Thinking about ultrafilters a question came to my mind. Suppose that we have $A$ and $B$ two Boolean algebras such that $A\subseteq B$ and $U$ be an ultrafilter composed by elements of $B$. Is it true that $U\cap A$ is an ultrafilter on $A$?
Clearly $U\cap A$ is a filter on $A$. Only remains to prove the maximality. For this I cannot prove it but I tried:
Suppose $U\cap A$ is not an ultrafilter on $A$. Then there exist $\mathcal{F}$ an ultrafilter on $A$ such that $U\cap A\subsetneq \mathcal{F}$ (proper contention). Then there exist $F\in\mathcal{F}$ such that $F\notin U\cap A$. But then $F\notin U$. From here I don't know how to continue. I think that we can use $F$ to construct a new filter that contains $U$ and $F$ to derive a contradiction with the maximality of $U$ in $B$. Any hint?
Your idea is correct. Consider the filter on $B$ generated by $U\cup\{F\}$. This will be a proper filter as long as there is no finite subset of $U\cup\{F\}$ whose meet is $0$. Since $U$ is closed under finite meets, this can only happen if some element $u\in U$ is disjoint from $F$. But then $u\leq \neg F$ so $\neg F\in U$. But $\neg F\in A$ so $\neg F\in U\cap A\subseteq\mathcal{F}$, which is a contradiction since $F\in\mathcal{F}$.
Or more simply: a filter is an ultrafilter iff for any $a$, exactly one of $a$ and $\neg a$ is in the filter. It is then trivial that if $U$ satisfies this condition as a filter on $B$, then $U\cap A$ satisfies it as a filter on $A$. (More generally, this characterization is typically the "right" way to think about ultrafilters; the fact that it is equivalent to being a maximal proper filter is somewhat of a coincidence.)