Extending a filter to an ultrafilter inside a coideal Assume the axiom of choice, every coideal contains an ultrafilter (which is equivalently a minimal coideal), and every filter may be extended to an ultrafilter.
Now suppose we have a coideal $\mathcal{H}$ and a filter $\mathcal{F} \subseteq \mathcal{H}$. Is it always possible to obtain an ultrafilter $\mathcal{U}$ such that $\mathcal{F} \subseteq \mathcal{U} \subseteq \mathcal{H}$?
 A: Yes. Applying Zorn's Lemma to the set of all filters containing $\mathcal{F}$ and contained in $\mathcal{H}$, we obtain a maximal such filter $\mathcal{U}$. I claim $\mathcal{U}$ is an ultrafilter.
Let $A$ be a set, and suppose for contradiction that neither $A$ nor its complement $A'$ is in $\mathcal{U}$. Let $\mathcal{U}[A]$ be the filter generated by $\mathcal{U}$ and $A$, and similarly for $\mathcal{U}[A']$. Both of these filters properly extend $\mathcal{U}$ and contain $\mathcal{F}$, so by maximality they must not be contained in $\mathcal{H}$.
So we can find $B\in \mathcal{U}[A]$ with $B\notin \mathcal{H}$ and $C\in \mathcal{U}[A']$ with $C\notin \mathcal{H}$. Since $\mathcal{H}$ is a coideal, $B\cup C\notin \mathcal{H}$. By the definition of $\mathcal{U}[A]$ and $\mathcal{U}[A']$, we have $B\supseteq X\cap A$ for some $X\in \mathcal{U}$ and $C\supseteq Y\cap A'$ for some $Y\in \mathcal{U}$. But then $$B\cup C \supseteq (X\cap A)\cup (Y\cap A') = (X\cup Y)\cap (X\cup A)\cap (Y\cup A') \cap (A\cup A')\in \mathcal{U},$$ so $B\cup C\in \mathcal{U}$. This contradicts $\mathcal{U}\subseteq \mathcal{H}$.

A slight generalization of this argument can be used to show that for any distributive lattice $(D,\land,\lor,\top,\bot)$, if $F\subseteq D$ is a filter and $I\subseteq D$ is an ideal such that $F\cap I = \varnothing$, then there is a prime filter containing $F$ and disjoint from $I$ (equivalently, a prime ideal containing $I$ and disjoint from $F$). This theorem is the key step in the Stone representation theorem for distributive lattices.
