For every cardinal $\kappa$ does there exist a Boolean algebra that has exactly $\kappa$ many ultrafilters?

Question

For every cardinal $\kappa$ does there exist a Boolean algebra that has exactly $\kappa$ many ultrafilters?

By Jech's Exercise I.7.25, I know that a Boolean algebra has at least as many ultrafilters in it as it has elements.

Is it easy to answer the question above? If not, what are good resources for this?

Answer 1

Let $FC(X)$ denote the Boolean algebra of finite and co-finite subsets of $X$.
Then the ultrafilters of $FC(X)$ are the principal filters $\uparrow\!\{x\}$, with $x \in X$ together with the free ultrafilter consisting of the co-finite subsets of $X$ (see here).
So $FC(X)$ has $\kappa$ many ultrafilters, whenever $|X| = \kappa$.

Answer 2

If $X$ is compact Hausdorff and zero-dimensional its clopen algebra Clop($X$) only has fixed ultrafilters by compactness, so exactly $|X|$ many.

I think that for every infinite cardinal $\kappa$, its successor ordinal $\kappa +1$ in the order topology is just such a space. So yes. (For finite cardinals $n$ just take $\mathscr{P}(n)$ of course).