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!