Birkhoff prime ideal theorem

Birkhoff prime ideal theorem

Birkhoff Prime Ideal Theorem

. Let

L

be a

distributive lattice

and

I

a proper

lattice ideal

of

L

. Pick any

element

a

∉

I

. Then there is a

prime ideal

P

in

L

such that

I

⊆

P

and

a

∉

P

.

Proof.

If

I

is prime, then we are done. Let

S

:=

{

J

∣

J

⁢

is an ideal in

⁢

L

⁢

, and

⁢

a

∉

J

}

. Then

I

∈

S

. Order

S

by

inclusion

. This turns

S

into a poset. Let

C

be a chain in

S

. Let

K

=

⋃

C

. If

x

,

y

∈

K

, then

x

∈

J

1

and

y

∈

J

2

for some ideals

J

1

,

J

2

∈

C

. Since

C

is a chain, we may assume that

J

1

⊆

J

2

, so that

x

∈

J

2

as well. This means

x

∨

y

∈

J

2

⊆

K

. Next, assume

x

∈

K

and

y

≤

x

. Then

x

∈

J

for some ideal

J

∈

C

, so that

y

∈

J

⊆

K

also. This shows that

K

is an ideal. If

a

∈

K

, then

a

∈

J

for some

J

∈

C

⊆

S

, contradicting the

definition

of

S

. So

a

∉

K

and

K

∈

S

also. This shows that every chain in

S

has an

upper bound

. We can now appeal to Zorn’s lemma, and conclude that

S

has a

maximal element

, say

P

.

We now want to show that

P

is the candidate that we are seeking:

P

is a prime ideal in

L

and

a

∉

P

. Since

P

∈

S

,

P

is an ideal such that

a

∉

P

. So the only thing left to prove is that

P

is prime. This amounts to showing that if

x

∧

y

∈

P

, then

x

∈

P

or

y

∈

P

. Suppose not:

x

,

y

∉

P

. Let

Q

1

be the

ideal generated by

elements of

P

and

x

, and

Q

2

the ideal generated by

P

and

y

. Since

Q

1

and

Q

2

properly contain

P

,

a

∈

Q

1

and

a

∈

Q

2

. Write

a

≤

p

1

∨

x

and

a

≤

p

2

∨

y

, where

p

1

,

p

2

∈

P

. Then

a

∨

p

2

≤

(

p

1

∨

p

2

)

∨

x

and

a

∨

p

1

≤

(

p

1

∨

p

2

)

∨

y

. Take the meet of these two

expressions

, and we obtain

(

a

∨

p

2

)

∧

(

a

∨

p

1

)

≤

(

(

p

1

∨

p

2

)

∨

x

)

∧

(

(

p

1

∨

p

2

)

∨

y

)

. Since

L

is

distributive

, on the left hand side, we get

a

∨

(

p

1

∧

p

2

)

. On the right hand side, we have

(

p

1

∨

p

2

)

∨

(

x

∧

y

)

∈

P

. As the left hand side is less than or equal to the right hand side, we get that

a

∨

(

p

1

∧

p

2

)

∈

P

. Since

a

≤

a

∨

(

p

1

∧

p

2

)

∈

P

,

a

∈

P

, a

contradiction

. Therefore,

P

is prime and the proof is

complete

.
∎

In the proof, we use the fact that, an element

a

∈

L

belongs to the ideal generated by ideals

I

k

iff

a

is less than or equal to a finite join of elements, each of which belongs to some

I

k

.

Remarks

.

1.

The

theorem

can be generalized: if we use a subset

S

∩

I

=

∅

instead of an element

a

∉

I

, there is a prime ideal

P

containing

I

but excluding

S

.

2.

Birkhoff’s prime ideal theorem has been shown to be

equivalent

to the

axiom of choice

, under ZF.

Title

Birkhoff prime ideal theorem

Canonical

name

BirkhoffPrimeIdealTheorem

Date of creation

2013-03-22 17:02:18

Last modified on

2013-03-22 17:02:18

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

11

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 06D05

Classification

msc 03E25

Related topic

DistributiveLattice

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