aarjaneiro/planetmath huggingface dataset:
maximal ideal

maximal ideal

Let

R

be a

ring with identity

. A proper left (right, two-sided) ideal

𝔪

⊊

R

is said to be

maximal

if

𝔪

is not a

proper subset

of any other proper left (right, two-sided) ideal of

R

.

One can prove:

•

A

left ideal

𝔪

is maximal if and only if

R

/

𝔪

is a simple left

R

-module.

•

A

right ideal

𝔪

is maximal if and only if

R

/

𝔪

is a simple right

R

-module.

•

A

two-sided ideal

𝔪

is maximal if and only if

R

/

𝔪

is a

simple ring

.

All

maximal ideals

are

prime ideals

. If

R

is

commutative

, an ideal

𝔪

⊂

R

is maximal if and only if the

quotient ring

R

/

𝔪

is a field.

Title

maximal ideal

Canonical name

MaximalIdeal

Date of creation

2013-03-22 11:50:57

Last modified on

2013-03-22 11:50:57

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Definition

Classification

msc 13A15

Classification

msc 16D25

Classification

msc 81R50

Classification

msc 46M20

Classification

msc 18B40

Classification

msc 22A22

Classification

msc 46L05

Related topic

ProperIdeal

Related topic

Module

Related topic

Comaximal

Related topic

PrimeIdeal

Related topic

EveryRingHasAMaximalIdeal

Generated on Fri Feb 9 17:21:58 2018 by

LaTeXML
multiplication ring

multiplication ring

Let

R

be a

commutative ring

with

non-zero unity

.  If

𝔞

and

𝔟

are two

fractional ideals

(

http://planetmath.org/FractionalIdealOfCommutativeRing

) of

R

with

a

⊆

b

and if

b

is invertible (

http://planetmath.org/FractionalIdealOfCommutativeRing

), then there is a

c

of

R

such that

a

=

b

⁢

c

(one can choose

c

=

b

-

1

⁢

a

).

Definition.

Let

R

be a commutative ring with non-zero unity and let

a

and

b

be ideals of

R

.  The ring

R

is a

multiplication ring

if

a

⊆

b

always implies that there exists a

c

of

R

such that

a

=

b

⁢

c

.

Theorem

.

Every

Dedekind domain

is a multiplication ring.  If a multiplication ring has no

zero divisors

, it is a Dedekind domain.

References

1

M. Larsen & P. McCarthy:

Multiplicative

theory of ideals

.  Academic Press. New York (1971).

Title

multiplication ring

Canonical name

MultiplicationRing

Date of creation

2013-03-22 14:27:02

Last modified on

2013-03-22 14:27:02

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

17

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 13A15

Related topic

PruferRing

Related topic

DedekindDomain

Related topic

DivisibilityInRings

Generated on Fri Feb 9 17:22:01 2018 by

LaTeXML
multiplication rule gives inverse ideal

multiplication rule gives inverse ideal

Theorem

.

Let

R

be a

commutative ring

with

non-zero unity

.  If an ideal

(

a

,

b

)

of

R

, with

a

or

b

regular

(

http://planetmath.org/RegularElement

), obeys the

multiplication

rule

(

a

,

b

)

⁢

(

c

,

d

)

=

(

a

⁢

c

,

a

⁢

d

+

b

⁢

c

,

b

⁢

d

)

(1)

with all ideals

(

c

,

d

)

of

R

, then

(

a

,

b

)

is an

invertible ideal

.

Proof.

The rule gives

(

a

,

b

)

2

=

(

a

,

-

b

)

⁢

(

a

,

b

)

=

(

a

2

,

a

⁢

b

-

b

⁢

a

,

b

2

)

=

(

a

2

,

b

2

)

.

Thus the

product

a

⁢

b

may be written in the form

a

⁢

b

=

u

⁢

a

2

+

v

⁢

b

2

,

where

u

and

v

are elements of

R

.  Let’s assume that e.g.

a

is regular.  Then

a

has the

multiplicative inverse

a

-

1

in the

total ring of fractions

R

.  Again applying the rule yields

(

a

,

b

)

⁢

(

v

⁢

a

,

a

-

v

⁢

b

)

⁢

(

a

-

2

)

=

(

v

⁢

a

2

,

a

2

-

v

⁢

a

⁢

b

+

v

⁢

a

⁢

b

,

a

⁢

b

-

v

⁢

b

2

)

⁢

(

a

-

2

)

=

(

v

⁢

a

2

,

a

2

,

u

⁢

a

2

)

⁢

(

a

-

2

)

=

(

v

,

1

,

u

)

=

R

.

Consequently the ideal

(

a

,

b

)

has an inverse ideal (which may be a

fractional ideal

(

http://planetmath.org/FractionalIdealOfCommutativeRing

)); this settles the proof.

Remark.

The rule (1) in the

theorem

may be replaced with the rule

(

a

,

b

)

⁢

(

c

,

d

)

=

(

a

⁢

c

,

(

a

+

b

)

⁢

(

c

+

d

)

,

b

⁢

d

)

(2)

as is seen from the identical

equation

(

a

+

b

)

⁢

(

c

+

d

)

-

a

⁢

c

-

b

⁢

d

=

a

⁢

d

+

b

⁢

c

.

Title

multiplication rule gives inverse ideal

Canonical

name

MultiplicationRuleGivesInverseIdeal

Date of creation

2013-03-22 15:24:16

Last modified on

2013-03-22 15:24:16

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 16D25

Related topic

PruferRing

Related topic

Characterization

Generated on Fri Feb 9 17:22:06 2018 by

LaTeXML
prime ideal factorization is unique

prime ideal factorization is unique

The following

theorem

shows that the decomposition of an (

integral

)

invertible ideal

into its

prime factors

is unique, if it exists. This applies to the

ring of integers

in a

number field

or, more generally, to any

Dedekind domain

, in which every nonzero ideal is

invertible

.

Theorem

.

Let

I

be an invertible ideal in an

integral domain

R

, and that

I

=

𝔭

1

⁢

𝔭

2

⁢

⋯

⁢

𝔭

m

=

𝔮

1

⁢

𝔮

2

⁢

⋯

⁢

𝔮

n

are two factorizations of

I

into a

product

of

prime ideals

. Then

m

=

n

and, up to reordering of the factors,

p

k

=

q

k

(

k

=

1

,

2

,

…

,

n

).

Here we allow the case where

m

or

n

is zero, in which case such an

empty product

is taken to be the full ring

R

.

Proof.

We use

induction

on

m

+

n

. First, the case with

m

+

n

=

0

is

trivial

, so suppose that

m

+

n

>

0

.
As the set of prime ideals

𝔭

k

,

𝔮

k

is partially ordered by

inclusion

, there must be a

minimal element

. After reordering, without loss of generality we may suppose that it is

𝔭

1

. Then

𝔮

1

⁢

𝔮

2

⁢

⋯

⁢

𝔮

n

⊆

𝔭

1

,

so

n

≥

1

. Furthermore, as

𝔭

1

is prime, this implies that

𝔮

k

⊆

𝔭

1

for some

k

. After reordering the factors, we can take

k

=

1

, so that

𝔮

1

⊆

𝔭

1

.

As

𝔭

1

is

minimal

among the prime factors, we have

𝔮

1

=

𝔭

1

. Also,

𝔭

1

is a factor of the invertible ideal

I

and so is itself invertible. Therefore, it can be cancelled from the products,

𝔭

2

⁢

⋯

⁢

𝔭

m

=

𝔮

2

⁢

⋯

⁢

𝔮

n

.

The

induction hypothesis

gives

m

=

n

and, after reordering,

𝔭

k

=

𝔮

k

for

k

=

2

,

…

,

n

.
∎

Title

prime ideal factorization is unique

Canonical

name

PrimeIdealFactorizationIsUnique

Date of creation

2013-03-22 18:34:24

Last modified on

2013-03-22 18:34:24

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

9

Author

gel (22282)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 13F05

Related topic

DedekindDomain

Related topic

FractionalIdeal

Related topic

PrimeIdeal

Related topic

FundamentalTheoremOfIdealTheory

Generated on Fri Feb 9 17:22:10 2018 by

LaTeXML
principal ideal

principal ideal

Let

R

be a ring and let

a

∈

R

. The principal left (resp. right, 2-sided) ideal of

a

is the smallest left (resp. right, 2-sided) ideal of

R

containing the element

a

.

When

R

is a

commutative ring

, the

principal ideal

of

a

is denoted

(

a

)

.

Title

principal ideal

Canonical name

PrincipalIdeal

Date of creation

2013-03-22 11:51:49

Last modified on

2013-03-22 11:51:49

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Definition

Classification

msc 13A15

Classification

msc 16D25

Classification

msc 81-00

Classification

msc 82-00

Classification

msc 83-00

Classification

msc 46L05

Generated on Fri Feb 9 17:22:13 2018 by

LaTeXML
proof of Dedekind-Mertens lemma

proof of Dedekind

-

Mertens lemma

Let

R

be

subring

of the

commutative ring

T

and

f

⁢

(

X

)

=

f

0

+

f

1

⁢

X

+

…

+

f

m

⁢

X

m

and

g

⁢

(

X

)

=

g

0

+

g

1

⁢

X

+

…

+

g

n

⁢

X

n

be arbitrary

polynomials

in

T

⁢

[

X

]

.  We will prove by induction on

n

that the

R

-

submodules

of

T

generated by the
coefficients of the polynomials

f

,

g

, and

f

⁢

g

satisfy

M

f

n

+

1

⁢

M

g

=

M

f

n

⁢

M

f

⁢

g

(1)

where the

product

modules are generated by the products of their

generators

.

The generators of the right hand side of (1) belong obviously to the left hand side,
whence only the containment

M

f

n

+

1

⁢

M

g

⊆

M

f

n

⁢

M

f

⁢

g

(2)

has to be proved.

Firstly, (2) is trivial in the case

n

=

0

.  Let now

n

>

0

.  Define

f

j

:=

0

for

j

<

0

or

j

>

m

and let

G

n

be the

R

-submodule of

T

generated by

g

0

,

g

1

,

…

,

g

n

-

1

.  We have

∑

i

<

n

f

k

-

i

⁢

g

i

=

h

k

-

f

k

-

n

⁢

g

n

∈

M

f

⁢

g

+

g

n

⁢

M

f

where

h

k

is the coefficient of

X

k

of the polynomial

f

⁢

g

, and thus by induction we
can write

M

f

n

⁢

G

n

⊆

M

f

n

-

1

⁢

(

M

f

⁢

g

+

g

n

⁢

M

f

)

⊆

M

f

n

-

1

⁢

M

f

⁢

g

+

M

f

n

⁢

g

n

.

This implies the containment

f

i

⁢

M

f

n

⁢

G

n

⊆

M

f

n

⁢

M

f

⁢

g

+

M

f

n

⁢

f

i

⁢

g

n

for every

i

.  In

addition

, we have

f

i

⁢

g

n

∈

M

f

⁢

g

+

f

i

+

1

⁢

G

n

+

M

i

+

2

⁢

G

n

+

…

+

f

n

⁢

G

n

,

whence

f

i

⁢

M

f

n

⁢

G

n

⊆

M

f

n

⁢

M

f

⁢

g

+

f

i

+

1

⁢

M

f

n

⁢

G

n

+

…

+

f

n

⁢

M

f

n

⁢

G

n

.

From this we infer that

f

i

⁢

M

f

n

⁢

G

n

⊆

M

f

n

⁢

M

f

⁢

g

is true for each

i

=

m

,

m

-

1

,

…

,

0

.  Thus also (2) is true.

References

1

J. Pahikkala:

“Some formulae for multiplying and inverting ideals”.  –

Ann. Univ. Turkuensis

183

(A) (1982).

2

J. Arnold & R. Gilmer:

“On the

contents of polynomials

”.  –

Proc. Amer. Math. Soc.

24

(1970).

3

T. Coquand

: “

A direct proof of
Dedekind–Mertens lemma

”. University of Gothenburg 2006. (Available

http://www.cse.chalmers.se/ coquand/mertens.pdf

here.)

Title

proof of Dedekind

-

Mertens lemma

Canonical name

ProofOfDedekindMertensLemma

Date of creation

2013-12-15 20:29:20

Last modified on

2013-12-15 20:29:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Proof

Classification

msc 13A15

Classification

msc 13M10

Classification

msc 16D10

Classification

msc 16D25

Generated on Fri Feb 9 17:22:23 2018 by

LaTeXML
proof of finite extensions of Dedekind domains are Dedekind

proof of finite extensions of Dedekind domains are Dedekind

Let

R

be a

Dedekind domain

with

field of fractions

K

. If

L

/

K

is a

finite extension

of fields and

A

is the

integral closure

of

R

in

L

, then we show that

A

is also a Dedekind domain.

We procede by splitting the proof up into the

separable

and

purely inseparable

cases. Letting

F

consist of all elements of

L

which are separable over

K

, then

F

/

K

is a

separable extension

and

L

/

F

is a

purely inseparable extension

.

First, the integral closure

B

of

R

in

F

is a Dedekind domain (see

proof of finite separable extensions of Dedekind domains are Dedekind

). Then, as

A

is

integrally closed

and contains

B

, it is equal to the integral closure of

B

in

L

and, therefore, is a Dedekind domain (see

proof of finite inseparable extensions of Dedekind domains are Dedekind

).

Title

proof of finite extensions of Dedekind domains are Dedekind

Canonical name

ProofOfFiniteExtensionsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:44

Last modified on

2013-03-22 18:35:44

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 13A15

Classification

msc 13F05

Generated on Fri Feb 9 17:22:26 2018 by

LaTeXML
proof of finite inseparable extensions of Dedekind domains are Dedekind

proof of finite inseparable extensions of Dedekind domains are Dedekind

Let

R

be a

Dedekind domain

with

field of fractions

K

and

L

/

K

be a

field extension

. We suppose that

K

has

characteristic

(

http://planetmath.org/characteristic

)

p

>

0

and that there is a

q

=

p

r

such that

x

q

∈

K

for all

x

∈

L

. In particular, this is satisfied if it is a

purely inseparable

and

finite extension

.

We show that the

integral closure

A

of

R

in

L

is a Dedekind domain.

We cannot apply the same method of proof as for the

proof of finite separable extensions of Dedekind domains are Dedekind

, because here

A

does not have to be

finitely generated

as an

R

-module.

We use the characterization of Dedekind domains as

integral domains

in which all nonzero ideals are

invertible

(see proof that a domain is Dedekind if its ideals are invertible).
Note that for any

x

∈

A

,

x

q

is in

K

and is integral over

R

so, by integral closure,

x

q

∈

R

.

So, let

𝔞

be a nonzero ideal in

A

, and let

𝔟

be the ideal of

R

generated by terms of the form

a

q

for

a

∈

𝔞

,

𝔟

=

(

a

q

:

a

∈

𝔞

)

R

.

Then, as

R

is a Dedekind domain, there is a

fractional ideal

𝔟

-

1

of

R

such that

𝔟

⁢

𝔟

-

1

=

R

, and write

𝔟

A

-

1

for the fractional ideal of

A

generated by

𝔟

-

1

. Then,

1

∈

R

=

𝔟

⁢

𝔟

-

1

⊆

𝔞

q

⁢

𝔟

A

-

1

.

(1)

On the other hand, if

a

1

,

…

,

a

q

∈

𝔞

and

b

∈

𝔟

-

1

then

(

a

1

⁢

⋯

⁢

a

q

⁢

b

)

q

=

(

a

1

q

⁢

b

)

⁢

⋯

⁢

(

a

q

q

⁢

b

)

∈

R

,

so

a

1

⁢

⋯

⁢

a

q

⁢

b

is integral over

R

and is in

A

. Therefore,

𝔞

q

⁢

𝔟

A

-

1

⊆

A

. Combining with (

1

) gives

𝔞

q

⁢

𝔟

A

-

1

=

A

, so

𝔞

is invertible with

inverse

𝔞

q

-

1

⁢

𝔟

A

-

1

.

Title

proof of finite inseparable extensions of Dedekind domains are Dedekind

Canonical name

ProofOfFiniteInseparableExtensionsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:42

Last modified on

2013-03-22 18:35:42

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 13A15

Classification

msc 13F05

Generated on Fri Feb 9 17:22:29 2018 by

LaTeXML
the set of prime ideals of a commutative ring with identity

the set of prime ideals of a commutative ring with identity

The set of all

prime ideals

of a

commutative ring

R

with

identity

is called the spectrum of the ring (

http://planetmath.org/PrimeSpectrum

) and is denoted

Spec

⁢

(

R

)

.

Title

the set of prime ideals of a commutative ring with identity

Canonical name

TheSetOfPrimeIdealsOfACommutativeRingWithIdentity

Date of creation

2013-03-22 14:01:47

Last modified on

2013-03-22 14:01:47

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

7

Author

bwebste (988)

Entry type

Definition

Classification

msc 13A15

Generated on Fri Feb 9 17:22:41 2018 by

LaTeXML
topics on ideals

topics on ideals

Below is a list of main

concepts

in the part of ring theory concerning ideals:

•

ideal (

http://planetmath.org/Ideal

)

•

zero ideal

•

proper ideal

•

ideal of elements with finite order

•

modular ideal

•

regular ideal

•

nil ideal

•

nilpotent ideal

•

primitive ideal

•

primary ideal

•

semiprime ideal

•

prime ideal

,

minimal prime ideal

•

characterization of prime ideals

•

large ideal

•

maximal ideal

•

maximal ideal is prime

•

comaximal ideal

•

principal ideal

•

finitely generated

ideal (

http://planetmath.org/EntriesOnFinitelyGeneratedIdeals

)

•

ideal generated by

•

sum of ideals

•

product of ideals

•

quotient of ideals

•

ideal

multiplication

laws

•

product

of finitely generated ideals

•

ideal included in union of prime ideals

•

ideals

contained

in a union of ideals

•

dense ideals

/subsets of a ring

•

cancellation ideal

•

radical ideal

•

fractional ideal of commutative ring

•

integral ideal

•

invertible ideal

•

invertibility of regularly generated ideal

•

inverse ideal

•

generators of inverse ideal

•

ideal

generators

in Prüfer ring

•

ideal inverting in Prüfer ring

•

fractional ideal

(

integral domain

)

•

image ideal of divisor

•

ideals in a

Dedekind domain

Title

topics on ideals

Canonical name

TopicsOnIdeals

Date of creation

2013-03-22 18:03:12

Last modified on

2013-03-22 18:03:12

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

16

Author

pahio (2872)

Entry type

Definition

Classification

msc 13A15

Classification

msc 16D25

Related topic

EntriesOnFinitelyGeneratedIdeals

Generated on Fri Feb 9 17:22:44 2018 by

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continuation of exponent

continuation of exponent

Theorem.

Let

K

/

k

be a

finite field extension

and

ν

an

exponent valuation

of the

extension field

K

.  Then there exists one and only one

positive integer

e

such that the function

(

1

)

ν

0

⁢

(

x

)

:=

{

∞

when

⁢

x

=

0

,

ν

⁢

(

x

)

e

⁢

when

⁢

x

≠

0

,

defined in the

base field

k

, is an exponent (

http://planetmath.org/ExponentValuation

) of

k

.

Proof.

The exponent

ν

of

K

attains in the set

k

∖

{

0

}

also non-zero values; otherwise

k

would be included in

𝒪

ν

, the ring of the exponent

ν

.  Since any element

ξ

of

K

are integral over

k

, it would then be also integral over

𝒪

ν

, which is

integrally closed

in its

quotient field

K

(see theorem 1 in

ring of exponent

); the situation would mean that

ξ

∈

𝒪

ν

and thus the whole

K

would be contained in

𝒪

ν

.  This is impossible, because an exponent of

K

attains also

negative

values.  So we infer that

ν

does not vanish in the whole

k

∖

{

0

}

.  Furthermore,

ν

attains in

k

∖

{

0

}

both negative and positive values, since

ν

⁢

(

a

)

+

ν

⁢

(

a

-

1

)

=

ν

⁢

(

a

⁢

a

-

1

)

=

ν

⁢

(

1

)

=

0

.

Let

p

be such an element of

k

on which

ν

attains as its value the least possible positive integer

e

in the field

k

and let

a

be an arbitrary non-zero element of

k

.  If

ν

(

a

)

=

m

=

q

e

+

r

(

q

,

r

∈

ℤ

,

0

≦

r

<

e

)

,

then

ν

⁢

(

a

⁢

p

-

q

)

=

m

-

q

⁢

e

=

r

,  and thus

r

=

0

on grounds of the choice of

p

.  This means that

ν

⁢

(

a

)

is always

divisible

by

e

, i.e. that the values of the function

ν

0

in

k

∖

{

0

}

are

integers

.  Because

ν

0

⁢

(

p

)

=

1

and

ν

0

⁢

(

p

l

)

=

l

,  the function attains in

k

every integer value.  Also the conditions

ν

0

⁢

(

a

⁢

b

)

=

ν

0

⁢

(

a

)

+

ν

0

⁢

(

b

)

,

ν

0

⁢

(

a

+

b

)

≧

min

⁡

{

ν

0

⁢

(

a

)

,

ν

0

⁢

(

b

)

}

are in , whence

ν

0

is an exponent of the field

k

.

Definition.

Let

K

/

k

be a finite field extension.  If the exponent

ν

0

of

k

is tied with the exponent

ν

of

K

via the condition (1), one says that

ν

induces

ν

0

to

k

and that

ν

is the

continuation

of

ν

0

to

K

.  The positive integer

e

, uniquely determined by (1), is the

ramification index

of

ν

with respect to

ν

0

(or with respect to the

subfield

k

).

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

continuation of exponent

Canonical name

ContinuationOfExponent

Date of creation

2013-03-22 17:59:49

Last modified on

2013-03-22 17:59:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Definition

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 11R99

Classification

msc 13F30

Synonym

prolongation of exponent

Defines

induce

Defines

continuation

Defines

continuation of the exponent

Defines

ramification index of the exponent

Generated on Fri Feb 9 17:22:49 2018 by

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divisor theory and exponent valuations

divisor theory and exponent valuations

A

divisor theory

𝒪

*

→

𝔇

of an

integral domain

𝒪

determines via its

prime divisors

a certain set

N

of

exponent valuations

on the

quotient field

of

𝒪

.  Assume to be known this set of

exponents

(

http://planetmath.org/ExponentValuation2

)

ν

𝔭

corresponding the prime divisors

𝔭

.  There is a

bijective

correspondence between the elements of

N

and of the set of all prime divisors.  The set of the prime divisors determines completely the of the

free monoid

𝔇

of all

divisors

in question. The

homomorphism

𝒪

*

→

𝔇

is then defined by the condition

α

↦

∏

i

𝔭

𝔦

ν

𝔭

i

⁢

(

α

)

=

(

α

)

,

Ã

⁢



⁢

Â

⁢

´

(1)

since for any

element

α

of

𝒪

*

there exists only a finite number of exponents

ν

𝔭

i

which do not vanish on

α

(corresponding the different prime divisor factors (

http://planetmath.org/DivisibilityInRings

) of the principal divisor

(

α

)

).

One can take the

concept

of exponent as

foundation

for divisor theory:

Theorem

.

Let

𝒪

be an integral domain with quotient field

K

and

N

a given set of exponents (

http://planetmath.org/ExponentValuation2

) of

K

.  The exponents in

N

determine, as in (1), a divisor theory of

𝒪

iff the following three conditions are in :

•

For every

α

∈

𝒪

there is at most a finite number of exponents

ν

∈

N

such that

ν

⁢

(

α

)

≠

0

.

•

An element

α

∈

K

belongs to

𝒪

if and only if

ν

⁢

(

α

)

≧

0

for each

ν

∈

N

.

•

For any

finite set

ν

1

,

…

,

ν

n

of distinct exponents in

N

and for the arbitrary set

k

1

,

…

,

k

n

of non-negative

integers

, there exists an element

α

of

𝒪

such that

ν

1

⁢

(

α

)

=

k

1

,

…

,

ν

n

⁢

(

α

)

=

k

n

.

For the proof of the theorem, we mention only how to

construct

the divisors when we have the exponent set

N

fulfilling the three conditions of the theorem.  We choose a

commutative monoid

𝔇

that allows

unique prime

factorisation and that may be mapped bijectively onto

N

.  The exponent in

N

which corresponds to arbitrary

prime element

𝔭

is denoted by

ν

𝔭

.  Then we obtain the homomorphism

α

↦

∏

ν

𝔭

ν

𝔭

⁢

(

α

)

:=

(

α

)

which can be seen to

satisfy

all required properties for a divisor theory

𝒪

*

→

𝔇

.

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

divisor theory and exponent valuations

Canonical name

DivisorTheoryAndExponentValuations

Date of creation

2013-03-22 17:59:34

Last modified on

2013-03-22 17:59:34

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Topic

Classification

msc 13A18

Classification

msc 12J20

Classification

msc 13A05

Synonym

divisors and exponents

Related topic

ExponentValuation2

Related topic

ImplicationsOfHavingDivisorTheory

Generated on Fri Feb 9 17:22:54 2018 by

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divisor theory in finite extension

divisor theory in finite extension

Theorem.

Let the

integral domain

𝒪

, with the

quotient field

k

, have the

divisor theory

𝒪

*

→

𝔇

, determined (see

divisors and exponents

) by the exponent (

http://planetmath.org/ExponentValuation2

) system

N

0

of

k

.  If

K

/

k

is a

finite extension

, then the exponent system

N

, consisting of the

continuations

(

http://planetmath.org/ContinuationOfExponent

) of all exponents in

N

0

to the field

K

, determines the divisor theory of the

integral closure

of

𝒪

in

K

.

Corollary.

In the

ring of integers

𝒪

of any

algebraic number field

ℚ

⁢

(

ϑ

)

, there is a divisor theory

𝒪

*

→

𝔇

, determined by the set of all

exponent valuations

of

ℚ

⁢

(

ϑ

)

.

References

1

S. Borewicz & I. Safarevic

:

Zahlentheorie

.  Birkhäuser Verlag. Basel und Stuttgart (1966).

Title

divisor theory in finite extension

Canonical name

DivisorTheoryInFiniteExtension

Date of creation

2013-03-22 17:59:59

Last modified on

2013-03-22 17:59:59

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A18

Classification

msc 13F05

Classification

msc 12J20

Classification

msc 13A05

Classification

msc 11A51

Related topic

FiniteExtensionsOfDedekindDomainsAreDedekind

Generated on Fri Feb 9 17:22:56 2018 by

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