unique factorization and ideals in ring of integers

unique factorization and ideals in ring of integers

Theorem.

Let

O

be the

maximal order

, i.e. the

ring of
integers

of an

algebraic number field

.  Then

O

is a

unique
factorization domain

if and only if

O

is a

principal ideal domain

.

Proof.

1

o

¯

. Suppose that

O

is a PID.

We first state, that any

prime number

π

of

O

generates

a

prime ideal

(

π

)

of

O

.  For if

(

π

)

=

𝔞

⁢

𝔟

,  then we have the

principal ideals

𝔞

=

(

α

)

and

𝔟

=

(

β

)

.  It follows that

(

π

)

=

(

α

⁢

β

)

, i.e.

π

=

λ

⁢

α

⁢

β

with some

λ

∈

O

,  and since

π

is prime, one of

α

and

β

must be a unit of

O

.  Thus one of

𝔞

and

𝔟

is the

unit ideal

O

, and accordingly

(

π

)

is a

maximal ideal

of

O

, so also a prime ideal.

Let a non-zero element

γ

of

O

be split to prime number factors

π

i

,

ϱ

j

in two ways:

γ

=

π

1

⁢

⋯

⁢

π

r

=

ϱ

1

⁢

⋯

⁢

ϱ

s

.  Then also the principal ideal

(

γ

)

splits to principal prime ideals in two ways:

(

γ

)

=

(

π

1

)

⁢

⋯

⁢

(

π

r

)

=

(

ϱ

1

)

⁢

⋯

⁢

(

ϱ

s

)

.  Since the

prime factorization

of ideals is unique, the

(

π

1

)

,

…

,

(

π

r

)

must be, up to the , identical with

(

ϱ

1

)

,

…

,

(

ϱ

s

)

(and

r

=

s

).  Let

(

π

1

)

=

(

ϱ

j

1

)

.  Then

π

1

and

ϱ

j

1

are

associates

of each other; the same may be said of all pairs

(

π

i

,

ϱ

j

i

)

.  So we have seen that the factorization in

O

is unique.

2

o

¯

. Suppose then that

O

is a UFD.

Consider any prime ideal

𝔭

of

O

.  Let

α

be a non-zero element of

𝔭

and let

α

have the prime factorization

π

1

⁢

⋯

⁢

π

n

.  Because

𝔭

is a prime ideal and divides the ideal product

(

π

1

)

⁢

⋯

⁢

(

π

n

)

,

𝔭

must divide one principal ideal

(

π

i

)

=

(

π

)

.  This means that

π

∈

𝔭

.  We write

(

π

)

=

𝔭

⁢

𝔞

, whence

π

∈

𝔭

and

π

∈

𝔞

.  Since

O

is a

Dedekind domain

, every its ideal can be generated by two elements, one of which may be chosen freely (see the

two-generator property

).  Therefore we can write

𝔭

=

(

π

,

γ

)

,

𝔞

=

(

π

,

δ

)

.

We multiply these, getting

𝔭

⁢

𝔞

=

(

π

2

,

π

⁢

γ

,

π

⁢

δ

,

γ

⁢

δ

)

,  and so

γ

⁢

δ

∈

𝔭

⁢

𝔞

=

(

π

)

.  Thus

γ

⁢

δ

=

λ

⁢

π

with some

λ

∈

O

.  According to the

unique factorization

, we have

π

|

γ

or

π

|

δ

.

The latter alternative means that

δ

=

δ

1

⁢

π

(with

δ

1

∈

O

),  whence

𝔞

=

(

π

,

δ

1

⁢

π

)

=

(

π

)

⁢

(

1

,

δ

1

)

=

(

π

)

⁢

(

1

)

=

(

π

)

;  thus we had

𝔭

⁢

𝔞

=

(

π

)

=

𝔭

⁢

(

π

)

which would imply the absurdity

𝔭

=

(

1

)

.  But the former alternative means that

γ

=

γ

1

⁢

π

(with

γ

1

∈

O

),  which shows that

𝔭

=

(

π

,

γ

1

⁢

π

)

=

(

π

)

⁢

(

1

,

γ

1

)

=

(

π

)

⁢

(

1

)

=

(

π

)

.

In other words, an arbitrary prime ideal

𝔭

of

O

is principal.  It follows that all ideals of

O

are principal. Q.E.D.

Title

unique factorization and ideals in ring of integers

Canonical name

UniqueFactorizationAndIdealsInRingOfIntegers

Date of creation

2015-05-06 15:32:53

Last modified on

2015-05-06 15:32:53

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13B22

Classification

msc 11R27

Synonym

equivalence of UFD and PID

Related topic

ProductOfFinitelyGeneratedIdeals

Related topic

PIDsAreUFDs

Related topic

NumberFieldThatIsNotNormEuclidean

Related topic

DivisorTheory

Related topic

FundamentalTheoremOfIdealTheory

Related topic

EquivalentDefinitionsForUFD

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