fundamental theorem of ideal theory

fundamental theorem of ideal theory

Theorem.

Every nonzero ideal of the

ring of integers

of an

algebraic number field

can be written as

product

(

http://planetmath.org/ProductOfIdeals

) of

prime ideals

of the ring.  The prime ideal of the factors (

http://planetmath.org/Product

).

In this entry we consider the ring

𝒪

of the

integers

of a

number field

ℚ

⁢

(

ϑ

)

.  We use as starting the fact that the ideals of

𝒪

are

finitely generated

submodules

of

𝒪

(cf.

basis of ideal in algebraic number field

) and that its prime ideals

𝔭

are

maximal ideals

, i.e. the only ideal factors of

𝔭

are

𝔭

itself and the

unit ideal

(

1

)

=

𝒪

.

For proving the above

fundamental theorem of ideal theory

, we present and prove some lemmata.

Lemma 1.

The

equation

𝔞

=

𝔟

⁢

𝔠

between the ideals of

𝒪

implies that

𝔞

⊆

𝔠

.

Proof.

Let

𝔟

=

(

β

1

,

…

,

β

s

)

and

𝔠

=

(

γ

1

,

…

,

γ

t

)

.  If

α

∈

𝔞

=

(

β

1

⁢

γ

1

,

…

,

β

i

⁢

γ

j

,

…

,

β

s

⁢

γ

t

)

,

then there are the elements

λ

i

⁢

j

of

𝒪

such that

α

=

∑

i

∑

j

λ

i

⁢

j

⁢

β

i

⁢

γ

j

=

∑

j

(

∑

i

λ

i

⁢

j

⁢

β

i

)

⁢

γ

j

.

But the of

γ

j

in the parentheses are elements of the ring

𝒪

, whence the last sum form of

α

shows that

α

∈

𝔠

.  Consequently,

𝔞

⊆

𝔠

.

Lemma 2.

Any nonzero element

α

of

𝒪

belongs only to a finite number of ideals of

𝒪

.

Proof.

Let

𝔞

=

(

α

1

,

…

,

α

r

)

be any ideal containing

α

and let

{

ϱ

1

,

…

,

ϱ

m

}

be a

complete residue system

modulo

α

(cf.

congruence in algebraic number field

).  Then

α

i

=

α

λ

i

+

ϱ

n

i

(

i

=

1

,

…

,

r

)

where the numbers

λ

i

belong to

𝒪

.  Since we have

𝔞

=

(

α

1

,

…

,

α

r

,

α

)

=

(

α

⁢

λ

1

+

ϱ

n

1

,

…

,

α

⁢

λ

r

+

ϱ

n

r

,

α

)

=

(

ϱ

n

1

,

…

,

ϱ

n

r

,

α

)

,

there can be different ideals

𝔞

only a finite number, at most

1

+

m

+

(

n

2

)

+

…

+

(

m

m

)

=

2

m

.

Lemma 3.

Each ideal

𝔞

of

𝒪

has only a finite number of ideal factors.

Proof.

If

𝔠

∣

𝔞

and

α

∈

𝔞

,  then by Lemma 1,

α

∈

𝔠

,  whence Lemma 2 implies that there is only a finite number of such factors

𝔠

.

Lemma 4.

All nonzero ideals of

𝒪

are

cancellative

(

http://planetmath.org/CancellationIdeal

), i.e. if

𝔞

⁢

𝔠

=

𝔞

⁢

𝔡

then

𝔠

=

𝔡

.

Proof.

The theorem of

http://planetmath.org/node/3154

Steinitz (1911) guarantees an ideal

𝔤

of

𝒪

such that the product

𝔤

⁢

𝔞

is a

principal ideal

(

ω

)

.  Then we may write

(

ω

)

⁢

𝔠

=

(

𝔤

⁢

𝔞

)

⁢

𝔠

=

𝔤

⁢

(

𝔞

⁢

𝔠

)

=

𝔤

⁢

(

𝔞

⁢

𝔡

)

=

(

𝔤

⁢

𝔞

)

⁢

𝔡

=

(

ω

)

⁢

𝔡

.

If

𝔠

=

(

γ

1

,

…

,

γ

s

)

and

𝔡

=

(

δ

1

,

…

,

δ

t

)

,  we thus have the equation

(

ω

⁢

γ

1

,

…

,

ω

⁢

γ

s

)

=

(

ω

⁢

δ

1

,

…

,

ω

⁢

δ

t

)

by which there must exist the elements

λ

i

⁢

1

,

…

,

λ

i

⁢

t

of

𝒪

such that

ω

⁢

γ

i

=

λ

i

⁢

1

⁢

ω

⁢

δ

1

+

…

+

λ

i

⁢

t

⁢

ω

⁢

δ

t

.

Consequently, the

http://planetmath.org/node/7040

generators

γ

i

=

λ

i

⁢

1

⁢

δ

1

+

…

+

λ

i

⁢

t

⁢

δ

t

of

𝔠

belong to the ideal

𝔡

, and therefore

𝔠

⊆

𝔡

.  Similarly one gets the reverse containment.

Lemma 5.

If

𝔞

=

𝔟

⁢

𝔠

and

𝔟

≠

(

1

)

,  then

𝔠

has less ideal factors than

𝔞

.

Proof.

Evidently, any factor of

𝔠

is a factor of

𝔞

.  But

𝔞

∣

𝔞

and

𝔞

∤

𝔠

, since otherwise we had

𝔠

=

𝔞

⁢

𝔡

=

𝔟

⁢

𝔠

⁢

𝔡

whence

(

1

)

=

𝔟

⁢

𝔡

which would, by Lemma 4, imply

𝔟

=

(

1

)

.

Lemma 6.

Any

proper ideal

𝔞

of

𝒪

has a prime ideal factor.

Proof.

Let

𝔠

be such a factor of

𝔞

that has as few factors as possible.  Then

𝔠

must be a prime ideal, because otherwise we had

𝔠

=

𝔠

1

⁢

𝔡

where

𝔠

1

and

𝔡

are proper ideals of

𝒪

and, by Lemma 5, the ideal

𝔠

1

would have less factors than

𝔠

; this however contradicts the fact

𝔠

1

∣

𝔞

.

Lemma 7.

Every nonzero proper ideal

𝔞

of

𝒪

can be written as a product

𝔭

1

⁢

⋯

⁢

𝔭

k

where

k

>

0

and the factors

𝔭

i

are prime ideals.

Proof.

If

𝔞

has only one factor

𝔭

distinct from

(

1

)

, then

𝔞

=

𝔭

is a prime ideal.

Induction hypothesis:  Lemma 7 is in always when

𝔞

has at most

n

factors.  Let

𝔞

now have

n

+

1

factors.  Lemma 6 implies that there is a prime ideal

𝔭

such that

𝔞

=

𝔭

⁢

𝔡

where

𝔡

≠

(

1

)

and

𝔡

has, by Lemma 5, at most

n

factors.  Hence,

𝔡

=

𝔭

1

⁢

⋯

⁢

𝔭

k

and therefore,

𝔞

=

𝔭

⁢

𝔭

1

⁢

⋯

⁢

𝔭

k

where all

𝔭

’s are prime ideals.

Lemma 8.

Any two

prime factor

𝔞

=

𝔭

1

⁢

⋯

⁢

𝔭

r

=

𝔮

1

⁢

⋯

⁢

𝔮

s

of a nonzero ideal

𝔞

of

𝒪

are identical, i.e.

r

=

s

and each prime factor

𝔭

i

is equal to a prime factor

𝔮

j

and vice versa.

Proof.

Any prime ideal has the property that if it divides a

product of ideals

, it divides one of the factors of the product; now these factors are prime ideals and therefore the prime ideal coincides with one of the factors.  Similarly as in the proof of the

fundamental theorem of arithmetics

, one sees the uniqueness of the prime factorisation of

𝔞

.

Title

fundamental theorem of ideal theory

Canonical name

FundamentalTheoremOfIdealTheory

Date of creation

2013-03-22 19:12:40

Last modified on

2013-03-22 19:12:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

27

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11R04

Synonym

principal theorem of ideal theory

Related topic

AlgebraicNumberTheory

Related topic

DedekindDomain

Related topic

EveryIdealInADedekindDomainIsAFactorOfAPrincipalIdeal

Related topic

PrimeIdealFactorizationIsUnique

Related topic

UniqueFactorizationAndIdealsInRingOfIntegers

Related topic

CancellativeSemigroup

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