aarjaneiro/planetmath huggingface dataset:
proof of Artin-Rees theorem

proof of Artin-Rees theorem

Define the

graded ring

Pow

𝔄

⁢

(

A

)

⊆

A

⁢

[

X

]

,
where

X

is an indeterminate by

Pow

𝔄

⁢

(

A

)

=

∐

n

≥

0

𝔄

n

⁢

X

n

=

{

z

0

+

z

1

X

+

⋯

+

z

r

X

r

∣

r

≥

0

,

z

j

∈

𝔄

j

}

.

Now,

M

gives rise to a

graded module

,

M

′

, over

Pow

𝔄

⁢

(

A

)

,
namely

M

′

=

∐

n

≥

0

𝔄

n

⁢

M

⁢

X

n

=

{

z

0

+

z

1

X

+

⋯

+

z

r

X

r

∣

r

≥

0

,

z

j

∈

𝔄

j

M

}

.

Observe that

Pow

𝔄

⁢

(

A

)

is a

noetherian ring

. For, if

α

1

,

…

,

α

q

generate

𝔄

in

A

, then the
elements of

𝔄

n

are sums of degree

n

monomials

in the

α

j

’s, i.e., if

Y

1

,

…

,

Y

q

are independent indeterminates
the map

A

⁢

[

Y

1

,

…

,

Y

q

]

⟶

Pow

𝔄

⁢

(

A

)

via

Y

j

↦

α

j

⁢

X

is

surjective

, and as

A

⁢

[

Y

1

,

…

,

Y

q

]

is

noetherian

, so is

Pow

𝔄

⁢

(

A

)

.

Let

m

1

,

…

,

m

t

generate

M

over

A

. Then,

m

1

,

…

,

m

t

generate

M

′

over

Pow

𝔄

⁢

(

A

)

. Therefore,

M

′

is
a

noetherian module

. Set

N

′

=

∐

n

≥

0

(

𝔄

n

⁢

M

∩

N

)

⁢

X

n

⊆

M

′

,

a

submodule

of

M

′

.
Moreover,

N

′

is a homogeneous submodule of

M

and it is f.g.
as

M

′

is noetherian.
Consequently,

N

′

possesses a finite number of

homogeneous

generators

:

u

1

⁢

X

n

1

,

…

,

u

s

⁢

X

n

s

, where

u

j

∈

𝔄

n

j

⁢

M

∩

N

. Let

k

=

max

⁡

{

n

1

,

…

,

n

s

}

.
Given any

n

≥

k

and any

z

∈

𝔄

n

⁢

M

∩

N

, look at

z

⁢

X

n

∈

N

n

′

. We have

z

⁢

X

n

=

∑

l

=

1

s

a

l

⁢

X

n

-

n

l

⁢

u

l

⁢

X

n

l

,

where

a

l

⁢

X

n

-

n

l

∈

(

Pow

𝔄

⁢

(

A

)

)

n

-

n

l

.
Thus,

a

l

∈

𝔄

n

-

n

l

=

𝔄

n

-

k

⁢

𝔄

k

-

n

l

and

a

l

⁢

u

l

∈

𝔄

n

-

k

⁢

(

𝔄

k

-

n

l

⁢

u

l

)

⊆

𝔄

n

-

k

⁢

(

𝔄

k

-

n

l

⁢

(

𝔄

n

l

⁢

M

∩

N

)

)

⊆

𝔄

n

-

k

⁢

(

𝔄

k

⁢

M

∩

N

)

.

It follows that

z

=

∑

l

=

1

s

a

l

⁢

u

l

∈

𝔄

n

-

k

⁢

(

𝔄

k

⁢

M

∩

N

)

, so

𝔄

n

⁢

M

∩

N

⊆

𝔄

n

-

k

⁢

(

𝔄

k

⁢

M

∩

N

)

.

Now, it is clear that the righthand side is

contained

in

𝔄

n

⁢

M

∩

N

, as

𝔄

n

-

k

⁢

N

⊆

N

.

□

Title

proof of Artin-Rees theorem

Canonical name

ProofOfArtinReesTheorem

Date of creation

2013-03-22 14:28:43

Last modified on

2013-03-22 14:28:43

Owner

mat_cross (707)

Last modified by

mat_cross (707)

Numerical id

4

Author

mat_cross (707)

Entry type

Proof

Classification

msc 13C99

Generated on Fri Feb 9 17:29:37 2018 by

LaTeXML
proof of Nakayama’s lemma

proof of Nakayama’s lemma

(This proof was taken from

[

1

]

.)

If

M

were not zero, it would have a simple quotient,

isomorphic

to

R

/

𝔪

for some

maximal ideal

𝔪

of

R

. Then we would have

𝔪

⁢

M

≠

M

, so that

𝔞

⁢

M

≠

M

as

𝔞

⊆

𝔪

.

References

1

Serre, J.-P.

Local Algebra.

Springer-Verlag, 2000.

Title

proof of Nakayama’s lemma

Canonical name

ProofOfNakayamasLemma

Date of creation

2013-03-22 13:16:50

Last modified on

2013-03-22 13:16:50

Owner

nerdy2 (62)

Last modified by

nerdy2 (62)

Numerical id

6

Author

nerdy2 (62)

Entry type

Proof

Classification

msc 13C99

Generated on Fri Feb 9 17:29:39 2018 by

LaTeXML
proof of Nakayama’s lemma

proof of Nakayama’s lemma

Let

X

=

{

x

1

,

x

2

,

…

,

x

n

}

be a

minimal set

of

generators

for

M

, in the sense that

M

is not generated by any

proper subset

of

X

.

Elements of

𝔞

⁢

M

can be written as

linear combinations

∑

a

i

⁢

x

i

, where

a

i

∈

𝔞

.

Suppose that

|

X

|

>

0

. Since

M

=

𝔞

⁢

M

, we can express

x

1

as a such a linear combination:

x

1

=

∑

a

i

⁢

x

i

.

Moving the term involving

a

1

to the left, we have

(

1

-

a

1

)

⁢

x

1

=

∑

i

>

1

a

i

⁢

x

i

.

But

a

1

∈

J

⁢

(

R

)

, so

1

-

a

1

is

invertible

, say with

inverse

b

.
Therefore,

x

1

=

∑

i

>

1

b

⁢

a

i

⁢

x

i

.

But this means that

x

1

is redundant as a generator of

M

, and so

M

is generated by the subset

{

x

2

,

x

3

,

…

,

x

n

}

. This contradicts the minimality of

X

.

We conclude that

|

X

|

=

0

and therefore

M

=

0

.

Title

proof of Nakayama’s lemma

Canonical name

ProofOfNakayamasLemma

Date of creation

2013-03-22 13:07:46

Last modified on

2013-03-22 13:07:46

Owner

mclase (549)

Last modified by

mclase (549)

Numerical id

5

Author

mclase (549)

Entry type

Proof

Classification

msc 13C99

Generated on Fri Feb 9 17:29:41 2018 by

LaTeXML
quotient ring modulo prime ideal

quotient ring modulo prime ideal

Theorem.

Let

R

be a

commutative ring

with

non-zero unity

1 and

𝔭

an ideal of

R

. The

quotient ring

R

/

𝔭

is an

integral domain

if and only if

𝔭

is a

prime ideal

.

Proof.

1

o

¯

. First, let

𝔭

be a prime ideal of

R

. Then

R

/

𝔭

is of course a commutative ring and has the unity

1

+

𝔭

. If the

product

(

r

+

𝔭

)

⁢

(

s

+

𝔭

)

of two

residue classes

vanishes, i.e. equals

𝔭

, then we have

r

⁢

s

+

𝔭

=

𝔭

,  and therefore

r

⁢

s

must belong to

𝔭

. Since

𝔭

is , either

r

or

s

belongs to

𝔭

, i.e.

r

+

𝔭

=

𝔭

or

s

+

𝔭

=

𝔭

.  Accordingly,

R

/

𝔭

has no

zero divisors

and is an integral domain.

2

o

¯

. Conversely, let

R

/

𝔭

be an integral domain and let the product

r

⁢

s

of two elements of

R

belong to

𝔭

. It follows that

(

r

+

𝔭

)

⁢

(

s

+

𝔭

)

=

r

⁢

s

+

𝔭

=

𝔭

. Since

R

/

𝔭

has no zero divisors,

r

+

𝔭

=

𝔭

or

s

+

𝔭

=

𝔭

. Thus,

r

or

s

belongs to

𝔭

, i.e.

𝔭

is a prime ideal.

Title

quotient ring modulo prime ideal

Canonical name

QuotientRingModuloPrimeIdeal

Date of creation

2013-03-22 17:37:09

Last modified on

2013-03-22 17:37:09

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 13C99

Related topic

CharacterisationOfPrimeIdeals

Related topic

QuotientRing

Generated on Fri Feb 9 17:29:43 2018 by

LaTeXML
scalar map

scalar map

Given a ring

R

, a left

R

-module

U

, a right

R

-module

V

and a two-sided

R

-module

W

then a map

b

:

U

×

V

→

W

is an

R

-

scalar map

if

1.

b

is biadditive, that is

b

⁢

(

u

+

u

′

,

v

)

=

b

⁢

(

u

,

v

)

+

b

⁢

(

u

′

,

v

)

and

b

⁢

(

u

,

v

+

v

′

)

=

b

⁢

(

u

,

v

)

+

b

⁢

(

u

,

v

′

)

for all

u

,

u

′

∈

U

and

v

,

v

′

∈

V

;

2.

b

⁢

(

r

⁢

u

,

v

)

=

r

⁢

b

⁢

(

u

,

v

)

and

b

⁢

(

u

,

v

⁢

r

)

=

b

⁢

(

u

,

v

)

⁢

r

for all

u

∈

U

,

v

∈

V

and

r

∈

R

.

Such maps can also be called

outer linear

.

Unlike

bilinear maps

, scalar maps do not force a

commutative

multiplication

on

R

even when the map is

non-degenerate

and the modules are

faithful

.
For example, if

A

is an

associative ring

then the multiplication of

A

,

b

:

A

×

A

→

A

is a

A

-outer linear:

b

⁢

(

x

⁢

y

,

z

)

=

(

x

⁢

y

)

⁢

z

=

x

⁢

(

y

⁢

z

)

=

x

⁢

b

⁢

(

y

,

z

)

and likewise

b

⁢

(

x

,

y

⁢

z

)

=

b

⁢

(

x

,

y

)

⁢

z

. Using a non-commutative ring

A

confirms
the claim.

It is immediate however that

⟨

b

⁢

(

U

,

V

)

⟩

is in fact an

R

-

bimodule

.
This is because:

s

⁢

(

b

⁢

(

u

,

v

)

⁢

r

)

=

s

⁢

b

⁢

(

u

,

v

⁢

r

)

=

b

⁢

(

s

⁢

u

,

v

⁢

r

)

=

s

⁢

b

⁢

(

u

,

v

⁢

r

)

=

(

s

⁢

b

⁢

(

u

,

v

)

)

⁢

r

for all

u

∈

U

,

v

∈

V

and

s

,

r

∈

R

. Therefore it is not uncommon to
require that indeed all of

W

be an

R

-bimodule.

Title

scalar map

Canonical name

ScalarMap

Date of creation

2013-03-22 17:24:22

Last modified on

2013-03-22 17:24:22

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

7

Author

Algeboy (12884)

Entry type

Definition

Classification

msc 13C99

Synonym

outer linear

Related topic

BilinearMap

Defines

scalar map

Generated on Fri Feb 9 17:29:45 2018 by

LaTeXML
second isomorphism theorem

second isomorphism theorem

Let

(

G

,

*

)

be a group. Let

H

be a

subgroup

of

G

and let

K

be a

normal subgroup

of

G

. Then

•

H

⁢

K

:=

{

h

*

k

∣

h

∈

H

,

k

∈

K

}

is a subgroup of

G

,

•

K

is a normal subgroup of

H

⁢

K

,

•

H

∩

K

is a normal subgroup of

H

,

•

There is a natural

group isomorphism

H

/

(

H

∩

K

)

=

H

⁢

K

/

K

.

The same statement also holds in the

category

of modules over a fixed ring (where normality is neither needed nor relevant), and indeed can be formulated so as to hold in any

abelian category

.

Title

second isomorphism theorem

Canonical name

SecondIsomorphismTheorem

Date of creation

2013-03-22 12:08:46

Last modified on

2013-03-22 12:08:46

Owner

djao (24)

Last modified by

djao (24)

Numerical id

9

Author

djao (24)

Entry type

Theorem

Classification

msc 13C99

Classification

msc 20A05

Generated on Fri Feb 9 17:29:47 2018 by

LaTeXML
sum of ideals

sum of ideals

Definition.

Let’s consider some set of ideals (left, right or two-sided) of a ring.  The

sum of the ideals

is the smallest ideal of the ring containing all those ideals.  The

sum of ideals

is denoted by using “+” and “

∑

” as usually.

It is not difficult to be persuaded of the following:

•

The sum of a finite amount of ideals is

𝔞

1

+

𝔞

2

+

⋯

+

𝔞

k

=

{

a

1

+

a

2

+

⋯

+

a

k

⋮

a

i

∈

𝔞

i

∀

i

}

.

•

The sum of any set of ideals consists of all finite sums

∑

j

a

j

where every

a

j

belongs to one

𝔞

j

of those ideals.

Thus, one can say that the sum ideal is

generated by

the set of all elements of the

individual

ideals; in fact it suffices to use all

generators

of these ideals.

Let

𝔞

+

𝔟

=

𝔡

in a ring

R

.  Because

𝔞

⊆

𝔡

and

𝔟

⊆

𝔡

,  we can say that

𝔡

is a of both

𝔞

and

𝔟

.

1

1

This may be motivated by the situation in

ℤ

:

(

n

)

⊆

(

m

)

iff

m

is a factor of

n

.

Moreover,

𝔡

is

contained

in every common factor

𝔠

of

𝔞

and

𝔟

by virtue of its minimality.  Hence,

𝔡

may be called the

greatest common divisor

of the ideals

𝔞

and

𝔟

.  The notations

𝔞

+

𝔟

=

gcd

⁡

(

𝔞

,

𝔟

)

=

(

𝔞

,

𝔟

)

are used, too.

In an analogous way, the

intersection

of ideals may be designated as the

least common

of the ideals.

The by “

⊆

”

partially ordered set

of all ideals of a ring forms a

lattice

, where the least

upper bound

of

𝔞

and

𝔟

is

𝔞

+

𝔟

and the

greatest lower bound

is

𝔞

∩

𝔟

.  See also the example 3 in

algebraic lattice

.

Title

sum of ideals

Canonical name

SumOfIdeals

Date of creation

2013-03-22 14:39:26

Last modified on

2013-03-22 14:39:26

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

22

Author

pahio (2872)

Entry type

Definition

Classification

msc 13C99

Classification

msc 16D25

Classification

msc 08A99

Synonym

greatest common divisor of ideals

Related topic

QuotientOfIdeals

Related topic

ProductOfIdeals

Related topic

LeastCommonMultiple

Related topic

TwoGeneratorProperty

Related topic

Submodule

Related topic

AlgebraicLattice

Related topic

LatticeOfIdeals

Related topic

MaximalIdealIsPrime

Related topic

AnyDivisorIsGcdOfTwoPrincipalDivisors

Related topic

GcdDomain

Defines

sum ideal

Defines

sum of the ideals

Defines

addition of ideals

Defines

factor of ideal

Defines

greatest common divisor of ideals

Defines

least common

multiple

of ideals

Generated on Fri Feb 9 17:29:49 2018 by

LaTeXML
global dimension

global dimension

For any ring

R

, the

left

global dimension

of

R

is defined to be the

supremum

of

projective dimensions

of

left modules

of

R

:

l

.

Gd

⁡

(

R

)

:=

sup

⁡

{

pd

R

⁡

(

M

)

∣

M

⁢

is a left R-module

}

.

Similarly, the

right global dimension

of

R

is:

r

.

Gd

⁡

(

R

)

:=

sup

⁡

{

pd

R

⁡

(

M

)

∣

M

⁢

is a right R-module

}

.

If

R

is

commutative

, then

l

.

Gd

⁡

(

R

)

=

r

.

Gd

⁡

(

R

)

and we may drop

l

and

r

and simply use

Gd

⁡

(

R

)

to mean the

global dimension

of

R

.

Remarks

.

1.

For a ring

R

,

l

.

Gd

⁡

(

R

)

=

0

iff

r

.

Gd

⁡

(

R

)

=

0

(see the first example below). However, in general,

l

.

Gd

⁡

(

R

)

is not necessarily the same as

r

.

Gd

⁡

(

R

)

.

2.

The left (right) global dimension of a ring can also be defined in terms of injective dimensions. For example, for right global dimension of

R

, we have:

r

.

Gd

⁡

(

R

)

=

sup

⁡

{

id

R

⁡

(

M

)

∣

M

⁢

is a right R-module

}

. This definition turns out to be

equivalent

to the one using projective dimensions.

Examples

.

1.

l

.

Gd

⁡

(

R

)

=

0

iff

R

is a

semisimple ring

iff

r

.

Gd

⁡

(

R

)

=

0

.

2.

r

.

Gd

⁡

(

R

)

=

1

iff

R

is a right

hereditary ring

that is not

semisimple

.

Title

global dimension

Canonical name

GlobalDimension

Date of creation

2013-03-22 14:51:51

Last modified on

2013-03-22 14:51:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13D05

Classification

msc 16E10

Classification

msc 18G20

Synonym

homological dimension

Generated on Fri Feb 9 17:29:51 2018 by

LaTeXML
global dimension of a subring

global dimension of a subring

Let

S

be a

ring with identity

and

R

⊂

S

a

subring

, such that

R

is

contained

in the center of

S

. In this case

S

is a (left)

R

-module via

multiplication

. Throughout by modules we will understand

left modules

and by

global dimension

we will understand left global dimension (we will denote it by

gl dim

⁢

(

S

)

).

Proposition

.

Assume that

gl dim

⁢

(

S

)

=

n

<

∞

. If

S

is free as a

R

-module, then

gl dim

⁢

(

R

)

≤

n

+

1

.

Proof.

Let

M

be a

R

-module. Then, there exists

exact sequence

0

→

K

→

P

n

→

⋯

→

P

0

→

M

→

0

,

of

R

-modules, where each

P

i

is projective (module

K

is just a kernel of a map

P

n

→

P

n

-

1

). We will show, that

K

is also projective (and since

M

is arbitrary, it will show that

gl dim

⁢

(

R

)

≤

n

+

1

).

Since

S

is free as a

R

-module, then the extension of scalars

(

-

⊗

R

S

)

is an

exact functor

from the cateogry of

R

-modules to the

category

of

S

-modules. Furthermore for any projective

R

-module

M

, the

S

-module

M

⊗

R

S

is projective (in the category of

S

-modules). Thus we have following exact sequence of

S

-modules

0

→

K

⊗

R

S

→

P

n

⊗

R

S

→

⋯

→

P

0

⊗

R

S

→

M

⊗

R

S

→

0

,

where each

P

i

⊗

R

S

is a projective

S

-module. But

projective dimension

of

M

⊗

R

S

is at most

n

(since

gl dim

⁢

(

S

)

=

n

). Thus

K

⊗

R

S

is a projective

S

-module (please, see this entry (

http://planetmath.org/ExactSequencesForModulesWithFiniteProjectiveDimension

) for more details).

Note that the restriction of scalars

functor

also maps projective

S

-modules into projective

R

-modules. Thus

K

⊗

R

S

is a projective

R

-module. But

S

is free

R

-module, so

S

≃

⊕

i

∈

I

R

,

for some

index set

I

. Finally we have

K

⊗

R

S

≃

K

⊗

R

(

⊕

i

∈

I

R

)

≃

⊕

i

∈

I

(

K

⊗

R

R

)

≃

⊕

i

∈

I

K

.

This shows, that

K

is a

direct summand

of a projective

R

-module

K

⊗

R

R

and therefore

K

is projective, which

completes

the proof.

□

Title

global dimension of a subring

Canonical name

GlobalDimensionOfASubring

Date of creation

2013-03-22 19:05:09

Last modified on

2013-03-22 19:05:09

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Theorem

Classification

msc 13D05

Classification

msc 16E10

Classification

msc 18G20

Generated on Fri Feb 9 17:29:53 2018 by

LaTeXML
criterion for a module to be noetherian

criterion for a module to be noetherian

Theorem 1

.

A module is

noetherian

if and only if all of its

submodules

and

quotients

(

http://planetmath.org/QuotientModule

) are noetherian.

Proof.

Suppose

M

is Noetherian (over a ring

R

), and

N

⊆

M

a submodule. Since any submodule of

M

is

finitely generated

, any submodule of

N

, being a submodule of

M

, is finitely generated as well. Next, if

A

/

N

is a submodule of

M

/

N

, and if

a

1

,

…

,

a

n

is a generating set for

A

⊆

M

, then

a

1

+

N

,

…

,

a

n

+

N

is a generating set for

A

/

N

.

Conversely

, if every submodule of

M

is Noetherian, then

M

, being a submodule itself, must be Noetherian.
∎

A weaker form of the

converse

is the following:

Theorem 2

.

If

N

⊆

M

is a submodule of

M

such that

N

and

M

/

N

are Noetherian, then

M

is Noetherian.

Proof.

Suppose

A

1

⊆

A

2

⊆

⋯

is an ascending chain of submodules of

M

. Let

B

i

=

A

i

∩

N

, then

B

1

⊆

B

2

⊆

⋯

is an ascending chain of submodules of

N

. Since

N

is Noetherian, the chain terminates at, say

B

n

. Let

C

i

=

(

A

i

+

N

)

/

N

, then

C

1

⊆

C

2

⊆

⋯

is an ascending chain of submodules of

M

/

N

. Since

M

/

N

is Noetherian, the chain stops at, say

C

m

. Let

k

=

max

⁡

(

m

,

n

)

. Then we have

B

k

=

B

k

+

1

and

C

k

=

C

k

+

1

. We want to show that

A

k

=

A

k

+

1

. Since

A

k

⊆

A

k

+

1

, we need the other

inclusion

. Pick

a

∈

A

k

+

1

. Then

a

+

N

=

b

+

N

, where

b

∈

A

k

. This means that

a

-

b

∈

N

. But

b

∈

A

k

+

1

as well, so

a

-

b

∈

N

∩

A

k

+

1

. Since

N

∩

A

k

=

N

∩

A

k

+

1

, this means that

a

-

b

∈

A

k

or

a

∈

A

k

.
∎

Title

criterion for a module to be noetherian

Canonical name

CriterionForAModuleToBeNoetherian

Date of creation

2013-03-22 15:28:46

Last modified on

2013-03-22 15:28:46

Owner

mps (409)

Last modified by

mps (409)

Numerical id

9

Author

mps (409)

Entry type

Theorem

Classification

msc 13E05

Generated on Fri Feb 9 17:29:55 2018 by

LaTeXML
Hilbert basis theorem

Hilbert basis theorem

Let

R

be a right (left)

Noetherian ring

. Then

R

⁢

[

x

]

is also right (left)

Noetherian

.

Title

Hilbert basis theorem

Canonical name

HilbertBasisTheorem

Date of creation

2013-03-22 11:44:55

Last modified on

2013-03-22 11:44:55

Owner

KimJ (5)

Last modified by

KimJ (5)

Numerical id

12

Author

KimJ (5)

Entry type

Theorem

Classification

msc 13E05

Classification

msc 16P40

Generated on Fri Feb 9 17:29:57 2018 by

LaTeXML
irreducible ideal

irreducible ideal

Let

R

be a ring. An ideal

I

in

R

is said to be if, whenever

I

is an

intersection

of two ideals:

I

=

J

∩

K

, then either

I

=

J

or

I

=

K

.

Irreducible ideals

are closely related to the notions of

irreducible elements

in a ring. In fact, the following holds:

Proposition 1

.

If

D

is a

gcd domain

, and

x

is an irreducible element, then

I

=

(

x

)

is an irreducible ideal.

Proof.

If

x

is a unit, then

I

=

D

and we are done. So we assume that

x

is not a unit for the

remainder

of the proof.

Let

I

=

J

∩

K

and suppose

a

∈

J

-

I

and

b

∈

K

-

I

. Then

a

⁢

b

=

x

n

for some

n

∈

ℕ

. Let

c

be a gcd of

a

and

x

. So

c

⁢

d

=

x

for some

d

∈

D

. Since

x

is

irreducible

, either

c

is a unit or

d

is. The proof now breaks down into two cases:

•

c

is a unit. Let

t

be a lcm of

a

and

x

. Then

t

⁢

c

is an

associate

of

a

⁢

x

. But

c

is a unit,

t

and

a

⁢

x

are associates, so that

a

⁢

x

is a lcm of

a

and

x

. As

a

⁢

b

=

x

n

, both

a

∣

a

b

and

x

∣

a

b

hold, which imply that

a

x

∣

a

b

. Write

a

⁢

x

⁢

r

=

a

⁢

b

, where

r

∈

D

. Then

b

=

x

⁢

r

∈

I

, which is impossible by

assumption

.

•

d

is a unit. So

c

is an associate of

x

. Because

c

divides

a

, we get that

x

∣

a

as well, or

a

∈

I

, which is again impossible by assumption.

Therefore, the assumption that

J

-

I

≠

∅

and

K

-

I

≠

∅

is false, which is the same as saying

J

⊆

I

or

K

⊆

I

. But

I

⊆

J

and

I

⊆

K

, either

I

=

J

or

I

=

K

, or

I

is irreducible.
∎

Remark

. In a

commutative

Noetherian ring

, the notion of an irreducible ideal can be used to prove the

Lasker-Noether theorem

: every ideal (in a Noetherian ring) has a

primary decomposition

.

References

1

D.G. Northcott,

Ideal Theory

, Cambridge University Press, 1953.

2

H. Matsumura,

Commutative Ring

Theory

, Cambridge University Press, 1989.

3

M. Reid,

Undergraduate

Commutative Algebra

, Cambridge University Press, 1996.

Title

irreducible ideal

Canonical

name

IrreducibleIdeal

Date of creation

2013-03-22 18:19:47

Last modified on

2013-03-22 18:19:47

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Definition

Classification

msc 13E05

Classification

msc 13A15

Classification

msc 16D25

Synonym

indecomposable ideal

Related topic

IrreducibleElement

Generated on Fri Feb 9 17:30:09 2018 by

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Krull intersection theorem

Krull intersection theorem

Given a

Noetherian ring

A

, an

A

-module

M

, and an ideal

I

inside the

radical

of

A

, we have that

M

is separated with respect to the

I

-adic topology.

Furthermore, if

A

is also an

integral domain

and

J

⊂

A

is a

proper ideal

, we have

⋂

n

>

0

J

n

=

(

0

)

Title

Krull intersection theorem

Canonical name

KrullIntersectionTheorem

Date of creation

2013-03-22 14:36:12

Last modified on

2013-03-22 14:36:12

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Theorem

Classification

msc 13E05

Generated on Fri Feb 9 17:30:11 2018 by

LaTeXML
Noetherian module

Noetherian module

\PMlinkescapephrase

generated by

\PMlinkescapephrase

left noetherian

\PMlinkescapephrase

right noetherian

A (left or right) module

M

over a ring

R

is said to be

Noetherian

if the following

equivalent

conditions hold:

1.

Every

submodule

of

M

is

finitely generated

over

R

.

2.

The

ascending chain condition

holds on submodules.

3.

Every nonempty family of submodules has a

maximal element

.

For example, the

ℤ

-module

ℚ

is not Noetherian,
as it is not finitely generated,
but the

ℤ

-module

ℤ

is Noetherian,
as every submodule is generated by a single

element

.

Observe that changing the ring can change whether a module is Noetherian or not:
for example, the

ℚ

-module

ℚ

is Noetherian,
since it is simple (

http://planetmath.org/SimpleModule

)
(has no nontrivial submodules).

There is also a notion of Noetherian for rings (

http://planetmath.org/Noetherian

):
a ring is left Noetherian if it is Noetherian as a

left module

over itself,
and right Noetherian if it is Noetherian as a right module over itself.
For

non-commutative

rings, these two notions can differ.

The corresponding property for groups is usually called the maximal condition.

Finally, there is the somewhat related notion of a

Noetherian topological space

(

http://planetmath.org/NoetherianTopologicalSpace

).

Title

Noetherian module

Canonical name

NoetherianModule

Date of creation

2013-03-22 11:44:57

Last modified on

2013-03-22 11:44:57

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

24

Author

yark (2760)

Entry type

Definition

Classification

msc 13E05

Classification

msc 33C75

Classification

msc 33E05

Classification

msc 14J27

Classification

msc 86A30

Classification

msc 14H52

Related topic

Noetherian

Defines

Noetherian

Defines

Noetherian left module

Defines

Noetherian right module

Defines

left Noetherian module

Defines

right Noetherian module

Generated on Fri Feb 9 17:30:14 2018 by

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