Lasker-Noether theorem

Lasker-Noether theorem

Theorem 1

(Lasker-Noether).

Let

R

be a

commutative

Noetherian ring

with

1

. Every ideal in

R

is

decomposable

(

http://planetmath.org/DecomposableIdeal

).

The

theorem

can be proved in two steps:

Proposition 1

.

Every ideal in

R

can be written as a finite

intersection

of

irreducible ideals

Proof.

Let

S

be the set of all ideals of a Noetherian ring

R

which can not be written as a finite intersection of irreducible ideals. Suppose

S

≠

∅

. Then any chain

I

1

⊆

I

2

⊆

⋯

in

S

must terminate in a finite number of steps, as

R

is

Noetherian

. Say

I

=

I

n

is the maxmimal element of this chain. Since

I

∈

S

,

I

itself can not be

irreducible

, so that

I

=

J

∩

K

where

J

and

K

are ideals strictly containing

I

. Now, if

J

∈

S

, then then

I

would not be

maximal

in the chain

I

1

⊆

I

2

⊆

⋯

. Therefore,

J

∉

S

. Similarly,

K

∉

S

. By the definition of

S

,

J

and

K

are both finite intersections of irreducible ideals. But this would imply that

I

∉

S

, a

contradiction

. So

S

=

∅

and we are done.
∎

Proposition 2

.

Every irreducible ideal in

R

is

primary

Proof.

Suppose

I

is irreducible and

a

⁢

b

∈

I

. We want to show that either

a

∈

I

, or some power

n

of

b

is in

I

. Define

J

i

=

I

:

(

b

i

)

, the

quotient of ideals

I

and

(

b

i

)

. Since

⋯

⊆

(

b

n

)

⊆

⋯

⊆

(

b

2

)

⊆

(

b

)

,

we have, by one of the rules on quotients of ideals, an ascending chain of ideals

J

1

⊆

J

2

⊆

⋯

⊆

J

n

⊆

⋯

Since

R

is Noetherian,

J

:=

J

n

=

J

m

for all

m

>

n

. Next, define

K

=

(

b

n

)

+

I

, the

sum of ideals

(

b

n

)

and

I

. We want to show that

I

=

J

∩

K

.

First, it is clear that

I

⊆

J

and

I

⊆

K

, which takes care of one of the

inclusions

. Now, suppose

r

∈

J

∩

K

. Then

r

=

s

+

t

⁢

b

n

, where

s

∈

I

and

t

∈

R

, and

r

⁢

b

n

∈

I

. So,

r

⁢

b

n

=

s

⁢

b

n

+

t

⁢

b

2

⁢

n

. Now,

t

∈

I

:

(

b

2

⁢

n

)

, so

t

∈

I

:

(

b

n

)

. But this means that

r

=

s

+

t

⁢

b

n

∈

I

, and this proves the other inclusion.

Since

I

is irreducible, either

I

=

J

or

I

=

K

. We analyze the two cases below:

•

If

I

=

J

=

I

:

(

b

n

)

, then

I

=

I

:

(

b

)

in particular, since

I

⊆

I

:

(

b

)

⊆

I

:

(

b

n

)

. As

a

⁢

b

∈

I

by

assumption

,

a

∈

I

:

(

b

)

=

I

.

•

If

I

=

K

=

(

b

n

)

+

I

, then

b

n

∈

I

.

This

completes

the proof.
∎

Remarks

.

•

The above theorem can be generalized to any

submodule

of a

finitely generated module

over a commutative Noetherian ring with 1.

•

A ring is said to be

Lasker

if every ideal is decomposable. The theorem above says that every commutative Noetherian ring with 1 is Lasker. There are

Lasker rings

that are not Noetherian.

Title

Lasker-Noether theorem

Canonical name

LaskerNoetherTheorem

Date of creation

2013-03-22 18:19:53

Last modified on

2013-03-22 18:19:53

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 13C99

Defines

Lasker ring

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