Prime ideal of a lattice

Remarks

A proper ideal $I$ is prime exactly when its complement $L \setminus I$ is a filter; such a filter is called a prime filter. Example: let $X$ be a set, $p \in X$, and $I = \set{A \subseteq X : p \notin A}$ in the power set of $X$. $I$ is an ideal, it is proper since $X \notin I$, and if $p \notin A \cap B$ then $p \notin A$ or $p \notin B$. Non-example: in the power set of $\set{1, 2, 3}$, the principal ideal $\set{\emptyset, \set{1}}$ is proper but not prime. Indeed $\set{2} \cap \set{3} = \emptyset$ lies in it, while neither $\set{2}$ nor $\set{3}$ does. In a distributive lattice, if an ideal $I$ and a filter $F$ are disjoint, then some prime ideal contains $I$ and is disjoint from $F$. This is the prime ideal theorem for distributive lattices, and its proof uses a form of the axiom of choice.