$\widetilde{\mid}\hspace{1mm}$-divisibility of ultrafilters

TL;DR

This paper explores a divisibility relation among ultrafilters on natural numbers, identifying prime ultrafilters, establishing a hierarchy, and analyzing the structure and positions of ultrafilters within this framework.

Contribution

It introduces a novel divisibility hierarchy for ultrafilters, characterizes prime ultrafilters, and analyzes their structural relationships and product positions.

Findings

Prime ultrafilters are characterized within the hierarchy.

A hierarchy of ultrafilters based on divisibility is established.

Ultrafilters with many immediate successors are constructed.

Abstract

We further investigate a divisibility relation on the set $\beta N$ of ultrafilters on the set of natural numbers. We single out prime ultrafilters (divisible only by 1 and themselves) and establish a hierarchy in which a position of every ultrafilter depends on the set of prime ultrafilters it is divisible by. We also construct ultrafilters with many immediate successors in this hierarchy and find positions of products of ultrafilters.

Equations

$$A\in x\cdot y\iff \{n\in N:A/n\in y\}\in x,$$

$$x\text{mathaccent}869\mid y$$

$$c(\{a_{1},a_{2},\dots,a_{n}\})=\left\{\begin{array}[]{ll}0,\mbox{if }a_{1}a_{2}\dots a_{n}\in S\\ 1,\mbox{otherwise.}\end{array}\right.$$

$$X_{n+1,k}\subseteq \{xa:x\in X_{n,k},a\in P\}$$

$$\text{mbox}foreveryarithmeticprogressions\text{mbox}oflength{2}^k+1$$

$$\text{mbox}therearea,b\in s\text{mbox}suchthatc(\{a,b\})=k.$$

$$\text{mbox}foreveryarithmeticprogressions\text{mbox}oflength{2}^k+n-1$$

$$\text{mbox}thereare{a}_1,a_2,\dots ,a_n\in s\text{mbox}suchthat{c}_n(\{a_1,a_2,\dots ,a_n\})=k.$$

$$(A_1^{k_1}{)}^{(n_1)}(A_2^{k_2}{)}^{(n_2)}\dots (A_m^{k_m}{)}^{(n_m)}=\{\prod _{i=1}^m\prod _{j=1}^{n_i}{a}_{i,j}^{k_i}$$

$$L_4=P^4\cup P^{3}P\cup (P^2{)}^{(2)}\cup P^2{P}^{(2)}\cup P^{(4)}.$$

$$u:=\sum _{k=m}^n{x}_k>v:=\sum _{k=m}^n{y}_k.$$

$$(A_{b_1}^{(n_1+n_1^{\prime })}{A}_{b_2}^{(n_2+n_2^{\prime })}\dots A_{b_m}^{(n_m+n_m^{\prime })})/a\supseteq B_{b_1}^{(n_1^{\prime })}\dots B_{b_m}^{(n_m^{\prime })},$$

$$\{a\in N:(A_{b_1}^{(n_1+n_1^{\prime })}{A}_{b_2}^{(n_2+n_2^{\prime })}\dots A_{b_m}^{(n_m+n_m^{\prime })})/a\in y\}\supseteq A_{b_1}^{(n_1)}{A}_{b_2}^{(n_2)}\dots A_{b_m}^{(n_m)}\in x,$$

Full text

2010 Mathematics Subject Classification: 03E20, 54D35, 54D80.

Key words and phrases: divisibility, Stone-Čech compactification, ultrafilter

1 Introduction

Let $N$ denote the set of natural numbers (without zero). The Stone-Čech compactification of the discrete space $N$ is the space $\beta N$ of all ultrafilters over $N$. For each $n\in N$ the principal ultrafilter $p_{n}=\{A\subseteq N:n\in A\}$ is identified with the respective element $n$. The topology on $\beta N$ is generated by (clopen) base sets of the form ${\bar{A}}=\{x\in\beta N:A\in x\}$.

A family ${\cal F}$ of subsets of $N$ has the finite intersection property (f.i.p.) if the intersection of every finitely many elements of ${\cal F}$ is nonempty. ${\cal F}$ has the uniform f.i.p. if the intersection of every finitely many elements of ${\cal F}$ is infinite. Every family with f.i.p. is contained in an ultrafilter, and every family with u.f.i.p. is contained in a nonprincipal ultrafilter. A family ${\cal F}$ with the f.i.p. generates a filter $F$ if for every $B\in F$ there are $A_{1},A_{2},\dots,A_{n}\in{\cal F}$ such that $A_{1}\cap A_{2}\cap\dots\cap A_{n}\subseteq B$.

If $f:N\rightarrow N$ is a function, the direct and inverse image of a set $A\subseteq N$ are $f[A]=\{f(a):a\in A\}$ and $f^{-1}[A]=\{b:f(b)\in A\}$. For every $f:N\rightarrow N$ there is unique continuous function $\mathaccent 869{f}:\beta N\rightarrow\beta N$ extending $f$. It is given by $\mathaccent 869{f}(x)=\{A\subseteq N:f^{-1}[A]\in x\}$ for $x\in\beta N$. $\mathaccent 869{f}(x)$ is also generated by sets $f[A]$ for $A\in x$. Since $\mathaccent 869{g}\circ\mathaccent 869{f}$ is a continuous extension of $g\circ f$, it follows that $\mathaccent 869{g}\circ\mathaccent 869{f}=\mathaccent 1373{g\circ f}$ for every two functions $f,g:N\rightarrow N$.

The multiplication $\cdot$ on $N$ can be extended to $\beta N$ (using the same notation $\cdot$ for the extension) in such way that $(\beta N,\cdot)$ is a compact Hausdorff right-topological semigroup: for $x,y\in\beta N$,

[TABLE]

where for $A\subseteq N$ and $n\in N$, $A/n=\{m\in N:mn\in A\}=\{\frac{a}{n}:a\in A,n\mid a\}$. The known properties if this structure are described in detail in [6].

We fix some more notation. Throughout the paper $P$ is the set of prime numbers. The complement of $A\subseteq N$ with respect to $N$ is $A^{c}=N\setminus A$, and the complement of $A\subseteq P$ with respect to $P$ is $A^{\prime}=P\setminus A$. $[X]^{k}$ is the set of subsets of $X$ of cardinality $k$. The set of functions $f:A\rightarrow B$ will be denoted by ${{}^{A}B}$.

For $x\in\beta N$, if $A\in x$ then $X\in x$ if and only if $X\cap A\in x$. Because of this we can identify each ultrafilter $x$ on $N$ containing $A$ with the ultrafilter $x\mathchar 13334\relax A=\{X\cap A:X\in x\}$ on $A$. Thus it is also common to think of $\overline{A}$ as a subspace of $\beta N$. Also, $A^{*}=\overline{A}\setminus A$ is the set of nonprincipal ultrafilters containing $A$.

The Rudin-Keisler preorder on $\beta N$ is defined as follows: $x\leq_{RK}y$ if and only of there is $f:N\rightarrow N$ such that $\mathaccent 869{f}(y)=x$.

An ultrafilter $x\in N^{*}$ is a P-point if, for every sequence $\langle A_{n}:n\in N\rangle$ of sets in $x$ such that $A_{m}\supseteq A_{n}$ for $m<n$ there is a set $B\in x$ (called a pseudointersection of sets $A_{n}$) such that $B\setminus A_{n}$ is finite for every $n\in N$.

A function $c:X\rightarrow\{0,1\}$ is called a $2$-coloring of $X$. An ultrafilter $x$ is called Ramsey (selective) if, for every $A\in x$, every $k\in N$ and every $2$-coloring $c$ of $[A]^{k}$ there is a set $M\in x$ that is monochromatic, i.e. such that $|c[[M]^{k}]|=1$. It is well-known that $x$ is a Ramsey ultrafilter if and only if $x$ is minimal in $\leq_{RK}$ (there are no nonprincipal ultrafilters $y\leq_{RK}x$).

If $A\subseteq N$, we will denote $A\uparrow=\{n\in N:\exists a\in A\;a\mid n\}$ and $A\downarrow=\{n\in N:\exists a\in A\;n\mid a\}$. ${\cal U}=\{A\subseteq N:A=A\uparrow\}$ and ${\cal V}=\{A\subseteq N:A=A\downarrow\}$ are the collections of subsets of $N$ upwards/downwards closed for divisibility.

In [8] four divisibility relations on $\beta N$ were introduced. In [8] and [9] some properties of these relations were investigated. In particular, the relation $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ was introduced as an extension of the usual divisibility relation $\mid$ on $N$ to $\beta N$ analogous to extensions of functions described above. It was proven that, for $x,y\in\beta N$,

[TABLE]

In this paper we will use these characterizations and under divisibility of ultrafilters we will understand $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-divisibility. For example, ”least upper bound for $x$ and $y$” will mean the $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-smallest ultrafilter $z$ (if such exists) such that $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptz$ and $y\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptz$. ”$x$ is below $y$” will mean $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$. $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ is a preorder and, if we define $x=_{\sim}y\Leftrightarrow x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty\land y\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$, then $=_{\sim}$ is an equivalence relation. We denote the equivalence class of $x$ by $[x]_{\sim}$ and think of $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ as an order on the set of these classes. Let us also call $x,y\in\beta N$ incompatible if there is no $z\in\beta N\setminus\{1\}$ such that $z\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$ and $z\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$.

The motivation for introducing divisibility of ultrafilters is to inspect the effects of existence of some infinite sets of natural numbers to ultrafilters and (hopefully) the opposite: to draw number-theoretic conclusions from the existence of certain ultrafilters. However, we also hope that some of the results in this paper will help to better understand the product (1) of ultrafilters by finding its place in the divisibility hierarchy.

2 Prime ultrafilters

Lemma 2.1

Let $x\in\beta N$, $A\in x$ and $f:N\rightarrow N$.

(a) If $f(a)\mid a$ for all $a\in A$, then $\mathaccent 869{f}(x)\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$.

(b) If $a\mid f(a)$ for all $a\in A$, then $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt\mathaccent 869{f}(x)$.

**Proof. **(a) If $B\in{\mathaccent 869{f}}(x)\cap{\cal U}$ then $f^{-1}[B]\in x$. But $f^{-1}[B]\cap A\subseteq B$ (because $B\in{\cal U}$) so $B\in x$. Hence we have ${\mathaccent 869{f}}(x)\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$.

(b) Analogously to (a) we prove that every $B\in{\mathaccent 869{f}}(x)\cap{\cal V}$ is also in $x$. $\Box$

If $A\in x$, in order to determine $\mathaccent 869{f}(x)$ it is enough to know values of $f(a)$ for $a\in A$. Hence, when using the lemma above we will sometimes define functions only on a set in $x$.

Lemma 2.2

For every $x\in\beta N\setminus\{1\}$ there is $p\in\overline{P}$ such that $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$.

**Proof. **We define a function $f:N\setminus\{1\}\rightarrow N$: let $f(n)$ be the smallest prime factor of $n$. Now ${\mathaccent 869{f}}(x)\in\overline{P}$ (for $x\in\beta N\setminus\{1\}$) and, by Lemma 2.1(a), ${\mathaccent 869{f}}(x)\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$. $\Box$

Clearly, 1 is the smallest element in $(\beta N,\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt)$. We will call $p\in\beta N\setminus\{1\}$ prime (or $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal) if it is divisible only by 1 and itself. We will reserve labels $p,q,r,\dots$ for prime ultrafilters and $x,y,z,\dots$ for ultrafilters in general.

Theorem 2.3

$p\in\beta N$* is prime if and only if $p\in\overline{P}$.*

**Proof. **Assume first that $p$ is prime but $p\in\overline{P^{c}}$. By Lemma 2.2 there is an element $q\in\overline{P}$ below $p$. But $q\neq_{\sim}p$ because the set of composite numbers $(P\cup\{1\})^{c}\in(p\cap{\cal U})\setminus q$; hence $q\neq p$, a contradiction.

Now assume $p\in\overline{P}$ and $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptp$, but $x\neq p$. By Lemma 2.2 there is $q\in\overline{P}$ such that $q\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$. It follows that $q\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptp$. If $A\in q\setminus p$, let $B=(A\cap P)\uparrow$. Then $B\in q\cap{\cal U}$, so $B\in p$ and $A\supseteq B\cap P\in p$, a contradiction. $\Box$

Comparing this with Lemmas 7.3 and 7.5 from [8] we conclude that $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal ultrafilters are also $\mid_{L}$-minimal, but not vice versa. Also, a corollary of this theorem is that there is a family of $2^{{c}}$ incompatible ultrafilters, which improves Theorem 3.8 from [9].

Let us also mention that, by Fact 6.1, there are ultrafilters with $2^{{c}}$-many divisors so not every divisibility can be established by means of Lemma 2.1(a). It follows that the Rudin-Keisler preorder is not stronger than $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$. On the other hand, neither is $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ stronger than $\leq_{RK}$, since ultrafilters containing $P$ that are not Ramsey are $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-minimal but nor $\leq_{RK}$-minimal.

Now we define levels of the $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-hierarchy.

Definition 2.4

Let $L_{0}=\{1\}$ and $L_{n}=\{a_{1}a_{2}\dots a_{n}:a_{1},a_{2},\dots,a_{n}\in P\}$ for $n\geq 1$. We will say that $x$ is of level $n$ if $x\in\overline{L_{n}}$.

In particular, $\overline{L_{1}}=\overline{P}$ is the set of prime ultrafilters.

In this paper we will mostly deal with ultrafilters on finite levels (belonging to $\overline{L_{n}}$ for some $n\in N$). In Section 6 we will see that there are also ultrafilters above all these finite levels.

3 Ultrafilters with one prime divisor

Definition 3.1

For $A\subseteq N$ and $n\in N\cup\{0\}$ we denote $A^{n}=\{a^{n}:a\in A\}$. If $pow_{n}:N\rightarrow N$ is defined by $pow_{n}(a)=a^{n}$ then, for $x\in N^{*}$, $\mathaccent 1373{pow_{n}}(x)$ is generated by sets $A^{n}$ for $A\in x$. We will denote $\mathaccent 1373{pow_{n}}(x)$ with $x^{n}$.

Of course, $A^{0}=\{1\}$, $x^{0}=1$ and $x^{1}=x$.

Lemma 3.2

If $p\in\overline{P}$, the only ultrafilters below $p^{n}$ are $p^{k}$ for $k\leq n$.

**Proof. **If $p\in P$, the lemma is obvious. So we prove it for $p\in P^{*}$.

Since $\bigcup_{k\leq n}P^{k}\in p^{n}\cap{\cal V}$, every ultrafilter $x\neq 1$ below $p^{n}$ must contain $P^{k}$ for some $k\leq n$.

If $x\in(P^{k})^{*}$ then for every $A\subseteq P$, if $A^{k}\in p^{k}$ then we have $\bigcup_{i\leq n}A^{i}\in p^{n}\cap{\cal V}$ so $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptp^{n}$ implies that $A^{k}=P^{k}\cap\bigcup_{i\leq n}A^{i}\in x$. Thus $x\mathchar 13334\relax P^{k}=p^{k}\mathchar 13334\relax P^{k}$, which means that $x=p^{k}$. $\Box$

Definition 3.3

For $A,B\subseteq N$ let $AB=\{ab:a\in A,b\in B\land gcd(a,b)=1\}$. In particular, $A^{(n)}=\underbrace{A\cdot A\cdot\dots\cdot A}_{n}$. If $p\in\overline{P}$ let $F_{n}^{p}=\{A^{(n)}:A\in p\mathchar 13334\relax P\}$.

Note the difference between $A^{(n)}$ and $A^{n}$ from Definition 3.1: elements of $A^{(n)}$ must be products of mutually prime numbers. $A$ will almost always be a subset of $P$ in which case ”mutually prime” will mean ”distinct”.

If $p\in P$ (a principal ultrafilter), $\{p\}^{(n)}=\emptyset\in F_{n}^{p}$, so there is no ultrafilter $x\supseteq F_{n}^{p}$.

Lemma 3.4

Let $p\in P^{*}$. For any ultrafilter $x\supseteq F_{n}^{p}$, $p$ is the only ultrafilter from $\overline{P}$ below $x$.

**Proof. **If $q\in\overline{P}$ is such that $q\neq p$, there is $A\subseteq P$ such that $A\in p\setminus q$. Then $\bigcup_{k\leq n}A^{(k)}\in x\cap{\cal V}\setminus q$, so $q\mathaccent 869{\mathchar 13613\relax}x$. $\Box$

Lemma 3.5

Let $p\in P^{*}$.

(a) $p\cdot p\supseteq F_{2}^{p}$.

(b) There are either finitely many or $2^{{c}}$ ultrafilters $x\supseteq F_{n}^{p}$.

**Proof. **(a) Let $A\in p\mathchar 13334\relax P$. Since $A^{(2)}/a=A\setminus\{a\}$ for $a\in A$, $\{a\in N:A^{(2)}/a\in p\}\supseteq A\in p$ and so $A^{(2)}\in p\cdot p$.

(b) The set of ultrafilters $x\supseteq F_{n}^{p}$ is actually $\bigcap_{X\in F_{n}^{p}}\overline{X}$ so it is closed. By [10], Theorem 3.3, closed subsets of $\beta N$ are either finite or of cardinality $2^{{c}}$. Hence, for every $p\in P^{*}$ there are either finitely many or $2^{{c}}$ ultrafilters containing $F_{n}^{p}$. $\Box$

Theorem 3.6

Let $p\in P^{*}$. There is unique ultrafilter $x\supseteq F_{n}^{p}$ if and only if $p$ is Ramsey.

**Proof. **Assume $p$ is Ramsey. To prove that $x$ is unique it suffices to show that for every set $S\subseteq P^{(n)}$ one of the sets $S$ and $P^{(n)}\setminus S$ contains a set from $F_{n}^{p}$. Define a coloring of $[P]^{n}$ as follows:

[TABLE]

Since $p$ is Ramsey there is a monochromatic set $M\in p$, say $c(\{a_{1},a_{2},\dots,a_{n}\})=0$ for all $a_{1},a_{2},\dots,a_{n}\in M$. This means that $M^{(n)}\subseteq S$.

Now assume $p$ is not Ramsey. Then there is a 2-coloring $c$ of $[P]^{n}$ such that $p$ does not contain a monochromatic subset. Let $S=\{a_{1}a_{2}\dots a_{n}:c(\{a_{1},a_{2},\dots,a_{n}\})=0\}$; then both $F_{n}^{p}\cup\{S\}$ and $F_{n}^{p}\cup\{P^{(n)}\setminus S\}$ have the f.i.p. so there are at least two ultrafilters containing $F_{n}^{p}$. $\Box$

It is well-known that, under CH, there are Ramsey ultrafilters in $N^{*}$ ([4], Theorem 9.19). Also, not all ultrafilters in $N^{*}$ are Ramsey. If $f:N\rightarrow P$ is any bijection, $\mathaccent 869{f}$ maps Ramsey to Ramsey, and non-Ramsey to non-Ramsey ultrafilters, so (under CH) in $P^{*}$ there are also ultrafilters of both types. On the other hand, each Ramsey ultrafilter is a P-point and Shelah proved that it is consistent with ZFC that there are no P-points in $N^{*}$ (the proof can be found in [7]).

Blass proved in [2] that (under CH) there is a non-Ramsey ultrafilter $p$ such that for every 3-coloring $c:[N]^{2}\rightarrow\{0,1,2\}$ there is $A\in p$ such that $|c[[A]^{2}]|\leq 2$. By a simple modification of the proof of Theorem 3.6 we get that, for such $p\in P^{*}$, there are exactly two ultrafilters $x\supseteq F_{2}^{p}$.

Lemma 3.7

Let $x\in\overline{P^{(2)}}$.

(a) If there are disjoint $A,B\subseteq P$ such that $AB\in x$ then $x$ is divisible by at least two ultrafilters from $\overline{P}$.

(b) For any two distinct $p,q\in\overline{P}$, $x$ is divisible by $p$ and $q$ if and only if for every two disjoint $A\in p$, $B\in q$ holds $AB\in x$.

**Proof. **(a) If $AB\in x$ for disjoint $A,B\subseteq P$ then we can define functions $f_{A}$ and $f_{B}$ on $AB$ such that $f_{A}(ab)=a$ and $f_{B}(ab)=b$ (for $a\in A$, $b\in B$) so, by Lemma 2.1(a), $\mathaccent 869{f_{A}}(x)\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$ and $\mathaccent 869{f_{B}}(x)\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$. Finally, $\mathaccent 869{f_{A}}(x)\neq\mathaccent 869{f_{B}}(x)$ because $A=f_{A}[AB]\in\mathaccent 869{f_{A}}(x)$ and $B=f_{B}[AB]\in\mathaccent 869{f_{B}}(x)$.

(b) First let $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$ and $q\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$. Let $A\in p\mathchar 13334\relax P$ and $B\in q\mathchar 13334\relax P$ be disjoint. Then $A\uparrow\in p\cap{\cal U}\subseteq x$ and $B\uparrow\in q\cap{\cal U}\subseteq x$, so $AB=A\uparrow\cap B\uparrow\cap P^{(2)}\in x$.

Now let $P=A\cup B$ be a partition of $P$ with $A\in p$, $B\in q$. As in (a) we define $f_{A}$ and $f_{B}$. Since $AB\in x$, $\mathaccent 869{f_{A}}(x)$ and $\mathaccent 869{f_{B}}(x)$ are ultrafilters below $x$. Assume $\mathaccent 869{f_{A}}(x)\neq p$. Let $A_{1}\in\mathaccent 869{f_{A}}(x)\setminus p$ be disjoint from $B$; by the first implication $A_{1}B\in x$. But $A_{1}B$ and $(P\setminus(A_{1}\cup B))B$ (belonging to $x$ by assumption) are disjoint so they can not both be in $x$, a contradiction. Thus $\mathaccent 869{f_{A}}(x)=p$, and in the same way we prove $\mathaccent 869{f_{B}}(x)=q$. $\Box$

Lemma 3.8

Let $p\in\overline{P}$. The ultrafilters such that their only proper divisors are 1 and $p$ are exactly $p^{2}$ and (for $p\in P^{*}$) ultrafilters containing $F_{2}^{p}$.

**Proof. **In Lemma 3.2 we proved that the only proper divisors of $p^{2}$ are 1 and $p$. The proof for $x\supseteq F_{2}^{p}$ is similar. Now assume that $x$ is any ultrafilter such that 1 and $p$ are its only proper divisors. $x$ belongs either to $\overline{P}$, $\overline{P^{2}}$, $\overline{P^{(2)}}$ or to $\overline{(P\cup P^{2}\cup P^{(2)}\cup\{1\})^{c}}$.

In the first case $x$ has only one proper divisor, 1.

In the second case, if we let $p=\{A\subseteq P:A^{2}\in x\}$, it is easy to prove $x=p^{2}$.

Let $x\in\overline{P^{(2)}}$ and $G=\{A\subseteq P:A^{(2)}\in x\}$. $G$ is closed for finite intersections and sets in $G$ are nonempty, so $G$ has the f.i.p. If $A\cup B$ is any partition of $P$, by Lemma 3.7 $AB\notin x$, so it follows that $A^{(2)}\in x$ or $B^{(2)}\in x$ (because $P^{(2)}=A^{(2)}\cup B^{(2)}\cup AB$), hence $A\in G$ or $B\in G$. This means that $G$ is an ultrafilter on $P$ and $x\supseteq F_{2}^{G}$. Since $p$ is the only element of $\overline{P}$ below $x$, we have $x\supseteq F_{2}^{p}$.

Finally, if $x\notin\overline{P\cup P^{2}\cup P^{(2)}\cup\{1\}}$, we can define the function $g:N\setminus(P\cup\{1\})\rightarrow N$ by $g(a_{1}a_{2}\dots a_{n})=a_{1}a_{2}$ (where $a_{1}\leq a_{2}\leq\dots\leq a_{n}$ are prime). Then $\mathaccent 869{g}(x)\neq x$ is an element of $P^{2}\cup P^{(2)}$ below $x$, so $x$ has more than two proper divisors. $\Box$

The following definition and lemmas will be used in the proof of Theorem 3.13.

Definition 3.9

Let $X\subseteq N$ and let $d=\{X_{k}:k\in N\}$ be a partition of $X^{(n)}$. A set $A\subseteq X$ is $d$-thick if for all $m\in N$ and all finite partitions $A=A_{1}\cup A_{2}\cup\dots\cup A_{m}$ there is $i\leq m$ such that for every $k\in N$ $A_{i}^{(n)}\cap X_{k}\neq\emptyset$.

The condition of $d$-thickness strengthens the condition that $A^{(n)}$ intersects every $X_{k}$. The idea of this strengthening is to satisfy (b) of the lemma below.

Lemma 3.10

Let $A\subseteq X$, $B\subseteq X$ and let $d=\{X_{k}:k\in N\}$ be a partition of $X^{(n)}$.

(a) If $A$ is $d$-thick and $A\subseteq B$, then $B$ is $d$-thick.

(b) If neither $A$ nor $B$ are $d$-thick, then $A\cup B$ is not $d$-thick.

**Proof. **(a) is obvious.

(b) By (a) we may assume without loss of generality that $A\cap B=\emptyset$. Let partitions $A=A_{1}\cup A_{2}\cup\dots\cup A_{m_{A}}$ and $B=B_{1}\cup B_{2}\cup\dots\cup B_{m_{B}}$ and $k_{i},l_{i}\in N$ be such that $A_{i}^{(n)}\cap X_{k_{i}}=\emptyset$ for all $i\leq m_{A}$ and $B_{i}^{(n)}\cap X_{l_{i}}=\emptyset$ for all $i\leq m_{B}$. Then the partition $A\cup B=A_{1}\cup A_{2}\cup\dots\cup A_{m_{A}}\cup B_{1}\cup B_{2}\cup\dots\cup B_{m_{B}}$ and the same $k_{i},l_{i}$ witness that $A\cup B$ is not $d$-thick. $\Box$

Lemma 3.11

If, for each $n\geq 2$, $d_{n}=\{X_{n,k}:k\in N\}$ is a partition of $X^{(n)}$ such that for all $k\in N$

[TABLE]

and $A\subseteq X$ is not $d_{n_{0}}$-thick then, for all $n\geq n_{0}$, $A$ is not $d_{n}$-thick.

**Proof. **It suffices to prove the theorem for $n=n_{0}+1$. Let the partition $A=A_{1}\cup A_{2}\cup\dots\cup A_{m}$ and $k_{i}\in N$ be such that $A_{i}^{(n_{0})}\cap X_{n_{0},k_{i}}=\emptyset$ for all $i\leq m$. Then the same partition witnesses that $A$ is not $d_{n_{0}+1}$-thick: if $xa\in A_{i}^{(n_{0}+1)}\cap X_{n_{0}+1,k_{i}}$ (for $x\in X_{n_{0},k}$, $a\in P$), then $x\in A_{i}^{(n_{0})}\cap X_{n_{0},k_{i}}$, a contradiction. $\Box$

Lemma 3.12

*(a) There is a coloring $c:[N]^{2}\rightarrow N$ such that for every $k\in N$:

$\displaystyle\mbox{for every arithmetic progression }s\mbox{ of length }2^{k}+1$

$\displaystyle\mbox{ there are }a,b\in s\mbox{ such that }c(\{a,b\})=k.$

(b) There are partitions $d_{n}=\{X_{n,k}:k\in N\}$ of $P^{(n)}$ (for $n\geq 2$) such that $P$ is $d_{n}$-thick for every $n\geq 2$ and (2) holds.

**Proof. **(a) We define sets $A^{0}_{i}=\{i\}$ (for $i\in N$) and, by recursion on $n$, $A^{n}_{i}=A^{n-1}_{2i-1}\cup A^{n-1}_{2i}$. Note that $a\in A_{2i-1}^{n-1}$ are exactly numbers with residue $1\leq r\leq 2^{n-1}$ modulo $2^{n}$.

First we color the pairs of numbers in the same $A^{1}_{i}$: $c(\{2i-1,2i\})=1$. By recursion on $n$, pairs of numbers $a,b\in A^{n}_{i}$ such that $a\in A^{n-1}_{2i-1}$ and $b\in A^{n-1}_{2i}$ are colored according to the difference $b-a$: if $2^{j}\leq b-a<2^{j+1}$ then $c(\{a,b\})=n-j$.

Let us prove (LABEL:progr). Let $s=\{a_{0}+md:0\leq m\leq 2^{k}\}$ and $j$ is such that $2^{j}\leq d<2^{j+1}$. At least one $a_{0}+md\in s$ (for $0\leq m<2^{k}$) has residue $2^{k+j-1}-d<r\leq 2^{k+j-1}$ modulo $2^{k+j}$, which means that $a_{0}+md\in A^{k+j-1}_{2i-1}$ and $a_{0}+(m+1)d\in A^{k+j-1}_{2i}$ for some $i\in N$. Then $c(\{a_{0}+md,a_{0}+(m+1)d\})=(k+j)-j=k$.

(b) Let $c$ be the coloring of $[N]^{2}$ defined in (a). We define $c_{n}:[N]^{n}\rightarrow N$ (for $n\geq 2$) by $c_{n}(\{a_{1},a_{2},\dots,a_{n}\})=c(\{a_{1},a_{2}\})$ for $a_{1}<a_{2}<\dots<a_{n}$. Then (LABEL:progr) implies that for every $n\geq 2$ and every $k\in N$

[TABLE]

[TABLE]

Now enumerate $P=\{p_{n}:n\in N\}$ in the increasing order. Define the partitions $d_{n}=\{X_{n,k}:k\in N\}$ of $P^{(n)}$: $X_{n,k}=\{p_{a_{1}}p_{a_{2}}\dots p_{a_{n}}:c_{n}(\{a_{1},a_{2},\dots,a_{n}\})=k\}$. Obviously, (2) holds.

To prove that $P$ is $d_{n}$-thick, let $P=A_{1}\cup A_{2}\cup\dots\cup A_{m}$ be any finite partition of $P$. If $B_{i}=\{a\in N:p_{a}\in A_{i}\}$, then $N=B_{1}\cup B_{2}\cup\dots\cup B_{m}$ is a partition of $N$. By the infinite Van der Waerden’s theorem there is $i\leq m$ such that $B_{i}$ contains arithmetic progressions of any length. So for every $k\in N$, there is a progression $s$ in $B_{i}$ of length $2^{k}+n-1$. By (3) there are $a_{1},a_{2},\dots,a_{n}\in s\subseteq B_{i}$ such that $c_{n}(\{a_{1},a_{2},\dots,a_{n}\})=k$ so $p_{a_{1}}p_{a_{2}}\dots p_{a_{n}}\in X_{n,k}\cap A_{i}^{(n)}$. $\Box$

Theorem 3.13

(CH) There is $p\in P^{*}$ such that for every $n\geq 2$ there are $2^{{c}}$ ultrafilters $x\supseteq F_{n}^{p}$.

**Proof. **Let, for $n\geq 2$, $d_{n}$ be the partition of $P^{(n)}$ given by Lemma 3.12(b). By the Continuum Hypothesis we can enumerate all subsets of $P$ as $\langle S_{\xi}:\xi<\omega_{1}\rangle$; recall that $S_{\xi}^{\prime}=P\setminus S_{\xi}$. By recursion on $\xi<\omega_{1}$ we define sets $A_{\xi}$ and families ${\cal F}_{\xi}$ such that:

(1ξ) ${\cal F}_{\xi}$ is a countable family of infinite subsets of $P$, closed for finite intersections;

(2ξ) ${\cal F}_{\zeta}\subseteq{\cal F}_{\xi}$ for $\zeta<\xi$;

(3ξ) ${\cal F}_{\xi}=\bigcup_{\zeta<\xi}{\cal F}_{\zeta}$ for $\xi$ a limit ordinal;

(4ξ) $A$ is $d_{n}$-thick for all $A\in{\cal F}_{\xi}$ and all $n\geq 2$;

(5ξ) either $A_{\xi}=S_{\xi}$ or $A_{\xi}=S_{\xi}^{\prime}$, and $A_{\xi}\in{\cal F}_{\xi+1}$.

First we let ${\cal F}_{0}=\{P\}$; by Lemma 3.12 $P$ is $d_{n}$-thick for all $n\geq 2$. Now for every $\xi<\omega_{1}$, assuming we have already defined ${\cal F}_{\xi}$ satisfying (1ξ)-(4ξ), we define $A_{\xi}$ and ${\cal F}_{\xi+1}$.

We first prove that for at least one of the possibilities $A_{\xi}=S_{\xi}$ and $A_{\xi}=S_{\xi}^{\prime}$ all sets $A_{\xi}\cap A$ for $A\in F_{\xi}$ are $d_{n}$-thick for all $n\geq 2$. Assume not: then there are $A,B\in{\cal F}_{\xi}$ and $m,n\in N$ such that $S_{\xi}\cap A$ is not $d_{m}$-thick and $S_{\xi}^{\prime}\cap B$ is not $d_{n}$-thick. If, say, $m\leq n$, by Lemma 3.11 both those sets are not $d_{n}$-thick. By Lemma 3.10(a), $S_{\xi}\cap(A\cap B)$ and $S_{\xi}^{\prime}\cap(A\cap B)$ are not $d_{n}$-thick and, finally, by (b) of the same lemma, their union $P\cap(A\cap B)=A\cap B$ is not $d_{n}$-thick, which is impossible by (4ξ) and (1ξ).

Hence we define $A_{\xi}=S_{\xi}$ if all $S_{\xi}\cap A$ are $d_{n}$-dense for all $n\geq 2$ and all $A\in{\cal F}_{\xi}$, and $A_{\xi}=S_{\xi}^{\prime}$ otherwise. Let ${\cal F}_{\xi+1}={\cal F}_{\xi}\cup\{A_{\xi}\cap A:A\in{\cal F}_{\xi}\}$. If $\xi$ is a limit ordinal, we define ${\cal F}_{\xi}$ as in (3ξ). Clearly, all the properties (1ξ)-(5ξ) are now satisfied.

In the end, by (1ξ) and (5ξ) $p:=\bigcup_{\xi<\omega_{1}}{\cal F}_{\xi}$ is an ultrafilter on $P$. Let $n\geq 2$. For every $k\in N$ the family $F_{n}^{p}\cup\{X_{n,k}\}$ has the f.i.p. (since every $A\in p\mathchar 13334\relax P$ is $d_{n}$-thick, every $A^{(n)}\in F_{n}^{p}$ intersects $X_{n,k}$), so there are ultrafilters $x_{k}\supseteq F_{n}^{p}\cup\{X_{n,k}\}$. But $X_{n,k}$ (for $k\in N$) are disjoint, so all $x_{k}$ are distinct ultrafilters. By Lemma 3.5(b) there are $2^{{c}}$ ultrafilters $x\supseteq F_{n}^{p}$. $\Box$

4 Ultrafilters with two prime divisors

Lemma 4.1

No ultrafilter $x\in\overline{P^{(2)}}$ is divisible by more than two prime ultrafilters.

**Proof. **Assume the opposite, that $x$ is divisible by $p_{1},p_{2},p_{3}\in\overline{P}$. Let $P=A_{1}\cup A_{2}\cup A_{3}$ be a partition such that $A_{1}\in p_{1}$, $A_{2}\in p_{2}$ and $A_{3}\in p_{3}$. We consider three cases. $1^{\circ}$ If $A_{1}P\notin x$, then clearly $A_{1}\uparrow\notin x$ (because $A_{1}\uparrow\cap P^{(2)}=A_{1}P$), so $p_{1}\hskip 2.84526pt\mathaccent 869{\mathchar 13613\relax}\hskip 2.84526ptx$. $2^{\circ}$ If $A_{1}(A_{1}\cup A_{3})\in x$, then $A_{2}P\notin x$ so, as in $1^{\circ}$, $p_{2}\hskip 2.84526pt\mathaccent 869{\mathchar 13613\relax}\hskip 2.84526ptx$. $3^{\circ}$ Finally, if $A_{1}A_{2}\in x$ then $A_{3}P\notin x$ so $p_{3}\hskip 2.84526pt\mathaccent 869{\mathchar 13613\relax}\hskip 2.84526ptx$. In each case we reach a contradiction. $\Box$

Definition 4.2

$F_{1,1}^{p,q}=\{AB:A\in p\mathchar 13334\relax P,B\in q\mathchar 13334\relax P,A\cap B=\emptyset\}$* for $p,q\in\overline{P}$.*

By Lemma 3.7 ultrafilters $x\supseteq F_{1,1}^{p,q}$ are exactly those ultrafilters in $P^{(2)}$ divisible by $p$ and $q$.

Lemma 4.3

For any distinct $p,q\in\overline{P}$:

(a) $p\cdot q\supseteq F_{1,1}^{p,q}$ and $q\cdot p\supseteq F_{1,1}^{p,q}$.

(b) there are either finitely many or $2^{{c}}$ ultrafilters $r\supseteq F_{1,1}^{p,q}$.

**Proof. **(a) Let $A\in p$ and $B\in q$ be disjoint. Since $AB/a=B$ for $a\in A$, $\{a\in N:AB/a\in q\}\supseteq A\in p$ and so $AB\in p\cdot q$. Analogously $AB\in q\cdot p$.

(b) Analogous to the proof of Lemma 3.5(b). $\Box$

If $n\in N$ then $nq=qn$ for all $q\in\beta N$ and it is not hard to see that this is the only ultrafilter containing $x\supseteq F_{1,1}^{n,q}$: by Lemma 5.1 of [8] every ultrafilter divisible by $n$ must contain the set $nN=\{na:a\in N\}$ so $nP\in x$; but $F_{1,1}^{n,q}=\{n(B\setminus\{n\}):B\in q\}$ generates an ultrafilter on $nP$.

On the other hand, by [6], Corollary 6.51, for every $p\in P^{*}$ there is $q\in P^{*}$ such that $pq\neq qp$, and we have at least two ultrafilters containing $F_{1,1}^{p,q}$. We will improve this in Theorem 4.5.

Theorem 4.4

Let $p,q\in P^{*}$. If there is unique $x\supseteq F_{1,1}^{p,q}$ then both $p$ and $q$ are P-points.

**Proof. **First let $X_{n}\in q$ for $n\in N$ and, without loss of generality, assume $X_{0}=P$ and $X_{m}\subseteq X_{n}$ for $m>n$. Let $Y=\{mn\in P^{(2)}:m>n\mbox{ and }m\in X_{n}\}$. Then, for every $n\in P$, $Y/n\supseteq X_{n}\setminus\{1,2,\dots,n\}\in q$ so $\{n\in N:Y/n\in q\}=P\in p$ and $Y\in p\cdot q$.

Since $F_{1,1}^{p,q}$ generates the unique ultrafilter, there is $AB\in F_{1,1}^{p,q}$ such that $AB\subseteq Y$ ($A\in p$, $B\in q$). To prove that $q$ is a P-point it suffices to show that $B$ is a pseudointersection of the sets $X_{n}$. For every $n\in N$ there is $a\in A$ such that $n\leq a$. If $b\in B$ is such that $b>a$, then $ab\in Y$ implies $b\in X_{a}\subseteq X_{n}$. Hence $B\setminus X_{n}$ is finite.

Since $p\cdot q$ and $q\cdot p$ both contain $F_{1,1}^{p,q}$, we have $p\cdot q=q\cdot p$, so by interchanging the roles of $p$ and $q$ we prove in the same way that $p$ is a P-point. $\Box$

Theorem 4.5

For every $p\in P^{*}$ there is an ultrafilter $q\in P^{*}$ such that there are $2^{{c}}$ ultrafilters $r\supseteq F_{1,1}^{p,q}$.

**Proof. **Let $p\in P^{*}$ be given. Let $f:P\rightarrow{{}^{N}P}$ be such that, if $f(i)=f_{i}$, then $\langle f_{i}:i\in P\rangle$ is the sequence of all eventually constant functions $f_{i}:N\rightarrow P$ (i.e. such that there are $n_{0}\in N$ and $a\in P$ so that $f_{i}(n)=a$ for $n\geq n_{0}$), ordered in such way that for all $n\in N$: $f_{i}(n)\leq i$ for $i\in\{2,3\}$ and $f_{i}(n)<i$ for $i>3$. The set $D=\{f_{i}:i\in P\}$ is dense in the space ${{}^{N}P}$ (with the usual Tychonoff topology). For $m,n\in N$ and $A\in p$ let $U_{n}(A)=\{x\in{{}^{N}P}:x(n)\in A\}$ and $V_{m,n}=\{x\in{{}^{N}P}:x(m)\neq x(n)\}$. Then the family $\{U_{n}(A):n\in N\land A\in p\}\cup\{V_{m,n}:m\neq n\}$ has the uniform f.i.p. so the family $F=\{f^{-1}[U_{n}(A)]:n\in N\land A\in p\}\cup\{f^{-1}[V_{m,n}]:m\neq n\}$ also has the uniform f.i.p. (because $f$ is one-to-one, all $U_{n}(A)$ and $V_{m,n}$ are open and $D$ is dense in ${{}^{N}P}$). Hence there is an ultrafilter $q\in P^{*}$ containing $F$.

Now, for $n\in N$, let $g_{n}:P\rightarrow N$ be defined by $g_{n}(i)=if_{i}(n)$, and let $r_{n}=\mathaccent 869{g_{n}}(q)$. Since $f_{i}(n)\neq i$ for all $i>3$ and all $n$, we conclude that $r_{n}\in\overline{P^{(2)}\cup\{4,9\}}$. Moreover, by Lemma 2.1(b) all $r_{n}$ are divisible by $q$, hence they are nonprincipal and $r_{n}\in(P^{(2)})^{*}$.

Let $h_{1},h_{2}:P^{(2)}\cup\{4,9\}\rightarrow P$ be defined by $h_{1}(ab)=a$ and $h_{2}(ab)=b$ for $a,b\in P$ and $a\leq b$. We prove that $\mathaccent 869{h_{1}}(r_{n})=p$ and $\mathaccent 869{h_{2}}(r_{n})=q$ for every $n\in N$. First, $\mathaccent 869{h_{2}}(r_{n})=\mathaccent 869{h_{2}}\circ\mathaccent 869{g_{n}}(q)=\mathaccent 1373{h_{2}\circ g_{n}}(q)=q$ (because $h_{2}\circ g_{n}$ is the identity function).

On the other hand, $h_{1}\circ g_{n}(i)=f_{i}(n)$ for all $i$. For any $A\in p$ and all $i\in f^{-1}[U_{n}(A)]$ we have $f_{i}(n)\in A$. Hence $h_{1}\circ g_{n}[f^{-1}[U_{n}(A)]]\subseteq A$ and so $A\in\mathaccent 1373{h_{1}\circ g_{n}}(q)$. Thus $\mathaccent 869{h_{1}}(r_{n})=\mathaccent 1373{h_{1}\circ g_{n}}(q)=p$.

By Lemma 2.1(a) $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptr_{n}$ and $q\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptr_{n}$ for each $n\in N$, so $r_{n}\in F_{1,1}^{p,q}$. It remains to prove that all $r_{n}$ are distinct so, by Lemma 4.3(b), the set of ultrafilters $r\supseteq F_{1,1}^{p,q}$ will be of cardinality $2^{{c}}$. Let $m<n$. We prove that the sets $g_{m}[f^{-1}[V_{m,n}]]$ and $g_{n}[f^{-1}[V_{m,n}]]$ are disjoint: if we assume that $g_{m}(i)=g_{n}(j)$ for some $i,j\in f^{-1}[V_{m,n}]$ then $if_{i}(m)=jf_{j}(n)$, so since $f_{i}(m)\leq i$, $f_{j}(n)\leq j$ and all of the numbers $i,j,f_{i}(m),f_{j}(n)$ are prime, we have $i=j$ and $f_{i}(m)=f_{i}(n)$, a contradiction with the fact $f_{i}\in V_{m,n}$. But $g_{m}[f^{-1}[V_{m,n}]]\in r_{m}$ and $g_{n}[f^{-1}[V_{m,n}]]\in r_{n}$, so $r_{m}\neq r_{n}$. $\Box$

5 The higher levels

Definition 5.1

We call ultrafilters of the form $p^{k}$ for some $p\in\overline{P}$ and $k\in N$ basic. Let ${\cal B}$ be the set of all basic ultrafilters, and let ${\cal A}$ be the set of all functions $\alpha:{\cal B}\rightarrow N\cup\{0\}$ with finite support (i.e. such that $\{b\in{\cal B}:\alpha(b)\neq 0\}$ is finite).

We will abuse notation and write $\alpha=\{(b_{1},n_{1}),(b_{2},n_{2}),\dots,(b_{m},n_{m})\}$ if $\alpha(b)=0$ for $b\notin\{b_{1},b_{2},\dots,b_{m}\}$ (allowing also some of the $n_{i}$ to be zeros).

Definition 5.2

Let $\alpha=\{(p_{1}^{k_{1}},n_{1}),(p_{2}^{k_{2}},n_{2}),\dots,(p_{m}^{k_{m}},n_{m})\}\in{\cal A}$ ($p_{i}\in\overline{P}$). With $F_{\alpha}$ we denote the family of all sets

[TABLE]

such that: (i) $A_{i}\in p_{i}\mathchar 13334\relax P$, (ii) $A_{i}=A_{j}$ if $p_{i}=p_{j}$ and $A_{i}\cap A_{j}=\emptyset$ otherwise.

Example 5.3

$F_{1,1}^{p,q}=F_{\alpha}$* for $\alpha=\{(p,1),(q,1)\}$ and $F_{2}^{p}=F_{\beta}$ for $\beta=\{(p,2)\}$. If $\gamma=\{(p^{2},1)\}$ then the set $F_{\gamma}=\{A^{2}:A\in p\mathchar 13334\relax P\}$ generates $p^{2}$.*

Definition 5.4

If $\alpha=\{(p_{1}^{k_{1}},n_{1}),(p_{2}^{k_{2}},n_{2}),\dots,(p_{m}^{k_{m}},n_{m})\}\in{\cal A}$ ($p_{i}\in\overline{P}$), we denote $\sigma(\alpha)=\sum_{i=1}^{m}(k_{i}n_{i})$.

Theorem 5.5

The $n$-th level $\overline{L_{n}}$ (for $n\in N$) consists precisely of ultrafilters containing $F_{\alpha}$ for some $\alpha\in{\cal A}$ such that $\sigma(\alpha)=n$.

**Proof. **To avoid cumbersome notation we prove the theorem only for $n=4$. This special case contains essentially all the ideas needed for the proof in general.

Ultrafilters in the 4th level contain $L_{4}=\{a_{1}a_{2}a_{3}a_{4}:a_{1},a_{2},a_{3},a_{4}\in P\}$. We partition $L_{4}$ as

[TABLE]

So for every ultrafilter $x\in\overline{L_{4}}$ we have 5 cases.

$1^{\circ}$ $x\ni P^{4}=\{a^{4}:a\in P\}$. Then $p:=\{A\subseteq P:A^{4}\in x\}$ is an ultrafilter in $\overline{P}$ and it is easy to see that $x=p^{4}$, so $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{4},1)\}$.

$2^{\circ}$ $x\ni P^{3}P=\{a^{3}b:a,b\in P,a\neq b\}$. We define functions $f_{1}(a^{3}b)=a^{3}$ and $f_{2}(a^{3}b)=b$ for $a^{3}b\in P^{3}P$. Let $p^{3}={\mathaccent 869{f}_{1}}(x)$ (clearly, ${\mathaccent 869{f}_{1}}(x)\in\overline{P^{3}}$) and $q={\mathaccent 869{f}_{2}}(x)$. We prove that $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{3},1),(q,1)\}$. Let $A^{3}B\in F_{\alpha}$ (it may be that $p=q$ - then $A=B$, or $p\neq q$ and $A\cap B=\emptyset$). Then $A^{3}B=f_{1}^{-1}[A^{3}]\cap f_{2}^{-1}[B]\in x$.

$3^{\circ}$ $x\ni(P^{2})^{(2)}=\{a^{2}b^{2}:a,b\in P,a\neq b\}$. We define functions $f_{1}(a^{2}b^{2})=a^{2}$ and $f_{2}(a^{2}b^{2})=b^{2}$ for $a^{2}b^{2}\in(P^{2})^{(2)}$ and $a<b$. Let $p^{2}={\mathaccent 869{f}_{1}}(x)$ and $q^{2}={\mathaccent 869{f}_{2}}(x)$. If $p\neq q$ then $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{2},1),(q^{2},1)\}$: for each disjoint $A\in p\mathchar 13334\relax P$ and $B\in q\mathchar 13334\relax P$, $A^{2}B^{2}\supseteq f_{1}^{-1}[A^{2}]\cap f_{2}^{-1}[B^{2}]\in x$. Otherwise, $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{2},2)\}$: for each $A\in p$, $(A^{2})^{(2)}=f_{1}^{-1}[A^{2}]\cap f_{2}^{-1}[A^{2}]\in x$.

$4^{\circ}$ $x\ni P^{2}P^{(2)}=\{a^{2}bc:a,b,c\in P,a\neq b\neq c\neq a\}$. We define functions $f_{1}(a^{2}bc)=a^{2}$, $f_{2}(a^{2}bc)=b$ and $f_{3}(a^{2}bc)=c$ for $a^{2}bc\in P^{2}P^{(2)}$ and $b<c$. Let $p^{2}={\mathaccent 869{f}_{1}}(x)$, $q={\mathaccent 869{f}_{2}}(x)$ and $r={\mathaccent 869{f}_{3}}(x)$. If $q=r$ then $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{2},1),(q,2)\}$, otherwise $x\supseteq F_{\alpha}$ for $\alpha=\{(p^{2},1),(q,1),(r,1)\}$.

$5^{\circ}$ $x\ni P^{(4)}=\{abcd:a,b,c,d\in P,\;a,b,c,d\mbox{ all distinct}\}$. Analogously to previous cases, $x\supseteq F_{\alpha}$ for one of the following: $\alpha=\{(p,4)\}$, $\alpha=\{(p,3),(q,1)\}$, $\alpha=\{(p,2),(q,1),(r,1)\}$, $\alpha=\{(p,2),(q,2)\}$ or $\alpha=\{(p,1),(q,1),(r,1),(s,1)\}$ for some $p,q,r,s\in\overline{P}$. $\Box$

Definition 5.6

For every $\alpha\in{\cal A}$ and every $p\in\overline{P}$ let $\alpha\mathchar 13334\relax p=\langle\alpha(p^{k}):k\in N\rangle$. Clearly, all such sequences have finitely many non-zero elements. If $\vec{x}=\langle x_{k}:k\in N\rangle$ and $\vec{y}=\langle y_{k}:k\in N\rangle$ are two such sequences in $N\cup\{0\}$ we say that $\vec{y}$ dominates $\vec{x}$ if for every $m\in N$, $\sum_{k\geq m}x_{k}\leq\sum_{k\geq m}y_{k}$.

We define an order on ${\cal A}$ as follows: $\alpha\leq\beta$ if for every $p\in\overline{P}$ $\beta\mathchar 13334\relax p$ dominates $\alpha\mathchar 13334\relax p$.

Example 5.7

If $\alpha=\{(p,2)\}$ and $\beta=\{(p,1),(p^{2},1)\}$ (for some $p\in\overline{P}$), then $\alpha\mathchar 13334\relax p=\langle 2,0,0,\dots\rangle$ and $\beta\mathchar 13334\relax p=\langle 1,1,0,0,\dots\rangle$ so $\beta\mathchar 13334\relax p$ dominates $\alpha\mathchar 13334\relax p$ and $\alpha\leq\beta$.

But if $\alpha=\{(p,2),(p^{2},2),(q,2)\}$ and $\beta=\{(p,1),(p^{3},3),(q^{2},1)\}$, then $\beta\mathchar 13334\relax p=\langle 1,0,3,0,\dots\rangle$ dominates $\alpha\mathchar 13334\relax p=\langle 2,2,0,0,\dots\rangle$, but $\beta\mathchar 13334\relax q=\langle 0,1,0,0,\dots\rangle$ does not dominate $\alpha\mathchar 13334\relax q=\langle 2,0,0,\dots\rangle$ so $\alpha\leq\beta$ does not hold.

It is not hard to see that $\alpha\leq\beta$ implies $\sigma(\alpha)\leq\sigma(\beta)$.

Theorem 5.8

Let $\alpha,\beta\in{\cal A}$.

(a) If $x\supseteq F_{\alpha}$, $y\supseteq F_{\beta}$ and $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$, then $\alpha\leq\beta$.

(b) There is no $x\supseteq F_{\alpha}\cup F_{\beta}$ for $\alpha\neq\beta$.

**Proof. **(a) Assume that $\alpha\leq\beta$ does not hold. This means that, for some prime $p$, $\beta\mathchar 13334\relax p=\langle y_{k}:k\in N\rangle$ does not dominate $\alpha\mathchar 13334\relax p=\langle x_{k}:k\in N\rangle$, i.e. there is $m\in N$ such that $\sum_{k\geq m}x_{k}>\sum_{k\geq m}y_{k}.$ Let $n=\max\{k:x_{k}>0\lor y_{k}>0\}$; then

[TABLE]

For any $A\in p$, if we denote $B=((A^{m})^{(x_{m})}(A^{m+1})^{(x_{m+1})}\dots(A^{n})^{(x_{n})})\uparrow$, we have $B\in x$, because $B\cap L_{\sigma(\alpha)}$ contains a set in $F_{\alpha}$. On the other hand, $B\notin y$, because $B$ is disjoint from any set in $F_{\beta}$: every element of $B$ has a divisor of the form $a_{m,1}^{m}\dots a_{m,x_{m}}^{m}a_{m+1,1}^{m+1}\dots a_{n,x_{n}}^{n}$ with $u$ prime factors from $A$ to powers $\geq m$ and, because of (5), no element from a set in $F_{\beta}$ does (they all have exactly $v$ prime factors from $A$ to powers $\geq m$). Thus, $B\in x\cap{\cal U}\setminus y$. This is a contradiction with $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$.

(b) Since $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$, by (a) $x\supseteq F_{\alpha}\cup F_{\beta}$ would imply $\alpha\leq\beta$ and $\beta\leq\alpha$. However, the relation $\leq$ on ${\cal A}$ is clearly antisymmetric, so $\alpha=\beta$. $\Box$

Example 5.9

The reverse of Theorem 5.8(a) does not hold: we will construct ultrafilters $x\supseteq F_{\alpha}$ and $y\supseteq F_{\beta}$ such that $\alpha\leq\beta$ but $x\hskip 2.84526pt\mathaccent 869{\mathchar 13613\relax}\hskip 2.84526pty$.

If $p\in P^{*}$ is not Ramsey, by Theorem 3.6 there is $A\subseteq P^{(2)}$ such that neither $A$ nor $P^{(2)}\setminus A$ contain a set in $F_{2}^{p}$. Exactly one of these two sets is in $p\cdot p$, say $A\in p\cdot p$, i.e. $\{a\in P:A/a\in p\}\in p$. Then for every $X\in p$ we can choose $a\in X$ such that $A/a\in p$. If $bc\notin A$ for all distinct $b,c\in X\cap A/a$, then $(X\cap A/a)^{(2)}\subseteq P^{(2)}\setminus A$, a contradiction with our choice of $A$. So for every $X\in p$ there are $a,b,c\in X$ such that $ab,ac,bc\in A$. This means that, if we denote $S=\{abc:ab,ac,bc\in A\}$, the family $F_{3}^{p}\cup\{S\}$ has the f.i.p. Hence we can find ultrafilters $x\supseteq F_{2}^{p}\cup\{P^{(2)}\setminus A\}$ and $y\supseteq F_{3}^{p}\cup\{S\}$, so $S\downarrow\in(y\cap{\cal V})\setminus x$ and thus $x\hskip 2.84526pt\mathaccent 869{\mathchar 13613\relax}\hskip 2.84526pty$.

Theorem 5.10

Let $\alpha,\beta\in{\cal A}$ be such that $\alpha\leq\beta$.

(a) If $x\supseteq F_{\alpha}$ then there is at least one $y\supseteq F_{\beta}$ such that $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$.

(b) If $y\supseteq F_{\beta}$ then there is at least one $x\supseteq F_{\alpha}$ such that $x\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$.

**Proof. **(a) Let $\alpha=\{(b_{1},n_{1}),(b_{2},n_{2}),\dots,(b_{m},n_{m})\}$ and $\beta=\{(b_{1},n_{1}^{\prime}),(b_{2},n_{2}^{\prime})$, $\dots,(b_{m},n_{m}^{\prime})\}$. We prove that the family $F_{\beta}\cup(x\cap{\cal U})$ has the f.i.p. Since every intersection of finitely many elements of $F_{\beta}$ contains an element of $F_{\beta}$ and $x\cap{\cal U}$ is closed for finite intersections, it suffices to prove that every set $B^{\prime}=A_{1}^{(n_{1}^{\prime})}A_{2}^{(n_{2}^{\prime})}\dots A_{m}^{(n_{m}^{\prime})}\in F_{\beta}$ intersects every $X\in x\cap{\cal U}$. But $X$ intersects $B=A_{1}^{(n_{1})}A_{2}^{(n_{2})}\dots A_{m}^{(n_{m})}\in F_{\alpha}$, say $l\in B\cap X$. It suffices to show that $B^{\prime}$ contains numbers divisible by $l$.

Let $p$ be any prime ultrafilter which is a divisor of $x$. Let $l=a_{1}^{k_{1}}a_{2}^{k_{2}}\dots a_{r}^{k_{r}}b$ be the factorization of $l$ such that $a_{1}^{k_{1}},a_{2}^{k_{2}},\dots$ (where $a_{i}\neq a_{j}$ for $i\neq j$) are elements of those sets $A_{1},A_{2},\dots$ that belong to a power of $p$, ordered so that $k_{1}\geq k_{2}\geq\dots$, and $b$ is the product of the other factors of $l$. Let $l^{\prime}=(a_{1}^{\prime})^{k_{1}^{\prime}}(a_{2}^{\prime})^{k_{2}^{\prime}}\dots(a_{s}^{\prime})^{k_{s}^{\prime}}b^{\prime}$ be the factorization of any element $l^{\prime}\in B^{\prime}$ obtained in the same way. Then the fact that $\beta\mathchar 13334\relax p$ dominates $\alpha\mathchar 13334\relax p$ implies that $s\geq r$ and $k_{i}^{\prime}\geq k_{i}$ for $i\leq r$. We let $l_{p}=a_{1}^{k_{1}^{\prime}}a_{2}^{k_{2}^{\prime}}\dots a_{r}^{k_{r}^{\prime}}(a_{r+1}^{\prime})^{k_{r+1}^{\prime}}\dots(a_{s}^{\prime})^{k_{s}^{\prime}}$; this element is clearly divisible by $a_{1}^{k_{1}}a_{2}^{k_{2}}\dots a_{r}^{k_{r}}$. In the same way we construct, for every prime $q\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$, a number $l_{q}$. The product of all such $l_{q}$ will be the desired element divisible by $l$.

(b) Analogously to (a), we can prove that $F_{\alpha}\cup(y\cap{\cal V})$ has the f.i.p. $\Box$

Theorem 3.6 shows that the ultrafilter $y$ from Theorem 5.10(a) need not be unique. The next example shows the same for $x$ from Theorem 5.10(b).

Example 5.11

As in Example 5.9, let $p\in P^{*}$ be non-Ramsey and let $A\subseteq P^{(2)}$ be such that both $F_{2}^{p}\cup\{A\}$ and $F_{2}^{p}\cup\{A"\}$ (where $A"=P^{(2)}\setminus A$) have the f.i.p. Then $F_{4}^{p}\cup\{AA"\}$ also has the f.i.p.: if $B^{(4)}\in F_{4}^{p}$ then $B^{(2)}$ intersects both $A$ and $A"$, say $a=b_{1}b_{2}\in B^{(2)}\cap A$ and $a"=b_{1}"b_{2}"\in B^{(2)}\cap A"$ (and without loss of generality $b_{1}",b_{2}"$ are different from $b_{1},b_{2}$). But then $aa"\in B^{(4)}\cap AA"$. Hence there is $y\supseteq F_{4}^{p}\cup\{AA"\}$ and, if we define $f_{1}(aa")=a$ and $f_{2}(aa")=a"$ for $a\in A$, $a"\in A"$, then $\mathaccent 869{f_{1}}(y)\supseteq F_{2}^{p}\cup\{A\}$ and $\mathaccent 869{f_{2}}(y)\supseteq F_{2}^{p}\cup\{A"\}$ are both divisors of $y$.

Theorem 5.12

If $x\supseteq F_{\alpha}$ and $y\supseteq F_{\beta}$, then $x\cdot y\supseteq F_{\alpha+\beta}$, where $\alpha+\beta=\{(b,n_{b}+n_{b}^{\prime}):(b,n_{b})\in x,(b,n_{b}^{\prime})\in y\}$.

**Proof. **First, the theorem clearly holds if one of the ultrafilters $x$, $y$ is in $N$. If $x=kx^{\prime}$ and $y=ly^{\prime}$ for some $k,l\in N$, $x^{\prime},y^{\prime}\in N^{*}$, then $x\cdot y=kl(x^{\prime}\cdot y^{\prime})$, so we can assume without loss of generality that $x$ and $y$ are not divisible by any elements of $N$.

Now let $\alpha=\{(b_{1},n_{1}),(b_{2},n_{2}),\dots,(b_{m},n_{m})\}$ and $\beta=\{(b_{1},n_{1}^{\prime}),(b_{2},n_{2}^{\prime}),\dots$, $(b_{m},n_{m}^{\prime})\}$. Then the sets in $F_{\alpha+\beta}$ are of the form $A_{b_{1}}^{(n_{1}+n_{1}^{\prime})}A_{b_{2}}^{(n_{2}+n_{2}^{\prime})}\dots A_{b_{m}}^{(n_{m}+n_{m}^{\prime})}$ (where $A_{p}\in p$ for prime divisors $p$ of $x$ or $y$, $A_{p}\cap A_{q}=\emptyset$ for $p\neq q$ and, for basic divisors of the form $p^{k}$, $A_{p^{k}}=A_{p}^{k}$). But if $a=a_{1,1}a_{1,2}\dots a_{1,n_{1}}a_{2,1}\dots a_{m,n_{m}}\in A_{b_{1}}^{(n_{1})}A_{b_{2}}^{(n_{2})}\dots A_{b_{m}}^{(n_{m})}$, then

[TABLE]

where $B_{b_{k}}=A_{b_{k}}\setminus\{a_{1,1},a_{1,2},\dots a_{m,n_{m}}\}$. For every $a\in A_{b_{1}}^{(n_{1})}A_{b_{2}}^{(n_{2})}\dots A_{b_{m}}^{(n_{m})}$ we have $B_{b_{1}}^{(n_{1}^{\prime})}\dots B_{b_{m}}^{(n_{m}^{\prime})}\in y$ so

[TABLE]

i.e. $A_{b_{1}}^{(n_{1}+n_{1}^{\prime})}A_{b_{2}}^{(n_{2}+n_{2}^{\prime})}\dots A_{b_{m}}^{(n_{m}+n_{m}^{\prime})}\in x\cdot y$. $\Box$

In particular, if $x\in\overline{L_{m}}$ and $y\in\overline{L_{n}}$ then $xy\in\overline{L_{m+n}}$. A corollary of this theorem is that levels $\overline{L_{n}}$ contain no idempotents (ultrafilters such that $x\cdot x=x$), since if $x\in\overline{L_{n}}$, then $x\cdot x\in\overline{L_{2n}}$. Another corollary is that, if $p\in\overline{P}$ and $x$ and $y$ are on finite levels, then $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx\cdot y$ implies $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx$ or $p\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pty$, which (partly) justifies our calling such ultrafilters prime.

Lemma 5.13

$|[x]_{\sim}|=1$* for every $n\in N$ and all ultrafilters $x\in\overline{L_{n}}$.*

**Proof. **Assume the opposite, that $y=_{\sim}x$. If $x\in F_{\alpha}$ and $y\in F_{\beta}$, then Theorem 5.8(a) implies that $\alpha=\beta$, so $y\in\overline{L_{n}}$ as well. Let $A\subseteq L_{n}$ be such that $A\in y$ and $L_{n}\setminus A\in x$. Then $A\uparrow\in y\cap{\cal U}$ and $A\uparrow\notin x$, a contradiction. $\Box$

All of the proofs in the next lemma are analogous to Lemma 3.5(b).

Lemma 5.14

(a) For every $\alpha\in{\cal A}$ there are either finitely many or $2^{{c}}$ ultrafilters containing $F_{\alpha}$.

(b) For every $x\in\beta N$ there are either finitely many or $2^{{c}}$ ultrafilters above $x$, and either finitely many or $2^{{c}}$ ultrafilters below $x$.

(c) For every $x\supseteq F_{\alpha}$ and all $\beta\leq\alpha\leq\gamma$ there are either finitely many or $2^{{c}}$ ultrafilters above $x$ in $F_{\gamma}$, and either finitely many or $2^{{c}}$ ultrafilters below $x$ in $F_{\beta}$.

6 Above all finite levels

Fact 6.1

([9], Theorem 4.1) There is the $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$-greatest class (consisting of ultrafilters divisible by all other ultrafilters).

Lemma 6.2

Let $\langle x_{n}:n\in N\rangle$ be a sequence of ultrafilters such that $x_{n}\in\overline{L_{n}}$ and $x_{m}\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526ptx_{n}$ for $m<n$. Then there is an ultrafilter divisible by all $x_{n}$ and not divisible by any ultrafilter which is not below any $x_{n}$.

**Proof. **Let $F_{1}=\{A\in{\cal U}:A\in x_{n}\mbox{ for some }n\in N\}$ and $F_{2}=\{B\in{\cal V}:B^{c}\notin x_{n}\mbox{ for all }n\in N\}$. We prove that $F_{1}\cup F_{2}$ has the f.i.p. First, both $F_{1}$ and $F_{2}$ are closed for finite intersections. So let $A\in F_{1}$ and $B\in F_{2}$. There is $n\in N$ such that $A\in x_{n}$. Since also $B^{c}\notin x_{n}$, we have $A\cap B\in x_{n}$, so $A\cap B\neq\emptyset$. $\Box$

In particular, for every countable set $S\subseteq\overline{P}$ there is an ultrafilter divisible by all $p\in S$ but not divisible by any prime $p\in\overline{P}\setminus S$. Since there are $2^{2^{{c}}}$ subsets of $\overline{P}$, this can not hold for all uncountable $S\subseteq\overline{P}$.

Let us also note that it was shown in [8] that $\hskip 2.84526pt\mathaccent 869{\mid}\hskip 2.84526pt$ is not antisymmetric, so there are ultrafilters above all finite levels such that $|[x]_{\sim}|>1$.

7 Final remarks

The ultrafilters containing $F_{2}^{p}$ and $F_{1,1}^{p,q}$ bear similarities with ultrafilters that are preimages under the natural map from $\beta(N\times N)$ to $\beta N\times\beta N$. Such ultrafilters were investigated, among other papers, in [1], [5] and [3]. Hence some of the proofs in this paper are modifications of ideas from these papers (most of which can also be found in Chapter 16 of [4]). Since some of these modifications were not quite trivial, and for the sake of completeness, we decided to include all the proofs here. Also, the proof of Lemma 3.12(b) is more general, different and (hopefully) more intuitive then in [5].