Is every p-point ultrafilter Ramsey? 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?
 A: 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)
A: Amit: 
The converse is not true, not even under MA. This is a result of Kunen, and the paper you want to look at is "Some points in $\beta{\mathbb N}$", Math. Proc. Cambridge Philos. Soc. 80 (1976), no. 3, 385–398. 
There is a related notion, called $q$-point. These are ultrafilters such that any finite-to-one $f:\omega\to\omega$ is injective on a set in the ultrafilter. A Ramsey ultrafilter is one that is simultaneously a $p$-point, and a $q$-point.
Miller proved ("There are no $Q$-points in Laver's model for the Borel conjecture", Proc. Amer. Math. Soc. 78 (1980), no. 1, 103–106) that it is consistent that there are no $q$-points. The consistency of the non-existence of $p$-points is significantly harder, and due to Shelah (see for example Chapter VI of his "Proper and improper forcing"). 
There is a fairly extensive literature on related results. You may want to start by looking at Blass' article in the Handbook of Set Theory, "Combinatorial Cardinal Characteristics of the Continuum". 
A: Another small addendum to Andres's and Andreas's answers.
It is also consistent that the answer to your question is yes.
Shelah has constructed a model of ZFC in which there exists (up to isomorphism) exactly one p-point -- and that p-point is, in fact, selective. This construction is Section XVIII.4 in  Shelah, Proper and Improper Forcing .
A: A few addenda to Andres Caicedo's answer: It was proved around 1970 by several people (Adrian Mathias was one of them) that the continuum hypothesis (CH) implies the existence of P-points that are not selective. (CH also implies the existence of selective ultrafilters and the existence of Q-points that are not selective.  ZFC alone suffices to prove the existence of ultrafilters that are neither P-points nor Q-points.)  The more difficult task of producing a model of set theory in which P-points exist but selective ultrafilters don't was achieved in Kunen's paper cited by Andres.  It is a famous open problem whether there are models of set theory in which neither P-points nor Q-points exist; it is known that in such a model the cardinal of the continuum must be at least $\aleph_3$.  (In contrast, there are models with $2^{\aleph_0}=\aleph_2$ with no P-points and others with no Q-points.)   
