Boolean Prime Ideal Theorem

Theorem
Let $\left({S, \le}\right)$ be a Boolean algebra.

Let $I$ be an ideal in $S$.

Let $F$ be a filter on $S$.

Let $I \cap F = \varnothing$.

Then there exists a prime ideal $P$ in $S$ such that:
 * $I \subseteq P$

and:
 * $P \cap F = \varnothing$

Proof from the Axiom of Choice
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 $\displaystyle U = \bigcup N$ is clearly disjoint from $F$ and contains $I$.

Let $x \in U$ and $y \le x$.

Then:
 * $\exists A \in N: x \in A$

By the definition of union:
 * $x \in U$

Let $x \in U$ and $y \in U$.

Then:
 * $\exists A, B \in N: x \in A, y \in B$

By the definition of a chain:
 * $A \subseteq B$

or:
 * $B \subseteq A$

Hence $U$ is also an ideal.

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:

Every Boolean algebra is a distributive lattice.

So by Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice, $M$ 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.