aarjaneiro/planetmath huggingface dataset:
proof of Hilbert basis theorem

proof of Hilbert basis theorem

Let

R

be a

noetherian ring

and let

f

⁢

(

x

)

=

a

n

⁢

x

n

+

a

n

-

1

⁢

x

n

-

1

+

…

+

a

1

⁢

x

+

a

0

∈

R

⁢

[

x

]

with

a

n

≠

0

. Then
call

a

n

the

initial

coefficient

of

f

.

Let

I

be an ideal in

R

⁢

[

x

]

. We will show

I

is

finitely
generated

, so that

R

⁢

[

x

]

is

noetherian

. Now let

f

0

be a

polynomial

of least degree in

I

, and if

f

0

,

f

1

,

…

,

f

k

have been chosen then choose

f

k

+

1

from

I

∖

(

f

0

,

f

1

,

…

,

f

k

)

of minimal degree. Continuing inductively
gives a

sequence

(

f

k

)

of elements of

I

.

Let

a

k

be the initial coefficient of

f

k

, and
consider the ideal

J

=

(

a

1

,

a

2

,

a

3

,

…

)

of initial
coefficients. Since

R

is noetherian,

J

=

(

a

0

,

…

,

a

N

)

for
some

N

.

Then

I

=

(

f

0

,

f

1

,

…

,

f

N

)

. For if not then

f

N

+

1

∈

I

∖

(

f

0

,

f

1

,

…

,

f

N

)

, and

a

N

+

1

=

∑

k

=

0

N

u

k

⁢

a

k

for some

u

1

,

u

2

,

…

,

u

N

∈

R

. Let

g

⁢

(

x

)

=

∑

k

=

0

N

u

k

⁢

f

k

⁢

x

ν

k

where

ν

k

=

deg

⁡

(

f

N

+

1

)

-

deg

⁡

(

f

k

)

.

Then

deg

⁡

(

f

N

+

1

-

g

)

<

deg

⁡

(

f

N

+

1

)

, and

f

N

+

1

-

g

∈

I

and

f

N

+

1

-

g

∉

(

f

0

,

f

1

,

…

,

f

N

)

. But this contradicts
minimality of

deg

⁡

(

f

N

+

1

)

.

Hence,

R

⁢

[

x

]

is noetherian.

□

Title

proof of Hilbert basis theorem

Canonical name

ProofOfHilbertBasisTheorem

Date of creation

2013-03-22 12:59:27

Last modified on

2013-03-22 12:59:27

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

6

Author

bwebste (988)

Entry type

Proof

Classification

msc 13E05

Generated on Fri Feb 9 17:30:17 2018 by

LaTeXML
free modules over a ring which is not a PID

free modules over a ring which is not a PID

Let

R

be a

unital ring

. In the following modules will be

left modules

.

We will say that

R

has

the free

submodule

property

if for any

free module

F

over

R

and any submodule

F

′

⊆

F

we have that

F

′

is also free. It is well known, that if

R

is a PID, then

R

has the free submodule property. One can ask whether the

converse

is also true? We will try to answer this question.

Proposition

.

If

R

is a

commutative ring

, which is not a PID, then

R

does not have the free submodule property.

Proof.

Assume that

R

is not a PID. Then there are two possibilities: either

R

is not a domain or there is an ideal

I

⊆

R

which is not principal. Assume that

R

is not a domain and let

a

,

b

∈

R

be two

zero divisors

, i.e.

a

≠

0

,

b

≠

0

and

a

⋅

b

=

0

. Let

(

b

)

⊆

R

be an

ideal generated by

b

. Then obviously

(

b

)

is a submodule of

R

(regarded as a

R

-module). Assume that

(

b

)

is free. In particular there exists

m

∈

(

b

)

,

m

≠

0

such that

r

⋅

m

=

0

if and only if

r

=

0

. But

m

is of the form

λ

⋅

b

and because

R

is

commutative

we have

a

⋅

m

=

a

⋅

(

λ

⋅

b

)

=

λ

⋅

(

a

⋅

b

)

=

0

.

Contradiction

, because

a

≠

0

. Thus

(

b

)

is not free although

(

b

)

is a submodule of a free module

R

.

Assume now that there is an ideal

I

⊆

R

which is not principal and assume that

I

is free as a

R

-module. Since

I

is not principal, then there exist

a

,

b

∈

I

such that

{

a

,

b

}

is

linearly independent

. On the other hand

a

,

b

∈

R

and

1

is a free

generator

of

R

. Thus

{

1

,

a

}

is linearly dependent, so

λ

⋅

1

+

α

⋅

a

=

0

for some nonzero

λ

,

α

∈

R

(note that in this case both

λ

,

α

are nonzero, more precisely

λ

=

a

and

α

=

-

1

). Multiply the

equation

by

b

. Thus we have

λ

⋅

b

+

(

α

⋅

b

)

⋅

a

=

0

.

Note that here we used commutativity of

R

. Since

{

a

,

b

}

is linearly independend (in

I

), then the last equation implies that

λ

=

0

. Contradiction.

□

Corollary.

Commutative ring is a PID if and only if it has the free submodule property.

Title

free modules over a ring which is not a PID

Canonical name

FreeModulesOverARingWhichIsNotAPID

Date of creation

2013-03-22 18:50:08

Last modified on

2013-03-22 18:50:08

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

5

Author

joking (16130)

Entry type

Definition

Classification

msc 13E15

Generated on Fri Feb 9 17:30:23 2018 by

LaTeXML
Dedekind domains with finitely many primes are PIDs

Dedekind domains with finitely many primes are PIDs

A

commutative ring

in which there are only finitely many

maximal ideals

is known as a

semi-local ring

. For such rings, the property of being a

Dedekind domain

and of being a

principal ideal domain

coincide.

Theorem

.

A Dedekind domain in which there are only finitely many

prime ideals

is a principal ideal domain.

This result is sometimes proven using the

chinese remainder theorem

or, alternatively, it follows directly from the fact that

invertible ideals in semi-local rings

are principal.

Suppose that

R

is a Dedekind domain such as the

ring of algebraic integers

in a

number field

. Although there are infinitely many prime ideals in such a ring, we can use the result that

localizations of Dedekind domains are Dedekind

and apply the above theorem to

localizations

of

R

.

In particular, if

𝔭

is a nonzero prime ideal, then

R

𝔭

≡

(

R

∖

𝔭

)

-

1

⁢

R

is a Dedekind domain with a unique nonzero prime ideal, so the theorem shows that it is a principal ideal domain.

Title

Dedekind domains with finitely many primes are PIDs

Canonical name

DedekindDomainsWithFinitelyManyPrimesArePIDs

Date of creation

2013-03-22 18:35:18

Last modified on

2013-03-22 18:35:18

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Theorem

Classification

msc 13F05

Classification

msc 11R04

Related topic

DivisorTheory

Generated on Fri Feb 9 17:30:25 2018 by

LaTeXML
finitely generated torsion-free modules over Prüfer domains

finitely generated torsion-free modules over Prüfer domains

Theorem

.

Let

M

be a

finitely generated

torsion-free module over a Prüfer domain

R

. Then,

M

is

isomorphic

to a

direct sum

(

http://planetmath.org/DirectSum

)

M

≅

𝔞

1

⊕

⋯

⊕

𝔞

n

of finitely generated ideals

a

1

,

…

,

a

n

.

As

invertible ideals are projective

and direct sums of

projective modules

are themselves projective, this

theorem

shows that

M

is also a projective module.

Conversely

, if every finitely generated torsion-free module over an

integral domain

R

is projective then, in particular, every finitely generated nonzero ideal of

R

will be projective and hence

invertible

. So, we get the following

characterization

of Prüfer domains.

Corollary

.

An integral domain

R

is Prüfer if and only if every finitely generated

torsion-free

R

-module is projective (

http://planetmath.org/ProjectiveModule

).

Title

finitely generated torsion-free modules over Prüfer domains

Canonical name

FinitelyGeneratedTorsionfreeModulesOverPruferDomains

Date of creation

2013-03-22 18:36:11

Last modified on

2013-03-22 18:36:11

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Theorem

Classification

msc 13F05

Classification

msc 13C10

Related topic

EquivalentCharacterizationsOfDedekindDomains

Generated on Fri Feb 9 17:30:29 2018 by

LaTeXML
Jaffard ring

Jaffard ring

Let

dim

denote

Krull dimension

. A

Jaffard

ring is a ring

A

for which

dim

⁡

(

A

⁢

[

x

]

)

=

dim

⁡

(

A

)

+

1

(compare to the bound on the Krull dimension of

polynomial rings

). Such a ring is said to be

Jaffardian

.

Since this condition holds for

Noetherian rings

, every Noetherian ring is Jaffardian.

Examples of rings

that are not Jaffardian are thus relatively difficult to come by, since we are already forced to search exclusively in the realm of non-Noetherian rings. The first example of a non-Jaffardian ring seems to have been found by A. Seidenberg

[

Seid

]

: the

subring

of

ℚ

¯

⁢

[

[

T

]

]

consisting of power series whose

constant term

is

rational

.

A

Jaffard domain

is a Jaffard ring which is also an

integral domain

.

References

Seid

A. Seidenberg,

A note on the dimension theory of rings.

Pacific J. of Mathematics, Volume 3 (1953), 505-512.

Title

Jaffard ring

Canonical name

JaffardRing

Date of creation

2013-03-22 15:22:15

Last modified on

2013-03-22 15:22:15

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

4

Author

mathcam (2727)

Entry type

Definition

Classification

msc 13F05

Defines

Jaffard domain

Defines

Jaffard

Generated on Fri Feb 9 17:30:31 2018 by

LaTeXML
proof of Dedekind domains with finitely many primes are PIDs

proof of Dedekind domains with finitely many primes are PIDs

Proof.

Let

𝔭

1

,

…

,

𝔭

k

be all the primes of a

Dedekind domain

R

. If

I

is any ideal of

R

, then by the

Weak Approximation Theorem

we can choose

x

∈

R

such that

ν

𝔭

i

⁢

(

(

x

)

)

=

ν

𝔭

i

⁢

(

I

)

for all

i

(where

ν

𝔭

is the

𝔭

-adic

valuation

). But since

R

is Dedekind, ideals have

unique factorization

; since

(

x

)

and

I

have identical factorizations, we must have

(

x

)

=

I

and

I

is principal.
∎

Title

proof of Dedekind domains with finitely many primes are PIDs

Canonical name

ProofOfDedekindDomainsWithFinitelyManyPrimesArePIDs

Date of creation

2013-03-22 18:35:24

Last modified on

2013-03-22 18:35:24

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Proof

Classification

msc 13F05

Classification

msc 11R04

Generated on Fri Feb 9 17:30:33 2018 by

LaTeXML
proof of finite separable extensions of Dedekind domains are Dedekind

proof of finite separable extensions of Dedekind domains are Dedekind

Let

R

be a

Dedekind domain

with

field of fractions

K

and

L

/

K

be a finite (

http://planetmath.org/FiniteExtension

)

separable extension

of fields. We show that the

integral closure

A

of

R

in

L

is also a Dedekind domain. That is,

A

is

Noetherian

(

http://planetmath.org/Noetherian

),

integrally closed

and every nonzero

prime ideal

is

maximal

(

http://planetmath.org/MaximalIdeal

).

First, as integral closures are themselves integrally closed,

A

is integrally closed. Second, as

integral closures in separable extensions are finitely generated

,

A

is

finitely generated

as an

R

-module. Then, any ideal

𝔞

of

A

is a

submodule

of

A

, so is finitely generated as an

R

-module and therefore as an

A

-module. So,

A

is Noetherian.

It only remains to show that a nonzero prime ideal

𝔭

of

A

is maximal. Choosing any

p

∈

𝔭

∖

{

0

}

there is a nonzero

polynomial

f

=

∑

k

=

0

n

c

k

⁢

X

k

for

c

k

∈

R

,

c

0

≠

0

and such that

f

⁢

(

p

)

=

0

. Then

c

0

=

-

p

⁢

∑

k

=

1

n

c

k

⁢

p

k

-

1

∈

𝔭

∩

R

,

so

𝔭

∩

R

is a nonzero prime ideal in

R

and is therefore a

maximal ideal

. So,

R

/

(

𝔭

∩

R

)

→

A

/

𝔭

gives an

algebraic extension

of the field

R

/

(

𝔭

∩

R

)

to the

integral domain

A

/

𝔭

. Therefore,

A

/

𝔭

is a field (see a

condition of algebraic extension

) and

𝔭

is a maximal ideal.

Title

proof of finite separable extensions of Dedekind domains are Dedekind

Canonical name

ProofOfFiniteSeparableExtensionsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:36

Last modified on

2013-03-22 18:35:36

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 13F05

Classification

msc 13A15

Generated on Fri Feb 9 17:30:35 2018 by

LaTeXML
proof of finitely generated torsion-free modules over Prüfer domains

proof of finitely generated torsion-free modules over Prüfer domains

Let

M

be a

finitely generated

torsion-free module over a Prüfer domain

R

with

field of fractions

k

. We show that

M

is

isomorphic

to a

direct sum

(

http://planetmath.org/DirectSum

) of finitely generated ideals in

R

.

We shall write

k

⊗

M

for the

vector space

over

k

generated by

M

. This is just the

localization

(

http://planetmath.org/LocalizationOfAModule

) of

M

at

R

∖

{

0

}

and, as

M

is

torsion-free

, the natural map

M

→

k

⊗

M

is

one-to-one

and we can regard

M

as a subset of

k

⊗

M

.

As

M

is finitely generated, the vector space

k

⊗

M

will finite dimensional (

http://planetmath.org/Dimension2

), and we use

induction

on its

dimension

n

.
Supposing that

n

>

0

, choose any basis

e

1

,

…

,

e

n

and define the

linear map

f

:

k

⊗

M

→

k

by

projection

(

http://planetmath.org/Projection

) onto the first

component

,

f

⁢

(

x

1

⁢

e

1

+

⋯

+

x

n

⁢

e

n

)

=

x

1

.

Restricting to

M

, this gives a nonzero map

M

→

k

. Furthermore, as

M

is finitely generated,

f

⁢

(

M

)

will be a finitely generated

fractional ideal

in

k

. Choosing any nonzero

c

∈

R

such that

𝔞

≡

c

⁢

f

⁢

(

M

)

⊆

R

,

g

:

M

→

𝔞

,

g

⁢

(

u

)

=

c

⁢

f

⁢

(

u

)

defines a homorphism from

M

onto the nonzero and finitely generated ideal

𝔞

. As

R

is Prüfer and

invertible ideals are projective

,

g

has a right-inverse

h

:

𝔞

→

M

.
Then

h

has the left-inverse

g

and is one-to-one, so defines an

isomorphism

between

𝔞

and its image (

http://planetmath.org/ImageOfALinearTransformation

). We decompose

M

as the direct sum of the kernel of

g

and the image of

h

,

M

=

ker

⁡

(

g

)

⊕

Im

⁡

(

h

)

≅

ker

⁡

(

g

)

⊕

𝔞

.

Projection from the finitely generated module

M

onto

ker

⁡

(

g

)

shows that it is finitely generated and,

dim

⁡

(

k

⊗

ker

⁡

(

g

)

)

=

dim

⁡

(

k

⊗

M

)

-

dim

⁡

(

k

⊗

𝔞

)

=

n

-

1

.

So, the result follows from applying the

induction hypothesis

to

ker

⁡

(

g

)

.

Title

proof of finitely generated torsion-free modules over Prüfer domains

Canonical name

ProofOfFinitelyGeneratedTorsionfreeModulesOverPruferDomains

Date of creation

2013-03-22 18:36:14

Last modified on

2013-03-22 18:36:14

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 13F05

Classification

msc 13C10

Generated on Fri Feb 9 17:30:37 2018 by

LaTeXML
proof that a domain is Dedekind if its ideals are products of maximals

proof that a domain is Dedekind if its ideals are products of maximals

Let

R

be an

integral domain

. We show that it is a Dedekind domain if and only if every nonzero

proper ideal

can be expressed as a

product

of

maximal ideals

.
To do this, we make use of the

characterization

of Dedekind domains as integral domains in which every nonzero

integral ideal

is

invertible

(proof that a domain is Dedekind if its ideals are invertible).

First, let us suppose that every nonzero proper ideal in

R

is a product of maximal ideals.
Let

𝔪

be a maximal ideal and choose a nonzero

x

∈

𝔪

. Then, by

assumption

,

(

x

)

=

𝔪

1

⁢

⋯

⁢

𝔪

𝔫

for some

n

≥

0

and maximal ideals

𝔪

k

. As

(

x

)

is a

principal ideal

, each of the factors

𝔪

k

is invertible.
Also,

𝔪

1

⁢

⋯

⁢

𝔪

𝔫

⊆

𝔪

.

As

𝔪

is prime, this gives

𝔪

k

⊆

𝔪

for some

k

. However,

𝔪

k

is maximal so must equal

𝔪

, showing that

𝔪

is indeed invertible.
Then, every nonzero proper ideal is a product of maximal, and hence invertible, ideals and so is invertible, and it follows that

R

is Dedekind.

We now show the reverse direction, so suppose that

R

is Dedekind.

Proof by contradiction

will be used to show that every nonzero ideal is a product of maximals, so suppose that this is not the case.
Then, as

R

is defined to be

Noetherian

(

http://planetmath.org/Noetherian

), there is an ideal

𝔞

maximal (

http://planetmath.org/MaximalElement

) (w.r.t. the

partial order

of

set inclusion

) among those proper ideals which are not a product of maximal ideals.
Then

𝔞

cannot be a maximal ideal itself, so is strictly

contained

in a maximal ideal

𝔪

and, as

𝔪

is invertible, we can write

𝔞

=

𝔪

⁢

𝔟

for an ideal

𝔟

.

Therefore

𝔞

⊆

𝔟

and we cannot have

equality

, otherwise cancelling

𝔞

from

𝔞

=

𝔪

⁢

𝔞

would give

𝔪

=

R

. So,

𝔟

is strictly larger than

𝔞

and, by the choice of

𝔞

, is therefore a product of maximal ideals. Finally,

𝔞

=

𝔪

⁢

𝔟

is then also a product of maximal ideals.

Title

proof that a domain is Dedekind if its ideals are products of maximals

Canonical name

ProofThatADomainIsDedekindIfItsIdealsAreProductsOfMaximals

Date of creation

2013-03-22 18:35:04

Last modified on

2013-03-22 18:35:04

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Proof

Classification

msc 13F05

Classification

msc 13A15

Related topic

DedekindDomain

Related topic

MaximalIdeal

Related topic

FractionalIdeal

Generated on Fri Feb 9 17:30:39 2018 by

LaTeXML
proof that a Noetherian domain is Dedekind if it is locally a PID

proof that a Noetherian domain is Dedekind if it is locally a PID

We show that for a

Noetherian

(

http://planetmath.org/Noetherian

) domain

R

with

field of fractions

k

, the following are

equivalent

.

1.

R

is Dedekind. That is, it is

integrally closed

and every nonzero

prime ideal

is maximal.

2.

for every

maximal ideal

𝔪

,

R

𝔪

≡

{

s

-

1

⁢

x

:

s

∈

R

∖

𝔪

,

x

∈

R

}

is a

principal ideal domain

.

For a given maximal ideal

𝔪

and ideal

𝔞

of

R

, we shall write

𝔞

¯

for the

ideal generated by

𝔞

in

R

𝔪

, which consists of the elements of the form

s

-

1

⁢

a

for

a

∈

𝔞

and

s

∈

R

∖

𝔪

. It is then easily seen that

𝔭

↦

𝔭

¯

gives a

bijection

between the prime ideals of

R

contained

in

𝔪

and the prime ideals of

R

𝔪

, with

inverse

𝔭

↦

R

∩

𝔭

. In particular

𝔪

¯

is the unique maximal ideal of

R

𝔪

, which is therefore a

local ring

.

Now suppose that

R

is Dedekind, then the

localization

R

𝔪

will be a

Dedekind domain

(

localizations of Dedekind domains are Dedekind

) with a unique maximal ideal, so it is a principal ideal domain (

Dedekind domains with finitely many primes are PIDs

).

Only the

converse

remains to be shown, so suppose that

R

is a Noetherian domain such that

R

𝔪

is a principal ideal domain for every maximal ideal

𝔪

. In particular,

R

𝔪

is integrally closed and every nonzero prime ideal is maximal, so it contains a unique nonzero prime ideal

𝔪

¯

.

We start by showing that every nonzero prime ideal

𝔭

of

R

is maximal. Choose a maximal ideal

𝔪

containing

𝔭

. Then,

𝔭

¯

is a nonzero prime ideal, so

𝔭

¯

=

𝔪

¯

and therefore

𝔭

=

𝔪

is maximal.

We finally show that

R

is integrally closed. So, choose any

x

integral over

R

and lying in its field of fractions. Let

𝔞

be the ideal

𝔞

=

{

a

∈

R

:

a

⁢

x

∈

R

}

.

We use

proof by contradiction

to show that

𝔞

is the whole of

R

. So, supposing that this is not the case, there exists a maximal ideal

𝔪

containing

𝔞

. Then

x

will be integral over the integrally closed ring

R

𝔪

and therefore

x

∈

R

𝔪

. So,

x

=

s

-

1

⁢

y

for some

s

∈

R

∖

𝔪

and

y

∈

R

. Then,

s

⁢

x

=

y

∈

R

so

s

∈

𝔞

⊆

𝔪

, which is the required

contradiction

. Therefore,

𝔞

=

R

and, in particular,

1

∈

𝔞

and

x

=

1

⁢

x

∈

R

, showing that

R

is integrally closed.

Title

proof that a Noetherian domain is Dedekind if it is locally a PID

Canonical name

ProofThatANoetherianDomainIsDedekindIfItIsLocallyAPID

Date of creation

2013-03-22 18:35:27

Last modified on

2013-03-22 18:35:27

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 13F05

Classification

msc 13A15

Generated on Fri Feb 9 17:30:41 2018 by

LaTeXML
weak approximation theorem

weak approximation theorem

The

weak approximation theorem

allows selection, in a

Dedekind ring

, of an

element

having specific

valuations

at a specific

finite set

of primes, and nonnegative valuations at all other primes. It is essentially a

generalization

of the

Chinese Remainder theorem

, as is evident from its proof.

Theorem 1

(Weak ).

Let

A

be a

Dedekind domain

with

fraction field

K

. Then for any finite set

p

1

,

…

,

p

k

of primes of

A

and

integers

a

1

,

…

,

a

k

, there is

x

∈

K

⋆

such that

ν

p

i

⁢

(

(

x

)

)

=

a

i

and for all other

prime ideals

p

,

ν

p

⁢

(

(

x

)

)

≥

0

. Here

ν

p

is the

p

-adic valuation associated with a prime ideal

p

.

Proof.

Assume first that all

a

i

≥

0

. By the Chinese Remainder Theorem,

A

/

𝔭

1

a

1

+

1

×

⋯

⁢

A

/

𝔭

k

a

k

+

1

≅

A

/

𝔭

1

a

1

+

1

⁢

⋯

⁢

𝔭

k

a

k

+

1

Thus the map

A

→

A

/

𝔭

1

a

1

+

1

×

⋯

⁢

A

/

𝔭

k

a

k

+

1

is

surjective

. Now choose

x

i

∈

p

i

a

i

,

x

i

∉

p

i

a

i

+

1

; this is possible since these two ideals are unequal by

unique factorization

. Choose

x

∈

A

with image

(

x

1

,

…

,

x

k

)

. Clearly

ν

𝔭

i

⁢

(

(

x

)

)

=

a

i

. But

x

∈

A

, so all other valuations are nonnegative.

In the general case, assume wlog that we are given a set

𝔭

1

,

…

,

𝔭

r

of primes of

A

and integers

a

1

,

…

,

a

r

≥

0

, and a set

𝔮

1

,

…

,

𝔮

t

of primes with integers

b

1

,

…

,

b

t

<

0

. First choose

y

∈

K

⋆

(using the case already proved above) so that

{

ν

𝔭

⁢

(

(

y

)

)

=

0

𝔭

=

𝔭

i

ν

𝔭

⁢

(

(

y

)

)

=

-

b

i

𝔭

=

𝔮

j

ν

𝔭

⁢

(

(

y

)

)

≥

0

otherwise

Now, there are only a finite number of primes

𝔭

k

′

such that

𝔭

k

′

is not the same as any of the

𝔮

j

and

ν

𝔭

k

′

⁢

(

(

y

)

)

>

0

. Let

ν

𝔭

k

′

⁢

(

(

y

)

)

=

c

k

>

0

. Again using the case proved above, choose

x

∈

K

⋆

such that

{

ν

𝔭

⁢

(

(

x

)

)

=

a

i

𝔭

=

𝔭

i

ν

𝔭

⁢

(

(

x

)

)

=

0

𝔭

=

𝔮

j

ν

𝔭

⁢

(

(

x

)

)

=

c

k

𝔭

=

𝔭

k

′

ν

𝔭

⁢

(

(

x

)

)

≥

0

otherwise

Then

x

/

y

is the required element.
∎

Title

weak approximation theorem

Canonical name

WeakApproximationTheorem

Date of creation

2013-03-22 18:35:21

Last modified on

2013-03-22 18:35:21

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 13F05

Classification

msc 11R04

Related topic

IndependenceOfTheValuations

Related topic

ChineseRemainderTheoremInTermsOfDivisorTheory

Generated on Fri Feb 9 17:30:45 2018 by

LaTeXML
Euclidean domain

Euclidean domain

A

Euclidean domain

is an

integral domain

on which a

Euclidean valuation

can be defined.

Every Euclidean domain is a

principal ideal domain

,
and therefore also a

unique factorization domain

.

Any two elements of a Euclidean domain have a

greatest common divisor

,
which can be computed using the

Euclidean algorithm

.

An example of a Euclidean domain is the ring

ℤ

.
Another example is the

polynomial ring

F

⁢

[

x

]

, where

F

is any field.
Every field is also a Euclidean domain.

Title

Euclidean domain

Canonical name

EuclideanDomain

Date of creation

2013-03-22 12:40:42

Last modified on

2013-03-22 12:40:42

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

13

Author

yark (2760)

Entry type

Definition

Classification

msc 13F07

Synonym

Euclidean ring

Related topic

PID

Related topic

UFD

Related topic

EuclidsAlgorithm

Related topic

Ring

Related topic

IntegralDomain

Related topic

EuclideanValuation

Related topic

WhyEuclideanDomains

Generated on Fri Feb 9 17:30:48 2018 by

LaTeXML