ideal class group is finite

ideal class group is finite

We give two proofs of the finiteness of the

class group

, one using the bound provided by Minkowski’s theorem (

http://planetmath.org/MinkowskisConstant

) and a second, more elementary, proof that does not provide the same computational benefits as Minkowski’s bound does. Both proofs rely on the following lemma:

Lemma 1

.

If

K

is an

algebraic extension

of

Q

and

0

<

d

∈

Z

, then there are only a finite number of ideals of norm

d

.

Proof.

The norm of a

prime ideal

𝔓

of

K

lying over a

rational prime

p

is

p

f

, where

f

is the

residue field

degree

[

𝒪

K

/

𝔓

⁢

𝒪

K

:

ℤ

/

p

⁢

ℤ

]

, and there are at most

[

K

:

ℚ

]

prime ideals lying over any given rational prime. There are thus only a finite number of possibilities for ideals with norm

d

- simply factor

d

into a product of prime powers and note that each prime power must correspond to one of a finite number of possibilities.

The finiteness of the class group now follows trivially from Minkowski’s theorem:

Theorem 1

.

If

K

is an algebraic extension of

Q

, then the class group of

K

, denoted

Cl

⁡

(

K

)

, is finite.

Proof.

Minkowski’s theorem guarantees that each

ideal class

contains a representative

integral ideal

whose norm is

bounded

by a constant depending only on the field, and the lemma shows that there are only a finite number of integral ideals with norm less than that constant.

Minkowski’s theorem gives enough information about the size of the class group to be computationally useful in some cases (see the topic on using Minkowski’s constant to find a

class number

). It does, however, require quite a bit of machinery. To see in a more elementary way that

Cl

⁡

(

K

)

is finite, one can proceed as follows:

Proof.

(alternate proof of theorem)

By the lemma, it suffices to show that there is

some

constant

C

, depending only on

K

, such that every class in

Cl

⁡

(

K

)

has a representative

I

⊂

𝒪

K

with

N

⁡

(

I

)

≤

C

.
For

u

∈

K

, denote by

T

u

the linear map left

multiplication

by

u

, and let

e

1

,

…

,

e

n

be a basis of

𝒪

K

as a

ℤ

-module (where

[

K

:

ℚ

]

=

n

). Then if

u

=

∑

b

i

⁢

(

u

)

⁢

e

i

, it follows that

N

⁡

(

u

)

=

det

⁡

(

T

u

)

=

det

⁡

(

∑

b

i

⁢

(

u

)

⁢

T

e

i

)

is a

polynomial

of total degree at most

n

in the

b

i

⁢

(

u

)

, whose coefficients are functions of the

T

e

i

and thus depend only on

K

(and not on

u

). Let

C

be the sum of the magnitude of those coefficients.

Let

c

be a class in

Cl

⁡

(

K

)

and let

J

⊂

𝒪

K

be a representative of the class

c

-

1

. Consider

S

=

{

∑

i

=

1

n

r

i

e

i

∣

1

≤

r

i

≤

⌊

N

(

J

)

1

/

n

+

1

⌋

}

. The cardinality of

S

is strictly greater than

N

⁡

(

J

)

, while

|

𝒪

K

/

J

|

=

N

⁡

(

J

)

. So by the pigenhole principle, two distinct elements of

S

are in the same

J

-coset of

𝒪

K

. Taking their

difference

, we get an element

0

≠

x

=

∑

b

i

⁢

(

x

)

⁢

e

i

∈

J

, where not all the

b

i

⁢

(

x

)

are zero, and

|

b

i

(

x

)

|

≤

N

(

J

)

1

/

n

.

Now, since

x

∈

J

, using

unique factorization

of ideals in the

Dedekind ring

𝒪

K

, we can construct an integral ideal

I

such that

I

⁢

J

=

(

x

)

, so that

I

is in the class

c

∈

Cl

⁡

(

K

)

. Finally,

N

⁡

(

I

)

⁢

N

⁡

(

J

)

=

N

⁡

(

x

)

≤

C

⁢

(

max

⁡

(

b

1

⁢

(

x

)

,

…

,

b

n

⁢

(

x

)

)

)

n

≤

C

⁢

N

⁡

(

J

)

so that

N

⁡

(

I

)

≤

C

.

Title

ideal class group is finite

Canonical name

IdealClassGroupIsFinite

Date of creation

2013-03-22 17:57:23

Last modified on

2013-03-22 17:57:23

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

7

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11R29

Related topic

IdealNorm

Generated on Fri Feb 9 15:44:09 2018 by

LaTeXML