the previous

example. These imply 4dy2 ∈ Z. So if y = rs with r, s coprime, r, s ∈ Z, then we

must have s2 | 4d. But d is square-free. So s = 1 or 2. So

x =u

2, y =

v

2

for some u, v ∈ Z. Then we know u2 − dv2 ∈ 4Z, i.e. u2 ≡ dv2 (mod 4). But weknow the squares mod 4 are always 0 and 1. So if d 6≡ 1 (mod 4), then u2 ≡ dv2

11


(mod 4) imply that u2 = v2 = 0 (mod 4), and hence u, v are even. So x, y ∈ Z,giving OL = Z[

√d].

On the other hand, if d ≡ 1 (mod 4), then u, v have the same parity mod 2,i.e. we can write x+ y

√d as a Z-combination of 1 and 1

2 (1 +√d).

As a sanity check, we find that the minimal polynomial of 12 (1 +

√d) is

x2 − x+ 14 (1− d) which is in Z if and only if d ≡ 1 (mod 4).

Field embeddings

Recall the following theorem from IID Galois Theory:

Theorem (Primitive element theorem). LetK ⊆ L be a separable field extension.Then there exists an α ∈ L such that K(α) = L.

For example, Q(√

2,√

3) = Q(√

2 +√

3).Since Q has characteristic zero, it follows that all number fields are separable

extensions. So any number field L/Q is of the form L = Q(α). This makes itmuch easier to study number fields, as the only extra “stuff” we have on top ofQ.

One particular thing we can do is to look at the number of ways we canembed L ↪→ C. For example, for Q(

√−1), there are two such embeddings — one

sends√−1 to i and the other sends

√−1 to −i.

Lemma. The degree [L : Q] = n of a number field is the number of fieldembeddings L ↪→ C.

Proof. Let α be a primitive element, and pα(x) ∈ Q[x] its minimal polynomial.Then by we have deg pα = [L : Q] = n, as 1, α, α2, · · · , αn−1 is a basis. Moreover,

Q[x]

(pα)∼= Q(α) = L.

Since L/Q is separable, we know pα has n distinct roots in C. Write

pα(x) = (x− α1) · · · (x− αn).

Now an embedding Q[x]/(pα) ↪→ C is uniquely determined by the image of x,and x must be sent to one of the roots of pα. So for each i, the map x 7→ αigives us a field embedding, and these are all. So there are n of them.

Using these field embeddings, we can come up with the following alternativeformula for the norm and trace.

Corollary. Let L/Q be a number field. If σ1, · · · , σn : L→ C are the differentfield embeddings and β ∈ L, then

trL/Q(β) =∑

σi(β), NL/Q(β) =∏i

σi(β).

We call σ1(β), · · · , σn(β) the conjugates of β in C.

Proof is in the Galois theory course.Using this characterization, we have the following very concrete test for when

something is a unit.

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Lemma. Let x ∈ OL. Then x is a unit if and only if NL/Q(x) = ±1.

Notation. Write O×L = {x ∈ OL : x−1 ∈ OL}, the units in OL.

Proof. (⇒) We know N(ab) = N(a)N(b). So if x ∈ O×L , then there is somey ∈ OL such that xy = 1. So N(x)N(y) = 1. So N(x) is a unit in Z, i.e. ±1.

(⇐) Let σ1, · · · , σn : L→ C be the n embeddings of L in C. For notationalconvenience, We suppose that L is already subfield of C, and σ1 is the inclusionmap. Then for each x ∈ OL, we have

N(x) = xσ2(x) · · ·σn(x).

Now if N(x) = ±1, then x−1 = ±σ2(x) · · ·σn(x). So we have x−1 ∈ OL, sincethis is a product of algebraic integers. So x is a unit in OL.

Corollary. If x ∈ OL is such that N(x) is prime, then x is irreducible.

Proof. If x = ab, then N(a)N(b) = N(x). Since N(x) is prime, either N(a) = ±1or N(b) = ±1. So a or b is a unit.

We can consider a more refined notion than just the number of field embed-dings.

Definition (r and s). We write r for the number of field embeddings L ↪→ R,and s the number of pairs of non-real field embeddings L ↪→ C. Then

n = r + 2s.

Alternatively, r is the number of real roots of pα, and s is the number of pairs ofcomplex conjugate roots.

The distinction between real embeddings and complex embeddings will beimportant in the second half of the course.

Discriminant

The final invariant we will look at in this chapter is the discriminant. It is basedon the following observation:

Proposition. Let L/K be a separable extension. Then a K-bilinear formL × L → K defined by (x, y) 7→ trL/K(xy) is non-degenerate. Equivalent, ifα1, · · · , αn are a K-basis for L, the Gram matrix (tr(αiαj))i,j=1,··· ,n has non-zerodeterminant.

Recall from Galois theory that if L/K is not separable, then trL/K = 0, andit is very very degenerate. Also, note that if K is of characteristic 0, then there isa quick and dirty proof of this fact — the trace map is non-degenerate, becausefor any x ∈ K, we have trL/K(x · x−1) = n 6= 0. This is really the only casewe care about, but in the proof of the general result, we will also find a usefulformula for the discriminant when the basis is 1, θ, θ2, . . . , θn−1.

We will use the following important notation:

Notation.∆(α1, · · · , αn) = det(trL/K(αiαj)).

13


Proof. Let σ1, · · · , σn : L → K be the n distinct K-linear field embeddingsL ↪→ K. Put

S = (σi(αj))i,j=1,··· ,n =

σ1(α1) · · · σ1(αn)...

. . ....

σn(α1) · · · σn(αn).

Then

STS =

(n∑k=1

σk(αi)σk(αj)

)i,j=1,···n

.

We know σk is a field homomorphism. So

n∑k=1

σk(αi)σk(αj) =

n∑k=1

σk(αiαj) = trL/K(αiαj).

SoSTS = (tr(αiαj))i,j=1,··· ,n.

So we have∆(α1, · · · , αn) = det(STS) = det(S)2.

Now we use the theorem of primitive elements to write L = K(θ) such that1, θ, · · · , θn−1 is a basis for L over K, with [L : K] = n. Now S is just

S =

1 σ1(θ) · · · σ1(θ)n−1

......

. . ....

1 σn(θ) · · · σn(θ)n−1

.

This is a Vandermonde matrix, and so

∆(1, θ, · · · , θn−1) = (detS)2 =∏i<j

(σi(θ)− σj(θ))2.

Since the field extension is separable, and hence σi 6= σj for all i, j, this impliesσi(θ) 6= σj(θ), since θ generates the field. So the product above is non-zero.

So we have this nice canonical bilinear map. However, this determinant isnot canonical. Recall that if α1, · · · , αn is a basis for L/K, and α′1, · · · , α′n isanother basis, then

α′i =∑

aijαj

for some A = (aij) ∈ GLn(K). So

∆(α′1, · · · , α′n) = (detA)2∆(α1, · · · , αn).

However, for number fields, we shall see that we can pick a “canonical” basis,and get a canonical value for ∆. We will call this the discriminant.

Definition (Integral basis). Let L/Q be a number field. Then a basis α1, · · · , αnof L is an integral basis if

OL =

{n∑i=1

miαi : mi ∈ Z

}=

n⊕1

Zαi.

In other words, it is simultaneously a basis for L over Q and OL over Z.

14


Note that integral bases are not unique, just as with usual bases. Given onebasis, you can get any other by acting by GLn(Z).

Example. Consider Q(√d) with d square-free, d 6= 0, 1. If d ∼= 1 (mod 4),

we’ve seen that 1, 12 (1 +

√λ) is an integral basis. Otherwise, if d ∼= 2, 3 (mod 4),

then 1,√d is an integral basis.

The important theorem is that an integral basis always exists.

Theorem. Let Q/L be a number field. Then there exists an integral basis forOL. In particular, OL ∼= Zn with n = [L : Q].

Proof. Let α1, · · · , αn be any basis of L over Q. We have proved that there issome ni ∈ Z such that niαi ∈ OL. So wlog α1, · · · , αn ∈ OL, and are an basis ofL over Q. Since αi are integral, so are αiαj , and so all these have integer trace,as we have previously shown. Hence ∆(α1, · · · , αn), being the determinant of amatrix with integer entries, is an integer.

Now choose a Q-basis α1, · · · , αn ∈ OL such that ∆(α1, · · · , αn) ∈ Z \ {0}has minimal absolute value. We will show that these are an integral basis.

Let x ∈ OL, and write

x =∑

λiαi

for some λi ∈ Q. These λi are necessarily unique since α1, · · · , αn is a basis.Suppose some λi 6∈ Z. wlog say λ1 6∈ Z. We write

λ1 = n1 + ε1,

for n1 ∈ Z and 0 < ε1 < 1. We put

α′1 = x− n1α1 = ε1α1 + λ2α2 + · · ·+ λnαn ∈ OL.

So α′1, α2, · · · , αn is still a basis for L/Q, and are still in OL. But then

∆(α′1, · · · , αn) = ε21 ·∆(α1, · · · , αn) < ∆(α1, · · · , αn).

This contradicts minimality. So we must have λi ∈ Z for all Z. So this is a basisfor OL.

Now if α′1, · · · , α′n is another integral basis of L over Q, then there is someg ∈ GLn(Z) such that gαi = α′i. Since det(g) is invertible in Z, it must be 1 or−1, and hence

det ∆(α′1, · · · , α′n) = det(g)2∆(α1, · · · , αn) = ∆(α1, · · · , αn)

and is independent of the choice of integral basis.

Definition (Discriminant). The discriminant DL of a number field L is definedas

DL = ∆(α1, · · · , αn)

for any integral basis α1, · · · , αn.

15


Example. Let L = Q(√d), where d 6= 0, 1 and d is square-free. If d ∼= 2, 3

(mod 4), then it has an integral basis 1,√d. So

DL = det

(1√d

1 −√d

)2

= 4d.

Otherwise, if d ∼= 1 (mod 4), then

DL = det

(1 1

2 (1 +√d)

1 12 (1−

√d)

)2

= d.

Recall that we have seen the word discriminant before, and let’s make surethese concepts are more-or-less consistent. Recall that the discriminant of apolynomial f(x) =

∏(x− αi) is defined as

disc(f) =∏i<j

(αi − αj)2 = (−1)n(n−1)/2∏i 6=j

(αi − αj).

If pθ(x) ∈ K[x] is the minimal polynomial of θ (where L = K[θ]), then the rootsof pθ are σi(θ). Hence we get

disc(pθ) =∏i<j

(σi(θ)− σj(θ))2.

In other words,disc(pθ) = ∆(1, θ, · · · , θn−1).

So this makes sense.

16

3 Multiplicative structure of ideals II Number Fields

3 Multiplicative structure of ideals

Again, let L/Q be a number field. It turns out that in general, the integralring OL is not too well-behaved as a ring. In particular, it fails to be a UFD ingeneral.

Example. Let L = Q(√

5). Then OL = Z[√−5]. Then we find

3 · 7 = (1 + 2√−5)(1− 2

√−5).

These have norms 9, 49, 21, 21. So none of 3, 7, 1 + 2√

5 are associates.Moreover, 3, 7, 1± 2

√−5 are all irreducibles. The proof is just a straightfor-

ward check on the norms.For example, to show that 3 is irreducible, if 3 = αβ, then 9 = N(3) =

N(α)N(β). Since none of the terms on the right are±1, we must haveN(α) = ±3.But there are no solutions to

x2 + 5y2 = ±3

where x, y are integers. So there is no α = x+ y√−5 such that N(α) = ±3.

So unique factorization fails.

Note that it is still possible to factor any element into irreducibles, just notuniquely — we induct on |N(α)|. If |N(α)| = 1, then α is a unit. Otherwise, αis either irreducible, or α = βγ. Since N(β)N(γ) = N(α), and none of them are±1, we must have |N(β)|, |N(γ)| < |N(α)|. So done by induction.

To fix the lack of unique factorization, we instead look at ideals in OL.This has a natural multiplicative structure — the product of two ideals a, b isgenerated by products ab, with a ∈ a, b ∈ b. The big theorem is that every idealcan be written uniquely as a product of prime ideals.

Definition (Ideal multiplication). Let a, bCOL be ideals. Then we define theproduct ab as

ab =

∑i,j

αiβj : αi ∈ a, βj ∈ b

.

We write a | b if there is some ideal c such that ac = b, and say a divides b.

The proof of unique factorization is the same as the proof that Z is a UFD.Usually, when we want to prove factorization is unique, we write an object as

a = x1x2 · · ·xm = y1y2 · · · yn.

We then use primality to argue that x1 must be equal to some of the yi, andthen cancel them from both sides. We can usually do this because we are workingwith an integral domain. However, we don’t have this luxury when working withideals.

Thus, what we are going to do is to find inverses for our ideals. Of course,given any ideal a, there is no ideal a−1 such that aa−1 = OL, as for any b,we know ab is contained in a. Thus we are going to consider more generalobjects known as fractional ideals, and then this will allow us to prove uniquefactorization.

17


Even better, we will show that a | b is equivalent to b ⊆ a. This is a veryuseful result, since it is often very easy to show that b ⊆ a, but it is usually veryhard to actually find the quotient a−1b.

We first look at some examples of multiplication and factorization of idealsto get a feel of what these things look like.

Example. We have

〈x1, · · · , xn〉〈y1, · · · , ym〉 = 〈xiyj : 1 ≤ i ≤ n, 1 ≤ j ≤ m〉.

In particular,〈x〉〈y〉 = 〈xy〉.

It is also an easy exercise to check (ab)c = a(bc).

Example. In Z[√−5], we claim that we have

〈3〉 = 〈3, 1 +√−5〉〈3, 1−

√−5〉.

So 〈3〉 is not irreducible.Indeed, we can compute

〈3, 1 +√−5〉〈3, 1−

√−5〉 = 〈9, 3(1 + 2

√−5), 3(1− 2

√−5), 21〉.

But we know gcd(9, 21) = 3. So 〈9, 21〉 = 〈3〉 by Euclid’s algorithm. So this is infact equal to 〈3〉.

Notice that when we worked with elements, the number 3 was irreducible, asthere is no element of norm 3. Thus, scenarios such as 2 ·3 = (1+

√−5)(1−

√−5)

could appear and mess up unique factorization. By passing on to ideals, we canfurther factorize 〈3〉 into a product of smaller ideals. Of course, these cannot beprincipal ideals, or else we would have obtained a factorization of 3 itself. Sowe can think of these ideals as “generalized elements” that allow us to furtherbreak elements down.

Indeed, given any element in α ∈ OL, we obtain an ideal 〈α〉 correspondingto α. This map is not injective — if two elements differ by a unit, i.e. they areassociates, then they would give us the same ideal. However, this is fine, as weusually think of associates as being “the same”.

We recall the following definition:

Definition (Prime ideal). Let R be a ring. An ideal p ⊆ R is prime if R/p isan integral domain. Alternatively, for all x, y ∈ R, xy ∈ p implies x ∈ p or y ∈ p.

In this course, we take the convention that a prime ideal is non-zero. This isnot standard, but it saves us from saying “non-zero” all the time.

It turns out that the ring of integers OL is a very special kind of rings, knownas Dedekind domains:

Definition (Dedekind domain). A ring R is a Dedekind domain if

(i) R is an integral domain.

(ii) R is a Noetherian ring.

18


(iii) R is integrally closed in FracR, i.e. if x ∈ FracR is integral over R, thenx ∈ R.

(iv) Every proper prime ideal is maximal.

This is a rather specific list of properties OL happens to satisfy, and it turnsout most interesting properties of OL can be extended to arbitrary Dedekinddomains. However, we will not do the general theory, and just study numberfields in particular.

The important result is, of course:

Proposition. Let L/Q be a number field, and OL be its ring of integers. ThenOL is a Dedekind domain.

The first three parts of the definition are just bookkeeping and not toointeresting. The last one is what we really want. This says that OL is “onedimensional”, if you know enough algebraic geometry.

Proof of (i) to (iii).

(i) Obvious, since OL ⊆ L.

(ii) We showed that as an abelian group, OL = Zn. So if a ≤ OL is an ideal,then a ≤ Zn as a subgroup. So it is finitely generated as an abelian group,and hence finitely generated as an ideal.

(iii) Note that FracOL = L. If x ∈ L is integral over OL, as OL is integralover Z, x is also integral over Z. So x ∈ OL, by definition of OL.

To prove the last part, we need the following lemma, which is also veryimportant on its own right.

Lemma. Let aCOL be a non-zero ideal. Then a ∩ Z 6= {0} and OL/a is finite.

Proof. Let α ∈ a and α 6= 0. Let

pα = xm + am−1xm−1 + · · ·+ a0

be its minimal polynomial. Then pα ∈ Z[x]. We know a0 6= 0 as pα is irreducible.Since pα(α) = 0, we know

a0 = −α(αm−1 + am−1αm−2 + · · ·+ a2α+ a1).

We know α ∈ a by assumption, and the mess in the brackets is in OL. So thewhole thing is in a. But a0 ∈ Z. So a0 ∈ Z ∩ a.

Thus, we know 〈a0〉 ⊆ a. Thus we get a surjection

OL〈a0〉

→ OLa.

Hence it suffices to show that OL/〈a0〉 is finite. But for every d ∈ Z, we know

OL〈d〉

=Zn

dZn=

(ZdZ

)n,

which is finite.

19


Finally, recall that a finite integral domain must be a field — let x ∈ R withx 6= 0. Then mx : y 7→ xy is injective, as R is an integral domain. So it is abijection, as R is finite. So there is some y ∈ R such that xy = 1.

This allows us to prove the last part

Proof of (iv). Let p be a prime ideal. Then OL/p is an integral domain. Sincethe lemma says OL/p is finite, we know OL/p is a field. So p is maximal.

We now continue on to prove a few more technical results.

Lemma. Let p be a prime ideal in a ring R. Then for a, b C R ideals, thenab ⊆ p implies a ⊆ p or b ⊆ p.

Once we’ve shown that inclusion of ideals is equivalent to divisibility, this ineffect says “prime ideals are primes”.

Proof. If not, then there is some a ∈ a \ p and b ∈ b \ p. Then ab ∈ ab ⊆ p. Butthen a ∈ p or b ∈ p. Contradiction.

Eventually, we will prove that every ideal is a product of prime ideals.However, we cannot prove that just yet. Instead, we will prove the following“weaker” version of that result:

Lemma. Let 0 6= aCOL a non-zero ideal. Then there is a subset of a that is aproduct of prime ideals.

The proof is some unenlightening abstract nonsense.

Proof. We are going to use the fact that OL is Noetherian. If this does not hold,then there must exist a maximal ideal a not containing a product of prime ideals(by which we mean any ideal greater than a contains a product of prime ideals,not that a is itself a maximal ideal). In particular, a is not prime. So there aresome x, y ∈ OL such that x, y 6∈ a but xy ∈ a.

Consider a + 〈x〉. This is an ideal, strictly bigger than a. So there existsprime ideals p1, · · · , pr such that p1 · · · pr ⊆ a + 〈x〉, by definition.

Similarly, there exists q1, · · · qs such that q1 · · · qs ⊆ a + 〈y〉.But then

p1 · · · prq1 · · · qs ⊆ (a + 〈x〉)(a + 〈y〉) ⊆ a + 〈xy〉 = a

So a contains a product of prime ideals. Contradiction.

Recall that for integers, we can multiply, but not divide. To make life easier,we would like to formally add inverses to the elements. If we do so, we obtainthings like 1

3 , and obtain the rationals.Now we have ideals. What can we do? We can formally add some inverse

and impose some nonsense rules to make sure it is consistent, but it is helpfulto actually construct something explicitly that acts as an inverse. We can thenunderstand what significance these inverses have in terms of the rings.

Proposition.

(i) Let 0 6= aCOL be an ideal. If x ∈ L has xa ⊆ a, then x ∈ OL.

20


(ii) Let 0 6= aCOL be a proper ideal. Then

{y ∈ L : ya ≤ OL}

contains elements that are not in OL. In other words,

{y ∈ L : ya ≤ OL}OL

6= 0.

We will see that the object {y ∈ L : ya ≤ OL} is in some sense an inverse toa.

Before we prove this, it is helpful to see what this means in a concrete setting.

Example. Consider OL = Z and a = 3Z. Then the first part says if ab ·3Z ⊆ 3Z,then a

b ∈ Z. The second says {ab

:a

b· 3 ∈ Z

}contains something not in Z, say 1

3 . These are both “obviously true”.

Proof.

(i) Let a ⊆ OL. Then since OL is Noetherian, we know a is finitely generated,say by α1, · · · , αm. We consider the multiplication-by-x map mx : a→ a,i.e. write

xαi =∑

aijαj ,

where A = (aij) is a matrix in OL. So we know

(xI −A)

α1

...αn

= 0.

By multiplying by the adjugate matrix, this implies det(xI −A) = 0. So xsatisfies a monic polynomial with coefficients in OL, i.e. x is integral overOL. Since OL is integrally closed, x ∈ OL.

(ii) It is clear that if the result is true for a, then it is true for all a′ ⊆ a. Soit is enough to prove this for a = p, a maximal, and in particular prime,ideal.

Let α ∈ p be non-zero. By the previous lemma, there exists prime idealsp1, · · · , pr such that p1 · · · pr ⊆ 〈α〉. We also have that 〈α〉 ⊆ p by definition.Assume r is minimal with this property. Since p is prime, there is some isuch that pi ⊆ p. wlog, we may as well assume i = 1, i.e. p1 ⊆ p. But p1 isa prime ideal, and hence maximal. So p1 = p.

Also, since r is minimal, we know p2 · · · pr 6⊆ 〈a〉.Pick β ∈ p2 · · · pr \ 〈a〉. Then

βp = βp1 ⊆ p1p2 · · · pr ⊆ 〈α〉.

Dividing by α, we get βαp ⊆ OL. But β 6∈ 〈α〉. So we know β

α 6∈ OL. Sodone.

21


What is this {x ∈ L : xa ≤ OL}? This is not an ideal, but it almost is. Theonly way in which it fails to be an ideal is that it is not contained inside OL. Bythis we mean it is closed under addition and multiplication by elements in OL.So it is an OL module, which is finitely generated (we will see this in a second),and a subset of L. We call this a “fractional ideal”.

Definition (Fractional ideal). A fractional ideal of OL is a subset of L that isalso an OL module and is finitely generated.

Definition (Integral/honest ideal). If we want to emphasize that aCOL is anideal, we say it is an integral or honest ideal. But we never use “ideal” to meanfractional ideal.

Note that the definition of fractional ideal makes sense only because OL isNoetherian. Otherwise, the non-finitely-generated honest ideals would not qualifyas fractional ideals, which is bad. Rather, in the general case, the followingcharacterization is more helpful:

Lemma. An OL module q ⊆ L is a fractional ideal if and only if there is somec ∈ L× such that cq is an ideal in OL. Moreover, we can pick c such that c ∈ Z.

In other words, each fractional ideal is of the form 1ca for some honest ideal

a and integer c.

Proof.

(⇐) We have to prove that q is finitely generated. If q ⊆ L×, c ∈ L non-zero,then cq ∼= q as an OL module. Since OL is Noetherian, every ideal isfinitely-generated. So cq, and hence q is finitely generated.

(⇒) Suppose x1, · · · , xn generate q as an OL-module. Write xi = yini

, withyi ∈ OL and ni ∈ Z, ni 6= 0, which we have previously shown is possible.

We let c = lcm(n1, · · · , nk). Then cq ⊆ OL, and is an OL-submodule ofOL, i.e. an ideal.

Corollary. Let q be a fractional ideal. Then as an abelian group, q ∼= Zn, wheren = [L : Q].

Proof. There is some c ∈ L× such that cqCOL as an ideal, and cq ∼= q as abeliangroups. So it suffices to show that any non-zero ideal q ≤ OL is isomorphic toZn. Since q ≤ OL ∼= Zn as abelian groups, we know q ∼= Zm for some m. Butalso there is some a0 ∈ Z ∩ q, and Zn ∼= 〈a0〉 ≤ q. So we must have n = m, andq ∼= Zn.

Corollary. Let a ≤ OL be a proper ideal. Then {x ∈ L : xa ≤ OL} is afractional ideal.

Proof. Pick a ∈ a. Then a · {x ∈ L : xa ≤ OL} ⊆ OL and is an ideal in OL.

Finally, we can state the proposition we want to prove, after all that nonsensework.

Definition (Invertible fractional ideal). A fractional ideal q is invertible if thereexists a fractional ideal r such that qr = OL = 〈1〉.

22


Notice we can multiply fractional ideals using the same definition as forintegral ideals.

Proposition. Every non-zero fractional ideal is invertible. The inverse of q is

{x ∈ L : xq ⊆ OL}.

This is good.Note that if q = 1

na and r = 1mb, and a, bCOL are integral ideals, then

qr =1

mnab = OL

if and only if ab = 〈mn〉. So the proposition is equivalent to the statement thatfor every aCOL, there exists an ideal bCOL such that ab is principal.

Proof. Note that for any n ∈ OL non-zero, we know q is invertible if and only ifnq is invertible. So if the proposition is false, there is an integral ideal aCOLwhich is not invertible. Moreover, as OL is Noetherian, we can assume a ismaximal with this property, i.e. if a < a′ < OL, then a′ is invertible.

Let b = {x ∈ L : xa ⊆ OL}, a fractional ideal. We clearly have OL ⊆ b, andby our previous proposition, we know this inclusion is strict.

As OL ⊆ b, we know a ⊆ ab. Again, this inclusion is strict — if ab = a, thenfor all x ∈ b, we have xa ⊆ a, and we have shown that this implies x ∈ OL, butwe cannot have b ⊆ OL.

So a ( ab. By assumption, we also have ab ⊆ OL, and since a is not invertible,this is strict. But then by definition of a, we know ab is invertible, which impliesa is invertible (if c is an inverse of ab, then bc is an inverse of a). This is acontradiction. So all fractional ideals must be invertible.

Finally, we have to show that the formula for the inverse holds. We write

c = {x ∈ L : xq ⊆ OL}.

Then by definition, we know q−1 ⊆ c. So

OL = qq−1 ⊆ qc ⊆ OL.

Hence we must have qc = OL, i.e. c = q−1.

We’re now done with the annoying commutative algebra, and can finallyprove something interesting.

Corollary. Let a, b, cCOL be ideals, c 6= 0. Then

(i) b ⊆ a if and only if bc ⊆ ac

(ii) a | b if and only if ac | bc

(iii) a | b if and only if b ⊆ a.

Proof.

(i) (⇒) is clear, and (⇐) is obtained by multiplying with c−1.

(ii) (⇒) is clear, and (⇐) is obtained by multiplying with c−1.

23


(iii) (⇒) is clear. For the other direction, we notice that the result is easy ifa = 〈α〉 is principal. Indeed, if b = 〈β1, · · · , βr〉, then b ⊆ 〈α〉 means thereare some β′a, · · · , β′r ∈ OL such that βi = β′iα. But this says

〈βa, · · · , βr〉 = 〈β′1, · · · , β′r〉〈α〉,

So 〈α〉 | b.

In general, suppose we have b ⊆ a. By the proposition, there exists anideal cCOL such that ac = 〈α〉 is principal with α ∈ OL, α 6= 0. Then

– b ⊆ a if and only if bc ⊆ 〈α〉 by (i); and

– a | b if and only if 〈α〉 | bc by (ii).

So the result follows.

Finally, we can prove the unique factorization of prime ideals:

Theorem. Let aCOL be an ideal, a 6= 0. Then a can be written uniquely as aproduct of prime ideals.

Proof. To show existence, if a is prime, then there is nothing to do. Otherwise,if a is not prime, then it is not maximal. So there is some b ) a with bCOL.Hence b | a, i.e. there is some cCOL with a = bc, and c ⊇ a. We can continuefactoring this way, and it must stop eventually, or else we have an infinite chainof strictly ascending ideals.

We prove uniqueness the usual way. We have shown p | ab implies p | a orp | b. So if p1 · · · pr = q1 · · · qs, with pi, qj prime, then we know p1 | q1 · · · qs,which implies p1 | qi for some i, and wlog i = 1. So q1 ⊆ p1. But q1 is primeand hence maximal. So p1 = q1.

Multiply the equation p1 · · · pr = q1 · · · qs by p−11 , and we get p2 · · · pr =

q2 · · · qs. Repeat, and we get r = s and pi = qi for all i (after renumbering).

Corollary. The non-zero fractional ideals form a group under multiplication.We denote this IL. This is a free abelian group generated by the prime ideals,i.e. any fractional ideal q can be written uniquely as pa11 · · · pqrr , with pi distinctprime ideals and ai ∈ Z.

Moreover, if q is an integral ideal, i.e. qCOL, then ai, · · · , ar ≥ 0.

Proof. We already have unique factorization of honest ideals. Now take anyfractional ideal, and write it as q = ab−1, with a, b ∈ OL (e.g. take b = 〈n〉 forsome n), and the result follows.

Unimportant side note: we have shown that there are two ways we canpartially order the ideals of OL — by inclusion and by division. We have shownthat these two orders are actually the same. Thus, it follows that the “leastcommon multiple” of two ideals a, b is their intersection a ∩ b, and the “greatestcommon divisor” of two ideals is their sum

a + b = {a+ b : a ∈ a, b ∈ b}.

Example. Again let [L : Q] = 2, i.e. L = Q(√d) with d 6= 0, 1 and square-free.

While we proved that every ideal can be factorized into prime ideals, we havecompletely no idea what prime ideals look like. We just used their very abstract

24


properties like being prime and maximal. So we would like to play with someactual ideals.

Recall we had the example

〈3, 1 + 2√−5〉〈3, 1− 2

√−5〉 = 〈3〉.

This is an example where we multiply two ideals together to get a principal ideal,and the key to this working is that 1 + 2

√−5 is conjugate to 1− 2

√−5. We will

use this idea to prove the previous result for number fields of this form.Let aCOL be a non-zero ideal. We want to find some bCOL such that ab

is principal.We know OL ∼= Z2, and a ≤ OL as a subgroup. Moreover, we have shown

that we must have a ∼= Z2 as abelian groups. So it is generated by 2 elements assubgroups of Z2. Since Z is a subring of OL, we know a must be generated byat most 2 elements as OL-modules, i.e. as ideals of OL. If it is generated by oneelement, then it is already principal. Otherwise, suppose a = 〈α, β〉 for someα, β ∈ OL.

Further, we claim that we can pick α, β such that β ∈ Z. We write α = a+b√d

and β = a′ + b′√d. Then let ` = gcd(b, b′) = mb + m′b′, with m,m′ ∈ Z (by

Euclid’s algorithm). We set

β′ = (mα+m′β) · −b′

`+ β

= (ma+m′a′ + `√d)−b′

`+ a′ + b′

√d

= (ma+m′a′)−b′

`+ a′ ∈ Z

using the fact that that − b′

` ∈ Z. Then 〈α, β′〉 = 〈α, β〉.So suppose a = 〈b, α〉, with b ∈ Z and α ∈ OL. We now claim

〈b, α〉〈b, α〉

is principal (where α = x+ y√d, α = x− y

√d). In particular, if aCOL, then

aa is principal, so the proposition is proved by hand.To show this, we can manually check

〈b, α〉〈b, α〉 = 〈b2, bα, bα, αα〉= 〈b2, bα, b tr(α), N(α)〉,

using the fact that tr(α) = α+ α and N(α) = αα. Now note that b2, b tr(α) andN(α) are all integers. So we can take the gcd c = gcd(b2, b tr(α), N(α)). Thenthis ideal is equal to

= 〈c, bα〉.

Finally, we claim that bα ∈ 〈c〉.Write bα = cx, with x ∈ L. Then trx = b

c trα ∈ Z since bc ∈ Z by definition,

and

N(x) = N

(bα

c

)=b2N(α)

c2=b2

c

N(α)

c∈ Z.

So x ∈ OL. So c | bα in OL. So 〈c, bα〉 = 〈c〉.

25


Finally, after all these results, we can get to the important definition of thecourse.

Definition (Class group). . The class group or ideal class group of a numberfield L is

clL = IL/PL,

where IL is the group of fractional ideals, and PL is the subgroup of principalfractional ideals.

If a ∈ IL, we write [a] for its equivalence class in clL. So [a] = [b] if and onlyif there is some γ ∈ L× such that γa = b.

The significance is that clL measures the failure of unique factorization:

Theorem. The following are equivalent:

(i) OL is a principal ideal domain

(ii) OL is a unique factorization domain

(iii) clL is trivial.

Proof. (i) and (iii) are equivalent by definition, while (i) implies (ii) is well-knownfrom IB Groups, Rings and Modules. So the real content is (ii) to (i), which isspecific to Dedekind domains.

If pCOL is prime, and x ∈ p \ {0}, we factor x = α1 · · ·αk such that αi isirreducible in OL. As p is prime, there is some αi ∈ p. But then 〈αi〉 ⊆ p, and〈αi〉 is prime as OL is a UFD. So we must have 〈αi〉 = p as prime ideals aremaximal. So p is principal.

In the next few chapters, we will come up with methods to explicitly computethe class group of any number field.

26

4 Norms of ideals II Number Fields

4 Norms of ideals

In the previous chapter, we defined the class group, and we know it is generatedby prime ideals of OL. So we now want to figure out what the prime ideals are.In the case of finding irreducible elements, one very handy tool was the norm

— we know an element of OL is a unit iff it has norm 1. So if x ∈ OL is notirreducible, then there must be some element whose norm strictly divides N(x).Similarly, we would want to come up with the notion of the norm of an ideal,which turns out to be an incredibly useful notion.

Definition (Norm of ideal). Let aCOL be an ideal. We define

|N(a)| = |OL/a| ∈ N.

Recall that we’ve already proved that |OL/a| is finite. So this definitionmakes sense. It is also clear that N(a) = 1 iff a = OL (i.e. a is a “unit”).

Example. Let d ∈ Z. Then since OL ∼= Zn, we have dOL ∼= dZn. So we have

N(〈d〉) = |Zn/(dZ)n| = |Z/dZ|n = dn.

We start with a simple observation:

Proposition. For any ideal a, we have N(a) ∈ a ∩ Z.

Proof. It suffices to show that N(a) ∈ a. Viewing OL/a as an additive group,the order of 1 is a factor of N(a). So N(a) = N(a) · 1 = 0 ∈ OL/a. HenceN(a) ∈ a.

The most important property of the norm is the following:

Proposition. Let a, bCOL be ideals. Then N(ab) = N(a)N(b).

We will provide two proofs of the result.

Proof. By the factorization into prime ideals, it suffices to prove this for b = pprime, i.e.

N(ap) = N(a)N(p).

In other words, we need to show that∣∣∣∣OLa∣∣∣∣ =

∣∣∣∣OLap∣∣∣∣/ ∣∣∣∣OLp

∣∣∣∣ .By the third isomorphism theorem, we already know that

OLa∼=(OLap

)/( a

ap

).

So it suffices to show that OL/p ∼= a/ap as abelian groups.In the case of the integers, we could have, say, p = 7Z, a = 12Z. We would

then simply define

Z7Z

12Z7 · 12Z

x 12x

27


However, in general, we do not know that a is principal, but it turns out itdoesn’t really matter. We can just pick an arbitrary element to multiply with.

By unique factorization, we know a 6= ap. So we can find some α ∈ a \ ap.We now claim that the homomorphism of abelian groups

OLp

a

ap

x+ p αx+ ap

is an isomorphism. We first check this is well-defined — if p ∈ p, then αp ∈ apsince α ∈ a. So the image of x+p and (x+p)+p are equal. So this is well-defined.

To prove our claim, we have to show injectivity and surjectivity. To showinjectivity, since 〈α〉 ⊆ a, we have a | 〈a〉, i.e. there is an ideal c ⊆ OL such thatac = 〈α〉. If x ∈ OL is in the kernel of the map, then αx ∈ ap. So

xac ⊆ ap.

Soxc ⊆ p.

As p is prime, either c ⊆ p or x ∈ p. But c ⊆ p implies 〈α〉 = ac ⊆ ap,contradicting the choice of α. So we must have x ∈ p, and the map is injective.

To show this is surjective, we notice that surjectivity means 〈α〉/ap = a/ap,or equivalently ap + 〈α〉 = a.

Using our knowledge of fractional ideals, this is equivalent to saying (ap +〈α〉)a−1 = OL. But we know

ap < ap + 〈α〉 ⊆ a.

We now multiply by a−1 to obtain

p < (ap + 〈α〉)a−1 = p + c ⊆ OL.

Since p is a prime, and hence maximal ideal, the last inclusion must be anequality. So ap + 〈α〉 = a, and we are done.

Now we provide the sketch of a proof that makes sense. The details are leftas an exercise in the second example sheet.

Proof. It is enough to show that N(pa11 · · · parr ) = N(p1)a1 · · ·N(pr)ar by unique

factorization.By the Chinese remainder theorem, we have

OLpa11 · · · p

arr

∼=OLpa11

× · · · × OLparr

where p1, · · · , pr are distinct prime ideals.Next, we show by hand that∣∣∣∣OLpr

∣∣∣∣ =

∣∣∣∣OLp∣∣∣∣× ∣∣∣∣ pp2

∣∣∣∣× · · · × ∣∣∣∣pr−1

pr

∣∣∣∣ =

∣∣∣∣OLp∣∣∣∣r ,

by showing that pk/pk+1 is a 1-dimensional vector space over the field OL/p.Then the result follows.

28


This is actually the same proof, but written in a much saner form. This isbetter because we are combining a general statement (the Chinese remaindertheorem), with a special property of the integral rings. In the first proof, whatwe really did was simultaneously proving two parts using algebraic magic.

We’ve taken an obvious invariant of an ideal, the size, and found it ismultiplicative. How does this relate to the other invariants?

Recall that

∆(α1, · · · , αn) = det(trL/Q(αiαj)) = det(σi(αj))2.

Proposition. Let aCOL be an ideal, n = [L : Q]. Then

(i) There exists α1, · · · , αn ∈ a such that

a ={∑

riαi : ri ∈ Z}

=

n⊕1

αiZ,

and α1, · · · , αn are a basis of L over Q. In particular, a is a free Z-moduleof n generators.

(ii) For any such α1, · · · , αn,

∆(α1, · · · , αn) = N(a)2DL.

The prove this, we recall the following lemma from IB Groups, Rings andModules:

Lemma. Let M be a Z-module (i.e. abelian group), and suppose M ≤ Zn.Then M ∼= Zr for some 0 ≤ r ≤ n.

Moreover, if r = n, then we can choose a basis v1, · · · , vn of M such that thechange of basis matrix A = (aij) ∈Mn×n(Z) is upper triangular, where

vj =∑

aijei,

where e1, · · · , en is the standard basis of Zn.In particular,

|Zn/M | = |a11a12 · · · ann| = |detA|.

Proof of proposition. Let d ∈ a ∩ Z, say d = N(α). Then dOL ⊆ a ⊆ OL. Asabelian groups, after picking an integral basis α′1, · · · , α′n of OL, we have

Zn ∼= dZn ≤ a ≤ Zn.

So a ∼= Zn. Then the lemma gives us a basis α1, · · · , αn of a as a Z-module. Asa Q-module, since the αi are obtained from linear combinations of α′i, by basiclinear algebra, α1, · · · , αn is also a basis of L over Q.

Moreover, we know that we have

∆(α1, · · · , αn) = det(A)2∆(α′1, · · · , α′n).

Since det(A)2 = |OL/a|2 = N(a) and DL = ∆(α′1, · · · , α′n) by definition, thesecond part follows.

29


This result is very useful for the following reason:

Corollary. Suppose aCOL has basis α1, · · · , αn, and ∆(α1, · · · , αn) is square-free. Then a = OL (and DL is square-free).

This is a nice trick, since it allows us to determine immediately whether aparticular basis is an integral basis.

Proof. Immediate, since this forces N(a)2 = 1.

Note that nothing above required a to be an actual ideal. It merely hadto be a subgroup of OL that is isomorphic to Zn, since the quotient OL/a iswell-defined as long as a is a subgroup. With this, we can have the followinguseful result:

Example. Let α be an algebraic integer and L = Q(α). Let n = [Q(α) : Q].Then a = Z[α] COL. We have

disc(pα) = ∆(1, α, α2, · · · , αn−1) = discriminant of minimal polynomial of α.

Thus if disc(pα) is square-free, then Z[α] = OL.Even if disc(pα) is not square-free, it still says something: let d2 | disc(pα)

be such that disc(pα)/d2 is square-free. Then |N(Z[α])| divides d.Let x ∈ OL. Then the order of x+ Z[α] ∈ OL/Z[α] divides N(Z[α]), hence

d. So d · x ∈ Z[α]. So x ∈ 1dZ[α]. Hence we have

Z[α] ⊆ OL ⊆1

dZ[α].

For example, if α =√a for some square-free a, then disc(

√a) is the discriminant

of x2 − a, which is 4a. So the d above is 2, and we have

Z[α] ⊆ OQ(√d) ⊆

1

2Z[α],

as we have previously seen.

We shall prove one more lemma, and start factoring things. Recall that wehad two different notions of norm. Given α ∈ OL, we can take the norm N(〈α〉),or NL/Q(α). It would be great if they are related, like if they are equal. However,that cannot possibly be true, since N(〈α〉) is always positive, but NL/Q(α) canbe negative. So we take the absolute value.

Lemma. If α ∈ OL, then

N(〈α〉) = |NL/Q(α)|.

Proof. Let α1, · · · , αn be an integral basis of OL. Then αα1, .., ααn is an integralbasis of 〈α〉. So by the previous lemma,

∆(αα1, · · · , ααn) = N(〈α〉)2DL.

30


But

∆(αα1, · · · , ααn) = det(σi(ααj)ij)2

= det(σi(α)σi(αj))2

=

(n∏i=1

σi(α)

)2

∆(α1, · · · , αn)

= NL/Q(α)2DL.

SoNL/Q(α)2 = N(〈α〉)2.

But N(〈α〉) is positive. So the result follows.

31

5 Structure of prime ideals II Number Fields

5 Structure of prime ideals

We can now move on to find all prime ideals of OL. We know that every idealfactors as a product of prime ideals, but we don’t know what the prime idealsare. The only obvious way we’ve had to obtain prime ideals is to take a usualprime, take its principal ideal and factor it in OL, and get the resultant primeideals.

It turns out this gives us all prime ideals.

Lemma. Let p C OL be a prime ideal. Then there exists a unique p ∈ Z, pprime, with p | 〈p〉. Moreover, N(p) = pf for some 1 ≤ f ≤ n.

This is not really too exciting, as soon as we realize that p | 〈p〉 is the sameas saying 〈p〉 ⊆ p, and we already know p ∩ Z is non-empty.

Proof. Well p ∩ Z is an ideal in Z, and hence principal. So p ∩ Z = pZ for somep ∈ Z.

We now claim p is a prime integer. If p = ab with ab ∈ Z. Then since p ∈ p,either a ∈ p or b ∈ p. So a ∈ p ∩ Z = pZ or b ∈ p ∩ Z = pZ. So p | a or p | b.

Since 〈p〉 ⊆ p, we know 〈p〉 = pa for some ideal a by factorization. Takingnorms, we get

pn = N(〈p〉) = N(p)N(a).

So the result follows.

This is all good. So all we have to do is to figure out how principal ideals 〈p〉factor into prime ideals.

We write〈p〉 = pe11 · · · pemm

for some distinct prime ideals pi, with N(pi) = pfi for some positive integers ei.Taking norms, we get

pn =∏

pfiei .

Son =

∑eifi.

We start by giving some names to the possible scenarios.

Definition (Ramification indices). Let 〈p〉 = pe11 · · · pemm be the factorizationinto prime ideals. Then e1, · · · , em are the ramification indices.

Definition (Ramified prime). We say p is ramified if some ei > 1.

Definition (Inert prime). We say p is inert if m = 1 and em = 1, i.e. 〈p〉remains prime.

Definition (Splitting prime). We say p splits completely if e1 = · · · = em = 1 =f1 = · · · = fm. So m = n.

Note that this does not exhaust all possibilities. The importance of theseterms, especially ramification, will become clear later.

So how do we actually compute pi and ei? In other words, how can we factorthe ideal 〈p〉 C OL into prime ideals? The answer is given very concretely byDedekind’s criterion.

32


Theorem (Dedekind’s criterion). Let α ∈ OL and g(x) ∈ Z[x] be its minimalpolynomial. Suppose Z[α] ⊆ OL has finite index, coprime to p (i.e. p - |OL/Z[α]|).We write

g(x) = g(x) (mod p),

so g(x) ∈ Fp[x]. We factor

g(x) = ϕe11 · · ·ϕemm

into distinct irreducibles in Fp[x]. We define the ideal

pi = 〈p, ϕi(α)〉COL,

generated by p and ϕi, where ϕi is any polynomial in Z[x] such that ϕi mod p =ϕi. Notice that if ϕ′ is another such polynomial, then p | (ϕi−ϕi), so 〈p, ϕ′(α)〉 =〈p, ϕ(α)〉.

Then the pi are prime, and

〈p〉 = pe11 · · · pemm .

Moreover, fi = degϕi, so N(a) = pdegϕi .

If we are lucky, we might just find an α such that Z[α] = OL. If not, wecan find something close, and as long as p is not involved, we are fine. Afterfinding α, we get its minimal polynomial, factor it, and immediately get theprime factorization of 〈p〉.

Example. Consider L = Q(√−11). We want to factor 〈5〉 in OL. We consider

Z[√−11] ⊆ OL. This has index 2, and (hopefully) 5 - 2. So this is good

enough. The minimal polynomial is x2 + 11. Taking mod 5, this reduces tox2 − 4 = (x− 2)(x+ 2). So Dedekind says

〈5〉 = 〈5,√−11 + 2〉〈5,

√−11− 2〉.

In general, consider L = Q(√d), d 6= 0, 1 and square-free, and p an odd

prime. Then Z[√d] ⊆ OL has index 1 or 2, both of which are coprime to p. So

Dedekind says factor x2 − d mod p. What are the possibilities?

(i) There are two distinct roots mod p, i.e. d is a square mod p, i.e.(dp

)= 1.

Thenx2 − d = (x+ r)(x− r) (mod p)

for some r. So Dedekind says

〈p〉 = p1p2,

wherep1 = 〈p,

√d− r〉, p2 = 〈p,

√d+ r〉,

and N(p1) = N(p2) = p. So p splits.

(ii) x2 − d is irreducible, i.e. d is not a square mod p, i.e.(dp

)= −1. Then

Dedekind says 〈p〉 = p is prime in OL. So p is inert.

33


(iii) x2−d has a repeated root mod p, i.e. p | d, or alternatively(dp

)= 0. Then

by Dedekind, we know〈p〉 = p2,

wherep = 〈p,

√d〉.

So p ramifies.

So in fact, we see that the Legendre symbol encodes the ramification behaviourof the primes.

What about the case where p = 2, for L = Q(√d)? How do we factor 〈2〉?

Lemma. In L = Q(√d),

(i) 2 splits in L if and only if d ≡ 1 (mod 8);

(ii) 2 is inert in L if and only if d ≡ 5 (mod 8);

(iii) 2 ramifies in L if d ≡ 2, 3 (mod 4).

Proof.

– If d ≡ 1 (mod 4), then then OL = Z[α], where α = 12 (1 +

√d). This has

minimal polynomial

x2 − x+1

4(1− d).

We reduce this mod 2.

◦ If d ≡ 1 (mod 8), we get x(x+ 1). So 2 splits.

◦ If d ≡ 5 (mod 8), then we get x2 + x+ 1, which is irreducible. So 〈2〉is prime, hence 2 is inert.

– If d ≡ 2, 3 (mod 4), thenOL = Z[√d], and x2−d is the minimal polynomial.

Taking mod 2, we get x2 or x2 +1 = (x+1)2. In both cases, 2 ramifies.

Note how important p - |OL/Z[α]| is. If we used Z[√d] when d ≡ 1 (mod 4),

we would have gotten the wrong answer.Recall

DL =

{4d d ≡ 2, 3 (mod 4)

d d ≡ 1 (mod 4)

The above computations show that p | DL if and only if p ramifies in L. Thishappens to be true in general. This starts to hint how important these invariantslike DL are.

Now we get to prove Dedekind’s theorem.

Proof of Dedekind’s criterion. The key claim is that

Claim. We haveOLpi∼=

Fp[x]

〈ϕi〉.

34


Suppose this is true. Then since ϕi is irreducible, we knowFp[x]〈ϕi〉 is a field.

So pi is maximal, hence prime.Next notice that

pe11 = 〈p, ϕi(α)〉ei ⊆ 〈p, ϕi(α)ei〉.

So we have

pe11 · · · pemm ⊆ 〈p, ϕ1(α)e1 · · · ϕm(α)em〉 = 〈p, g(α)〉 = 〈p〉,

using the fact that g(α) = 0.So to prove equality, we notice that if we put fi = degϕi, then N(pi) = pfi ,

andN(pe11 · · · pemm ) = N(p1)e1 · · ·N(pm)em = p

∑eifi = pdeg g.

Since N(〈p〉) = pn, it suffices to show that deg g = n. Since Z[α] ⊆ OL has finiteindex, we know Z[α] ∼= Zn. So 1, α, · · · , αn−1 are independent over Z, hence Q.So deg g = [Q(α) : Q] = n = [L : Q], and we are done.

So it remains to prove that

OLpi∼=

Z[α]

pi ∩ Z[α]∼=

Fp[x]

〈ϕi〉.

The second isomorphism is clear, since

Z[α]

〈p, ϕi(α)〉∼=

Z[x]

〈p, ϕi(x), g(x)〉∼=

Fp[x]

〈ϕi(x), g(x)〉=

Fp[x]

〈ϕi(x), g(x)〉=

Fp[x]

〈ϕi〉.

To prove the first isomorphism, it suffices to show that the following map is anisomorphism:

Z[α]

pZ[α]→ OL

pOL(∗)

x+ pZ[α] 7→ x+ pOL

If this is true, then quotienting further by ϕi gives the desired isomorphism.To prove the claim, we consider a slightly different map. We notice p -

|OL/Z[α]| means the “multiplication by p” map

OLZ[α]

OLZ[α]

p(†)

is injective. But OL/Z[α] is a finite abelian group. So the map is an isomorphism.By injectivity of (†), we have Z[α] ∩ pOL = pZ[α]. By surjectivity, we have

Z[α] + pOL = OL. It thus follows that (∗) is injective and surjective respectively.So it is an isomorphism. We have basically applied the snake lemma to thediagram

0 Z[α] OL OL

Z[α] 0

0 Z[α] OL OL

Z[α] 0

p p p

35


Corollary. If p is prime and p < n = [L : Q], and Z[α] ⊆ OL has finite indexcoprime to p, then p does not split completely in OL.

Proof. By Dedekind’s theorem, if g(x) is the minimal polynomial of α, then thefactorization of g(x) = g(x) mod p determines the factorization of 〈p〉 into primeideals. In particular, p splits completely if and only if g factors into distinctlinear factors, i.e.

g(x) = (x− α1) · · · (x− αn),

where αi ∈ Fp and αi are distinct. But if p < n, then there aren’t n distinctelements of Fp!

Example. Let L = Q(α), where α has minimal polynomial x3 − x2 − 2x − 8.This is the case where n = 3 > 2 = p. On example sheet 2, you will see that 2splits completely, i.e. OL/2OL = F2 × F2 × F2. But then this corollary showsthat for all β ∈ OL, Z[β] ⊆ OL has even index, i.e. there does not exist anβ ∈ OL with |OL/Z[β]| odd.

As we previously alluded to, the following is true:

Theorem. p | DL if and only if p ramifies in OL.

We will not prove this.

36

6 Minkowski bound and finiteness of class group II Number Fields

6 Minkowski bound and finiteness of class group

Dedekind’s criterion allowed us to find all prime factors of 〈p〉, but if we wantto figure out if, say, the class group of a number field is trivial, or even finite,we still have no idea how to do so, because we cannot go and check every singleprime p and see what happens.

What we are now going to do is the following — we are going to use purelygeometric arguments to reason about ideals, and figure that each element ofthe class group clL = IL/PL has a representative whose norm is bounded bysome number cL, which we will find rather explicitly. After finding the cL, tounderstand the class group, we just need to factor all prime numbers less thancL and see what they look like.

We are first going to do the case of quadratic extensions explicitly, since2-dimensional pictures are easier to draw. We will then do the full general caseafterwards.

Quadratic extensions

Consider again the case L = Q(√d), where d < 0. Then OL = Z[α], where

α =

{√d d ≡ 2, 3 (mod 4)

12 (1 +

√d) d ≡ 1 (mod 4)

We can embed this as a subfield L ⊆ C. We can then plot the points on thecomplex plane. For example, if d ≡ 2, 3 (mod 4), then the points look like this:

0

√d

1

1 +√d

2√d

Then an ideal of OL, say a = 〈2,√d〉, would then be the sub-lattice given by

the blue crosses.

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

37


We always get this picture, since any ideal of OL is isomorphic to Z2 as anabelian group.

If we are in the case where d ≡ 1 (mod 4), then the lattice is hexagonal:

0

√d

1

12 (1 +

√d)

The key result is the following purely geometric lemma:

Lemma (Minskowski’s lemma). Let Λ = Zv1 + Zv2 ⊆ R2 be a lattice, withv1,v2 linearly independent in R (i.e. Rv1+Rv2 = R2). We write vi = aie1+bie2.Then let

A(Λ) = area of fundamental parallelogram =

∣∣∣∣det

(a1 a2

b1 b2

)∣∣∣∣ ,where the fundamental parallelogram is the following:

v1

v2

v1 + v2

Then a closed disc S around 0 contains a non-zero point of Λ if

area(S) ≥ 4A(Λ).

In particular, there exists an α ∈ Λ with α 6= 0, such that

0 < |α|2 ≤ 4A(Λ)

π.

This is just an easy piece of geometry. What is remarkable is that the radiusof the disc needed depends only on the area of the fundamental parallelogram,and not its shape.

Proof. We will prove a general result in any dimensions later.

We now apply this to ideals a ≤ OL, regarded as a subset of C = R2 viasome embedding L ↪→ C. The following proposition gives us the areas of therelevant lattices:

Proposition.

(i) If α = a+ b√λ, then as a complex number,

|α|2 = (a+ b√λ)(a− b

√λ) = N(α).

38


(ii) For OL, we have

A(OL) =1

2

√|DL|.

(iii) In general, we have

A(a) =1

2

√|∆(α1, α2)|,

where α1, α2 are the integral basis of a.

(iv) We haveA(a) = N(a)A(OL).

Proof.

(i) This is clear.

(ii) We know OL has basis 1, α, where again

α =

{√d d ≡ 2, 3 (mod 4)

12 (1 +

√d) d ≡ 1 (mod 4)

.

So we can just look at the picture of the lattice, and compute to get

A(OL) =

{√|d| d ≡ 2, 3 (mod 4)

12

√|d| d ≡ 1 (mod 4)

=1

2

√|DL|.

(iii) If α1, α2 are the integral basis of a, then the lattice of a is in fact spannedby the vectors α1 = a+ bi, α2 = a′ + b′i. This has area

A(a) = det

(a ba′ b′

),

whereas we have

∆(α1, α2) = det

(α1 α1

α2 α2

)2

= (α1α2 − α2α1)2

= Im(2α1α2)2

= 4(a′b− ab′)2

= 4A(a)2.

(iv) This follows from (ii) and (iii), as

∆(α1, · · · , αn) = N(a)2DL

in general.

Now what does Minkowski’s lemma tell us? We know there is an α ∈ a suchthat

N(α) ≤ 4A(a)

π= N(a)cL,

39


where

cL =2√|DL|π

.

But α ∈ a implies 〈α〉 ⊆ a, which implies 〈α〉 = ab for some ideal b. So

|N(α)| = N(〈α〉) = N(a)N(b).

So this implies

N(b) ≤ cL =2√|DL|π

.

Recall that the class group is clL = IL/PL, the fractional ideals quotiented byprincipal ideals, and we write [a] for the class of a in clL. Then if ab = 〈α〉, thenwe have

[b] = [a−1]

in clL. So we have just shown,

Proposition (Minkowski bound). For all [a] ∈ clL, there is a representative bof [a] (i.e. an ideal b ≤ OL such that [b] = [a]) such that

N(b) ≤ cL =2√|DL|π

.

Proof. Find the b such that [b] = [(a−1)−1] and N(b) ≤ cL.

Combining this with the following easy lemma, we know that the class groupis finite!

Lemma. For every n ∈ Z, there are only finitely many ideals a ≤ OL withN(a) = m.

Proof. If N(a) = m, then by definition |OL/a| = m. So m ∈ a by Lagrange’stheorem. So 〈m〉 ⊆ a, i.e. a | 〈m〉. Hence a is a factor of 〈m〉. By uniquefactorization of prime ideals, there are only finitely many such ideals.

Another proof is as follows:

Proof. Each ideal bijects with an ideal in OL/mOL = (Z/m)n. So there areonly finitely many.

Thus, we have proved

Theorem. The class group clL is a finite group, and the divisors of ideals ofthe form 〈p〉 for p ∈ Z, p a prime, and 0 < p < cL, collectively generate clL.

Proof.

(i) Each element is represented by an ideal of norm less than 2√|DL|/π, and

there are only finitely many ideals of each norm.

(ii) Given any element of clL, we pick a representative a such that N(a) < cL.We factorize

a = pe11 · · · perr .Then

N(pi) ≤ N(a) < cL.

Suppose pi | 〈p〉. Then N(p) is a power of p, and is thus at least p. Sop < cL.

40


We now try to work with some explicit examples, utilizing Dedekind’s criterionand the Minkowski bound.

Example. Consider d = −7. So Q(√−7) = L, and DL = −7. Then we have

1 < cL =2√

7

π< 2.

So clL = {1}, since there are no primes p < cL. So OL is a UFD.Similarly, if d = −1,−2,−3, then OL is a UFD.

Example. Let d = −5. Then DL = −20. We have

2 < cL =4√

5

π< 3.

So clL is generated by primes dividing 〈2〉.Recall that Dirichlet’s theorem implies

〈2〉 = 〈2, 1 +√−5〉2 = p2.

Also, p = 〈2, 1 +√−5〉 is not principal. If it were, then p = 〈β〉, with β =

x+ y√−5, and N(β) = 2. But there are no solutions in Z of x2 + 5y2 = 2. So

clL = 〈p〉 = Z/2.

Example. Consider d = −17 ≡ 3 (mod 4). So cL ≈ 5.3. So clL is generated byprimes dividing by 〈2〉, 〈3〉, 〈5〉. We factor

x2 + 17 ≡ x2 + 1 ≡ (x+ 1)2 (mod 2).

So〈2〉 = p2 = 〈2, 1 +

√d〉2.

Doing this mod 3, we have

x2 + 17 ≡ x2 − 1 ≡ (x− 1)(x+ 1) (mod 3).

So we have〈3〉 = qq = 〈3, 1 +

√d〉〈3, 1−

√d〉.

Finally, mod 5, we have

x2 + 17 ≡ x2 + 2 (mod 5).

So 5 is inert, and [〈5〉] = 1 in clL. So

clL = 〈[p], [q]〉,

and we need to compute what this is. We can just compute powers q2, q3, · · · ,pq, pq2, · · · , and see what happens.

But a faster way is to look for principal ideals with small norms that aremultiples of 2 and 3. For example,

N(〈1 +√d〉) = 18 = 2 · 32.

41


But we have1 +√d ∈ p, q.

So p, q | 〈1 +√d〉. Thus we know pq | 〈1 +

√d〉. We have N(pq) = 2 · 3 = 6. So

there is another factor of 3 to account for. In fact, we have

〈1 +√d〉 = pq2,

which we can show by either thinking hard or expanding it out. So we must have

[p] = [q]−2

in clL. So we have clL = 〈[q]〉. Also, [q]−2 = [p] 6= 1 in clL, as if it did, then p isprincipal, i.e. p = 〈x+ y

√d〉, but 2 = N(p) = x2 + 7y2 has no solution in the

integers. Also, we know [p]2 = [1]. So we know

clL = Z/4Z.

In fact, we have

Theorem. Let L = Q(√d) with d < 0. Then OL is a UFD if

−d ∈ {1, 2, 3, 7, 11, 19, 43, 67, 163}.

Moreover, this is actually an “if and only if”.

The first part is a straightforward generalization of what we have been doing,but the proof that no other d’s work is hard.

General case

Now we want to extend these ideas to higher dimensions. We are really justdoing the same thing, but we need a bit harder geometry and proper definitions.

Definition (Discrete subset). A subset X ⊆ Rn is discrete if for every x ∈ X,there is some ε > 0 such that Bε(x) ∩X = {x}. This is true if and only if forevery compact K ⊆ Rn, K ∩X is finite.

We have the following very useful characterization of discrete subgroups ofRn:

Proposition. Suppose Λ ⊆ Rn is a subgroup. Then Λ is a discrete subgroup of(Rn,+) if and only if

Λ =

{m∑1

nixi : ni ∈ Z

}for some x1, · · · ,xm linearly independent over R.

Note that linear independence is important. For example, Z√

2 + Z√

3 ⊆ Ris not discrete. On the other hand, if Λ = aCOL is an ideal, where L = Q(

√d)

and d < 0, then this is discrete.

42


Proof. Suppose Λ is generated by x1, · · · ,xm. By linear independence, thereis some g ∈ GLn(R) such that gxi = ei for all 1 ≤ i ≤ m, where e1, · · · , en isthe standard basis. We know acting by g preserves discreteness, since it is ahomeomorphism, and gΛ = Zm ⊆ Rm × Rn−m is clearly discrete (take ε = 1

2 ).So this direction follows.

For the other direction, suppose Λ is discrete. We pick y1, · · · ,ym ∈ Λwhich are linearly independent over R, with m maximal (so m ≤ n). Then bymaximality, we know{

m∑i=1

λiyi : λi ∈ R

}=

{m∑1

λizi : λi ∈ R, zi ∈ Λ

},

and this is the smallest vector subspace of Rn containing Λ. We now let

X =

{m∑i=1

λiyi : λi ∈ [0, 1]

}∼= [0, 1]m.

This is closed and bounded, and hence compact. So X ∩ Λ is finite.Also, we know ⊕

Zyi = Zm ⊆ Λ,

and if γ is any element of Λ, we can write it as γ = γ0 + γ1, where γ0 ∈ X andγ1 ∈ Zm. So ∣∣∣∣ Λ

Zm

∣∣∣∣ ≤ |X ∩ Λ| <∞.

So let d = |Λ/Zm|. Then dΛ ⊆ Zm, i.e. Λ ⊆ 1dZ

m. So

Zm ⊆ Λ ⊆ 1

dZm.

So Λ is a free abelian group of rank m. So there exists x1, · · · ,xm ∈ 1dZ

m whichis an integral basis of Λ and are linearly independent over R.

Definition (Lattice). If rank Λ = n = dimRn, then Λ is a lattice in Rn.

Definition (Covolume and fundamental domain). Let Λ ⊆ Rn be a lattice, andx1, · · · ,xn be a basis of Λ, then let

P =

{n∑i=1

λixi : λi ∈ [0, 1]

},

and define the covolume of Λ to be

covol(Λ) = vol(P ) = |detA|,

where A is the matrix such that xi =∑aijej .

We say P is a fundamental domain for the action of Λ on Rn, i.e.

Rn =⋃γ∈Λ

(γ + P ),

and(γ + P ) ∩ (µ+ P ) ⊆ ∂(γ + P ).

In particular, the intersection has zero volume.

43


This is called the covolume since if we consider the space Rn/Λ, which is ann-dimensional torus, then this has volume covol(Λ).

Observe now that if x′1, · · · ,x′n is a different basis of Λ, then the transitionmatrix x′i =

∑bijxj has B ∈ GLn(Z). So we have detB = ±1, and covol(Λ) is

independent of the basis choice.With these notations, we can now state Minkowski’s theorem.

Theorem (Minkowski’s theorem). Let Λ ⊆ Rn be a lattice, and P be a funda-mental domain. We let S ⊆ Rn be a measurable set, i.e. one for which vol(S) isdefined.

(i) Suppose vol(S) > covol(Λ). Then there exists distinct x,y ∈ S such thatx− y ∈ Λ.

(ii) Suppose 0 ∈ S, and S is symmetric around 0, i.e. s ∈ S if and only if−s ∈ S, and S is convex, i.e. for all x,y ∈ S and λ ∈ [0, 1], then

λx + (1− λ)y ∈ S.

Then suppose either

(a) vol(S) > 2n covol(Λ); or

(b) vol(S) ≥ 2n covol(Λ) and S is closed.

Then S contains a γ ∈ Λ with γ 6= 0.

Note that for n = 2, this is what we used for quadratic fields.By considering Λ = Zn ⊆ Rn and S = [−1, 1]n, we know the bounds are

sharp.

Proof.

(i) Suppose vol(S) > covol(Λ) = vol(P ). Since P ⊆ Rn is a fundamentaldomain, we have

vol(S) = vol(S ∩ Rn) = vol

S ∩∑γ∈Λ

(P + γ)

=∑γ∈Λ

vol(S ∩ (P + γ)).

Also, we know

vol(S ∩ (P + γ)) = vol((S − γ) ∩ P ),

as volume is translation invariant. We now claim the sets (S − γ) ∩ P forγ ∈ Λ are not pairwise disjoint. If they were, then

vol(P ) ≥∑γ∈Λ

vol((S − γ) ∩ P ) =∑γ∈Λ

vol(S ∩ (P + γ)) = vol(S),

contradicting our assumption.

Then in particular, there are some distinct γ and µ such that (S − γ) and(S − µ) are not disjoint. In other words, there are x,y ∈ S such thatx− γ = y − µ, i.e. x− y = γ − µ ∈ Λ 6= 0.

44


(ii) We now let

S′ =1

2S =

{1

2s : s ∈ S

}.

So we havevol(S′) = 2−n vol(S) > covol(Λ),

by assumption.

(a) So there exists some distinct y, z ∈ S′ such that y − z ∈ Λ \ {0}. Wenow write

y − z =1

2(2y + (−2z)),

Since 2z ∈ S implies −2z ∈ S by symmetry around 0, so we knowy − z ∈ S by convexity.

(b) We apply the previous part to Sm =(1 + 1

m

)S for all m ∈ N, m > 0.

So we get a non-zero γm ∈ Sm ∩ Λ.

By convexity, we know Sm ⊆ S1 = 2S for all m. So γ1, γ2, · · · ∈ S1∩Λ.But S1 is compact set. So S1 ∩Λ is finite. So there exists γ such thatγm is γ infinitely often. So

γ ∈⋂m≥0

Sm = S.

So γ ∈ S.

We are now going to use this to mimic our previous proof that the classgroup of an imaginary quadratic field is finite.

To begin with, we need to produce lattices from ideals of OL. Let L be anumber field, and [L : Q] = n. We let σ1, · · · , σr : L→ R be the real embeddings,and σr+1, · · · , σr+s, σr+1, · · · , σr+s : L→ C be the complex embeddings (notethat which embedding is σr+i and which is σr+i is an arbitrary choice).

Then this defines an embedding

σ = (σ1, σ2, · · · , σr, σr+1, · · · , σr+s) : L ↪→ Rr × Cs ∼= Rr × R2s = Rr+2s = Rn,

under the isomorphism C→ R2 by x+ iy 7→ (x, y).Just as we did for quadratic fields, we can relate the norm of ideals to their

covolume.

Lemma.

(i) σ(OL) is a lattice in Rn of covolume 2−s|DL|12 .

(ii) More generally, if aCOL is an ideal, then σ(a) is a lattice and the covolume

covol(σ(a)) = 2−s|DL|12N(a).

Proof. Obviously (ii) implies (i). So we just prove (ii). Recall that a has anintegral basis γ1, · · · , γn. Then a is the integer span of the vectors

(σ1(γi), σ2(γi), · · · , σr+s(γi))

45


for i = 1, · · · , n, and they are independent as we will soon see when we computethe determinant. So it is a lattice.

We also know that

∆(γ1, · · · , γn) = det(σi(γj))2 = N(a)2DL,

where the σi run over all σ1, · · · , σr, σr+1, · · · , σr+s, σr+1, · · · σr+s.So we know

|det(σi(γj))| = N(a)|DL|12 .

So what we have to do is to relate det(σi(γj)) to the covolume of σ(a). Butthese two expressions are very similar.

In the σi(γj) matrix, we have columns that look like(σr+i(γj) σr+i(γj)

)=(z z

).

On the other hand, the matrix of σ(γ) has corresponding entries

(Re(z) Im(z)

)=(

12 (z + z) i

2 (z − z))

=1

2

(1 1i −i

)(zz

)

We call the last matrix A =1

2

(1 1i −i

). We can compute the determinant as

|detA| =∣∣∣∣det

1

2

(1 1i −i

)∣∣∣∣ =1

2.

Hence the change of basis matrix from (σi(γj)) to σ(γ) is s diagonal copies of A,so has determinant 2−s. So this proves the lemma.

Proposition. Let aCOL be an ideal. Then there exists an α ∈ a with α 6= 0such that

|N(α)| ≤ cLN(a),

where

cL =

(4

π

)sn!

nn|DL|

12 .

This is the Minkowski bound .

Proof. Let

Br,s(t) ={

(y1, · · · , yr, z1, · · · , zs) ∈ Rr × Cs :∑|yi|+ 2

∑|zi| ≤ t

}.

This

(i) is closed and bounded;

(ii) is measurable (it is defined by polynomial inequalities);

(iii) has volume

vol(Br,s(t)) = 2r(π

2

)s tnn!

;

(iv) is convex and symmetric about 0.

46


Only (iii) requires proof, and it is on the second example sheet, i.e. we are notdoing it here. It is just doing the integral.

We now choose t so that

volBr,s(t) = 2n covol(σ(a)).

Explicitly, we let

tn =

(4

π

)sn!|DL|1/2N(a).

Then by Minkowski’s lemma, there is some α ∈ a non-zero such that σ(α) ∈Br,s(t). We write

σ(α) = (y1, · · · , yr, z1, · · · , zs).

Then we observe

N(α) = y1 · · · yrz1z1z2z2 · · · zszs =∏

yi∏|zj |2.

By the AM-GM inequality, we know

|N(α)|1/n ≤ 1

n

(∑yi + 2

∑|zj |)≤ t

n,

as we know σ(a) ∈ Br,s(t). So we get

|N(α)| ≤ tn

nn= cLN(a).

Corollary. Every [a] ∈ clL has a representative a ∈ OL with N(a) ≤ cL.

Theorem (Dirichlet). The class group clL is finite, and is generated by primeideals of norm ≤ cL.

Proof. Just as the case for imaginary quadratic fields.

47

7 Dirichlet’s unit theorem II Number Fields

7 Dirichlet’s unit theorem

We have previously characterized the units on OL as the elements with unitnorm, i.e. α ∈ OL is a unit if and only if |N(α)| = 1. However, this doesn’t tellus much about how many units there are, and how they are distributed. Theanswer to this question is given by Dirichlet’s unit theorem.

Theorem (Dirichlet unit theorem). We have the isomorphism

O×L ∼= µL × Zr+s−1,

whereµL = {α ∈ L : αN = 1 for some N > 0}

is the group of roots of unity in L, and is a finite cyclic group.

Just as in the finiteness of the class group, we do it for an example first, orelse it will be utterly incomprehensible.

We do the example of real quadratic fields, L = Q(√d), where d > 1 is

square-free. So r = 2, s = 0, and L ⊆ R implies µL = {±1}. So

O×L ∼= {±1} × Z.

Also, we know that

N(x+ y√d) = (x+ y

√d)(x− y

√d) = x2 − dy2.

So Dirichlet’s theorem is saying that there are infinitely many solutions ofx2 − dy2 = ±1, and are all (plus or minus) the powers of one single element.

Theorem (Pell’s equation). There are infinitely many x+ y√d ∈ OL such that

x2 − dy2 = ±1.

You might have seen this in IIC Number Theory, where we proved it directlyby continued fractions. We will provide a totally unconstructive proof here, sincethis is more easily generalized to arbitrary number fields.

This is actually just half of Dirichlet’s theorem. The remaining part is toshow that they are all powers of the same element.

Proof. Recall that σ : OL → R2 sends

α = x+ y√d 7→ (σ1(α), σ2(α)) = (x+ y

√d, x− y

√d).

(in the domain,√d is a formal symbol, while in the codomain, it is a real number,

namely the positive square root of d)Also, we know

covol(σ(OL)) = |DL|12 .

48


(1, 1)

Z[X]

N(α) = 1

Consider

st =

{(y1, y2) ∈ R2 : |y1| ≤ t, |y2| ≤

|DL|1/2

t

}.

Sovol(st) = 4|DL|

12 = 2n covol(OL),

as n = [L : Q] = 2. Now Minkowski implies there is an α ∈ OL non-zero suchthat σ(α) ∈ st. Also, if we write

σ(α) = (y1, y2),

thenN(α) = y1y2.

So such an α will satisfy

1 ≤ |N(α)| ≤ |DL|1/2.

This is not quite what we want, since we need |N(α)| = 1 exactly. Nevertheless,this is a good start. So let’s try to find infinitely such elements.

First notice that no points on the lattice (apart from the origin) hits the xor y axis, since any such point must satisfy x± y

√d = 0, but

√d is not rational.

Also, st is compact. So st ∩ σ(OL) contains finitely many points. So we canfind a t2 such that for each (y1, y2) ∈ st ∩ OL, we have |y1| > t2. In particular,st2 does not contain any point in st ∩ σ(OL). So we get a new set of pointsα ∈ st2 ∩ OL such that 1 ≤ |N(α)| ≤ |DL|1/2.

s1

s2

49


We can do the same thing for st2 and get a new t3. In general, given t1 > · · · > tn,pick tn+1 be such that

0 < tn+1 < min

{|y1| : (y1, y2) ∈

n⋃i=1

sti ∩ σ(OL)

},

and the minimum is finite since st is compact and hence contains finitely manylattice points on σ(OL).

Then we get an infinite sequence of ti such that sti ∩ σ(OL) are disjoint fordifferent i. Since each must contain at least one point, we have got infinitelymany points in OL satisfying 1 ≤ |N(α)| ≤ |DL|1/2.

Since there are only finitely many integers between 1 and |DL|1/2, we canapply the pigeonhole principle, and get that there is some integer satisfying1 ≤ |m| ≤ |DL|1/2 such that there exists infinitely many α ∈ OL with N(α) = m.

This is not quite good enough. We consider

OL/mOL ∼= (Z/mZ)[L:Q],

another finite set. We notice that each α ∈ OL must fall into one of finitelymany the cosets of mOL in OL. In particular, each α such that N(α) = m mustbelong to one of these cosets.

So again by the pigeonhole principle, there exists a β ∈ OL with N(β) = m,and infinitely many α ∈ OL with N(α) = m and α = β (mod mOL).

Now of course α and β are not necessarily units, if m 6= 1. However, we willshow that α/β is. The hard part is of course showing that it is in OL itself,because it is clear that α/β has norm 1 (alternatively, by symmetry, β/α is inOL, so an inverse exists).

Hence all it remains is to prove the general fact that if

α = β +mγ,

where α, β, γ ∈ OL and N(α) = N(β) = m, then α/β ∈ OL.To show this, we just have to compute

α

β= 1 +

m

βγ = 1 +

N(β)

βγ = 1 + βγ ∈ OL,

since N(β) = ββ. So done.

We have thus constructed infinitely many units.We now prove the remaining part

Theorem (Dirchlet’s unit theorem for real quadratic fields). Let L = Q(√d).

Then there is some ε0 ∈ O×L such that

O×L = {±εn0 : n ∈ Z}.

We call such an ε0 a fundamental unit (which is not unique). So

O×L ∼= {±1} × Z.

50


Proof. We have just proved the really powerful theorem that there are infinitelymany ε with N(ε) = 1. We are not going to need the full theorem. All we needis that there are three — in particular, something that is not ±1.

We pick some ε ∈ O×L with ε 6= ±1. This exists by what we just proved.Then we know

|σ1(ε)| 6= 1,

as |σ1(ε)| = 1 if and only if ε = ±1. Replacing by ε−1 if necessary, we wlogE = |σ1(ε)| > 1. Now consider

{α ∈ OL : N(α) = ±1, 1 ≤ |σ1(α)| ≤ E}.

This is again finite, since it is specified by a compact subset of the OL-lattice.So we pick ε0 in this set with ε0 6= ±1 and |σ1(ε0)| minimal (> 1). Replacing ε0

by −ε0 if necessary, we can assume σ1(ε) > 1.Finally, we claim that if ε ∈ O×L and σ1(ε) > 0, then ε = εN0 for some N ∈ Z.

This is obvious if we have addition instead of multiplication. So we take logs.Suppose

log ε

log ε0= N + γ,

where N ∈ Z and 0 ≤ γ < 1. Then we know

εε−N0 = εγ0 ∈ O×L ,

but |εγ0 | = |ε0|γ < |ε0|, as |ε0| > 1. By our choice of ε0, we must have γ = 0. Sodone.

Now we get to prove the Dirichlet unit theorem in its full glory.

Theorem (Dirichlet unit theorem). We have the isomorphism

O×L ∼= µL × Zr+s−1,

whereµL = {α ∈ L : αN = 1 for some N > 0}

is the group of roots of unity in L, and is a finite cyclic group.

Proof. We do the proof in the opposite order. We throw in the logarithm at thevery beginning. We define

` : O×L → Rr+s

by

x 7→ (log |σ1(x)|, · · · , log |σr(x)|, 2 log |σr+1(x)|, · · · , 2 log |σr+s(x)|).

Note that |σr+i(x)| = |σr+`(x)|. So this is independent of the choice of one ofσr+i, σr+i.

Claim. We now claim that im ` is a discrete group in Rr+s and ker ` = µL is afinite cyclic group.

51


We note thatlog |ab| = log |a|+ log |b|.

So this is a group homomorphism, and the image is a subgroup. To prove thefirst part, it suffices to show that im ` ∩ [−A,A]r+s is finite for all A > 0. Wenotice ` factors as

O×L OL Rr × Cs Rr+s.σ j

where σ maps α 7→ (σ1(α), · · · , σr+s(α)), and

j : (y1, · · · , yr, z1, · · · , zs) 7→ (log |y1|, · · · , log |yr|, 2 log |z1|, · · · , 2 log |z2|).

We see

j−1([−A,A]r+s) = {(yi, zj) : e−A ≤ |yi| ≤ eA, e−A ≤ 2|zj | ≤ eA}

is a compact set, and σ(OL) is a lattice, in particular discrete. So σ(OL) ∩j−1([−A,A]r+s) is finite. This also shows the kernel is finite, since the kernel isthe inverse image of a compact set.

Now as ker ` is finite, all elements are of finite order. So ker ` ⊆ µL. Con-versely, it is clear that µL ⊆ ker `. So it remains to show that µL is cyclic. SinceL embeds in C, we know µL is contained in the roots of unity in C. Since µL isfinite, we know L is generated by a root of unity with the smallest argument(from, say, IA Groups).

Claim. We claim that

im ` ⊆{

(y1, · · · , yr+s) :∑

yi = 0}∼= Rr+s−1.

To show this, note that if α ∈ O×L , then

N(α) =

n∏i=1

σi(α)

s∏`=1

σr+`(α)σr+` = ±1.

Taking the log of the absolute values, we get

0 =∑

log |σi(α)|+ 2∑

log |σr+i(α)|.

So we know im ` ⊆ Rr+s−1 as a discrete subgroup. So it is isomorphic to Zafor some a ≤ r + s− 1. Then what we want to show is that im ` ⊆ Rr+s−1 is alattice, i.e. it is congruent to Zr+s−1.

Note that so far what we have done is the second part of what we did for thereal quadratic fields. We took the logarithm to show that these form a discretesubgroup. Next, we want to find r + s− 1 independent elements to show it is alattice.

Claim. Fix a k such that 1 ≤ k ≤ r + s and α ∈ OL with α 6= 0. Then thereexists a β ∈ OL such that

|N(β)| ≤(

2

π

)s|DL|1/2,

52


and moreover if we write

`(α) = (a1, · · · , ar+s)`(β) = (b1, · · · , br+s),

then we have bi < ai for all i 6= k.

We can apply Minkowski to the region

S = {(y1, · · · , yr, z1, · · · , zs) ∈ Rr × Cs : |yi| ≤ ci, |zj | ≤ cr+j}

(we will decide what values of ci to take later). Then this has volume

vol(S) = 2rπsc1 · · · cr+s.

We notice S is convex and symmetric around 0. So if we choose 0 < ci < eai fori 6= k, and choose

ck =

(2

π

)s|DL|1/2

1

c1 · · · ck · · · cr+s.

Then Minkowski gives β ∈ σ(OL) ∩ S, satisfying the two conditions above.

Claim. For any k = 1, · · · , r + s, there is a unit uk ∈ O×L such that if `(uk) =(y1, · · · , yr+s, then yi < 0 for all i 6= k (and hence yk > 0 since

∑yi = 0).

This is just as in the proof for the real quadratic case. We can repeatedlyapply the previous claim to get a sequence α1, α2, · · · ∈ OL such that N(αt)is bounded for all t, and for all i 6= k, the ith coordinate of `(α1), `(α2), · · ·is strictly decreasing. But then as with real quadratic fields, the pigeonholeprinciple implies we can find t, t′ such that

N(αt) = N(αt′) = m,

say, andαt ≡ αt′ (mod mOL),

i.e. αt = αt′ in OL/mOL. Hence for each k, we get a unit uk = αt/αt′ such that

`(uk) = `(αt)− `(α′t) = (y1, · · · , yr+s)

has yi < 0 if i 6= k (and hence yk > 0, since∑yi = 0). We need a final trick to

show the following:

Claim. The units u1, · · · , ur+s−1 are linearly independent in Rr+s−1. Hencethe rank of `(O×L ) = r + s− 1, and Dirichlet’s theorem is proved.

We let A be the (r + s)× (r + s) matrix whose jth row is `(uj), and applythe following lemma:

Claim. Let A ∈ Matm(R) be such that aii > 0 for all i and aij < 0 for all i 6= j,and

∑j aij ≥ 0 for each i. Then rank(A) ≥ m− 1.

53


To show this, we let vi be the ith column of A. We show that v1, · · · ,vm−1

are linearly independent. If not, there exists a sequence ti ∈ R such that

m−1∑i=1

tivi = 0, (∗)

with not all of the ti non-zero. We choose k so that |tk| is maximal among thet1, · · · , tm−1’s. We divide the whole equation by tk. So we can wlog assumetk = 1, ti ≤ 1 for all i.

Now consider the kth row of (∗). We get

0 =

m−1∑i=1

tiaki ≥m−1∑i=1

aki,

as a < 0 and t ≤ 1 implies at ≥ a. Moreover, we know ami > 0 strictly. So weget

0 >

m∑i=1

aki ≥ 0.

This is a contradiction. So done.

You should not expect this to be examinable.We make a quick definition that we will need later.

Definition (Regulator). The regulator of a number field L is

RL = covol(`(O×L ) ⊆ Rr+s−1).

More concretely, we pick fundamental units ε1, · · · , εr+s−1 ∈ O×L so that

O×L = µL × {εn11 · · · ε

nr+s−1

r+s−1 : ni ∈ Z}.

We take any (r+ s− 1)(r+ s− 1) subminor of the matrix(`(ε1) · · · `(εr+s)

).

Their determinants all have the same absolute value, and

|det(subminor)| = RL.

This is a definition we will need later.We quickly look at some examples with quadratic fields. Consider L = Q(

√d),

where d 6= 0, 1 square-free.

Example. If d < 0, then r = 0 and s = 1. So r + s− 1 = 0. So O×L = µL is afinite group. So RL = 1.

Lemma.

(i) If d = −1, then Z[i]× = {±1,±i} = Z/4Z.

(ii) If d = −3, then let ω = 12 (1 +

√d), and we have ω6 = 1. So Z[ω]× =

{1, ω, · · · , ω5} ∼= Z/6Z.

(iii) For any other d < 0, we have O×L = {±1}.

54


Proof. This is just a direct check.If d ≡ 2, 3 (mod 4), then by looking at the solution of x2 − dy2 = ±1 in the

integers, we get (i) and (iii).

If d ≡ 1 (mod 4), then by looking at the solutions to(x+ y

2

)2 − d4y

2 = ±1in the integers, we get (ii) and (iii).

Now if d > 0, then RL = | log |ε||, where ε is a fundamental unit. So how dowe find a fundamental unit? In general, there is no good algorithm for finding thefundamental unit of a fundamental field. The best algorithm takes exponentialtime. We do have a good algorithm for quadratic fields using continued fractions,but we are not allowed to use that.

Instead, we could just guess a solution — we find a unit by guessing, andthen show there is no smaller one by direct check.

Example. Consider the field Q(√

2). We can try ε = 1 +√

2. We haveN(ε) = 1 − 2 = −1. So this is a unit. We claim this is fundamental. If not,there there exists u = a+ b

√2, where a, b ∈ Z and 1 < u < ε (as real numbers).

Then we haveu = a− b

√2

has uu = ±1. Since u > 1, we know |u| < 1. Then we must have u± u > 0. Sowe need a, b > 0. We know can only be finitely many possibilities for

1 < a+ b√

2 < 1 +√

2,

where a, b are positive integers. But there actually are none. So done.

Example. Consider Q(√

11). We guess ε = 10 − 3√

11 is a unit. We cancompute N(ε) = 100− 99 = 1. Note that ε < 1 and ε−1 > 1.

Suppose this is not fundamental. Then we have some u such that

1 < u = a+ b√

11 < 10 + 3√

11 = ε−1 < 20. (∗)

We can check all the cases, but there is a faster way.We must have N(u) = ±1. If N(u) = −1, then a2 − 11b2 = −1. But −1 is

not a square mod 11.So there we must have N(u) = 1. Then u−1 = u. We get 0 < ε < u−1 = u < 1

also. So−1 < −a+ b

√11 < 0.

Adding this to (∗), we get

0 < 2b√

11 < 10 + 3√

11 < 7√

11.

So b = 1, 2 or 3, but 11b2 + 1 is not a square for each of these. So done.

55

8 L-functions, Dirichlet series* II Number Fields

8 L-functions, Dirichlet series*

This section is non-examinable.We start by proving the exciting fact that there are infinitely many primes.

Theorem (Euclid). There are infinitely many primes.

Proof. Consider the function

∏p primes

(1− 1

p

)−1

=∏

p prime

(1 +

1

p+

1

p2+ · · ·

)=∑n>0

1

n.

This is since every n = pe11 · · · perr factors uniquely as a product of primes, andeach such product appears exactly once in this. If there were finitely many

primes, as∑

1pn converges to

(1− 1

p

)−1

, the sum

∑n≥1

1

n=

∏p prime

(1− 1

p

)must be finite. But the harmonic series diverges. This is a contradiction.

We all knew that. What we now want to prove is something more interesting.

Theorem (Dirichlet’s theorem). Let a, q ∈ Z be coprime. Then there existsinfinitely many primes in the sequence

a, a+ q, a+ 2q, · · · ,

i.e. there are infinitely many primes in any such arithmetic progression.

We want to imitate the Euler proof, but then that would amount to showingthat ∏

p≡a mod qp prime

(1− 1

p

)−1

is divergent, and there is no nice expression for this. So it will be a lot morework.

To begin with, we define the Riemann zeta function.

Definition (Riemann zeta function). The Riemann zeta function is defined as

ζ(s) =∑n≥1

n−s

for s ∈ C.

There are some properties we will show (or assert):

Proposition.

(i) The Riemann zeta function ζ(s) converges for Re(s) > 1.

56


(ii) The function

ζ(s)− 1

s− 1

extends to a holomorphic function when Re(s) > 0.

In other words, ζ(s) extends to a meromorphic function on Re(s) > 0 witha simple pole at 1 with residue 1.

(iii) We have the expression

ζ(s) =∏

p prime

(1− 1

ps

)−1

for Re(s) > 1, and the product is absolutely convergent. This is the Eulerproduct.

The first part follows from the following general fact about Dirichlet series.

Definition (Dirichlet series). A Dirichlet series is a series of the form∑ann

−s,where a1, a2, · · · ∈ C.

Lemma. If there is a real number r ∈ R such that

a1 + · · ·+ aN = O(Nr),

then ∑ann

−s

converges for Re(s) > r, and is a holomorphic function there.

Then (i) is immediate by picking r = 1, since in the Riemann zeta function,a1 = a2 = · · · = 1.

Recall that xs = es log x has

|xs| = |xRe(s)|

if x ∈ R, x > 0.

Proof. This is just IA Analysis. Suppose Re(s) > r. Then we can write

N∑n=1

ann−s = a1(1−s − 2−s) + (a1 + a2)(2−s − 3−s) + · · ·

+ (a1 + · · ·+ aN−1)((N − 1)−s −N−s) +RN ,

where

RN =a1 + · · ·+ aN

Ns.

This is getting annoying, so let’s write

T (N) = a1 + · · ·+ aN .

We know ∣∣∣∣T (N)

Ns

∣∣∣∣ =

∣∣∣∣T (N)

Nr

∣∣∣∣ 1

NRe(s)−r → 0

57


as N →∞, by assumption. Thus we have∑n≥1

ann−s =

∑n≥1

T (n)(n−s − (n+ 1)−s)

if Re(s) > r. But again by assumption, T (n) ≤ B · nr for some constant B andall n. So it is enough to show that∑

n

nr(n−s − (n+ 1)−s)

converges. But

n−s − (n+ 1)−s =

∫ n+1

n

s

xs+1dx,

and if x ∈ [n, n+ 1], then nr ≤ xr. So we have

nr(n−s − (n+ 1)−s) ≤∫ n+1

n

xrs

xs+1dx = s

∫ n+1

n

dx

xs+1−r .

It thus suffices to show that ∫ n

1

dx

xs+1−r

converges, which it does (to ss−r ).

We omit the proof of (ii). The idea is to write

1

s− 1=

∞∑n=1

∫ n+1

n

dx

xs,

and show that∑φn is uniformly convergent when Re(s) > 0, where

φn = n−s −∫ n+1

n

dx

xs.

For (iii), consider the first r primes p1, · · · , pr, and

r∏i=1

(1− p−si )−1 =∑

n−s,

where the sum is over the positive integers n whose prime divisors are amongp1, · · · , pr. Notice that 1, · · · , r are certainly in the set.

So ∣∣∣∣∣ζ(s)−r∏i=1

(1− p−si )−1

∣∣∣∣∣ ≤∑n≥r

|n−s| =∑n≥r

n−Re(s).

But∑n≥r n

−Re(s) → 0 as r → ∞, proving the result, if we also show that itconverges absolutely. We omit this proof, but it follows from the fact that∑

p prime

p−s ≤∑n

n−s.

and the latter converges absolutely, plus the fact that∏

(1 − an) converges ifand only if

∑an converges, by IA Analysis I.

This is good, but not what we want. Let’s mimic this definition for anarbitrary number field!

58


Definition (Zeta function). Let L ⊇ Q be a number field, and [L : Q] = n. Wedefine the zeta function of L by

ζL(s) =∑

aCOL

N(a)−s.

It is clear that if L = Q and OL = Z, then this is just the Riemann zetafunction.

Theorem.

(i) ζL(s) converges to a holomorphic function if Re(s) > 1.

(ii) Analytic class number formula: ζL(s) is a meromorphic function if Re(s) >1− 1

n and has a simple pole at s = 1 with residue

| clL |2r(2π)sRL|DL|1/2|µL|

,

where clL is the class group, r and s are the number of real and complexembeddings, you know what π is, RL is the regulator, DL is the discriminantand µL is the roots of unity in L.

(iii)

ζL(s) =∏

pCOL prime ideal

(1−N(p)−s)−1.

This is again known as the Euler product.

We will not prove this, but the proof does not actually require any new ideas.Note that ∑

aCOL

N(a)−s =∏

pCOL,p prime

(1−N(p)−s)−1

holds “formally”, as in the terms match up when you expand, as an immediateconsequence of the unique factorization of ideals into a product of prime ideals.The issue is to study convergence of

∑N(a)−s, and this comes down to estimating

the number of ideals of fixed norm geometrically, and that is where all the factorsin the pole come it.

Example. We try to compute ζL(s), where L = Q(√d). This has discriminant

D, which may be d or 4d. We first look at the prime ideals.If p is a prime ideal in OL, then p | 〈p〉 for a unique p. So let’s enumerate

the factors of ηL controlled by p ∈ Z.Now if p | |DL|, then 〈p〉 = p2 ramifies, and N(p) = p. So this contributes a

factor of (1− p−s)−1.Now if p remains prime, then we have N(〈p〉) = p2. So we get a factor of

(1− p−2s)−1 = (1− p−s)−1(1 + p−s)−1.

If p splits completely, then〈p〉 = p1p2.

SoN(pi) = p,

59


and so we get a factor of

(1− p−s)−1(1− p−s)−1.

So we find thatζL(s) = ζ(s)L(χD, s),

where we define

Definition (L-function). We define the L-function by

L(χ, s) =∏

p prime

(1− χ(p)p−s)−1.

In our case, χ is given by

χD(p) =

0 p | D−1 p remains prime

1 p splits

=

{(Dp

)p is odd

depends on d mod 8 p = 2.

Example. If L = Q(√−1), then we know(−4

p

)=

(−1

p

)= (−1)

p−12 if p 6= 2,

and χD(2) = 0 as 2 ramifies. We then have

L(χD, s) =∏

p>2 prime

(1− (−1)p−12 p−s)−1 = 1− 1

3s+

1

5s− 1

7s+ · · · .

Note that χD was defined for primes only, but we can extend it to a functionχD : Z→ C by imposing

χD(nm) = χD(n)χD(m),

i.e. we defineχD(pe11 · · · perr ) = χD(p1)e1 · · ·χD(pr)

er .

Example. Let L = Q(√−1). Then

χ−4(m) =

{(−1)

m−12 m odd

0 m even.

It is an exercise to show that this is really the extension, i.e.

χ−4(mn) = χ−4(m)χ−4(n).

Notice that this has the property that

χ−4(m− 4) = χ−4(m).

We give these some special names

60


Definition (Dirichlet character). A function χ : Z→ C is a Dirichlet characterof modulus D if there exists a group homomorphism

w :

(ZDZ

)×→ C×

such that

χ(m) =

{w(m mod D) gcd(m,D) = 1

0 otherwise.

We say χ is non-trivial if ω is non-trivial.

Example. χ−4 is a Dirichlet character of modulus 4.

Note thatχ(mn) = χ(m)χ(n)

for such Dirichlet characters, and so

L(χ, s) =∏

p prime

(1− χ(p)p−s)−1 =∑n≥1

χ(n)

ns

for such χ.

Proposition. χD, as defined for L = Q(√d) is a Dirichlet character of modulus

D.

Note that this is a very special Dirichlet character, as it only takes values0,±1. We call this a quadratic Dirichlet character.

Proof. We must show that

χD(p+Da) = χD(p)

for all p, a.

(i) If d ≡ 3 (mod 4), then D = 4d. Then

χD(2) = 0,

as (2) ramifies. So χD(even) = 0. For p > 2, we have

χD(p) =

(D

p

)=

(d

p

)=(pd

)(−1)

p−12

as d−12 ≡ 1 (mod 2), by quadratic reciprocity. So

χD(p+Da) =

(p+Da

d

)(−1)

p−12 (−1)4da/2 = χD(p).

(ii) If d ≡ 1, 2 (mod 4), see example sheet.

Lemma. Let χ be any non-trivial Dirichlet character. Then L(χ, s) is holomor-phic for Re(s) > 0.

61


Proof. By our lemma on convergence of Dirichlet series, we have to show that

N∑i=1

χ(i) = O(1),

i.e. it is bounded. Recall from Representation Theory that distinct irreduciblecharacters of a finite group G are orthogonal, i.e.

1

|G|∑g∈G

χ1(g)χ2(g) =

{1 χ1 = χ2

0 otherwise.

We apply this to G = (Z/DZ)×, where χ1 is trivial and χ2 = χ. So orthogonalitygives ∑

aD<i≤(a+1)D

χ(i) =∑

i∈(Z/DZ)×

χ(i) = 0,

using that χ(i) = 0 if i is not coprime to D. So we are done.

Corollary. For quadratic characters χD, we have

L(χD, 1) 6= 0.

For example, if D < 0, then

L(χD, 1) =2π| clQ(

√d) |

|D|1/2|µQ(√d)|.

Proof. We have shown that

ζQ(√d)(s) = ζQ(s)L(χD, s).

Note that ζQ(√d)(s) and ζQ(s) have simple poles at s = 1, while L(χD, s) is

holomorphic at s = 1.Since the residue of ζQ(s) at s = 1 is 1, while the residue of ζQ(

√D) at s = 1

is non-zero by the analytic class number formula. So L(χD, 1) is non-zero, andgiven by the analytic class number formula.

Example. If L = Q(√−1), then

1− 1

3+

1

5− 1

7+ · · · = 2π · 1

2 · 4=π

4.

In general, for any field whose class number we know, we can get a seriesexpansion for π. And it converges incredibly slow.

Note that this corollary required two things — the analytic input for theanalytic class number formula, and quadratic reciprocity (to show that χD is aDirichlet character).

More ambitiously, we now compute the zeta function of a cyclotomic field,L = Q(ωq), where ωq is the primitive qth root of unity and q ∈ N. We need toknow the following facts about cyclotomic extensions:

Proposition.

62


(i) We have [L : Q] = ϕ(q), where

ϕ(q) = |(Z/qZ)×|.

(ii) L ⊇ Q is a Galois extension, with

Gal(L/Q) = (Z/qZ)×,

where if r ∈ (Z/qZ)×, then r acts on Q(wQ) by sending ωq 7→ ωrq . This iswhat plays the role of quadratic reciprocity for cyclotomic fields.

(iii) The ring of integers is

OL = Z[ωq] = Z[x]/Φq(x),

where

Φq(x) =xq − 1∏

d|q,d6=q Φd(x)

is the qth cyclotomic polynomial.

(iv) Let p be a prime. Then p ramifies in OL if and only if p | DL, if and onlyif p | q. So while D might be messy, the prime factors of D are the primefactors of q.

(v) Let p be a prime and p - q. Then 〈p〉 factors as a product of ϕ(q)/f distinctprime ideals, each of norm pf , where f is the order of p in (Z/qZ)×.

Proof.

(i) In the Galois theory course.

(ii) In the Galois theory course.

(iii) In the example sheet.

(iv) In the example sheet.

(v) Requires proof, but is easy Galois theory, and is omitted.

Example. Let q = 8. Then

Φ8 =x8 − 1

(x+ 1)(x− 1)(x2 + 1)=x8 − 1

x4 − 1= x4 + 1.

So given a prime p (that is not 2), we need to understand

OL/p =Fp[x]

Φ8,

i.e. how Φ8 factors factors mod p (Dedekind’s criterion). We have

(Z/8)× = {1, 3, 5, 7} = {1, 3,−3,−1} = Z/2× Z/2.

Then (v) says if p = 17, then x4 factors into 4 linear factors, which it does.If p = 3, then (v) says x4 factors into 2 quadratic factors. Indeed, we have

(x2 − x− 1)(x2 + x− 1) = (x2 − 1)2 − x2 = x4 + 1.

63


Given all of these, let’s compute the zeta function! Recall that

ζQ(ωq)(s) =∏p

(1−N(p)−s)−1.

We consider the prime ideals p dividing 〈p〉, where p is a fixed integer primenumber. If p - q, then (v) says this contributes a factor of

(1− p−fs)−ϕ(q)/f ,

to the zeta function, where f is the order of p in (Z/qZ)×. We observe that thisthing factors, since

1− tf =∏γ∈µf

(1− γt),

withµf = {γ ∈ C : γf = 1},

and we can put t = p−s.We let

ω1, · · · , ωϕ(q) : (Z/qZ)× → C×

be the distinct irreducible (one-dimensional) representations of (Z/qZ)×, withω1 being the trivial representation, i.e. ω1(a) = 1 for all a ∈ (Z/qZ)×.

The claim is that ω1(p), · · · , ωϕ(q)(p) are fth roots of 1, each repeated ϕ(q)/ftimes. We either say this is obvious, or we can use some representation theory.

We know p generates a cyclic subgroup 〈p〉 of (Z/qZ)× of order f , by definitionof f . So this is equivalent to saying the restrictions of ω1, · · · , ωϕ(q) to p are thef distinct irreducible characters of 〈p〉 ∼= Z/f , each repeated ϕ(q)/f times.

Equivalently, note that

Res(Z/qZ)×

〈p〉 (ω1 ⊕ · · · ⊕ ωϕ(q)) = Res(Z/qZ)×

〈p〉 (regular representation of (Z/qZ)×).

So this claims that

Res(Z/qZ)×

〈p〉 (regular rep. of (Z/qZ)×) =ϕ(q)

f(regular rep. of Z/f).

But this is true for any group, since

ResGH CG = |G/H|CH,

as the character of both sides is |G|δe.So we have

(1− p−fs)−ϕ(q)/f =

ϕ(q)∏i=1

(1− ωi(p)p−s)−1.

So we let

χi(n) =

{wi(n mod q) gcd(n, q) = 1

0 otherwise

be the corresponding Dirichlet characters. So we have just shown that

64


Proposition. We have

ζQ(ωq)(s) =

ϕ(q)∏i=1

L(χi, s) · (corr. factor) = ζQ(s)

ϕ(q)∏i=2

L(χi, s) · (corr. factor)

where the correction factor is a finite product coming from the primes that divideq.

By defining the L functions in a slightly more clever way, we can hide thecorrection factors into the L(χ, s), and then the ζ function is just the product ofthese L-functions.

Proof. Our analysis covered all primes p - q, and the correction factor is just toinclude the terms with p | q. The second part is just saying that

ζQ(s) = L(χ1, s)∏p|q

(1− p−s)−1.

This allows us to improve our result on the non-vanishing of L(χ, 1) to allDirichlet characters, and not just quadratic Dirichlet characters.

Corollary. If χ is any non-trivial Dirichlet character, then L(χ, 1) 6= 0.

Proof. By definition, Dirichlet characters come from representations of some(Z/qZ)×, so they appear in the formula of the ζ function of some cyclotomicextension.

Consider the formula

ζQ(ωq)(s) = ζQ(s)

ϕ(q)∏i=2

L(χi, s) · (corr. factor)

at s = 1. We know that the L(χi, s) are all holomorphic at s = 1. Moreover,both ζQ(ωq) and ζQ have a simple pole at 0. Since the correction terms are finite,it must be the case that all L(χi, s) are non-zero.

Theorem (Dirichlet, 1839). Let a, q ∈ N be coprime, i.e. gcd(a, q) = 1. Thenthere are infinitely many primes in the arithmetic progression

a, a+ q, a+ 2q, a+ 3q, · · · .

Proof. As before, let

ω1, · · · , ωϕ(q) : (Z/qZ)× → C×

be the irreducible characters, and let

χ1, · · · , χϕ(q) : Z→ C

be the corresponding Dirichlet character, with ω1 the trivial one.Recall the orthogonality of columns of the character table, which says that if

gcd(p, q) = 1, then

1

ϕ(q)

∑i

ωi(a)ωi(p) =

{1 a ≡ p (mod q)

0 otherwise.

65


Hence we know

1

ϕ(q)

∑i

χi(a)χi(p) =

{1 a ≡ p (mod q)

0 otherwise,

even if gcd(p, q) 6= 1, as then χi(p) = 0. So∑p≡a mod qp prime

p−s =1

ϕ(q)

∑i

χi(a)∑

all primes p

χi(p)p−s. (‡)

We want to show this has a pole at s = 1, as in Euclid’s proof.To do so, we show that

∑p χi(p)p

−s is “essentially” logL(χi, s), up to somebounded terms. We Taylor expand

logL(χ, s) = −∑

log(1− χ(p)p−s) =∑n≥1

p prime

χ(p)n

npns=

∑n≥1

p prime

χ(pn)

npns.

What we care about is the n = 1 term. So we claim that∑n≥2,p prime

χ(pn)

npns

converges at s = 1. This follows from the geometric sum∣∣∣∣∣∣∑p

∑n≥2

χ(pn)

npns

∣∣∣∣∣∣ ≤∑p

∑n≥2

p−ns =∑

p prime

1

ps(ps − 1)≤∑n≥2

1

ns(ns − 1)<∞.

Hence we know

logL(χ, s) =∑p

χi(p)p−s + bounded stuff

near s = 1.So at s = 1, we have

(‡) ∼ 1

ϕ(q)

∑i

χi(a) logL(χi, s).

and we have to show that the right hand side has a pole at s = 1.We know that for i 6= 1, i.e. χi non-trivial, L(χi, s) is holomorphic and

non-zero at s = 1. So we just have to show that logL(χ1, s) has a pole. Notethat L(χ1, s) is essentially ζQ(s). Precisely, we have

L(χ1, s) = ζQ(s)∏p|q

(1− p−s).

Moreover, we already know that ζQ(s) blows up at s = 1. We have

ζQ(s) =1

s− 1+ holomorphic function

=1

s− 1(1 + (s− 1)(holomorphic function)).

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So we know

logL(χ1, s) ∼ log ζQ(s) ∼ log

(1

s− 1

),

and this does blow up at s = 1.

So far, we have been working with abelian extensions over Q, i.e. extensionsL/Q whose Galois group is abelian. By the Kronecker–Weber theorem, everyabelian extension of Q is contained within some cyclotomic extension. So insome sense, we have considered the “most general” abelian extension.

Can we move on to consider more complicated number fields? In general,suppose L/Q is Galois, and G = Gal(L/Q). We can still make sense of the ζfunctions, and it turns out it always factors as

ζL(s) =∏ρ

L(ρ, s)dim ρ,

where ρ ranges over all the irreducible representations of G, and L(ρ, s) is theArtin L-function. It takes some effort to define the Artin L-function, and weshall not do so here. However, it is worth noting that L(1, s) is just ζQ(s), andfor ρ 6= 1, we still have a factorization of the form

L(ρ, s) =∏

p prime

Lp(ρ, s).

This Lp(ρ, s) is known as the Euler factor .One can show that L(ρ, s) is always a meromorphic function of s, and is

conjectured to be holomorphic for all s (if ρ 6= 1, of course).If ρ is one-dimensional, then L(ρ, s) is a Dirichlet series L(χ, s) for some χ.

Recall that to establish this fact for quadratic fields, we had to use quadraticreciprocity. In general given a ρ, finding χ is a higher version of “quadraticreciprocity”. This area is known as class field theory . If dim ρ > 1, then this is“non-abelian class field theory”, known as Langlands programme.

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Index II Number Fields

Index

DL, 15IL, 24L-function, 60clL, 26ζ, 56ζL, 59pα, 8r, 13s, 13

addition of ideals, 24algebraic integer, 4analytic class number formula, 59Artin L-function, 67associates, 18

class field theory, 67class group, 26

finiteness, 47conjugate, 12covolume, 43

Dedekind domain, 18Dedekind’s criterion, 33degree, 4Dirichlet character, 61Dirichlet series, 57Dirichlet unit theorem, 48Dirichlet’s theorem, 56Dirichlet’s theorem on primes in

AP, 65Dirichlet’s unit theorem, 50, 51discrete subset, 42discriminant, 15divides, 17

Euler factor, 67

field extension, 4finite extension, 4finitely-generated, 5finiteness of class group, 47fractional ideal, 22

invertible, 22fundamental domain, 43

honest ideal, 22

ideal

fractional, 22honest, 22integral, 22invertible, 22multiplication, 17norm, 27prime, 18sum, 24unique factorization, 24

ideal class group, 26inert prime, 32integral, 5integral basis, 14integral ideal, 22invertible fractional ideal, 22

Kronecker–Weber theorem, 67

Langlands programme, 67lattice, 43

minimal polynomial, 8Minkowski bound, 40, 46Minkowski’s lemma, 38Minkowski’s theorem, 44multiplication of ideals, 17

norm, 10of ideal, 27

number field, 4

Pell’s equation, 48prime ideal, 18primitive element theorem, 12

quadratic Dirichlet character, 61

ramification index, 32ramified prime, 32regulator, 54Riemann zeta function, 56

splitting prime, 32sum of ideals, 24

trace, 10

unique factorization of ideals, 24

zeta function, 59

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part ii - number fields - srcf · 1 number elds ii number fields this is a re nement of the idea of...

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