\(\widetilde{|}\)-divisibility of ultrafilters (Q2003920)

In [\textit{B. Šobot}, Rep. Math. Logic 50, 53--66 (2015; Zbl 1379.54020)] the author defined four ways in which an ultrafilter~\(u\) on~\(\mathbb{N}\) can be said to divide another ultrafilter~\(v\). In this paper he studies one of these in depth: \(u\mathrel{\tilde|}v\) means that for every set~\(U\) in~\(u\) the set~\(V\) of multiples of members of~\(U\), that is \(V=\bigl\{n:(\exists m\in U)(m\mathrel|n)\bigr\}\), belongs to~\(v\).\par The ultrafilters that contain the set~\(P\) of primes are exactly the primes of~\(\beta\mathbb{N}\): they are minimal with respect to this divisibility relation. The present paper deals with ultrafilters on finite levels, by which the author means the following: let \(L_o=\{1\}\) and for \(n\ge1\) let \(L_n\) be the set of natural numbers that are products of exactly \(n\)~man prime numbers, not necessarily distinct (so \(P=L_1\)). An ultrafilter is at level~\(n\) if it contains~\(L_n\). The author characterizes the ultrafilters that are divisible by~\(1\) and one prime ultrafilter~\(p\). If the prime ultrafilter~\(p\) is Ramsey there is just one such ultrafilter, and the Continuum Hypothesis ensures there is a prime ultrafilter~\(p\) for which there are \(2^\mathfrak{c}\) such ultrafilters. If \(p\) and~\(q\) are non-principal prime ultrafilters and there is exactly one ultrafilter divisible by both \(p\) and \(q\) then \(p\) and~\(q\) are both \(P\)-points. On the other hand for any \(p\in P^*\) there is a \(q\in P^*\) such that there are \(2^\mathfrak{c}\) many ultrafilters divisible by both \(p\) and \(q\). There are also some general, but less conclusive, results about ultrafilters at the higher finite levels.\par The author announces that in [``More about divisibility in \(\beta\mathbb{N}\)'', to appear] he shows that the real line embeds in the divisibility order and the prime ultrafilters are truly prime: if \(p\) divides \(x\cdot y\) then it divides~\(x\) or~\(y\).