II Number Fields - Minkowski bound and finiteness of class group

6Minkowski bound and finiteness of class group

II Number Fields

6 Minkowski bound and finiteness of class group

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

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.