aarjaneiro/planetmath huggingface dataset:
Krull valuation

Krull valuation

Definition.

The mapping

|

.

|

:

K

→

G

,  where

K

is a field and

G

an

ordered group equipped with zero

, is a

Krull valuation

of

K

, if it has the properties

1.

|

x

|

=

0

⇔

x

=

0

;

2.

|

x

⁢

y

|

=

|

x

|

⋅

|

y

|

;

3.

|

x

+

y

|

≦

max

⁡

{

|

x

|

,

|

y

|

}

.

Thus the Krull valuation is more general than the usual

valuation

(

http://planetmath.org/Valuation

), which is also characterized as and which has real values.  The image

|

K

∖

{

0

}

|

is called the

value group

of the Krull valuation; it is

abelian

.  In general, the

rank of Krull valuation

the rank (

http://planetmath.org/IsolatedSubgroup

) of the value group.

We may say that a Krull valuation is non-archimedean (

http://planetmath.org/Valuation

).

Some values

•

|

1

|

=

1

because the Krull valuation is a

group homomorphism

from the

multiplicative group

of

K

to the ordered group.

•

|

-

1

|

=

1

because

1

=

|

(

-

1

)

2

|

=

|

-

1

|

2

and 1 is the only element of the ordered group being its own

inverse

(

S

∩

S

-

1

=

∅

).

•

|

-

x

|

=

|

(

-

1

)

⁢

x

|

=

|

-

1

|

⋅

|

x

|

=

|

x

|

References

1

Emil Artin

:

Theory of

Algebraic Numbers

.  Lecture notes.  Mathematisches Institut, Göttingen (1959).

2

P. Jaffard

:

Les systèmes d’idéaux

.  Dunod, Paris (1960).

Title

Krull valuation

Canonical name

KrullValuation

Date of creation

2013-03-22 14:54:39

Last modified on

2013-03-22 14:54:39

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

19

Author

pahio (2872)

Entry type

Definition

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Related topic

OrderedGroup

Related topic

TrivialValuation

Related topic

IsolatedSubgroup

Related topic

ValueGroupOfCompletion

Related topic

PlaceOfField

Related topic

OrderValuation

Related topic

AlternativeDefinitionOfValuation2

Related topic

UniquenessOfDivisionAlgorithmInEuclideanDomain

Defines

value group

Defines

rank of Krull valuation

Defines

rank of valuation

Generated on Fri Feb 9 17:31:51 2018 by

LaTeXML
Krull valuation domain

Krull valuation domain

Theorem

.

Any

Krull valuation

|

⋅

|

of a field

K

determines a unique

valuation domain

R

=

{

a

∈

K

:

|

x

|

≦

1

}

, whose

field of fraction

is

K

.

Proof.

We first see that

1

∈

R

since

|

1

|

=

1

.  Let then

a

,

b

be any two elements of

R

.  The

non-archimedean triangle inequality

shows that

|

a

-

b

|

≦

max

⁡

{

|

a

|

,

|

b

|

}

≦

1

,  i.e. that the

difference

a

-

b

belongs to

R

.  Using the

multiplication

rule (

http://planetmath.org/OrderedGroup

) 4 of

inequalities

we obtain

|

a

⁢

b

|

=

|

a

|

⋅

|

b

|

≦

1

⋅

1

=

1

which shows that also the product

a

⁢

b

is element of

R

.  Thus,

R

is a

subring

of the field

K

, and so an

integral domain

.  Let now

c

be an arbitrary element of

K

not belonging to

R

.  This implies that

1

<

|

c

|

,  whence

|

c

-

1

|

=

|

c

|

-

1

<

1

(see the

inverse

rule (

http://planetmath.org/OrderedGroup

) 5).  Consequently, the inverse

c

-

1

belongs to

R

, and we conclude that

R

is a valuation domain.   The

a

=

a

1

and

c

=

1

c

-

1

make evident that

K

is the field of fractions of

R

.

Title

Krull valuation domain

Canonical name

KrullValuationDomain

Date of creation

2013-03-22 14:55:01

Last modified on

2013-03-22 14:55:01

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Related topic

ValuationDeterminedByValuationDomain

Generated on Fri Feb 9 17:31:53 2018 by

LaTeXML
place of field

place of field

Let

F

be a field and

∞

an

element

not belonging to

F

.  The

mapping

φ

:

k

→

F

∪

{

∞

}

,

where

k

is a field, is called a

place of the field

k

, if it

satisfies

the following conditions.

•

The

preimage

φ

-

1

⁢

(

F

)

=

𝔬

is a

subring

of

k

.

•

The

restriction

φ

|

𝔬

is a

ring homomorphism

from

𝔬

to

F

.

•

If

φ

⁢

(

a

)

=

∞

,  then

φ

⁢

(

a

-

1

)

=

0

.

It is easy to see that the subring

𝔬

of the field

k

is a

valuation domain

; so any place of a field determines a unique valuation domain in the field.

Conversely

, every valuation domain

𝔬

with

field of fractions

k

determines a place of

k

:

Theorem

.

Let

𝔬

be a valuation domain with field of fractions

k

and

𝔭

the

maximal ideal

of

𝔬

, consisting of the non-units of

𝔬

.  Then the mapping

φ

:

k

→

𝔬

/

𝔭

∪

{

∞

}

defined by

φ

⁢

(

x

)

:=

{

x

+

𝔭

when

⁢

x

∈

𝔬

,

∞

when

⁢

x

∈

k

∖

𝔬

,

is a place of the field

k

.

Proof.

Apparently,

φ

-

1

⁢

(

𝔬

/

𝔭

)

=

𝔬

and the restriction

φ

|

𝔬

is the

canonical homomorphism

from the ring

𝔬

onto the

residue-class ring

𝔬

/

𝔭

.  Moreover, if

φ

⁢

(

x

)

=

∞

,  then

x

does not belong to the valuation domain

𝔬

and thus the inverse element

x

-

1

must belong to it without being its unit.  Hence

x

-

1

belongs to the ideal

𝔭

which is the kernel of the

homomorphism

φ

|

𝔬

.  So we see that

φ

⁢

(

x

-

1

)

=

0

.

References

1

Emil Artin: .  Lecture notes.  Mathematisches Institut, Göttingen (1959).

Title

place of field

Canonical name

PlaceOfField

Date of creation

2013-03-22 14:56:51

Last modified on

2013-03-22 14:56:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12E99

Synonym

place

Synonym

spot of field

Related topic

KrullValuation

Related topic

ValuationDeterminedByValuationDomain

Related topic

IntegrityCharacterizedByPlaces

Related topic

RamificationOfArchimedeanPlaces

Defines

place of field

Generated on Fri Feb 9 17:31:57 2018 by

LaTeXML
ring of exponent

ring of exponent

Definition

.

Let

ν

be an

exponent valuation

of the field

K

.
The

subring

𝒪

ν

:=

{

α

∈

K

⋮

ν

(

α

)

≧

0

}

of

K

is called the

ν

.  It is, naturally, an

integral domain

.  Its elements are called

ν

.

Theorem

1.

The ring of the exponent

ν

of the field

K

is

integrally closed

in

K

.

Theorem 2.

The ring

𝒪

ν

only one

prime element

π

, when one does not regard associated elements as different.  Any non-zero element

α

can be represented uniquely with a

π

in the form

α

=

ε

⁢

π

m

,

where

ε

is a unit of

𝒪

ν

and

m

=

ν

⁢

(

α

)

≧

0

.  This means that

𝒪

is a UFD.

Remark 1.

The prime elements

π

of the ring

𝒪

ν

are characterised by the equation

ν

⁢

(

π

)

=

1

and the units

ε

the equation

ν

⁢

(

ε

)

=

0

.

Remark 2.

In an algebraically closed field

Ω

, there are no exponents (

http://planetmath.org/ExponentValuation

).  In fact, if there were an exponent

ν

of

Ω

and if

π

were a prime element of the ring of the exponent, then, since the equation

x

2

-

π

=

0

would have a root (

http://planetmath.org/Equation

)

ϱ

in

Ω

, we would obtain

2

⁢

ν

⁢

(

ϱ

)

=

ν

⁢

(

ϱ

2

)

=

ν

⁢

(

π

)

=

1

;  this is however impossible, because an exponent attains only integer values.

Theorem 3.

Let

𝔒

1

,

…

,

𝔒

r

be the rings of the different exponent valuations

ν

1

,

…

,

ν

r

of the field

K

.  Then also the intersection

𝔒

:=

⋂

i

=

1

r

𝔒

i

is a subring of

K

with unique factorisation (

http://planetmath.org/UFD

).  To be precise, any non-zero element

α

of

𝔒

may be uniquely represented in the form

α

=

ε

⁢

π

1

n

1

⁢

⋯

⁢

π

r

n

r

,

in which

ε

is a unit of

𝔒

,  the

integers

n

1

,

…

,

n

r

are nonnegative and

π

1

,

…

,

π

r

are

coprime

prime elements of

𝔒

satisfying

ν

i

⁢

(

π

j

)

=

δ

i

⁢

j

=

{

1

⁢

for

⁢

i

=

j

,

0

⁢

for

⁢

i

≠

j

.

Title

ring of exponent

Canonical

name

RingOfExponent

Date of creation

2013-03-22 17:59:43

Last modified on

2013-03-22 17:59:43

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Definition

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Related topic

DiscreteValuationRing

Related topic

ValuationRingOfAField

Related topic

LocalRing

Defines

ring of an exponent

Defines

ring of the exponent

Defines

integral

with respect to an exponent

Generated on Fri Feb 9 17:32:03 2018 by

LaTeXML
valuation determined by valuation domain

valuation determined by valuation domain

Theorem

.

Every

valuation domain

determines a

Krull valuation

of the

field of fractions

.

Proof.

Let

R

be a valuation domain,

K

its field of fractions and

E

the

group of units

of

R

. Then

E

is a

normal subgroup

of the

multiplicative group

K

*

=

K

∖

{

0

}

.  So we can form the

factor group

K

*

/

E

, consisting of all cosets

a

⁢

E

where

a

∈

K

*

,  and attach to it the additional “coset”

0

⁢

E

getting thus a multiplicative group

K

/

E

equipped with zero.  If

𝔪

=

R

∖

E

is the

maximal ideal

of

R

(any valuation domain has a unique maximal ideal
— cf.

valuation domain is local

), then we denote

𝔪

*

=

𝔪

∖

{

0

}

and

S

=

𝔪

*

/

E

=

{

a

⁢

E

:

a

∈

𝔪

*

}

.  Then the

subsemigroup

S

of

K

/

E

makes

K

/

E

an

ordered group equipped with zero

.  It is not hard to check that the mapping

x

↦

|

x

|

:=

x

⁢

E

from

K

to

K

/

E

is a Krull valuation of the field

K

.

Title

valuation determined by valuation domain

Canonical name

ValuationDeterminedByValuationDomain

Date of creation

2013-03-22 14:54:58

Last modified on

2013-03-22 14:54:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Related topic

ValuationDomainIsLocal

Related topic

KrullValuationDomain

Related topic

PlaceOfField

Generated on Fri Feb 9 17:32:11 2018 by

LaTeXML
valuation domain is local

valuation domain is local

Theorem

.

Every

valuation domain

is a

local ring

.

Proof.

Let

R

be a valuation domain and

K

its

field of fractions

.  We shall show that the set of all non-units of

R

is the only

maximal ideal

of

R

.

Let

a

and

b

first be such elements of

R

that

a

-

b

is a unit of

R

; we may suppose that

a

⁢

b

≠

0

since otherwise one of

a

and

b

is instantly stated to be a unit.  Because

R

is a valuation domain in

K

, therefore e.g.

a

b

∈

R

.  Because now

a

-

b

b

=

1

-

a

b

and

(

a

-

b

)

-

1

belong to

R

, so does also the

product

a

-

b

b

⋅

(

a

-

b

)

-

1

=

1

b

,  i.e.

b

is a unit of

R

.  We can conclude that the difference

a

-

b

must be a non-unit whenever

a

and

b

are non-units.

Let

a

and

b

then be such elements of

R

that

a

⁢

b

is its unit, i.e.

a

-

1

⁢

b

-

1

∈

R

.  Now we see that

a

-

1

=

b

⋅

a

-

1

⁢

b

-

1

∈

R

,

b

-

1

=

a

⋅

a

-

1

⁢

b

-

1

∈

R

,

and consequently

a

and

b

both are units.  So we conclude that the product

a

⁢

b

must be a non-unit whenever

a

is an element of

R

and

b

is a non-unit.

Thus the non-units form an ideal

𝔪

.  Suppose now that there is another ideal

𝔫

of

R

such that

𝔪

⊂

𝔫

⊆

R

.  Since

𝔪

contains all non-units, we can take a unit

ε

in

𝔫

.  Thus also the product

ε

-

1

⁢

ε

, i.e. 1, belongs to

𝔫

, or

R

⊆

𝔫

.  So we see that

𝔪

is a maximal ideal.  On the other hand, any maximal ideal of

R

contains no units and hence is contained in

𝔪

; therefore

𝔪

is the only maximal ideal.

Title

valuation domain is local

Canonical name

ValuationDomainIsLocal

Date of creation

2013-03-22 14:54:49

Last modified on

2013-03-22 14:54:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13G05

Classification

msc 16U10

Related topic

ValuationRing

Related topic

ValuationDeterminedByValuationDomain

Related topic

HenselianField

Generated on Fri Feb 9 17:32:13 2018 by

LaTeXML
value group of completion

value group of completion

Let

k

be a field and

|

⋅

|

its

non-archimedean valuation

of rank one (

http://planetmath.org/KrullValuation

).  Then its

value group

|

k

∖

{

0

}

|

may be considered to be a

subgroup

of the

multiplicative group

of

ℝ

.  In the

completion

K

of the valued field

k

, the

extension

of the

valuation

is defined by

|

x

|

=

:

lim

n

→

∞

|

x

n

|

,

when the

Cauchy sequence

x

1

,

x

2

,

…

,

x

n

,

…

of elements of

k

determines the element

x

of

K

.

Theorem

.

The

non-archimedean field

k

and its completion

K

have the same value group.

Proof.

Of course,

|

k

|

⊆

|

K

|

.  Let

x

=

lim

n

→

∞

⁡

x

n

be any non-zero element of

K

, where

x

j

’s form a Cauchy sequence in

k

.  Then there exists a positive number

n

0

such that

|

x

n

-

x

|

<

|

x

|

for all

n

>

n

0

.  For all these values of

n

we have

|

x

n

|

=

|

x

+

(

x

n

-

x

)

|

=

|

x

|

according to the

ultrametric triangle inequality

.  Thus we see that

|

K

|

⊆

|

k

|

.

Title

value group of completion

Canonical name

ValueGroupOfCompletion

Date of creation

2013-03-22 14:58:14

Last modified on

2013-03-22 14:58:14

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13F30

Classification

msc 13J10

Classification

msc 13A18

Classification

msc 12J20

Related topic

KrullValuation

Related topic

ExtensionOfValuationFromCompleteBaseField

Defines

value group of the completion

Generated on Fri Feb 9 17:32:15 2018 by

LaTeXML
a finite integral domain is a field

a finite integral domain is a field

A finite

integral domain

is a field.

Proof:

Let

R

be a finite integral domain. Let

a

be nonzero

element of

R

.

Define a

function

φ

:

R

→

R

by

φ

⁢

(

r

)

=

a

⁢

r

.

Suppose

φ

⁢

(

r

)

=

φ

⁢

(

s

)

for some

r

,

s

∈

R

. Then

a

⁢

r

=

a

⁢

s

, which implies

a

⁢

(

r

-

s

)

=

0

. Since

a

≠

0

and

R

is a

cancellation ring

, we have

r

-

s

=

0

. So

r

=

s

, and hence

φ

is

injective

.

Since

R

is finite and

φ

is injective, by the

pigeonhole principle

we see that

φ

is also

surjective

. Thus there exists some

b

∈

R

such that

φ

⁢

(

b

)

=

a

⁢

b

=

1

R

, and thus

a

is a unit.

Thus

R

is a finite division ring. Since it is commutative, it is also a field.

Note:

A more general result is that an

Artinian

integral domain is a field.

Title

a finite integral domain is a field

Canonical name

AFiniteIntegralDomainIsAField

Date of creation

2013-03-22 12:50:02

Last modified on

2013-03-22 12:50:02

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

11

Author

yark (2760)

Entry type

Theorem

Classification

msc 13G05

Related topic

FiniteRingHasNoProperOverrings

Generated on Fri Feb 9 17:32:17 2018 by

LaTeXML
alternative proof that a finite integral domain is a field

alternative proof that a finite integral domain is a field

Proof.

Let

R

be a finite

integral domain

and

a

∈

R

with

a

≠

0

. Since

R

is finite, there exist

positive integers

j

and

k

with

j

<

k

such that

a

j

=

a

k

. Thus,

a

k

-

a

j

=

0

. Since

j

<

k

and

j

and

k

are positive integers,

k

-

j

is a positive integer. Therefore,

a

j

⁢

(

a

k

-

j

-

1

)

=

0

. Since

a

≠

0

and

R

is an integral domain,

a

j

≠

0

. Thus,

a

k

-

j

-

1

=

0

. Hence,

a

k

-

j

=

1

. Since

k

-

j

is a positive integer,

k

-

j

-

1

is a

nonnegative integer

. Thus,

a

k

-

j

-

1

∈

R

. Note that

a

⋅

a

k

-

j

-

1

=

a

k

-

j

=

1

. Hence,

a

has a

multiplicative inverse

in

R

. It follows that

R

is a field.
∎

Title

alternative proof that a finite integral domain is a field

Canonical name

AlternativeProofThatAFiniteIntegralDomainIsAField

Date of creation

2013-03-22 16:21:54

Last modified on

2013-03-22 16:21:54

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 13G05

Generated on Fri Feb 9 17:32:19 2018 by

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Bezout domain

Bezout domain

A

Bezout domain

D

is an

integral domain

such that every

finitely generated

ideal of

D

is principal (

http://planetmath.org/PID

).

Remarks

.

•

A PID is obviously a Bezout domain.

•

Furthermore, a Bezout domain is a

gcd domain

. To see this, suppose

D

is a Bezout domain with

a

,

b

∈

D

. By definition, there is a

d

∈

D

such that

(

d

)

=

(

a

,

b

)

, the

ideal generated by

a

and

b

. So

a

∈

(

d

)

and

b

∈

(

d

)

and therefore,

d

∣

a

and

d

∣

b

. Next, suppose

c

∈

D

and that

c

∣

a

and

c

∣

b

. Then both

a

,

b

∈

(

c

)

and so

d

∈

(

c

)

. This means that

c

∣

d

and we are done.

•

From the discussion above, we see in a Bezout domain

D

, a

greatest common divisor

exists for every pair of elements. Furthermore, if

gcd

⁡

(

a

,

b

)

denotes one such greatest common divisor between

a

,

b

∈

D

, then for some

r

,

s

∈

D

:

gcd

⁡

(

a

,

b

)

=

r

⁢

a

+

s

⁢

b

.

The above

equation

is known as the

Bezout identity

, or Bezout’s Lemma.

Title

Bezout domain

Canonical name

BezoutDomain

Date of creation

2013-03-22 14:19:53

Last modified on

2013-03-22 14:19:53

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13G05

Synonym

Bézout domain

Related topic

GcdDomain

Related topic

DivisibilityByProduct

Defines

Bezout identity

Generated on Fri Feb 9 17:32:32 2018 by

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Dedekind-Hasse valuation

Dedekind-Hasse valuation

If

D

is an

integral domain

then it is a PID iff it has a

Dedekind-Hasse valuation

, that is, a function

ν

:

D

-

{

0

}

→

ℤ

+

such that for any

a

,

b

∈

D

-

{

0

}

either

•

a

∈

(

b

)

or

•

∃

α

∈

(

a

)

∃

β

∈

(

b

)

[

0

<

ν

(

α

+

β

)

<

ν

(

b

)

]

Proof:

First, let

ν

be a Dedekind-Hasse valuation and let

I

be an ideal of an integral domain

D

. Take some

b

∈

I

with

ν

⁢

(

b

)

minimal (this exists because the integers are

well-ordered

) and some

a

∈

I

such that

a

≠

0

.

I

must contain both

(

a

)

and

(

b

)

, and since it is closed under

addition

,

α

+

β

∈

I

for any

α

∈

(

a

)

,

β

∈

(

b

)

.

Since

ν

⁢

(

b

)

is minimal, the second possibility above is ruled out, so it follows that

a

∈

(

b

)

. But this holds for any

a

∈

I

, so

I

=

(

b

)

, and therefore every ideal is princple.

For the converse, let

D

be a PID. Then define

ν

⁢

(

u

)

=

1

for any unit. Any non-zero, non-unit can be factored into a finite

product

of

irreducibles

(since

http://planetmath.org/node/PIDsareUFDs

every PID is a UFD), and every such factorization of

a

is of the same length,

r

. So for

a

∈

D

, a non-zero non-unit, let

ν

⁢

(

a

)

=

r

+

1

. Obviously

r

∈

ℤ

+

.

Then take any

a

,

b

∈

D

-

{

0

}

and suppose

a

∉

(

b

)

. Then take the ideal of elements of the form

{

α

+

β

|

α

∈

(

a

)

,

β

∈

(

b

)

}

. Since this is a PID, it is a

principal ideal

(

c

)

for some

r

∈

D

-

{

0

}

, and since

0

+

b

=

b

∈

(

c

)

, there is some non-unit

x

∈

D

such that

x

⁢

c

=

b

. Then

N

⁢

(

b

)

=

N

⁢

(

x

⁢

r

)

. But since

x

is not a unit, the factorization of

b

must be longer than the factorization of

c

, so

ν

⁢

(

b

)

>

ν

⁢

(

c

)

, so

ν

is a Dedekind-Hasse valuation.

Title

Dedekind-Hasse valuation

Canonical

name

DedekindHasseValuation

Date of creation

2013-03-22 12:51:16

Last modified on

2013-03-22 12:51:16

Owner

Henry (455)

Last modified by

Henry (455)

Numerical id

5

Author

Henry (455)

Entry type

Definition

Classification

msc 13G05

Related topic

EuclideanValuation

Defines

Dedekind-Hasse norm

Defines

Dedekind-Hasse valuation

Generated on Fri Feb 9 17:32:34 2018 by

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equivalent definitions for UFD

equivalent definitions for UFD

Let

R

be an

integral domain

. Define

T

=

{

u

∈

R

|

u

⁢

is invertible

}

∪

{

p

1

⁢

⋯

⁢

p

n

∈

R

|

p

i

⁢

is prime

}

.

Of course

0

∉

T

and

T

is a

multiplicative subset

(recall that a

prime element

multiplied by an invertible element is again prime). Furthermore

R

is a UFD if and only if

T

=

R

\

{

0

}

(see the parent object for more details).

Lemma.

If

a

,

b

∈

R

are such that

a

⁢

b

∈

T

, then both

a

,

b

∈

T

.

Proof.

If

a

⁢

b

is

invertible

, then (since

R

is

commutative

) both

a

,

b

are invertible and thus they belong to

T

. Therefore assume that

a

⁢

b

is not invertible. Then

a

⁢

b

=

p

1

⁢

⋯

⁢

p

k

for some prime elements

p

i

∈

R

. We can group these prime elements in such way that

p

1

⁢

⋯

⁢

p

n

divides

a

and

p

n

+

1

⁢

⋯

⁢

p

k

divides

b

. Thus

a

=

α

⁢

p

1

⁢

⋯

⁢

p

n

and

b

=

β

⁢

p

n

+

1

⁢

⋯

⁢

p

k

for some

α

,

β

∈

R

. Since

R

is an integral domain we conclude that

α

⁢

β

=

1

, which means that both

α

,

β

are invertible in

R

. Therefore (for example)

α

⁢

p

1

is prime and thus

a

∈

T

. Analogously

b

∈

T

, which

completes

the proof.

□

Theorem

. (Kaplansky)

An integral domain

R

is a UFD if and only if every nonzero

prime ideal

in

R

contains prime element.

Proof.

Without loss of generality we may assume that

R

is not a field, because the thesis trivialy holds for fields. In this case

R

always contains nonzero prime ideal (just take a

maximal ideal

).

,,

⇒

” Let

P

be a nonzero prime ideal. In particular

P

is proper, thus there is nonzero

x

∈

P

which is not invertible. By

assumption

x

∈

T

and since

x

is not invertible, then there are prime elements

p

1

,

…

,

p

k

∈

R

such that

x

=

p

1

⁢

⋯

⁢

p

k

∈

P

. But

P

is prime, therefore there is

i

∈

{

1

,

…

,

k

}

such that

p

i

∈

P

, which completes this part.

,,

⇐

” Assume that

R

is not a UFD. Thus there is a nonzero

x

∈

R

such that

x

∉

T

. Consider an ideal

(

x

)

. We will show, that

(

x

)

∩

T

=

∅

. Assume that there is

r

∈

R

such that

r

⁢

x

∈

T

. It follows that

x

∈

T

(by lemma).

Contradiction

.

Since

(

x

)

∩

T

=

∅

and

T

is a multiplicative subset, then there is a prime ideal

P

in

R

such that

(

x

)

⊆

P

and

P

∩

T

=

∅

(please, see this entry (

http://planetmath.org/MultiplicativeSetsInRingsAndPrimeIdeals

) for more details). But we assumed that every nonzero prime ideal contains prime element (and

P

is nonzero, since

x

∈

P

). Obtained contradiction completes the proof.

□

Title

equivalent definitions for UFD

Canonical name

EquivalentDefinitionsForUFD

Date of creation

2013-03-22 19:04:04

Last modified on

2013-03-22 19:04:04

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 13G05

Related topic

UniqueFactorizationAndIdealsInRingOfIntegers

Generated on Fri Feb 9 17:32:36 2018 by

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