aarjaneiro/planetmath huggingface dataset:
Chinese remainder theorem for rings, noncommutative case

Chinese remainder theorem for rings, noncommutative case

Theorem 1

.

(

Chinese Remainder Theorem

) Let

R

be a ring and

I

1

,

I

2

,

…

,

I

n

pairwise comaximal (

http://planetmath.org/Comaximal

) ideals such that

R

=

I

j

+

R

2

for all

j

. The

homomorphism

:

f

:

R

→

R

/

I

1

×

R

/

I

2

×

…

×

R

/

I

n

f

⁢

(

a

)

=

(

a

+

I

1

,

a

+

I

2

,

…

,

a

+

I

n

)

is

surjective

and

k

⁢

e

⁢

r

⁢

f

=

I

1

∩

I

2

∩

⋯

∩

I

n

.

Proof.

Clearly

f

is a homomorphism with kernel

I

1

∩

I

2

∩

⋯

∩

I

n

. It remains to show the surjectivity.

We have:

R

=

I

1

+

R

2

=

I

1

+

(

I

1

+

I

2

)

⁢

(

I

1

+

I

3

)

⊆

I

1

+

I

1

2

+

I

1

⁢

I

3

+

I

2

⁢

I

1

+

I

2

⁢

I

3

⊆

I

1

+

(

I

2

∩

I

3

)

.

Moreover,

R

=

I

1

+

R

2

=

I

1

+

(

I

1

+

I

2

∩

I

3

)

⁢

(

I

1

+

I

4

)

=

I

1

+

I

1

⁢

I

4

+

(

I

2

∩

I

3

)

⁢

I

1

+

(

I

2

∩

I

3

)

⁢

I

4

⊆

I

1

+

(

I

2

∩

I

3

∩

I

4

)

.

Continuing, we obtain that

R

=

I

1

+

⋂

j

≠

1

I

j

. We show similarly that:

R

=

I

2

+

⋂

j

≠

2

I

j

=

I

3

+

⋂

j

≠

3

I

j

=

⋯

=

I

n

+

⋂

j

≠

n

I

j

.

Given

elements

a

1

,

a

2

,

…

,

a

n

, we can find

x

j

∈

I

j

and

y

j

∈

⋂

j

≠

k

I

k

such that

a

j

=

x

j

+

y

j

.

Take

a

:=

∑

i

=

1

n

x

i

=

a

j

(

mod

I

j

)

.

Hence

f

⁢

(

a

)

=

(

a

1

+

I

1

,

a

2

+

I

2

,

…

,

a

n

+

I

n

)

,

and we conclude that

f

is surjective as required.∎

Notes

1.The

relation

R

=

I

j

+

R

2

is satisfied when

R

is

ring with unity

. In that case

R

2

=

R

.

2. The Chinese Remainder Theorem (

http://planetmath.org/ChineseRemainderTheorem

) case for

integers

is obtained from the above result. For this, take

R

=

ℤ

and

I

j

=

(

p

j

)

=

p

j

⁢

ℤ

. The fact that two

solutions

of the set of

congruences

must

x

=

x

0

(

mod

p

1

⁢

…

⁢

p

n

)

is a

consequence

of:

I

1

∩

I

2

∩

⋯

∩

I

n

=

(

p

1

)

∩

(

p

2

)

∩

⋯

∩

(

p

n

)

=

(

p

1

⁢

p

2

⁢

…

⁢

p

n

)

⁢

ℤ

.

Title

Chinese remainder theorem for rings,

noncommutative

case

Canonical name

ChineseRemainderTheoremForRingsNoncommutativeCase

Date of creation

2013-03-22 16:53:45

Last modified on

2013-03-22 16:53:45

Owner

polarbear (3475)

Last modified by

polarbear (3475)

Numerical id

16

Author

polarbear (3475)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 11D79

Synonym

chinese remainder theorem

Generated on Fri Feb 9 17:20:55 2018 by

LaTeXML
contracted ideal

contracted ideal

Let

f

:

A

→

B

be a

ring homomorphism

. Let

𝔟

be an ideal in

B

. Then it is easy to show that the

inverse image

of

𝔟

, that is

f

-

1

⁢

(

𝔟

)

, is an ideal in

A

, and we call it a

contracted ideal

. A common notation for the contracted ideal in this case is

𝔟

c

.

Title

contracted ideal

Canonical name

ContractedIdeal

Date of creation

2013-03-22 12:55:31

Last modified on

2013-03-22 12:55:31

Owner

drini (3)

Last modified by

drini (3)

Numerical id

5

Author

drini (3)

Entry type

Definition

Classification

msc 13A15

Classification

msc 14K99

Classification

msc 16D25

Related topic

ExtendedIdeal

Generated on Fri Feb 9 17:20:59 2018 by

LaTeXML
entries on finitely generated ideals

entries on finitely generated ideals

This entry lists PlanetMath entries concerning

finitely generated

ideals of

commutative rings

.

One

1.

principal ideal

2.

zero ideal

3.

unit ideal

4.

principal ideal ring

5.

principal ideal domain

6.

Bézout ring

7.

divisibility by product

Two

1.

two-generator property

2.

multiplication rule gives inverse ideal

3.

invertibility of regularly generated ideal

4.

unique factorization and ideals in ring of integers

n

1.

invertible ideal is finitely generated

2.

generators of inverse ideal

3.

invertibility of regularly generated ideal (

http://planetmath.org/InvertibilityOfRegularlyGeneratedIdeal

)

4.

Prüfer ring (

http://planetmath.org/pruferring

)

5.

Prüfer domain (

http://planetmath.org/pruferdomain

)

6.

ideal

generators

in Prüfer ring

7.

finitely generated modules over a principal ideal domain

Formulae

1.

well-definedness of product of finitely generated ideals

2.

product of finitely generated ideals

3.

cancellation ideal

4.

ideal decomposition in Dedekind domain

5.

ideal inverting in Prüfer ring

6.

least common

multiple

7.

quotient of ideals

8.

generators of a

quotient

polynomial ring

Title

entries on finitely generated ideals

Canonical name

EntriesOnFinitelyGeneratedIdeals

Date of creation

2015-05-06 15:39:55

Last modified on

2015-05-06 15:39:55

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Topic

Classification

msc 13A15

Classification

msc 13C05

Classification

msc 16D25

Related topic

ProductOfFinitelyGeneratedIdeals

Related topic

TopicsOnIdeals

Generated on Fri Feb 9 17:21:02 2018 by

LaTeXML
extended ideal

extended ideal

Let

f

:

A

→

B

be a ring map. We can look at the

ideal generated by

the image of

𝔞

, which is called an

extended ideal

and is denoted by

𝔞

e

.

It is not true in general that if

𝔞

is an ideal in

A

, the image of

𝔞

under

f

will be an ideal in

B

. (For example, consider the embedding

f

:

ℤ

→

ℚ

. The image of the ideal

(

2

)

⊂

ℤ

is

not

an ideal in

ℚ

, since the only ideals in

ℚ

are

{

0

}

and all of

ℚ

.)

Title

extended ideal

Canonical name

ExtendedIdeal

Date of creation

2013-03-22 12:55:34

Last modified on

2013-03-22 12:55:34

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Definition

Classification

msc 13A15

Classification

msc 14K99

Related topic

ContractedIdeal

Generated on Fri Feb 9 17:21:08 2018 by

LaTeXML
finite extensions of Dedekind domains are Dedekind

finite extensions of Dedekind domains are Dedekind

Theorem

.

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

A

is also a Dedekind domain.

For example, a

number field

K

is a finite extension of

ℚ

and its

ring of integers

is denoted by

𝒪

K

. Although such rings can fail to be

unique factorization domains

, the above theorem shows that they are always Dedekind domains and therefore

unique factorization

of ideals (

http://planetmath.org/IdealDecompositionInDedekindDomain

) is satisfied.

Title

finite extensions of Dedekind domains are Dedekind

Canonical name

FiniteExtensionsOfDedekindDomainsAreDedekind

Date of creation

2013-03-22 18:35:30

Last modified on

2013-03-22 18:35:30

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 13F05

Related topic

FiniteExtension

Related topic

DivisorTheoryInFiniteExtension

Generated on Fri Feb 9 17:21:11 2018 by

LaTeXML
first isomorphism theorem

first isomorphism theorem

If

f

:

G

→

H

is a

homomorphism of groups

(or rings, or modules), then it induces an

isomorphism

G

/

ker

⁡

f

≅

im

⁢

f

.

Title

first isomorphism theorem

Canonical name

FirstIsomorphismTheorem

Date of creation

2013-03-22 12:03:39

Last modified on

2013-03-22 12:03:39

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 20A05

Generated on Fri Feb 9 17:21:15 2018 by

LaTeXML
fractional ideal

fractional ideal

1

Basics

Let

A

be an

integral domain

with

field of fractions

K

. Then

K

is
an

A

–module, and we define a

fractional ideal

of

A

to be a

submodule

of

K

which is

finitely generated

as an

A

–module.

The product of two fractional ideals

𝔞

and

𝔟

of

A

is defined
to be the submodule of

K

generated by all the products

x

⋅

y

∈

K

, for

x

∈

𝔞

and

y

∈

𝔟

. This product is denoted

𝔞

⋅

𝔟

, and it is always a fractional ideal of

A

as well. Note
that, if

A

itself is considered as a fractional ideal of

A

, then

𝔞

⋅

A

=

𝔞

. Accordingly, the set of fractional ideals is always
a monoid under this product operation, with

identity element

A

.

We say that a fractional ideal

𝔞

is

invertible

if there
exists a fractional ideal

𝔞

′

such that

𝔞

⋅

𝔞

′

=

A

. It can
be shown that if

𝔞

is invertible, then its

inverse

must be

𝔞

′

=

(

A

:

𝔞

)

, the

annihilator

1

1

In general, for any fractional
ideals

𝔞

and

𝔟

, the annihilator of

𝔟

in

𝔞

is the
fractional ideal

(

𝔞

:

𝔟

)

consisting of all

x

∈

K

such that

x

⋅

𝔟

⊂

𝔞

.

of

𝔞

in

A

.

2

Fractional ideals in Dedekind domains

We now suppose that

A

is a

Dedekind domain

. In this case, every
nonzero fractional ideal is invertible, and consequently the nonzero
fractional ideals in

A

form a group under ideal

multiplication

,
called the

ideal group

of

A

.

The

unique factorization

of ideals

theorem states that every
fractional ideal in

A

factors uniquely into a finite product of

prime ideals

of

A

and their (fractional ideal) inverses. It follows
that the ideal group of

A

is

freely generated

as an

abelian group

by
the nonzero prime ideals of

A

.

A fractional ideal of

A

is said to be

principal

if it is
generated as an

A

–module by a single element. The set of nonzero
principal fractional ideals is a

subgroup

of the ideal group of

A

,
and the

quotient group

of the ideal group of

A

by the subgroup of
principal fractional ideals is nothing other than the

ideal class
group

of

A

.

Title

fractional ideal

Canonical name

FractionalIdeal

Date of creation

2013-03-22 12:42:38

Last modified on

2013-03-22 12:42:38

Owner

djao (24)

Last modified by

djao (24)

Numerical id

5

Author

djao (24)

Entry type

Definition

Classification

msc 13A15

Classification

msc 13F05

Related topic

IdealClassGroup

Defines

ideal group

Generated on Fri Feb 9 17:21:18 2018 by

LaTeXML
generators of inverse ideal

generators of inverse ideal

Theorem.

Let

R

be a

commutative ring

with

non-zero
unity

and let

T

be the

total ring of fractions

of

R

.  If

𝔞

=

(

a

1

,

…

,

a

n

)

is an

invertible

fractional ideal

(

http://planetmath.org/FractionalIdealOfCommutativeRing

) of

R

with

𝔞

⁢

𝔟

=

R

,  then also the inverse
ideal

𝔟

can be generated by

n

elements of

T

.

Proof.

The

equation

𝔞

⁢

𝔟

=

(

1

)

implies the existence of the elements

a

i

′

of

𝔞

and

b

i

′

of

𝔟

(

i

=

1

,

…

,

m

)

such that

a

1

′

⁢

b

1

′

+

⋯

+

a

m

′

⁢

b

m

′

=

1

.  Because the

a

i

′

’s are
in

𝔞

, they may be expressed as

a

i

′

=

∑

j

=

1

n

r

i

⁢

j

a

j

(

i

=

1

,

…

,

m

)

,

where the

r

i

⁢

j

’s are some elements of

R

.  Now the unity acquires the
form

1

=

∑

i

=

1

m

a

i

′

⁢

b

i

′

=

∑

i

=

1

m

∑

j

=

1

n

r

i

⁢

j

⁢

a

j

⁢

b

i

′

=

∑

j

=

1

n

a

j

⁢

∑

i

=

1

m

r

i

⁢

j

⁢

b

i

′

=

∑

j

=

1

n

a

j

⁢

b

j

,

in which

b

j

=

∑

i

=

1

m

r

i

⁢

j

b

i

′

∈

R

𝔟

=

𝔟

(

j

=

1

,

…

,

n

)

.

Thus an arbitrary element

b

of the

𝔟

satisfies the condition

b

=

b

⋅

1

=

∑

j

=

1

n

(

a

j

⁢

b

)

⁢

b

j

∈

R

⁢

b

1

+

⋯

+

R

⁢

b

n

=

(

b

1

,

…

,

b

n

)

.

Consequently,

𝔟

⊆

(

b

1

,

…

,

b

n

)

.  Since the

inverse

inclusion is apparent, we have the equality

𝔞

-

1

=

𝔟

=

(

b

1

,

…

,

b

n

)

.

Title

generators of inverse ideal

Canonical name

GeneratorsOfInverseIdeal

Date of creation

2015-05-06 14:52:30

Last modified on

2015-05-06 14:52:30

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

18

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A15

Related topic

FractionalIdealOfCommutativeRing

Related topic

IdealGeneratedByASet

Related topic

PruferRing

Generated on Fri Feb 9 17:21:23 2018 by

LaTeXML
homogeneous ideal

homogeneous ideal

Let

R

=

⊕

g

∈

G

R

g

be a

graded ring

. Then an element

r

of

R

is said to be

homogeneous

if it is an element of some

R

g

. An ideal

I

of

R

is said to be homogeneous if it can be generated by a set of homogeneous elements, or equivalently if it is the

ideal generated by

the set of elements

⋃

g

∈

G

I

∩

R

g

.

One observes that if

I

is a

homogeneous ideal

and

r

=

∑

i

r

g

i

is the sum of homogeneous elements

r

g

i

for distinct

g

i

, then each

r

g

i

must be in

I

.

To see some examples, let

k

be a field, and take

R

=

k

⁢

[

X

1

,

X

2

,

X

3

]

with the usual

grading

by total degree. Then the ideal generated by

X

1

n

+

X

2

n

-

X

3

n

is a homogeneous ideal. It is also a

radical ideal

. One reason homogeneous ideals in

k

⁢

[

X

1

,

…

,

X

n

]

are of

interest

is because (if they are

radical

) they define

projective varieties

; in this case the projective variety is the Fermat (

http://planetmath.org/FermatsLastTheorem

) curve. For contrast, the ideal generated by

X

1

+

X

2

2

is not homogeneous.

Title

homogeneous ideal

Canonical

name

HomogeneousIdeal

Date of creation

2013-03-22 11:45:00

Last modified on

2013-03-22 11:45:00

Owner

archibal (4430)

Last modified by

archibal (4430)

Numerical id

11

Author

archibal (4430)

Entry type

Definition

Classification

msc 13A15

Classification

msc 33C75

Classification

msc 33E05

Classification

msc 86A30

Classification

msc 14H52

Classification

msc 14J27

Related topic

GradedRing

Related topic

ProjectiveVariety

Related topic

HomogeneousElementsOfAGradedRing

Related topic

HomogeneousPolynomial

Defines

homogeneous

Defines

homogeneous element

Generated on Fri Feb 9 17:21:27 2018 by

LaTeXML
ideals contained in a union of ideals

ideals contained in a union of ideals

Assume that

R

is a commutative ring.

Lemma.

Let

A

,

B

,

C

be ideals in

R

such that

A

⊆

B

∪

C

. Then

A

⊆

B

or

A

⊆

C

.

Proof.

Assume that this is not true. Then there are

x

,

y

∈

A

such that

x

∈

B

,

y

∈

C

and

x

∉

C

,

y

∉

B

. Obviously

x

+

y

∈

A

⊆

B

∪

C

and without loss of generality we may assume that

x

+

y

∈

B

. Then

y

=

(

x

+

y

)

-

x

∈

B

.

Contradiction

.

□

Remark.

This lemma is also true if we exchange ring with a group and ideals with

subgroups

(because we didn’t use

multiplication

and

commutativity

of

addition

in proof).

Proposition

.

Let

I

,

P

1

,

…

,

P

n

be ideals in

R

such that each

P

i

is prime. If

I

⊆

P

1

∪

⋯

∪

P

n

, then there exists

i

∈

{

1

,

…

,

n

}

such that

I

⊆

P

i

.

Proof.

We will use the

induction

on

n

. For

n

=

2

our lemma applies. Let

n

>

2

. Assume that

I

⊈

P

1

∪

⋯

∪

P

n

. For

i

∈

{

1

,

…

,

n

}

define

P

i

¯

=

P

1

∪

⋯

∪

P

i

-

1

∪

P

i

+

1

∪

⋯

∪

P

n

.

By our

assumption

(and

induction hypothesis

)

I

⊈

P

i

¯

for any

i

∈

{

1

,

…

,

n

}

. Thus for any

i

there is

x

i

∈

I

such that

x

i

∉

P

i

¯

.

Now for any

i

∈

{

1

,

…

,

n

}

define

x

i

¯

=

x

1

⁢

⋯

⁢

x

i

-

1

⁢

x

i

+

1

⁢

⋯

⁢

x

n

∈

I

. Then we have

x

1

¯

+

⋯

+

x

n

¯

∈

I

and thus there is

j

∈

{

1

,

…

,

n

}

such that

x

1

¯

+

⋯

+

x

n

¯

∈

P

j

. Since

x

i

¯

∈

P

j

for any

i

≠

j

, then we have that

x

j

¯

∈

P

j

.

But

P

j

is prime, so there is

k

≠

j

such that

x

k

∈

P

j

⊆

P

k

¯

. Contradiction.

□

Counterexample

.

We will show, that if

P

i

’s are not prime, then the thesis no longer hold, even when

n

=

3

. Consider the ring of polynomials in two

variables

over a simple field of order

2

, i.e.

ℤ

2

⁢

[

X

,

Y

]

. Let

R

=

ℤ

2

⁢

[

X

,

Y

]

/

(

X

2

,

X

⁢

Y

,

Y

2

)

. For

W

⁢

(

X

,

Y

)

∈

ℤ

2

⁢

[

X

,

Y

]

we shall write

W

⁢

(

X

,

Y

)

¯

=

W

⁢

(

X

,

Y

)

+

(

X

2

,

X

⁢

Y

,

Y

2

)

∈

R

. Then it is easy to see, that

R

=

{

0

¯

,

1

¯

,

X

¯

,

Y

¯

,

X

¯

+

Y

¯

,

X

¯

+

1

¯

,

Y

¯

+

1

¯

,

X

¯

+

Y

¯

+

1

¯

}

.

Let

I

=

{

0

¯

,

X

¯

,

Y

¯

,

X

¯

+

Y

¯

}

;

A

1

=

{

0

¯

,

X

¯

}

;

A

2

=

{

0

¯

,

Y

¯

}

;

A

3

=

{

0

¯

,

X

¯

+

Y

¯

}

.

It can be easily checked, that

I

,

A

1

,

A

2

,

A

3

are all ideals and

I

⊆

A

1

∪

A

2

∪

A

3

but obviously

I

⊈

A

i

for any

i

=

1

,

2

,

3

.

□

Title

ideals

contained

in a union of ideals

Canonical

name

IdealsContainedInAUnionOfIdeals

Date of creation

2013-03-22 19:03:55

Last modified on

2013-03-22 19:03:55

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 13A15

Related topic

IdealIncludedInUnionOfPrimeIdeals

Generated on Fri Feb 9 17:21:33 2018 by

LaTeXML
ideals contained in a union of radical ideals

ideals contained in a union of radical ideals

Let

R

be a

commutative ring

and

I

⊆

R

an ideal. Recall that

the

radical

of

I

is defined as

r

⁢

(

I

)

=

{

x

∈

R

|

∃

n

∈

ℕ

x

n

∈

I

}

.

It can be shown, that

r

⁢

(

I

)

is again an ideal and

I

⊆

r

⁢

(

I

)

. Let

V

⁢

(

I

)

=

{

P

⊆

R

|

P

⁢

is a prime ideal and

⁢

I

⊆

P

}

.

Of course

V

⁢

(

I

)

≠

∅

(because

I

is

contained

in at least one

maximal ideal

) and it can be shown, that

r

⁢

(

I

)

=

⋂

P

∈

V

⁢

(

I

)

P

.

Finaly, recall that an ideal

I

is called

radical

, if

I

=

r

⁢

(

I

)

.

Proposition

.

Let

I

,

R

1

,

…

,

R

n

be ideals in

R

, such that each

R

i

is radical. If

I

⊆

R

1

∪

⋯

∪

R

n

,

then there exists

i

∈

{

1

,

…

,

n

}

such that

I

⊆

R

i

.

Proof.

Assume that this not true, i.e. for every

i

we have

I

⊈

R

i

. Then for any

i

∈

{

1

,

…

,

n

}

there exists

P

i

∈

V

⁢

(

R

i

)

such that

I

⊈

P

i

(this follows from the fact, that

R

i

=

r

⁢

(

R

i

)

and the

characterization

of radicals via

prime ideals

). But for any

i

we have

R

i

⊆

P

i

and thus

I

⊆

P

1

∪

⋯

∪

P

n

.

Contradiction

, since each

P

i

is prime (see the parent object for details).

□

Title

ideals contained in a union of radical ideals

Canonical

name

IdealsContainedInAUnionOfRadicalIdeals

Date of creation

2013-03-22 19:04:23

Last modified on

2013-03-22 19:04:23

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Corollary

Classification

msc 13A15

Generated on Fri Feb 9 17:21:37 2018 by

LaTeXML
image ideal of divisor

image ideal of divisor

Theorem

.

If an

integral domain

𝒪

has a

divisor theory

𝒪

*

→

𝔇

,  then the subset

[

𝔞

]

of

𝒪

, consisting of 0 and all elements

divisible

by a

divisor

𝔞

, is an ideal of

𝒪

.  The mapping

𝔞

↦

[

𝔞

]

from the set

𝔇

of divisors into the set of ideals of

𝒪

is injective and maps any principal divisor

(

α

)

to the

principal ideal

(

α

)

.

Proof.

Let

α

,

β

∈

[

𝔞

]

and

λ

∈

𝒪

.  Then, by the postulate 2 of divisor theory (

http://planetmath.org/DivisorTheory

),

α

-

β

is divisible by

𝔞

or is 0, and in both cases belongs to

[

𝔞

]

.  When

λ

⁢

α

≠

0

,  we can write

(

α

)

=

𝔞

⁢

𝔠

with

𝔠

a divisor.  According to the homomorphicity of the mapping

𝒪

*

→

𝔇

,  we have

(

λ

⁢

α

)

=

(

λ

)

⁢

(

α

)

=

(

λ

)

⁢

𝔞

⁢

𝔠

,

and therefore the element

λ

⁢

α

is divisible by

𝔞

, i.e.

λ

⁢

α

∈

[

𝔞

]

.  Thus,

[

𝔞

]

is an ideal of

𝒪

.

The injectivity of the mapping

𝔞

↦

[

𝔞

]

follows from the postulate 3 of divisor theory (

http://planetmath.org/DivisorTheory

).

The ideal

[

𝔞

]

may be called the

image ideal

of

𝔞

or the

ideal determined by the divisor

𝔞

.

Remark.

There are integral domains

𝒪

having a divisor theory but also having ideals which are not of the form

[

𝔞

]

(for example a

polynomial ring

in two

indeterminates

and its ideal formed by the polynomials without constant term).  Such rings have ‘‘too many ideals’’.  On the other hand, in some integral domains the monoid of principal ideals cannot be embedded into a free monoid; thus those rings cannot have a divisor theory.

References

1

М. М. Постников:

Введение  в  теорию  алгебраических
чисел

.  Издательство
‘‘Наука’’. Москва (1982).

Title

image ideal of divisor

Canonical

name

ImageIdealOfDivisor

Date of creation

2013-03-22 18:02:40

Last modified on

2013-03-22 18:02:40

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 13A05

Classification

msc 11A51

Defines

image ideal

Defines

ideal determined by the divisor

Generated on Fri Feb 9 17:21:48 2018 by

LaTeXML
invertibility of regularly generated ideal

invertibility of regularly generated ideal

Lemma.

Let

R

be a

commutative ring

containing

regular elements

.  If

𝔞

,

𝔟

and

𝔠

are three ideals of

R

such that

𝔟

+

𝔠

,

𝔠

+

𝔞

and

𝔞

+

𝔟

are

invertible

(

http://planetmath.org/FractionalIdealOfCommutativeRing

), then also their

sum ideal

𝔞

+

𝔟

+

𝔠

is .

Proof.

We may assume that

R

has a unity, therefore the

product

of an ideal and its

inverse

(

http://planetmath.org/FractionalIdealOfCommutativeRing

) is always

R

.  Now, the ideals

𝔟

+

𝔠

,

𝔠

+

𝔞

and

𝔞

+

𝔟

have the

𝔣

1

,

𝔣

2

and

𝔣

3

, respectively, so that

(

𝔟

+

𝔠

)

⁢

𝔣

1

=

(

𝔠

+

𝔞

)

⁢

𝔣

2

=

(

𝔞

+

𝔟

)

⁢

𝔣

3

=

R

.

Because

𝔞

⁢

𝔣

2

⊆

R

and

𝔠

⁢

𝔣

1

⊆

R

,  we obtain

(

𝔞

+

𝔟

+

𝔠

)

⁢

(

𝔞

⁢

𝔣

2

⁢

𝔣

3

+

𝔠

⁢

𝔣

1

⁢

𝔣

2

)

=

(

𝔞

+

𝔟

)

⁢

𝔞

⁢

𝔣

2

⁢

𝔣

3

+

𝔠

⁢

(

𝔞

⁢

𝔣

2

)

⁢

𝔣

3

+

𝔞

⁢

(

𝔠

⁢

𝔣

1

)

⁢

𝔣

2

+

(

𝔟

+

𝔠

)

⁢

𝔠

⁢

𝔣

1

⁢

𝔣

2

=

𝔞

⁢

𝔣

2

+

𝔠

⁢

𝔣

2

=

(

𝔠

+

𝔞

)

⁢

𝔣

2

=

R

.

Theorem

.

Let

R

be a commutative ring containing regular
elements.  If every ideal of

R

generated by two regular elements is , then in

R

also every

ideal generated by

a

finite set

of regular elements is .

Proof.

We use

induction

on

n

, the number of the regular elements of the

generating set

.  We thus assume that every ideal of

R

generated by

n

regular elements  (

n

≧

2

)

is .  Let

{

r

1

,

r

2

,

…

,

r

n

+

1

}

be any set of regular elements of

R

.  Denote

𝔞

=

:

(

r

1

)

,

𝔟

=

:

(

r

2

,

…

,

r

n

)

,

𝔠

=

:

(

r

n

+

1

)

.

The sums

𝔟

+

𝔠

,

𝔠

+

𝔞

and

𝔞

+

𝔟

are, by the

assumptions

, .  Then the ideal

(

r

1

,

r

2

,

…

,

r

n

,

r

n

+

1

)

=

𝔞

+

𝔟

+

𝔠

is, by the lemma, , and the induction
proof is

complete

.

References

1

R. Gilmer:

Multiplicative

ideal theory

.  Queens University Press. Kingston, Ontario (1968).

Title

invertibility of regularly generated ideal

Canonical name

InvertibilityOfRegularlyGeneratedIdeal

Date of creation

2015-05-06 15:27:47

Last modified on

2015-05-06 15:27:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

17

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13A15

Classification

msc 11R04

Related topic

IdealMultiplicationLaws

Related topic

PruferRing

Related topic

InvertibleIdealIsFinitelyGenerated

Generated on Fri Feb 9 17:21:54 2018 by

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