# Nontriviality of rings of integral-valued polynomials

Giulio Peruginelli111Department of Mathematics “Tullio Levi-Civita” University
of Padova, Via Trieste, 63 35121 Padova, Italy<EMAIL_ADDRESS>Nicholas J. Werner222Department of Mathematics, Computer and Information
Science, State University of New York at Old Westbury, Old Westbury, NY 11560,
USA<EMAIL_ADDRESS>
###### Abstract

Let $S$ be a subset of $\overline{\mathbb{Z}}$, the ring of all algebraic
integers. A polynomial $f\in\mathbb{Q}[X]$ is said to be integral-valued on
$S$ if $f(s)\in\overline{\mathbb{Z}}$ for all $s\in S$. The set
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ of all integral-
valued polynomials on $S$ forms a subring of $\mathbb{Q}[X]$ containing
$\mathbb{Z}[X]$. We say that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial if
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$, and
nontrivial otherwise. We give a collection of necessary and sufficient
conditions on $S$ in order
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to be nontrivial. Our
characterizations involve, variously, topological conditions on $S$ with
respect to fixed extensions of the $p$-adic valuations to
$\overline{\mathbb{Q}}$; pseudo-monotone sequences contained in $S$;
ramification indices and residue field degrees; and the polynomial closure of
$S$ in $\overline{\mathbb{Z}}$.

Keywords: integer-valued polynomial, algebraic integers, pseudo-monotone
sequences, polynomial closure.

MSC Primary 12J20, 13F30, 13A18, 13B25, 13F20.

##  1 Introduction and Motivation

Let $\overline{\mathbb{Q}}$ be a fixed algebraic closure of the field of
rational numbers, and let $\overline{\mathbb{Z}}$ be the integral closure of
$\mathbb{Z}$ in $\overline{\mathbb{Q}}$, which is the ring of all algebraic
integers. For a subset $S\subseteq\overline{\mathbb{Z}}$, a polynomial
$f\in\mathbb{Q}[X]$ is said to be integral-valued if
$f(s)\in\overline{\mathbb{Z}}$ for all $s\in S$. Sets of integral-valued
polynomials often form rings with interesting properties, and these rings have
been of interest in recent years [7, 13, 15, 18, 20, 26].

For each positive integer $n$, let $\mathcal{A}_{n}$ be the set of algebraic
integers of degree at most $n$. In [15] (see also [20] and [26]) the following
ring of integral-valued polynomials was introduced in order to provide an
example of a Prüfer domain strictly contained between $\mathbb{Z}[X]$ and the
classical ring of integer-valued polynomials
$\textnormal{Int}(\mathbb{Z})=\\{f\in\mathbb{Q}[X]\mid
f(\mathbb{Z})\subseteq\mathbb{Z}\\}$:

$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n}):=\\{f\in\mathbb{Q}[X]\mid
f(\mathcal{A}_{n})\subseteq\mathcal{A}_{n}\\}.$

Note that for $n=1$, we have $\textnormal{Int}(\mathbb{Z})$. Clearly,
$\mathcal{A}_{n}\subsetneq\mathcal{A}_{n+1}$ for each $n$. If $\alpha$ is an
algebraic number of degree at most $n$ and $f\in\mathbb{Q}[X]$, then the
degree of $f(\alpha)$ is also bounded by $n$. As we will later show in Lemma
4.2, there exist polynomials in
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n+1})$ that are not in
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$. Thus, we have the following
chain of strict inclusions (see Lemma 4.2):

$\ldots\subsetneq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n+1})\subsetneq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})\subsetneq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n-1})\subsetneq\ldots\subsetneq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{1})=\textnormal{Int}(\mathbb{Z}).$
(1.1)

By [15, Theorem 3.9], for each $n$ the ring
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$ is a Prüfer domain. In
particular, this means that the polynomial ring $\mathbb{Z}[X]$ is strictly
contained in $\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$ for each $n\geq
1$. Thus, for $n\geq 2$, these rings provide examples of Prüfer domains
strictly contained between $\mathbb{Z}[X]$ and $\textnormal{Int}(\mathbb{Z})$.

More generally, for $S\subseteq\overline{\mathbb{Z}}$ one may then consider
the ring

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}}):=\\{f\in\mathbb{Q}[X]\mid
f(S)\subseteq\overline{\mathbb{Z}}\\}.$

Note that
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})=\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n},\overline{\mathbb{Z}})$.
It is stated in [15, p. 2482] that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ lies properly between
$\mathbb{Z}[X]$ and $\textnormal{Int}(\mathbb{Z})$ for every subset $S$ of
$\overline{\mathbb{Z}}$ properly containing $\mathbb{Z}$. The containment
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})\subsetneq\textnormal{Int}(\mathbb{Z})$
holds whenever $\mathbb{Z}\subsetneq S$ (see [13]), but we demonstrate below
in Example 1.3 that $\mathbb{Z}[X]$ may equal
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$. Hence, the
aforementioned statement from [15, p. 2482] is false.333Thankfully, the
fallacious claim from [15] was merely an expository remark, and does not
affect the results of that paper.

######  Definition 1.2.

For any $\alpha\in\overline{\mathbb{Z}}$, let $O_{\mathbb{Q}(\alpha)}$ be the
ring of integers of $\mathbb{Q}(\alpha)$. A subset
$S\subseteq\overline{\mathbb{Z}}$ is said to have unbounded degree if the set
$\\{[\mathbb{Q}(\alpha):\mathbb{Q}]\mid\alpha\in S\\}$ is unbounded. For
$\alpha\in\overline{\mathbb{Z}}$, the index of $\alpha$ is
$\iota_{\alpha}:=[O_{\mathbb{Q}(\alpha)}:\mathbb{Z}[\alpha]]$.

######  Example 1.3.

Let $S\subseteq\overline{\mathbb{Z}}$ be such that $S$ contains a sequence
$\\{\alpha_{n}\\}_{n\in\mathbb{N}}$ of unbounded degree and
$\iota_{\alpha_{n}}=1$ for all $n$. We will show that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$.

Certainly,
$\mathbb{Z}[X]\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$.
Let $f(X)\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ of degree
$d$, and write $f(X)=\sum_{i=0}^{d}a_{i}X^{i}$, where each
$a_{i}\in\mathbb{Q}$. Take an element $\alpha\in S$ of degree $n>d$, and
consider $f(\alpha)$. By assumption $f(\alpha)$ is integral over $\mathbb{Z}$
and so it belongs to $O_{\mathbb{Q}(\alpha)}=\mathbb{Z}[\alpha]$. Thus,

$f(\alpha)=a_{0}+a_{1}\alpha+\ldots+a_{d}\alpha^{d}\in\mathbb{Z}[\alpha].$

As a $\mathbb{Z}$-module, $\mathbb{Z}[\alpha]$ is a free with basis given by
$1,\alpha,\ldots,\alpha^{n-1}$, so we also have

$f(\alpha)=b_{0}+b_{1}\alpha+\ldots+b_{n-1}\alpha^{n-1}$

for some integers $b_{0},\ldots,b_{n-1}$ that are uniquely determined by
$f(\alpha)$. Since $n>d$ and every element of the $\mathbb{Q}$-vector space
$\mathbb{Q}(\alpha)$ can be written uniquely as a $\mathbb{Q}$-linear
combination of $1,\alpha,\ldots,\alpha^{n-1}$, each coefficients $a_{i}$ must
be in $\mathbb{Z}$. Thus, $f(X)\in\mathbb{Z}[X]$ and consequently
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$.

In particular, Example 1.3 applies when $S=\overline{\mathbb{Z}}$ or when
$S=\\{\zeta_{n}\\}_{n\in\mathbb{N}}$, where $\zeta_{n}$ is a primitive
$n^{\text{th}}$ root of unity. Moreover, since
$\textnormal{Int}_{\mathbb{Q}}(\overline{\mathbb{Z}},\overline{\mathbb{Z}})=\bigcap_{n\in\mathbb{N}}\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$,
Example 1.3 demonstrates that the intersection of all the rings in (1.1) is
equal to $\mathbb{Z}[X]$, which answers a question posed by David Dobbs to the
first author in 2014.

The goal of this paper is to characterize those subsets
$S\subseteq\overline{\mathbb{Z}}$ such that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})\neq\mathbb{Z}[X]$. In
pursuit of this problem, we will deal with more general rings of integer-
valued polynomials, and describe when these rings are trivial, in the sense
that they are equal to ordinary rings of polynomials. We refer to the papers
[25, 28] for studies on related problems.

######  Definition 1.4.

Let $D$ be an integral domain, let $L$ be a field containing $D$, and let $F$
be a subfield of $L$. For each subset $S\subseteq D$, we define the following
ring of integer-valued polynomials:

$\textnormal{Int}_{F}(S,D):=\\{f\in F[X]\mid f(S)\subseteq D\\}.$

In the literature, when $F$ is the fraction field of $D$, the above ring is
denoted by $\textnormal{Int}(S,D)$. Because our work will often involve
changing the field of coefficients of these polynomials, we will always
include the subscript $F$ in $\textnormal{Int}_{F}(S,D)$ for the sake of
clarity, even in the case where $F$ is the fraction field of $D$.

It is clear that $(D\cap F)[X]\subseteq\textnormal{Int}_{F}(S,D)$. We say that
$\textnormal{Int}_{F}(S,D)$ is trivial if $(D\cap
F)[X]=\textnormal{Int}_{F}(S,D)$, and nontrivial otherwise.

We seek to describe those subsets $S\subseteq\overline{\mathbb{Z}}$ for which
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial. If $S$
is of bounded degree $n$, then $S\subseteq\mathcal{A}_{n}$, and we have
$\mathbb{Z}[X]\subsetneq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\mathbb{Z})$.
Thus, the sets of interest in this problem are all of unbounded degree.
Example 1.3 shows that if $S$ has unbounded degree and $\iota_{s}=1$ for all
$s\in S$, then $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is
trivial. However, if we relax the condition on the indices $\iota_{s}$, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ could be trivial or
nontrivial.

######  Example 1.5.

Let
$S=2\overline{\mathbb{Z}}=\\{2\alpha\mid\alpha\in\overline{\mathbb{Z}}\\}$.
Then, $S$ has unbounded degree and not all of the indices $\iota_{s}$ are
equal to 1. For this $S$,
$X/2\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, and hence
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial. It is
also possible to construct a set $S$ that has unbounded degree and with not
all $\iota_{s}$ equal to 1, but for which
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial; see
Example 4.11.

######  Example 1.6.

Fix a prime $p$. For each $k\in\mathbb{N}$ let $a_{k}=1-\tfrac{1}{2^{k}}$, and
take
$S=\\{p^{a_{k}}\\}_{k\in\mathbb{N}}=\\{p^{1/2},p^{3/4},p^{7/8},\ldots\\}$.
Then, $S$ has unbounded degree, but
$f(X)=X^{2}/p\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$
because $f(p^{a_{k}})=p^{a_{k-1}}$ for all $k\geq 2$.

This example can be generalized. Once again fix $p\in\mathbb{P}$, but also fix
a positive integer $n\geq 2$. For each $k\in\mathbb{N}$, let
$b_{k}=\frac{1}{n-1}(1-\frac{1}{n^{k}})$. Take
$S=\\{p^{b_{k}}\\}_{k\in\mathbb{N}}$. Then,
$f(X)=X^{n}/p\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$,
because $f(p^{b_{1}})=1$ and $f(p^{b_{k}})=p^{b_{k-1}}$ for all $k\geq 2$.

In this paper, we provide both local and global characterizations of subsets
$S\subseteq\overline{\mathbb{Z}}$ for which
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial. We
begin our study by considering integer-valued polynomial rings over subsets of
valuation domains. Let $V$ be a valuation domain with fraction field $K$, and
suppose $S\subseteq V$. In Section 2, we review the concept of pseudo-monotone
sequences (as defined in [6]), and we use these sequences to provide necessary
and sufficient conditions on $S$ in order for $\textnormal{Int}_{K}(S,V)$ to
be nontrivial (Theorem 2.6). In Section 3, we return to the case where
$S\subseteq\overline{\mathbb{Z}}$ and relate
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to rings of integer-
valued polynomials over valuation domains in $p$-adic fields. In this way,
Theorem 2.6 can be applied to
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ (see Theorem 3.2 and
Corollary 3.3), and provides several characterizations of when
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial.

In Section 4, we examine global conditions that imply that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial or
nontrivial. We also consider how the degrees and indices of the algebraic
integers in $S$ can influence
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, present variations
on Examples 1.3 and 1.6. Section 5 examines similar questions with regard to
ramification indices and residue field degrees in number fields generated by
elements of $S$. We produce a new descending chain of Prüfer domains between
$\mathbb{Z}[X]$ and $\textnormal{Int}(\mathbb{Z})$ by considering for each
$n\in\mathbb{N}$ the compositum $\mathbb{Q}^{(n)}=\mathbb{Q}(\mathcal{A}_{n})$
of all the number fields of degree bounded by $n$ and the ring
$\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(n)}})$ of polynomials integral-
valued over the ring of integers $O_{\mathbb{Q}^{(n)}}$ of this infinite
algebraic extension (which is a non-Noetherian almost Dedekind domain with
finite residue fields). Moreover, we review some classical examples due to
Gilmer [12, Example 14] and Chabert [5, Example 6.2] of an infinite algebraic
extension $K$ of $\mathbb{Q}$ such that the integral closure $D$ in $K$ of
$\mathbb{Z}_{(p)}$ for some prime $p\in\mathbb{Z}$ is an almost Dedekind
domain with finite residue fields such that either the set of the residue
field degrees (ramification indexes, respectively) of all the prime ideals of
$D$ over $\mathbb{Z}_{(p)}$ are unbounded and
$\textnormal{Int}_{\mathbb{Q}}(D)$ is trivial. By means of the double
boundedness condition of Loper [14], in Theorem 5.18 we prove that if $D$ is
the integral closure of $\mathbb{Z}_{(p)}$ in some algebraic extension of
$\mathbb{Q}$, then $\textnormal{Int}_{\mathbb{Q}}(D)$ is nontrivial if and
only if $\textnormal{Int}_{\mathbb{Q}}(D)$ is Prüfer.

The final section of the paper considers a broad generalization of Example
1.5, describes the polynomial closure in $\overline{\mathbb{Z}}$ of a subset
$S$ of the ring of all algebraic integers, and relates this concept to the
nontriviality of $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$.

##  2 Nontriviality over Valuation Domains

Throughout, $V$ is a valuation domain with valuation $v$, value group
$\Gamma_{v}$, maximal ideal $M$, and fraction field $K$. We begin by recalling
the definitions of pseudo-monotone sequences and related objects, which were
introduced in [6] in order to study the polynomial closure of subsets of a
rank one valuation domain and studied further in [21, 23]. After proving some
basic properties of such sequences, we use them to characterize when
$\textnormal{Int}_{K}(S,V)$ is nontrivial (see Theorem 2.6).

######  Definition 2.1.

Let $\Lambda$ be an index set, and let $E=\\{s_{i}\\}_{i\in\Lambda}$ be a
sequence in $K$. Then, $E$ is said to be

  1. (i)

pseudo-convergent if $v(s_{i}-s_{j})<v(s_{j}-s_{i})$ for all
$i<j<k\in\Lambda$;

  2. (ii)

pseudo-divergent if $v(s_{i}-s_{j})>v(s_{j}-s_{k})$ for all $i<j<k\in\Lambda$;

  3. (iii)

pseudo-stationary if $v(s_{i}-s_{j})=v(s_{k}-s_{\ell})$ for all $i\neq
j\in\Lambda$, $k\neq\ell\in\Lambda$.

A sequence that satisfies any one of these three properties is said to be
pseudo-monotone.

######  Definition 2.2.

Let $E=\\{s_{i}\\}_{i\in\Lambda}$ be a pseudo-monotone sequence in $K$. We
define the _gauge_ of $E$ as the following sequence
$\\{\delta_{i}\\}_{i\in\Lambda}$ of $\Gamma_{v}$:

  1. (i)

if $E$ is pseudo-convergent, for each $i\in\Lambda$ we set
$\delta_{i}=v(s_{i}-s_{j})$, with $j\in\Lambda$ and $j>i$;

  2. (ii)

if $E$ is pseudo-divergent, for each $i\in\Lambda$ we set
$\delta_{i}=v(s_{i}-s_{j})$, with $j\in\Lambda$ and $j<i$;

  3. (iii)

if $E$ is pseudo-stationary, we let $\delta=v(s_{i}-s_{j})$ for any
$i,j\in\Lambda$, $i\neq j$.

######  Definition 2.3.

Each pseudo-monotone sequence $E=\\{s_{i}\\}_{i\in\Lambda}$ in $K$ has an
associated breadth ideal, which we denote by $\textnormal{Br}(E)$. The
definition of $\textnormal{Br}(E)$ depends on which one the three properties
from Definition 2.1 that $E$ satisfies.

  1. (i)

If $E$ is pseudo-convergent, then $\textnormal{Br}(E)=\\{x\in K\mid
v(x)>\delta_{i}\text{ for all }i\in\Lambda\\}$;

  2. (ii)

If $E$ is pseudo-divergent, then $\textnormal{Br}(E)=\\{x\in K\mid
v(x)>\delta_{i}\text{ for some }i\in\Lambda\\}$;

  3. (iii)

If $E$ is pseudo-stationary, then $\textnormal{Br}(E)=\\{x\in K\mid
v(x)\geq\delta\\}$.

For a pseudo-divergent sequence $E$, it is possible that
$\textnormal{Br}(E)=K$ (this occurs when the gauge of $E$ is coinitial in
$\Gamma_{v}$). In all the other cases, $\textnormal{Br}(E)$ is a fractional
$V$-ideal. When $V$ has rank one and $E=\\{s_{n}\\}_{n\in\mathbb{N}}$ is a
pseudo-monotone sequence, the breadth
$\delta_{E}\in\mathbb{R}\cup\\{\infty\\}$ of $E$ is
$\delta_{E}:=\lim_{n\to\infty}v(s_{n+1}-s_{n})$. In this case, $\delta_{i}$
can be replaced with $\delta_{E}$ in each of the above definitions for
$\textnormal{Br}(E)$.

Note that if $E$ is pseudo-stationary and $\delta=v(c)$ for some $c\in K$,
then $\textnormal{Br}(E)=cV$ is a principal (fractional) ideal. Also, if
$E\subseteq V$ is any pseudo-monotone sequence, then
$\textnormal{Br}(E)\subseteq V$, and $\textnormal{Br}(E)\subseteq M$ if $E$ is
pseudo-divergent.

Finally, we recall that, given $a\in K$ and $\gamma\in\Gamma_{v}$, the ball of
center $a$ and radius $\gamma$ is $B(a,\delta)=\\{x\in K\mid
v(x-a)\geq\delta\\}$. For each $b\in B(a,\delta)$ it is well known that
$B(a,\delta)=B(b,\delta)$.

######  Lemma 2.4.

Let $S\subseteq V$. Assume there exist a finite subset $T\subseteq S$ and
$b\in M$ such that for all $s\in S$, there exists $t\in T$ such that
$v(s-t)\geq v(b)$.

  1. (1)

If $S$ contains a pseudo-divergent sequence $E$, then
$\textnormal{Br}(E)\subseteq bM\subsetneq M$.

  2. (2)

If $S$ contains a pseudo-stationary sequence $E$, then
$\textnormal{Br}(E)\subseteq bV\subsetneq V$.

Note that the assumption in this lemma is equivalent to saying that $S$ is
covered by a finite union of balls $B(t,\delta)$, where $\delta>0$ is fixed
and each $t\in T$.

###### Proof.

Let $\delta=v(b)>0$. Note that if either one of the two conditions is
satisfied, then $S$ is infinite. In particular, $b\not=0$, because otherwise
$S$ would be finite. Since $T$ is finite, there exists $t\in T$ such that
$B(t,\delta)$ contains an infinite subsequence
$E^{\prime}=\\{s_{i}\\}_{i\in\Lambda^{\prime}}$ of $E$. If $E$ is pseudo-
divergent (respectively, pseudo-stationary), then $E^{\prime}$ is also pseudo-
divergent (resp., pseudo-stationary), and
$\textnormal{Br}(E^{\prime})=\textnormal{Br}(E)$. For any distinct
$i,j\in\Lambda^{\prime}$, we have
$v(s_{i}-s_{j})=v((s_{i}-t)+(t-s_{j}))\geq\delta$.

When $E^{\prime}$ is pseudo-divergent, the gauge of $E^{\prime}$ is strictly
decreasing, and the above inequality implies that $v(s_{i}-s_{j})>\delta$. It
follows that $\textnormal{Br}(E^{\prime})\subseteq bM\subsetneq M$. If instead
$E^{\prime}$ is pseudo-stationary, then $v(s_{i}-s_{j})$ is constant for all
$i,j\in\Lambda^{\prime}$. In this case, we have $v(s_{i}-s_{j})\geq\delta>0$,
which means that $\textnormal{Br}(E^{\prime})\subseteq bV\subsetneq V$. ∎

######  Lemma 2.5.

Let $S\subseteq V$. The following are equivalent.

  1. (1)

There exist a finite subset $T\subseteq S$ and
$\delta\in\Gamma_{v}\cup\\{\infty\\}$ such that $\delta>0$ and for all $s\in
S$, there exists $t\in T$ such that $v(s-t)\geq\delta$.

  2. (2)

There exists $b\in M$ such that $S/bV$ is finite.

###### Proof.

Note that $\delta=\infty$ in (1) corresponds to $b=0$ in (2). In this case,
the set $S$ is finite.

(1) $\Rightarrow$ (2) Assume that (1) holds. Choose $b\in M$, such that
$v(b)=\delta$. Let $T=\\{t_{1},\ldots,t_{m}\\}$. We claim that
$S/bV=\\{t_{i}+bV\mid 1\leq i\leq m\\}$. Indeed, given $s\in S$, there exists
$i$ such that $v(s-t_{i})\geq\delta=v(b)$. So, $s-t_{i}\in bV$ and hence
$s+bV=t_{i}+bV$.

(2) $\Rightarrow$ (1) Assume that (2) holds. Let $t_{1},\ldots,t_{m}\in S$ be
such that $S/bV=\\{t_{i}+bV\mid 1\leq i\leq m\\}$. Take
$T=\\{t_{1},\ldots,t_{m}\\}$ and $\delta=v(b)>0$. Then, for any $s\in S$,
there exists $i$ such that $s+bV=t_{i}+bV$, and for this $i$ we have
$v(s-t_{i})\geq v(b)=\delta$. ∎

######  Theorem 2.6.

Let $S\subseteq V$. The following are equivalent.

  1. (1)

$\textnormal{Int}_{K}(S,V)$ is nontrivial.

  2. (2)

There exist a finite subset $T\subseteq S$ and
$\delta\in\Gamma_{v}\cup\\{\infty\\}$ such that $\delta>0$ and for each $s\in
S$, there exists $t\in T$ such that $v(s-t)\geq\delta$.

  3. (3)

There exists $b\in M$ such that $S/bV$ is finite.

  4. (4)

$S$ contains neither a pseudo-divergent sequence $E$ with
$\textnormal{Br}(E)=M$, nor a pseudo-stationary sequence $E$ with
$\textnormal{Br}(E)=V$.

###### Proof.

The theorem is trivial if $S$ is finite (which is equivalent to having
$\delta=\infty$ in (2), or $b=0$ in (3)), so we will assume throughout that
$S$ is infinite.

(1) $\Rightarrow$ (2) Assume $V[X]\subsetneq\textnormal{Int}_{K}(S,V)$. Among
all polynomials in $\textnormal{Int}_{K}(S,V)\setminus V[X]$, choose $f(X)$ of
minimal degree $m$. Then, $m\geq 1$. Write $f(X)=g(X)/d$, where $g(X)\in V[X]$
and $d\in M$. Let $a$ be the leading coefficient of $g$.

We claim that $v(a)<v(d)$. If $v(a)\geq v(d)$, then $aX^{m}/d\in V[X]$, and
$f(X)-aX^{m}/d$ would be either an element of $V[X]$, or an element of
$\textnormal{Int}_{K}(S,V)\setminus V[X]$ of degree less than $m$. We reach a
contradiction in either case. So, $v(a)<v(d)$.

Next, let $L/K$ be the splitting field of $g(X)$, and let $w$ be an extension
of $v$ to $L$. In $L$, we have $g(X)=a(X-\alpha_{1})\cdots(X-\alpha_{m})$ for
some $\alpha_{1},\ldots,\alpha_{m}\in L$. Let $\delta=(v(d)-v(a))/m$, which is
greater than 0. Then, for all $s\in S$, there exists $1\leq i\leq m$ such that
$w(s-\alpha_{i})\geq\delta$. Indeed, if this is not the case, then for some
$s\in S$,

$v(f(s))=w(f(s))<v(a)+m\delta-v(d)=0,$

and $f\notin\textnormal{Int}_{K}(S,V)$.

For each $1\leq i\leq m$, let $B_{w}(\alpha_{i},\delta)=\\{x\in L\mid
w(x-\alpha_{i})\geq\delta\\}$ and let $B_{i}=B_{w}(\alpha_{i},\delta)\cap V$.
Whenever $B_{i}$ has nonempty intersection with $S$, we may choose $t_{i}\in
B_{i}\cap S$, and then $B_{i}=B(t_{i},\delta)$. Take $T=\\{t_{i}\mid 1\leq
i\leq m\text{ and }B_{i}\cap S\neq\varnothing\\}$. Then (2) holds.

(2) $\Leftrightarrow$ (3) This is the content of Lemma 2.5.

(3) $\Rightarrow$ (4) This follows from Lemma 2.4.

(4) $\Rightarrow$ (1) We prove the contrapositive. Assume that
$V[X]=\textnormal{Int}_{K}(S,V)$. Then, for all $m\geq 1$ and all
$s_{1},\ldots,s_{m}\in S$, and all $d\in M$, the polynomial
$(X-s_{1})\cdots(X-s_{m})/d$ is not in $\textnormal{Int}_{K}(S,V)$.
Consequently,

for all $s_{1},\ldots,s_{m}\in S$, and all $\gamma\in\Gamma_{v}$ with
$\gamma>0$, (2.7) there exists $t\in S$ such that
$v(\textstyle\prod_{i=1}^{m}(t-s_{i}))<\gamma$.

Suppose first that $v$ is discrete. By (2.7), given any $s_{1},\ldots,s_{m}\in
S$, there exists $s_{m+1}\in S$ such that
$v(\prod_{i=1}^{m}(s_{m+1}-s_{i}))=0$. So, we can construct a sequence
$E=\\{s_{1},s_{2},\ldots\\}$ such that $v(s_{i}-s_{j})=0$ for all $i\neq j$.
This sequence is pseudo-stationary with $\textnormal{Br}(E)=V$.

For the remainder of the proof, we will assume that $v$ is not discrete. If
$S$ contains a pseudo-stationary sequence $E$ with $\textnormal{Br}(E)=V$,
then we are done. So, we will further assume that this is not the case. We
will demonstrate how to construct a pseudo-divergent sequence $E\subseteq S$
with $\textnormal{Br}(E)=M$.

Choose $s_{1}\in S$. For each $n\geq 1$, choose—if possible—$s_{n+1}\in S$
such that $v(\prod_{i=1}^{n}(s_{n+1}-s_{i}))=0$. We cannot do this
indefinitely, since then $\\{s_{n}\\}_{n\in\mathbb{N}}$ would be pseudo-
stationary with breadth ideal equal to $V$. So, there exists $N\geq 1$ such
that

$v(s_{i}-s_{j})=0\text{ for all }1\leq i<j\leq N,\text{ and
}v(\textstyle\prod_{i=1}^{N}(s-s_{i}))>0\text{ for all }s\in S.$ (2.8)

In other words, for each $s\in S$, there exists a unique
$i\in\\{1,\ldots,N\\}$ such that $v(s-s_{i})>0$ and for every
$j\in\\{1,\ldots,N\\}$, $j\not=i$, we have $v(s-s_{j})=0$.

Let $\delta_{1}\in\Gamma_{v},\delta_{1}>0$. By (2.7) and (2.8), there exists
$s_{N+1}\in S$ such that $0<v(\prod_{i=1}^{N}(s_{N+1}-s_{i}))<\delta_{1}$. By
the previous paragraph, we have $v(s_{N+1}-s_{i_{1}})>0$ for a unique
$i_{1}\in\\{1,\ldots,N\\}$ and $v(s_{N+1}-s_{j})=0$ for every
$j\in\\{1,\ldots,N\\}\setminus\\{i_{1}\\}$. In particular,
$0<v(s_{N+1}-s_{i_{1}})<\delta_{1}$. We choose now $\delta_{2}\in\Gamma_{v}$
such that $0<\delta_{2}<v(s_{N+1}-s_{i_{1}})$ and correspondingly, by (2.7)
and (2.8), we can find $s_{N+2}\in S$ such that
$0<v(\prod_{i=1}^{N+1}(s_{N+2}-s_{i}))<\delta_{2}$. As before, there exists a
unique $i_{2}\in\\{1,\ldots,N\\}$ such that $v(s_{N+2}-s_{i_{2}})>0$ and
$v(s_{N+2}-s_{j})=0$ for every $j\in\\{1,\ldots,N\\}\setminus\\{i_{2}\\}$.

At this point, we have

$0<v(s_{N+2}-s_{i_{2}})<\delta_{2}<v(s_{N+1}-s_{i_{1}})<\delta_{1}.$

If we continue in this way, we can produce a sequence
$E=\\{s_{N+k}\\}_{k\in\mathbb{N}}\subseteq S$ and a sequence
$\\{\delta_{k}\\}_{k\in\mathbb{N}}\subseteq\\{\gamma\in\Gamma_{v}\mid\gamma>0\\}$
such that $\\{\delta_{k}\\}_{k\in\mathbb{N}}$ decreases to 0 and for each
$k,\ell\in\mathbb{N}$, $k<\ell$ we have

$0<v(s_{N+\ell}-s_{i_{\ell}})<\delta_{\ell}<v(s_{N+k}-s_{i_{k}})<\delta_{k}$
(2.9)

where $i_{k},i_{\ell}\in\\{1,\ldots,N\\}$. For each $1\leq i\leq N$, define

$E_{i}:=\\{s_{N+k}\mid k\in\mathbb{N}\text{ and }v(s_{N+k}-s_{i})>0\\}.$

Note that if $s_{N+k}\in E_{i}$, then $0<v(s_{N+k}-s_{i})<\delta_{k}$ and
$v(s_{N+k}-s_{j})=0$ for every $j\in\\{1,\ldots,N\\}\setminus\\{i\\}$.

Now, at least one of the sets $E_{i}$, $i=1,\ldots,N$, must be infinite.
Without loss of generality, assume that $E_{1}$ is infinite. We will prove
that $E_{1}$ is pseudo-divergent with $\textnormal{Br}(E_{1})=M$. Write
$E_{1}=\\{s_{k_{0}},s_{k_{1}},s_{k_{2}},\ldots\\}$, where $s_{k_{0}}=s_{1}$
and $k_{1}<k_{2}<\cdots$. By (2.9), for $k_{j}>k_{i}$ we have

$0<v(s_{k_{j}}-s_{1})<\delta_{k_{j}}<v(s_{k_{i}}-s_{1})<\delta_{k_{i}}$

which shows that $E_{1}$ is pseudo-divergent with pseudo-limit $s_{1}$ (see
for example [6, Remark 4.7]). Moreover, $\textnormal{Br}(E_{1})=M$ because the
sequence $\\{\delta_{k}\\}_{k\in\mathbb{N}}$ decreases to 0. ∎

##  3 Local Case

In this section, we show how Theorem 2.6 can be used to decide whether or not
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial. Let
$\mathbb{P}$ be the set of all prime numbers. For $p\in\mathbb{P}$, $v_{p}$ is
the standard $p$-adic valuation, $\mathbb{Z}_{p}$ denotes the ring of $p$-adic
integers, $\mathbb{Q}_{p}$ is the field of $p$-adic numbers, and
$\mathbf{F}_{p}$ is the finite field with $p$ elements. We let
$\overline{\mathbb{Q}_{p}}$ be an algebraic closure of $\mathbb{Q}_{p}$, and
let $\overline{\mathbb{Z}_{p}}$ be the absolute integral closure of
$\overline{\mathbb{Z}_{p}}$. Note that $\overline{\mathbb{Z}_{p}}$ is a rank
one non-discrete valuation domain with residue field equal to an algebraic
closure of $\mathbb{F}_{p}$.

######  Definition 3.1.

Let $p\in\mathbb{P}$ and let $S\subseteq\overline{\mathbb{Z}_{(p)}}$. Let
$\mathcal{P}(S)\subseteq\mathbb{Z}_{(p)}[X]$ be set of minimal polynomials
over $\mathbb{Q}$ of all the elements of $S$. We define $\Sigma_{p}(S)$ to be
the set of roots in $\overline{\mathbb{Z}_{p}}$ of the polynomials in
$\mathcal{P}(S)$.

######  Theorem 3.2.

Let $S\subseteq\overline{\mathbb{Z}}$. The following are equivalent.

  1. (1)

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial.

  2. (2)

There exists $p\in\mathbb{P}$ such that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ is nontrivial.

  3. (3)

There exists $p\in\mathbb{P}$ such that
$\textnormal{Int}_{\mathbb{Q}_{p}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial.

  4. (4)

There exists $p\in\mathbb{P}$ such that
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial.

Since $\overline{\mathbb{Z}_{p}}$ is a valuation domain with fraction field
$\overline{\mathbb{Q}_{p}}$, combining Theorem 2.6 and Theorem 3.2 yields the
following corollary.

######  Corollary 3.3.

Let $S\subseteq\overline{\mathbb{Z}}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial if and
only if there exists $p\in\mathbb{P}$ such that
$\Sigma_{p}(S)\subseteq\overline{\mathbb{Z}_{p}}$ satisfies one of the
equivalent conditions of Theorem 2.6.

The proof of Theorem 3.2 is quite long and requires numerous intermediate
results. To keep things organized, we will prove the equivalences in the
theorem one at a time.

### Equivalence of 3.2(1) and 3.2(2)

First, we show that our rings of integer-valued polynomials are well-behaved
with respect to localization at primes of $\mathbb{Z}$.

######  Lemma 3.4.

Let $p\in\mathbb{P}$. Then,
$\overline{\mathbb{Z}_{(p)}}=\overline{\mathbb{Z}}_{(p)}$. Moreover,
$(\mathcal{A}_{n})_{(p)}=\\{\tfrac{\alpha}{t}\mid\alpha\in\mathcal{A}_{n},t\in\mathbb{Z}\setminus
p\mathbb{Z}\\},$ and so
$\bigcap_{p\in\mathbb{P}}(\mathcal{A}_{n})_{(p)}=\mathcal{A}_{n}$.

###### Proof.

This follows from the standard principle (see e.g. [1, Proposition 5.12]) that
integral closure and localization commute with one another. ∎

Since
$\overline{\mathbb{Z}}=\bigcap_{p\in\mathbb{P}}\overline{\mathbb{Z}_{(p)}}$,
we clearly have

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\bigcap_{p\in\mathbb{P}}\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}}).$
(3.5)

The next proposition shows that for each $p\in\mathbb{P}$,
$\textnormal{Int}(S,\overline{\mathbb{Z}_{(p)}})$ is the localization of
$\textnormal{Int}(S,\overline{\mathbb{Z}})$ at the multiplicative set
$\mathbb{Z}\setminus p\mathbb{Z}$.

######  Proposition 3.6.

Let $S\subseteq\overline{\mathbb{Z}}$ and let $p\in\mathbb{P}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})_{(p)}=\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$.

###### Proof.

First, let let $f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$
and $n\in\mathbb{Z}\setminus p\mathbb{Z}$. Then, for each $s\in S$, we have
that $\frac{f(s)}{n}$ is integral over $\mathbb{Z}_{(p)}$ by Lemma 3.4. Thus,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})_{(p)}\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$.

The proof of the other containment is similar to the proof of [8, (7)]. Let
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$. There
exists $d\in\mathbb{Z},d\not=0$ such that $df(X)\in\mathbb{Z}[X]$. Suppose
that $d=p^{a}t$, for some $a\geq 0$ and $t\in\mathbb{Z}$ not divisible by $p$.
Let $g(X)=tf(X)$. Then
$g\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ because $t$
is invertible in $\mathbb{Z}_{(p)}$. Moreover, for each prime $q\not=p$,
$g\in\mathbb{Z}_{(q)}[X]\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(q)}})$.
Now, we have

$g\in\bigcap_{p\in\mathbb{P}}\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$

so that $f=\frac{g}{t}$ is in
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})_{(p)}$. ∎

######  Proposition 3.7.

Let $S\subseteq\overline{\mathbb{Z}}$. Then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial if and
only if, for each $p\in\mathbb{P}$,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ is trivial.

###### Proof.

The ‘if’ direction follows by (3.5) and the fact that
$\mathbb{Z}[X]=\bigcap_{p\in\mathbb{P}}\mathbb{Z}_{(p)}[X]$. Conversely,
suppose that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$.
Localizing at any $p\in\mathbb{P}$, by Proposition 3.6 we get
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})_{(p)}=\mathbb{Z}_{(p)}[X]=\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$.
∎

### Equivalence of 3.2(2) and 3.2(3)

######  Proposition 3.8.

Let $p\in\mathbb{P}$ and let $S\subseteq\overline{\mathbb{Z}_{(p)}}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$.

###### Proof.

First, assume
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ and let
$\alpha\in\Sigma_{p}(S)$. Then, there exists $\beta\in S$ such that $\alpha$
and $\beta$ have the same minimal polynomial over $\mathbb{Q}$. Moreover,
$\alpha=\sigma(\beta^{\prime})$, where $\beta^{\prime}$ is a conjugate of
$\beta$ over $\mathbb{Q}$ and
$\sigma:\mathbb{Q}(\beta^{\prime})\to\mathbb{Q}_{p}(\alpha)$ is a
$\mathbb{Q}$-embedding (which corresponds to some prime ideal $P$ of
$O_{\mathbb{Q}(\beta^{\prime})}$ above $p$). By assumption, $f(\beta)$ is
integral over $\mathbb{Z}_{(p)}$; but, $f$ is also integral-valued on
$\beta^{\prime}$, even if $\beta^{\prime}$ does not lie in $S$. Hence,
$f(\alpha)=\sigma(f(\beta^{\prime}))\in
O_{\mathbb{Q}_{p}(\alpha)}\subseteq\overline{\mathbb{Z}_{p}}$.

Conversely, suppose
$f\in\textnormal{Int}_{\mathbb{Q}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
and let $\beta\in S$. For each $\mathbb{Q}$-embedding
$\sigma:\mathbb{Q}(\beta)\to\overline{\mathbb{Q}_{p}}$, we have
$\sigma(\beta)\in\Sigma_{p}(S)$. Thus, $f(\sigma(\beta))\in
O_{\mathbb{Q}_{p}(\sigma(\beta))}$. Denote the localization of
$O_{\mathbb{Q}(\beta)}$ at a prime $P$ by $O_{\mathbb{Q}(\beta),P}$. Then,

$f(\beta)\in\bigcap_{P|p}O_{\mathbb{Q}(\beta),P}=O_{\mathbb{Q}(\beta),(p)}.$

Since $O_{\mathbb{Q}(\beta),(p)}\subseteq\overline{\mathbb{Z}_{(p)}}$, we
obtain the desired result. ∎

######  Proposition 3.9.

Let $p\in\mathbb{P}$ and $S\subseteq\overline{\mathbb{Z}_{(p)}}$. Then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ is nontrivial
if and only if
$\textnormal{Int}_{\mathbb{Q}_{p}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial.

###### Proof.

Note that by Proposition 3.8, we have
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
so that

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}_{p}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})\cap\mathbb{Q}[X].$

Now, $\overline{\mathbb{Z}_{p}}\cap\mathbb{Q}_{p}=\mathbb{Z}_{p}$ and
$\mathbb{Z}_{p}\cap\mathbb{Q}=\mathbb{Z}_{(p)}$. So, if
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ is nontrivial,
then
$\textnormal{Int}_{\mathbb{Q}_{p}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial.

Conversely, suppose that there exists
$f\in\textnormal{Int}_{\mathbb{Q}_{p}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})\setminus\mathbb{Z}_{p}[X]$.
Since $\mathbb{Q}_{p}=\mathbb{Z}_{p}[\frac{1}{p}]$, we may write

$f(X)=\frac{F(X)}{p^{n}}=\frac{\sum_{i=0}^{d}\alpha_{i}X^{i}}{p^{n}}$

where $F(X)=\sum_{i}\alpha_{i}X^{i}\in\mathbb{Z}_{p}[X]$ and $n\geq 1$. For
each $i=0,\ldots,d$, choose $a_{i}\in\mathbb{Z}_{(p)}$ such that
$v_{p}(a_{i}-\alpha_{i})\geq n$. We consider the polynomial

$g(X)=\frac{G(X)}{p^{n}}=\frac{\sum_{i=0}^{d}a_{i}X^{i}}{p^{n}}.$

By construction, $g-f\in\mathbb{Z}_{p}[X]$, so
$g\in\mathbb{Q}[X]\setminus\mathbb{Z}_{(p)}[X]$. We will show that
$g\in\textnormal{Int}_{\mathbb{Q}}(\Sigma_{p}(S),\overline{\mathbb{Z}_{p}})$.
Let $s\in\Sigma_{p}(S)$. Then,

$g(s)=\frac{G(s)-F(s)+F(s)}{p^{n}}=\frac{G(s)-F(s)}{p^{n}}+f(s)$

For each $0\leq i\leq d$, we have $v_{p}(a_{i}-\alpha_{i})+iv(s)\geq n$, so
$\frac{G(s)-F(s)}{p^{n}}$ has non-negative valuation. Thus,
$g(s)\in\overline{\mathbb{Z}}_{p}$. ∎

### Equivalence of 3.2(3) and 3.2(4)

For a prime $p$, we denote by $G_{p}$ the absolute Galois group of
$\mathbb{Q}_{p}$, that is,
$G_{p}=\text{Gal}(\overline{\mathbb{Q}_{p}}/\mathbb{Q}_{p})$. For a subset $S$
of $\overline{\mathbb{Z}_{p}}$ we set

$G_{p}(S)=\\{\sigma(s)\mid s\in S,\sigma\in G_{p}\\},$

which is a subset of $\overline{\mathbb{Z}_{p}}$ consisting of the union of
all the conjugates of the elements in $S$. Note that we have the equality
$\textnormal{Int}_{\mathbb{Q}_{p}}(S,\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}_{p}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$.

######  Proposition 3.10.

Let $p\in\mathbb{P}$ and let $S\subseteq\overline{\mathbb{Z}_{p}}$. The
following are equivalent.

  1. (1)

$\textnormal{Int}_{\mathbb{Q}_{p}}(S,\overline{\mathbb{Z}_{p}})$ is
nontrivial.

  2. (2)

$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial.

  3. (3)

$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(S,\overline{\mathbb{Z}_{p}})$ is
nontrivial.

###### Proof.

(1) $\Leftrightarrow$ (2) By [22, Lemma 2.20], the integral closure of
$\textnormal{Int}_{\mathbb{Q}_{p}}(S,\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}_{p}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$
in $\overline{\mathbb{Q}_{p}}(X)$ is equal to
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$.
In particular,
$\textnormal{Int}_{\mathbb{Q}_{p}}(S,\overline{\mathbb{Z}_{p}})$ strictly
contains $\mathbb{Z}_{p}[X]$ if and only if
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$
strictly contains $\overline{\mathbb{Z}_{p}}[X]$.

(2) $\Rightarrow$ (3) Since $S\subseteq G_{p}(S)$, we have
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(G_{p}(S),\overline{\mathbb{Z}_{p}})\subseteq\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(S,\overline{\mathbb{Z}_{p}})$.
So, (2) implies (3).

(3) $\Rightarrow$ (2) Assume
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(S,\overline{\mathbb{Z}_{p}})$ is
nontrivial. By Theorem 2.6, there exists a finite set $T\subseteq S$ and
$\delta>0$ such that for each $s\in S$, there exists $t\in T$ with
$v(s-t)\geq\delta$. Note that since $T$ is finite, so is the set $G_{p}(T)$ of
all images of $T$ under $G_{p}$. Let $\sigma(s)\in G_{p}(S)$, where $\sigma\in
G_{p}$ and $s\in S$. Then, there exists $t\in T$ with $v(s-t)\geq\delta$.
Since $\overline{\mathbb{Q}}_{p}$ is Henselian, we have
$v(\sigma(\alpha))=v(\alpha)$ for all $\alpha\in\overline{\mathbb{Q}}_{p}$.
So,

$v(\sigma(s)-\sigma(t))=v(\sigma(s-t))=v(s-t)\geq\delta.$

Thus, condition (2) of Theorem 2.6 holds for $G_{p}(S)$, and we conclude that
$\textnormal{Int}_{\overline{\mathbb{Q}}_{p}}(G_{p}(S),\overline{\mathbb{Z}_{p}})$
is nontrivial. ∎

##  4 Global Conditions and Examples

Here, we examine global conditions that can be used to decide when
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial. Our
first theorem of this type relates
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to some distinguished
polynomials that lie in the rings
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$.

######  Definition 4.1.

For each $p\in\mathbb{P}$ and each positive integer $n\geq 1$, let

$\Psi_{p,n}(X)=(X^{p^{n}}-X)(X^{p^{n-1}}-X)\cdots(X^{p}-X).$

Recall that when $\alpha\in\overline{\mathbb{Z}}$, the index of $\alpha$ is
$\iota_{\alpha}=[O_{\mathbb{Q}(\alpha)}:\mathbb{Z}[\alpha]]$. The lemma below
summarizes the basic relationships among $\Psi_{p,n}$, $\iota_{\alpha}$, and
integer-valued polynomials.

######  Lemma 4.2.

Let $n\in\mathbb{N}$ and $p\in\mathbb{P}$.

  1. (1)

Modulo $p$, $\Psi_{p,n}$ is the monic least common multiple of all polynomials
in $\mathbb{F}_{p}[X]$ of degree at most $n$.

  2. (2)

For each prime $p$,
$\Psi_{p,n}/p\in\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})\setminus\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n+1})$.

  3. (3)

Let $\alpha\in\overline{\mathbb{Z}}$ and let
$f\in\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\overline{\mathbb{Z}})$. If
$\deg f<[\mathbb{Q}(\alpha):\mathbb{Q}]$, then
$\iota_{\alpha}f\in\mathbb{Z}[X]$. In particular, $f\in\mathbb{Z}_{(q)}[X]$
for each prime $q$ not dividing $\iota_{\alpha}$.

###### Proof.

Part (1) is shown in [2, Theorem 3]. For (2), if $\alpha\in\mathcal{A}_{n}$
with minimal polynomial $m_{\alpha}(X)$, then $m_{\alpha}\mid\Psi_{p,n}$
modulo $p$. Hence, $\Psi_{p,n}(\alpha)\in pO_{\mathbb{Q}(\alpha)}$. To show
that $\Psi_{p,n}/p\notin\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n+1})$, let
$f\in\mathbb{F}_{p}[X]$ be monic and irreducible of degree $n+1$. Then,
$f\nmid\Psi_{p,n}$ mod $p$. Next, let $F\in\mathbb{Z}[X]$ be monic,
irreducible, and such that $F\equiv f$ mod $p$ (the existence of $F$ follows
from Perron’s Criterion for irreducibility [27, Theorem 2.2.5]). Let
$\beta\in\overline{\mathbb{Z}}$ be a root of $F$, and let
$O=O_{\mathbb{Q}(\beta)}$; note that $\beta\in\mathcal{A}_{n+1}$. Suppose that
$\Psi_{p,n}(\beta)/p\in\overline{\mathbb{Z}}$. Then, $\Psi_{p,n}(\beta)\in
pO$, and hence $\Psi_{p,n}(\beta)\equiv 0$ mod $p$. Since the image of $\beta$
in $O/pO$ is a root of $f$, this means that $f\mid\Psi_{p,n}$ mod $p$, which
is a contradiction.

For (3), let $N=[\mathbb{Q}(\alpha):\mathbb{Q}]$ and assume that
$f(X)=\sum_{i=0}^{d}a_{i}X^{i}$, where each $a_{i}\in\mathbb{Q}$ and $d<N$.
Then,
$f(\alpha)\in\overline{\mathbb{Z}}\cap\mathbb{Q}(\alpha)=O_{\mathbb{Q}(\alpha)}$.
Now, $\iota_{\alpha}O_{\mathbb{Q}(\alpha)}\subseteq\mathbb{Z}[\alpha]$, so
$\sum_{i=0}^{d}\iota_{\alpha}a_{i}\alpha^{i}\in\mathbb{Z}[\alpha]$. Since
$\mathbb{Z}[\alpha]$ is a free $\mathbb{Z}$-module with basis
$1,\alpha,\ldots,\alpha^{N-1}$ and $d<N$, each
$\iota_{\alpha}a_{i}\in\mathbb{Z}$, as required. ∎

In the examples later in this section, we will sometimes use extensions of
$p$-adic valuations to $\overline{\mathbb{Q}}$ in order to prove that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial. The next
two lemmas demonstrate how to do this.

######  Lemma 4.3.

Let $S\subseteq\overline{\mathbb{Z}}$.

  1. (1)

Let $\sigma\in G=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ and $U$ be a
valuation of $\overline{\mathbb{Q}}$ extending $\mathbb{Z}_{(p)}$ for some
$p\in\mathbb{P}$. Then
$\textnormal{Int}_{\mathbb{Q}}(S,U)=\textnormal{Int}_{\mathbb{Q}}(\sigma(S),\sigma(U))$.
In particular, if $S$ is $G$-invariant (i.e., $\sigma(S)=S,\forall\sigma\in
G$) and $U,U^{\prime}$ are two valuation domains extending $\mathbb{Z}_{(p)}$
then
$\textnormal{Int}_{\mathbb{Q}}(S,U)=\textnormal{Int}_{\mathbb{Q}}(S,U^{\prime})$
and so
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(S,U)$
for any valuation domain $U$ extending $\mathbb{Z}_{(p)}$.

  2. (2)

Let $p\in\mathbb{P}$. If there exists an extension $u$ of $v_{p}$ to
$\overline{\mathbb{Q}}$ with associated valuation domain $U$ such that
$\textnormal{Int}_{\overline{\mathbb{Q}}}(S,U)$ is trivial, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$.

  3. (3)

If, for every $p\in\mathbb{P}$, there exists an extension $u_{p}$ of $v_{p}$
to $\overline{\mathbb{Q}}$ with associated valuation domain $U_{p}$ such that
$\textnormal{Int}_{\overline{\mathbb{Q}}}(S,U_{p})$ is trivial, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

###### Proof.

(1) Let $f\in\textnormal{Int}_{\mathbb{Q}}(S,U)$. Since $f\in\mathbb{Q}[X]$,
$\sigma(f(s))=f(\sigma(s))\in\sigma(U)$ for each $s\in S$, so
$f\in\textnormal{Int}_{\mathbb{Q}}(\sigma(S),\sigma(U))$. The other
containment is proved in the same way by applying $\sigma^{-1}$ to the
previous relation. Now, suppose that $S$ is $G$-invariant. The valuation
domains $U,U^{\prime}$ are conjugate by some $\sigma\in G$, so
$U^{\prime}=\sigma(U)$ and
$\textnormal{Int}_{\mathbb{Q}}(S,U)=\textnormal{Int}_{\mathbb{Q}}(S,U^{\prime})$.
For the final claim, note that
$\overline{\mathbb{Z}_{(p)}}=\bigcap_{u|v_{p}}U$, where the the intersection
is taken over all extensions $u$ of $v_{p}$ to $\overline{\mathbb{Q}}$, and
$U$ is the valuation domain corresponding to $u$. We have,

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\bigcap_{u|v_{p}}\textnormal{Int}_{\mathbb{Q}}(S,U)$
(4.4)

and so if $S$ is $G$-invariant, we have
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(S,U)$
for each $u$ extending $v_{p}$ to $\overline{\mathbb{Q}}$.

(2) Assume the desired $u$ and $U$ exist. Then,
$\textnormal{Int}_{\overline{\mathbb{Q}}}(S,U)=U[X]$, so

$\textnormal{Int}_{\mathbb{Q}}(S,U)=\textnormal{Int}_{\overline{\mathbb{Q}}}(S,U)\cap\mathbb{Q}[X]=\mathbb{Z}_{(p)}[X].$

By (4.4), in order for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ to be trivial
it is sufficient that $\textnormal{Int}_{\mathbb{Q}}(S,U)=\mathbb{Z}_{(p)}[X]$
for only one extension $u$ of $v_{p}$. As shown above, this holds for $u$, so
we conclude that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$.

(3) By (2), we have
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$
for each $p$, and then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$ by
(3.5). ∎

######  Remark 4.5.

We stress that the condition appearing in item (2) of Lemma 4.3 is sufficient
but not necessary, if $S$ is not $G$-invariant ((1) of the same Lemma). We
will see in Example 5.9 (due to Gilmer, [12, Example 14]) and in Example 5.14
(due to Chabert, [5, Example 6.2]) that there are subsets
$S\subset\overline{\mathbb{Z}_{(p)}}$ such that
$\textnormal{Int}_{\mathbb{Q}}(S,U)$ is non trivial for each extension $U$ of
$\mathbb{Z}_{(p)}$ to $\overline{\mathbb{Q}}$ but their intersection
$\bigcap_{U|\mathbb{Z}_{(p)}}\textnormal{Int}_{\mathbb{Q}}(S,U)=\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$
(4.4) is trivial (see Propositions 5.13 and 5.15).

######  Lemma 4.6.

Let $S=\\{s_{i}\\}_{i\in\mathbb{N}}\subseteq\overline{\mathbb{Z}}$. Assume
that $p\in\mathbb{P}$ and $u$ is an extension of $v_{p}$ to
$\overline{\mathbb{Q}}$ such that one of the following two conditions holds:

  1. (i)

$\\{u(s_{i})\\}_{i\in\mathbb{N}}$ is a strictly decreasing sequence with limit
0.

  2. (ii)

$u(s_{i}-s_{j})=0$ for all distinct $i,j\in\mathbb{N}$.

Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$.
Moreover, if for every $p\in\mathbb{P}$ there exists an extension $u$ of
$v_{p}$ satisfying either (i) or (ii), then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

###### Proof.

Let $U\subseteq\overline{\mathbb{Q}}$ be the valuation domain associated to
$u$, and let $M$ be the maximal ideal of $U$. If (i) holds, then with respect
to $u$, $S$ is a pseudo-divergent sequence with breadth ideal $M$; and if (ii)
holds, then $S$ is pseudo-stationary with respect to $u$, and the breadth
ideal is $U$. In either case, $\textnormal{Int}_{\overline{\mathbb{Q}}}(S,U)$
is trivial by Theorem 2.6. By Lemma 4.3,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$,
and if this is true for every prime $p$, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial by (3.5).
∎

Next, we will explore examples of unbounded sets
$S\subseteq\overline{\mathbb{Z}}$ such that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial. Later in
Sections 5 and 6, we construct unbounded sets for which
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial. As
demonstrated in Example 1.3, the indices of algebraic integers can be used to
give a sufficient condition for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to be trivial. We now
restate that example, and give a slight variation that does not require each
index $\iota_{s}$ to be 1.

######  Proposition 4.7.

Let $S\subseteq\overline{\mathbb{Z}}$.

  1. (1)

Assume that $S$ contains a sequence $\\{s_{i}\\}_{i\in\mathbb{N}}\subseteq S$
of unbounded degree and such that $\iota_{s_{i}}=1$ for all $i\in\mathbb{N}$.
Then, $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

  2. (2)

Assume that for each integer $n\geq 1$, there exists a finite subset
$\\{s_{1},\ldots,s_{m}\\}\subseteq S$ such that
$[\mathbb{Q}(s_{i}):\mathbb{Q}]>n$ for each $1\leq i\leq m$ and
$\gcd(\iota_{s_{1}},\ldots,\iota_{s_{m}})=1$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

###### Proof.

Part (1) is Example 1.3. For (2), let
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ of degree $d$ and
let $s_{1},\ldots,s_{m}$ be the elements of $S$ with the stated properties.
For each $i$, we have $[\mathbb{Q}(s_{i}):\mathbb{Q}]>d$, so
$\iota_{s_{i}}f\in\mathbb{Z}[X]$ by Lemma 4.2(3). The condition
$\gcd(\iota_{s_{1}},\ldots,\iota_{s_{m}})=1$ then implies that
$f\in\mathbb{Z}[X]$. ∎

In Example 4.11, we will show that the conditions on indices in Proposition
4.7 are not necessary for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to be trivial. This
example requires Dedekind’s Index Theorem, which describes the primes $p$ such
that $p\mid\iota_{\alpha}$. For this theorem, we will follow the treatment
given in [10]. An equivalent statement is available in [9, Theorem 6.1.4].

######  Theorem 4.8.

(Dedekind Index Theorem) Let $\alpha$ be an algebraic integer with minimal
polynomial $f\in\mathbb{Z}[X]$. For a prime $p$, use a bar to denote reduction
mod $p$. Factor $f$ mod $p$ as

$\overline{f}=\overline{\pi_{1}}^{e_{1}}\cdots\overline{\pi_{k}}^{e_{k}},$

where each $\overline{\pi_{i}}\in\mathbb{F}_{p}[X]$ is monic and irreducible,
and each $e_{i}\geq 1$. For each $i$, let $\pi_{i}$ be a monic lift of
$\overline{\pi_{i}}$ to $\mathbb{Z}[X]$ and let $F\in\mathbb{Z}[X]$ be such
that

$f=\pi_{1}^{e_{1}}\cdots\pi_{k}^{e_{k}}+pF.$

Then, $p\mid\iota_{\alpha}$ if and only if
$\overline{\pi_{j}}\mid\overline{F}$ in $\mathbb{F}_{p}[X]$ for some $1\leq
j\leq k$ such that $e_{j}\geq 2$.

######  Definition 4.9.

Let $c,n\in\mathbb{Z}$ be such that $c\geq 2$ and $n\geq 2$. We define

$f_{c,n}(X):=X^{n}+c^{3}X^{n-1}+c^{2}$

and $\iota_{c,n}:=\iota_{\alpha}$, where $\alpha$ is a root of $f_{c,n}$.

######  Lemma 4.10.

Let $c,n\in\mathbb{Z}$ be such that $c\geq 2$ and $n\geq 2$.

  1. (1)

The polynomial $f_{c,n}$ is irreducible over $\mathbb{Q}$.

  2. (2)

Let $\alpha\in\overline{\mathbb{Q}}$ be a root of $f_{c,n}$ and $u$ an
extension of the $p$-adic valuation $v_{p}$ to $\overline{\mathbb{Q}}$. If $p$
is a prime and $p\mid c$, then $u(\alpha)=2v_{p}(c)/n$.

  3. (3)

If $p$ is a prime and $p\mid c$, then $p\mid\iota_{c,n}$.

  4. (4)

If $q$ is a prime and $q\nmid c$ but $q\mid n$, then $q\nmid\iota_{c,n}$.

  5. (5)

If $m\in\mathbb{N}$ and $q$ is a prime such that $q\nmid c$, then there exists
a prime $l>m$ such that $q\nmid\iota_{c,l}$.

###### Proof.

(1) This is true by from Perron’s Criterion for irreducibility [27, Theorem
2.2.5].

(2) We have $\alpha^{n}=-c^{2}(c\alpha^{n-1}+1)$. If $p$ divides $c$, then
$u(\alpha^{n})=v_{p}(c^{2})$, and so $u(\alpha)=2v_{p}(c)/n$.

(3) Assume $p\mid c$, and use a bar to denote reduction mod $p$. Then,
$\overline{f_{c,n}}(X)=X^{n}$, and $f_{c,n}(X)=X^{n}+pF_{c,n}(X)$, where
$F_{c,n}(X)=\tfrac{c^{3}}{p}X^{n-1}+\tfrac{c^{2}}{p}$. Note that
$F_{c,n}\in\mathbb{Z}[x]$ and $\overline{F}=0$. By Theorem 4.8,
$p\mid\iota_{p,n}$.

(4) Assume $q\nmid c$ and $q\mid n$. When reduced mod $q$, the derivative of
$f_{c,n}(X)$ equals $(n-1)c^{3}X^{n-2}$, which is either constant or has 0 as
its only root. Since $q\nmid f_{c,n}(0)$, we see that $f_{c,n}$ mod $q$ is
separable. Hence, when applying the Dedekind Index Theorem, each exponent
$e_{j}$ will equal 1. We conclude that $q\nmid\iota_{c,n}$.

(5) We have $f_{c,n}^{\prime}(X)=X^{n-2}((n-1)c^{3}+nX)$. By the Dirichlet’s
theorem on arithmetic progressions, there exists a prime $l$ of the form
$qk+1$, $l>m$; in particular, $v_{q}(l-1)>0$ and so, modulo $q$, the only root
of $f_{c,l}^{\prime}(X)$ is $0$ (note that $v_{q}(l)=0$) which is not a root
of $f_{c,l}$ modulo $q$ because we are assuming $v_{q}(c)=0$. Hence, $f_{c,l}$
is separable modulo $q$ so that $q\nmid\iota_{c,l}$ by Theorem 4.8. ∎

######  Example 4.11.

It is possible that the indices of the elements of an unbounded sequence
$S=\\{s_{i}\\}_{i\in\mathbb{N}}\subseteq\overline{\mathbb{Z}}$ can all share a
common prime factor, yet the corresponding ring of integral-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

Fix a prime $p$, and let $q_{1}<q_{2}<\ldots$ be an ordering of all the primes
not equal to $p$. For each $i\in\mathbb{N}$, let $f_{p,q_{i}}$ be as in
Definition 4.9, and let $s_{i}\in\overline{\mathbb{Q}}$ be a root of
$f_{p,q_{i}}$. Take $S=\\{s_{i}\\}_{i\in\mathbb{N}}$. Then,
$\\{[\mathbb{Q}(s_{i}):\mathbb{Q}]\\}_{i\in\mathbb{N}}$ is unbounded, and by
Lemma 4.10 each $\iota_{s_{i}}$ is divisible by $p$ and not divisible by
$q_{i}$. We will show that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

By Lemma 4.10(2), $u_{p}(s_{i})=2/q_{i}$ for each $i$, so
$\\{u_{p}(s_{i})\\}_{i\in\mathbb{N}}$ is a strictly decreasing sequence with
limit $0$. By Lemma 4.6,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$.

Let now $q$ be a prime different from $p$. Given
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, by Lemma
4.10(5), there exists a prime $q_{i}$ such that
$[\mathbb{Q}(s_{i}):\mathbb{Q}]=q_{i}>\deg(f)$ and $q\nmid\iota_{s_{i}}$. By
Lemma 4.2(3), $\iota_{s_{i}}\cdot f\in\mathbb{Z}[X]$ so that $f$ belongs to
$\mathbb{Z}_{(q)}[X]$. This argument shows that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(q)}})=\mathbb{Z}_{(q)}[X]$
for every prime $q\not=p$.

By (3.5), we conclude that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]$.

In the following examples, for each $p\in\mathbb{P}$, we fix an extension
$u_{p}$ of $v_{p}$ to $\overline{\mathbb{Q}}$ with corresponding valuation
domain $U_{p}$ having maximal ideal $\overline{M_{p}}$. For each positive
integer $n$, let $\zeta_{n}$ be a primitive $n^{\text{th}}$ root of unity.
Sequences of these algebraic integers can lead to interesting examples of sets
$S$ for which $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is
trivial. In the next three examples, we will construct a sequence
$S\subseteq\overline{\mathbb{Z}}$ that is pseudo-divergent with
$\textnormal{Br}(S)=\overline{M_{p}}$ with respect to $u_{p}$ for a single
prime $p$, but pseudo-stationary with $\textnormal{Br}(S)=U_{q}$ at all other
primes $q$; a sequence $S$ that is pseudo-stationary with
$\textnormal{Br}(S)=U_{p}$ with respect to all primes; and a sequence $S$
which is eventually pseudo-divergent with
$\textnormal{Br}(S)=\overline{M_{p}}$ with respect to each prime.

######  Example 4.12.

Let $p\in\mathbb{P}$ and take $S=\\{\zeta_{p^{k}}\\}_{k\in\mathbb{N}}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial by
Proposition 4.7(1). We claim that $S$ is a pseudo-divergent sequence with
$\textnormal{Br}(S)=M_{p}$ with respect to $u_{p}$ and it is pseudo-stationary
with $\textnormal{Br}(S)=U_{q}$ with respect to $u_{q}$ for every prime
$q\not=p$.

Whenever $j<k$, we have
$\zeta_{p^{k}}-\zeta_{p^{j}}=\zeta_{p^{k}}(1-\zeta_{p^{k}}^{m})$, where
$m=p^{k-j}-1$. Since $m$ and $p^{k}$ are coprime, $\zeta_{p^{k}}^{m}$ is also
a primitive $p^{k}$-th root of unity. So,

$u_{q}(\zeta_{p^{k}}-\zeta_{p^{j}})=u_{q}(1-\zeta_{p^{k}}^{m})=u_{q}(1-\zeta_{p^{k}}).$

We recall that the prime $p$ is totally ramified in
$\mathbb{Q}(\zeta_{p^{k}})$ for every $k\in\mathbb{N}$ (see for example [16,
Theorem 26]), and $u_{p}(1-\zeta_{p^{k}})=(p^{k-1}(p-1))^{-1}$ by [29, Chapter
2, p. 9]. Moreover, if $q$ is a prime different from $p$ then
$u_{q}(1-\zeta_{p^{k}})=0$. These calculations prove the claims about $S$.

######  Example 4.13.

Let $S=\\{\zeta_{p}\\}_{p\in\mathbb{P}}$. Once again,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial by
Proposition 4.7. We will show that $S$ is pseudo-stationary with
$\textnormal{Br}(E)=U_{p}$ with respect to $u_{p}$ for every prime $p$.

Given primes $p<q$, we have
$\zeta_{q}-\zeta_{p}=\zeta_{q}(1-\zeta_{pq}^{q-p})$, and $\zeta_{pq}^{q-p}$ is
a primitive $pq$-th root of unity because $q-p$ is coprime to $pq$. So, for
any prime $l$,

$u_{l}(\zeta_{q}-\zeta_{p})=u_{l}(1-\zeta_{pq}^{q-p})=u_{l}(1-\zeta_{pq}).$

By [29, Proposition 2.8], $1-\zeta_{pq}$ is a unit of
$\mathbb{Z}[\zeta_{pq}]$. Hence, $u_{l}(\zeta_{q}-\zeta_{p})=0$, and so for
every prime $l$, $S$ is pseudo-stationary with respect to $u_{l}$ with breadth
ideal $U_{l}$.

Note that in both Example 4.12 and Example 4.13, one could use Lemma 4.6
instead of Proposition 4.7 to conclude that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

It is not possible for a sequence
$S=\\{s_{k}\\}_{k\in\mathbb{N}}\subseteq\overline{\mathbb{Z}}$ to be pseudo-
divergent with respect to all primes. If that were the case, then for all
$i\neq j$, the difference $s_{i}-s_{j}$ would have finite positive value at
each integral prime, which is impossible. However, we can construct a sequence
$S$ that is eventually pseudo-divergent with respect to each prime.

######  Example 4.14.

Let $\mathbb{P}=\\{p_{1},p_{2},\ldots\\}$. Define $s_{1}=p_{1}$,
$s_{2}=(p_{1}p_{2})^{1/2}$, $s_{3}=(p_{1}p_{2}p_{3})^{1/3}$, and in general
$s_{k}=(p_{1}\cdots p_{k})^{1/k}$ for each $k\in\mathbb{N}$. Then, for each
$n\in\mathbb{N}$,

$u_{p_{n}}(s_{k})=\begin{cases}0,&k<n\\\ \tfrac{1}{k},&k\geq n.\end{cases}$

Thus, for every prime $p$, the sequence $\\{u_{p}(s_{k})\\}_{k\in\mathbb{N}}$
eventually strictly decreases to 0. Hence, $S$ is eventually pseudo-divergent
with breadth ideal equal to $\overline{M_{p}}$ with respect to every prime
$p$. As in the prior two examples,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is trivial.

##  5 Ramification Indices and Residue Field Degrees

We first show that having bounds on both the ramification indices and residue
field degrees of elements of $S$ is sufficient for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to be nontrivial.
Similar conditions appear in [14] in a theorem which classifies the almost
Dedekind domains $D$ with finite residue fields such that
$\textnormal{Int}(D)$ is a Prüfer domain.

######  Lemma 5.1.

Let $S\subseteq\overline{\mathbb{Z}}$. Assume that there exists
$p\in\mathbb{P}$ such that $S$ has both bounded ramification indices and
bounded residue field degrees at $p$. That is, assume that there exist
$e_{0},f_{0}\in\mathbb{N}$ such that for all $s\in S$ and every prime $P_{s}$
of $O_{\mathbb{Q}(s)}$ above $p$, we have $e(P_{s}|p)\leq e_{0}$ and
$f(P_{s}|p)\leq f_{0}$. Then, both
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ and
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ are nontrivial.

###### Proof.

Fix $s\in S$, and let $P_{s}$ be a prime of $O_{\mathbb{Q}(s)}$ above $p$. Let
$q=p^{f_{0}!}$. Then, $O_{\mathbb{Q}(s)}/P_{s}$ is a subfield of
$\mathbb{F}_{q}$, so $X^{q}-X$ maps $O_{\mathbb{Q}(s)}$ into $P_{s}$ under
evaluation for every $s\in S$. Since all ramification indices over $p$ are
bounded above by $e_{0}$, it follows that $(X^{q}-X)^{e_{0}}$ sends
$O_{\mathbb{Q}(s)}$ into $pO_{\mathbb{Q}(s)}$. Because the values of $q$ and
$e_{0}$ are independent of $s$, this is true for all elements of $S$. Thus,
the polynomial $((X^{q}-X)^{e_{0}})/p$ is in
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$,
and both rings are nontrivial. ∎

The hypothesis of Lemma 5.1 is clearly met when $S$ is of bounded degree.
However, there exist sets $S$ of unbounded degree for which Lemma 5.1 can be
applied.

######  Example 5.2.

Let $\mathbb{Q}^{(n)}=\mathbb{Q}(\mathcal{A}_{n})$ be the compositum in
$\overline{\mathbb{Q}}$ of all number fields of degree bounded by $n$ and let
$O_{\mathbb{Q}^{(n)}}$ be the ring of integers of $\mathbb{Q}^{(n)}$. It is
known that there exists a global bound on both ramification indexes and
residue field degrees of valuations of $\mathbb{Q}^{(n)}$. That is, there
exists $N\in\mathbb{N}$ such that if $u_{p}$ is a valuation of
$\mathbb{Q}^{(n)}$ extending some $v_{p}$, $p\in\mathbb{P}$, then
$e(u_{p}|v_{p})\leq N$ and $f(u_{p}|v_{p})\leq N$ [3, Proposition 4.5.3, p.
118]; we stress that $N$ is independent from $p$. So, any subset $S\subseteq
O_{\mathbb{Q}^{(n)}}$ meets the conditions of Lemma 5.1, and hence
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial.
Furthermore, subsets of $O_{\mathbb{Q}^{(n)}}$ may have unbounded degree. For
an explicit example, let $\mathbb{P}=\\{p_{k}\\}_{k\in\mathbb{N}}$. For each
$k\in\mathbb{N}$, let $s_{k}=\sum_{i=1}^{k}\sqrt{p_{i}}$, and take
$S=\\{s_{k}\\}_{k\in\mathbb{N}}$. Then, $S\subseteq O_{\mathbb{Q}^{(2)}}$ and
for each $k$, $[\mathbb{Q}(s_{k}):\mathbb{Q}]=2^{k}$.

As mentioned earlier, the double-boundedness condition of Lemma 5.1 was used
in [14] to classify the almost Dedekind domains $D$ with finite residue fields
such that $\textnormal{Int}(D)$ is Prüfer. We do not know whether this
condition guarantees that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is Prüfer. However,
we can prove that this is the case when $S\subseteq O_{\mathbb{Q}^{(n)}}$.

######  Proposition 5.3.

Let $n\in\mathbb{N}$. Define $\mathbb{Q}^{(n)}$ and $O_{\mathbb{Q}^{(n)}}$ as
in Example 5.2, and let $S\subseteq O_{\mathbb{Q}^{(n)}}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is a Prüfer domain.

###### Proof.

For readability, let $K=\mathbb{Q}^{(n)}$ and $O=O_{\mathbb{Q}^{(n)}}$. As
usual, we let $\textnormal{Int}(O)=\\{f\in K[X]\mid f(O)\subseteq O\\}$ and
$\textnormal{Int}_{\mathbb{Q}}(O)=\textnormal{Int}(O)\cap\mathbb{Q}[X]$. Then,

$\textnormal{Int}_{\mathbb{Q}}(O)=\textnormal{Int}_{\mathbb{Q}}(O,\overline{\mathbb{Z}})\subseteq\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}}).$

Since overrings of Prüfer domains are Prüfer, it suffices to prove that
$\textnormal{Int}_{\mathbb{Q}}(O)$ is Prüfer. By the properties of
ramification indexes and residue field degree mentioned in Example 5.2, it
follows that $O$ is an almost Dedekind domain with finite residue fields
satisfying the double-boundedness condition of [14, Theorem 2.5]. Hence,
$\textnormal{Int}(O)$ is Prüfer. By [22, Lemma 2.20], $\textnormal{Int}(O)$ is
integral over $\textnormal{Int}_{\mathbb{Q}}(O)$, so
$\textnormal{Int}_{\mathbb{Q}}(O)$ is Prüfer by [11, Theorem 22.4]. ∎

######  Remark 5.4.

For each $n\in\mathbb{N}$, we have
$\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(n)}})\subseteq\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n})$.
As in (1.1), we have a descending chain of Prüfer domains:

$\ldots\subseteq\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(n+1)}})\subseteq\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(n)}})\subseteq\ldots\subseteq\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(1)}})=\textnormal{Int}(\mathbb{Z}).$

It is not known whether each containment in this train is strict. Regardless,
as was shown in the Introduction for the rings
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n}),n\in\mathbb{N}$, we have
$\bigcap_{n\in\mathbb{N}}\textnormal{Int}_{\mathbb{Q}}(O_{\mathbb{Q}^{(n)}})=\textnormal{Int}_{\mathbb{Q}}(\overline{\mathbb{Z}})=\mathbb{Z}[X]$.

While bounds on both ramification indices and residue field degrees are
sufficient to conclude that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial by
Lemma 5.1, neither condition is necessary for nontriviality. The remainder of
this section is devoted to examples that illustrate this. We also consider how
to interpret the presence of either unbounded residue field degrees
(respectively, ramification degrees) or in terms of pseudo-stationary
(respectively, pseudo-divergent) sequences in suitable images of $S$ in
$\overline{\mathbb{Z}_{p}}$.

######  Example 5.5.

Let $S\subseteq\overline{\mathbb{Z}}$ and $p\in\mathbb{P}$. Let $I$ be the set
of all ramification indices $e(P_{s}|p)$, where $s$ runs through $S$ and
$P_{s}$ runs through the prime ideals of $O_{\mathbb{Q}(s)}$ above $p$. We
give two examples to show that if $I$ is unbounded, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ may or may not be
trivial.

First, let $S$ be as in Example 1.6. The associated set of ramification
indices is $I=\\{2^{k}\mid k\in\mathbb{N}\\}$, which is unbounded. However,
$X^{2}/p\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, so the
ring of integral-valued polynomials is nontrivial. By contrast, if $S$ is the
set defined in Example 4.14, then $S$ still exhibits unbounded ramification
indices, but $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is
trivial.

As for the situation of ramification indices, we can show that the bound of
the residue field degrees is not necessary for the
$\textnormal{Int}_{\mathbb{Q}}$-ring to be nontrivial.

######  Example 5.6.

For a fixed prime $p\in\mathbb{P}$, we consider the set $S$ of all $m$-th root
of unity in $\overline{\mathbb{Z}}$ with $(p,m)=1$; we recall that $p$ is not
ramified in $\mathbb{Q}(\zeta_{m})$ [16, Corollary of Theorem 26]. The set of
residue field degrees of the set of prime ideals of the finite extensions
generated by the elements of the set $\\{p\cdot\zeta_{m}\mid
m\in\mathbb{N},(p,m)=1\\}$ is unbounded but $u_{p}(p\cdot\zeta_{m})>0$ for
each of the relevant $m$’s. Clearly,
$\frac{X}{p}\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$. The
important condition here is that the residue set $S/\overline{P}$, for each
prime ideal $\overline{P}\subset\overline{\mathbb{Z}}$ above $p$, is finite.

######  Notation 5.7.

For $p\in\mathbb{P}$, we denote by $\mathcal{P}_{p}$ the set of prime ideals
$\overline{P}$ of $\overline{\mathbb{Z}}$ lying above $p$.

######  Lemma 5.8.

Let $S\subseteq\overline{\mathbb{Z}_{(p)}}$ for some $p\in\mathbb{P}$. Suppose
there exists $n\in\mathbb{N}$ such that either one of the following conditions
is satisfied:

  * (i)

there exists some $\overline{P}\in\mathcal{P}_{p}$ such that
$\\#S/\overline{P}>n$.

  * (ii)

there exist some $\overline{P}\in\mathcal{P}_{p}$ and $s\in S$ such that
$0<v_{\overline{P}}(s)<\frac{1}{n}$.

Then, given
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})\setminus\mathbb{Z}_{(p)}[X]$,
we have $\deg(f)>n$. In particular, if for every $n\in\mathbb{N}$ either one
of these conditions hold, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\mathbb{Z}_{(p)}[X]$.

###### Proof.

Let $f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ be of
degree $d\leq n$ and suppose that $f=\frac{g}{p}$ for some monic
$g\in\mathbb{Z}_{(p)}[X]$ (see the arguments of the proof of Proposition 6.4
for the monic assumption). (i) By assumption, there exists a prime ideal
$\overline{P}\in\mathcal{P}_{p}$ such that $\\#S/\overline{P}>n$, that is,
there exist $s_{0},\ldots,s_{n}\in S$ such that
$u_{\overline{P}}(s_{i}-s_{j})=0$ for each $0\leq i<j\leq n$. In particular,
$\overline{s_{i}}=s_{i}\pmod{\overline{P}}$ are $n+1$ distinct elements of
$\overline{\mathbb{F}_{p}}$. Then $g(s_{i})\equiv 0\pmod{\overline{P}}$, for
$i=0,\ldots,n$, that is, $\overline{g}(\overline{s_{i}})=0$ in
$\overline{\mathbb{F}_{p}}$, where $\overline{g}\in\mathbb{F}_{p}[X]$ is the
reduction modulo $p$ of $g$. It follows that
$\overline{g}\in\mathbb{F}_{p}[X]$ is a polynomial of degree $\leq n$ which
has $n+1$ distinct roots in $\overline{\mathbb{F}_{p}}$, a contradiction.

(ii) By assumption, there exists a prime ideal
$\overline{P}\in\mathcal{P}_{p}$ and $s\in S$ such that
$v_{\overline{P}}(s)<\frac{1}{n}$. Let $g(X)=\prod_{i=1}^{d}(X-\alpha_{i})$
over $\overline{\mathbb{Q}}$ and let $i\in\\{1,\ldots,n\\}$. If
$v_{\overline{P}}(\alpha_{i})$ is not $0$ then it is equal to $\frac{a}{n}>0$,
for some $a\in\mathbb{N}$ (not necessarily coprime with $n$) since
$e(\overline{P}\cap O_{\mathbb{Q}(\alpha_{i})}\mid
p)\leq[\mathbb{Q}(\alpha_{i}):\mathbb{Q}]\leq n$. If this second case occurs,
then $0<v_{\overline{P}}(s)<\frac{1}{n}\leq v_{\overline{P}}(\alpha_{i})$. In
either case we have:

$v_{\overline{P}}(s-\alpha_{i})=\min\\{v_{\overline{P}}(\alpha_{i}),v_{\overline{P}}(s)\\}=\begin{cases}0,&\text{
if }v_{\overline{P}}(\alpha_{i})=0\\\ v_{\overline{P}}(s),&\text{
otherwise}.\end{cases}$

Hence, $v_{\overline{P}}(g(s))\leq\sum_{i=1}^{n}v_{\overline{P}}(s)\leq
nv_{\overline{P}}(s)<1=v_{\overline{P}}(p)$ which contradicts the fact that
$f(s)\in\overline{\mathbb{Z}_{(p)}}\subset\overline{\mathbb{Z}}_{\overline{P}}$.
∎

######  Example 5.9.

For an explicit example of a subset $S$ of $\overline{\mathbb{Z}}$ for which
$S/\overline{P}$ is finite for each maximal ideal
$\overline{P}\in\mathcal{P}_{p}$ for some prime $p\in\mathbb{Z}$ but
$\\{\\#S/\overline{P}\mid\overline{P}\in\mathcal{P}_{p}\\}$ is unbounded, let
$S$ be equal to the almost Dedekind domain $D$ with finite residue fields of
[12, Example 14] (see also [4, Example VI.4.18]). In that example, $D$ is the
integral closure of $\mathbb{Z}_{(p)}$ in a suitable infinite algebraic
extension of $\mathbb{Q}$ constructed as the union of a tower of finite
algebraic extensions, so by Lemma 5.8 (see also [4, Lemma VI.4.2]) applied to
that example shows that
$\textnormal{Int}_{\mathbb{Q}}(D,\overline{\mathbb{Z}})=\textnormal{Int}_{\mathbb{Q}}(D)=\mathbb{Z}_{(p)}[X]$.

Note that in Example 5.9 there are neither pseudo-stationary sequence nor
pseudo-divergent sequences in $D$ with respect to every possible extension of
$v_{p}$ to $\overline{\mathbb{Q}}$ because $D$ is locally a DVR with finite
residue field. Nonetheless, $\textnormal{Int}_{\mathbb{Q}}(D)$ is trivial. The
reason for this is that if we work with all of the $\mathbb{Q}$-embeddings of
$\overline{\mathbb{Q}}$ into $\overline{\mathbb{Q}_{p}}$, and consider the
union of all of the images of $D$ under these embeddings, then we obtain the
same $\textnormal{Int}_{\mathbb{Q}}$-ring (Lemma 5.11) and this union contains
a pseudo-stationary sequence (see Proposition 5.13).

######  Definition 5.10.

For each $\overline{P}\in\mathcal{P}_{p}$, let $u_{\overline{P}}$ be the
valuation associated to $\overline{\mathbb{Z}}_{\overline{P}}$. By [19,
Chapter 2, §8, Theorem 8.1], for each $\overline{P}\in\mathcal{P}_{p}$, the
extension $u_{\overline{P}}$ of $v_{p}$ to $\overline{\mathbb{Q}}$ is equal to
$v_{p}\circ\tau_{\overline{P}}$, for some $\mathbb{Q}$-embedding
$\tau_{\overline{P}}:\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}_{p}}$,
where $v_{p}$ is the unique extension to $\overline{\mathbb{Q}_{p}}$ of the
$p$-adic valuation. Given $S\subseteq\overline{\mathbb{Z}}$, we define the
following subset of $\overline{\mathbb{Z}_{p}}$:

$\mathcal{S}_{p}=\bigcup_{\overline{P}\in\mathcal{P}_{p}}\tau_{\overline{P}}(S).$

The following lemma is analogous to Proposition 3.8.

######  Lemma 5.11.

Let $p\in\mathbb{P}$, $S\subseteq\overline{\mathbb{Z}_{(p)}}$ and
$\overline{P}\in\mathcal{P}_{p}$. Then, we have

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}}_{\overline{P}})=\textnormal{Int}_{\mathbb{Q}}(\tau_{\overline{P}}(S),\overline{\mathbb{Z}_{p}}).$

In particular,

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(\mathcal{S}_{p},\overline{\mathbb{Z}_{p}}).$

###### Proof.

$(\subseteq)$ Let
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}}_{\overline{P}})$
and $\tau_{\overline{P}}(s)\in\tau_{\overline{P}}(S)$, for some $s\in S$. Then
$f(s)\in\overline{\mathbb{Z}}_{\overline{P}}$, so $u_{\overline{P}}(f(s))\geq
0$. Hence,

$v_{p}\circ\tau_{\overline{P}}(f(s))=v_{p}(f(\tau_{\overline{P}}(s)))\geq 0$

so that $f(\tau_{\overline{P}}(s))\in\overline{\mathbb{Z}_{p}}$. It follows
that
$f\in\textnormal{Int}_{\mathbb{Q}}(\tau_{\overline{P}}(S),\overline{\mathbb{Z}_{p}})$.

$(\supseteq)$ Let
$f\in\textnormal{Int}_{\mathbb{Q}}(\tau_{\overline{P}}(S),\overline{\mathbb{Z}_{p}})$
and $s\in S$. By definition we have
$f(\tau_{\overline{P}}(s))\in\overline{\mathbb{Z}_{p}}$. Hence, applying
$\tau_{\overline{P}}^{-1}$ to the previous relation we get
$f(s)\in\overline{\mathbb{Z}}_{\overline{P}}$, and so
$f\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}}_{\overline{P}})$.

The final claim follows from the facts that
$\overline{\mathbb{Z}_{(p)}}=\bigcap_{\overline{P}\in\mathcal{P}_{p}}\overline{\mathbb{Z}}_{\overline{P}}$
and
$\bigcap_{\overline{P}\in\mathcal{P}_{p}}\textnormal{Int}_{\mathbb{Q}}(\tau_{\overline{P}}(S),\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}}(\mathcal{S}_{p},\overline{\mathbb{Z}_{p}})$.
∎

######  Remark 5.12.

Recall the set $\Sigma_{p}(S)$ that was introduced in Definition 3.1. One can
show that $\Sigma_{p}(S)$ is equal to the
$\text{Gal}(\overline{\mathbb{Q}_{p}}/\mathbb{Q}_{p})$-closure of
$\mathcal{S}_{p}$. Consequently, many of the results in Section 3 can be
stated in terms of $\mathcal{S}_{p}$ rather than $\Sigma_{p}(S)$. In
particular, the following equivalences hold:

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})\text{ is
nontrivial
}\stackrel{{\scriptstyle\text{Prop}3.9}}{{\Longleftrightarrow}}\textnormal{Int}_{\mathbb{Q}}(\mathcal{S}_{p},\overline{\mathbb{Z}_{p}})\text{
is nontrivial
}\stackrel{{\scriptstyle\text{Prop}3.10}}{{\Longleftrightarrow}}\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(\mathcal{S}_{p},\overline{\mathbb{Z}_{p}})\text{
is nontrivial }$

The proof of each equivalence is identical to that of the cited statement. One
can then apply Theorem 2.6 to the ring
$\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(\mathcal{S}_{p},\overline{\mathbb{Z}_{p}})$
to establish whether it is nontrivial or not.

######  Proposition 5.13.

Let $S\subseteq\overline{\mathbb{Z}_{(p)}}$ and let $\overline{M_{p}}$ be the
maximal ideal of $\overline{\mathbb{Z}_{p}}$. The following are equivalent.

  1. (1)

$\\{\\#S/\overline{P}\mid\overline{P}\in\mathcal{P}_{p}\\}$ is unbounded.

  2. (2)

$\mathcal{S}_{p}/\overline{M_{p}}$ is infinite.

  3. (3)

$\mathcal{S}_{p}$ contains a pseudo-stationary sequence $E$ (with respect to
$v_{p}$, the unique valuation of $\overline{\mathbb{Z}_{p}}$) such that
$\textnormal{Br}(E)=\overline{\mathbb{Z}_{p}}$.

Furthermore, if any of these conditions holds, then
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}_{(p)}})$ is trivial.

###### Proof.

(1) $\Rightarrow$ (2). Suppose that
$\\{\\#S/\overline{P}\mid\overline{P}\in\mathcal{P}_{p}\\}$ is unbounded, but
$\\#\mathcal{S}_{p}/\overline{M_{p}}=n$ for some $n\in\mathbb{N}$. By
assumption, there exists some $\overline{P}\in\mathcal{P}_{p}$ such that
$\\#S/\overline{P}>n$. Thus, there exist $s_{0},\ldots,s_{n}\in S$ such that
$u_{\overline{P}}(s_{i}-s_{j})=0$ for each $0\leq i<j\leq n$. Applying the
$\mathbb{Q}$-embedding $\tau_{\overline{P}}$ to each $s_{i}$, we obtain $n+1$
elements
$\tau_{\overline{P}}(s_{0}),\ldots,\tau_{\overline{P}}(s_{n})\in\mathcal{S}_{p}$
such that

$v_{p}(\tau_{\overline{P}}(s_{i})-\tau_{\overline{P}}(s_{j}))=v_{p}\circ\tau_{\overline{P}}(s_{i}-s_{j})=u_{\overline{P}}(s_{i}-s_{j})=0$

for each $0\leq i<j\leq n$. This is impossible, since some two of the
$\tau_{\overline{P}}(s_{i})$ must lie in the same residue class modulo
$\overline{M_{p}}$. We conclude that $\mathcal{S}_{p}/\overline{M_{p}}$ is
infinite.

(2) $\Rightarrow$ (1). Assume that $\mathcal{S}_{p}/\overline{M_{p}}$ is
infinite. Clearly, we have the identities

$\frac{\phantom{x}\mathcal{S}_{p}\phantom{x}}{\overline{M_{p}}}=\frac{\bigcup_{\overline{P}\in\mathcal{P}_{p}}\tau_{\overline{P}}(S)}{\overline{M_{p}}}=\bigcup_{\overline{P}\in\mathcal{P}_{p}}\frac{\tau_{\overline{P}}(S)}{\overline{M_{p}}}.$

Moreover, for each $\overline{P}\in\mathcal{P}_{p}$, $\tau_{\overline{P}}$
establishes a bijection between $S/\overline{P}$ and
$\tau_{\overline{P}}(S)/\overline{M_{p}}$. If
$\tau_{\overline{P}}(S)/\overline{M_{p}}$ is infinite for some
$\overline{P}\in\mathcal{P}_{p}$, then considering the pullback via
$\tau_{\overline{P}}^{-1}$ we have that $S/\overline{P}$ is infinite. If
instead $\tau_{\overline{P}}(S)/\overline{M_{p}}$ is finite for each
$\overline{P}\in\mathcal{P}_{p}$, since $\mathcal{S}_{p}/\overline{M_{p}}$ is
infinite, this means that there exists a sequence
$\\{\overline{P}_{n}\\}_{n\in\mathbb{N}}\subseteq\mathcal{P}_{p}$ such that
$\\#\tau_{\overline{P}_{n}}(S)/\overline{M_{p}}=d_{n}<\infty$ is diverging to
infinity. This amounts to say via the previous one-to-one correspondence that
$\\#S/\overline{P}_{n}=d_{n}$ is diverging to infinity, which means that
$\\{\\#S/\overline{P}\mid\overline{P}\in\mathcal{P}_{p}\\}$ is unbounded.

(2) $\Leftrightarrow$ (3) If $\mathcal{S}_{p}/\overline{M_{p}}$ is infinite,
then we can choose a sequence
$E=\\{t_{i}\\}_{i\in\mathbb{N}}\subseteq\mathcal{S}_{p}$ such that each
$t_{i}$ lies in a distinct residue class modulo $\overline{M_{p}}$. We then
have $v_{p}(t_{i}-t_{j})=0$ whenever $i\neq j$, so $E$ is pseudo-stationary
and $\textnormal{Br}(E)=\overline{\mathbb{Z}_{p}}$. Conversely, if
$\mathcal{S}_{p}/\overline{M_{p}}$ is finite, then no such pseudo-stationary
sequence can exist in $S$.

For the last claim, apply Theorem 2.6 and the equivalences noted in Remark
5.12. ∎

We conjecture that there is an analogous result linking unbounded ramification
indices of $S$ at $p$ to pseudo-divergent sequences in $\mathcal{S}_{p}$.
However, determining the correct conditions for such a relation to hold in
general has proven to be difficult. Nevertheless, we can provide one example
showing that such a connection is possible, at least in the case where $S$ is
a ring.

######  Example 5.14.

Following [5, Example 6.2] (see also [4, Example VI.4.17]), one may construct
an almost Dedekind domain $D$ with the following properties:

  * •

$D$ has finite residue fields.

  * •

$D$ is the integral closure of $\mathbb{Z}_{(p)}$ in a (suitably chosen)
infinite algebraic extension $K$ of $\mathbb{Q}$.

  * •

The set $\\{e(\overline{P}\cap D\mid p)\mid\overline{P}\in\mathcal{P}_{p}\\}$
is unbounded.

For this domain $D$, every valuation overring is a DVR, and every residue
field $D$ is equal to $\mathbb{F}_{p}$. These conditions guarantee that the
ramification indices of $D$ are bounded. Furthermore, $D$ cannot contain any
pseudo-divergent sequence with respect to any of its maximal prime ideals
$\overline{P}\cap D$, because $D_{\overline{P}\cap D}$ is a DVR. We will show
that
$\mathcal{D}_{p}^{\prime}:=\bigcup_{\overline{P}\in\mathcal{P}_{p}}\tau_{\overline{P}}(D_{\overline{P}\cap
D})\subseteq\overline{\mathbb{Z}_{p}}$ contains a pseudo-divergent sequence
and by an argument similar to Lemma 5.11, the ring
$\textnormal{Int}_{\mathbb{Q}}(D,\overline{\mathbb{Z}_{(p)}})$ must equal
$\mathbb{Z}_{(p)}[X]$.

######  Proposition 5.15.

Let $D\subseteq\overline{\mathbb{Z}_{(p)}}$ be a ring containing
$\mathbb{Z}_{(p)}$ and define $\mathcal{D}_{p}^{\prime}$ as above. The
following are equivalent.

  1. (1)

$\\{e(P\mid p)\mid p\in P\subset D\\}$ is unbounded.

  2. (2)

For each $n\in\mathbb{N}$, there exist $P_{n}\subset D$, $p\in P_{n}$ and
$s_{n}\in D_{P_{n}}$ such that $0<v_{P_{n}}(s_{n})<\frac{1}{n}$.

  3. (3)

There exists a pseudo-divergent sequence $E\subset\mathcal{D}_{p}^{\prime}$
with $\textnormal{Br}(E)=\overline{M_{p}}$.

Moreover, if any one of these conditions holds, then
$\textnormal{Int}_{\mathbb{Q}}(D)=\mathbb{Z}_{(p)}[X]$, and thus is trivial.

###### Proof.

(1) $\Rightarrow$ (2) If for some prime ideal $P$ of $D$ containing $p$ we
have $e(P\mid p)=\infty$, then it is easy to see that for each
$n\in\mathbb{N}$, there exists $s_{n}\in D_{P}$ with
$0<v_{P}(s_{n})<\frac{1}{n}$. Suppose instead that exists a sequence
$\\{P_{n}\\}_{n\in\mathbb{N}}$ of prime ideals of $D$ containing $p$ such that
$e(P_{n}\mid p)=e_{n}>n$ for each $n\in\mathbb{N}$. We claim that there exists
$s_{n}\in D_{P_{n}}$ such that $v_{P_{n}}(s_{n})=\frac{1}{e_{n}}<\frac{1}{n}$
which is precisely condition (2). Indeed, for each $n\in\mathbb{N}$ there
exists $\alpha_{n}\in D$ such that
$v_{P_{n}}(\alpha_{n})=\frac{a_{n}}{e_{n}}$, for some $a_{n}\in\mathbb{Z}$
coprime with $e_{n}$. Now, if $aa_{n}+be_{n}=1$ for some $a,b\in\mathbb{Z}$,
then $s_{n}=\alpha_{n}^{a}\cdot p^{b}$ is an element of $D_{P_{n}}$ with the
desired value.

(2) $\Rightarrow$ (3) Suppose (2) holds. Then for each $n\in\mathbb{N}$ we
have
$0<v_{P_{n}}(s_{n})=v_{p}(\tau_{P_{n}}(s_{n}))=v_{p}(\tilde{s}_{n})<\frac{1}{n}$,
where $\tilde{s}_{n}\in\mathcal{D}_{p}^{\prime}$. We remark that $P_{n}$ could
be the same prime ideal for infinitely many $n$’s; this corresponds to the
case $e(P\mid p)=\infty$. Suppose that for some $n\in\mathbb{N}$, we have
$\tilde{s}_{1},\ldots,\tilde{s}_{n}$ so that
$0<v_{p}(\tilde{s}_{i+1})<v_{p}(\tilde{s}_{i})$ for all $i=1,\ldots,n-1$. If
we take $m\in\mathbb{N}$ so that $0<\frac{1}{m}\leq v_{p}(\tilde{s}_{n})$,
then by assumption there exists $\tilde{s}_{n+1}\in\mathcal{D}_{p}^{\prime}$
such that $0<v_{p}(\tilde{s}_{n+1})<\frac{1}{m}\leq v_{p}(\tilde{s}_{n})$. In
this way we can extract a subsequence $E$ from
$\\{\tilde{s}_{n}\\}_{n\in\mathbb{N}}\subset\mathcal{D}_{p}^{\prime}$ which is
pseudo-divergent (i.e., $0<v_{p}(\tilde{s}_{n+1})<v_{p}(\tilde{s}_{n})$ for
each $n\in\mathbb{N}$) and has $\textnormal{Br}(E)=\overline{M_{p}}$.

(3) $\Rightarrow$ (1) Assume that (3) holds. Then, there exists a pseudo-
divergent sequence
$E=\\{\tilde{s}_{n}\\}_{n\in\mathbb{N}}\subset\mathcal{D}_{p}^{\prime}$ with
$\textnormal{Br}(E)=\overline{M_{p}}$. Hence, $v_{p}(\tilde{s}_{n})$ strictly
decreases to $0$ and, up to taking a subsequence, we may assume that
$0<v_{p}(\tilde{s}_{n})<\frac{1}{n}$. Since
$\tilde{s}_{n}=\tau_{P_{n}}(s_{n})$ for some $P_{n}\subset D$, $p\in D$ and
$s_{n}\in D_{P_{n}}$, we have
$0<v_{P_{n}}(s_{n})=v_{p}(\tau_{P_{n}}(s_{n}))<\frac{1}{n}$ which implies that
$n<e(P_{n}\mid p)$. Thus $\\{e(P_{n}\mid p)\\}_{n\in\mathbb{N}}$ is unbounded.
Note that, as in the proof of (2) $\Rightarrow$ (3), if $P_{n}$ is the same
prime ideal $P$ for infinitely many $n$’s, then $e(P\mid p)=\infty$.

The last claim follows from Lemma 5.8. ∎

We end this section by showing that when $S$ is an integrally closed subring
of $\overline{\mathbb{Z}_{(p)}}$ containing $\mathbb{Z}_{(p)}$, then the
conditions of Lemma 5.1 are both necessary and sufficient for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ to be nontrivial.

######  Notation 5.16.

For an algebraic extension $K$ of $\mathbb{Q}$ and a prime $p\in\mathbb{P}$,
let $D$ be the integral closure of $\mathbb{Z}_{(p)}$ in $K$ (note that
$D=O_{K,(p)}$, the localization of the integral closure $O_{K}$ of
$\mathbb{Z}$ in $K$ with respect to $p$). We set

$E_{D,p}:=\\{e(P|p)\mid P\subset D\\}\quad\text{ and }\quad
F_{D,p}:=\\{f(P|p)\mid P\subset D\\}.$

######  Lemma 5.17.

Let $K$ be an infinite algebraic extension of $\mathbb{Q}$ and let
$K^{\prime}$ be the Galois closure of $K$ over $\mathbb{Q}$. Let
$p\in\mathbb{P}$ and let $D$ and $D^{\prime}$ be the integral closures of
$\mathbb{Z}_{(p)}$ in $K$ and $K^{\prime}$, respectively. If $E_{D,p}$ and
$F_{D,p}$ are bounded, then $E_{D^{\prime},p}$ and $F_{D^{\prime},p}$ are
bounded.

###### Proof.

Let $\sigma\in G=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Note that
$E_{\sigma(D),p}=E_{D,p}$ and $F_{\sigma(D),p}=F_{D,p}$. Since
$\mathbb{Q}_{p}$ has only finitely many extensions of bounded degree, there
exists a finite extension $L$ of $\mathbb{Q}_{p}$ such that, for every
$\sigma\in G$, $L$ contains the completion of $\sigma(K)$ with respect to any
prime ideal of $\sigma(D)$. In particular, there are only finitely many such
completions.

Note that $K^{\prime}$ is equal to the compositum of the fields
$\sigma(K),\sigma\in G$. Let $Q$ be a prime ideal of $D^{\prime}$. We claim
that the completion $\widehat{K^{\prime}}$ of $K^{\prime}$ with respect to $Q$
is equal to the compositum of the completions $\widehat{\sigma(K)}$ of
$\sigma(K)$ with respect to $Q\cap\sigma(D)$ (which by the above are finite in
number). Indeed, $\widehat{K^{\prime}}$ contains $\widehat{\sigma(K)}$ for
every $\sigma\in G$ and so $\widehat{K^{\prime}}$ contains their compositum;
conversely, since $\widehat{K^{\prime}}$ is the completion of $K^{\prime}$, it
is contained in the compositum of the (finitely many) fields
$\widehat{\sigma(K)},\sigma\in G$, because the latter is complete and contains
$K^{\prime}$.

Now, if $L^{\prime}$ is the Galois closure of $L$ over $\mathbb{Q}_{p}$, then
it is clear that $\widehat{K^{\prime}}$ is contained in $L^{\prime}$. Since
this holds for every prime ideal $Q$ of $D^{\prime}$ (and $L^{\prime}$ is
independent from the choice of the prime ideal $Q$), it follows that
$E_{D^{\prime},p}$ and $F_{D^{\prime},p}$ are bounded. ∎

######  Theorem 5.18.

Let $D$ be an integrally closed subring of $\overline{\mathbb{Z}_{(p)}}$
containing $\mathbb{Z}_{(p)}$. The following conditions are equivalent.

  1. (1)

$\textnormal{Int}_{\mathbb{Q}}(D)$ is Prüfer.

  2. (2)

$\textnormal{Int}_{\mathbb{Q}}(D)$ is nontrivial (i.e.,
$\mathbb{Z}_{(p)}[X]\subsetneq\textnormal{Int}_{\mathbb{Q}}(D)$).

  3. (3)

The sets $F_{D,p}$ and $E_{D,p}$ are bounded.

###### Proof.

Clearly, (1) implies (2).

(2) $\Rightarrow$ (3) If $F_{p}$ is unbounded then by Proposition 5.13 and
Remark 5.12, $\textnormal{Int}_{\mathbb{Q}}(D)$ is trivial. Similarly, if
$E_{p}$ is unbounded the same conclusion holds by Proposition 5.15.

(3) $\Rightarrow$ (1) Let $K^{\prime}$ be the Galois closure of $K$ and let
$D^{\prime}$ be the integral closure of $D$ in $K^{\prime}$. By Lemma 5.17,
$E_{D^{\prime},p}$ and $F_{D^{\prime},p}$ are bounded. Therefore,
$\textnormal{Int}_{K^{\prime}}(D^{\prime})$ is Prüfer by [14, Theorem 2.5]. We
consider now the set $G(D)=\bigcup_{\sigma\in G}\sigma(D)\subset D^{\prime}$.
Since
$\textnormal{Int}_{K^{\prime}}(D^{\prime})\subseteq\textnormal{Int}_{K^{\prime}}(G(D),D^{\prime})$,
the latter ring is Prüfer, too. By [22, Lemma 2.20],
$\textnormal{Int}_{K^{\prime}}(G(D),D^{\prime})$ is the integral closure in
$K^{\prime}(X)$ of
$\textnormal{Int}_{\mathbb{Q}}(D)=\textnormal{Int}_{\mathbb{Q}}(G(D),D^{\prime})$.
It follows by [11, Theorem 22.4] that
$\textnormal{Int}_{\mathbb{Q}}(D)=\textnormal{Int}_{K^{\prime}}(G(D),D^{\prime})\cap\mathbb{Q}(X)$
is Prüfer. ∎

######  Corollary 5.19.

Let $D\subseteq\overline{\mathbb{Z}}$ be an integrally closed subring.

  1. (1)

$\textnormal{Int}_{\mathbb{Q}}(D)$ is nontrivial if and only if there exists
some $p\in\mathbb{P}$ such that both $E_{D,p}$ and $E_{F,p}$ are bounded.

  2. (2)

$\textnormal{Int}_{\mathbb{Q}}(D)$ is Prüfer if and only if for each
$p\in\mathbb{P}$ the sets $E_{D,p}$ and $E_{D,p}$ are bounded.

###### Proof.

(1) This follows by Proposition 3.7 and Theorem 5.18.

(2) It is easy to show that $\textnormal{Int}_{\mathbb{Q}}(D)$ is Prüfer if
and only if for each $p\in\mathbb{P}$, the localization
$\textnormal{Int}_{\mathbb{Q}}(D)_{(p)}$ is Prüfer. By Proposition 3.6, the
latter ring is equal to
$\textnormal{Int}_{\mathbb{Q}}(D,\overline{\mathbb{Z}_{(p)}})$, and since
$\overline{\mathbb{Z}_{(p)}}\cap K=D_{(p)}$, we have
$\textnormal{Int}_{\mathbb{Q}}(D,\overline{\mathbb{Z}_{(p)}})=\textnormal{Int}_{\mathbb{Q}}(D,D_{(p)})$.
Finally,
$\textnormal{Int}_{\mathbb{Q}}(D,D_{(p)})=\textnormal{Int}_{\mathbb{Q}}(D_{(p)})$
by [4, Corollary I.2.6]. The conclusion then follows by Theorem 5.18. ∎

##  6 Polynomial closure of subsets of algebraic integers

Recall that by Example 1.5, when
$S=2\overline{\mathbb{Z}}=\\{2\alpha\mid\alpha\in\overline{\mathbb{Z}}\\}$,
the ring $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is
nontrivial because it contains $X/2$. This example admits a rather strong
generalization.

######  Definition 6.1.

Let $f\in\mathbb{Z}[X]$ be nonconstant and monic. Let $d\in\mathbb{Z},d\geq
1$. For each $\alpha\in\overline{\mathbb{Z}}$, let
$\mathcal{Z}(f(X)-d\alpha)\subseteq\overline{\mathbb{Z}}$ be the set of all
the roots of $f(X)-d\alpha$. Define $\mathscr{S}(f,d)$ to be the following
subset of $\overline{\mathbb{Z}}$:

$\mathscr{S}(f,d):=\bigcup_{\alpha\in\overline{\mathbb{Z}}}\mathcal{Z}(f(X)-d\alpha).$

Note that if $d=1$, then $\mathscr{S}(f,1)=\overline{\mathbb{Z}}$.

With this notation, $2\overline{\mathbb{Z}}=\mathscr{S}(X,2)$. The sets
$\mathscr{S}(f,d)$ provide many examples of subsets of $\overline{\mathbb{Z}}$
for which the associated integral-valued polynomial ring is nontrivial.

######  Example 6.2.

Let $f\in\mathbb{Z}[X]$ be nonconstant and monic, and let $d\geq 2$. Then,
$\mathscr{S}(f,d)$ has unbounded degree, and
$\textnormal{Int}_{\mathbb{Q}}(\mathscr{S}(f,d),\overline{\mathbb{Z}})$
contains $f(X)/d$, hence is nontrivial.

There is more to say about the relationship between the sets
$\mathscr{S}(f,d)$ and
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$.

######  Definition 6.3.

Given a subset $S\subseteq\overline{\mathbb{Z}}$, the polynomial closure of
$S$ in $\overline{\mathbb{Z}}$ is the largest subset
$T\subseteq\overline{\mathbb{Z}}$ containing $S$ such that
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\textnormal{Int}_{\mathbb{Q}}(T,\overline{\mathbb{Z}})$.
We say that $S$ is _polynomially closed in $\overline{\mathbb{Z}}$_ if $S=T$.
In the literature [6, 24], the polynomial closure of $S$ is often denoted by
$\overline{S}$. We will not employ this notation in order to avoid confusion
with integral closures or algebraic closures.

In Theorem 6.5 below, we prove that each polynomially closed subset of
$\overline{\mathbb{Z}}$ is equal to an intersection of $\mathscr{S}(f,d)$
sets. In particular, this means that if
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial, then
$S\subseteq\mathscr{S}(f,d)$ for some $f$ and some $d\geq 2$.

We recall now the definitions of characteristic ideals and regular bases, as
discussed in [4, Section II.1]. For each integer $n\geq 0$, let
$\mathfrak{I}_{n}(S)$ be the $n^{\text{th}}$ characteristic ideal of
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ of degree $n$, which
consists of 0 and all the leading coefficients of polynomials in
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ of degree $n$. For
each $n\geq 0$, $\mathfrak{I}_{n}(S)$ is a principal fractional ideal of
$\mathbb{Z}$ that contains $\mathbb{Z}$. Let $R$ be a ring such that
$\mathbb{Z}[X]\subseteq R\subseteq\mathbb{Q}[X]$. Then, a regular basis for
$R$ is a $\mathbb{Z}$-module basis $\\{b_{n}\\}_{n\geq 0}$ for $R$ such that
each $b_{n}$ has degree $n$. This happens precisely when $\mathfrak{I}_{n}(S)$
is principal, generated by the leading coefficient of $b_{n}$, for each
$n\in\mathbb{N}$ [4, Theorem II.1.4].

######  Proposition 6.4.

Let $S\subseteq\overline{\mathbb{Z}}$. Then,
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ has a regular basis
of the form $\\{f_{n}(X)/d_{n}\\}_{n\geq 0}$ such that for each $n\geq 0$,
$f_{n}\in\mathbb{Z}[X]$ is monic and $d_{n}$ is a positive integer.

###### Proof.

Note that in the case $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$
is trivial, a regular basis is given by $\\{X^{n}\\}_{n\in\mathbb{N}}$.

For each $n\geq 0$, the characteristic ideal $\mathfrak{I}_{n}(S)$ is a
principal fractional ideal of $\mathbb{Z}$. Fix $n$, and assume that
$\mathfrak{I}_{n}(S)$ is generated by $\tfrac{a}{b}\in\mathbb{Q}$, where $a$
and $b$ are coprime integers and $b>0$. Let $x,y\in\mathbb{Z}$ such that
$ax+by=1$. Since $\mathbb{Z}\subseteq\mathfrak{I}_{n}(S)$, we have
$\tfrac{1}{b}=x(\tfrac{a}{b})+y\in\mathfrak{I}_{n}(S)$. Thus, we may assume
without loss of generality that each $\mathfrak{I}_{n}(S)$ is generated by
$\tfrac{1}{d_{n}}$, where $d_{n}\in\mathbb{Z}$ and $d_{n}\geq 1$.

Now, by [4, Proposition II.1.4],
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ has a regular basis
$\\{g_{n}\\}_{n\geq 0}\subseteq\mathbb{Q}[X]$ such that for each $n$, $g_{n}$
has degree $n$ and the leading coefficient of $g_{n}$ generates
$\mathfrak{I}_{n}(S)$. Moreover, by [4, Proposition II.1.7], each coefficient
of $g_{n}$ is an element of $\mathfrak{I}_{n}(S)$. In light of the previous
paragraph, we see that there are integers $c_{0},\ldots,c_{n-1}$ such that

$g_{n}(X)=\tfrac{1}{d_{n}}X^{n}+\sum_{i=0}^{n-1}\tfrac{c_{i}}{d_{n}}X^{i}.$

Taking $f_{n}(X)=X^{n}+\sum_{i=0}^{n-1}c_{i}X^{i}$, we achieve the desired
regular basis. ∎

######  Theorem 6.5.

Let $S\subseteq\overline{\mathbb{Z}}$. Let $\\{f_{n}(X)/d_{n}\\}_{n\geq 0}$ be
a regular basis for $\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$
such that each $f_{n}\in\mathbb{Z}[X]$ is monic and each $d_{n}$ is a positive
integer. Let $\mathscr{F}=\\{f_{n}(X)/d_{n}\mid d_{n}\geq 2\\}$.

  1. (1)

$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$ is nontrivial if and
only if $\mathscr{F}\neq\varnothing$.

  2. (2)

The polynomial closure of $S$ in $\overline{\mathbb{Z}}$ is
$\displaystyle\bigcap_{f(X)/d\in\mathscr{F}}\mathscr{S}(f,d)$.

###### Proof.

Part (1) is clear. For (2), let $T$ be the polynomial closure of $S$ in
$\overline{\mathbb{Z}}$. If $\mathscr{F}$ is empty, then
$T=\overline{\mathbb{Z}}$ because
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\mathbb{Z}[X]=\textnormal{Int}_{\mathbb{Q}}(\overline{\mathbb{Z}},\overline{\mathbb{Z}})$,
and an intersection of an empty collection of subsets of
$\overline{\mathbb{Z}}$ is also equal to $\overline{\mathbb{Z}}$. So, assume
that $\mathscr{F}\neq\varnothing$. Let $\alpha\in T$, and let
$f(X)/d\in\mathscr{F}$. Then,
$f(X)/d\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})=\textnormal{Int}_{\mathbb{Q}}(T,\overline{\mathbb{Z}})$,
so $f(\alpha)/d\in\overline{\mathbb{Z}}$ and $\alpha\in\mathscr{S}(f,d)$.
Thus,

$T\subseteq\bigcap_{f(X)/d\in\mathscr{F}}\mathscr{S}(f,d).$

Conversely, let $\beta\in\bigcap_{f(X)/d\in\mathscr{F}}\mathscr{S}(f,d)$, let
$g\in\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, and let $m=\deg
g$. Since $\\{f_{n}(X)/d_{n}\\}_{n\geq 0}$ is a regular basis for
$\textnormal{Int}_{\mathbb{Q}}(S,\overline{\mathbb{Z}})$, there exist
$c_{0},\ldots,c_{m}\in\mathbb{Z}$ such that

$g(X)=\sum_{i=0}^{m}\dfrac{c_{i}f_{i}(X)}{d_{i}}.$

For each $0\leq i\leq m$, if $d_{i}=1$ then clearly
$f_{i}(\beta)/d_{i}\in\overline{\mathbb{Z}}$. If $d_{i}\geq 2$, then
$f_{i}(X)/d_{i}\in\mathscr{F}$, so $\beta\in\mathscr{S}(f_{i},d_{i})$, which
means that $f_{i}(\beta)/d_{i}\in\overline{\mathbb{Z}}$. It follows that
$g(\beta)\in\overline{\mathbb{Z}}$. Thus, $\beta\in T$, and therefore
$T=\bigcap_{f(X)/d\in\mathscr{F}}\mathscr{S}(f,d)$. ∎

We close the paper by using a theorem of McQuillan to prove that
$\mathcal{A}_{1}=\mathbb{Z}$ is polynomially closed in
$\overline{\mathbb{Z}}$. It is an open problem to determine whether
$\mathcal{A}_{n}$ is polynomially closed in $\overline{\mathbb{Z}}$ when
$n\geq 2$, although we suspect that this is the case.

######  Theorem 6.6.

[17, Theorem] Let $\alpha\in\overline{\mathbb{Z}}$ and let
$\textnormal{Int}(\mathbb{Z})(\alpha)=\\{f(\alpha)\mid
f\in\textnormal{Int}(\mathbb{Z})\\}$. Then,

$\textnormal{Int}(\mathbb{Z})(\alpha)=\bigcap_{P\in\mathcal{S}_{1}}O_{\mathbb{Q}(\alpha),P}$

where the intersection is over the family $\mathcal{S}_{1}$ of prime ideals
$P$ of $O_{\mathbb{Q}(\alpha)}$ which are totally split over $\mathbb{Z}$,
that is $e(P\mid P\cap\mathbb{Z})f(P\mid P\cap\mathbb{Z})=1$.

######  Proposition 6.7.

$\mathbb{Z}$ is polynomially closed in $\overline{\mathbb{Z}}$.

###### Proof.

Given $\alpha\in\overline{\mathbb{Z}}\setminus\mathbb{Z}$, define
$\textnormal{Int}(\mathbb{Z})(\alpha)$ as in Theorem 6.6. Since not all the
prime ideals of $O_{\mathbb{Q}(\alpha)}$ are totally split over $\mathbb{Z}$,
by Theorem 6.6 $\textnormal{Int}(\mathbb{Z})(\alpha)$ is a proper overring of
$O_{\mathbb{Q}(\alpha)}$. Thus, there exists
$f\in\textnormal{Int}(\mathbb{Z})$ such that $f(\alpha)\notin
O_{\mathbb{Q}(\alpha)}$. This is equivalent to having
$f(\alpha)\notin\overline{\mathbb{Z}}$, so we conclude that $\alpha$ is not in
the polynomial closure of $\mathbb{Z}$ in $\overline{\mathbb{Z}}$. ∎

######  Conjecture 6.8.

For each $n\geq 1$, $\mathcal{A}_{n}$ is polynomially closed in
$\overline{\mathbb{Z}}$.

Acknowledgements: The first author wishes to thank D. Dobbs for raising
questions about the chain of Prüfer domains
$\textnormal{Int}_{\mathbb{Q}}(\mathcal{A}_{n}),n\in\mathbb{N}$ which are the
original motivations of this paper. He also wishes to thank S. Mulay for the
hospitality during his visit at the Department of Mathematics of the
University of Tenneessee in Knoxville, USA in 2014. He thanks also S. Checcoli
for stimulating discussions around Lemma 5.17.

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