I thought I had heard or read somewhere that the existence of a non-principal ultrafilter on $\omega$ was equivalent to some common weakening of AC. As I searched around, I read that this is not the case: neither countable choice nor dependent choice are strong enough.


This leads me to two questions:


Where would I find a proof that DC is not strong enough to prove the existence of a non-principal ultrafilter on $\omega$?


Is the assumption that there exists a non-principal ultrafilter on $\omega$ strong enough to show DC or countable choice? I.e. is "there exists a non-principal ultrafilter on $\omega$" stronger than either of countable or dependent choice?



Non-principal ultrafilters on ω
See the [Prime ideal theorem](http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem). 


The existence of ultrafilters on every Boolean algebra (which implies non-principal ultrafilters on ω, since these come from ultrafilters on the Boolean algebra P(ω)/Fin) is a set-theoretic principle that follows from AC and is not provable in ZF (if ZF is consistent), but which does not imply full AC. Thus, it is an intermediate weaker choice principle. 


Your statement about ultrafilters on ω appears to be even weaker, since it is such a special case of the Prime Ideal Theorem. 


Nevertheless, I believe that the method of forcing shows that it is consistent with ZF that there are no non-principal ultrafilters on ω. I believe that some of the standard models of ¬AC, built by using symmetric names for adding Cohen reals, have DC, and hence also ACω, but still have no nonprincipal ultrafilters on ω. In this case, neither DC nor ACω would imply the existence of such ultrafilters.


I'm less sure about finding models that have ultrafilters on ω, but not on all Boolean algebras. But I believe that this is likely the case. These models would show that your principle is strictly weaker even than the Prime Ideal Theorem.


