# Ultrafilter on a boolean algebra containing $u$ but not $v$

I was reading stone's representation theorem and a part of the proof was omitted and i wanted to check if i got it right:

So assume that $B$ is a boolean algebra and $u,v \in B \wedge u\neq v$ and we want to prove the existence of an ultrafilter that contains exactly one of $u$ or $v$. Since $u \neq v$ then $u.(-v) \neq 0$ and the set $\{ u, -v\}$ has the finite intersection property and there is a filter $F$ extending it. The dual of $F$ is an ideal $I$, such that $\{-u, v\} \subset I$ and according to the prime ideal theorem there is a prime ideal $P$ extending $I$ and the dual of $P$ is an ultrafilter $U$ such that $\{u, -v\} \subset U$ and we are done. Is this correct? Is there a faster way? Do we have tarski's ultrafilter theorem's equivalent for ultrafilters in boolean algebras?

This isn't quite correct. $u \neq v$ does not imply that $u \cdot (-v) \neq 0$. (E.g. if you start with a poset $(\mathbb P; \preceq)$ and $u,v \in \mathbb P$ are such that $u \prec v$, then in the Boolean completion of $\mathbb P$ we will have $u \cdot (-v) = \emptyset$.
A direct corollary of the prime ideal theorem is that every filter is contained in an ultrafilter (by the proof you outlined implicitly: Given a filter $F$ consider its associated ideal $I$ obtained by taking pointwise complements. Extend $I$ to a prime ideal $P$ and consider its associated filer $U$ -- again by taking pointwise complements. This will be an ultrafilter such that $F \subseteq U$).
• But then i think $v.(-u) \neq 0$ and we can switch their places, right? – Shervin Sorouri Jun 18 '18 at 9:55
• @ShervinSorouri Sorry, there was a typo. But you're right in that you can always find an ultrafilter containing only one of $u,v$. (In case that $u \prec v$ you cannot, however, find an ultrafilter containing $u$ but not $v$.) – Stefan Mesken Jun 18 '18 at 9:59