Do Ramsey idempotent ultrafilters exist?

I was studying idempotent ultrafilters when I saw that no principal ultrafilter could ever be idempotent, because $\left\langle n \right\rangle \oplus \left\langle n \right\rangle = \left\langle 2n \right\rangle$.

A Ramsey ultrafilter is, by definition, never principal, so the question that arose to me was: do Ramsey idempotent ultrafilters exist?

First some explanation:

One of the characterisations of an ultrafilter to be Ramsey is:
For every partition $\{X_i \mid i \in \mathbb{N} \}$ where $X_i \notin \mathcal{F}$ for every $i \in \mathbb{N} $, there exists $X \in \mathcal{F}$ such that $\lvert X \cap X_i \rvert \leq 1$ for every $ i \in \mathbb{N} $.

An ultrafilter is called idempotent if $U \oplus U = U$, where
$U \oplus V = \{ A \subset \mathbb{N} \mid \{ m \in \mathbb{N} \mid A – m \in V \} \in U \}$ and $A – m = \{ n \in \mathbb{N} \mid m + n \in A \}$

(I was trying to construct a partition of $\mathbb{N}$ of infinitely many finite pieces, such that if I get an $ A \in \mathcal{U}$ that satisfies $\lvert X \cap X_i \rvert \leq 1$ for every $ i \in \mathbb{N} $, $A$ can not be an element of $ U \oplus U$ but I did not succeed).

I have a strong feeling that a Ramsey ultrafilter cannot be idempotent, but ‘a strong feeling’ is not a proof. Maybe someone can help me thinking about the subject? Do there exists proofs of the form I was trying to make it? Or maybe what I want to prove is not true? Can someone recommend literature about this subject or is it just irrelevant what I’m trying to do?

Thanks in advance!

Solutions Collecting From Web of "Do Ramsey idempotent ultrafilters exist?"

The question seems (at least to some extent) to be answered in comments, so I am posting what was said in comments as a CW-answer.


From the above comment by Shawn Henry:
Ramsey ultrafilters are not idempotent. A nice topological proof can be put together from the material in section 5 of this paper of Blass: http://www.math.lsa.umich.edu/~ablass/ufdyn.pdf

First since $\mathcal U+\mathcal V=\operatorname{{\mathcal U}-lim}_n (n+\mathcal V)$, while P-points are, in view of their topological description, never limit points of a countable set of other ultrafilters, it follows that no P-point can be of the form $\mathcal U+\mathcal V$ with non-principal $\mathcal U$ and $\mathcal V$. In particular, no P-point can be recurrent. So the family of P-points and, a fortiori, the subfamily of selective ultrafilters are disjoint from the families of ultrafilters studied in the preceding sections (recurrent, idempotent, etc.).