Idempotent ultrafilters and the Rudin-Keisler ordering

Question

Short version: what can we say about the place of idempotent ultrafilters in the Rudin-Keisler ordering?

Longer version:

If $U$, $V$ are (nonprincipal) ultrafilters on $\omega$, then we write $U\ge_{RK}V$ in case there is some function $f:\omega\rightarrow\omega$ such that $$ \forall X\subseteq\omega,\quad f^{-1}(X)\in U\iff X\in V.$$

An ultrafilter $U$ is Ramsey if given any two-coloring of pairs $c: [\omega]^2\rightarrow 2$, there is a homogeneous set for $c$ in $U$. Equivalently, $U$ is Ramsey if whenever $\lbrace C_n\rbrace_{n\in\omega}$ is a partition of $\omega$ with each $C_n\not\in U$, there is some $H\in U$ such that for all $n\in\omega$, $$\vert H\cap C_n\vert=1.$$ An ultrafilter $U$ is idempotent if $U\oplus U=U$, where $$ V\oplus W=\lbrace X: \lbrace x: \lbrace y: x+y\in X \rbrace \in W \rbrace \in V\rbrace.$$ Idempotent ultrafilters can be proved to exist in $ZFC$: this amounts to showing that $\oplus$ is left-continuous and associative on the compact space $\beta\mathbb{N}$, and then applying Ellis' theorem that every left-continuous semigroup on a compact space has an idempotent element. By contrast, Ramsey ultrafilters cannot be shown to exist in $ZFC$, although their existence is equiconsistent with $ZFC$ (in particular, if the Continuum Hypothesis - or weaker statements - holds, then there are Ramsey ultrafilters).

Now, an easy argument shows that no Ramsey ultrafilter is idempotent. On the other hand, the Ramsey ultrafilters enjoy a special property with respect to the RK-ordering: they are precisely the RK-minimal ultrafilters. So combining these facts shows that no idempotent ultrafilter can be RK-minimal.

My question is, what else can be said about the idempotent ultrafilters in terms of RK-reducibility? For example, can we have an idempotent ultrafilter U with exactly one RK-class of (necessarily, Ramsey) ultrafilters strictly RK-below U? This seems clearly impossible, but I don't see how to prove it.

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Motivation: a few weeks ago, I taught a one-week course on the proof of Hindman's theorem from additive combinatorics using idempotent ultrafilters. On the last day, I talked a bit about other types of ultrafilters, and spent a bit of time defining Ramsey ultrafilters and explaining (not proving) their place in the RK-ordering. One of my students asked whether anything similar could be said about idempotent ultrafilters; besides the obvious, I couldn't come up with anything, so I'm asking here.

Answer 1

The idempotent ultrafilters closest to being Ramsey are the stable ordered-union ultrafilters. These are officially defined as certain ultrafilters on the set $\mathbb F$ of finite subsets of $\omega$, but they can be transferred to $\omega$ via the "binary expansion" map $\mathbb F\to\omega: s\mapsto\sum_{n\in s}2^n$. The image on $\omega$ of a stable ordered-union ultrafilter is idempotent and has has exactly three non-isomorphic non-principal ultrafilters RK-below it. Two of these are the ones Todd Eisworth mentioned in his comment; the third is the "pairing" of these two, i.e, the image of the idempotent under the map to $\omega^2$ given by $n\mapsto($position of leftmost $1,\,$position of rightmost $1)$. For details, see my paper "Ultrafilters related to Hindman's finite-unions theorem and its extensions" ["Logic and Combinatorics", Contemporary Math 65 (1987) 89-124] also available at http://www.math.lsa.umich.edu/~ablass/uf-hindman.pdf (this is a scanned picture and therefore not searchable; the stable ordered-union stuff starts on page 113).