representing a distributive lattice by ring of sets
In this entry, we present the proof of a fundamental fact that every distributive lattice^{} is lattice^{} isomorphic to a ring of sets, originally proved by Birkhoff and Stone in the 1930’s. The proof uses the prime ideal theorem of Birkhoff (http://planetmath.org/BirkhoffPrimeIdealTheorem). First, a simple results from the prime ideal theorem:
Lemma 1.
Let $L$ be a distributive lattice and $a\mathrm{,}b\mathrm{\in}L$ with $a\mathrm{\ne}b$. Then there is a prime ideal^{} containing one or the other.
Proof.
Let $I=\u27e8a\u27e9$ and $J=\u27e8b\u27e9$, the principal ideals^{} generated by $a,b$ respectively. If $I=J$, then $b\le a$ and $a\le b$, or $a=b$, contradicting the assumption^{}. So $I\ne J$, which means either $a\notin J$ or $b\notin I$. In either case, apply the prime ideal theorem to obtain a prime ideal containing $I$ (or $J$) not containing $b$ (or $a$). ∎
Definition. Let $L$ be a distributive lattice, and $X$ the set of all prime ideals of $L$. Define $F:L\to P(X)$, the powerset of $X$, by
$$F(a):=\{P\mid a\notin P\}.$$ 
Proposition 1.
$F$ is an injection.
Proof.
If $a\ne b$, then by the lemma there is a prime ideal $P$ containing one but not another, say $a\in P$ and $b\notin P$. Then $P\notin F(a)$ and $P\in F(b)$, so that $F(a)\ne F(b)$. ∎
Proposition 2.
$F$ is a lattice homomorphism^{}.
Proof.
There are two things to show:

•
$F$ preserves $\wedge $: If $P\in F(a\wedge b)$, then $a\wedge b\notin P$, so that $a\notin P$ and $b\notin P$, since $P$ is a sublattice. So $P\in F(a)$ and $P\in F(b)$ as a result. On the other hand, if $P\in F(a)\cap F(b)$, then $a\notin P$ and $b\notin P$. Since $P$ is prime, $a\wedge b\notin P$, so that $P\in F(a\wedge b)$. Therefore, $F(a\wedge b)=F(a)\cap F(b)$.

•
$F$ preserves $\vee $: If $P\in F(a\vee b)$, then $a\vee b\notin P$, which implies that $a\notin P$ or $b\notin P$, since $P$ is a sublattice of $L$. So $P\in F(a)\cup F(b)$. On the other hand, if $P\in F(a)\cup F(b)$, then $a\vee b\notin P$, since $P$ is a lattice ideal. Hence $F(a\vee b)=F(a)\cup F(b)$.
Therefore, $F$ is a lattice homomorphism. ∎
Theorem 1.
Every distributive lattice is isomorphic to a ring of sets.
Proof.
Let $L,X,F$ be as above. Since $F:L\to P(X)$ is an embedding, $L$ is lattice isomorphic to $F(L)$, which is a ring of sets. ∎
Remark. Using the result above, one can show that if $L$ is a Boolean algebra^{}, then $L$ is isomorphic to a field of sets. See link below for more detail.
Title  representing a distributive lattice by ring of sets 

Canonical name  RepresentingADistributiveLatticeByRingOfSets 
Date of creation  20130322 19:08:24 
Last modified on  20130322 19:08:24 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 06D99 
Classification  msc 06D05 
Related topic  RingOfSets 
Related topic  RepresentingABooleanLatticeByFieldOfSets 