# Exponential ultrafilters and patterns in Ramsey theory

Sayan Goswami
<EMAIL_ADDRESS>Institute of Mathematical Sciences, X6VW+FP7,
4th Cross Street, CIT Campus, Tharamani, Chennai, Tamil Nadu 600113. Lorenzo
Luperi Baglini
<EMAIL_ADDRESS>222Dipartimento di Matematica, Università di Milano,
Via Saldini 50, 20133 Milano, Italy. Sourav Kanti Patra
<EMAIL_ADDRESS>Department of Mathematics, Centre for Distance
and Online Education, University of Burdwan, Burdwan-713104,West Bengal.

###### Abstract

In this article, we show that the partition regularity of exponential triples
admits a simple proof by means of the Hales-Jewett theorem. By adapting this
proof, we extend in several directions (including polynomial extension) many
celebrated results about the partition regularity of finite and infinite
exponential configurations. This also leads to the introduction of two
exponentiations among ultrafilters, whose combinatorial properties are studied
in detail.

Mathematics subject classification 2020: 05D10, 05C55, 22A15, 54D35.
Keywords: Exponential patterns, Hales-Jewett theorem, Stone-Čech
compactification.

## 1 Introduction

In Arithmetic Ramsey Theory, problems regard the so-called partition regular
arithmetic configurations and equations on $\mathbb{N}=\\{1,2,\dots\\}$.

######  Definition 1.1.

Let
$f_{1}\left(x_{1},\dots,x_{n}\right),\dots,f_{m}\left(x_{1},\dots,x_{n}\right):\mathbb{N}^{n}\rightarrow\mathbb{N}$.
We say that the configuration
$\\{f_{1}\left(x_{1},\dots,x_{n}\right),\dots,f_{m}\left(x_{1},\dots,x_{n}\right)\\}$
is partition regular on $\mathbb{N}$ (PR from now on) if for all finite
partitions $\mathbb{N}=A_{1}\cup\dots\cup A_{k}$ there are $i\leq k$ and
$x_{1},\dots,x_{n}\in\mathbb{N}$ such that for all $j\leq m$
$f_{j}\left(x_{1},\dots,x_{n}\right)\in A_{i}$.

Analogously, we say that the system $\sigma\left(x_{1},\dots,x_{n}\right)$
defined by

$\begin{cases}f_{1}\left(x_{1},\dots,x_{n}\right)=0,\\\ \ \ \ \ \ \ \ \ \
\vdots\\\ f_{m}\left(x_{1},\dots,x_{n}\right)=0\end{cases}$

is PR if for all finite partitions $\mathbb{N}=A_{1}\cup\dots\cup A_{k}$ there
are $i\leq k$ and $x_{1},\dots,x_{n}\in A_{i}$ such that
$\sigma\left(x_{1},\dots,x_{n}\right)=0$.

In [21], Schur proved one of the first and most important results in
Arithmetic Ramsey Theory.

######  Theorem 1.2 (Additive Schur Theorem).

The configuration $\\{x,y,x+y\\}$ is PR.

Equivalently, the additive Schur Theorem states that the equation $x+y=z$ is
PR.

Applying transformations to partitions allows, sometimes, to easily deduce new
results by just considering the shape of transformed partition regular
configurations. A useful trick, that sometimes allows turning additive
structures in multiplicative structures, and multiplicative structures in
exponential ones, is that of considering $\log_{2}$-base partitions: for
$B\subseteq\mathbb{N}$, let $2^{-B}=\\{n\in\mathbb{N}\mid 2^{n}\in B\\}$ and,
given a partition $\mathbb{N}=A_{1}\cup\dots\cup A_{k}$, let the associated
$\log_{2}$-base partition be given by $\mathbb{N}=2^{-A_{1}}\cup\dots\cup
2^{-A_{k}}$. For example, given any finite partition of $\mathbb{N}$, by
Theorem 1.2 in the $\log_{2}$-base partition we find monochromatic $x,y,z$
with $x+y=z$; hence, $2^{x},2^{y},2^{z}=2^{x+y}=2^{x}\cdot 2^{y}$ were
monochromatic with respect to the original partition. This proves the
multiplicative version of Schur.

######  Theorem 1.3 (Multiplicative Schur Theorem).

The configuration $\\{x,y,x\cdot y\\}$ is PR.

Equivalently, the multiplicative Schur Theorem states that the equation $xy=z$
is PR.

Recently, in [23], the question about the partition regularity of
$\\{x,y,x^{y}\\}$, which we will call an exponential triple, was posed and
partially answered, by using ultrafilters, by Sisto; its full solution was
first given in [20] by Sahasrabudhe who, among other generalizations, proved
the exponential Schur Theorem444At the best of our knowledge, this theorem has
not been called explicitly like this in the literature before.:

######  Theorem 1.4 (Exponential Schur Theorem).

The configuration $\\{x,y,x^{y}\\}$ is PR.

Again, the above theorem can be rephrased in terms of equations: it states
that $x^{y}=z$ is PR. Sahasrabudhe’s proof used repeated applications of van
der Waerden theorem and the compactness property on the convergence of
subsequences in the space of $k$-partitions of $\mathbb{N}$. One of the key
ingredients in his proof was the use of the function $f(n,m):=2^{n}m$,
combined with associated $\log_{2}$-base partitions. Starting from
observations on this function, in [9, Theorem 1.1] Di Nasso and Ragosta found
a much simpler proof of the partition regularity of the configuration
$\\{x,y,x2^{y}\\}$, from which the exponential Schur’s theorem can be
immediately deduced by the same $\log_{2}$-partition trick that allows proving
the multiplicative version of Schur Theorem from the additive one (see [9,
Corollary 1.3]). In their proof, a clever trick based on tensor products of
ultrafilters555Through the paper we assume the reader to have a basic
knowledge of the main properties of $\beta\mathbb{N}$; we refer to the
monograph [17] for an expensive presentation of ultrafilters and their
applications to combinatorics. and the extension to
$\left(\beta\mathbb{N}\right)^{2}$ of the same function $f(n,m)=2^{n}m$ used
by Sahasrabudhe is applied, which allows getting a much simpler proof of the
partition regularity of exponential triples.

In this paper, we have two primary goals.

Our first goal is to show that, actually, the partition regularity of
exponential triples, as well as of many more complicated exponential
configurations, can be deduced with rather elementary methods just by applying
the Hales-Jewett theorem. Moreover, our results allow us to have a good
qualitative control on the kind of sets that witness the partition regularity
of exponential configuration. Explicitly, when focused on exponential triples
our main result will be the following:

######  Theorem 1.5.

Let $n\in\mathbb{N},n>1$ and let $A\subseteq\mathbb{N}$ be multiplicatively
piecewise syndetic666We will recall the definition of multiplicatively
piecewise syndetic sets in Section Hales-Jewett Theorem.. Then there exists
$x,y\in A$ such that $xn^{y}\in A$.

Notice that, in particular, $n^{A}$ contains the exponential triple
$n^{x},n^{y},x^{xn^{y}}=\left(n^{x}\right)^{\left(n^{y}\right)}$, and this
suffices to deduce straightforwardly the exponential Schur Theorem just by
using the same $\log_{2}$-associated partition trick discussed above and the
fact that multiplicatively piecewise syndetic sets are partition regular.

Several questions that we study in this paper are related to our main goal. In
particular, in Section 3 we will show the partition regularity of some
exponential patterns with additional structure, whilst in Section 4 we will
consider infinite exponential configurations, providing a polynomial extension
of a result of Di Nasso and Ragosta about exponential towers, namely [9,
Theorem 3.1], as well as showing that a famous result of Bergelson and
Hindman, namely the partition regularity of combined infinite multiplicative
and additive structures, can be extended to include exponentiation as well.

Our second main goal is to start a systematic study of ultrafilters related to
exponential triples. The starting observation is that, motivated by Theorem
1.5, if for for any $n\in\mathbb{N}$ and $p\in\beta\mathbb{N}$ we let

$n^{p}=\\{A\in\mathbb{N}\mid\\{m:n^{m}\in A\\}\in p\\},$

(equivalently, $n^{p}$ is the image of $p$ with respect to the extension to
$\beta\mathbb{N}$ of the function that maps any $x\in\mathbb{N}$ to $n^{x}$),
then we can use the well-known fact that ultrafilters in
$\overline{K(\beta\mathbb{N},\odot)}$ are made by multiplicatively piecewise
syndetic sets to rephrase Theorem 1.5 as follows:

######  Theorem 1.6.

Let $n\in\mathbb{N}$ and $\mathcal{U}\in\overline{K(\beta\mathbb{N},\odot)}$.
Then every set $A\in n^{\mathcal{U}}$ contains exponential triples.

Theorem 5.8 leads naturally to consider the exponentiation of ultrafilters;
since exponentiation is not associative, this will force us to introduce two
different operations between ultrafilters.

We will start a detailed study of these operations, discussing in particular
their relationships with exponential triples and the existence of idempotents
in Section 5. This will lead also to consider a generalization of the
operation $\star$, which first appeared in [9], defined as $p\star
q=2^{p}\odot q$.

Finally, in Section 7 we present and discuss some open questions that arise as
consequences of our results.

## 2 Exponential consequences of the Hales-Jewett Theorem

As shortly discussed in the introduction, so far all proofs777At least, all
proofs that we are aware of. of the partition regularity of exponential
triples and related configurations use some kind of non-elementary method:
topological methods in [20], tensor products of ultrafilters in [9]. In this
Section, we show that a rather straightforward application of the 60-year-old
Hales-Jewett theorem actually suffices to produce several partition regular
exponential configurations, and to connect them with multiplicatively
piecewise syndetic sets.

Before starting to explore our Hales-Jewett exponential consequences, to the
readers interested solely in exponential triples we offer a straight proof of
Theorem 1.5 which uses basic facts about MPS sets, ultrafilters in
$K(\beta\mathbb{N},\odot)$ and the Pigeonhole principle.

###### Direct proof of Theorem 1.5.

As $A$ is a multiplicative piecewise syndetic set, there exists $p\in
K(\beta\mathbb{N},\odot)$ such that $A\in p$. Hence, by [17, Theorem 4.39] the
set $S=\\{x:x^{-1}A\in p\\}$ is syndetic. Let $\mathbb{N}=\bigcup_{t\in
F}t^{-1}S$, and $|F|=r.$ As $A$ is a piecewise syndetic set, it follows from
the Lemma 2.2 that it must contain an $IP_{r+1}$, say $FS(\langle
b_{i}\rangle_{i=1}^{r+1})\subset A.$

As
$\\{n^{b_{1}},n^{b_{1}+b_{2}},\ldots,n^{b_{1}+\cdots+b_{r+1}}\\}\subset\bigcup_{t\in
F}t^{-1}S$ and $|F|=r$, by Pigeonhole principle there exists $t\in F$ and
$m,n\in[1,r+1]$ such that $\\{tn^{b_{1}+\cdots+b_{m}},tn^{b_{1}+\cdots
b_{n}}\\}\subset S.$ Let $m<n$, $b=n^{b_{1}+\cdots+b_{m}}$ and
$c=b_{m+1}+\cdots b_{n}$. So, $\\{tb,tbn^{c}\\}\subset S$. Choose $a\in
A\cap(tb)^{-1}A\cap(tbn^{c})^{-1}A\in p.$ If $x=atb\in A$ and $y=c\in A$, by
construction $xn^{y}\in A$.∎

###  The Hales-Jewett trick

As we are going to repeatedly use the Hales-Jewett theorem, we start this
section by recalling it, taking in the meantime the chance to fix some
notation.

Let $\mathbb{A}$ be a nonempty set. A word of lenght $n\in\mathbb{N}$ in the
alphabet $\mathbb{A}$ is a sequence $w=a_{1}a_{2}\ldots a_{n}$ with $n\geq 1$
and $a_{1},\dots,a_{n}\in\mathbb{A}$. We let $\mathbb{A}^{<\omega}$ to be the
set of words of finite length in the alphabet $\mathbb{A}$. If
$x\notin\mathbb{A}$, a variable word over $\mathbb{A}$ is a word in
$\left(\mathbb{A}\cup\\{x\\}\right)^{<\omega}$ containing at least one
occurrence of $x$. In this setting, in [13] in 1963 Hales-Jewett proved the
following fundamental theorem.

######  Hales-Jewett Theorem.

For all $n,m\in\mathbb{N}$, there exists a number $\text{HJ}\left(n,m\right)$
such that, if $N\geq\text{HJ}\left(n,m\right)$ and
$\\{1,\dots,m\\}^{N}=A_{1}\cup\dots\cup A_{n}$ is a partition, then there
exists a variable word $w$ and an index $j\leq n$ such that
$\left\\{w\left(a\right):a\in\\{1,\dots,m\\}\right\\}\subseteq A_{j}$.

Before proceeding, we recall also a few well-known classes of sets that are
relevant for our aims.

######  Definition 2.1.

Let $(S,\cdot)$ be a semigroup, let $n\in\mathbb{N}$ and let $A\subseteq S$.
We say that

  * •

$A$ is a thick set if for any finite subset $F\subset S$, there exists an
element $x\in S$ such that $Fx=\\{fx:f\in F\\}\subset A$;

  * •

$A$ is a syndetic set if there exists a finite set $F\subset S$ such that
$S=\bigcup_{x\in F}x^{-1}A$, where $x^{-1}A=\\{y:xy\in A\\}$;

  * •

$A$ is piecewise syndetic if there exists a finite set $F\subset S$ such that
$\bigcup_{x\in F}x^{-1}A$ is a thick set;

  * •

in particular, piecewise syndetic sets in the semigroup $(\mathbb{N},\cdot)$
will be called multiplicative piecewise syndetic sets, abbreviated as MPS from
now on;

  * •

$A$ is an $IP_{n}$-set if there exists $F\subseteq A$,
$F=\\{x_{1},\dots,x_{n}\\}$ such that for all nonempty
$I\subseteq\\{1,\dots,n\\}$ one has that $\sum_{i\in I}x_{i}\in A$;

  * •

$A$ is an $IP_{n}^{\ast}$-set if $A\cap B\neq\emptyset$ for all $IP_{n}$-set
$B$;

  * •

$A$ is central if it belongs to a minimal idempotent in $\beta S$.

The following is a widely known result in the literature, which admits several
different proofs. We refer, for example, to [2, Theorem 7.1.].

######  Lemma 2.2.

Let $A\subseteq\mathbb{N}$ be a MPS set. Then, for each $n\in\mathbb{N}$, $A$
is an $IP_{n}$-set.

We are in the position to prove Theorem 1.5; actually, we will prove it in the
following much stronger version.

######  Theorem 2.3.

Let $A\subset\mathbb{N}$ be a MPS set and let $F\subset\mathbb{N}$ be a finite
set. Then there exists $a,b$ such that $\left\\{a,b,a\cdot n^{b}\mid n\in
F\right\\}\subset A.$

###### Proof.

Without loss of generality, assume that $F=\\{1,\dots,k\\}$.

Let $G\subseteq\mathbb{N}$ be a finite set such that $\bigcup_{t\in G}t^{-1}A$
is a multiplicative thick set, and let $r=|G|$. Let
$\mathbb{A}=F\cup\left\\{0\right\\}$ and $N=\text{HJ}\left(r,k+1\right).$

Let $x_{1},\dots,x_{N}$ be mutually distinct elements in $\mathbb{N}$. To any
word $w=a_{1}a_{2}\cdots a_{N}$ of length $N$ over $\mathbb{A}$ we associate
the number $e\left(w\right)=\prod_{i=1}^{N}a_{i}^{x_{i}}$.

Now consider the finite set $M=\left\\{e\left(w\right)\mid
w\in\mathbb{A}^{N}\right\\}.$ As $\bigcup_{t\in G}t^{-1}A$ is multiplicatively
thick, there exists $z\in\mathbb{N}$ such that $M\cdot z\subset\bigcup_{t\in
G}t^{-1}A$. We use this fact to construct an $r$-coloring
$\Xi:\mathbb{A}^{N}\rightarrow G$ as follows:

$\Xi(w)=t\ \text{if and only if}\ t=\min\\{g\in G\mid e\left(w\right)\cdot
z\cdot g\in A\\}.$

As $|G|=r$ and $N=\text{HJ}\left(r,k+1\right),$ by the Hales-Jewett Theorem
there exists a monochromatic combinatorial line in $\mathbb{A}^{N}$; namely,
there exists a variable word $w(x)=b_{1}\dots
b_{N},b_{i}\in\mathcal{A}\cup\\{x\\}$ such that
$\Xi(w(0))=\Xi(w(1))=\cdots=\Xi(w(k))$. Call $s$ this common color, let
$H=\\{i\leq N\mid b_{i}=x\\}$ (which is nonempty as $w(x)$ is a variable word)
and

$y=\begin{cases}1,&\text{if}\ H=\\{1,\dots,N\\};\\\ \prod_{i\notin
H}b_{i},&\text{otherwise.}\end{cases}$

By our definitions, by letting $a=yzs,b=\sum_{i\in H}x_{i}$ we have that
$a,a\cdot n^{\sum_{i\in H}x_{i}}\in A$ for all $n\in\\{1,2,\ldots,k\\}.$

In particular $b\in FS\left(\langle x_{i}\rangle_{i=1}^{N}\right)$ so, by the
arbitrarity of $x_{1},\dots,x_{N}$, we have that

$\left\\{b\in\mathbb{N}\mid\ \text{there exists some $a$ satisfying
}\left\\{a,an^{b}:n\in F\right\\}\subset A\right\\}$

is an $IP_{N}^{\ast}$ set. By Lemma 2.2, $A$ is an $IP_{N}$-set, hence

$C=\left\\{y\mid\text{there exists $x$ satisfying }\left\\{x,xn^{y}:n\in
F\right\\}\subset A\right\\}\cap A\neq\emptyset.$

Our thesis now follows just by taking any $b\in C\subseteq A$ and $a\in A$ so
that $\left\\{a,a\cdot n^{b}:n\in F\right\\}\subset A$.∎

The above result can be used to show that any MPS set $A$, and sets of the
form $n^{A}$ consequently, contain several interesting configurations.

######  Example 2.4.

Let $A$ be a MPS set and $n,m_{1},\dots,m_{k}\in\mathbb{N}$. Then

  * •

$\exists x,y\in A$ such that $x,y,xn^{y}\in A$; as discussed in the
introduction, this proves that $n^{A}$ contains exponential triples, which are
hence partition regular;

  * •

$\exists x,y\in A$ such that $\\{x,y,xm_{i}^{jy}\mid i\leq k,j\leq n\\}\in A$;
therefore $n^{A}$ contains arbitrarily long exponential progressions of the
form $\\{x,y,x^{y},x^{(y^{2})},\dots,x^{(y^{k})}\\}$. Arguing as above, this
proves the exponential analogue of van der Waerden Theorem.

Theorem 2.3 can be generalized in several ways. A first direct generalization
can be obtained by using basically the same proof, but appealing to the
multidimensional Hales-Jewett theorem. To state it we need to recall two
definitions from [11] first.

######  Definition 2.5.

Given $k,d,m\in{\mathbb{N}}$ , a $d$-dimensional $m$-variable word on $k$
letters is an element of the set
$\left([k]\cup\\{\ast_{1},...,\ast_{m}\\}\right)^{d}$ where each symbol
$\ast_{i}$ appears at least once. A $m$-dimensional combinatorial subspace of
$[k]^{d}$ is the image of a $m$-variable word.

We can now recall the needed generalization of the Hales-Jewett Theorem proven
in [11, Theorem 7, page 40].

######  Multidimensional Hales-Jewett Theorem.

For each $k,r,m\in{\mathbb{N}}$ there exists some $N=MHJ(r,k,m)$ such that any
for any $r$-coloring of $[k]^{N}$, there is a monochromatic $m$-dimensional
combinatorial subspace.

By using the multidimensional Hales-Jewett Theorem instead of the classical
one, we can modify the proof of Theorem 2.3 to obtain even richer exponential
structures.

######  Theorem 2.6.

Let $m\in\mathbb{N}$, $A\subset\mathbb{N}$ be a MPS set and let
$F\subset\mathbb{N}$ be a finite set. Then there exists $a\in\mathbb{N}$ and a
finite sequence $\langle b_{i}\rangle_{i=1}^{m}$ such that $\left\\{a,b,a\cdot
n^{b}\mid n\in F,\,b\in FS(\langle b_{i}\rangle_{i=1}^{m})\right\\}\subset A.$

###### Proof.

Without loss of generality, assume that $F=\\{1,\dots,k\\}$.

Let $G\subseteq\mathbb{N}$ be a finite set such that $\bigcup_{t\in G}t^{-1}A$
is a multiplicative thick set, and let $r=|G|$. Let
$\mathbb{A}=F\cup\left\\{0\right\\}$ and $N=\text{MHJ}\left(r,k+1,m\right).$

Let $x_{1},\dots,x_{N}$ be mutually distinct elements in $A$ such that
$FS(\langle x_{i}\rangle_{i=1}^{N})\subset A$. To any word $w=a_{1}a_{2}\cdots
a_{N}$ of length $N$ over $\mathbb{A}$ we associate the number
$e\left(w\right)=\prod_{i=1}^{N}a_{i}^{x_{i}}$.

Now consider the finite set $M=\left\\{e\left(w\right)\mid
w\in\mathbb{A}^{N}\right\\}.$ As $\bigcup_{t\in G}t^{-1}A$ is multiplicatively
thick, there exists $z\in\mathbb{N}$ such that $M\cdot z\subset\bigcup_{t\in
G}t^{-1}A$. We use this fact to construct an $r$-coloring
$\Xi:\mathbb{A}^{N}\rightarrow G$ as follows:

$\Xi(w)=t\ \text{if and only if}\ t=\min\\{g\in G\mid e\left(w\right)\cdot
z\cdot g\in A\\}.$

As $|G|=r$ and $N=\text{MHJ}\left(r,k+1,m\right),$ by the multidimensional
Hales-Jewett Theorem there exists a $m$-dimensional monochromatic
combinatorial line in $\mathbb{A}^{N}$; namely, there exists a variable word
$w(\ast_{1},\ast_{2},\cdots,\ast_{m})=w_{1}\dots w_{N},$ such that
$\Xi(w(i_{1},i_{2},\ldots,i_{m}))=\Xi(w(j_{1},j_{2},\ldots,j_{m}))$ for all
$(i_{1},i_{2},\ldots,i_{m}),(j_{1},j_{2},\ldots,j_{m})\in\mathbb{A}^{m}$.

Call $s$ this common color. For each $j\in\\{1,2,\ldots,m\\},$ let
$H_{j}=\\{i\leq N\mid w_{i}=\ast_{j}\\}$ (which is nonempty as
$w(\ast_{1},\ast_{2},\cdots,\ast_{m})$ is a variable word) and

$y=\begin{cases}1,&\text{if}\ \bigcup_{j=1}^{m}H_{j}=\\{1,\dots,N\\};\\\
\prod_{i\notin\bigcup_{j=1}^{m}H_{j}}b_{i},&\text{otherwise.}\end{cases}$

By our definitions, by letting $a=yzs,b_{j}=\sum_{i\in H_{j}}x_{i}$ (for
$j\in\\{1,2,\cdots,m\\}$) we have that for any $H\subseteq\\{1,2,\cdots,m\\}$,
$\\{a,a\cdot n^{\sum_{i\in H}b_{i}}\\}\subset A$ for all
$n\in\\{1,2,\ldots,k\\}$.

As $b\in FS\left(\langle x_{i}\rangle_{i=1}^{N}\right)$, we have
$\left\\{a,b,a\cdot n^{b}\mid n\in F,\,b\in FS(\langle
b_{i}\rangle_{i=1}^{m})\right\\}\subset A.$∎

######  Example 2.7.

Let $A$ be a MPS set and $n,m\in\mathbb{N}$. Then $n^{A}$ contains
configurations of the form $\\{x,y,x^{y}:y\in FP(\langle
y_{i}\rangle_{i=1}^{m})\\}$ for some $y_{1},\dots,y_{m}\in\mathbb{N}$.

A different kind of generalization of Theorems 2.3 and 2.6 can be obtained if
one uses ultrafilters. In here, we will use solely a very simple trick that
was introduced in [8] by Di Nasso and Luperi Baglini; the relationship between
exponential configurations and ultrafilters will be studied more in detail in
Section 5.

It is long known that partition regular configurations and ultrafilters are
closely connected. In fact, the following result holds, which is just a
reformulation of [17, Theorem 57], holds.

######  Theorem 2.8.

A configuration
$\\{f_{1}\left(x_{1},\dots,x_{n}\right),\dots,f_{m}\left(x_{1},\dots,x_{n}\right)\\}$
is PR if and only if there exists an ultrafilter $p\in\beta\mathbb{N}$ such
that for all $A\in p$ there exists $a_{1},\dots,a_{n}\in\mathbb{N}$ with

$f_{1}\left(a_{1},\dots,a_{n}\right),\dots,f_{m}\left(a_{1},\dots,a_{n}\right)\in
A.$

The above characterization of PR leads to the introduction of the following
useful definition.

######  Definition 2.9.

Let $p\in\beta\mathbb{N}$. We say that $p$ witnesses the partition regularity
of the configuration
$\\{f_{1}\left(x_{1},\dots,x_{n}\right),\dots,f_{m}\left(x_{1},\dots,x_{n}\right)\\}$
if for all $A\in p$ there exists $a_{1},\dots,a_{n}\in\mathbb{N}$ such that
$\\{f_{1}\left(a_{1},\dots,a_{n}\right),\dots,f_{m}\left(a_{1},\dots,a_{n}\right)\\}\in
p$.

We will use an analogous definition when talking about partition regular
systems of equations. The simple, but useful, result we need to extend Theorem
2.3 is a reformulation of a theorem that first appeared in [8, Lemma 2.1], to
which we refer for an elementary proof.

######  Theorem 2.10.

Let $\mathcal{U}\in\beta\mathbb{N}$, and let
$f\left(x_{1},\dots,x_{n}\right),g\left(y_{1},\dots,y_{m}\right)$ be two
functions such that
$\mathcal{U}\models\left(f_{1}\left(x_{1},\dots,x_{n}\right)=0\right)\wedge\left(g\left(y_{1},\dots,y_{m}\right)=0\right)$.
Then
$\mathcal{U}\models\left(f_{1}\left(x_{1},\dots,x_{n}\right)=0\right)\wedge\left(g\left(y_{1},\dots,y_{m}\right)=0\right)\wedge\left(x_{1}=y_{1}\right)$.

The above result allows us to recover the partition regularity of many
equations and configurations.

######  Theorem 2.11.

The equation $xy=a^{b}$ is partition regular.

More generally, any equation that can be obtained starting from $x=y$ and
operating a finite amount of admissible substitutions is partition regular,
where by admissible substitution we mean any substitution of a variable $v$ by
either a product $v_{1}v_{2}$ or an exponentiation $v_{1}^{v_{2}}$ with
$v_{1},v_{2}$ not already appearing in the equation.

###### Proof.

For the first claim: take any $p\in\overline{K(\beta\mathbb{N},\odot)}$. As
ultrafilters in $\overline{K(\beta\mathbb{N},\odot)}$ are witnesses for
arbitrary partition regular homogeneous systems, $p\models x+y=z$, henceforth
$2^{p}\models xy=z$ and, by Theorem 5.8, $2^{p}$ also models $a^{b}=c$.
Therefore, by Theorem 2.10 we can impose $z=c$ to get the desired
conclusion888Note that we could also have applied the same method to any
ultrafilter witnessing the partition regularity of $\\{x,y,x^{y},xy\\}$, which
was proved by Sahasrabudhe in [20]. For the general claim: just do the same
proof inductively. ∎

In the statement of Theorem 2.11 we decided to highlight the equation
$xy=a^{b}$ as it resembles an exponential analog of the equation $x+y=ab$,
whose partition regularity has been an open problem for a few years999Its
first solution was given by Bergelson in [1] using the algebra of
$\beta\mathbb{N}$; a very simple proof can be given using Theorem 2.10.
Several generalizations have appeared since..

Theorem 2.11 shows the partition regularity of several other interesting
configurations.

######  Example 2.12.

The following configurations are partition regular and witnessed by any
ultrafilter of the form $n^{p}$ for $p\in\overline{K(\beta\mathbb{N},\odot)}$,
as a consequence of Theorem 2.11:

  * •

$\\{x,y_{1},y_{2},xy_{1}^{y_{2}}\\}$;

  * •

$\left\\{x,y_{1},y_{2},\left(x^{y_{1}}\right)^{y_{2}}\right\\}$;

  * •

$\\{x,y_{1},y_{2},x^{\left(y_{1}^{y_{2}}\right)}\\}$;

  * •

$\left\\{x_{1},\dots,x_{n},x_{i}^{x_{i+1}^{\mathinner{\mkern 1.0mu\raise
1.0pt\vbox{\kern 7.0pt\hbox{.}}\mkern 2.0mu\raise 4.0pt\hbox{.}\mkern
2.0mu\raise 7.0pt\hbox{.}\mkern 1.0mu}^{x_{j}}}}\mid j<i\right\\}$;

  * •

more in general, the same as the previous example but where the iterated
exponentiation are associated in any possible way;

  * •

$\left\\{x_{1,1},\dots,x_{1,n_{1}},\dots,x_{m,1},\dots,x_{m,n_{m}},x_{1,1}^{x_{1,2}^{\mathinner{\mkern
1.0mu\raise 1.0pt\vbox{\kern 7.0pt\hbox{.}}\mkern 2.0mu\raise
4.0pt\hbox{.}\mkern 2.0mu\raise 7.0pt\hbox{.}\mkern
1.0mu}^{x_{1,n_{1}}}}}\cdot\ \dots\ \cdot x_{m,1}^{x_{m,2}^{\mathinner{\mkern
1.0mu\raise 1.0pt\vbox{\kern 7.0pt\hbox{.}}\mkern 2.0mu\raise
4.0pt\hbox{.}\mkern 2.0mu\raise 7.0pt\hbox{.}\mkern
1.0mu}^{x_{m,n_{m}}}}}\right\\}$;

  * •

again, more in general, the same as the previous example but where each
iterated exponentiation is associated in any possible way.

## 3 New mixed exponential structures

It follows from Hindman’s theorem [14] that for every finite coloring of
$\mathbb{N}$, there exists a monochromatic pattern of the form
$A,B,A+B=\\{a+b:a\in A,b\in B\\}$, where $A,B$ can be chosen finite or
infinite sets. It is a routine exercise to check that if we choose $p\in
E\left(K(\beta\mathbb{N},\oplus)\right),$ then every member of $p$ contains
patterns of the form $A+B,$ where $A,B$ are sets containing arithmetic
progressions of a given length. Passing to the map $n\rightarrow 2^{n},$ one
can immediately shows that members of $2^{p}$ contain patterns of the form
$A\cdot B$, where $A,B$ are geometric progressions of a given length. An
immediate question appears: is there an exponential analogue to these results?
This is by no means trivial. To answer this question affirmatively, we will
introduce a new shadow alphabet argument.

Before we proceed let us define the notion of MPSC good and centrally
exponential good family, which will be necessary for our work.

######  Definition 3.1.

Let $\mathcal{F}$ be a family of finite subsets of $\mathbb{N}$.

  * •

We will call it dilation invariant family, that is for each $n\in\mathbb{N}$,
and $F\in\mathcal{F}$, we have $n\cdot F\in\mathcal{F}$.

  * •

A dilation invariant family will be called MPSC good if every multiplicative
central sets contains an element $F\in\mathcal{F}$.

  * •

A family $\mathcal{A}$ is said to be centrally exponential good if there
exists a MPSC good family $\mathcal{F}$ such that
$2^{\mathcal{F}}=\\{2^{F}|F\in\mathcal{F}\\}\subseteq\mathcal{A}$, where
$2^{F}=\\{2^{f}:f\in F\\}.$

The set of all arithmetic progressions (geometric progressions resp.) is a
MPSC good (exponential good resp.) family.

###  Mixed patterns

In this section we will prove that for any finite coloring of $\mathbb{N}$,
there exist two “sufficiently rich” sets 101010for example sets containing
geometric progressions, or sets of the form $FP(\langle
x_{n}\rangle_{n=1}^{r})$ etc. $A,B$ of given length such that
$A,B,A^{B}=\\{a^{b}:a\in A,b\in B\\}$ are monochromatic. To prove our main
theorem 3.3, we need the following lemma.

######  Lemma 3.2.

Let $p$ be any idempotent in $(\beta\mathbb{N},\odot)$, then
$(2^{p})^{p}=2^{p}.$

###### Proof.

As $p$ is an idempotent, we have $A\in 2^{p}=2^{p\odot p}$. Hence
$\\{x:2^{x}\in A\\}\in p\odot p=p.$ In other words $\\{y:\\{z:2^{yz}\in
A\\}\in p\\}\in p.$ That is $\\{y:\\{z:(2^{y})^{z}\in A\\}\in p\\}\in p$,
which implies $\\{m:\\{z:m^{z}\in A\\}\in p\\}\in 2^{p}.$ Now by definition
5.1, $A\in 2^{p\odot p}$, that is $2^{p\odot p}\subseteq(2^{p})^{p}$. Now from
the maximality of ultrafilters we have $(2^{p})^{p}=2^{p}.$ ∎

Now we are in the position to prove our main theorem.

######  Theorem 3.3.

Let $n,M\in\mathbb{N}$, $\mathcal{A}\subset\mathbb{N}^{M}$ be an centrally
exponential good family and $\mathcal{B}\subset\mathbb{N}^{n}$ be a MPSC good
family. Let $p\in E\left(K\left(\beta\mathbb{N},\odot\right)\right)$ and $A\in
2^{p}$. Then there exists $C\in\mathcal{A}$, $D\in\mathcal{B}$ and an element
$a\in\mathbb{N}$ such that

$C\cup\left\\{c^{a},c^{x_{1}},\ldots,c^{x_{n}},(b^{a})^{c^{x_{1}}},(b^{a})^{c^{x_{2}}},\ldots,(b^{a})^{c^{x_{n}}}:x_{i}\in
D,\,\forall i\in\\{1,2,\ldots,n\\}\text{ and }b,c\in C\right\\}\subset A.$

###### Proof.

Let $p\in K\left(\beta\mathbb{N},\odot\right)$ be an idempotent, and $A\in
2^{p}.$ From Lemma 3.2, we have $A\in 2^{p\odot p}.$ Hence
$\\{2^{m}:\\{n:(2^{m})^{n}\in A\\}\in p\\}\in 2^{p}.$ Let
$A^{\times}=A\cap\\{2^{m}:\\{n:(2^{m})^{n}\in A\\}\in p\\}\in 2^{p}.$ As
$\\{n:2^{n}\in A^{\times}\\}\in p$ is a MPSC set, there exists
$C\in\mathcal{A}$ such that $C\subset A^{\times}$. For each $c\in C$, let
$B_{c}=\\{n:c^{n}\in A\\}\in p$. Define $B=\bigcap_{c\in C}B_{c}.$ Then $B\in
p.$ Let $G$ be a finite set such that $\cup_{s\in G}s^{-1}B$ is thick. Let
$|G|=r$ and $N=HJ(r,n\cdot M+1)$ be the Hales-Jewett number coming from
Theorem Hales-Jewett Theorem. As $B\in p,$ from [17, Theorem 4.39], we have
$S=\\{x:x^{-1}B\in p\\}$ is a syndetic set. Now from Lemma 2.2, we have an
$IP_{N}$ set $FS\left(\langle z_{i}\rangle_{i=1}^{N}\right)\subset S.$ Hence
$E=B\cap\bigcap_{t\in FS\left(\langle z_{i}\rangle_{i=1}^{N}\right)}t^{-1}B\in
p.$ As $E\in p$, we have an element $D\in\mathcal{B}$ such that $D\subset E.$
Let $D=\\{y_{1},y_{2},\ldots,y_{n}\\}.$

Now we will introduce the new shadow alphabet argument. The main trick is to
choose a finite number of shadows of an alphabet and consider a new alphabet
set as the union of these shadows.

Let $C=\\{c_{1},c_{2},\ldots,c_{M}\\}$. For each $j\in[1,M],$ let
$\mathbb{A}_{j}=\\{y_{1,j},y_{2,j},\ldots,y_{n,j}\\}$. Define
$\mathbb{A}=\\{0\\}\cup\bigcup_{j=1}^{M}\mathbb{A}_{j}$. Let
$\mathcal{O}:\mathbb{A}\rightarrow\\{0\\}\cup\\{y_{1},y_{2},\ldots,y_{n}\\}$
be a function defined by $\mathcal{O}(y_{i,j})=y_{i}$ if
$y_{i,j}\in\mathbb{A}_{j}$ and $\mathcal{O}(0)=0$. Define
$f:\mathbb{A}\rightarrow C\cup\\{1\\}$ (w.l.g. we will assume $1\notin C$) by
$f(y_{i,j})=c_{j}$ if $y_{i,j}\neq 0$ and $f(0)=1.$ For each
$w=w_{1}w_{2}\cdots w_{N}\in\mathbb{A}^{N}$, let
$e(w)=\prod_{i=1}^{N}f(w_{i})^{z_{i}\cdot\mathcal{O}(w_{i})}$.

Now from the Hales-Jewett theorem, we have $a\in\mathbb{N}$ and $l\in
FS\left(\langle z_{i}\rangle_{i=1}^{N}\right)$ such that $\\{a,ac^{lb}:c\in
C,b\in\\{y_{1},y_{2},\ldots,y_{n}\\}\\}\subset B.$ Again
$\\{lb:b\in\\{y_{1},y_{2},\ldots,y_{n}\\}\\}\subset B$. Let $x_{i}=ly_{i}$ for
each $i\in\\{1,2,\ldots,n\\}.$ As, $\\{a,x_{1},x_{2},\ldots,x_{n}\\}\subset
B,$ we have $\\{c^{x_{i}}:c\in C,i\in\\{1,2,\ldots,n\\}\\}\subset A,$ and
$\\{c^{a}:c\in C\\}\subset A.$ Again
$\\{ac^{x_{i}}:i\in\\{1,2,\ldots,n\\}\\}\subset B$, hence we have
$\\{(b^{a})^{c^{x_{i}}}:i\in\\{1,2,\ldots,n\\}\text{ and }b,c\in C\\}\subset
A.$ This completes the proof. ∎

The following examples are completely new.

######  Example 3.4.

Let $N\in\mathbb{N}$ and $A$ be a set as in theorem 3.3. We can choose
$\mathcal{B}$ to be the set of all A.P. of length $N$. Now choose
$\mathcal{A}$ to be the set of all G.P. of length $N$. Then from theorem 3.3,
we have two sets of G.P. $F,G$ such that

$\\{f,g,g^{f}|f\in F\,,g\in G\\}\subset A.$

######  Example 3.5.

In the above example, we can choose $G$ to be a set of the form
$FP\left(\langle b_{i}\rangle_{i=1}^{N}\right)$ instead of a geometric
progression.

######  Example 3.6.

Let $N\in\mathbb{N}$ and $A$ be a set as in theorem 3.3. Let $n=2^{N}-1$, and
choose $D=\langle x_{m}\rangle_{m=1}^{n}=FS(\langle y_{i}\rangle_{i=1}^{N})$
for some sequence $\langle y_{i}\rangle_{i=1}^{N}$. Let $C$ be a geometric
progression of length $N.$ Now for each $i\in\\{1,2,\ldots,n\\}$
$C^{x_{i}}=\\{c^{x_{i}}:c\in C\\}\subset A$ is a geometric progression of
length $N$. Choose $a\in\mathbb{N}$ as in theorem 3.3. Let $G=C^{a}\subset A.$
For each nonempty $H\subseteq\\{1,2,\ldots,N\\}$, let $F_{H}=C^{\sum_{i\in
H}y_{i}}.$ Clearly $F_{H}\subset A$ and it is a geometric progression. Hence
we can conclude that, for each $N$ and $n=2^{N}-1$ we have geometric
progressions $G,\langle F_{i}\rangle_{i=1}^{n}$ each of length $N$ such that

$\\{f,g,g^{f}:g\in G,f\in\cup_{i=1}^{n}F_{i}\\}\subset A.$

## 4 Infinite exponential configurations

For any sequence $\langle x_{n}\rangle_{n}$, let
$\left\\{x_{i_{n}}^{x_{i_{n-1}}^{\cdot^{\cdot^{\cdot^{x_{i_{1}}}}}}}:i_{1}<\cdots<i_{n-1}<i_{n},n\in\mathbb{N}\right\\}$
be the Hindman tower generated by the sequence $\langle x_{n}\rangle_{n}$. In
[10], Di Nasso and Ragosta proved that for every finite coloring of
$\mathbb{N}$ there must exist a sequence $\langle x_{n}\rangle_{n}$ such that
the Hindman tower generated by $\langle x_{n}\rangle_{n}$ will be in a single
partition. Taking into account the non commutativity of exponentiation, this
can be seen as an exponential analogue of the Hindman theorem. In their
approach, they used a groupoid operation $\star$ over $\beta\mathbb{N}$
defined as follows: first, consider the operation
$f:\mathbb{N}^{2}\rightarrow\mathbb{N}$ defined as $f(n,m)=2^{n}m$. Let
$\overline{f}:\beta\left(\mathbb{N}^{2}\right)\rightarrow\beta\mathbb{N}$ be
its continuous extension. $\star$ is the restriction of $\overline{f}$ to
tensor pairs, namely: for all $p,q\in\beta\mathbb{N}$

$p\star q=\overline{f}(p\otimes q)=2^{p}\odot q.$

We will discuss $\star$ in detail in section 5.3. The following, which is
Theorem 3.1 in [10] is, at the best of our knowledge, the most general result
known for infinitary exponential configurations known in the literature,
namely Theorem 3.1 in [10].

######  Di Nasso-Ragosta.

Let $p\in E\left(K(\beta\mathbb{N},\oplus)\right)$ and $A\in p\star p$. For
every $N\in\mathbb{N}$ and for every sequence $\varPhi=\left(f_{n}:n\geq
2\right)$ of functions $f_{n}:\mathbb{N}\rightarrow\mathbb{N},$ there exists a
sequence $\left(a_{k}\right)_{k\in\mathbb{N}}$ such that

$\mathcal{F}_{k,\varPhi}\left(a_{k}\right)_{k\in\mathbb{N}}=\left\\{a_{k}\cdot
2^{\sum_{i=1}^{k-1}\lambda_{i}a_{i}}:k\in\mathbb{N},0\leq\lambda_{1}\leq
N,\text{ and }0\leq\lambda_{i}\leq f_{i}\left(a_{i-1}\right)\text{ for }2\leq
i\leq k-1\right\\}\subset A.$

Let $N=1$, and choose the function $f_{2}(x)=2^{x}.$ Inductively assume that
we have defined the function $f_{n}(x)$ for some $n\in\mathbb{N}$. Now define
$f_{n+1}(x)=(2^{x})^{f_{n}(x)}$. Now choosing $\varPhi=\left(f_{n}:n\geq
2\right)$, theorem Di Nasso-Ragosta immediately implies the existence of
monochromatic Hindman tower. For details see [10].

This section is aimed to find out the polynomial extension of Theorem Di
Nasso-Ragosta. As a byproduct, this will allow us to prove that a variation of
[20, Question 25] is true.

###  Polynomial extension of Di Nasso-Ragosta Theorem

The polynomial extension of the van der Waerden’s theorem was proved in [5].
To proof the polynomial extension of Di Nasso-Ragosta Theorem we will use the
recurrence property of the polynomial van der Waerden’s theorem. Our version
of the polynomial van der Waerden’s theorem is a direct consequence of the
polynomial Hales-Jewett theorem, proved in [6].

######  $IP_{r}^{\star}$ polynomial van der Waerden’s theorem.

Let $P$ be a finite collection of polynomials with integer coefficient, and
having no constant term, and let $A$ be an additively piecewise syndetic set.
Then

$\left\\{d\in\mathbb{N}\mid\text{ there exists }a\in\mathbb{N}\text{ such that
}\left\\{a,a+p(d):p\in P\right\\}\subset A\right\\}$

is an $IP_{r}^{\star}$ set for some $r\in\mathbb{N}$.

For any set $X,$ let $\mathcal{P}_{f}(X)$ be the collection of all nonempty
finite subsets of $X.$ Let $\mathbb{P}$ be the collection of all polynomials
with integer coefficients with no constant term. For every $P\in\mathbb{P},$
$deg(P)$ is the degree of the polynomial $P,$ and $coef(P)=\max\\{|c|:c\text{
is a coefficient of }P\\}.$ Define a new operation $\star_{n}$ on $\mathbb{N}$
by $a\star_{n}b=n^{a}b.$ The above result is all we need to prove the
polynomial version of Di Nasso-Ragosta Theorem.

######  Theorem 4.1.

Let $n\in\mathbb{N}$, $p\in E\left(K(\beta\mathbb{N},\oplus)\right)$ and
$q\in\beta\mathbb{N}$ be such that for every $N\in\mathbb{N}$, each element of
$q$ contains an $IP_{N}$ sets. Let $n\in\mathbb{N}_{>1},$ and $A\in
p\star_{n}q$ and $F_{1}\in\mathcal{P}_{f}(\mathbb{P})$. Let
$\varPhi=\left(f_{n}:n\geq 2\right)$ be a sequence of functions
$f_{n}:\mathbb{N}\rightarrow\mathbb{N}.$ For every $n(>1),x\in\mathbb{N}$, let
$F_{n,x}=\\{P\in\mathbb{P}:deg(P)\leq f_{n}(x)\text{ and }coef(P)\leq
f_{n}(x)\\}.$

Under this hypothesis we have a sequence $\left(x_{k}\right)_{k\in\mathbb{N}}$
such that

$\mathcal{PF}^{n}_{k,\varPhi}\left(x_{k}\right)_{k\in\mathbb{N}}=\left\\{x_{k}\cdot
n^{\sum_{i=1}^{k-1}p_{i}\left(x_{i}\right)}:k\in\mathbb{N},p_{1}\in
F_{1},\text{ and }p_{i}\in F_{i,f_{i}\left(x_{i-1}\right)}\text{ for }2\leq
i\leq k-1\right\\}\subset A.$

###### Proof.

To avoid the complexity in the calculation we will prove this theorem up to
$k=3.$

The rest of the part can be proved iteratively or inductively. The technique
is verbatim.

For any $p\in E\left(\beta\mathbb{N},\oplus\right)$ and any $A\in p,$ let
$A^{\star}=\\{n\in A:-n+A\in p\\}.$ As $p$ is an idempotent, we have
$A^{\star}\in p$.

Let $p\in E\left(K(\beta\mathbb{N},\oplus)\right)$ and $A\in
p\star_{n}q=E_{1}(n,p)\odot q.$ Then
$B=\left\\{m_{0}:\left\\{n_{0}:n^{m_{0}}n_{0}\in A\right\\}\in q\right\\}\in
p.$ Let $m_{1}\in B^{\star}$ and let
$C_{1}=\left\\{n^{\prime}:n^{m_{1}}n^{\prime}\in A\right\\}\in q.$ Let $F_{1}$
be the finite collection of polynomials. Then from theorem $IP_{r}^{\star}$
polynomial van der Waerden’s theorem

$D_{1}=\left\\{d:\text{ there exists }a_{1}\text{ such that
}\left\\{a_{1},a_{1}+p_{1}\left(d\right):p_{1}\in F_{1}\right\\}\subset
B^{\star}\right\\}$

is an $IP_{r}^{\ast}$ set for some $r\in\mathbb{N}$. So
$\frac{1}{n^{m_{1}}}D_{1}\cap D_{1}$ is an $IP_{s}^{\star}$ set for some
$s\in\mathbb{N}$. Hence there exists an element
$n_{1}\in\frac{1}{n^{m_{1}}}D_{1}\cap D_{1}\cap C_{1}\in q.$ Let
$x_{1}=n^{m_{1}}n_{1}\in A$ and again $x_{1}\in D.$ So there exists $a_{1}\in
B^{\star}$ such that $\left\\{a_{1},a_{1}+p_{1}\left(x_{1}\right):p_{1}\in
F_{1}\right\\}\subset B^{\star}.$

So,

$B_{2}=B^{\star}\cap\left(-a_{1}+B^{\star}\right)\cap\bigcap_{p_{1}\in
F_{1}}-\left(a_{1}+p_{1}\left(x_{1}\right)\right)+B^{\star}\in p.$

Let

$C_{2}=C_{1}\cap\bigcap_{p_{1}\in
F_{1}}\left\\{n^{\prime}:n^{a_{1}+p_{1}\left(x_{1}\right)}n^{\prime}\in
A\right\\}\cap\left\\{n^{\prime}:n^{a_{1}}n^{\prime}\in A\right\\}\in q$

and

$D_{2}=\left\\{d:\text{ there exists }a_{2}\text{ such that
}\left\\{a_{2},a_{2}+p_{2}\left(d\right):p_{2}\in
F_{2,f_{2}\left(x_{1}\right)}\right\\}\subset B_{2}^{\star}\right\\}$

is an $IP_{r}^{\star}$ set for some $r\in\mathbb{N}$.

Choose $n_{2}\in\frac{1}{n^{a_{1}}}D_{2}\cap D_{2}\cap D_{1}\cap C_{2}\in q.$
Let $n^{a_{1}}n_{2}=x_{2}\in A$ and so we have

$n^{a_{1}+p_{1}\left(x_{1}\right)}n_{2}=x_{2}n^{p_{1}\left(x_{1}\right)}\in A$

for all $p_{1}\in F_{1}.$

Now from the choice of $x_{2},$ we have $a_{2}\in B_{2}^{\star}$ such that
$\left\\{a_{2},a_{2}+p_{2}\left(d\right):p_{2}\in
F_{2,f_{2}\left(x_{1}\right)}\right\\}\subset B_{2}^{\star}.$ Letting
$c=a_{1}+a_{2}$, we have

$\left\\{c+p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right):p_{1}\in
F_{1},p_{2}\in F_{2,f_{2}\left(x_{1}\right)}\right\\}\subset B^{\star}.$

Let

$C_{3}=C_{1}\cap\bigcap_{p_{1}\in F_{1}}\bigcap_{p_{2}\in
F_{2,f_{2}\left(x_{1}\right)}}\left\\{n^{\prime}:n^{c+p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right)}n^{\prime}\in
A\right\\}\in q$

and

$B_{3}^{\star}=B^{\star}\cap\bigcap_{p_{1}\in F_{1}}\bigcap_{p_{2}\in
F_{2,f_{2}\left(x_{1}\right)}}-(c+p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right))+B^{\star}\in
p.$

As

$D_{3}=\left\\{d:\text{ there exists }a_{3}\text{ such that
}\left\\{a,a+p_{3}\left(d\right):p_{3}\in
F_{3,f_{3}\left(x_{2}\right)}\right\\}\subset B_{3}^{\star}\right\\}$

is an $IP_{r}^{\star}$ set for some $r\in\mathbb{N}$, we have

$n_{3}\in D_{3}\cap\bigcap_{p_{1}\in F_{1}}\bigcap_{p_{2}\in
F_{2,f_{2}\left(x_{1}\right)}}\frac{1}{n^{c+p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right)}}D_{3}\cap
C_{3}\in q.$

Let $n^{c}n_{3}=x_{3}\in A$ and

$n^{c+p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right)}n_{3}=x_{3}n^{p_{1}\left(x_{1}\right)+p_{2}\left(x_{2}\right)}\in
A$

for all $p_{1}\in F_{1},\text{ and }p_{2}\in F_{2,f_{2}\left(x_{1}\right)}.$

Now iterating this argument we have an infinite sequence $\langle
x_{n}\rangle_{n\in\mathbb{N}}$ such that the desired result is true. ∎

In [20, Question 25], Sahasrabudhe asked if $\\{a,b,a^{b},a^{b^{b}}\\}$ is PR
or not? As a corollary of theorem 4.1, we will show that a variation of this
question has an affirmative answer.

######  Corollary 4.2.

The family $\\{a,b,a^{b},a^{b^{log_{2}b}}:a,b\in\mathbb{N}\\}$ is PR.

###### Proof.

Let $p_{1},p_{2}:\mathbb{N}\rightarrow\mathbb{N}$ be defined by $p_{1}(n)=n$
and $p_{2}(n)=n^{2}.$ Now from theorem 4.1 we have $a_{1},a_{2}\in\mathbb{N}$
such that $\\{a_{1},a_{2},a_{2}2^{a_{1}},a_{2}2^{a_{1}^{2}}\\}$ is
monochromatic. Let $b=2^{a_{1}}$ and $a=2^{a_{2}}.$ Now
$(2^{a_{2}})^{(2^{a_{1}})}=a^{b}$ and
$(2^{a_{2}})^{(2^{a_{1}^{2}})}=a^{(2^{a_{1}})^{a_{1}}}=a^{b^{log_{2}b}}.$ This
completes the proof. ∎

###  Combined additive, exponential and multiplicative structure

In [3, Theorem 2.4.], Bergelson and Hindman proved the following combined
extension of the additive and multiplicative Hindman theorem.

######  Theorem 4.3.

If $\mathbb{N}$ is finitely colored, then there exist two sequences $\langle
x_{n}\rangle_{n}$ and $\langle y_{n}\rangle_{n}$ such that $FS(\langle
x_{n}\rangle_{n})\cup FP(\langle y_{n}\rangle_{n})$ is in the same color.

From Theorem 4.1 the immediate question that arises is if it would be possible
to extend Theorem 4.3 so to have in the same color also another sequence
$\langle z_{n}\rangle_{n}$ such that
$\mathcal{F}_{n,\varPhi}\left(z_{n}\right)_{n\in\mathbb{N}}$. As we are going
to show, this is actually the case. Our proof is based on the observation that
Theorem 4.1 could be proved for multiples of ultrafilters of the form
$p\star_{n}p$, if $p$ is chosen properly.

Let $\mathcal{DR}$ be the collection of those ultrafilters $p$ such that each
member of $p$ satisfies the conclusion of theorem 4.1, i.e.
$\mathcal{DR}=\\{p:p\text{ satisfies conclusion of theorem }\ref{poly}\\}.$

######  Lemma 4.4.

Let $p\in\mathcal{DR}$. Then for every $q\in\beta\mathbb{N}$, we have $q\odot
p\in\mathcal{DR}.$

###### Proof.

If $A\in q\odot p$, then
$B=\\{m^{\prime}:\\{n^{\prime}:m^{\prime}n^{\prime}\in A\\}\in p\\}\in q.$ Let
$a\in B$, and so $a^{-1}A\in p.$ Let $F_{1}$ be a finite set of polynomials
with no constant term and $\varPhi=\left(f_{n}:n\geq 2\right)$ be a given
sequence of functions. Let $F_{n,x}$ be a finite collection of polynomials
with no constant term and having the degree and the absolute value of each of
the coefficients is less or equal to $f_{n}\left(x\right).$

Let $G_{1}\in\mathcal{P}_{f}(\mathbb{P})$ defined as follows:

$g\in G_{1}\Leftrightarrow\exists f\in F_{1}\ \forall x\in\mathbb{N}\
g(x)=f(ax).$

Define a new sequence of functions $\varPhi^{\star}=\left(g_{m}:m\geq
2\right)$ by $g_{m}(x)=f_{m}(ax)$ for all $x\in\mathbb{N}$. As $a^{-1}A\in p,$
we have a sequence $\langle x_{n}\rangle_{n=1}^{\infty}$, such that

$\mathcal{PF}^{n}_{k,\varPhi^{\star}}\left(x_{k}\right)_{k\in\mathbb{N}}=\left\\{x_{k}\cdot
n^{\sum_{i=1}^{k-1}q_{i}\left(x_{i}\right)}:k\in\mathbb{N},q_{1}\in
G_{1},\text{ and }q_{i}\in G_{i,g_{i}\left(x_{i-1}\right)}\text{ for }2\leq
i\leq k-1\right\\}\subset a^{-1}A.$

In other words,

$\mathcal{PF}^{n}_{k,\varPhi}\left((ax)_{k}\right)_{k\in\mathbb{N}}=\left\\{ax_{k}\cdot
n^{\sum_{i=1}^{k-1}p_{i}\left(ax_{i}\right)}:k\in\mathbb{N},p_{1}\in
F_{1},\text{ and }p_{i}\in F_{i,f_{i}\left(x_{i-1}\right)}\text{ for }2\leq
i\leq k-1\right\\}\subset A.$

This completes the proof. ∎

Now we are in the position to extend Theorem 4.3 to include exponential
configurations.

######  Theorem 4.5.

Let $n\in\mathbb{N},$ $F_{1}\in\mathcal{P}_{f}(\mathbb{P})$. For every
sequence $\varPhi=\left(f_{n}:n\geq 2\right)$ of functions
$f_{n}:\mathbb{N}\rightarrow\mathbb{N},$ let $F_{n,x}$ be a finite collection
of polynomials defined as in Theorem 4.1. Then for every finite partition of
$\mathbb{N}$, there exist three sequences $\langle
x_{k}\rangle_{k=1}^{\infty},\langle y_{k}\rangle_{k=1}^{\infty},\langle
z_{k}\rangle_{k=1}^{\infty}$ such that

$\mathcal{PF}^{n}_{k,\varPhi}\left(x_{k}\right)_{k=1}^{\infty}\cup FS(\langle
y_{k}\rangle_{k=1}^{\infty})\cup FP(\langle z_{k}\rangle_{k=1}^{\infty})$

is in the same partition.

###### Proof.

Let $p\in E\left(K(\beta\mathbb{N},\oplus)\right)$ and
$L=\beta\mathbb{N}\odot(p\star_{n}p).$ Now $L$ is a left ideal of
$(\beta\mathbb{N},\odot).$ So it contains an idempotent say
$q\odot(p\star_{n}p).$ Let $A\in q\odot(p\star_{n}p)=(q\odot n^{p})\odot p.$
Now $q\odot(p\star_{n}p)$ is an idempotent, hence there exists a sequence
$\langle z_{k}\rangle_{k=1}^{\infty}$ such that $FP(\langle
z_{k}\rangle_{k=1}^{\infty})\subset A.$ As $p\in
E\left(K(\beta\mathbb{N},\oplus)\right)\subset\overline{E\left(K(\beta\mathbb{N},\oplus)\right)}$,
and $\overline{E\left(K(\beta\mathbb{N},\oplus)\right)}$ is a left ideal of
$(\beta\mathbb{N},\odot)$, we have $(q\odot n^{p})\odot
p\in\overline{E\left(K(\beta\mathbb{N},\oplus)\right)}$. So $A$ contains
$FS(\langle y_{k}\rangle_{k=1}^{\infty})$ for some sequence $\langle
y_{k}\rangle_{k=1}^{\infty}.$ And from lemma 4.4, we have a sequence $\langle
x_{n}\rangle_{n=1}^{\infty}$ such that
$\mathcal{PF}^{n}_{k,\varPhi}\left(x_{k}\right)_{k=1}^{\infty}\subset A.$

This completes the proof. ∎

## 5 Exponentiations of ultrafilters

Ultrafilters and partition regularity are closely related, as proven by an
enormous literature on the subject (instead of citing here hundreds of papers,
we refer to [17] and its bibliography). When it comes to exponential
structures, ultrafilters have been used already by Sisto in [23], although in
an implicit way. In [9], Di Nasso and Ragosta managed to produce explicit
witnesses of the partition regularity of exponential triples and related
structure, by considering the operation $\star$ that maps a pair
$(p,q)\in\left(\beta\mathbb{N}\right)^{2}$ to $2^{p}\odot q$. In this section,
we introduce two direct notions of exponentiation of ultrafilters and we show
their basic properties and their relationship with exponential triples. Then,
we consider the problem of the existence of idempotents with respect to these
notions and for $\star$, proving some partial negative result.

###  Two extensions of exponentiation to ultrafilters

As it is well known, there are two natural ways to extend the sum from
$\mathbb{N}$ to $\beta\mathbb{N}$: the nowadays usual one $(p,q)\rightarrow
p\oplus q$ defined by letting, for every $A\in\mathbb{N}$,

$A\in p\oplus q\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid
n+m\in A\\}\in q\\}\in p,$

and the reverse sum $(p,q)\rightarrow q\oplus p$. Whilst different, these two
extensions have the same properties, modulo properly translating their meaning
by switching ”left” with ”right” notions. This is a consequence of the sum
being commutative on $\mathbb{N}$.

Exponentiation, on the other hand, is not commutative. Hence, when extending
it to $\beta\mathbb{N}$, we have at least two natural and truly distinct
possible definitions to consider111111Not including those that could be
obtained trivially by reversing the order of the variables in the operations;
of course, this discussion could be made from the general point of view of
extending operations from $\mathbb{N}^{2}$ to $\beta\mathbb{N}^{2}$, but we
prefer to give explicit set-theoretical definitions here..

######  Definition 5.1.

Given $p,q\in\beta\mathbb{N}$ we let

$E_{1}(p,q):=\\{A\subseteq\mathbb{N}\mid\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid
n^{m}\in q\\}\in p\\},$

and

$E_{2}(p,q):=\\{A\subseteq\mathbb{N}\mid\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid
m^{n}\in q\\}\in p\\}.$

Equivalently, $E_{1}$ can be identified with the restriction to tensor pairs
of the extension
$\overline{e_{1}}:\beta\left(\mathbb{N}^{2}\right)\rightarrow\beta\mathbb{N}$
of the function $e_{1}:\mathbb{N}^{2}\rightarrow\mathbb{N}$ that maps $(n,m)$
to $n^{m}$. Analogously, $E_{2}$ can be identified with the restriction to
tensor pairs of the extension
$\overline{e_{2}}:\beta\left(\mathbb{N}^{2}\right)\rightarrow\beta\mathbb{N}$
of the function $e_{2}:\mathbb{N}^{2}\rightarrow\mathbb{N}$ that maps $(n,m)$
to $m^{n}$.

As usual, to simplify notations we identify each natural number
$n\in\mathbb{N}$ with the principal idempotent it generates, so that
$\mathbb{N}\subseteq\beta\mathbb{N}$. For $n,m\in\mathbb{N}$, trivially we
have that $E_{1}(n,m)=E_{2}(m,n);$ moreover, for
$n\in\mathbb{N},p\in\beta\mathbb{N}$ we have that $E_{1}(n,p)=E_{2}(p,n)$ and
$E_{1}(p,n)=E_{2}(n,p)$ as, for all $A\subseteq\mathbb{N}$

$\displaystyle A\in E_{1}(n,p)\Leftrightarrow\\{m\in\mathbb{N}\mid n^{m}\in
A\\}\in p\Leftrightarrow A\in E_{2}(p,n),$ $\displaystyle A\in
E_{1}(p,n)\Leftrightarrow\\{m\in\mathbb{N}\mid m^{n}\in A\\}\in
p\Leftrightarrow A\in E_{2}(n,p).$

Notation: we will continue to use the simplified notation $n^{p}$ to denote
any of $E_{1}(n,p),E_{2}(p,n)$, and we will use the simplified notation
$p^{n}$ to denote any of $E_{1}(p,n),E_{2}(n,p)$.

At first sight, it might look like the equality $E_{1}(p,q)=E_{2}(q,p)$ could
be true for all $p,q\in\beta\mathbb{N}$; however, this is not the case. In
fact, for example, let both $p,q$ contain the set $P$ of primes. Then

$P_{1}=\\{a^{b}\mid a,b\in P,a<b\\}\in E_{1}(p,q),P_{2}=\\{a^{b}\mid a,b\in
P,a>b\\}\in E_{2}(p,q)$

but $P_{1}\cap P_{2}=\emptyset$, so $E_{1}(p,q)\neq E_{2}(q,p)$. We will
provide some examples of nonprincipal ultrafilters $p,q$ satisfying
$E_{1}(p,q)=E_{2}(q,p)$ after Proposition 5.2.

$E_{1}$ and $E_{2}$ allow to extend, in a proper way, several algebraical
properties of exponentiation from natural numbers to ultrafilters. We list
here those that we will need in the rest of the paper. To avoid cumbersome
reformulations, some of the properties proven in the following proposition are
stated for rational numbers and ultrafilters on the rationals, to which the
exponentiation functions $E_{1},E_{2}$ can be extended trivially.

######  Proposition 5.2.

Let
$p,q,r\in\beta\mathbb{N}\setminus\\{0,1\\},n,m\in\mathbb{N}\setminus\\{1\\},a\in\mathbb{Q}\setminus\\{0\\}$.
The following properties hold:

  1. 1.

$E_{1}\left(p^{a},q\right)=E_{1}\left(p,aq\right)$ and
$E_{2}\left(q,p^{a}\right)=E_{2}\left(aq,p\right)$;

  2. 2.

$n^{p}=n^{q}\Leftrightarrow p=q$ and $p^{n}=q^{n}\Leftrightarrow p=q$;

  3. 3.

$n^{p}=m^{p}\Leftrightarrow n=m$ and $p^{n}=p^{m}\Leftrightarrow n=m$;

  4. 4.

$n^{p}\odot n^{q}=n^{p\oplus q}$, but $E_{1}(r,p)\odot E_{1}(r,q)\neq
E_{1}(r,p\oplus q)$ and $E_{2}(p,r)\odot E_{2}(q,r)\neq E_{2}(p\oplus q,r)$ in
general;

  5. 5.

$E_{1}\left(p,q\odot r\right)=E_{1}\left(E_{1}(p,q),r\right)$; in particular,
if $q$ is multiplicatively idempotent,
$E_{1}(p,q)=E_{1}\left(E_{1}(p,q),q\right)$;

  6. 6.

$E_{2}\left(p,E_{2}(q,r)\right)=E_{2}\left(p\odot q,r\right)$; in particular,
if $p$ is multiplicatively idempotent,
$E_{2}\left(p,E_{2}(p,q)\right)=E_{2}(p,q)$;

###### Proof.

All these proofs consist, basically, of straightforward checks on the
definitions.

$(1)$ Let $A\subseteq\mathbb{N}$. Then

$\displaystyle A\in
E_{1}\left(p^{a},q\right)\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid
n^{m}\in A\\}\in q\\}\in p^{a}\Leftrightarrow$
$\displaystyle\\{n\in\mathbb{N}\mid
n^{a}\in\\{l\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid l^{m}\in A\\}\in q\\}\\}\in
p\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid\left(n^{a}\right)^{m}\in
A\\}\in q\\}\in p\Leftrightarrow$
$\displaystyle\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid\left(n^{am}\right)\in
A\\}\in q\\}\in
p\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid\left(n^{m}\right)\in
A\\}\in aq\\}\in p\Leftrightarrow$ $\displaystyle A\in
E_{1}\left(p,aq\right).$

The $E_{2}$-case is analogous.

$(2)$ If $p=q$, both equivalences are trivial.

If $n^{p}=n^{q}$ it means that for all $A\subseteq\mathbb{N}$, if
$B=\\{a\in\mathbb{N}\mid n^{a}\in A\\}$ then $B\in p\Leftrightarrow B\in q$,
which shows that if we let $P_{n}=\\{a\in\mathbb{N}\mid\exists b\in\mathbb{N}\
a=b^{n}\\}$ and $log_{n}:P_{n}\rightarrow\mathbb{N}$ be the $n$-base logarithm
and $\overline{log_{n}}$ be its extension to ultrafilters containing $P_{n}$
then $p=\log_{n}\left(n^{p}\right)=\log_{n}\left(n^{q}\right)=q$.

The other equivalence can be proven similarly.

$(3)$ If $n=m$, both equivalences are trivial.

If $n^{p}=m^{p}$, first we observe that $n^{\mathbb{N}}=\\{n^{k}\mid
k\in\mathbb{N}\\}\in n^{p}$ and $m^{\mathbb{N}}=\\{m^{k}\mid
k\in\mathbb{N}\\}\in m^{p}$. Therefore, if $n^{p}=m^{p}$ we have that
$n^{\mathbb{N}}\cap m^{\mathbb{N}}\in n^{p}$. Hence $n^{\mathbb{N}}\cap
m^{\mathbb{N}}\neq\emptyset$, which forces $n^{a}=m$ for some
$a\in\mathbb{Q}$, so $n^{p}=n^{ap}$ from $(1)$, which from $(2)$ forces
$p=ap$, which forces $a=1$, so $n=m$.

The other equivalence is similar.

$(4)$ First, let us show that $n^{p}\odot n^{q}=n^{p\oplus q}$: let
$A\subseteq\mathbb{N}$. Then

$\displaystyle A\in n^{p}\odot
n^{q}\Leftrightarrow\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid ab\in A\\}\in
n^{q}\\}\in n^{p}\Leftrightarrow$ $\displaystyle\\{d\in\mathbb{N}\mid
n^{d}\in\\{a\in\mathbb{N}\mid\\{c\in\mathbb{N}\mid
n^{c}\in\\{b\in\mathbb{N}\mid ab\in A\\}\in q\\}\\}\\}\in p\Leftrightarrow$
$\displaystyle\\{d\in\mathbb{N}\mid\\{c\in\mathbb{N}\mid n^{c}\cdot n^{d}\in
A\\}\in q\\}\in p\Leftrightarrow\\{d\in\mathbb{N}\mid\\{c\in\mathbb{N}\mid
n^{c+d}\in A\\}\in q\\}\in p\Leftrightarrow$
$\displaystyle\\{e\in\mathbb{N}\mid n^{e}\in A\\}\in p\oplus q\Leftrightarrow
A\in n^{p\oplus q}.$

For the other claims: let $P=\\{n\in\mathbb{N}\mid n\ \text{is a prime}\\}$
and let $P\in r$. Let $p,q$ be nonprincipal. Then a direct computation shows
that

$I_{1}=\\{n\in\mathbb{N}\mid\exists p_{1},p_{2}\in P,a,b\in\mathbb{N}\
\text{s.t.}\ p_{1}<a<p_{2}<b\ \text{and}\ p_{1}^{a}p_{2}^{b}=n\\}\in
E_{1}(r,p)\odot E_{1}(r,q),$ $I_{2}=\\{n\in\mathbb{N}\mid\exists
p_{1},p_{2}\in P,a,b\in\mathbb{N}\ \text{s.t.}\ p_{1}>a>p_{2}>b\ \text{and}\
p_{1}^{a}p_{2}^{b}=n\\}\in E_{2}(p,r)\odot E_{2}(q,r),$

whilst

$J=\\{n\in\mathbb{N}\mid\exists p\in P,a\in\mathbb{N}\ \text{s.t.}\
p^{a}=n\\}\in E_{1}(r,p\oplus q)\cap E_{2}(p\oplus q,r),$

and we conclude as $I_{1}\cap J=I_{2}\cap J=\emptyset.$

$(5)$ Let $A\subseteq\mathbb{N}$. Then

$\displaystyle A\in E_{1}(p,q\odot
r)\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid n^{m}\in A\\}\in
q\odot r\\}\in p\Leftrightarrow$
$\displaystyle\\{n\in\mathbb{N}\mid\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid
n^{ab}\in A\\}\in r\\}\in q\\}\in
p\Leftrightarrow\\{n\in\mathbb{N}\mid\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid\left(n^{a}\right)^{b}\in
A\\}\in r\\}\in q\\}\in p\Leftrightarrow$
$\displaystyle\\{n\in\mathbb{N}\mid\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid\left(n^{a}\right)^{b}\in
A\\}\in r\\}\in q\\}\in
p\Leftrightarrow\\{c\in\mathbb{N}\mid\\{d\in\mathbb{N}\mid c^{d}\in A\\}\in
r\\}\in E_{1}\left(p,q\right)\Leftrightarrow$ $\displaystyle A\in
E_{1}\left(E_{1}\left(p,q\right),r\right).$

$(6)$ This is analogous to $(5)$. Let $A\subseteq\mathbb{N}$. Then

$\displaystyle A\in
E_{2}\left(p,E_{2}(q,r)\right)\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid
m^{n}\in A\\}\in E_{2}(q,r)\\}\in p\Leftrightarrow$
$\displaystyle\\{n\in\mathbb{N}\mid\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid\left(b^{a}\right)^{n}\in
A\\}\in r\\}\in q\\}\in
p\Leftrightarrow\\{n\in\mathbb{N}\mid\\{a\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid
b^{an}\in A\\}\in r\\}\in q\\}\in p\Leftrightarrow$
$\displaystyle\\{c\in\mathbb{N}\mid\\{b\in\mathbb{N}\mid b^{c}\in A\\}\in
r\\}\in p\odot q.\qed$

######  Example 5.3.

Let $n\in\mathbb{N},n>1$ and $p,q\in\beta\mathbb{N}\setminus\\{0\\}$ be such
that $p\odot q=q\odot p$. Then by Proposition 5.2.we have that

$E_{1}\left(n^{p},q\right)=n^{p\odot q}=n^{q\odot
p}=E_{2}\left(q,n^{p}\right).$

In particular, this shows the existence of infinitely many pairs of
nonprincipal ultrafilters solving $E_{1}(p,q)=E_{2}(q,p)$ (just take, e.g.,
$p=2^{q}$).

Their definition should make evident that $E_{1},E_{2}$ are ultrafilters
operations with relationships with exponential configurations.

######  Example 5.4.

Let $p\in\beta\mathbb{N}\setminus\mathbb{N}$ and let $q=E_{1}(p,p)$. Let $A\in
q$. Then

$B_{0}=\\{m\in\mathbb{N}\mid\\{n\in\mathbb{N}\mid\\{m^{n}\in A\\}\in p\\}\in
p.$

Now take $F_{1}\in\mathcal{P}_{f}(B_{0})$ and let

$B_{1}=\bigcap_{t\in F_{1}}\\{n\in\mathbb{N}\mid t^{n}\in A\\}\cap B_{0}\in
p.$

Proceeding inductively like above, we construct a sequence $\langle
F_{i}\rangle_{i=1}^{\infty}$ of finite subsets of $B_{0}$ with the property
that for all $j<i$ one has that

$F_{j}^{F_{i}}=\\{m^{n}\mid m\in F_{j},n\in F_{i}\\}\in A.$

For example, if $F_{i}=\\{a_{i}\\}$ for all $i$, we obtain an infinite subset
$X=\\{a_{i}\mid i\in\mathbb{N}\\}$ of $B_{0}$ such that $\\{a_{j}^{a_{i}}\mid
a_{i},a_{j}\in X,j<i\\}\subseteq A$.

Moreover, by choosing $p$ with additional properties, one could get more
complicated configurations. For example, by taking $p$ such that each set in
$p$ contains arbitrarily long arithmetic progressions, we can take all sets
$F_{i}$ to be arithmetic progressions of length $i$ in $B$.

######  Example 5.5.

If we argue as above with $E_{2}$ instead of $E_{1}$, we get analogous results
with the order relationship between $i$ and $j$ switched. So, in the same
notations, we get a sequence $\langle F_{i}\rangle_{i=1}^{\infty}$ of finite
subsets of $B_{0}$ with the property that for all $j<i$ one has that

$F_{j}^{F_{i}}=\\{m^{n}\mid m\in F_{j},n\in F_{i}\\}\in A.$

For example, if $F_{i}=\\{a_{i}\\}$ for all $i$, we obtain an infinite subset
$X=\\{a_{i}\mid i\in\mathbb{N}\\}$ of $B_{0}$ such that $\\{a_{j}^{a_{i}}\mid
a_{i},a_{j}\in X,j>i\\}\subseteq A$.

The reader with a basic knowledge of ultrafilters will recognize that a
trivial way to strengthen the above examples would be to take $p$ such that
$E_{1}(p,p)=p$ or $E_{2}(p,p)=p$, namely some kind of exponential idempotent.
We devote Section 5.2 to the study of the existence of such ultrafilters.

To conclude this Section, we want to show that the operations $E_{1},E_{2}$
are related to exponential triples. This should not come as a surprise as,
even if using a different language, Di Nasso and Ragosta in [9] to prove the
exponential Schur theorem employed ultrafilters $u$ of the form $u=2^{q}$
where $q$ had the form $q=2^{p}\odot p$ for some $p$ witness of van der
Waerden Theorem. This can be generalized. Let us first fix a definition.

######  Definition 5.6.

We let

$EVdW=\\{p\in\beta\mathbb{N}\mid\forall A\in p,\forall
k\in\mathbb{N}\,\exists\,x,y\in A\ \text{s.t.}\
x^{y},\dots,x^{\left(y^{k}\right)}\in A\\}.$

It is well known (see e.g. [17, Theorem 4.40]) that MPS sets and
$\overline{K\left(\beta\mathbb{N},\odot\right)}$ are strictly related: every
set belonging to an ultrafilter in
$\overline{K\left(\beta\mathbb{N},\odot\right)}$ is MPS, and for any MPS set
$A$ there exists $p\in\overline{K\left(\beta\mathbb{N},\odot\right)}$ with
$A\in p$. Hence, the discussion in Example 2.4 can be rephrased as follows:

######  Theorem 5.7.

For all $\mathcal{U}\in\overline{K(\beta\mathbb{N},\odot)}$, for all
$n\in\mathbb{N}_{>1}$, $n^{\mathcal{U}}\in EVdW$.

Actually, Theorem 5.7 can be seen as a very particular case of a much more
general phenomenon involving $E_{1},E_{2}$.

######  Theorem 5.8.

Let $p,q\in\beta\mathbb{N}\setminus\\{0\\}$. Then:

  1. 1.

If $q\in\overline{K(\beta\mathbb{N},\odot)}$ then $E_{1}\left(p,q\right)\in
EVdW$. In particular, $E_{1}\left(p,q\right)$ witnesses the partition
regularity of exponential triples;

  2. 2.

If $q\in EVdW$ then $E_{2}(p,q)\in EVdW$. In particular, $E_{2}(p,q)$
witnesses the partition regularity of exponential triples.

###### Proof.

(1) Let $k\in\mathbb{N}$ and let $A\in E_{1}(p,q)$. We will show that $A$
contains a configuration of the form $x,y,x^{y},\dots,x^{\left(y^{k}\right)}$.

By definition, $I=\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid n^{m}\in A\\}\in
q\\}\in p$. Let $n\in I$ and $J_{n}=\\{m\in\mathbb{N}\mid n^{m}\in A\\}$. Take
$a,b\in J_{n}$ with $a,b,an^{b},\dots,an^{kb}\in J_{n}$, whose existence is
ensured by Theorem 2.3 since, as $J_{n}\in
q\in\overline{K(\beta\mathbb{N},\odot)}$, $J_{n}$ is MPS. Therefore,

$n^{a},n^{b},n^{an^{b}}=\left(n^{a}\right)^{\left(n^{b}\right)},\dots,n^{an^{kb}}=\left(n^{a}\right)^{\left(\left(n^{b}\right)^{k}\right)}\in
A,$

and we conclude by letting $x=n^{a},y=n^{b}$.

(2) Let $q\in EVdW$, let $k\in\mathbb{N}$ be given and let $A\in E_{2}(p,q)$.
By definition, $I=\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid m^{n}\in A\\}\in
q\\}\in p$. Let $n\in I$ and let $J_{n}=\\{m\in\mathbb{N}\mid m^{n}\in A\\}$.
Take $a,b\in J_{n}$ with
$a,b,a^{\left(b^{n}\right)},\dots,a^{\left(b^{kn}\right)}\in J_{n}$. Hence

$a^{n},b^{n},\left(a^{\left(b^{n}\right)}\right)^{n}=a^{nb^{n}}=\left(a^{n}\right)^{\left(b^{n}\right)},\dots,\left(a^{\left(b^{kn}\right)}\right)^{n}=a^{nb^{kn}}=\left(a^{n}\right)^{\left(\left(b^{n}\right)^{k}\right)}\in
A,$

and we conclude by letting $x=a^{n},y=b^{n}$.∎

As a consequence of [20, Theorem 6], if
$Sc_{+}=\\{p\in\beta\mathbb{N}\mid\forall A\in p\ \exists a,b\in A\ a+b\in
A\\}$, then $Sc_{+}\cap EVdW=\emptyset$; in particular,

$\overline{K(\beta\mathbb{N},\odot)}\cap
EVdW=\emptyset,\overline{E(\oplus)}\cap EVdW=\emptyset,$

as $\overline{K(\beta\mathbb{N},\odot)}\cup\overline{E(\oplus)}\subseteq
Sc_{+}$. Then, n particular, by Theorem 5.7 we also deduce that for all $n>1$,
for all $p\in\overline{K(\beta\mathbb{N},\odot)}$,
$n^{p}\notin\overline{K(\beta\mathbb{N},\odot)}$.

###  Non existence of exponential idempotents

As it is nowadays well-known, idempotents are often key ingredients in the
search for infinite partition regular structures, the most famous example
being the relationship between idempotent ultrafilters with respect to the sum
and Hindman Theorem (see [17, Chapter 5]). As Examples 5.4, 5.5 suggest,
having introduced the binary operations $E_{1},E_{2}$, it makes sense to study
if they admit idempotents:

######  Definition 5.9.

We say that $p\in\beta\mathbb{N}$ is $E_{1}$-idempotent (resp.
$E_{2}$-idempotent) if $E_{1}(p,p)=p$ (resp. if $E_{2}(p,p)=p$.

First of all, since $E_{1},E_{2}$ are not associative, we cannot approach the
existence of idempotents problem via the usual straightforward application of
Ellis’ Theorem (see [17, Theorem 2.5.] for details). Moreover, since $1$ is
the only principal $E_{1}-$idempotent, and the only $E_{2}$-idempotent as
well, from now on we will use ”idempotent” to actually mean ”nonprincipal
idempotent”.

The following result shows that the existence of idempotents for $E_{2}$ is
related to that of $\star_{n}$-idempotents, whilst the existence of
idempotents for $E_{1}$ is related to that of
$\widetilde{\star_{n}}$-idempotents, where $\widetilde{\star_{n}}$ is the
switch of $\star_{n}$ defined, for all $p,q\in\beta\mathbb{N}$, as
$p\widetilde{\star_{n}}q=p\odot n^{q}$.

######  Proposition 5.10.

If $n\in\mathbb{N}_{>1}$, then $(1)\Rightarrow(2)\Rightarrow(3)$, where

  1. (i)

there exists $p\in\beta\mathbb{N}$ such that $p\oplus n^{p}=p$;

  2. (ii)

there exists $p\in\beta\mathbb{N}$ such that $p\widetilde{\star_{n}}p=p$;

  3. (iii)

there exists $p\in\beta\mathbb{N}$ such that $E_{1}(p,p)=p$.

Analogously, $(4)\Rightarrow(5)\Rightarrow(6)$, where

  1. (iv)

there exists $p\in\beta\mathbb{N}$ such that $n^{p}\oplus p=p$;

  2. (v)

there exists $p\in\beta\mathbb{N}$ such that $p\widetilde{\star_{n}}p=p$;

  3. (vi)

there exists $p\in\beta\mathbb{N}$ such that $E_{2}(p,p)=p$.

###### Proof.

All the implications follows by taking $q$ as given by the hypothesis, by
letting $p=n^{q}$ and by applying Proposition 5.2. For this reason, we prove
only that $(v)$ entails $(vi)$, as the other implications are very similar.

$(v)\Rightarrow(vi)$ Let $q\in\beta\mathbb{N}$ be such that
$q\widetilde{\star_{n}}q=q$ and let $p=n^{q}$. Then
$E_{2}(p,p)=E_{2}\left(n^{q},n^{q}\right)$. As $n^{q}=E_{2}(q,n)$, by
Proposition 5.2.(6) we have that

$E_{2}\left(n^{q},n^{q}\right)=E_{2}\left(n^{q},E_{2}\left(q,n\right)\right)=E_{2}\left(n^{q}\odot
q,n\right)=E_{2}\left(q,n\right)=p.\qed$

Moreover, in the search of $E_{1}$\- and $E_{2}$-idempotent ultrafilters, we
make the following observation:

######  Proposition 5.11.

Assume that there exists $p,q\in\beta\mathbb{N}\setminus\\{0,1\\}$ such that
$p=p\odot E_{1}(q,p)$. Then $E_{1}(q,p)$ is $E_{1}$-idempotent.

Analogously, if there exists $p,q\in\beta\mathbb{N}\setminus\\{0,1\\}$ such
that $p=E_{2}(p,q)\odot p$ then $E_{2}(p,q)$ is $E_{2}$-idempotent.

###### Proof.

We just have to apply Proposition 5.2.(5)-(6).

In fact, if $p=p\odot E_{1}(q,p)$ then

$E_{1}(q,p)=E_{1}(q,p)=E_{1}(q,p\odot
E_{1}(q,p))=E_{1}\left(E_{1}(q,p),E_{1}(q,p)\right)$

and, if $p=E_{2}(p,q)\odot p$ then

$E_{2}(p,q)=E_{2}\left(E_{2}(p,q)\odot
p,q\right)=E_{2}\left(E_{2}(p,q),E_{2}(p,q)\right).\qed$

All that has been said so far might point to the actual existence of
exponential idempotents. Actually, as we are going to show, this is not the
case for $E_{1}$, and at least not trivially the case for $E_{2}$.

As for $E_{1}$, the non existence of idempotents follows from an even stronger
fact, that we now prove.

######  Theorem 5.12.

For all $p\in\beta\mathbb{N},q\in\beta\mathbb{N}\setminus\mathbb{N}$ we have
that121212Notice, by contrast, that in Proposition 5.2 we proved that the
apparently similar equation $E_{1}(p,q)=p$ has infinitely many solutions.
$E_{1}(p,q)\neq q$; in particular, there are no $E_{1}-$idempotents
ultrafilters in $\beta\mathbb{N}\setminus\mathbb{N}$.

###### Proof.

By [20, Lemma 23], there is a $4$-coloring $\mathbb{N}=A_{1}\cup\dots\cup
A_{4}$ of $\mathbb{N}$ such that there is no monochromatic pair $b,a^{b}$ with
$\log_{2}a\leq b$ and $a,b\geq 1$.

Now assume that $E_{1}(p,q)=q$; by definition, this means that for all
$A\subseteq\mathbb{N}$ we have that

$A\in q\Leftrightarrow\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid n^{m}\in
A\\}\in q\\}\in p.$

Now take $A_{i}\in\\{A_{1},\dots,A_{4}\\}$ such that $A_{i}\in q$. In
particular, $B=\\{n\in\mathbb{N}\mid\\{m\in\mathbb{N}\mid n^{m}\in A_{i}\\}\in
q\\}\neq\emptyset$, so we can pick $n\in B$. Let $I_{n}:=\\{m\in\mathbb{N}\mid
n^{m}\in A_{i}\\}$. As $q$ is non principal, $I_{n}$ must contain some
$m>\log_{2}(n)$. But $m,n^{m}\in A_{i}$ with $\log_{2}(n)\leq m$, which is
absurd. ∎

Putting together Theorem 5.12 and Propositions 5.10, 5.11, we get the
following:

######  Corollary 5.13.

For all $n>1\in\mathbb{N}$ and $p,q\in\beta\mathbb{N}\setminus\\{0,1\\}$ we
have that $p\oplus n^{p}\neq p$ and $p\odot E_{1}(q,p)\neq p$.

Notice that, in particular, also $p\odot n^{p}=p$ has no solution in
$\beta\mathbb{N}$ if $n>1$.

Naively, from Theorem 5.12 it would be natural to imagine that some similar
argument could be used to prove the following conjecture:

######  Conjecture 5.14.

$E_{2}$-idempotent ultrafilters do not exist.

Unfortunately, we have not been able to prove the above conjecture; we can,
however, show a partial result towards its solution. The basic observation is
the following: by Proposition 5.2, if $p$ is $E_{2}$-idempotent, we have that

$p=E_{2}(p,p)=E_{2}(p,E_{2}(p,p))=E_{2}(p\odot p,p).$

As, for all $n,m\in\mathbb{N},p\in\beta\mathbb{N}\setminus\mathbb{N}$ one has
that $E_{2}(n,p)=E_{2}(m,p)\Leftrightarrow n=m$, it becomes natural to ask if
it is possible to have $p\in\beta\mathbb{N}\setminus\mathbb{N}$ such that
$p=p\odot p=E_{2}(p,p)$; the answer is no.

######  Theorem 5.15.

There is no ultrafilter $p\in\beta\mathbb{N}\setminus\mathbb{N}$ that is
multiplicatively and $E_{2}$-idempotent.

###### Proof.

By contrast, let $p$ be such an ultrafilter. Let
$S:\mathbb{N}\rightarrow\mathbb{N}$ be the function that maps
$n=p_{1}^{a_{1}}\cdot\dots\cdot p_{k}^{a_{k}}$ to $\sum_{i\leq k}a_{i}$, where
$n=p_{1}^{a_{1}}\cdot\dots\cdot p_{k}^{a_{k}}$ is the prime decomposition of
$n$. Observe that $S(E_{2}(p_{1},p_{2}))=p_{1}\odot S(p_{2})$ and
$S(p_{1}\odot p_{2})=S(p_{1})\oplus S(p_{2})$ for all
$p_{1},p_{2}\in\beta\mathbb{N}$. In particular, $S(p)=S(E_{2}(p,p))=p\odot
S(p)$ and $S(p)=S(p\odot p)=S(p)\oplus S(p)$, therefore $S(p)$ is additively
idempotent and $p\odot S(p)=S(p)\oplus S(p)$. If $S(p)\in\mathbb{N}$, then by
additive idempotency we would get that $S(p)=0$, from which $p=1$, which is
absurd as $p\in\beta\mathbb{N}\setminus\mathbb{N}$. So $S(p)$ is additively
idempotent and nonprincipal, so $n\mathbb{N}\in S(p)$ for all
$n\in\mathbb{N}$. Therefore we can apply [17, Theorem 13.14] to deduce that it
must be $p\odot S(p)\neq S(p)\oplus S(p)$, reaching the desired contrast. ∎

###  Properties of $\star_{n}$-idempotents

As said before, in [9] the authors introduced131313To be precise, they
considered the extension to $\beta\left(\mathbb{N}^{2}\right)$ of the function
$f:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$ that maps $(n,m)$ to
$2^{n}m$, restricted to tensor products of the form $p\otimes p$ where $p$ has
additional combinatorial properties; we are rephrasing this here in a language
closer to that we used in section 5.2. the operation $\star$ defined on
$(\beta\mathbb{N})^{2}$ as follows: $p\star q=2^{p}\odot q$. This was then
used to prove the existence of exponential triples; in Proposition 5.10 we
showed how idempotents for $\widetilde{\star_{n}}$, (where
$n\in\mathbb{N}_{>1}$) and for the similar function
$p\widetilde{\star_{n}}q=p\odot n^{q}$, are related, respectively, to
$E_{2}$\- and $\widetilde{\star_{n}}$-idempotents. In particular, by Theorem
5.12 it follows that $\beta\mathbb{N}$ admits no $E_{1}$-idempotents; in this
section, we want to study what can be said about $\star_{n}$-idempotents for
each $n\in\mathbb{N}$. Such ultrafilters would be very useful to produce
partition regular exponential configurations; for example, if one would be
able to find $p\in\beta\mathbb{N}$ such that $p\star_{n}p=p\oplus p=p$ for
some $n(>1)\in\mathbb{N}$, then every $A\in p$ would contain an element
$\\{a,b,a+b,an^{b}\\}$, so that every set in $n^{p}$ would contain a
monochromatic configuration141414By different methods, the partition
regularity of configurations of the form $\\{x,y,xy,x^{y}\\}$ was proved by
Sahasrabudhe in [20], although with no control on the kind of ultrafilter
witnesses of this partition regularity. of the form $\\{x,y,xy,x^{y}\\}$. As
done with $E_{2}$, we are not able to provide a definitive answer, but we have
some partial result that leads us to formulate this conjecture:

######  Conjecture 5.16.

For any $n\in\mathbb{N}_{>1},$ $\star_{n}$-idempotent ultrafilters do not
exist.

The material of this Section, particularly the proof of Theorem 5.20, has been
influenced by discussions with Prof. Dona Strauss. We are thankful for her
help.

Before we proceed to our main theorem, we will prove three lemmas. Note that
if $(X,\mathcal{U})$ is a uniform space, then there is a topological
compactification $(\phi,\gamma_{u}X)$ of $X$ such that it is precisely the
uniformly continuous functions in $C_{\mathbb{R}}(X)$ which have continuous
extensions to $\gamma_{u}X.$ This $\gamma_{u}X$ will be called the uniform
compactification of $X$. For details, we refer to [17, Chapter 21].

######  Lemma 5.17.

Let $x\in[1,\infty)$ and $q\in U$, where $U$ is the uniform compactification
of $([0,\infty),+)$. Then the equation $x+q=q$ can not hold.

###### Proof.

Choose an integer $n>x$. Now the map
$\varphi:[0,\infty)\longrightarrow\mathbb{T}$ given by
$\varphi(t)=exp\frac{2\pi it}{n}$ is a uniformly continuous homomorphism,
where $\mathbb{T}=\\{z\in\mathbb{C}:\|z\|=1\\}$ is the compact circle group.
So it extends to a continuous homomorphism
$\widetilde{\varphi}:U\longrightarrow\mathbb{T}$. Now if $x+q=q$, then
$\varphi(x)=1$, a contradiction. ∎

If $q$ is an idempotent in $(\beta\mathbb{N},\star_{n})$ for some
$n\in\mathbb{N}$, then $q=n^{q}\odot q$. This implies that
$q\in\mathbb{H}_{n}$, where
$\mathbb{H}_{n}=\bigcap_{m\in\mathbb{N}}\overline{n^{m}\mathbb{N}}$ is a
subsemigroup of $(\beta\mathbb{N},\oplus)$.

######  Lemma 5.18.

If $p$ is an idempotent in $(\beta\mathbb{N},\star_{n})$ for some
$n\in\mathbb{N}_{>1}$, then there is no $u\in\beta\mathbb{N}$ for which
$p=u\oplus p$. In addition, for any $p,q\text{ and }u\in\beta\mathbb{N}$, the
equation $q\star_{n}p=p=u\oplus p$ is not solvable.

###### Proof.

Let $U$ be the uniform compactification of $([0,\infty),+)$ and
$f:\beta\mathbb{N}\longrightarrow U$ be the continuous extension to
$\beta\mathbb{N}$ of the function $\log_{n}:\mathbb{N}\rightarrow[0,\infty)$.
$f$ has the property that $f(x\odot y)=f(x)\oplus f(y)$ for every
$x,y\in\beta\mathbb{N}$ and $f(x\oplus y)=f(y)$ for every
$x,y\in\beta\mathbb{N}$ and for every
$y\in\mathbb{N^{*}}=\beta\mathbb{N}\setminus\mathbb{N}$. By contrast, assume
that $p=q\star_{n}p=u\oplus p$ for some $u\in\beta\mathbb{N}$.

Claim: There exists at most one $m\in\mathbb{N}$ such that $m\oplus
p\in\mathbb{H}_{n}$.

To prove the claim, let $a\oplus p,b\oplus p\in\mathbb{H}_{n}$ for distinct
$a,b\in\mathbb{N}$ and choose $m\in\mathbb{N}$ such that
$n^{m}>\max\\{a,b\\}$. Let
$h:\beta\mathbb{N}\longrightarrow\mathbb{Z}_{n^{m}}$ denote the extension to
$\beta\mathbb{N}$ of the natural homomorphism that maps $\mathbb{N}$ onto
$\mathbb{Z}_{n^{m}}$. Then $h(a)+h(p)=h(b)+h(p)$. So $h(a)=h(b)$, which
implies that $n^{m}|a-b$, a contradiction.

Having proven the Claim, now observe that $p\in cl\\{n^{m}\odot
p:m\in\mathbb{N}\\}\cap cl\\{m\oplus p:m\in\mathbb{N},m\oplus
p\notin\mathbb{H}_{n}\\}$, because $p=n^{q}\odot p=u\oplus p$ for some
$q,u\in\beta\mathbb{N}$. Hence from [17, Corollary 3.42],

there exists $x\in\beta\mathbb{N}$ and $m\in\mathbb{N}$ with $m\oplus
p\notin\mathbb{H}_{n}$, such that $n^{x}\odot p=m\oplus p$ or, $n^{m}\odot
p=x\oplus p$. The first possibility cannot hold because $m\oplus
p\notin\mathbb{H}_{n}$ but $n^{x}\odot p\in\mathbb{H}_{n}$. The second
possibility also cannot hold because it implies that $m\oplus f(p)=f(p)$,
which makes a contradiction in Lemma 5.17.

∎

The following is a basic observation that we prove in detail for completeness.

######  Lemma 5.19.

If $p\in K(\beta\mathbb{N},\oplus)$, then $u\oplus p=p$ for some $u\in
K(\beta\mathbb{N},\oplus)$.

###### Proof.

Choose a minimal left ideal $L$ of $(\beta\mathbb{N},\oplus)$ such that $p\in
L$. Then $L\subseteq K(\beta\mathbb{N},\oplus)$. Now $L\oplus p$ is also a
left ideal of $(\beta\mathbb{N},\oplus)$ and $L\oplus p\subseteq L$. By the
minimality of $L$, we have that $L\oplus p=L$. Thus $p\in L\oplus p$. Take
$u\in L\subseteq K(\beta\mathbb{N},\oplus)$ such that $p=u\oplus p$, as
required. ∎

The three Lemmas above allow to deduce that $\star_{n}$-idempotents, if they
exist, cannot belong to $K\left(\beta\mathbb{N},\oplus\right)$.

######  Theorem 5.20.

For any $n\in\mathbb{N}_{>1}$, there is no idempotent $p$ of
$(\beta\mathbb{N},\star_{n})$ such that $p\in K(\beta\mathbb{N},\oplus)\cup
E(\beta\mathbb{N},\oplus)$.

###### Proof.

If $p\in K(\beta\mathbb{N},\oplus)$, then there exists $u\in
K(\beta\mathbb{N},\oplus)$ such that $u\oplus p=p$ by Lemma 5.19. Now by Lemma
5.18, $p$ can not be an idempotent of $(\beta\mathbb{N},\star_{n})$.

If $p\in E(\beta\mathbb{N},\oplus)$, then $p\oplus p=p.$ Now by Lemma 5.18,
$p$ can not be an idempotent of $(\beta\mathbb{N},\star_{n})$. ∎

Putting together the results of this section, we have the limitation that
$\star_{n}$-idempotents (for any $n\in\mathbb{N}_{>1}$), if they exist, must
belong to $\mathbb{H}_{n}\setminus\left(K(\beta\mathbb{N},\oplus)\cup
E(\beta\mathbb{N},\oplus)\right)$. In particular, this tells us that the
ultrafilters used by Di Nasso and Ragosta in [10] to produce a Hindman-like
infinitary extension of the partition regularity of exponential triples,
namely ultrafilters of the form $p\star_{n}p$ for $p$ minimal additive
idempotent, are not $\star_{n}$-idempontents.

We have seen in Lemma 4.4 that $\mathcal{DR}$ is the left ideal of
$(\beta\mathbb{N},\odot).$ From Theorem [14, Theorem 4.48], if $A$ is a
multiplicative syndetic set, then $\overline{A}\cap\mathcal{DR}\neq\emptyset.$
Hence the following corollary is an immediate application of Lemma 4.4.

######  Corollary 5.21.

Every multiplicatively syndetic set witness the conclusion of theorem 4.1.

In Theorem 5.20, we proved that $\star_{n}$ idempotents do not exist for a
large class of ultrafilters. Our next lemma shows that $p\star_{n}q=q$ is
solvable in $\beta\mathbb{N}$. Here we can choose $p,q\in\beta\mathbb{N}$
witnessing rich combinatorial structures. As a consequence we will prove that
$\overline{K\left(\beta\mathbb{N},\odot\right)}\subseteq\mathcal{DR}.$ This
immediately strengthen the above corollary.

Let $X$ be a compact space and $(S,\cdot)$ be a semigroup. For each $s\in S,$
let $T_{s}:X\rightarrow X$ be a continuous map and $T_{st}=T_{s}\circ T_{t}.$
Then $(X,\langle T_{s}\rangle_{s\in S})$ is called a dynamical system. A point
$x\in X$ is called an uniformly recurrent point if for every open set
$U\subseteq X,$ the return times set $\\{s:T_{s}x\in U\\}$ is a syndetic set.

######  Lemma 5.22.

Let $n\in\mathbb{N}_{>1}$, and $L^{a}$ and $L^{m}$ be both left ideals of
$(\beta\mathbb{N},\oplus)$ and $(\beta\mathbb{N},\odot).$ Then there exist
$p\in L^{a}$ and $q\in L^{m}$ such that $p\star_{n}q=q.$

###### Proof.

Fix $n\in\mathbb{N}_{>1}$. For each $s\in\mathbb{N},$ let
$T_{s}:\beta\mathbb{N}\rightarrow\beta\mathbb{N}$ be the continuous extension
of the map $f_{s}:\mathbb{N}\rightarrow\mathbb{N}$ given by $f_{s}(t)=n^{s}t$.
Clearly $(\beta\mathbb{N},\langle T_{s}\rangle_{s\in\mathbb{N}})$ is a
topological dynamical system. Since $L^{m}$ is a left ideal of
$(\beta\mathbb{N},\odot)$, we have $T_{s}[L^{m}]\subseteq L^{m}$ for all
$s\in\mathbb{N}.$ Hence $(L^{m},\langle T_{s}\rangle_{s\in\mathbb{N}})$ is
also a topological dynamical system. Let $q$ be an uniformly recurrent point
of the dynamical system $(L^{m},\langle T_{s}\rangle_{s\in\mathbb{N}})$. From
[18, Lemma 2.1], there exists $p\in L^{a}$ such that $T_{p}(q)=p\star_{n}q=q.$
Also from [18, Lemma 2.1], we can choose $p$ as an idempotent of $L^{a}.$ ∎

The following corollary shows that every multiplicative piecewise syndetic set
witness polynomial Di Nasso-Ragosta configuration.

######  Corollary 5.23.

$\overline{K\left(\beta\mathbb{N},\odot\right)}\subseteq\mathcal{DR}.$

###### Proof.

Let $p\in\overline{K\left(\beta\mathbb{N},\odot\right)},$ and $A\in p.$ Then
$\overline{A}\cap K\left(\beta\mathbb{N},\odot\right)\neq\emptyset.$ Hence it
will be sufficient to prove
$K\left(\beta\mathbb{N},\odot\right)\subseteq\mathcal{DR}.$

From [14, Theorem 1.51] we know that
$K\left(\beta\mathbb{N},\odot\right)=\bigcup\\{L:L\text{ is a minimal left
ideal of }(\beta\mathbb{N},\odot)\\}.$ Choose any minimal left ideal $L$ of
$(\beta\mathbb{N},\odot).$ From Lemma 5.22, there exists $p\in L$ such that
$p\in\mathcal{DR}.$ Again from Theorem 4.4, we have $\beta\mathbb{N}\odot
p\subseteq\mathcal{DR}.$ But $L$ is minimal. Hence $\beta\mathbb{N}\odot p=L.$
Hence $L\subseteq\mathcal{DR}.$ As $L$ is an arbitrary minimal left ideal, we
have $K\left(\beta\mathbb{N},\odot\right)\subseteq\mathcal{DR}.$ ∎

## 6 Very rich sets with no exponential triples

Theorem 2.3 connects power sets constructed starting from MPS sets and
exponential triples; from Sahasrabudhe work [20], we know that we cannot do
the same by working directly with MPS sets, as the fact that the configuration
$\\{a,b,a+b,x,y,x^{y}\\}$ is not partition regular forces that all
ultrafilters in $\overline{K(\beta\mathbb{N},\odot)}$ cannot be witnesses of
the partition regularity of exponential triples, as they are already witnesses
of the partition regularity of $\\{a,b,a+b\\}$; in particular, there must be
infinitely many MPS sets that do not contain exponential triples.

In this section, we want to (slightly) improve this result to prove that
containing exponential triples and having a rich additive or multiplicative
structure are very different properties. In fact, we have the following
result:

######  Theorem 6.1.

There exists a set $A$ which is both additively and multiplicatively thick
that does not contain any exponential triple.

###### Proof.

We construct inductively longer and longer intervals
$I_{n}=\left[x_{n},y_{n}\right]$ so that $A=\bigcup_{n\in\mathbb{N}}I_{n}$ has
the desired properties.

For $n=1$, just take $x_{1}=y_{1}>2$.

Assuming to have defined $I_{n}=\left[x_{n},y_{n}\right]$, pick
$x_{n+1},y_{n+1}$ such that

  1. 1.

$y_{n}^{y_{n}}<x_{n+1}<y_{n+1}<x_{n+1}^{2}<2^{x_{n+1}}$ and

  2. 2.

$y_{n}=x_{n}\cdot y_{n-1}$.

Let $A=\bigcup_{n\in\mathbb{N}}[x_{n},y_{n}]$. Then:

  1. 1.

$A$ is additively thick as condition (2) above entails that
$|I_{n}|\rightarrow\infty$ for $n\rightarrow\infty$;

  2. 2.

$A$ is multiplicatively thick, as for any $n\in\mathbb{N}$, if we take $y_{p}$
with $y_{p}>n$ we have that $x_{p},2x_{p},\dots,nx_{p}\in I_{p}\subseteq A$;

  3. 3.

$A$ does not contain any exponential triple. In fact, let $x,y\in A$. If
$x<y$, let $y\in I_{n}$. Then

$y_{n}<2^{x_{n}}<x^{x_{n}}<x^{y}<y^{y}\leq y_{n}^{y_{n}}<x_{n+1}.$

Hence $x^{y}\notin A.$

If $x>y$, analogously let $x\in I_{n}$. Then

$y_{n}<x_{n}^{2}<x_{n}^{y}<x^{y}<y_{n}^{y}<y_{n}^{y_{n}}<x_{n+1}.$

So $x^{y}\notin A.$

This completes the proof. ∎

In Corollary 5.23 we proved that
$K\left(\beta\mathbb{N},\odot\right)\subseteq\mathcal{DR}.$ The following
construction shows that even an additively thick set may not contain Di Nasso-
Ragosta configuration.

######  Theorem 6.2.

For any $m\in\mathbb{N}$, there exists an additive thick set that does not
contain configuration of the form $\\{x,y,x\cdot m^{y}\\}$.

###### Proof.

First choose $x_{1}=y_{1}>m$. Now for each $n\geq 2,$ choose $x_{n},y_{n}$
inductively satisfying the following conditions

  1. 1.

$y_{n-1}\cdot m^{y_{n-1}}<x_{n}<y_{n}<x_{n}\cdot m^{2}<m^{x_{n}+1}$ (again
$x_{n}>>y_{n-1}\cdot m^{y_{n-1}}$) and

  2. 2.

$y_{n}-x_{n}\rightarrow\infty$.

Let $A=\bigcup_{n\in\mathbb{N}}[x_{n},y_{n}].$ Then $A$ is a thick set.
Assuming $x,y\in A$, $x<y$, and $y\in[x_{n},y_{n}],$ we have

$y_{n}<m^{x_{n}+1}<m\cdot m^{x_{n}}<x\cdot m^{x_{n}}<x\cdot m^{y}<y\cdot
m^{y}<y_{n}\cdot m^{y_{n}}<x_{n+1}.$

Hence $x\cdot m^{y}\notin A.$ Again if $x>y$ and $x\in[x_{m},y_{m}],$ we have

$y_{n}<x_{n}\cdot m^{2}<x_{n}\cdot m^{y}<x\cdot m^{y}<x\cdot m^{x}<y_{n}\cdot
m^{y_{n}}<x_{n+1}.$

Hence $x\cdot m^{y}\notin A.$ This completes the proof. ∎

## 7 Open questions

In this section, we summarize some questions which we could not solve, as well
as some possible directions we believe it could be interesting to pursue in
the future.

The first question originates from the fact that, in [20], Sahasrabudhe proved
the partition regularity of the configuration $\\{x,y,xy,x^{y}\\}$. We tried
to recover a more direct proof of this fact, along the lines of what we
developed in Section 2, but we failed. Our guess is that, by choosing $p$ in a
reasonable way, the partition regularity of $\\{x,y,xy,x^{y}\\}$ should be
witnessed by $2^{p}$. We conjecture that a family of ultrafilters $p$ that
should make our claim true is the family of those $p\in
E\left(K(\beta\mathbb{N},\odot)\right)$ such that each member of $p$ is also
additive central. Summarizing, our first question is the following:

######  Question 7.1.

Let $p\in E\left(K(\beta\mathbb{N},\odot)\right)$ be such that each member of
$p$ is also additive central. Is it true or not that each $A\in p$ contains a
configuration of the form $\\{m,n,m+n,m2^{n}\\}$?

Notice that an affirmative answer to the above question would immediately
entail, by using the same methods adopted in this paper, that each member of
$2^{p}$ contains patterns of the form $\\{x,y,xy,x^{y}\\}$.

The following question regards several technical properties of the exponential
operations $E_{1},E_{2}$. We proved the existence and non-existence of
ultrafilters satisfying certain equalities; for some other similar ones, we
did not manage to find a definitive answer.

######  Question 7.2.

Are there $p,q\in\beta\mathbb{N}\setminus\mathbb{N}$ that solve any of the
following equations

  1. 1.

$E_{1}(p,q)=E_{1}(q,p)$;

  2. 2.

$E_{2}(p,q)=E_{2}(q,p)$;

  3. 3.

$E_{1}(p,q)=E_{2}(q,p)$;

  4. 4.

$E_{1}(q,p)\odot p=p$;

  5. 5.

$E_{1}(q,p)\odot q=q$;

  6. 6.

$E_{2}(q,p)\odot p=p$;

  7. 7.

$E_{2}(q,p)\odot q=q$.

Of course, as discussed in the paper, more relevant questions are the
following:

######  Question 7.3.

Does it exist $p$ that is $E_{2}$-idempotent? Does it exist $p$ that is
$\star_{n}$-idempotent?

Finally, as a general direction connected to all the above questions, we
believe that it would be interesting to further study the properties of
exponentially rich ultrafilters; of course, this rather vague idea could be
made precise in several different ways. We believe it would be interesting to
start by studying the following:

######  Question 7.4.

What are the algebraical properties of $EVdW$, besides Theorem 5.7?

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