Talk:Boolean prime ideal theorem

ultrafilter lemma

"Apparently, the ultrafilter lemma also implies BPI, such that both statements are equivalent -- please confirm if this is known to you." I managed to work out a rather convoluted proof of this, showing that ultrafilter lemma-->compactness theorem-->BPI for free Boolean algebras-->BPI. But I get the feeling there ought to be a more direct proof, and although I was very careful, I might have tacitly used some aspect of the axiom of choice at some point in my proof. --—Preceding unsigned comment added by 70.245.244.82 (talk • contribs)

The compactness theorem (even for sentential logic) directly implies the BPI:

Fix a Boolean algebra B. For every b in B let p_b be a propositional variable. Let T be the set of all formulas of the following form:

1. not p_b iff p_{not b}
2. (p_b and p_c) iff p_{b and c}
3. (p_b or p_c) iff p_{b or c}

where "not", "and", "or" on the left side is a logical symbol, and "not", "and", "or" on the right side are the boolean operations. T is easily seen to be finitely consistent, and every truth assignment to the propositional variables that satisfies T will induce an ultrafilter on B.

--Aleph4 22:20, 2 May 2006 (UTC)

often?

"the Boolean prime ideal theorem is often taken as an axiom of set theory". Is it really? Ever? That sounds like there are (many) people who actually work in the set theory "ZF + BPI". I doubt that. Of course there are few people who investigate the strength of BPI, but I doubt that they consider BPI an "axiom of set theory".

--Aleph4 18:21, 24 September 2005 (UTC)

merged ultrafilter lemma

I merged ultrafilter lemma here as a new section. The old talk for that page is very brief; you can find it at Talk:Ultrafilter lemma if you are interested. CMummert 02:58, 18 July 2006 (UTC)

Weak BPI for Boolean algebras

Weak and strong PIT are equivalent for Boolean algebras via quotient algebras. The same should hold for distributive lattices, but I am not absolutely certain. TexD 19:12, 23 October 2006 (UTC)

Is article too technical?

I think the article is too technical and it would be possible to simplify it, especially in the lead. Absolutelypuremilk (talk) 14:08, 4 December 2016 (UTC)

Which part of the lead is "too technical"? It appears to be a straightforward and direct statement. As the lead says, these appear to be "simple and intuitive", so its hard to imagine how to write the lead so that its more-simpler and more-intuitive. One could try to jam a mini-definition of "ideal" into the lead, or maybe just "prime ideal" but why? Ideals occur in pretty much all branches of algebra; yet another definition won't save the reader if this is a new concept to them. 67.198.37.16 (talk) 17:51, 26 February 2018 (UTC)

Slanted article towards order-related topics

• It is announced that “This article focuses on prime ideal theorems from order theory.”

What about maximal/prime ideals from classical algebra? For example Krull's theorem is hardly mentioned; the converse –(Krull ⇒ AC) in ZF– (due to Wilfrid Hodges) is missing.
Likewise BPI is not only a consequence of “In every commutative unitary ring, every proper ideal is included in some prime ideal”, but both statements are in fact equivalent (Dana Scott). Shouldn't these results (which AFAIK don't seem to stem from poset considerations) be discussed in this article?

• The section named Applications appears misleading since, along with the ultrafilter lemma, the cited results are not only consequences but equivalent to BPI, and a much longer list of them should appear there.

The Stone–Čech compactification theorem, the Compactness theorem for 1st-order logic, the Banach–Alaoglu theorem from functional analysis, the compactness of any $\{0,1\}^{X}$, among others.

• Also the article asserts that (Ultrafilter lemma ⇒ BPI) in ZF alone by use of the Stone's representation theorem but… this latter theorem is not provable in ZF; actually ZF ⊦ (BPI ⟺ Stone's theorem). It remains true however that (Ultrafilter lemma ⟺ BPI) in ZF.--ヒナゲシさん (talk) 05:42, 8 November 2021 (UTC)
  
• I am baffled by the following statement: “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 ideal is replaced by filter”.

In ZF (weak BPI) $\Rightarrow$ (strong BPI); but the weak Ultrafilter theorem (there exists a nonprincipal ultrafilter on any infinite set) $\nRightarrow$ the strong form (every filter on an infinite set can be extended to a nonprincipal ultrafilter).

Quoting Thomas Jech (The Axiom of Choice, 1973, section 8.3): “There is a model of ZF in which the Ultrafilter Theorem (see Section 2.3) fails but every infinite set has a nontrivial ultrafilter.”--ヒナゲシさん (talk) 21:20, 8 November 2021 (UTC)