Question about ultrafilters in a distributive lattice

Question

Let $p$ and $q$ be distinct ultrafilters (maximal proper filters) in a distributive bounded lattice $L$. Is possible to find $x\in p\setminus q$ and $y\in q\setminus p$ such that $x\vee y = \mathbb{1}$?

Answer 1

If $L$ 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 $x$ and $y$ as desired. But this observation does not answer the following interpretation of the question: Suppose $L$ HAS distinct maximal ideals $p$ and $q$. Is it possible to find $x\in p\setminus q$ and $y\in q\setminus p$ such that $x\vee y = 1$?

The answer to the revised question is also "No" (or, "not necessarily"). Let $L$ be the lattice of finite subsets of the natural numbers with a new top element called $\top$ adjoined. For each $n\in \mathbb N$, the set $F_n$ of all subsets $S\subseteq \mathbb N$ which contain $n$ is a maximal (hence prime) filter. Let $p=F_3$ and $q=F_7$. If $x\in p\setminus q$ and $y\in q\setminus p$, then $x$ and $y$ are finite, so $x\vee y\neq \top$.