Two ultrafilters equal over a subalgebra I want to prove this theorem:
Let $\mathfrak{B}$ be a Boolean algebra and $\mathfrak A\subsetneq \mathfrak B$ be a subalgebra, then for any $x\in \mathfrak B\setminus \mathfrak A$ there exists ultrafilters $U$ and $V$ of $\mathfrak B$ such that $U\cap \mathfrak A=V\cap \mathfrak A$ and $x\in U$ and $\lnot x\in V$.
This is what I was thinking until I got stuck: Let $x\in U$. $\frak A$ is a subalgebra, so $U\cap \frak A$ is an ultrafilter in $\frak A$, since for any $a\in \frak A$ we have $a\in U$ iff $\lnot a \notin U$. I wanted to prove that $(U\cap{\frak A})\cup\{\lnot x\land a:a\in U\cap \frak A\}$ does not contain $0$, so that it can be extended to an ultrafilter $V$ on $\frak B$ that makes the theorem true, because $V$ is already decided on all elements of $\frak A$. But I don't know how to prove this, or if it is true.
 A: This is very straightforward using Stone duality.  Let $X$ be the Stone space of $\mathfrak{A}$ and $Y$ be the Stone space of $\mathfrak{B}$, so the inclusion $\mathfrak{A}\to\mathfrak{B}$ induces a continuous surjection $f:Y\to X$.  An element $x\in\mathfrak{B}\setminus\mathfrak{A}$ corresponds to a clopen subset $C$ of $Y$ that is not the preimage of any clopen subset of $X$.  The ultrafilters $U$ and $V$ we wish to find correspond to points $p,q\in Y$ such that $f(p)=f(q)$, $p\in C$, and $q\not\in C$.
So, suppose no such points $p$ and $q$ exist.  That means that the sets $f(C)$ and $f(Y\setminus C)$ are disjoint.  Since $C$ and $Y\setminus C$ are both compact, $f(C)$ and $f(Y\setminus C)$ are also compact and hence closed.  Since $f$ is surjective, they cover all of $X$ and so are complements of each other.  Thus $f(C)$ is in fact clopen.  But $f^{-1}(f(C))=C$ (since $f(C)$ and $f(Y\setminus C)$ are disjoint), so this contradicts our assumption that $C$ was not the preimage of a clopen subset of $X$.

Here is a purely algebraic translation of the argument.  Let $F=\{y\in\mathfrak{A}:y\geq x\}$ and $G=\{y\in\mathfrak{A}:y\geq\neg x\}$ (these are the filters corresponding to $f(C)$ and $f(Y\setminus C)$ in the topological argument).  Note that since $x\not\in\mathfrak{A}$, every element of $F$ intersects $\neg x$ and thus intersects every element of $G$, so $F\cup G$ generates a proper filter.  Extend it to an ultrafilter $W$ on $\mathfrak{A}$.  Since $W$ contains both $F$ and $G$, it does not contain any elements that are disjoint from either $x$ or $\neg x$.  Thus both $W\cup \{x\}$ and $W\cup\{\neg x\}$ can be extended to ultrafilters $U$ and $V$ on $\mathfrak{B}$, which will satisfy $U\cap\mathfrak{A}=V\cap\mathfrak{A}=W$.
(Note that your approach of starting with $U$ containing $x$ is doomed--for an arbitrary $U$ containing $x$, there might not exist any $V$ that works with it.  In terms of the topological picture, the quotient map $f$ must identify together two points from $C$ and $Y\setminus C$, but it may not identify every point of $C$ with a point of $Y\setminus$.  For instance, you could just pick one point in each of $C$ and $Y\setminus C$ and take $X$ to be the quotient of $Y$ that identifies those two points.)
