Can someone give me an example of an ultrafilter which is not principal?
closed as off topic by HJRW, Mark Sapir, Simon Thomas, Andreas Blass, Gerald Edgar Sep 27 '11 at 14:50
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
Do you mean on a set or on a Boolean algebra? You need some fragment of the axiom of choice to construct a non-principal ultrafilter on the natural numbers, for example. (See the link in Igor Rivin's answer.)
However, the Boolean algebra consisting of finite and cofinite subsets of $\mathbb N$ has a non-principal ultrafilter, namely the collection of all cofinite sets.
As far as I know, the existence of non principal (= free) ultrafilters is independent on ZF. It follows from AC, but it's weaker: Ultrafilter's lemma, which says exactly that non principal ultrafilter exist, by using AC in the form of Zorn's lemma, does not imply AC.
So, an explicit construction cannot be given, since otherwise you would have proved the Axiom of Choice. If you want a non-constructive construction, you can also follow the proof of Ultrafilter's lemma: take the family of all filters and apply Zorn's lemma (see this discussion http://everything2.com/title/Proof+that+any+filter+can+be+extended+to+an+ultrafilter for some apparently right details).
In some cases one can find quite explicit examples, as suggested above for Boolean algebras. But this more difficult for ultrafilters on a set: the family of cofinite (the completementary is finite) subsets of a countable set is NOT an ultrafilter!