Boolean Prime Ideal Theorem/Proof 1

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Theorem

Let $\struct {S, \le}$ be a Boolean algebra.

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

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

Let $I \cap F = \O$.

Then there exists a prime ideal $P$ in $S$ such that:

$I \subseteq P$

and:

$P \cap F = \O$

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 $\ds U = \bigcup N$ 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.

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Without loss of generality, suppose $A \subseteq B$.

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.

Axiom of Choice

This proof depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.