Let pp and qq be distinct ultrafilters (maximal proper filters) in a distributive bounded lattice LL. Is possible to find x∈p∖qx\in p\setminus q and y∈q∖py\in q\setminus p such that x∨y=1x\vee y = \mathbb{1}?

=================

Yes, I will correct.

– user34870

Oct 20 at 19:43

3

No, just consider a chain.

– Keith Kearnes

Oct 20 at 19:46

3

Usually, the term “ultrafilter” is used in the context of Boolean algebras, where there are (at least) two equivalent definitions for it: (1) maximal proper filter and (2) prime filter. In more general distributive lattices, these are not equivalent, so you should say which definition you meant. If it’s (2), then the comment by @KeithKearnes answers your question.

– Andreas Blass

Oct 20 at 20:12

=================

1 Answer

1

=================

If LL is a bounded chain, then it does not have distinct maximal ideals, and it does not have incomparable prime ideals. Thus, whether “ultrafilter” means “maximal filter” or “prime filter”, one cannot choose xx and yy as desired. But this observation does not answer the following interpretation of the question: Suppose LL HAS distinct maximal ideals pp and qq. Is it possible to find x∈p∖qx\in p\setminus q and y∈q∖py\in q\setminus p such that x∨y=1x\vee y = 1?

The answer to the revised question is also “No” (or, “not necessarily”). Let LL be the lattice of finite subsets of the natural numbers with a new top element called ⊤\top adjoined. For each n∈Nn\in \mathbb N, the set FnF_n of all subsets S⊆NS\subseteq \mathbb N which contain nn is a maximal (hence prime) filter. Let p=F3p=F_3 and q=F7q=F_7. If x∈p∖qx\in p\setminus q and y∈q∖py\in q\setminus p, then xx and yy are finite, so x∨y≠⊤x\vee y\neq \top.