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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. First, a simple results from the prime ideal theorem:
Lemma 1.
Let $L$ be a distributive lattice and $a,b\in L$ with $a\neq b$. Then there is a prime ideal containing one or the other.
Proof.
Let $I=\langle a\rangle$ and $J=\langle b\rangle$, the principal ideals generated by $a,b$ respectively. If $I=J$, then $b\leq a$ and $a\leq b$, or $a=b$, contradicting the assumption. So $I\neq J$, 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 $a$). ∎
Before proving the theorem, we have one more concept to introduce:
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 $a\neq b$, 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 $F(a)\neq F(b)$. ∎
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.
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