Cardinality of the set of ultrafilters on an infinite Boolean algebra

Question

Let $\mathfrak B$ be a Boolean algebra with an infinite power $\kappa$. My question is how many ultrafilters does it have? $\kappa$ or $2^\kappa$? Or even smaller?

Answer 1

It can be at least those two options:

Example I:

Consider the algebra $\{A\subseteq\mathbb R\mid A\text{ finite, or }\mathbb R\setminus A\text{ is finite}\}$, that is finite co-finite subsets of $\mathbb R$. It is not hard to verify that this is indeed a Boolean algebra of size $2^{\aleph_0}$.

Suppose that $U$ is an ultrafilter over this Boolean algebra, if it does not contain any singleton then it has to be the collection of co-finite sets; if it contains a singleton then it is principal. We have continuum many principal ultrafilters and one free. Therefore a Boolean algebra of size continuum with continuum many ultrafilters.

Example II: On the other hand consider $\mathcal P(\mathbb N)$. We know that there are $2^{2^{\aleph_0}}$ many ultrafilters over this Boolean algebra. So we have a Boolean algebra of size $2^{\aleph_0}$ with $2^{2^{\aleph_0}}$ many ultrafilters.

For a further discussion on this sort of example see: The set of ultrafilters on an infinite set is uncountable