proof of basis of ideal in algebraic number field

proof of basis of ideal in algebraic number field

Although it is stated in a

number field

context

this

theorem

is about

ℤ

-modules.

𝒪

K

is an

integer

ring, that is, it is the

integral closure

of

ℤ

in

K

.

𝒪

K

is naturally endowed with a

structure

of

ℤ

-module and so are all its ideals

𝔞

. Therefore the situation is that we have a

ℤ

-module (namely

𝔞

) that is embedded in a finite dimensional

vector space

over

ℚ

, namely

K

. It is a well-known fact that

discrete

ℤ

-modules of finite dimensional vector spaces over

ℚ

are

free modules

with

finite rank

(ie. they have a finite basis). This is exactly the claim of the theorem.

Therefore to prove the theorem we only need prove that

𝔞

is discrete (

K

is by definition a finite dimensional vector space over

ℚ

as it is an

algebraic number field

).
Let

E

:=

ℚ

⁢

𝔞

, it is a finite dimensional

ℚ

-

vector subspace

of

K

. Let

n

be the

dimension

of

K

over

ℚ

, and

k

the dimension of

E

over

ℚ

.
To say that

𝔞

⊂

K

≅

ℚ

n

is a discrete

ℤ

-module of

K

is

equivalent

to say that every

sequence

of elements of

𝔞

that

converges

in

𝔞

for the usual Euclidian norm (

http://planetmath.org/NormedVectorSpace

) (ie. for

x

=

(

x

1

,

…

,

x

n

)

⁢

ℝ

n

,

||

x

||

2

=

∑

i

=

1

n

x

i

2

) is ultimately

constant

. It suffices to prove this for sequences that converges to

0

(instead of an arbitrary element of

𝔞

) because

𝔞

is stable subset under

addition

(one can transform a sequence converging to any element of

𝔞

into a sequence converging to

0

by subtracting that

element

to the sequence).

Suppose there is a sequence of

𝔞

that converges to

0

, we want to prove that this sequence is ultimately constant with value

0

. The elements of

𝔞

can be seen as

ℚ

-linear

endomorphisms

of

K

, and their

characteristic polynomials

have

coefficients

in

ℤ

as they are in the integral closure of

ℤ

.
In particular the

determinant

of these endomorphisms (which is called norm in this context) is an integer (the determinant is the constant coefficient of the characteristic polynomial), and if it were possible to find a sequence of elements of

𝔞

that converges to

0

the determinant would also converge to

0

as it is a

continuous function

. But it has just been said that the the determinant is an integer for any element of

𝔞

, therefore for the determinants of the elements of the sequence to converge to

0

they have to be ultimately

0

, that is from a certain point on, the sequence is constantly equal to

0

.
Since

K

is a field all the

mappings

but the

0

mapping, are

injective

and therefore have non-zero norm. Therefore if the norm of the elements is

0

, it means that the elements themselves were

0

. Hence we have proved that the sequence that converged to

0

was ultimately evenly

0

. We have thus proved that

𝔞

is a discrete

ℤ

-module of

K

.

As a reminder, here is a proof of the afore-mentioned ”well-known fact” that a discrete

submodule

of a finite dimensional vector space over

ℚ

has a basis:

First, we prove that can find a

finite set

of generators

𝔅

of

E

whose elements are in

𝔞

.
This is a straightforward

induction

on the dimension of

Vect

ℚ

⁢

(

𝔅

)

: start with

𝔅

=

∅

, if there is element in

E

that is not in

Vect

ℚ

⁢

(

𝔅

)

, then there is an element of

𝔞

that lies outside

Vect

ℚ

⁢

(

𝔅

)

, add that element to

𝔅

and keep on doing this until

dim

⁡

Vect

ℚ

⁢

(

𝔅

)

=

dim

⁡

E

.

At that point

E

/

Vect

ℚ

⁢

(

𝔅

)

is a finite set: the

quotient

E

/

Vect

ℚ

⁢

(

𝔅

)

can be represented as the subset of

ℚ

dim

⁡

E

of elements whose

projections

to the element of

𝔅

lies in

[

0

,

1

]

.
In other words

E

/

Vect

ℚ

⁢

(

𝔅

)

is

isomorphic

to the torus

∑

i

=

1

i

=

dim

⁡

E

[

0

,

1

]

⁢

𝔅

i

. This is a

compact set

, and therefore as

𝔞

is discrete there are only finitely many elements of

𝔞

that lie in the torus. Therefore by adding the element of

𝔞

that lie in the torus to

𝔅

, one obtains a finite set of generators of

𝔞

.

As

ℤ

is a

principal ideal ring

, it is again well-known that modules with finite rank (ie. that admit a finite set of generators) over a principal ideal ring can be represented as the

product

of a free module times a

torsion module

(with finite rank). Here there is no torsion as it would mean there is an element of

𝔞

that is sent to

0

by

multiplication

by an integer, and this is impossible as integers are elements of

K

and

K

is a field. Therefore

𝔞

itself is a free module (with finite rank).

The

discriminant

property can be seen intrinsically. Given an

algebra

of

linear maps

, here

𝔞

, one can define the

symmetric

bilinear map

a

,

b

∈

E

↦

Tr

⁢

(

a

⁢

b

)

. If the algebra happens to have a basis, which we have just proved in our case, then the determinant of that map can be computed using the basis and this is what is called discriminant. But of course, the determinant of that map is not dependent on the basis…

The minimality property is in fact a property of

Gram matrices

for

scalar products

. The elements of

𝔞

can be represented as elements of

End

⁢

(

E

,

E

)

≅

ℝ

(

dim

⁡

E

)

2

as they are linear endomorphisms of

E

. The bilinear map

a

,

b

∈

E

↦

Tr

⁢

(

a

⁢

b

)

is then no more than the Gram matrix associated to the vectors of the basis

α

1

,

…

,

α

k

of

𝔞

. Indeed, taking the trace of the product of two matrices it no more than taking the sum of the pairwise products of the entries of the two matrices.
The determinant of the Gram matrix is the square of the scalar factor by which volume of the

unit ball

is multiplied when taking its image through

ϕ

α

:

(

λ

1

,

…

,

λ

k

)

∈

ℝ

d

⁢

i

⁢

m

⁢

E

↦

∑

i

λ

i

⁢

α

i

∈

End

⁢

(

E

,

E

)

≅

ℝ

(

d

⁢

i

⁢

m

⁢

E

)

2

.
When computing the discriminant of the

β

i

, we look at the multiplication factor introduced by the map

ϕ

β

. But if

M

is the map that

associates

to the

β

i

their expression in terms of the

α

i

then

ϕ

β

=

ϕ

α

∘

M

.
Therefore

Δ

⁢

(

β

i

)

=

d

⁢

e

⁢

t

⁢

(

M

)

2

⁢

Δ

⁢

(

α

i

)

. If the

β

1

,

…

,

β

n

are not

linearly independent

then the multiplication factor is

0

(the ball is flattened), if they are linearly independent det(M) is an integer (the

β

i

are

linear combination

of the

α

i

with integer coefficients), therefore

Δ

⁢

(

β

i

)

≥

Δ

⁢

(

α

i

)

. Finally

Δ

⁢

(

β

i

)

=

Δ

⁢

(

α

i

)

is equivalent to

det

⁡

(

M

)

=

1

which in turn is equivalent to

M

is

invertible

(

1

is the only

positive

invertible element

of

ℤ

), which exactly means that

β

i

is a basis

iff

Δ

⁢

(

β

i

)

=

Δ

⁢

(

α

i

)

.

Title

proof of basis of ideal in algebraic number field

Canonical name

ProofOfBasisOfIdealInAlgebraicNumberField

Date of creation

2013-03-22 17:56:43

Last modified on

2013-03-22 17:56:43

Owner

lalberti (18937)

Last modified by

lalberti (18937)

Numerical id

18

Author

lalberti (18937)

Entry type

Proof

Classification

msc 12F05

Classification

msc 11R04

Classification

msc 06B10

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