Prime ideal theorem for distributive lattices

Remarks

The usual proof applies Zorn's lemma to the set of ideals that contain $I$ and are disjoint from $F$, ordered by inclusion. Let $P$ be a maximal one, and suppose $x \wedge y \in P$ with $x \notin P$ and $y \notin P$. By maximality, the ideals generated by $P \cup \set{x}$ and by $P \cup \set{y}$ meet $F$, so $f \le p \vee x$ and $g \le q \vee y$ for some $f, g \in F$ and $p, q \in P$. Then $f \wedge g \le (p \vee q) \vee (x \wedge y)$ by distributivity, so $f \wedge g \in P \cap F$, a contradiction.

Example: in the lattice of divisors of $12$, with $\gcd$ as meet and $\lcm$ as join, take $I = \set{1, 2}$ and $F = \set{3, 6, 12}$. The set $P = \set{1, 2, 4}$ of divisors not divisible by $3$ is a prime ideal with $I \subseteq P$ and $P \cap F = \emptyset$.

Distributivity is necessary. In the diamond lattice M3 with middle elements $a, b, c$, the proper ideals are $\set{0}$ and $\set{0, x}$ for a middle element $x$. None of them is prime: for instance $a \wedge b = 0$, while neither $a$ nor $b$ lies in $\set{0, c}$. So the ideal $\set{0}$ and the filter $\set{1}$ admit no separating prime ideal.

The Boolean prime ideal theorem is the special case of Boolean algebras. Over ZF set theory, the two statements are equivalent, and both are strictly weaker than the axiom of choice.

Going deeper

For a student

The theorem is the key step in representing a distributive lattice as a lattice of sets: map $a$ to the set of prime filters that contain it. Separating $a \not\le b$ by a prime ideal containing $b$ and missing $a$ shows that this map is injective. This is the basis of Stone's representation theorem for distributive lattices and of Priestley duality. It needs a form of choice, and over ZF it is equivalent to the Boolean prime ideal theorem.