II Number Fields (Full)

Part II — Number Fields

Based on lectures by I. Grojnowski

Notes taken by Dexter Chua

Lent 2016

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) 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 and

discriminants. [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 statement

of 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]

Contents

0 Introduction

1 Number fields

2 Norm, trace, discriminant, numbers

3 Multiplicative structure of ideals

4 Norms of ideals

5 Structure of prime ideals

6 Minkowski bound and finiteness of class group

7 Dirichlet’s unit theorem

8 L-functions, Dirichlet series*

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 we

will not refrain from pointing out the correspondence. If you have not learnt

Galois Theory, then you can ignore them.

1 Number fields

The focus of this course is, unsurprisingly, number fields. Before we define what

number fields are, we look at some motivating examples. Suppose we wanted to

find all numbers of the form

, where

x, y ∈ Z

. For example, if

a, b

can

both be written in this form, does it follow that ab can?

In IB Groups, Rings and Modules, we did the clever thing of working with

[

]. The integers of the form

are exactly the norms of integers in

[

where the norm of x + iy is

N(x + iy) = |x + iy|

= x

+ y

Then the previous result is obvious — if

(

) and

(

), then

N(zw). So ab is of the form x

+ y

Similarly, in the IB Groups, Rings and Modules example sheet, we found

all solutions to the equation

+ 2 =

by working in

[

√

−2

]. This is a very

general technique — working with these rings, and corresponding fields

(

√

−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

is a vector

space over K, and the degree of the field extension is

[L : K] = dim

(L).

Definition

(Finite extension)

A finite field extension is a field extension with

finite 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 one

and the whole field itself. Thus, if we want to do something interesting with

number fields algebraically, we need to come up with something more interesting.

In the case of

itself, one interesting thing to talk about is the integers

It turns out the right generalization to number fields is algebraic integers.

Definition

(Algebraic integer)

Let

be a number field. An algebraic integer

is an

α ∈ L

such that there is some monic

f ∈ Z

[

] with

(

) = 0. We write

for the set of algebraic integers in L.

Example.

It is a fact that if

(

), then

[

]. We will prove this in

the next chapter after we have the necessary tools.

These are in fact the main objects of study in this course. Since we say this

is a generalization of Z ⊆ Q, the following had better be true:

Lemma. O

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

Proof. If α ∈ Z, then x − α ∈ Z[x] is a monic polynomial. So α ∈ O

On the other hand, let

α ∈ Q

. Then there is some coprime

r, s ∈ Z

such that

α =

. If it is an algebraic integer, then there is some

f(x) = x

+ a

n−1

+ ··· + a

with a

∈ Z such that f(α) = 0. Substituting in and multiplying by s

, we get

+ a

n−1

s + ···+ a

| {z }

divisible by s

= 0,

s | r

. But if

s 6

= 1, there is a prime

such that

p | s

, and hence

p | r

. Thus

p | r

. So

is a common factor of

and

. This is a contradiction. So

= 1, and

α is an integer.

How else is this a generalization of

? We know

is a ring. So perhaps

also is.

Theorem. O

is a ring, i.e. if α, β ∈ O

, then so is α ± β and αβ.

Note that in general O

is not a field. For example, Z = O

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 a

field extension with

α, β ∈ L

algebraic over

, then so is

αβ

and

α ± β

, as well

if α 6= 0.

To prove this, we notice that

α ∈ K

is algebraic if and only if

[

] is a finite

extension — if α is algebraic, with

f(α) = a

+ ··· + a

= 0, a

6= 0

then

[

] has degree at most

, since

(and similarly

−1

) can be written as

a linear combination of 1

, α, ··· , α

n−1

, and thus these generate

[

]. On the

other hand, if

[

] is finite, say of degree

, then 1

, α, ··· , α

are independent,

hence some linear combination of them vanishes, and this gives a polynomial

for which

is a root. Moreover, by the same proof, if

is any finite extension

over K, then any element in K

is algebraic.

Thus, to prove the result, notice that if

[

] is generated by 1

, α, ··· , α

and

[

] is generated by 1

, β, ··· , β

, then

[

α, β

] is generated by

{α

}

for

≤ i ≤ n,

≤ j ≤ m

. Hence

[

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 consider

as a ring extension of

. We will formulate the general notion of “being an

algebraic integer” in general ring extensions:

Definition

(Integrality)

Let

R ⊆ S

be rings. We say

α ∈ S

is integral over

if 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

is finitely-generated over

if there

exists elements

, ··· , α

∈ S

such that the function

→ S

defined by

(

, ··· , r

)

7→

is surjective, i.e. every element of

can be written as a

linear combination of elements

, ··· , α

. In other words,

is finitely-generated

as an R-module.

This is a refinement of the idea of being algebraic. We allow the use of rings

and restrict to monic polynomials. In Galois theory, we showed that finiteness

and algebraicity “are the same thing”. We will generalize this to integrality of

rings.

Example.

In a number field

Z ⊆ Q ⊆ L

α ∈ L

is an algebraic integer if and

only if α is integral over Z by definition, and O

is integral over Z.

Notation.

, ··· , α

∈ S

, we write

[

, ··· , α

] for the subring of

generated by

R, α

, ··· , α

. In other words, it is the image of the homomorphism

from the polynomial ring R[x

, ··· , x

] → S given by x

7→ α

Proposition.

(i)

Let

R ⊆ S

be rings. If

[

] and

is integral over

, then

finitely-generated over R.

(ii)

[

, ··· , s

] with

integral over

, then

is finitely-generated

over R.

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

Proof.

(i)

We know

is spanned by 1

, s, s

, ···

over

. However, since

is integral,

there exists a

, ··· , a

∈ R such that

= a

+ a

s + ··· + a

n−1

So the

-submodule generated by 1

, s, ··· , s

n−1

is stable under multiplica-

tion by s. So it contains s

, s

n+1

, s

n+2

, ···. So it is S.

(ii)

Let

[

, ··· , s

]. So

i−1

[

]. Since

is integral over

, it is

integral over

i−1

. By the previous part,

is finitely-generated over

i−1

To finish, it suffices to show that being finitely-generated is transitive.

More precisely, if

A ⊆ B ⊆ C

are rings,

is finitely generated over

and

is finitely generated over

, then

is finitely generated over

. This

is not hard to see, since if

, ··· , x

generate

over

, and

, ··· , y

generate

over

, then

is generated by

}

1≤i≤n,1≤j≤m

over

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 monic

polynomial which it satisfies. In Galois theory, we have fields and vector spaces,

and the proof is easy. We can just consider 1

, s, s

, ···

, and linear dependence

kicks in and gives us a relation. But even if this worked in our case, there is no

way we can make this polynomial monic.

Instead, consider the multiplication-by-

map:

S → S

γ 7→ sγ

. If

were a finite-dimensional vector space over

, then Cayley-Hamilton tells us

and thus

, satisfies its characteristic polynomial, which is monic. Even though

is not a finite-dimensional vector space, the proof of Cayley-Hamilton will

work.

Proof.

Let

, ··· , α

generate

as an

-module. wlog take

= 1

∈ S

. For

any s ∈ S, write

sα

for some

∈ R

. We write

= (

). This is the “matrix of multiplication by

S”. By construction, we have

(sI − B)













= 0. (∗)

Now recall for any matrix

, we have

adj

(

)

= (

det X

)

, where the

i, j

entry of

adj

(

) is given by the determinant of the matrix obtained by removing

the ith row and jth column of X.

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

det(sI − B)













= 0

In particular,

det

(

sI −B

)

= 0. Since we picked

= 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 O

is a ring.

Proof.

α, β ∈ O

, then

[

α, β

] is finitely-generated by the proposition. But

then

[

α, β

] is integral over

, 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

is finitely-generated over

. For example, if

is the set of all algebraic

integers in

, and

, then by definition

is an integral extension of

, but

S is not finitely generated over Z.

Thus the following corollary isn’t as trivial as the case with “integral” replaced

by “finitely generated”:

Corollary.

A ⊆ B ⊆ C

be ring extensions such that

over

and

over

are 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 polynomials

witnessing integrality.

Proof. If c ∈ C, let

f(x) =

i=0

∈ B[x]

be a monic polynomial such that

(

) = 0. Let

[

, ··· , b

] and let

[

]. Then

is finitely generated as

, ··· , b

are integral over

Also,

is finitely-generated over

, since

is integral over

. Hence

finitely-generated over

. So

is integral over

. Since

was arbitrary, we know

C is integral over A.

Now how do we recognize algebraic integers? If we want to show something

is an algebraic integer, we just have to exhibit a monic polynomial that vanishes

on the number. However, if we want to show that something is not an algebraic

integer, we have to make sure no monic polynomial kills the number. How can

we do so?

It turns out to check if something is an algebraic integer, we don’t have to

check all monic polynomials. We just have to check one. Recall that if

K ⊆ L

is a field extensions with

α ∈ L

, then the minimal polynomial is the monic

polynomial p

(x) ∈ K[x] of minimal degree such that p

(α) = 0.

Note that we can always make the polynomial monic. It’s just that the

coefficients 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 p

| f .

Proof. Write f = p

h + r, with r ∈ K[x] and deg(r) < deg(p

). 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 p

In particular, this lemma implies

is unique. One nice application of this

result is the following:

Proposition.

Let

be a number field. Then

α ∈ O

if and only if the minimal

polynomial p

(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 condition

for whether something is algebraic.

Proof. (⇐) is trivial, since this is just the definition of an algebraic integer.

(

⇒

) Let

α ∈ O

and

∈ Q

[

] be the minimal polynomial of

, and

(

)

∈ Z

[

] be a monic polynomial which

satisfies. The idea is to use

show that the coefficients of p

are algebraic, thus in fact integers.

Now there exists a bigger field M ⊇ L such that

(x) = (x − α

) ···(x − α

)

factors in

[

]. But by our lemma,

| h

. So

(

) = 0 for all

. So

∈ O

is an algebraic integer. But

is a ring, i.e. sums and products of the

’s are

still algebraic integers. So the coefficients of

are algebraic integers (in

But they are also in Q. Thus the coefficients must be integers.

Alternatively, we can deduce this proposition from the previous lemma plus

Gauss’ lemma.

Another relation between

and

is that

is the fraction field of

. This

is true for general number fields

Lemma. We have

Frac O



: α, β ∈ O

, β 6= 0



= L.

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

Proof.

α ∈ L

, let

(

)

∈ Q

[

] be its monic minimal polynomial. Then there

exists

n ∈ Z

non-zero such that

(

)

∈ Z

[

] (pick

to be the least common

multiple of the denominators of the coefficients of

(

)). Now the magic is to

put

h(x) = n

deg(g)





Then this is a monic polynomial with integral coefficients — in effect, we have

just multiplied the coefficient of

deg(g)−i

! Then

(

nα

) = 0. So

nα

integral.

2 Norm, trace, discriminant, numbers

Recall that in our motivating example of

[

], one important tool was the norm

of an algebraic integer

, given by

(

) =

. This can be

generalized to arbitrary number fields, and will prove itself to be a very useful

notion to consider. Apart from the norm, we will also consider a number known

as the trace, which is also useful. We will also study numbers associated with

the number field itself, rather than particular elements of the field, and it turns

out 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

. We

write

L → L

for the map

 7→ α

. Viewing this as a linear map of

vector

spaces, we define the norm of α to be

L/K

(α) = det m

and the trace to be

L/K

(α) = tr m

The following property is immediate:

Proposition.

For a field extension

L/K

and

a, b ∈ L

, we have

(

) =

N(a)N(b) and tr(a + b) = tr(a) + tr(b).

We can alternatively define the norm and trace as follows:

Proposition.

Let

∈ K

[

] be the minimal polynomial of

. Then the

characteristic polynomial of m

det(xI − m

) = p

[L:K(α)]

Hence if p

(x) splits in some field L

⊇ K(α), say

(x) = (x − α

) ···(x − α

then

K(α)/K

(α) =

, tr

K(α)/K

(α) =

and hence

L/K

(α) =





[L:K(α)]

, tr

L/K

= [L : K(α)]





This was proved in the IID Galois Theory course, and we will just use it

without proving.

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

(i) α ∈ O

(ii) The minimal polynomial p

is in Z[x]

(iii) The characteristic polynomial of m

is in Z[x].

This in particular implies N

L/Q

(α) ∈ Z and tr

L/Q

(α) ∈ Z.

Proof.

The equivalence between the first two was already proven. For the

equivalence between (ii) and (iii), if

∈ Z

[

], then

α ∈ O

since it vanishes

on a monic polynomial in

[

]. On the other hand, if

∈ Z

[

], then so is the

characteristic polynomial, since it is just p

The final implication comes from the fact that the norm and trace are just

coefficients of the characteristic polynomial.

It would be nice if the last implication is an if and only if. This is in general

not true, but it occurs, obviously, when the characteristic polynomial is quadratic,

since the norm and trace would be the only coefficients.

Example.

Let

(

√

) =

[

]

(

−d

), where

is not a square in

. As a

vector space over

, we can take 1

√

as our basis. So every

can be written

α = x + y

√

Hence the matrix of multiplication by α is



x dy

y x



So the trace and norm are given by

L/K

(x + y

√

d) = 2x = (x + y

√

d) + (x − y

√

L/K

(x + y

√

d) = x

− dy

= (x + y

√

d)(x − y

√

We can also obtain this by consider the roots of the minimal polynomial of

α = x + y

√

d, namely (α − x)

− y

d = 0, which has roots x ± y

√

In particular, if

(

√

), 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

(

√

d] d ≡ 2 or 3 (mod 4)

(1 +

√

d ≡ 1 (mod 4)

Proof.

We know

√

λ ∈ O

if and only if 2

x, x

− dy

∈ Z

by the previous

example. These imply 4

∈ Z

. So if

with

r, s

coprime,

r, s ∈ Z

, then we

must have s

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

x =

, y =

for some

u, v ∈ Z

. Then we know

−dv

∈

, i.e.

≡ dv

(

mod

4). But we

know the squares mod 4 are always 0 and 1. So if

d 6≡

1 (

mod

4), then

≡ dv

(

mod

4) imply that

= 0 (

mod

4), and hence

u, v

are even. So

x, y ∈ Z

giving O

= 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 +

√

d).

As a sanity check, we find that the minimal polynomial of

(1 +

√

) is

− x +

(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)

Let

K ⊆ L

be a separable field extension.

Then there exists an α ∈ L such that K(α) = L.

For example, Q(

√

3) = Q(

√

2 +

√

3).

Since

has characteristic zero, it follows that all number fields are separable

extensions. So any number field

L/Q

is of the form

(

). This makes it

much easier to study number fields, as the only extra “stuff” we have on top of

One particular thing we can do is to look at the number of ways we can

embed

L → C

. For example, for

(

√

−1

), there are two such embeddings — one

sends

√

−1 to i and the other sends

√

−1 to −i.

Lemma.

The degree [

] =

of a number field is the number of field

embeddings L → C.

Proof.

Let

be a primitive element, and

(

)

∈ Q

[

] its minimal polynomial.

Then by we have

deg p

= [

] =

, as 1

, α, α

, ··· , α

n−1

is a basis. Moreover,

Q[x]

)

∼

Q(α) = L.

Since L/Q is separable, we know p

has n distinct roots in C. Write

(x) = (x − α

) ···(x − α

Now an embedding

[

]

(

)

→ C

is uniquely determined by the image of

and

must be sent to one of the roots of

. So for each

, the map

x 7→ α

gives 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 alternative

formula for the norm and trace.

Corollary.

Let

L/Q

be a number field. If

, ··· , σ

L → C

are the different

field embeddings and β ∈ L, then

L/Q

(β) =

(β), N

L/Q

(β) =

(β).

We call σ

(β), ··· , σ

(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.

Lemma. Let x ∈ O

. Then x is a unit if and only if N

L/Q

(x) = ±1.

Notation. Write O

= {x ∈ O

: x

−1

∈ O

}, the units in O

Proof.

(

⇒

) We know

(

) =

(

)

(

). So if

x ∈ O

, then there is some

y ∈ O

such that xy = 1. So N(x)N(y) = 1. So N(x) is a unit in Z, i.e. ±1.

(

⇐

) Let

, ··· , σ

L → C

be the

embeddings of

. For notational

convenience, We suppose that

is already subfield of

, and

is the inclusion

map. Then for each x ∈ O

, we have

N(x) = xσ

(x) ···σ

(x).

Now if

(

) =

1, then

−1

±σ

(

)

···σ

(

). So we have

−1

∈ O

, since

this is a product of algebraic integers. So x is a unit in O

Corollary. If x ∈ O

is such that N(x) is prime, then x is irreducible.

Proof.

, then

(

)

(

) =

(

). Since

(

) is prime, either

(

) =

or 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

(

and

)

We write

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,

is the number of real roots of

, and

is the number of pairs of

complex conjugate roots.

The distinction between real embeddings and complex embeddings will be

important in the second half of the course.

Discriminant

The final invariant we will look at in this chapter is the discriminant. It is based

on the following observation:

Proposition.

Let

L/K

be a separable extension. Then a

-bilinear form

L × L → K

defined by (

x, y

)

7→ tr

L/K

(

) is non-degenerate. Equivalent, if

, ··· , α

are a

-basis for

, the Gram matrix (

(

))

i,j=1,···,n

has non-zero

determinant.

Recall from Galois theory that if

L/K

is not separable, then

L/K

= 0, and

it is very very degenerate. Also, note that if

is of characteristic 0, then there is

a quick and dirty proof of this fact — the trace map is non-degenerate, because

for any

x ∈ K

, we have

L/K

(

x · x

−1

) =

n 6

= 0. This is really the only case

we care about, but in the proof of the general result, we will also find a useful

formula for the discriminant when the basis is 1, θ, θ

, . . . , θ

n−1

We will use the following important notation:

Notation.

∆(α

, ··· , α

) = det(tr

L/K

(α

)).

Proof.

Let

, ··· , σ

L →

be the

distinct

-linear field embeddings

L →

K. Put

S = (σ

(α

))

i,j=1,···,n







(α

) ··· σ

(α

)

(α

) ··· σ

(α







Then

S =

k=1

(α

)σ

(α

)

i,j=1,···n

We know σ

is a field homomorphism. So

k=1

(α

)σ

(α

) =

k=1

(α

) = tr

L/K

(α

S = (tr(α

))

i,j=1,···,n

So we have

∆(α

, ··· , α

) = det(S

S) = det(S)

Now we use the theorem of primitive elements to write

(

) such that

1, θ, ··· , θ

n−1

is a basis for L over K, with [L : K] = n. Now S is just

S =







1 σ

(θ) ··· σ

(θ)

n−1

1 σ

(θ) ··· σ

(θ)

n−1







This is a Vandermonde matrix, and so

∆(1, θ, ··· , θ

n−1

) = (det S)

i<j

(σ

(θ) − σ

(θ))

Since the field extension is separable, and hence

for all

i, j

, this implies

(

)

(

), since

generates the field. So the product above is non-zero.

So we have this nice canonical bilinear map. However, this determinant is

not canonical. Recall that if

, ··· , α

is a basis for

L/K

, and

, ··· , α

another basis, then

for some A = (a

) ∈ GL

(K). So

∆(α

, ··· , α

) = (det A)

∆(α

, ··· , α

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

, ··· , α

of L is an integral basis if

(

i=1

: m

∈ Z

)

Zα

In other words, it is simultaneously a basis for L over Q and O

over Z.

Note that integral bases are not unique, just as with usual bases. Given one

basis, you can get any other by acting by GL

(Z).

Example.

Consider

(

√

) with

square-free,

d 6

= 0

1. If

∼

1 (

mod

4),

we’ve seen that 1

(1 +

√

) is an integral basis. Otherwise, if

∼

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 for

. In particular, O

∼

with n = [L : Q].

Proof.

Let

, ··· , α

be any basis of

over

. We have proved that there is

some

∈ Z

such that

∈ O

. So wlog

, ··· , α

∈ O

, and are an basis of

over

. Since

are integral, so are

, and so all these have integer trace,

as we have previously shown. Hence ∆(

, ··· , α

), being the determinant of a

matrix with integer entries, is an integer.

Now choose a

-basis

, ··· , α

∈ O

such that ∆(

, ··· , α

)

∈ Z \ {

}

has minimal absolute value. We will show that these are an integral basis.

Let x ∈ O

, and write

x =

for some λ

∈ Q. These λ

are necessarily unique since α

, ··· , α

is a basis.

Suppose some λ

6∈ Z. wlog say λ

6∈ Z. We write

= n

+ ε

for n

∈ Z and 0 < ε

< 1. We put

= x − n

= ε

+ λ

+ ··· + λ

∈ O

So α

, α

, ··· , α

is still a basis for L/Q, and are still in O

. But then

∆(α

, ··· , α

) = ε

· ∆(α

, ··· , α

) < ∆(α

, ··· , α

This contradicts minimality. So we must have

∈ Z

for all

. So this is a basis

for O

Now if

, ··· , α

is another integral basis of

over

, then there is some

g ∈ GL

(

) such that

gα

. Since

det

(

) is invertible in

, it must be 1 or

−1, and hence

det ∆(α

, ··· , α

) = det(g)

∆(α

, ··· , α

) = ∆(α

, ··· , α

)

and is independent of the choice of integral basis.

Definition

(Discriminant)

The discriminant

of a number field

is defined

= ∆(α

, ··· , α

)

for any integral basis α

, ··· , α

Example.

Let

(

√

), where

d 6

= 0

1 and

is square-free. If

∼

(mod 4), then it has an integral basis 1,

√

d. So

= det



√

1 −

√



= 4d.

Otherwise, if d

∼

1 (mod 4), then

= det



(1 +

√

(1 −

√



= d.

Recall that we have seen the word discriminant before, and let’s make sure

these concepts are more-or-less consistent. Recall that the discriminant of a

polynomial f(x) =

(x − α

) is defined as

disc(f) =

i<j

(α

− α

)

= (−1)

n(n−1)/2

i6=j

(α

− α

(

)

∈ K

[

] is the minimal polynomial of

(where

[

]), then the roots

of p

are σ

(θ). Hence we get

disc(p

) =

i<j

(σ

(θ) − σ

(θ))

In other words,

disc(p

) = ∆(1, θ, ··· , θ

n−1

So this makes sense.

3 Multiplicative structure of ideals

Again, let

L/Q

be a number field. It turns out that in general, the integral

ring

is not too well-behaved as a ring. In particular, it fails to be a UFD in

general.

Example. Let L = Q(

√

5). Then O

= 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

√

−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 =

(3) =

(

)

(

). Since none of the terms on the right are

1, we must have

(

) =

But there are no solutions to

+ 5y

= ±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 not

uniquely — we induct on

(

)

. If

(

)

= 1, then

is a unit. Otherwise,

is either irreducible, or

βγ

. Since

(

)

(

) =

(

), 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

This has a natural multiplicative structure — the product of two ideals

a, b

generated by products

, with

a ∈ a, b ∈ b

. The big theorem is that every ideal

can be written uniquely as a product of prime ideals.

Definition

(Ideal multiplication)

Let

a, b C O

be ideals. Then we define the

product ab as

ab =







i,j

: α

∈ a, β

∈ 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

is a UFD.

Usually, when we want to prove factorization is unique, we write an object as

a = x

···x

= y

···y

We then use primality to argue that

must be equal to some of the

, and

then cancel them from both sides. We can usually do this because we are working

with an integral domain. However, we don’t have this luxury when working with

ideals.

Thus, what we are going to do is to find inverses for our ideals. Of course,

given any ideal

, there is no ideal

−1

such that

−1

, as for any

we know

is contained in

. Thus we are going to consider more general

objects known as fractional ideals, and then this will allow us to prove unique

factorization.

Even better, we will show that

a | b

is equivalent to

b ⊆ a

. This is a very

useful result, since it is often very easy to show that

b ⊆ a

, but it is usually very

hard to actually find the quotient a

−1

We first look at some examples of multiplication and factorization of ideals

to get a feel of what these things look like.

Example. We have

, ··· , x

ihy

, ··· , y

i = hx

: 1 ≤ i ≤ n, 1 ≤ j ≤ mi.

In particular,

hxihyi = hxyi.

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

Example. In Z[

√

−5], we claim that we have

h3i = h3, 1 +

√

−5ih3, 1 −

√

−5i.

So h3i is not irreducible.

Indeed, we can compute

h3, 1 +

√

−5ih3, 1 −

√

−5i = h9, 3(1 + 2

√

−5), 3(1 − 2

√

−5), 21i.

But we know

gcd

21) = 3. So

by Euclid’s algorithm. So this is in

fact equal to h3i.

Notice that when we worked with elements, the number 3 was irreducible, as

there 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 can

further factorize

into a product of smaller ideals. Of course, these cannot be

principal ideals, or else we would have obtained a factorization of 3 itself. So

we can think of these ideals as “generalized elements” that allow us to further

break elements down.

Indeed, given any element in

α ∈ O

, we obtain an ideal

hαi

corresponding

to α. This map is not injective — if two elements differ by a unit, i.e. they are

associates, then they would give us the same ideal. However, this is fine, as we

usually think of associates as being “the same”.

We recall the following definition:

Definition

(Prime ideal)

Let

be a ring. An ideal

p ⊆ R

is prime if

R/p

an integral domain. Alternatively, for all

x, y ∈ R

xy ∈ p

implies

x ∈ p

y ∈ p

In this course, we take the convention that a prime ideal is non-zero. This is

not standard, but it saves us from saying “non-zero” all the time.

It turns out that the ring of integers

is a very special kind of rings, known

as 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.

(iii) R

is integrally closed in

Frac R

, i.e. if

x ∈ Frac R

is integral over

, then

x ∈ R.

(iv) Every proper prime ideal is maximal.

This is a rather specific list of properties

happens to satisfy, and it turns

out most interesting properties of

can be extended to arbitrary Dedekind

domains. However, we will not do the general theory, and just study number

fields in particular.

The important result is, of course:

Proposition.

Let

L/Q

be a number field, and

be its ring of integers. Then

is a Dedekind domain.

The first three parts of the definition are just bookkeeping and not too

interesting. The last one is what we really want. This says that

is “one

dimensional”, if you know enough algebraic geometry.

Proof of (i) to (iii).

(i) Obvious, since O

⊆ L.

(ii)

We showed that as an abelian group,

. So if

a ≤ O

is an ideal,

then

a ≤ Z

as a subgroup. So it is finitely generated as an abelian group,

and hence finitely generated as an ideal.

(iii)

Note that

Frac O

. If

x ∈ L

is integral over

, as

is integral

over Z, x is also integral over Z. So x ∈ O

, by definition of O

To prove the last part, we need the following lemma, which is also very

important on its own right.

Lemma.

Let

a C O

be a non-zero ideal. Then

a ∩ Z 6

{

}

and

is finite.

Proof. Let α ∈ a and α 6= 0. Let

= x

+ a

m−1

+ ··· + a

be its minimal polynomial. Then

∈ Z

[

]. We know

= 0 as

is irreducible.

Since p

(α) = 0, we know

= −α(α

m−1

+ a

m−1

m−2

+ ··· + a

α + a

We know

α ∈ a

by assumption, and the mess in the brackets is in

. So the

whole thing is in a. But a

∈ Z. So a

∈ Z ∩ a.

Thus, we know ha

i ⊆ a. Thus we get a surjection

→

Hence it suffices to show that O

/ha

i is finite. But for every d ∈ Z, we know

hdi





which is finite.

Finally, recall that a finite integral domain must be a field — let

x ∈ R

with

x 6

= 0. Then

y 7→ xy

is injective, as

is an integral domain. So it is a

bijection, 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

be a prime ideal. Then

is an integral domain. Since

the lemma says O

/p is finite, we know O

/p is a field. So p is maximal.

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

Lemma.

Let

be a prime ideal in a ring

. Then for

a, b C R

ideals, then

ab ⊆ p implies a ⊆ p or b ⊆ p.

Once we’ve shown that inclusion of ideals is equivalent to divisibility, this in

effect says “prime ideals are primes”.

Proof.

If not, then there is some

a ∈ a \ p

and

b ∈ b \ p

. Then

ab ∈ ab ⊆ p

. But

then 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

a C O

a non-zero ideal. Then there is a subset of

that is a

product of prime ideals.

The proof is some unenlightening abstract nonsense.

Proof.

We are going to use the fact that

is Noetherian. If this does not hold,

then there must exist a maximal ideal

not containing a product of prime ideals

(by which we mean any ideal greater than

contains a product of prime ideals,

not that

is itself a maximal ideal). In particular,

is not prime. So there are

some x, y ∈ O

such that x, y 6∈ a but xy ∈ a.

Consider

hxi

. This is an ideal, strictly bigger than

. So there exists

prime ideals p

, ··· , p

such that p

···p

⊆ a + hxi, by definition.

Similarly, there exists q

, ···q

such that q

···q

⊆ a + hyi.

But then

···p

···q

⊆ (a + hxi)(a + hyi) ⊆ a + hxyi = 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 obtain

things like

, 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 helpful

to actually construct something explicitly that acts as an inverse. We can then

understand what significance these inverses have in terms of the rings.

Proposition.

(i) Let 0 6= a C O

be an ideal. If x ∈ L has xa ⊆ a, then x ∈ O

(ii) Let 0 6= a C O

be a proper ideal. Then

{y ∈ L : ya ≤ O

}

contains elements that are not in O

. In other words,

{y ∈ L : ya ≤ O

}

6= 0.

We will see that the object

{y ∈ L

ya ≤ O

}

is in some sense an inverse to

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

Example.

Consider

and

= 3

. Then the first part says if

Z ⊆

then

∈ Z. The second says

· 3 ∈ Z

contains something not in Z, say

. These are both “obviously true”.

Proof.

(i)

Let

a ⊆ O

. Then since

is Noetherian, we know

is finitely generated,

say by

, ··· , α

. We consider the multiplication-by-

map

a → a

i.e. write

xα

where A = (a

) is a matrix in O

. So we know

(xI − A)













= 0.

By multiplying by the adjugate matrix, this implies

det

(

xI −A

) = 0. So

satisfies a monic polynomial with coefficients in

, i.e.

is integral over

. Since O

is integrally closed, x ∈ O

(ii)

It is clear that if the result is true for

, then it is true for all

⊆ a

. So

it is enough to prove this for

, a maximal, and in particular prime,

ideal.

Let

α ∈ p

be non-zero. By the previous lemma, there exists prime ideals

, ··· , p

such that

···p

⊆ hαi

. We also have that

hαi ⊆ p

by definition.

Assume

is minimal with this property. Since

is prime, there is some

such that

⊆ p

. wlog, we may as well assume

= 1, i.e.

⊆ p

. But

a prime ideal, and hence maximal. So p

= p.

Also, since r is minimal, we know p

···p

6⊆ hai.

Pick β ∈ p

···p

\ hai. Then

βp = βp

⊆ p

···p

⊆ hαi.

Dividing by

, we get

p ⊆ O

. But

β 6∈ hαi

. So we know

6∈ O

. So

done.

What is this

{x ∈ L

xa ≤ O

}

? This is not an ideal, but it almost is. The

only way in which it fails to be an ideal is that it is not contained inside

. By

this we mean it is closed under addition and multiplication by elements in

So it is an

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

is a subset of

that is

also an O

module and is finitely generated.

Definition

(Integral/honest ideal)

If we want to emphasize that

a C O

is an

ideal, we say it is an integral or honest ideal. But we never use “ideal” to mean

fractional ideal.

Note that the definition of fractional ideal makes sense only because

Noetherian. Otherwise, the non-finitely-generated honest ideals would not qualify

as fractional ideals, which is bad. Rather, in the general case, the following

characterization is more helpful:

Lemma.

module

q ⊆ L

is a fractional ideal if and only if there is some

c ∈ L

such that

is an ideal in

. Moreover, we can pick

such that

c ∈ Z

In other words, each fractional ideal is of the form

for some honest ideal

a and integer c.

Proof.

(⇐)

We have to prove that

is finitely generated. If

q ⊆ L

c ∈ L

non-zero,

then

∼

as an

module. Since

is Noetherian, every ideal is

finitely-generated. So cq, and hence q is finitely generated.

(⇒)

Suppose

, ··· , x

generate

as an

-module. Write

, with

∈ O

and n

∈ Z, n

6= 0, which we have previously shown is possible.

We let

lcm

(

, ··· , n

). Then

cq ⊆ O

, and is an

-submodule of

, i.e. an ideal.

Corollary.

Let

be a fractional ideal. Then as an abelian group,

∼

, where

n = [L : Q].

Proof.

There is some

c ∈ L

such that

cqC O

as an ideal, and

∼

as abelian

groups. So it suffices to show that any non-zero ideal

q ≤ O

is isomorphic to

. Since

q ≤ O

∼

as abelian groups, we know

∼

for some

. But

also there is some

∈ Z ∩ q

, and

∼

i ≤ q

. So we must have

, and

∼

Corollary.

Let

a ≤ O

be a proper ideal. Then

{x ∈ L

xa ≤ O

}

is a

fractional ideal.

Proof.

Pick

a ∈ a

. Then

a · {x ∈ L

xa ≤ O

} ⊆ O

and is an ideal in

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

work.

Definition

(Invertible fractional ideal)

A fractional ideal

is invertible if there

exists a fractional ideal r such that qr = O

= h1i.

Notice we can multiply fractional ideals using the same definition as for

integral ideals.

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

{x ∈ L : xq ⊆ O

This is good.

Note that if q =

a and r =

b, and a, b C O

are integral ideals, then

qr =

ab = O

if and only if

hmni

. So the proposition is equivalent to the statement that

for every a C O

, there exists an ideal b C O

such that ab is principal.

Proof.

Note that for any

n ∈ O

non-zero, we know

is invertible if and only if

is invertible. So if the proposition is false, there is an integral ideal

a C O

which is not invertible. Moreover, as

is Noetherian, we can assume

maximal with this property, i.e. if a < a

< O

, then a

is invertible.

Let

{x ∈ L

xa ⊆ O

}

, a fractional ideal. We clearly have

⊆ b

, and

by our previous proposition, we know this inclusion is strict.

⊆ b

, we know

a ⊆ ab

. Again, this inclusion is strict — if

, then

for all

x ∈ b

, we have

xa ⊆ a

, and we have shown that this implies

x ∈ O

, but

we cannot have b ⊆ O

a ( ab

. By assumption, we also have

ab ⊆ O

, and since

is not invertible,

this is strict. But then by definition of

, we know

is invertible, which implies

is invertible (if

is an inverse of

, then

is an inverse of

). This is a

contradiction. 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 ⊆ O

Then by definition, we know q

−1

⊆ c. So

= qq

−1

⊆ qc ⊆ O

Hence we must have qc = O

, i.e. c = q

−1

We’re now done with the annoying commutative algebra, and can finally

prove something interesting.

Corollary. Let a, b, c C O

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

(iii)

(

⇒

) is clear. For the other direction, we notice that the result is easy if

hαi

is principal. Indeed, if

hβ

, ··· , β

, then

b ⊆ hαi

means there

are some β

, ··· , β

∈ O

such that β

= β

α. But this says

hβ

, ··· , β

i = hβ

, ··· , β

ihαi,

So hαi | b.

In general, suppose we have

b ⊆ a

. By the proposition, there exists an

ideal c C O

such that ac = hαi is principal with α ∈ O

, α 6= 0. Then

– b ⊆ a if and only if bc ⊆ hαi by (i); and

– a | b if and only if hαi | bc by (ii).

So the result follows.

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

Theorem.

Let

a C O

be an ideal,

a 6

= 0. Then

can be written uniquely as a

product of prime ideals.

Proof.

To show existence, if

is prime, then there is nothing to do. Otherwise,

is not prime, then it is not maximal. So there is some

b ) a

with

b C O

Hence

b | a

, i.e. there is some

c C O

with

, and

c ⊇ a

. We can continue

factoring this way, and it must stop eventually, or else we have an infinite chain

of strictly ascending ideals.

We prove uniqueness the usual way. We have shown

p | ab

implies

p | a

p | b

. So if

···p

···q

, with

, q

prime, then we know

| q

···q

which implies

| q

for some

, and wlog

= 1. So

⊆ p

. But

is prime

and hence maximal. So p

= q

Multiply the equation

···p

···q

−1

, and we get

···p

···q

. Repeat, and we get

and

for all

(after renumbering).

Corollary.

The non-zero fractional ideals form a group under multiplication.

We denote this

. This is a free abelian group generated by the prime ideals,

i.e. any fractional ideal

can be written uniquely as

···p

, with

distinct

prime ideals and a

∈ Z.

Moreover, if q is an integral ideal, i.e. q C O

, then a

, ··· , a

≥ 0.

Proof.

We already have unique factorization of honest ideals. Now take any

fractional ideal, and write it as

−1

, with

a, b ∈ O

(e.g. take

hni

for

some n), and the result follows.

Unimportant side note: we have shown that there are two ways we can

partially order the ideals of

— by inclusion and by division. We have shown

that these two orders are actually the same. Thus, it follows that the “least

common multiple” of two ideals

a, b

is their intersection

a ∩ b

, and the “greatest

common 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 have

completely no idea what prime ideals look like. We just used their very abstract

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

actual ideals.

Recall we had the example

h3, 1 + 2

√

−5ih3, 1 − 2

√

−5i = h3i.

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

−

√

−5

. We will

use this idea to prove the previous result for number fields of this form.

Let

a C O

be a non-zero ideal. We want to find some

b C O

such that

is principal.

We know

∼

, and

a ≤ O

as a subgroup. Moreover, we have shown

that we must have

∼

as abelian groups. So it is generated by 2 elements as

subgroups of

. Since

is a subring of

, we know

must be generated by

at most 2 elements as

-modules, i.e. as ideals of

. If it is generated by one

element, then it is already principal. Otherwise, suppose

hα, βi

for some

α, β ∈ O

Further, we claim that we can pick

α, β

such that

β ∈ Z

. We write

√

and

√

. Then let



gcd

(

b, b

) =

, with

m, m

∈ Z

(by

Euclid’s algorithm). We set

= (mα + m

β) ·

−b



+ β

= (ma + m

+ 

√

−b



+ a

+ b

√

= (ma + m

)

−b



+ a

∈ Z

using the fact that that −

∈ Z. Then hα, β

i = hα, βi.

So suppose a = hb, αi, with b ∈ Z and α ∈ O

. We now claim

hb, αihb, ¯αi

is principal (where

√

¯α

x − y

√

). In particular, if

a C O

, then

a is principal, so the proposition is proved by hand.

To show this, we can manually check

hb, αihb, ¯αi = hb

, bα, b¯α, α¯αi

= hb

, bα, b tr(α), N(α)i,

using the fact that

(

) =

¯α

and

(

) =

α¯α

. Now note that

, b tr

(

) and

(

) are all integers. So we can take the gcd

gcd

(

, b tr

(

)

, N

(

)). Then

this ideal is equal to

= hc, bαi.

Finally, we claim that bα ∈ hci.

Write

bα

, with

x ∈ L

. Then

tr x

tr α ∈ Z

since

∈ Z

by definition,

and

N(x) = N



bα



N(α)

∈ Z.

So x ∈ O

. So c | bα in O

. So hc, bαi = hci.

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

course.

Definition

(Class group)

. The class group or ideal class group of a number

field L is

= I

where

is the group of fractional ideals, and

is the subgroup of principal

fractional ideals.

a ∈ I

, we write [

] for its equivalence class in

. So [

] = [

] if and only

if there is some γ ∈ L

such that γa = b.

The significance is that cl

measures the failure of unique factorization:

Theorem. The following are equivalent:

(i) O

is a principal ideal domain

(ii) O

is a unique factorization domain

(iii) cl

is trivial.

Proof.

(i) and (iii) are equivalent by definition, while (i) implies (ii) is well-known

from IB Groups, Rings and Modules. So the real content is (ii) to (i), which is

specific to Dedekind domains.

p C O

is prime, and

x ∈ p \ {

}

, we factor

···α

such that

irreducible in

. As

is prime, there is some

∈ p

. But then

hα

i ⊆ p

, and

hα

is prime as

is a UFD. So we must have

hα

as prime ideals are

maximal. So p is principal.

In the next few chapters, we will come up with methods to explicitly compute

the class group of any number field.

4 Norms of ideals

In the previous chapter, we defined the class group, and we know it is generated

by prime ideals of

. 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

is a unit iff it has norm 1. So if

x ∈ O

is not

irreducible, then there must be some element whose norm strictly divides

(

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 a C O

be an ideal. We define

|N(a)| = |O

/a| ∈ N.

Recall that we’ve already proved that

/a|

is finite. So this definition

makes sense. It is also clear that N(a) = 1 iff a = O

(i.e. a is a “unit”).

Example. Let d ∈ Z. Then since O

∼

, we have dO

∼

. So we have

N(hdi) = |Z

/(dZ)

| = |Z/dZ|

= d

We start with a simple observation:

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

Proof.

It suffices to show that

(

)

∈ a

. Viewing

as an additive group,

the order of 1 is a factor of

(

). So

(

) =

(

)

1 = 0

∈ O

. Hence

N(a) ∈ a.

The most important property of the norm is the following:

Proposition. Let a, b C O

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

prime, i.e.

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

In other words, we need to show that



By the third isomorphism theorem, we already know that

∼











So it suffices to show that O

∼

a/ap as abelian groups.

In the case of the integers, we could have, say,

= 7

= 12

. We would

then simply define

12Z

7 · 12Z

x 12x

However, in general, we do not know that

is principal, but it turns out it

doesn’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

x + p αx + ap

is an isomorphism. We first check this is well-defined — if

p ∈ p

, then

αp ∈ ap

since

α ∈ a

. So the image of

and (

are equal. So this is well-defined.

To prove our claim, we have to show injectivity and surjectivity. To show

injectivity, since

hαi ⊆ a

, we have

a | hai

, i.e. there is an ideal

c ⊆ O

such that

ac = hαi. If x ∈ O

is in the kernel of the map, then αx ∈ ap. So

xac ⊆ ap.

xc ⊆ p.

is prime, either

c ⊆ p

x ∈ p

. But

c ⊆ p

implies

hαi

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

hαi/ap

a/ap

or equivalently ap + hαi = a.

Using our knowledge of fractional ideals, this is equivalent to saying (

hαi)a

−1

= O

. But we know

ap < ap + hαi ⊆ a.

We now multiply by a

−1

to obtain

p < (ap + hαi)a

−1

= p + c ⊆ O

Since

is a prime, and hence maximal ideal, the last inclusion must be an

equality. So ap + hαi = a, and we are done.

Now we provide the sketch of a proof that makes sense. The details are left

as an exercise in the second example sheet.

Proof.

It is enough to show that

(

···p

) =

(

)

···N

(

)

by unique

factorization.

By the Chinese remainder theorem, we have

···p

∼

× ··· ×

where p

, ··· , p

are distinct prime ideals.

Next, we show by hand that



× ··· ×



r−1



by showing that

k+1

is a 1-dimensional vector space over the field

Then the result follows.

This is actually the same proof, but written in a much saner form. This is

better because we are combining a general statement (the Chinese remainder

theorem), with a special property of the integral rings. In the first proof, what

we really did was simultaneously proving two parts using algebraic magic.

We’ve taken an obvious invariant of an ideal, the size, and found it is

multiplicative. How does this relate to the other invariants?

Recall that

∆(α

, ··· , α

) = det(tr

L/Q

(α

)) = det(σ

(α

))

Proposition. Let a C O

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

(i) There exists α

, ··· , α

∈ a such that

a =

: r

∈ Z

and

, ··· , α

are a basis of

over

. In particular,

is a free

-module

of n generators.

(ii) For any such α

, ··· , α

∆(α

, ··· , α

) = N(a)

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

Modules:

Lemma.

Let

be a

-module (i.e. abelian group), and suppose

M ≤ Z

Then M

∼

for some 0 ≤ r ≤ n.

Moreover, if

, then we can choose a basis

, ··· , v

such that the

change of basis matrix A = (a

) ∈ M

n×n

(Z) is upper triangular, where

where e

, ··· , e

is the standard basis of Z

In particular,

/M | = |a

···a

| = |det A|.

Proof of proposition.

Let

d ∈ a ∩ Z

, say

(

). Then

⊆ a ⊆ O

. As

abelian groups, after picking an integral basis α

, ··· , α

of O

, we have

∼

≤ a ≤ Z

∼

. Then the lemma gives us a basis

, ··· , α

as a

-module. As

-module, since the

are obtained from linear combinations of

, by basic

linear algebra, α

, ··· , α

is also a basis of L over Q.

Moreover, we know that we have

∆(α

, ··· , α

) = det(A)

∆(α

, ··· , α

Since

det

(

)

/a|

(

) and

= ∆(

, ··· , α

) by definition, the

second part follows.

This result is very useful for the following reason:

Corollary.

Suppose

a C O

has basis

, ··· , α

, and ∆(

, ··· , α

) is square-

free. Then a = O

(and D

is square-free).

This is a nice trick, since it allows us to determine immediately whether a

particular basis is an integral basis.

Proof. Immediate, since this forces N(a)

= 1.

Note that nothing above required

to be an actual ideal. It merely had

to be a subgroup of

that is isomorphic to

, since the quotient

well-defined as long as

is a subgroup. With this, we can have the following

useful result:

Example.

Let

be an algebraic integer and

(

). Let

= [

(

) :

Then a = Z[α] C O

. We have

disc(p

) = ∆(1, α, α

, ··· , α

n−1

) = discriminant of minimal polynomial of α.

Thus if disc(p

) is square-free, then Z[α] = O

Even if

disc

(

) is not square-free, it still says something: let

| disc

(

)

be such that disc(p

)/d

is square-free. Then |N(Z[α])| divides d.

Let

x ∈ O

. Then the order of

[

]

∈ O

[

] divides

(

[

]), hence

d. So d · x ∈ Z[α]. So x ∈

Z[α]. Hence we have

Z[α] ⊆ O

⊆

Z[α].

For example, if

√

for some square-free

, then

disc

(

√

) is the discriminant

of x

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

Z[α] ⊆ O

√

⊆

Z[α],

as we have previously seen.

We shall prove one more lemma, and start factoring things. Recall that we

had two different notions of norm. Given

α ∈ O

, we can take the norm

(

hαi

L/Q

(

). It would be great if they are related, like if they are equal. However,

that cannot possibly be true, since

(

hαi

) is always positive, but

L/Q

(

) can

be negative. So we take the absolute value.

Lemma. If α ∈ O

, then

N(hαi) = |N

L/Q

(α)|.

Proof.

Let

, ··· , α

be an integral basis of

. Then

αα

, .., αα

is an integral

basis of hαi. So by the previous lemma,

∆(αα

, ··· , αα

) = N(hαi)

But

∆(αα

, ··· , αα

) = det(σ

(αα

)

= det(σ

(α)σ

(α

))

i=1

(α)

∆(α

, ··· , α

)

= N

L/Q

(α)

L/Q

(α)

= N(hαi)

But N (hαi) is positive. So the result follows.

5 Structure of prime ideals

We can now move on to find all prime ideals of

. We know that every ideal

factors as a product of prime ideals, but we don’t know what the prime ideals

are. The only obvious way we’ve had to obtain prime ideals is to take a usual

prime, take its principal ideal and factor it in

, and get the resultant prime

ideals.

It turns out this gives us all prime ideals.

Lemma.

Let

p C O

be a prime ideal. Then there exists a unique

p ∈ Z

prime, with p | hpi. Moreover, N (p) = p

for some 1 ≤ f ≤ n.

This is not really too exciting, as soon as we realize that

p | hpi

is the same

as saying hpi ⊆ p, and we already know p ∩ Z is non-empty.

Proof.

Well

p ∩ Z

is an ideal in

, and hence principal. So

p ∩ Z

for some

p ∈ Z.

We now claim

is a prime integer. If

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

hpi ⊆ p

, we know

hpi

for some ideal

by factorization. Taking

norms, we get

= N(hpi) = 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

hpi

factor into prime ideals.

We write

hpi = p

···p

for some distinct prime ideals

, with

(

) =

for some positive integers

Taking norms, we get

n =

We start by giving some names to the possible scenarios.

Definition

(Ramification indices)

Let

hpi

···p

be the factorization

into prime ideals. Then e

, ··· , e

are the ramification indices.

Definition (Ramified prime). We say p is ramified if some e

> 1.

Definition

(Inert prime)

We say

is inert if

= 1 and

= 1, i.e.

hpi

remains prime.

Definition

(Splitting prime)

We say

splits completely if

···

= 1 =

= ··· = f

. So m = n.

Note that this does not exhaust all possibilities. The importance of these

terms, especially ramification, will become clear later.

So how do we actually compute

and

? In other words, how can we factor

the ideal

hpi C O

into prime ideals? The answer is given very concretely by

Dedekind’s criterion.

Theorem

(Dedekind’s criterion)

Let

α ∈ O

and

(

)

∈ Z

[

] be its minimal

polynomial. Suppose

[

]

⊆ O

has finite index, coprime to

(i.e.

p - |O

[

]

We write

¯g(x) = g(x) (mod p),

so ¯g(x) ∈ F

[x]. We factor

¯g(x) = ϕ

···ϕ

into distinct irreducibles in F

[x]. We define the ideal

= hp, ˜ϕ

(α)i C O

generated by

and

˜ϕ

, where

˜ϕ

is any polynomial in

[

] such that

˜ϕ

mod p

. Notice that if

˜ϕ

is another such polynomial, then

p |

(

˜ϕ

−ϕ

), so

hp, ˜ϕ

(

)

hp, ˜ϕ(α)i.

Then the p

are prime, and

hpi = p

···p

Moreover, f

= deg ϕ

, so N (a) = p

deg ϕ

If we are lucky, we might just find an

such that

[

] =

. If not, we

can find something close, and as long as

is not involved, we are fine. After

finding

, we get its minimal polynomial, factor it, and immediately get the

prime factorization of hpi.

Example.

Consider

(

√

−11

). We want to factor

. We consider

[

√

−11

]

⊆ O

. This has index 2, and (hopefully) 5

2. So this is good

enough. The minimal polynomial is

+ 11. Taking mod 5, this reduces to

− 4 = (x − 2)(x + 2). So Dedekind says

h5i = h5,

√

−11 + 2ih5,

√

−11 − 2i.

In general, consider

(

√

d 6

= 0

1 and square-free, and

an odd

prime. Then

[

√

]

⊆ O

has index 1 or 2, both of which are coprime to

. So

Dedekind says factor x

− d mod p. What are the possibilities?

(i)

There are two distinct roots mod

, i.e.

is a square mod

, i.e.





= 1.

Then

− d = (x + r)(x − r) (mod p)

for some r. So Dedekind says

hpi = p

where

= hp,

√

d − ri, p

= hp,

√

d + ri,

and N (p

) = N(p

) = p. So p splits.

(ii) x

− d

is irreducible, i.e.

is not a square mod

, i.e.





−

1. Then

Dedekind says hpi = p is prime in O

. So p is inert.

(iii) x

−d

has a repeated root mod

, i.e.

p | d

, or alternatively





= 0. Then

by Dedekind, we know

hpi = p

where

p = hp,

√

di.

So p ramifies.

So in fact, we see that the Legendre symbol encodes the ramification behaviour

of the primes.

What about the case where p = 2, for L = Q(

√

d)? How do we factor h2i?

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.

–

d ≡

1 (

mod

4), then then

[

], where

(1 +

√

). This has

minimal polynomial

− x +

(1 − d).

We reduce this mod 2.

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

◦

d ≡

5 (

mod

8), then we get

+ 1, which is irreducible. So

is prime, hence 2 is inert.

–

d ≡

3 (

mod

4), then

[

√

], and

−d

is the minimal polynomial.

Taking mod 2, we get

+ 1 = (

+ 1)

. In both cases, 2 ramifies.

Note how important

p - |O

[

]

is. If we used

[

√

] when

d ≡

1 (

mod

4),

we would have gotten the wrong answer.

Recall

(

4d d ≡ 2, 3 (mod 4)

d d ≡ 1 (mod 4)

The above computations show that

p | D

if and only if

ramifies in

. This

happens to be true in general. This starts to hint how important these invariants

like D

are.

Now we get to prove Dedekind’s theorem.

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

Claim. We have

∼

[x]

hϕ

Suppose this is true. Then since

is irreducible, we know

[x]

hϕ

is a field.

So p

is maximal, hence prime.

Next notice that

= hp, ˜ϕ

(α)i

⊆ hp, ˜ϕ

(α)

So we have

···p

⊆ hp, ˜ϕ

(α)

··· ˜ϕ

(α)

i = hp, g(α)i = hpi,

using the fact that g(α) = 0.

So to prove equality, we notice that if we put

deg ϕ

, then

(

) =

and

N(p

···p

) = N(p

)

···N(p

)

= p

deg g

Since

(

hpi

) =

, it suffices to show that

deg g

. Since

[

]

⊆ O

has finite

index, we know Z[α]

∼

. So 1, α, ··· , α

n−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

∼

Z[α]

∩ Z[α]

∼

[x]

hϕ

The second isomorphism is clear, since

Z[α]

hp, ˜ϕ

(α)i

∼

Z[x]

hp, ˜ϕ

(x), g(x)i

∼

[x]

h˜ϕ

(x), g(x)i

[x]

hϕ

(x), ¯g(x)i

[x]

hϕ

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

isomorphism:

Z[α]

pZ[α]

→

(∗)

x + pZ[α] 7→ x + pO

If this is true, then quotienting further by ˜ϕ

gives the desired isomorphism.

To prove the claim, we consider a slightly different map. We notice

p -

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

Z[α]

(†)

is injective. But

[

] is a finite abelian group. So the map is an isomorphism.

By injectivity of (

†

), we have

[

]

∩ pO

[

]. By surjectivity, we have

[

] +

. It thus follows that (

∗

) is injective and surjective respectively.

So it is an isomorphism. We have basically applied the snake lemma to the

diagram

0 Z[α] O

Z[α]

0 Z[α] O

Z[α]

Corollary.

is prime and

p < n

= [

], and

[

]

⊆ O

has finite index

coprime to p, then p does not split completely in O

Proof.

By Dedekind’s theorem, if

(

) is the minimal polynomial of

, then the

factorization of

¯g

(

) =

(

)

mod p

determines the factorization of

hpi

into prime

ideals. In particular,

splits completely if and only if

¯g

factors into distinct

linear factors, i.e.

¯g(x) = (x − α

) ···(x − α

where

∈ F

and

are distinct. But if

p < n

, then there aren’t

distinct

elements of F

Example.

Let

(

), where

has minimal polynomial

− x

−

x −

This is the case where

= 3

2 =

. On example sheet 2, you will see that 2

splits completely, i.e.

× F

. But then this corollary shows

that for all

β ∈ O

[

]

⊆ O

has even index, i.e. there does not exist an

β ∈ O

with |O

/Z[β]| odd.

As we previously alluded to, the following is true:

Theorem. p | D

if and only if p ramifies in O

We will not prove this.

6 Minkowski bound and finiteness of class group

Dedekind’s criterion allowed us to find all prime factors of

hpi

, but if we want

to 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 single

prime p and see what happens.

What we are now going to do is the following — we are going to use purely

geometric arguments to reason about ideals, and figure that each element of

the class group

has a representative whose norm is bounded by

some number

, which we will find rather explicitly. After finding the

, to

understand the class group, we just need to factor all prime numbers less than

and see what they look like.

We are first going to do the case of quadratic extensions explicitly, since

2-dimensional pictures are easier to draw. We will then do the full general case

afterwards.

Quadratic extensions

Consider again the case L = Q(

√

d), where d < 0. Then O

= Z[α], where

α =

(

√

d d ≡ 2, 3 (mod 4)

(1 +

√

d) d ≡ 1 (mod 4)

We can embed this as a subfield

L ⊆ C

. We can then plot the points on the

complex plane. For example, if

d ≡

3 (

mod

4), then the points look like this:

√

1 +

√

Then an ideal of

, say

√

, would then be the sub-lattice given by

the blue crosses.

We always get this picture, since any ideal of

is isomorphic to

as an

abelian group.

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

√

(1 +

√

The key result is the following purely geometric lemma:

Lemma

(Minskowski’s lemma)

Let Λ =

⊆ R

be a lattice, with

, v

linearly independent in

(i.e.

). We write

Then let

A(Λ) = area of fundamental parallelogram =



det







where the fundamental parallelogram is the following:

+ v

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|

≤

4A(Λ)

This is just an easy piece of geometry. What is remarkable is that the radius

of 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 ≤ O

, regarded as a subset of

via

some embedding

L → C

. The following proposition gives us the areas of the

relevant lattices:

Proposition.

(i) If α = a + b

√

λ, then as a complex number,

|α|

= (a + b

√

λ)(a − b

√

λ) = N(α).

(ii) For O

, we have

A(O

) =

(iii) In general, we have

A(a) =

|∆(α

, α

)|,

where α

, α

are the integral basis of a.

(iv) We have

A(a) = N(a)A(O

Proof.

(i) This is clear.

(ii) We know O

has basis 1, α, where again

α =

(

√

d d ≡ 2, 3 (mod 4)

(1 +

√

d) d ≡ 1 (mod 4)

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

A(O

) =

(

|d| d ≡ 2, 3 (mod 4)

|d| d ≡ 1 (mod 4)

(iii)

, α

are the integral basis of

, then the lattice of

is in fact spanned

by the vectors α

= a + bi, α

= a

+ b

i. This has area

A(a) = det



a b



whereas we have

∆(α

, α

) = det



¯α



= (α

¯α

− α

¯α

)

= Im(2α

¯α

)

= 4(a

b − ab

)

= 4A(a)

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

∆(α

, ··· , α

) = N(a)

in general.

Now what does Minkowski’s lemma tell us? We know there is an

α ∈ a

such

that

N(α) ≤

4A(a)

= N(a)c

where

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

|N(α)| = N (hαi) = N(a)N(b).

So this implies

N(b) ≤ c

Recall that the class group is

, the fractional ideals quotiented by

principal ideals, and we write [

] for the class of

. Then if

hαi

, then

we have

[b] = [a

−1

]

in cl

. So we have just shown,

Proposition

(Minkowski bound)

For all [

]

∈ cl

, there is a representative

of [a] (i.e. an ideal b ≤ O

such that [b] = [a]) such that

N(b) ≤ c

Proof. Find the b such that [b] = [(a

−1

)

−1

] and N (b) ≤ c

Combining this with the following easy lemma, we know that the class group

is finite!

Lemma.

For every

n ∈ Z

, there are only finitely many ideals

a ≤ O

with

N(a) = m.

Proof.

(

) =

, then by definition

/a|

. So

m ∈ a

by Lagrange’s

theorem. So

hmi ⊆ a

, i.e.

a | hmi

. Hence

is a factor of

hmi

. By unique

factorization of prime ideals, there are only finitely many such ideals.

Another proof is as follows:

Proof.

Each ideal bijects with an ideal in

/mO

= (

Z/m

)

. So there are

only finitely many.

Thus, we have proved

Theorem.

The class group

is a finite group, and the divisors of ideals of

the form hpi for p ∈ Z, p a prime, and 0 < p < c

, collectively generate cl

Proof.

(i)

Each element is represented by an ideal of norm less than 2

|/π

, and

there are only finitely many ideals of each norm.

(ii)

Given any element of

, we pick a representative

such that

(

)

< c

We factorize

a = p

···p

Then

N(p

) ≤ N (a) < c

Suppose

| hpi

. Then

(

) is a power of

, and is thus at least

. So

p < c

We now try to work with some explicit examples, utilizing Dedekind’s criterion

and the Minkowski bound.

Example. Consider d = −7. So Q(

√

−7) = L, and D

= −7. Then we have

1 < c

√

< 2.

So cl

= {1}, since there are no primes p < c

. So O

is a UFD.

Similarly, if d = −1, −2, −3, then O

is a UFD.

Example. Let d = −5. Then D

= −20. We have

2 < c

√

< 3.

So cl

is generated by primes dividing h2i.

Recall that Dirichlet’s theorem implies

h2i = h2, 1 +

√

−5i

= p

Also,

1 +

√

−5i

is not principal. If it were, then

hβi

, with

√

−5

, and

(

) = 2. But there are no solutions in

+ 5

= 2. So

= hpi = Z/2.

Example.

Consider

−

≡

3 (

mod

4). So

≈

3. So

is generated by

primes dividing by h2i, h3i, h5i. We factor

+ 17 ≡ x

+ 1 ≡ (x + 1)

(mod 2).

h2i = p

= h2, 1 +

√

Doing this mod 3, we have

+ 17 ≡ x

− 1 ≡ (x − 1)(x + 1) (mod 3).

So we have

h3i = q

q = h3, 1 +

√

dih3, 1 −

√

di.

Finally, mod 5, we have

+ 17 ≡ x

+ 2 (mod 5).

So 5 is inert, and [h5i] = 1 in cl

. So

= h[p], [q]i,

and we need to compute what this is. We can just compute powers

, q

, ···

pq, pq

, ···, and see what happens.

But a faster way is to look for principal ideals with small norms that are

multiples of 2 and 3. For example,

N(h1 +

√

di) = 18 = 2 · 3

But we have

1 +

√

d ∈ p, q.

p, q | h

1 +

√

. Thus we know

pq | h

1 +

√

. We have

(

) = 2

3 = 6. So

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

h1 +

√

di = pq

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

[p] = [q]

−2

. So we have

[

]

. Also, [

]

−2

= [

]

= 1 in

, as if it did, then

principal, i.e.

√

, but 2 =

(

) =

+ 7

has no solution in the

integers. Also, we know [p]

= [1]. So we know

= Z/4Z.

In fact, we have

Theorem. Let L = Q(

√

d) with d < 0. Then O

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 just

doing the same thing, but we need a bit harder geometry and proper definitions.

Definition

(Discrete subset)

A subset

X ⊆ R

is discrete if for every

x ∈ X

there is some

ε >

0 such that

(

)

∩ X

{x}

. This is true if and only if for

every compact K ⊆ R

, K ∩ X is finite.

We have the following very useful characterization of discrete subgroups of

Proposition.

Suppose Λ

⊆ R

is a subgroup. Then Λ is a discrete subgroup of

, +) if and only if

Λ =

(

: n

∈ Z

)

for some x

, ··· , x

linearly independent over R.

Note that linear independence is important. For example,

√

3 ⊆ R

is not discrete. On the other hand, if Λ =

a C O

is an ideal, where

(

√

)

and d < 0, then this is discrete.

Proof.

Suppose Λ is generated by

, ··· , x

. By linear independence, there

is some

g ∈ GL

(

) such that

for all 1

≤ i ≤ m

, where

, ··· , e

the standard basis. We know acting by

preserves discreteness, since it is a

homeomorphism, and

Λ =

⊆ R

× R

n−m

is clearly discrete (take

So this direction follows.

For the other direction, suppose Λ is discrete. We pick

, ··· , y

∈

which are linearly independent over

, with

maximal (so

m ≤ n

). Then by

maximality, we know

(

i=1

: λ

∈ R

)

(

: λ

∈ R, z

∈ Λ

)

and this is the smallest vector subspace of R

containing Λ. We now let

X =

(

i=1

: λ

∈ [0, 1]

)

∼

[0, 1]

This is closed and bounded, and hence compact. So X ∩ Λ is finite.

Also, we know

= Z

⊆ Λ,

and if γ is any element of Λ, we can write it as γ = γ

+ γ

, where γ

∈ X and

∈ Z

. So



≤ |X ∩ Λ| < ∞.

So let d = |Λ/Z

|. Then dΛ ⊆ Z

, i.e. Λ ⊆

. So

⊆ Λ ⊆

So Λ is a free abelian group of rank

. So there exists

, ··· , x

∈

which

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

Definition (Lattice). If rank Λ = n = dim R

, then Λ is a lattice in R

Definition

(Covolume and fundamental domain)

Let Λ

⊆ R

be a lattice, and

, ··· , x

be a basis of Λ, then let

P =

(

i=1

: λ

∈ [0, 1]

)

and define the covolume of Λ to be

covol(Λ) = vol(P ) = |det A|,

where A is the matrix such that x

We say P is a fundamental domain for the action of Λ on R

, i.e.

[

γ∈Λ

(γ + P ),

and

(γ + P ) ∩ (µ + P ) ⊆ ∂(γ + P ).

In particular, the intersection has zero volume.

This is called the covolume since if we consider the space

Λ, which is an

n-dimensional torus, then this has volume covol(Λ).

Observe now that if

, ··· , x

is a different basis of Λ, then the transition

matrix

has

B ∈ GL

(

). So we have

det B

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

⊆ R

be a lattice, and

be a funda-

mental domain. We let

S ⊆ R

be a measurable set, i.e. one for which

vol

(

) is

defined.

(i)

Suppose

vol

(

)

> covol

(Λ). Then there exists distinct

x, y ∈ S

such that

x − y ∈ Λ.

(ii)

Suppose

0 ∈ S

, and

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) > 2

covol(Λ); or

(b) vol(S) ≥ 2

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 =

⊆ R

and

= [

−

, we know the bounds are

sharp.

Proof.

(i)

Suppose

vol

(

)

> covol

(Λ) =

vol

(

). Since

P ⊆ R

is a fundamental

domain, we have

vol(S) = vol(S ∩ R

) = 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

for

γ ∈ 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

and

such that (

S − γ

) and

(

S − µ

) are not disjoint. In other words, there are

x, y ∈ S

such that

x − γ = y − µ, i.e. x − y = γ − µ ∈ Λ 6= 0.

(ii) We now let

S =



s : s ∈ S



So we have

vol(S

) = 2

−n

vol(S) > covol(Λ),

by assumption.

(a)

So there exists some distinct

y, z ∈ S

such that

y − z ∈

\ {

}

. We

now write

y − z =

(2y + (−2z)),

Since 2

z ∈ S

implies

−

z ∈ S

by symmetry around

, so we know

y − z ∈ S by convexity.

(b)

We apply the previous part to



1 +



for all

m ∈ N

m >

So we get a non-zero γ

∈ S

∩ Λ.

By convexity, we know

⊆ S

= 2

for all

. So

, γ

, ··· ∈ S

∩

Λ.

But

is compact set. So

∩

Λ is finite. So there exists

such that

is γ infinitely often. So

γ ∈

m≥0

= S.

So γ ∈ S.

We are now going to use this to mimic our previous proof that the class

group of an imaginary quadratic field is finite.

To begin with, we need to produce lattices from ideals of

. Let

be a

number field, and [

] =

. We let

, ··· , σ

L → R

be the real embeddings,

and

r+1

, ··· , σ

r+s

, ¯σ

r+1

, ··· , ¯σ

r+s

L → C

be the complex embeddings (note

that which embedding is σ

r+i

and which is ¯σ

r+i

is an arbitrary choice).

Then this defines an embedding

σ = (σ

, σ

, ··· , σ

, σ

r+1

, ··· , σ

r+s

) : L → R

× C

∼

× R

= R

r+2s

= R

under the isomorphism C → R

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(O

) is a lattice in R

of covolume 2

−s

(ii)

More generally, if

aC O

is an ideal, then

(

) is a lattice and the covolume

covol(σ(a)) = 2

−s

N(a).

Proof.

Obviously (ii) implies (i). So we just prove (ii). Recall that

has an

integral basis γ

, ··· , γ

. Then a is the integer span of the vectors

(σ

(γ

), σ

(γ

), ··· , σ

r+s

(γ

))

for

= 1

, ··· , n

, and they are independent as we will soon see when we compute

the determinant. So it is a lattice.

We also know that

∆(γ

, ··· , γ

) = det(σ

(γ

))

= N(a)

where the σ

run over all σ

, ··· , σ

, σ

r+1

, ··· , σ

r+s

, ¯σ

r+1

, ··· ¯σ

r+s

So we know

|det(σ

(γ

))| = N (a)|D

So what we have to do is to relate

det

(

)) to the covolume of

(

). But

these two expressions are very similar.

In the σ

(γ

) matrix, we have columns that look like



r+i

(γ

) ¯σ

r+i

(γ

)





z ¯z



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



Re(z) Im(z)





(z + ¯z)

(¯z − z)





1 1

i −i



¯z



We call the last matrix A =



1 1

i −i



. We can compute the determinant as

|det A| =



det



1 1

i −i





Hence the change of basis matrix from (

(

)) to

(

) is

diagonal copies of

so has determinant 2

−s

. So this proves the lemma.

Proposition.

Let

a C O

be an ideal. Then there exists an

α ∈ a

with

α 6

= 0

such that

|N(α)| ≤ c

N(a),

where





This is the Minkowski bound.

Proof. Let

r,s

(t) =

, ··· , y

, z

, ··· , z

) ∈ R

× C

| + 2

| ≤ t

This

(i) is closed and bounded;

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

(iii) has volume

vol(B

r,s

(t)) = 2





;

(iv) is convex and symmetric about 0.

Only (iii) requires proof, and it is on the second example sheet, i.e. we are not

doing it here. It is just doing the integral.

We now choose t so that

vol B

r,s

(t) = 2

covol(σ(a)).

Explicitly, we let





n!|D

1/2

N(a).

Then by Minkowski’s lemma, there is some

α ∈ a

non-zero such that

(

)

∈

r,s

(t). We write

σ(α) = (y

, ··· , y

, z

, ··· , z

Then we observe

N(α) = y

···y

¯z

···z

¯z

By the AM-GM inequality, we know

|N(α)|

1/n

≤



+ 2



≤

as we know σ(a) ∈ B

r,s

(t). So we get

|N(α)| ≤

= c

N(a).

Corollary. Every [a] ∈ cl

has a representative a ∈ O

with N (a) ≤ c

Theorem

(Dirichlet)

The class group

is finite, and is generated by prime

ideals of norm ≤ c

Proof. Just as the case for imaginary quadratic fields.

7 Dirichlet’s unit theorem

We have previously characterized the units on

as the elements with unit

norm, i.e.

α ∈ O

is a unit if and only if

(

)

= 1. However, this doesn’t tell

us much about how many units there are, and how they are distributed. The

answer to this question is given by Dirichlet’s unit theorem.

Theorem (Dirichlet unit theorem). We have the isomorphism

∼

× Z

r+s−1

where

= {α ∈ L : α

= 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, or

else it will be utterly incomprehensible.

We do the example of real quadratic fields,

(

√

), where

d >

1 is

square-free. So r = 2, s = 0, and L ⊆ R implies µ

= {±1}. So

∼

{±1} × Z.

Also, we know that

N(x + y

√

d) = (x + y

√

d)(x − y

√

d) = x

− dy

So Dirichlet’s theorem is saying that there are infinitely many solutions of

− dy

= ±1, and are all (plus or minus) the powers of one single element.

Theorem

(Pell’s equation)

There are infinitely many

√

d ∈ O

such that

− dy

= ±1.

You might have seen this in IIC Number Theory, where we proved it directly

by continued fractions. We will provide a totally unconstructive proof here, since

this is more easily generalized to arbitrary number fields.

This is actually just half of Dirichlet’s theorem. The remaining part is to

show that they are all powers of the same element.

Proof. Recall that σ : O

→ R

sends

α = x + y

√

d 7→ (σ

(α), σ

(α)) = (x + y

√

d, x − y

√

d).

(in the domain,

√

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(O

)) = |D

(1, 1)

Z[X]

N(α) = 1

Consider



, y

) ∈ R

: |y

| ≤ t, |y

| ≤

1/2



vol(s

) = 4|D

= 2

covol(O

= [

] = 2. Now Minkowski implies there is an

α ∈ O

non-zero such

that σ(α) ∈ s

. Also, if we write

σ(α) = (y

, y

then

N(α) = y

So such an α will satisfy

1 ≤ |N (α)| ≤ |D

1/2

This is not quite what we want, since we need

(

)

= 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

axis, since any such point must satisfy

x ± y

√

= 0, but

√

is not rational.

Also,

is compact. So

∩ σ

(

) contains finitely many points. So we can

find a

such that for each (

, y

)

∈ s

∩ O

, we have

| > t

. In particular,

does not contain any point in

∩ σ

(

). So we get a new set of points

α ∈ s

∩ O

such that 1 ≤ |N(α)| ≤ |D

1/2

We can do the same thing for

and get a new

. In general, given

> ··· > t

pick t

n+1

be such that

0 < t

n+1

< min

(

| : (y

, y

) ∈

[

i=1

∩ σ(O

)

and the minimum is finite since

is compact and hence contains finitely many

lattice points on σ(O

Then we get an infinite sequence of

such that

∩ σ

(

) are disjoint for

different

. Since each must contain at least one point, we have got infinitely

many points in O

satisfying 1 ≤ |N (α)| ≤ |D

1/2

Since there are only finitely many integers between 1 and

1/2

, we can

apply the pigeonhole principle, and get that there is some integer satisfying

≤ |m| ≤ |D

1/2

such that there exists infinitely many

α ∈ O

with

(

) =

This is not quite good enough. We consider

/mO

∼

(Z/mZ)

[L:Q]

another finite set. We notice that each

α ∈ O

must fall into one of finitely

many the cosets of

. In particular, each

such that

(

) =

must

belong to one of these cosets.

So again by the pigeonhole principle, there exists a

β ∈ O

with

(

) =

and infinitely many α ∈ O

with N (α) = m and α = β (mod mO

Now of course

and

are not necessarily units, if

m 6

= 1. However, we will

show that

α/β

is. The hard part is of course showing that it is in

itself,

because it is clear that

α/β

has norm 1 (alternatively, by symmetry,

β/α

is in

, so an inverse exists).

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

α = β + mγ,

where α, β, γ ∈ O

and N (α) = N(β) = m, then α/β ∈ O

To show this, we just have to compute

= 1 +

γ = 1 +

N(β)

γ = 1 +

βγ ∈ O

since N (β) = β

β. 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

(

√

Then there is some ε

∈ O

such that

= {±ε

: n ∈ Z}.

We call such an ε

a fundamental unit (which is not unique). So

∼

{±1} × Z.

Proof.

We have just proved the really powerful theorem that there are infinitely

many

with

(

) = 1. We are not going to need the full theorem. All we need

is that there are three — in particular, something that is not ±1.

We pick some

ε ∈ O

with

ε 6

1. This exists by what we just proved.

Then we know

|σ

(ε)| 6= 1,

|σ

(

)

= 1 if and only if

1. Replacing by

−1

if necessary, we wlog

E = |σ

(ε)| > 1. Now consider

{α ∈ O

: N(α) = ±1, 1 ≤ |σ

(α)| ≤ E}.

This is again finite, since it is specified by a compact subset of the

-lattice.

So we pick

in this set with

1 and

|σ

(

)

minimal (

1). Replacing

by −ε

if necessary, we can assume σ

(ε) > 1.

Finally, we claim that if

ε ∈ O

and

(

)

0, then

for some

N ∈ Z

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

Suppose

log ε

= N + γ,

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

εε

−N

= ε

∈ O

but

|ε

< |ε

, as

|ε

| >

1. By our choice of

, we must have

= 0. So

done.

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

Theorem (Dirichlet unit theorem). We have the isomorphism

∼

× Z

r+s−1

where

= {α ∈ L : α

= 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 the

very beginning. We define

 : O

→ R

r+s

x 7→ (log |σ

(x)|, ··· , log |σ

(x)|, 2 log |σ

r+1

(x)|, ··· , 2 log |σ

r+s

(x)|).

Note that

|σ

r+i

(

)

|σ

r+`

(x)|

. So this is independent of the choice of one of

r+i

, ¯σ

r+i

Claim.

We now claim that

im 

is a discrete group in

r+s

and

ker 

is a

finite cyclic group.

We note that

log |ab| = log |a| + log |b|.

So this is a group homomorphism, and the image is a subgroup. To prove the

first part, it suffices to show that

im  ∩

[

−A, A

]

r+s

is finite for all

A >

0. We

notice  factors as

× C

r+s

where σ maps α 7→ (σ

(α), ··· , σ

r+s

(α)), and

j : (y

, ··· , y

, z

, ··· , z

) 7→ (log |y

|, ··· , log |y

|, 2 log |z

|, ··· , 2 log |z

|).

We see

−1

([−A, A]

r+s

) = {(y

, z

) : e

−A

≤ |y

| ≤ e

, e

−A

≤ 2|z

| ≤ e

}

is a compact set, and

(

) is a lattice, in particular discrete. So

(

)

∩

−1

([

−A, A

]

r+s

) is finite. This also shows the kernel is finite, since the kernel is

the inverse image of a compact set.

Now as

ker 

is finite, all elements are of finite order. So

ker  ⊆ µ

. Con-

versely, it is clear that

⊆ ker 

. So it remains to show that

is cyclic. Since

embeds in

, we know

is contained in the roots of unity in

. Since

finite, we know

is generated by a root of unity with the smallest argument

(from, say, IA Groups).

Claim. We claim that

im  ⊆

, ··· , y

r+s

) :

= 0

∼

r+s−1

To show this, note that if α ∈ O

, then

N(α) =

i=1

(α)

`=1

r+`

(α)¯σ

r+`

= ±1.

Taking the log of the absolute values, we get

0 =

log |σ

(α)| + 2

log |σ

r+i

(α)|.

So we know

im  ⊆ R

r+s−1

as a discrete subgroup. So it is isomorphic to

for some

a ≤ r

s −

1. Then what we want to show is that

im  ⊆ R

r+s−1

is a

lattice, i.e. it is congruent to Z

r+s−1

Note that so far what we have done is the second part of what we did for the

real quadratic fields. We took the logarithm to show that these form a discrete

subgroup. Next, we want to find

s −

1 independent elements to show it is a

lattice.

Claim.

Fix a

such that 1

≤ k ≤ r

and

α ∈ O

with

α 6

= 0. Then there

exists a β ∈ O

such that

|N(β)| ≤





1/2

and moreover if we write

(α) = (a

, ··· , a

r+s

)

(β) = (b

, ··· , b

r+s

then we have b

< a

for all i 6= k.

We can apply Minkowski to the region

S = {(y

, ··· , y

, z

, ··· , z

) ∈ R

× C

: |y

| ≤ c

, |z

| ≤ c

r+j

}

(we will decide what values of c

to take later). Then this has volume

vol(S) = 2

···c

r+s

We notice

is convex and symmetric around 0. So if we choose 0

< c

< e

for

i 6= k, and choose





1/2

···ˆc

···c

r+s

Then Minkowski gives β ∈ σ(O

) ∩ S, satisfying the two conditions above.

Claim.

For any

= 1

, ··· , r

, there is a unit

∈ O

such that if



(

) =

, ··· , y

r+s

, then y

< 0 for all i 6= k (and hence y

> 0 since

= 0).

This is just as in the proof for the real quadratic case. We can repeatedly

apply the previous claim to get a sequence

, α

, ··· ∈ O

such that

(

)

is bounded for all

, and for all

i 6

, the

th coordinate of



(

)

, 

(

)

, ···

is strictly decreasing. But then as with real quadratic fields, the pigeonhole

principle implies we can find t, t

such that

N(α

) = N(α

) = m,

say, and

≡ α

(mod mO

i.e.

/mO

. Hence for each

, we get a unit

/α

such that

(u

) = (α

) − (α

) = (y

, ··· , y

r+s

)

has

0 if

i 6

(and hence

0, since

= 0). We need a final trick to

show the following:

Claim.

The units

, ··· , u

r+s−1

are linearly independent in

r+s−1

. Hence

the rank of (O

) = r + s − 1, and Dirichlet’s theorem is proved.

We let

be the (

)

(

) matrix whose

th row is



(

), and apply

the following lemma:

Claim.

Let

A ∈ Mat

(

) be such that

0 for all

and

0 for all

i 6

and

≥ 0 for each i. Then rank(A) ≥ m − 1.

To show this, we let

be the

th column of

. We show that

, ··· , v

m−1

are linearly independent. If not, there exists a sequence t

∈ R such that

m−1

i=1

= 0, (∗)

with not all of the

non-zero. We choose

so that

is maximal among the

, ··· , t

m−1

’s. We divide the whole equation by

. So we can wlog assume

= 1, t

≤ 1 for all i.

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

0 =

m−1

i=1

≥

m−1

i=1

a <

0 and

t ≤

1 implies

at ≥ a

. Moreover, we know

0 strictly. So we

get

0 >

i=1

≥ 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

= covol((O

) ⊆ R

r+s−1

More concretely, we pick fundamental units ε

, ··· , ε

r+s−1

∈ O

so that

= µ

× {ε

···ε

r+s−1

: n

∈ Z}.

We take any (

s −

1)(

s −

1) subminor of the matrix



(ε

) ··· (ε

r+s

)



Their determinants all have the same absolute value, and

|det(subminor)| = R

This is a definition we will need later.

We quickly look at some examples with quadratic fields. Consider

(

√

where d 6= 0, 1 square-free.

Example.

d <

0, then

= 0 and

= 1. So

s −

1 = 0. So

is a

finite group. So R

= 1.

Lemma.

(i) If d = −1, then Z[i]

= {±1, ±i} = Z/4Z.

(ii)

−

3, then let

(1 +

√

), and we have

= 1. So

[

]

{1, ω, ··· , ω

}

∼

Z/6Z.

(iii) For any other d < 0, we have O

= {±1}.

Proof. This is just a direct check.

d ≡

3 (

mod

4), then by looking at the solution of

− dy

1 in the

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

d ≡

1 (

mod

4), then by looking at the solutions to



x +



−

in the integers, we get (ii) and (iii).

Now if

d >

0, then

|log |ε||

, where

is a fundamental unit. So how do

we find a fundamental unit? In general, there is no good algorithm for finding the

fundamental unit of a fundamental field. The best algorithm takes exponential

time. 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, and

then show there is no smaller one by direct check.

Example.

Consider the field

(

√

). We can try

= 1 +

√

. We have

(

) = 1

−

2 =

−

1. So this is a unit. We claim this is fundamental. If not,

there there exists

√

, where

a, b ∈ Z

and 1

< u < ε

(as real numbers).

Then we have

¯u = a − b

√

has u¯u = ±1. Since u > 1, we know |¯u| < 1. Then we must have u ± ¯u > 0. So

we need a, b > 0. We know can only be finitely many possibilities for

1 < a + b

√

2 < 1 +

√

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

Example.

Consider

(

√

). We guess

= 10

−

√

is a unit. We can

compute 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

(

) =

1. If

(

) =

−

1, then

−

1. But

−

1 is

not a square mod 11.

So there we must have

(

) = 1. Then

−1

¯u

. We get 0

< ε < u

−1

¯u <

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 11b

+ 1 is not a square for each of these. So done.

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 prime



1 +

+ ···



n>0

This is since every

···p

factors uniquely as a product of primes, and

each such product appears exactly once in this. If there were finitely many

primes, as

converges to



1 −



−1

, the sum

n≥1

p prime



1 −



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 exists

infinitely 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 showing

that

p≡a mod q

p prime



1 −



−1

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

work.

To begin with, we define the Riemann zeta function.

Definition

(Riemann zeta function)

The Riemann zeta function is defined as

ζ(s) =

n≥1

−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.

(ii) The function

ζ(s) −

s − 1

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

In other words,

(

) extends to a meromorphic function on

(

)

0 with

a simple pole at 1 with residue 1.

(iii) We have the expression

ζ(s) =

p prime



1 −



−1

for

(

)

1, and the product is absolutely convergent. This is the Euler

product.

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

Definition

(Dirichlet series)

A Dirichlet series is a series of the form

−s

where a

, a

, ··· ∈ C.

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

+ ··· + a

= O(N

then

−s

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

Then (i) is immediate by picking

= 1, since in the Riemann zeta function,

= a

= ··· = 1.

Recall that x

= e

s log x

has

| = |x

Re(s)

if x ∈ R, x > 0.

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

n=1

−s

= a

−s

− 2

−s

) + (a

+ a

)(2

−s

− 3

−s

) + ···

+ (a

+ ··· + a

N−1

)((N − 1)

−s

− N

−s

) + R

where

+ ··· + a

This is getting annoying, so let’s write

T (N) = a

+ ··· + a

We know



T (N)



T (N)



Re(s)−r

→ 0

as N → ∞, by assumption. Thus we have

n≥1

−s

n≥1

T (n)(n

−s

− (n + 1)

−s

)

(

)

> r

. But again by assumption,

(

)

≤ B · n

for some constant

and

all n. So it is enough to show that

−s

− (n + 1)

−s

)

converges. But

−s

− (n + 1)

−s

n+1

s+1

dx,

and if x ∈ [n, n + 1], then n

≤ x

. So we have

−s

− (n + 1)

−s

) ≤

n+1

s+1

dx = s

n+1

s+1−r

It thus suffices to show that

s+1−r

converges, which it does (to

s−r

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

s − 1

∞

n=1

n+1

and show that

is uniformly convergent when Re(s) > 0, where

= n

−s

−

n+1

For (iii), consider the first r primes p

, ··· , p

, and

i=1

(1 − p

−s

)

−1

−s

where the sum is over the positive integers

whose prime divisors are among

, ··· , p

. Notice that 1, ··· , r are certainly in the set.



ζ(s) −

i=1

(1 − p

−s

)

−1



≤

n≥r

−s

| =

n≥r

−Re(s)

But

n≥r

−Re(s)

→

0 as

r → ∞

, proving the result, if we also show that it

converges absolutely. We omit this proof, but it follows from the fact that

p prime

−s

≤

−s

and the latter converges absolutely, plus the fact that

− a

) converges if

and only if

converges, by IA Analysis I.

This is good, but not what we want. Let’s mimic this definition for an

arbitrary number field!

Definition

(Zeta function)

Let

L ⊇ Q

be a number field, and [

] =

. We

define the zeta function of L by

(s) =

aCO

N(a)

−s

It is clear that if

and

, then this is just the Riemann zeta

function.

Theorem.

(i) ζ

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

(ii)

Analytic class number formula:

(

) is a meromorphic function if

(

)

1 −

and has a simple pole at s = 1 with residue

|cl

(2π)

1/2

|µ

where

is the class group,

and

are the number of real and complex

embeddings, you know what

is,

is the regulator,

is the discriminant

and µ

is the roots of unity in L.

(iii)

(s) =

pCO

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

aCO

N(a)

−s

pCO

,p prime

(1 − N(p)

−s

)

−1

holds “formally”, as in the terms match up when you expand, as an immediate

consequence of the unique factorization of ideals into a product of prime ideals.

The issue is to study convergence of

(

)

−s

, and this comes down to estimating

the number of ideals of fixed norm geometrically, and that is where all the factors

in the pole come it.

Example.

We try to compute

(

), where

(

√

). This has discriminant

D, which may be d or 4d. We first look at the prime ideals.

is a prime ideal in

, then

p | hpi

for a unique

. So let’s enumerate

the factors of η

controlled by p ∈ Z.

Now if

p | |D

, then

hpi

ramifies, and

(

) =

. So this contributes a

factor of (1 − p

−s

)

−1

Now if p remains prime, then we have N (hpi) = p

. So we get a factor of

(1 − p

−2s

)

−1

= (1 − p

−s

)

−1

(1 + p

−s

)

−1

If p splits completely, then

hpi = p

N(p

) = p,

and so we get a factor of

(1 − p

−s

)

−1

(1 − p

−s

)

−1

So we find that

(s) = ζ(s)L(χ

, 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

(p) =











0 p | D

−1 p remains prime

1 p splits

(





p is odd

depends on d mod 8 p = 2

Example. If L = Q(

√

−1), then we know



−4





−1



= (−1)

p−1

if p 6= 2,

and χ

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

L(χ

, s) =

p>2 prime

(1 − (−1)

p−1

−s

)

−1

= 1 −

−

+ ··· .

Note that

was defined for primes only, but we can extend it to a function

: Z → C by imposing

(nm) = χ

(n)χ

(m),

i.e. we define

···p

) = χ

)

···χ

)

Example. Let L = Q(

√

−1). Then

−4

(m) =

(

(−1)

m−1

m odd

0 m even.

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

−4

(mn) = χ

−4

(m)χ

−4

(n).

Notice that this has the property that

−4

(m − 4) = χ

−4

(m).

We give these some special names

Definition

(Dirichlet character)

A function

Z → C

is a Dirichlet character

of modulus D if there exists a group homomorphism

w :





→ 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 that

χ(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)

for such χ.

Proposition. χ

, as defined for

(

√

) is a Dirichlet character of modulus

Note that this is a very special Dirichlet character, as it only takes values

0, ±1. We call this a quadratic Dirichlet character.

Proof. We must show that

(p + Da) = χ

(p)

for all p, a.

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

(2) = 0,

as (2) ramifies. So χ

(even) = 0. For p > 2, we have

(p) =













(−1)

p−1

d−1

≡ 1 (mod 2), by quadratic reciprocity. So

(p + Da) =



p + Da



(−1)

p−1

(−1)

4da/2

= χ

(p).

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

Lemma.

Let

be any non-trivial Dirichlet character. Then

(

χ, s

) is holomor-

phic for Re(s) > 0.

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

i=1

χ(i) = O(1),

i.e. it is bounded. Recall from Representation Theory that distinct irreducible

characters of a finite group G are orthogonal, i.e.

|G|

g∈G

(g)χ

(g) =

(

1 χ

= χ

0 otherwise

We apply this to

= (

Z/DZ

)

, where

is trivial and

. So orthogonality

gives

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 χ

, we have

L(χ

, 1) 6= 0.

For example, if D < 0, then

L(χ

, 1) =

2π|cl

√

|D|

1/2

|µ

√

Proof. We have shown that

√

(s) = ζ

(s)L(χ

, s).

Note that

√

(

) and

(

) have simple poles at

= 1, while

(

, s

) is

holomorphic at s = 1.

Since the residue of

(

) at

= 1 is 1, while the residue of

√

= 1

is non-zero by the analytic class number formula. So

(

1) is non-zero, and

given by the analytic class number formula.

Example. If L = Q(

√

−1), then

1 −

−

+ ··· =

2π · 1

2 · 4

In general, for any field whose class number we know, we can get a series

expansion for π. And it converges incredibly slow.

Note that this corollary required two things — the analytic input for the

analytic class number formula, and quadratic reciprocity (to show that

is a

Dirichlet character).

More ambitiously, we now compute the zeta function of a cyclotomic field,

(

), where

is the primitive

th root of unity and

q ∈ N

. We need to

know the following facts about cyclotomic extensions:

Proposition.

(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

acts on

(

) by sending

7→ ω

. This is

what plays the role of quadratic reciprocity for cyclotomic fields.

(iii) The ring of integers is

= Z[ω

] = Z[x]/Φ

(x),

where

(x) =

− 1

d|q,d6=q

(x)

is the qth cyclotomic polynomial.

(iv)

Let

be a prime. Then

ramifies in

if and only if

p | D

, if and only

p | q

. So while

might be messy, the prime factors of

are the prime

factors of q.

(v)

Let

be a prime and

p - q

. Then

hpi

factors as a product of

(

)

distinct

prime ideals, each of norm p

, 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

− 1

(x + 1)(x − 1)(x

+ 1)

− 1

= x

+ 1.

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

/p =

[x]

i.e. how Φ

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 x

factors into 4 linear factors, which it does.

If p = 3, then (v) says x

factors into 2 quadratic factors. Indeed, we have

− x − 1)(x

+ x − 1) = (x

− 1)

− x

= x

+ 1.

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

Q(ω

)

(s) =

(1 − N(p)

−s

)

−1

We consider the prime ideals

dividing

hpi

, where

is a fixed integer prime

number. If p - q, then (v) says this contributes a factor of

(1 − p

−fs

)

−ϕ(q)/f

to the zeta function, where

is the order of

in (

Z/qZ

)

. We observe that this

thing factors, since

1 − t

γ∈µ

(1 − γt),

with

= {γ ∈ C : γ

= 1},

and we can put t = p

−s

We let

, ··· , ω

ϕ(q)

: (Z/qZ)

→ C

be the distinct irreducible (one-dimensional) representations of (

Z/qZ

)

, with

being the trivial representation, i.e. ω

(a) = 1 for all a ∈ (Z/qZ)

The claim is that

(

)

, ··· , ω

ϕ(q)

(

) are

th roots of 1, each repeated

(

)

times. We either say this is obvious, or we can use some representation theory.

We know

generates a cyclic subgroup

hpi

of (

Z/qZ

)

of order

, by definition

. So this is equivalent to saying the restrictions of

, ··· , ω

ϕ(q)

are the

f distinct irreducible characters of hpi

∼

Z/f, each repeated ϕ(q)/f times.

Equivalently, note that

Res

(Z/qZ)

hpi

(ω

⊕ ··· ⊕ ω

ϕ(q)

) = Res

(Z/qZ)

hpi

(regular representation of (Z/qZ)

So this claims that

Res

(Z/qZ)

hpi

(regular rep. of (Z/qZ)

) =

ϕ(q)

(regular rep. of Z/f).

But this is true for any group, since

Res

CG = |G/H|CH,

as the character of both sides is |G|δ

So we have

(1 − p

−fs

)

−ϕ(q)/f

ϕ(q)

i=1

(1 − ω

(p)p

−s

)

−1

So we let

(n) =

(

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

0 otherwise

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

Proposition. We have

Q(ω

)

(s) =

ϕ(q)

i=1

L(χ

, s) · (corr. factor) = ζ

(s)

ϕ(q)

i=2

L(χ

, s) · (corr. factor)

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

By defining the

functions in a slightly more clever way, we can hide the

correction factors into the

(

χ, s

), and then the

function is just the product of

these L-functions.

Proof.

Our analysis covered all primes

p - q

, and the correction factor is just to

include the terms with p | q. The second part is just saying that

(s) = L(χ

, s)

p|q

(1 − p

−s

)

−1

This allows us to improve our result on the non-vanishing of

(

χ,

1) to all

Dirichlet 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

function of some cyclotomic

extension.

Consider the formula

Q(ω

)

(s) = ζ

(s)

ϕ(q)

i=2

L(χ

, s) · (corr. factor)

= 1. We know that the

(

, s

) are all holomorphic at

= 1. Moreover,

both

Q(ω

)

and

have a simple pole at 0. Since the correction terms are finite,

it must be the case that all L(χ

, s) are non-zero.

Theorem

(Dirichlet, 1839)

Let

a, q ∈ N

be coprime, i.e.

gcd

(

a, q

) = 1. Then

there are infinitely many primes in the arithmetic progression

a, a + q, a + 2q, a + 3q, ··· .

Proof. As before, let

, ··· , ω

ϕ(q)

: (Z/qZ)

→ C

be the irreducible characters, and let

, ··· , χ

ϕ(q)

: Z → C

be the corresponding Dirichlet character, with ω

the trivial one.

Recall the orthogonality of columns of the character table, which says that if

gcd(p, q) = 1, then

ϕ(q)

(a)ω

(p) =

(

1 a ≡ p (mod q)

0 otherwise

Hence we know

ϕ(q)

(a)χ

(p) =

(

1 a ≡ p (mod q)

0 otherwise

even if gcd(p, q) 6= 1, as then χ

(p) = 0. So

p≡a mod q

p prime

−s

ϕ(q)

(a)

all primes p

(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

(

)

−s

is “essentially”

log L

(

, s

), up to some

bounded terms. We Taylor expand

log L(χ, s) = −

log(1 − χ(p)p

−s

) =

n≥1

p prime

χ(p)

n≥1

p prime

χ(p

)

What we care about is the n = 1 term. So we claim that

n≥2,p prime

χ(p

)

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



n≥2

χ(p

)



≤

n≥2

−ns

p prime

− 1)

≤

n≥2

− 1)

< ∞.

Hence we know

log L(χ, s) =

(p)p

−s

+ bounded stuff

near s = 1.

So at s = 1, we have

(‡) ∼

ϕ(q)

(a) log L(χ

, 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.

non-trivial,

(

, s

) is holomorphic and

non-zero at

= 1. So we just have to show that

log L

(

, s

) has a pole. Note

that L(χ

, s) is essentially ζ

(s). Precisely, we have

L(χ

, s) = ζ

(s)

p|q

(1 − p

−s

Moreover, we already know that ζ

(s) blows up at s = 1. We have

(s) =

s − 1

+ holomorphic function

s − 1

(1 + (s − 1)(holomorphic function)).

So we know

log L(χ

, s) ∼ log ζ

(s) ∼ log



s − 1



and this does blow up at s = 1.

So far, we have been working with abelian extensions over

, i.e. extensions

L/Q

whose Galois group is abelian. By the Kronecker–Weber theorem, every

abelian extension of

is contained within some cyclotomic extension. So in

some 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

Gal

(

L/Q

). We can still make sense of the

functions, and it turns out it always factors as

(s) =

L(ρ, s)

dim ρ

where

ranges over all the irreducible representations of

, and

(

ρ, s

) is the

Artin

-function. It takes some effort to define the Artin

-function, and we

shall not do so here. However, it is worth noting that

, s

) is just

(

), and

for ρ 6= 1, we still have a factorization of the form

L(ρ, s) =

p prime

(ρ, s).

This L

(ρ, s) is known as the Euler factor .

One can show that

(

ρ, s

) is always a meromorphic function of

, and is

conjectured to be holomorphic for all s (if ρ 6= 1, of course).

is one-dimensional, then

(

ρ, s

) is a Dirichlet series

(

χ, s

) for some

Recall that to establish this fact for quadratic fields, we had to use quadratic

reciprocity. In general given a

, finding

is a higher version of “quadratic

reciprocity”. This area is known as class field theory. If dim ρ > 1, then this is

“non-abelian class field theory”, known as Langlands programme.