TITLE: Is every p-point ultrafilter Ramsey?
QUESTION [16 upvotes]: A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is a p-point (or weakly selective) iff for every partition $\omega = \bigsqcup _{n < \omega} Z_n$ into null sets, i.e each $Z_n \not \in \mathcal{U}$, there exists a measure one set $S \in \mathcal{U}$ such that $S \cap Z_n$ is finite for each $n$.
A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is Ramsey (or selective) iff for every partition as above, there exists a measure one set $S$ such that $|S \cap Z_n| = 1$ for each $n$.
Clearly, every Ramsey ultrafilter is a p-point.  What is known about the converse?
I couldn't find anything, not even a consistency result, in any searches I've done or sources I've checked.  Is very little known/published about the converse?

REPLY [9 votes]: Another small and slightly trivial addendum:
If there are no p-points, then every p-point is a Ramsey ultrafilter. (Duh!)
As Andreas Blass remarked above, this situation is consistent, which is easier to prove than the consistency of a 
unique p-point. ("It is usually significantly harder to prove there is a unique object than to prove there is none". See Shelah's  Proper and improper forcing VI.5)