A choice of an ultrafilter in a Rudin-Keisler equivalence class

Question

Given a Rudin-Keisler equivalence class of ultrafilters, is it possible to construct a member of this class without using axiom of choice?

Answer 1

In the case of nonprincipal ultrafilters on the natural numbers, you cannot in general define an ultrafilter in a given Rudin-Keisler equivalence class. Specifically, suppose you begin with Solovay's model for "all sets are Lebesgue measurable" (a model that has no nonprincipal ultrafilters on the natural numbers) and adjoin an ultrafilter $U$ by forcing with the set of infinite subsets of $\mathbb N$ ordered by inclusion. Then, using the language of set theory plus a symbol for the RK-equivalence class of $U$, it is not possible to define $U$ itself or any other member of this equivalence class.

On the other hand, there are situations where remarkably many nonprincipal ultrafilters are definable. For example, under the assumption of the axiom of determinacy (AD), there are no nonprincipal ultrafilters on the natural numbers but there are lots of nonprincipal ultrafilters on other sets, and here there are lots of definable ultrafilters. For example, the closed unbounded subsets of $\omega_1$ generate an ultrafilter under AD. More generally, any RK-equivalence class contains at most one normal ultrafilter, so if an RK-equivalence class contains a normal ultrafilter then that's a canonical representative of the equivalence class. It is known that AD implies the existence of normal ultrafilters on many cardinals, so there are many RK-equivalence classes with canonical representatives.