Boolean Prime Ideal Theorem

Theorem
If $(S, \le)$ is a Boolean algebra, $I$ is an ideal in $S$, $F$ is a filter on $S$, and $I \cap F = \varnothing$, then

there exists a prime ideal $P$ in $S$ such that $I \subseteq P$ and $I \cap F = \varnothing$.

Proof
Let $T$ be the set of ideals in $S$ that contain $I$ and are disjoint from $F$, ordered by inclusion.

Let $N$ be a chain in $T$.

Then $U=\bigcup N$ is clearly disjoint from $F$ and contains $I$. It is also an ideal:


 * Suppose $x\in U$ and $y \le x$.


 * Then for some $A \in N$, $x \in A$.


 * By the definition of union, $x \in U$.


 * Suppose $x \in U$ and $y \in U$.


 * Then for some $A,\,B \in N$, $x \in A$ and $y\in B$.


 * By the definition of a chain, $A \subseteq B$ or $B \subseteq A$.

Suppose without loss of generality that $A \subseteq B$.


 * Then $x \in B$. Since $y$ is also in $B$, and $B$ is an ideal,


 * $x \vee y \in U$.

By Zorn's Lemma, $T$ has a maximal element, $M$.

It remains to show that $M$ is a prime ideal.

Significance
The Boolean Prime Ideal Theorem is weaker than the Axiom of Choice, but is similarly independent of ZF theory. It is sufficient to prove a number of important theorems, although such proofs are often more involved than ones relying on the Axiom of Choice.