Editing Boolean prime ideal theorem (section)

==Boolean prime ideal theorem==
The Boolean prime ideal theorem is the strong prime ideal theorem for Boolean algebras. Thus the formal statement is:

: Let ''B'' be a Boolean algebra, let {{mvar|I}} be an ideal and let ''F'' be a filter of ''B'', such that {{mvar|I}} and ''F'' are [[disjoint set|disjoint]]. Then {{mvar|I}} is contained in some prime ideal of ''B'' that is disjoint from ''F''.

The weak prime ideal theorem for Boolean algebras simply states:

: Every Boolean algebra contains a prime ideal.

We refer to these statements as the weak and strong ''BPI''. The two are equivalent, as the strong BPI clearly implies the weak BPI, and the reverse implication can be achieved by using the weak BPI to find prime ideals in the appropriate quotient algebra.

The BPI can be expressed in various ways. For this purpose, recall the following theorem:

For any ideal {{mvar|I}} of a Boolean algebra ''B'', the following are equivalent:
* {{mvar|I}} is a prime ideal.
* {{mvar|I}} is a maximal ideal, i.e. for any proper ideal {{mvar|J}}, if {{mvar|I}} is contained in {{mvar|J}} then {{math|1={{var|I}} = {{var|J}}}}.
* For every element ''a'' of ''B'', {{mvar|I}} contains exactly one of {''a'', ¬''a''}.
This theorem is a well-known fact for Boolean algebras. Its dual establishes the equivalence of prime filters and ultrafilters. Note that the last property is in fact self-dual—only the prior assumption that {{mvar|I}} is an ideal gives the full characterization. All of the implications within this theorem can be proven in ZF.

Thus the following (strong) maximal ideal theorem (MIT) for Boolean algebras is equivalent to BPI:

:Let ''B'' be a Boolean algebra, let {{mvar|I}} be an ideal and let ''F'' be a filter of ''B'', such that {{mvar|I}} and ''F'' are disjoint. Then {{mvar|I}} is contained in some maximal ideal of ''B'' that is disjoint from ''F''.

Note that one requires "global" maximality, not just maximality with respect to being disjoint from ''F''. Yet, this variation yields another equivalent characterization of BPI:

:Let ''B'' be a Boolean algebra, let {{mvar|I}} be an ideal and let ''F'' be a filter of ''B'', such that {{mvar|I}} and ''F'' are disjoint. Then {{mvar|I}} is contained in some ideal of ''B'' that is maximal among all ideals disjoint from ''F''.

The fact that this statement is equivalent to BPI is easily established by noting the following theorem: For any [[distributive lattice]] ''L'', if an ideal {{mvar|I}} is maximal among all ideals of ''L'' that are disjoint to a given filter ''F'', then {{mvar|I}} is a prime ideal. The proof for this statement (which can again be carried out in ZF set theory) is included in the article on ideals. Since any Boolean algebra is a distributive lattice, this shows the desired implication.

All of the above statements are now easily seen to be equivalent. Going even further, one can exploit the fact the dual orders of Boolean algebras are exactly the Boolean algebras themselves. Hence, when taking the equivalent duals of all former statements, one ends up with a number of theorems that equally apply to Boolean algebras, but where every occurrence of {{em|ideal}} is replaced by {{em|filter}}{{Citation needed|reason=(weak Ultrafilter ⇏ strong Ultrafilter), whereas (weak BPI ⇒ strong BPI)|date=November 2021}}. It is worth noting that for the special case where the Boolean algebra under consideration is a [[powerset]] with the [[subset]] ordering, the "maximal filter theorem" is called the ultrafilter lemma.

Summing up, for Boolean algebras, the weak and strong MIT, the weak and strong PIT, and these statements with filters in place of ideals are all equivalent. It is known that all of these statements are consequences of the [[Axiom of Choice]], ''AC'', (the easy proof makes use of [[Zorn's lemma]]), but cannot be proven in [[Zermelo–Fraenkel set theory|ZF]] (Zermelo-Fraenkel set theory without ''AC''), if ZF is [[consistent]]. Yet, the BPI is strictly weaker than the axiom of choice, though the proof of this statement, due to J. D. Halpern and [[Azriel Lévy]] is rather non-trivial.