The last sentence of the section "Types and existence of ultrafilters" of Wikipedia's article on ultrafilter (current revision) says:
In ZF without the axiom of choice, it is possible that every ultrafilter is principal.{see p.316, [Halbeisen, L.J.] "Combinatorial Set Theory", Springer 2012}
This statement is incorrect isn't it? Shouldn't "ultrafilter" here be replaced by "ultrafilter on $\mathscr{P}(\omega)$ (the power set of the set of natural numbers)"?
I think that for some Boolean algebras, one can show the existence of a nonprincipal ultrafilter without assuming the axiom of choice (or the ultrafilter lemma or the Boolean prime ideal theorem). Let $B$ be the set of all finite and cofinite subsets of the set of natural numbers, and let $U$ be the set of all cofinite subsets of the set of natural numbers. Then $U$ is a nonprincipal ultrafilter on the Boolean algebra $B$, isn't it?
I would also like to ask the following more general question: for what kind of Boolean algebra, can one show the existence of a nonprincipal ultrafilter without assuming some choice?
EDIT: I am not asking whether there is an infinite Boolean algebra with a unique ultrafilter (unless "nonprincipal ultrafilter that can be found without choice" and "unique filter" should mean the same thing, which I fail to see). I am first asking whether my example is a counterexample to the statement from Wikipedia. Assuming that I am right about this first point, I am then asking in what cases you can find nonprincipal ultrafilters without using choice. I actually don't know what $\mathcal{P}(\omega)/\text{fin}$ denotes in the suggested question. In any case, the answer given there is only an example and cannot be an answer to my question. I would appreciate if you could explain how my question is a duplicate or reopen my question so that people could provide answers that I would be able to understand.