# Polynomial Dedekind domains with finite residue fields

Giulio Peruginelli111Department of Mathematics ”Tullio Levi-Civita”,
University of Padova, Via Trieste, 63, 35121 Padova, Italy. E-mail:
<EMAIL_ADDRESS>
###### Abstract

We show that every Dedekind domain $R$ with finite residue fields lying
between the polynomial rings $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ is equal to a
generalized ring of integer-valued polynomials, that is, for each prime
$p\in\mathbb{Z}$ there exists a finite subset $E_{p}$ of the absolute integral
closure $\overline{\mathbb{Z}_{p}}$ of the ring of $p$-adic integers such that
$R=\\{f\in\mathbb{Q}[X]\mid
f(E_{p})\subseteq\overline{\mathbb{Z}_{p}},\forall\text{ prime
}p\in\mathbb{Z}\\}$. Moreover, we prove that the class group of $R$ is
isomorphic to a direct sum of a countable family of finitely generated abelian
groups. Conversely, any group of this kind is the class group of a Dedekind
domain $R$ between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$.

Keywords: Dedekind domain, Class group, Integer-valued polynomials

MSC Classification codes: 13F20, 13A15, 11S05, 13B25, 20K99

##  1 Introduction

Given a Dedekind domain $D$, the class group of $D$ measures how far $D$ is
from being a UFD and it is therefore an important object in the study of
factorization problems in the ring $D$. It is well-known that the class group
of the ring of integers of a number field is a finite abelian group. In
contrast with this result, Claborn proved in [7] the groundbreaking result
that every abelian class group occurs as the class group of a suitable
Dedekind domain.

After a few years, Eakin and Heinzer showed in [8] that every finitely
generated abelian group is the class group of a Dedekind domain between
$\mathbb{Z}[X]$ and $\mathbb{Q}[X]$. More generally, they proved that if
$V_{1},\ldots,V_{n}$ are distinct DVRs with same quotient field $K$ and for
each $i=1,\ldots,n$, $\\{V_{i,j}\\}_{j=1}^{g_{i}}$ is a finite collection of
DVRs extending $V_{i}$ to $K(X)$, each of which is residually algebraic over
$V_{i}$ (i.e., the extension of the residue fields is algebraic), then the
following is a Dedekind domain:

$D=\bigcap_{i,j}V_{i,j}\cap K[X]$

Moreover, they also give an explicit description of the class group of such a
domain $D$. By means of this result, they were able to show the aforementioned
result considering suitable residually algebraic extensions of a finite set of
DVRs of $\mathbb{Q}$ to $\mathbb{Q}(X)$.

Actually, if we suppose that each residue field extension of $V_{i,j}$ over
$V_{i}$ to be finite, a ring $D$ constructed as above can be represented as a
ring of integer-valued polynomials in the following way. For each $i,j$, by
[19, Theorem 2.5 & Proposition 2.2], there exists an element $\alpha_{i,j}$ in
the algebraic closure $\overline{\widehat{K}_{i}}$ of the $V_{i}$-adic
completion $\widehat{K}_{i}$ of $K$, $\alpha_{i,j}$ transcendental over $K$,
such that $V_{i,j}=V_{i,\alpha_{i,j}}=\\{\varphi\in
K(X)\mid\varphi(\alpha_{i,j})\in\overline{\widehat{V_{i}}}\\}$, where
$\overline{\widehat{V_{i}}}$ is the absolute integral closure of
$\widehat{V_{i}}$, the completion of $V_{i}$. Hence, the above ring $D$ can be
represented as $D=\\{f\in K[X]\mid
f(\alpha_{i,j})\in\overline{\widehat{V_{i}}},\forall i,j\\}$ (for more
details, see the following section and also [19, Remark 2.8]).

More recently, Glivický and S̆aroch in [14] investigated a family of quasi-
euclidean subrings of $\mathbb{Q}[X]$ depending on a parameter
$\alpha\in\widehat{\mathbb{Z}}$, the profinite completion of $\mathbb{Z}$. A
ring of this family is always a Bézout domain (i.e., finitely generated ideals
are principal) and might be a PID or not, according to the finiteness of some
set of primes depending on $\alpha$ and the set of polynomials in
$\mathbb{Z}[X]$. It has been observed in [15] that these rings can be realized
as overrings of the classical ring of integer-valued polynomials
$\textnormal{Int}(\mathbb{Z})=\\{f\in\mathbb{Q}[X]\mid
f(\mathbb{Z})\subseteq\mathbb{Z}\\}$; such overrings have been completely
characterized in [4]. We will review this representation in the following
section.

In the same area, Chang generalized Eakin and Heinzer’s result in [6] proving
that there exists an almost Dedekind domain $D$ (i.e., $D_{M}$ is a DVR for
each maximal ideal $M$ of $D$) which is not Noetherian, lies between
$\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ and with class group isomorphic to a
direct sum $G$ of a prescribed countable family of finitely generated abelian
groups. As before, assuming the finiteness of the residue field extensions of
the involved DVRs, Chang’s construction falls in the class of integer-valued
polynomial rings that we consider in this paper. It is worth mentioning that
the Picard group of $\textnormal{Int}(\mathbb{Z})$ is a free abelian group of
countably infinite rank ([13]).

The paper is organized as follows. We first set the notation we will use
throughout the paper and introduce the class of _generalized ring of integer-
valued polynomials_ , which are subrings of $\mathbb{Q}[X]$ formed by
polynomials which are simultaneously integer-valued over different subsets of
integral elements over $\mathbb{Z}_{p}$, the ring of $p$-adic integers, for
$p$ running over the set of integer primes. In section 2, we review Loper and
Werner’s construction of Prüfer domains in [16] and show that it falls into
the class of generalized rings of integer-valued polynomials. We then
characterize when a ring of their construction is a Dedekind domain. In order
to accomplish this objective, we introduce the definition of _polynomially
factorizable_ subsets $\underline{E}$ of
$\overline{\widehat{\mathbb{Z}}}=\prod_{p}\overline{\mathbb{Z}_{p}}$ (we refer
to §1.1 for the unexplained notation), which turns out to be the key
assumption for such a ring to be of finite character (hence, a noetherian
Prüfer domain, thus Dedekind). Furthermore, we show that every Dedekind domain
with finite residue fields between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ is
equal to a generalized ring of integer-valued polynomials with class group
equal to a direct sum of a countable family of finitely generated abelian
groups. Among other things, we will also characterize the PIDs among these
class of domains, generalizing the aforementioned work of [14] (see also
[15]). We will also give a criteria for when two such generalized rings of
integer-valued polynomials are equal. Finally, in section 3, by means of a
suitable modification of Chang’s construction, given a group $G$ which is the
direct sum of a countable family of finitely genereated abelian groups, we
prove that there exists a Dedekind domain $D$, $\mathbb{Z}[X]\subset
D\subseteq\mathbb{Q}[X]$, with class group $G$ thus giving a positive answer
to a question raised by Chang in [6]. Note that, by the previous result, such
a domain is a generalized ring of integer-valued polynomials.

###  1.1 Notation

All the generalized rings of integer-valued polynomials considered in this
paper falls into the class of integer-valued polynomials on algebras (see for
example [9, 21]), which encompasses also the classical definition of ring of
integer-valued polynomials. Let $D$ be an integral domain with quotient field
$K$ and $A$ a torsion-free $D$-algebra such that $A\cap K=D$. We may evaluate
polynomials $f\in K[X]$ at any element $a\in A$ inside the extended algebra
$A\otimes_{D}K$. The $D$-algebra $A$ clearly embeds into $A\otimes_{D}K$ and
if $f(a)\in A$ we say that $f$ is integer-valued at $a$. In general, given a
subset $S$ of $A$, we define the following ring of integer-valued polynomials
over $S$:

$\textnormal{Int}_{K}(S,A)=\\{f\in K[X]\mid f(s)\in A,\forall s\in S\\}$

Note that when $A=D$ we get the usual definition of ring of integer-valued
polynomials on $D$, and in that case we omit the subscript $K$.

For an integral domain $D$, we define the Picard group of $D$, denoted by
$\text{Pic}(D)$, as the quotient of the abelian group of the invertible
fractional ideals of $D$ by the subgroup generated by the nonzero principal
fractional ideals (see [2, §VIII.1]). If $D$ is a Dedekind domain, then
$\text{Pic}(D)$ is the usual class group of $D$.

Let $p\in\mathbb{Z}$ be a fixed prime and $\mathbb{P}$ the set of all prime
numbers. We adopt the following notations:

  * -

$\mathbb{Z}_{(p)}$ the localization of $\mathbb{Z}$ at $p\mathbb{Z}$.

  * -

$\mathbb{Z}_{p},\mathbb{Q}_{p}$, the ring of $p$-adic integers and the field
of $p$-adic numbers, respectively.

  * -

$\overline{\mathbb{Q}_{p}},\overline{\mathbb{Z}_{p}}$, a fixed algebraic
closure of $\mathbb{Q}_{p}$ and the absolute integral closure of
$\mathbb{Z}_{p}$, respectively.

  * -

for a finite extension $K$ of $\mathbb{Q}_{p}$, we denote by $O_{K}$ the ring
of integers of $K$.

  * -

$v_{p}$ denotes the unique extension of the $p$-adic valuation on
$\mathbb{Q}_{p}$ to $\overline{\mathbb{Q}_{p}}$.

  * -

$\widehat{\mathbb{Z}}=\prod_{p\in\mathbb{P}}\mathbb{Z}_{p}$ the profinite
completion of $\mathbb{Z}$.

  * -

$\overline{\widehat{\mathbb{Z}}}=\prod_{p\in\mathbb{P}}\overline{\mathbb{Z}_{p}}$

  * -

for $\alpha\in\overline{\mathbb{Q}_{p}}$, we set

$V_{p,\alpha}=\\{\varphi\in\mathbb{Q}(X)\mid\varphi(\alpha)\in\overline{\mathbb{Z}_{p}}\\}.$

Clearly, $V_{p,\alpha}$ is a valuation domain of $\mathbb{Q}(X)$ extending
$\mathbb{Z}_{(p)}$ with maximal ideal equal to $M_{p,\alpha}=\\{\varphi\in
V_{p,\alpha}\mid v_{p}(\varphi(\alpha))>0\\}$. Moreover, $V_{p,\alpha}$ is a
DVR if $\alpha$ is transcendental over $\mathbb{Q}$ and it has rank $2$
otherwise. In the former case, the ramification index
$e(V_{p,\alpha}\mid\mathbb{Z}_{(p)})$ is equal to the ramification index
$e(\mathbb{Q}_{p}(\alpha)\mid\mathbb{Q}_{p})$. For simplicity, we denote this
integer by $e_{\alpha}$. In either case, let $O_{\alpha}$ and $M_{\alpha}$ be
the valuation domain and maximal ideal of $\mathbb{Q}_{p}(\alpha)$,
respectively. Then, the residue field of $V_{p,\alpha}$ is equal to
$O_{\alpha}/M_{\alpha}$ and $pO_{\mathbb{Q}_{p}(\alpha)}=M_{\alpha}^{e}$, for
some integer $e$, which is equal to $e_{\alpha}$ ([19, Proposition 2.2 &
Theorem 2.5]).

The following result mentioned in the introduction characterizes residually
algebraic extensions of $\mathbb{Z}_{(p)}$ to $\mathbb{Q}(X)$ of a certain
kind:

######  Theorem 1.1.

[19, Theorem 2.5] Let $W\subset\mathbb{Q}(X)$ be a valuation domain with
maximal ideal $M$ extending $\mathbb{Z}_{(p)}$ for some $p\in\mathbb{P}$. If
$pW=M^{e}$ for some $e\geq 1$ and $W/M\supseteq\mathbb{Z}/p\mathbb{Z}$ is a
finite extension, then there exists $\alpha\in\overline{\mathbb{Q}_{p}}$ such
that $W=V_{p,\alpha}$.

Given $f\in\mathbb{Q}[X]$, the evaluation of $f(X)$ at an element
$\alpha=(\alpha_{p})\in\overline{\widehat{\mathbb{Z}}}$ is done componentwise:

$f(\alpha)=(f(\alpha_{p}))\in\prod_{p\in\mathbb{P}}\overline{\mathbb{Q}_{p}}$

We say that $f$ is _integer-valued_ at $\alpha$ if
$f(\alpha)\in\overline{\widehat{\mathbb{Z}}}\Leftrightarrow f\in
V_{p,\alpha_{p}}$ for all $p\in\mathbb{P}$.

######  Definition 1.2.

Given a subset $\underline{E}$ of $\overline{\widehat{\mathbb{Z}}}$, we define
the _generalized ring of integer-valued polynomials on $\underline{E}$_ as:

$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\\{f\in\mathbb{Q}[X]\mid
f(\alpha)\in\overline{\widehat{\mathbb{Z}}},\forall\alpha\in\underline{E}\\}$

If $\underline{E}=\widehat{\mathbb{Z}}$, then it is easy to show that
$\textnormal{Int}_{\mathbb{Q}}(\widehat{\mathbb{Z}},\overline{\widehat{\mathbb{Z}}})=\textnormal{Int}(\mathbb{Z})$.
We recall that the family of overrings of $\textnormal{Int}(\mathbb{Z})$ which
are contained in $\mathbb{Q}[X]$ is formed exactly by the rings
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\widehat{\mathbb{Z}})$, as
$\underline{E}$ ranges through the subsets of $\widehat{\mathbb{Z}}$ of the
form $\prod_{p\in\mathbb{P}}E_{p}$, where for each prime $p$, $E_{p}$ is a
closed (possibly empty) subset of $\mathbb{Z}_{p}$ (see [4, Theorem 6.2]). In
the study of a generalized ring of integer-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\widehat{\mathbb{Z}})$ without
loss of generality we may suppose that the subsets $\underline{E}$ of
$\overline{\widehat{\mathbb{Z}}}$ are of the form
$\underline{E}=\prod_{p\in\mathbb{P}}E_{p}$ (see the arguments given in [4,
Remark 6.3]). Note that we allow each component $E_{p}$ of $\underline{E}$ to
be equal to the empty set.

##  2 Polynomial Dedekind domain with finite residue fields

Loper and Werner exhibit a construction of Prüfer domains between
$\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ in [16] in order to show the existence of
a Prüfer domain strictly contained in $\textnormal{Int}(\mathbb{Z})$. As in
the work of Eakin and Heinzer [8] mentioned in the Introduction, their
construction is obtained by intersecting a suitable family of valuation
domains indexed by the set $\mathbb{P}$ of prime numbers with $\mathbb{Q}[X]$.
A valuation domain of this family is equal to $V_{p,\alpha}$, for some
$\alpha\in\overline{\mathbb{Z}_{p}}$, by Theorem 1.1. As it has already been
observed in [19, Remark 2.8], a ring of the Loper and Werner’s construction
can be represented as a generalized ring of integer-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$,
for a suitable subset $\underline{E}$ of $\overline{\widehat{\mathbb{Z}}}$
which satisfies the following definition.

######  Definition 2.1.

Let
$\underline{E}=\prod_{p\in\mathbb{P}}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$.
We say that $\underline{E}$ is locally bounded, if, for each prime $p$,
$E_{p}$ is a subset of $\overline{\mathbb{Z}_{p}}$ of bounded degree, that is,
$\\{[\mathbb{Q}_{p}(\alpha):\mathbb{Q}_{p}]\mid\alpha\in E_{p}\\}$ is bounded.

Note that some of the components $E_{p}$ of $\underline{E}$ may be equal to
the empty set. Since $\mathbb{Q}_{p}$ has at most finitely many extensions of
degree bounded by some fixed positive integer, if
$E_{p}\subset\overline{\mathbb{Z}_{p}}$ has bounded degree then $E_{p}$ is
contained in a finite extension of $\mathbb{Q}_{p}$.

By Theorem 1.1, a Prüfer domain constructed in [16] can be represented as the
following intersection of valuation domains (see also [4]), where
$\underline{E}=\prod_{p\in\mathbb{P}}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$
is locally bounded:

$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\bigcap_{p\in\mathbb{P}}\bigcap_{\alpha_{p}\in
E_{p}}V_{p,\alpha_{p}}\cap\bigcap_{q\in\mathcal{P}^{\text{irr}}}\mathbb{Q}[X]_{(q)}$
(2.2)

Here, $\mathcal{P}^{\text{irr}}$ denotes the set of irreducible polynomials in
$\mathbb{Q}[X]$. Similarly for
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\\{f\in\mathbb{Q}[X]\mid
f(E_{p})\subseteq\overline{\mathbb{Z}_{p}}\\}$ we have:

$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\bigcap_{\alpha_{p}\in
E_{p}}V_{p,\alpha_{p}}\cap\bigcap_{q\in\mathcal{P}^{\text{irr}}}\mathbb{Q}[X]_{(q)}$
(2.3)

In particular, note that
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\bigcap_{p\in\mathbb{P}}\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$.

The main result of Loper and Werner ([16, Corollary 2.12]) by means of the
representation (2.2) can now be restated as follows:

######  Theorem 2.4.

Let $\underline{E}\subset\overline{\widehat{\mathbb{Z}}}$ be locally bounded.
Then the ring
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a Prüfer domain.

We now characterize when such a Prüfer domain is a Dedekind domain. In order
to do so, we need to describe the prime spectrum of
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$.
Throughout this section, we assume that
$\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ is
locally bounded. It is customary for ring of integer-valued polynomials to
distinguish the prime ideals into two different kinds, and we do the same here
in our setting: given a prime ideal $P$ of
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$,
we say that $P$ is a _non-unitary_ if $P\cap\mathbb{Z}=(0)$ and that $P$ is
_unitary_ if $P\cap\mathbb{Z}=p\mathbb{Z}$ for some $p\in\mathbb{P}$.

It is a classical result that each non-unitary prime ideal of $R$ is equal to:

$\mathfrak{P}_{q}=q(X)\mathbb{Q}[X]\cap R$

for some $q\in\mathcal{P}^{\text{irr}}$ (see for example [2, Corollary
V.1.2]).

If $P\cap\mathbb{Z}=p\mathbb{Z}$ and $\alpha\in E_{p}$, the following is a
unitary prime ideal of $R$:

$\mathfrak{M}_{p,\alpha}=\\{f\in R\mid v_{p}(f(\alpha))>0\\}$

If $E_{p}$ is a closed subset of $\overline{\mathbb{Z}_{p}}$ for each prime
$p$, we are going to show that each unitary prime ideal of $R$ is equal to
$\mathfrak{M}_{p,\alpha}$, for some $p\in\mathbb{P}$ and $\alpha\in E_{p}$.

######  Lemma 2.5.

Let $P$ be a finite subset of $\mathbb{P}$ and $S$ the multiplicative subset
of $\mathbb{Z}$ generated by $\mathbb{P}\setminus P$. Then
$S^{-1}\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\bigcap_{p\in
P}\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$.

In particular, for each $p\in\mathbb{P}$, $(\mathbb{Z}\setminus
p\mathbb{Z})^{-1}\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$.

###### Proof.

The proof follows by an argument similar to the one of [5, Proposition 4.2].
Let
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
and $R_{p}=\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$,
for each $p\in P$. The containment $S^{-1}R\subseteq\bigcap_{p\in P}R_{p}$ is
clear, since $R\subseteq R_{p}$ and for every $d\in S$, $d$ is a unit in
$R_{p}$, for each $p\in P$. Conversely, let $f\in\bigcap_{p\in P}R_{p}$. Let
$d\in\mathbb{Z}$ be such that $df\in\mathbb{Z}[X]$ and let $d=t\prod_{p\in
P}p^{a_{p}}$, $a_{p}\geq 0$ and $t\in\mathbb{Z}$ not divisible by any $p\in
P$. Then, letting $g=tf$, we have that $g$ is in $\mathbb{Z}_{(q)}[X]\subset
R_{q}$ for each $q\notin P$ and $g$ is in $R_{p}$ for each $p\in P$ because
$t$ is a unit in $\mathbb{Z}_{(p)}$, $\forall p\in P$. Hence,
$f=\frac{g}{t}\in S^{-1}R$, as wanted. ∎

######  Proposition 2.6.

Let $\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ be
locally bounded and closed. If
$M\subset\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a unitary prime ideal such that $M\cap\mathbb{Z}=p\mathbb{Z}$ for some
$p\in\mathbb{P}$, then $M$ is maximal and there exists $\alpha\in E_{p}$ such
that $M=\mathfrak{M}_{p,\alpha}$.

###### Proof.

Let
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$.
We use the fact that $R$ is a Prüfer domain by Theorem 2.4.

Let $M$ be a maximal unitary ideal of $R$ and let $V=R_{M}$. Then, by Lemma
2.5, we have
$R_{p}=\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})\subset
V$, since $(\mathbb{Z}\setminus p\mathbb{Z})^{-1}V=V$. Let $M^{\prime}$ be the
center of $V$ on $R_{p}$. Since $M^{\prime}\cap R=M$, it is sufficient to show
that $M^{\prime}=\mathfrak{M}_{p,\alpha}=\\{f\in R_{p}\mid
v_{p}(f(\alpha))>0\\}$, for some $\alpha\in E_{p}$ (with a slight abuse of
notation, we denotes the unitary prime ideals of $R$ and $R_{p}$ in the same
way). Let $f\in R_{p}$. Let $K$ be a finite extension of $\mathbb{Q}_{p}$ such
that $O_{K}$ contains $E_{p}$ and let $i_{0},\ldots,i_{q-1}\in O_{K}$ be a set
of representatives for $O_{K}/\pi O_{K}\cong\mathbb{F}_{q}$, where $\pi$ is a
uniformizer of $O_{K}$ (i.e., a generator of the maximal ideal of $O_{K}$).
For each $\alpha\in E_{p}$, there exists some $j\in\\{0,\ldots,q-1\\}$ such
that $f(\alpha)-i_{j}\in\pi O_{K}$. In particular,
$\prod_{j=0}^{q-1}(f(\alpha)-i_{j})\in\pi O_{K}$ for each $\alpha\in E_{p}$.
Note that the polynomials $X^{q}-X$ and $\prod_{j=0}^{q-1}(X-i_{j})$ coincide
modulo $\pi$, so in particular $f(\alpha)^{q}-f(\alpha)\in\pi O_{K}$. If
$e=e(O_{K}\mid\mathbb{Q}_{p})$, we have $(f(\alpha)^{q}-f(\alpha))^{e}\in
pO_{K}$. Equivalently, $(f^{q}-f)^{e}\in pR_{p}$, which is contained in
$M^{\prime}$. Since $M^{\prime}$ is a prime ideal, it follows that $f^{q}-f\in
M^{\prime}$, so modulo $M^{\prime}$, $f$ satisfies the equation $X^{q}-X=0$.
This shows that $R_{p}/M^{\prime}$ is contained in the finite field
$\mathbb{F}_{q}$, so it is a finite domain, hence a field. This proves that
$M^{\prime}$ is maximal.

Since $R_{p}$ is countable, $M^{\prime}$ is countably generated, say
$M^{\prime}=\bigcup_{n\in\mathbb{N}}I_{n}$, where
$I_{n}=(p,f_{1},\ldots,f_{n})$ for each $n\in\mathbb{N}$. By [12, Proposition
1.4], for each $n\in\mathbb{N}$, there exists $\alpha_{n}\in E_{p}$ such that
$I_{n}\subset\mathfrak{M}_{p,\alpha_{n}}$ (we may exclude the non-unitary
prime ideals of $R_{p}$ because they do not contain $p$, hence neither $I_{n}$
for every $n$). Suppose first that $E_{p}$ is finite. Then there exists
$\alpha\in E_{p}$ such that the set $J=\\{n\in\mathbb{N}\mid
I_{n}\subset\mathfrak{M}_{p,\alpha}\\}$ is a cofinal subset of $\mathbb{N}$.
Hence, for each $f\in M^{\prime}$, there exists $n\in J$ such that $f\in
I_{n}\subset\mathfrak{M}_{p,\alpha}$, so that
$M^{\prime}\subseteq\mathfrak{M}_{p,\alpha}$ and therefore equality holds
since $M^{\prime}$ is maximal. If $E_{p}$ is infinite, since it is a closed
subset (because $\underline{E}$ is closed) contained in a finite extension of
$\mathbb{Q}_{p}$, by compactness we may extract a sequence
$\\{\alpha_{n}\\}_{n\in\mathbb{N}}$ from $E_{p}$ converging to some element
$\alpha\in E_{p}$. Without loss of generality we suppose that
$\alpha_{n}\to\alpha$. Now, for each $f\in M^{\prime}$, $f\in
I_{n}\subset\mathfrak{M}_{p,\alpha_{n}}$ for some $n$. Since $I_{n}\subseteq
I_{n+1}$ for each $n\in\mathbb{N}$, $f\in\mathfrak{M}_{p,\alpha_{m}}$ for each
$m\geq n$, that is $v_{p}(f(\alpha_{m}))>0$. By continuity we get that
$v_{p}(f(\alpha))>0$, that is, $f\in\mathfrak{M}_{p,\alpha}$. Therefore as
before we conclude that $M^{\prime}=\mathfrak{M}_{p,\alpha}$. ∎

In particular, since
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a Prüfer domain, given a maximal unitary ideal $\mathfrak{M}_{p,\alpha}$,
$p\in\mathbb{P}$ and $\alpha\in E_{p}$, we have

$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})_{\mathfrak{M}_{p,\alpha}}=V_{p,\alpha}.$
(2.7)

Similarly, for $q\in\mathcal{P}^{\text{irr}}$, we have

$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})_{\mathfrak{P}_{q}}=\mathbb{Q}[X]_{(q)}.$
(2.8)

We call the valuation domains $V_{p,\alpha}$ unitary, and the others
$\mathbb{Q}[X]_{(q)}$ non-unitary. Similar equations hold for the Prüfer
domain $\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$.

###  2.1 Local case

For a fixed prime $p\in\mathbb{Z}$, we characterize in this section the
subsets $E_{p}$ of $\overline{\mathbb{Z}_{p}}$ for which the corresponding
ring of integer-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is a Dedekind
domain.

######  Lemma 2.9.

Let $p\in\mathbb{P}$ and $W$ a valuation domain of $\mathbb{Q}_{p}(X)$ such
that $W\cap\mathbb{Q}(X)=V_{p,\alpha}$ for some
$\alpha\in\overline{\mathbb{Z}_{p}}$. Then
$W=\widehat{V}_{p,\alpha}=\\{\psi\in\mathbb{Q}_{p}(X)\mid\psi(\alpha)\in\overline{\mathbb{Z}_{p}}\\}$.

###### Proof.

Let $w$ be a valuation associated to $W$. We note first that since $X\in
V_{p,\alpha}$ because $\alpha\in\overline{\mathbb{Z}_{p}}$, it follows that
$w(X)\geq 0$.

Let $f\in W\cap\mathbb{Q}_{p}[X]$, say $f(X)=\sum_{i=0}^{d}\alpha_{i}X^{i}$.
Then for $g(X)=\sum_{i=0}^{d}a_{i}X^{i}\in\mathbb{Q}[X]$ we have

$w(f-g)\geq\min_{0\leq i\leq d}\\{v_{p}(\alpha_{i}-a_{i})+iw(X)\\}.$

Therefore, if we choose $a_{i}\in\mathbb{Q}$ sufficiently $v_{p}$-adically
close to $\alpha_{i}$ for each $i=0,\ldots,d$, we have $w(f-g)\geq 0$. In
particular, $g\in W\cap\mathbb{Q}(X)=V_{p,\alpha}$. The polynomial $h=f-g$ is
in $p^{n}\mathbb{Z}_{p}[X]$ for some $n\geq 1$; therefore
$f(\alpha)=h(\alpha)+g(\alpha)\in\overline{\mathbb{Z}_{p}}$, so that
$f\in\widehat{V}_{p,\alpha}$. Therefore
$W\cap\mathbb{Q}_{p}[X]\subseteq\widehat{V}_{p,\alpha}\cap\mathbb{Q}_{p}[X]$.
Similarly, one can easily show that the other containment holds, so
$W\cap\mathbb{Q}_{p}[X]=\widehat{V}_{p,\alpha}\cap\mathbb{Q}_{p}[X]$. In the
same way,
$M_{W}\cap\mathbb{Q}_{p}[X]=\widehat{M}_{p,\alpha}\cap\mathbb{Q}_{p}[X]$,
where $M_{W}$ and $\widehat{M}_{p,\alpha}$ are the maximal ideals of $W$ and
$\widehat{V}_{p,\alpha}$, respectively.

Let now $\psi=\frac{f}{g}\in\widehat{V}_{p,\alpha}$, where
$f,g\in\mathbb{Q}_{p}[X]$. We have $v_{p}(f(\alpha))\geq v_{p}(g(\alpha))$ by
assumption. There exists $c\in\mathbb{Q}_{p}$, $c\not=0$, such that
$v_{p}(c)+v_{p}(g(\alpha))=0$. Then, if we let $f_{1}=cf$ and $g_{1}=cg$ we
have $\psi=\frac{f_{1}}{g_{1}}$ with
$f_{1}\in\widehat{V}_{p,\alpha}\cap\mathbb{Q}_{p}[X]=W\cap\mathbb{Q}_{p}[X]$
and
$g_{1}\in\widehat{V}_{p,\alpha}^{*}\cap\mathbb{Q}_{p}[X]=W^{*}\cap\mathbb{Q}_{p}[X]$.
Then, we have $v_{p,\alpha}(f_{1})\geq v_{p,\alpha}(g_{1})=0$ and also that

$w(f_{1})\geq 0=w(g_{1})$

which proves that $\psi\in W$. Hence, $\widehat{V}_{p,\alpha}\subseteq W$ and
the equality follows because both rings are extensions of $\mathbb{Z}_{p}$ to
$\mathbb{Q}_{p}(X)$ and the one-dimensional valuation overring of
$\widehat{V}_{p,\alpha}$ is non-unitary (i.e., $\mathbb{Q}_{p}[X]_{(p)}$,
where $p\in\mathbb{Q}_{p}[X]$ is the minimal polynomial of $\alpha$). ∎

The following lemma provides a generalization of [6, Theorem 4.3 (2)].

######  Proposition 2.10.

Let $E_{p}$ be a subset of $\overline{\mathbb{Z}_{p}}$. Then
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is a Dedekind
domain with finite residue fields if and only if $E_{p}$ is a finite subset of
transcendental elements over $\mathbb{Q}$.

Moreover, if $E_{p}=\\{\alpha_{1},\ldots,\alpha_{n}\\}$, then the class group
of $\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is
isomorphic to $\mathbb{Z}/e\mathbb{Z}\oplus\mathbb{Z}^{n-1}$, where
$e=\text{g.c.d.}\\{e_{\alpha_{i}}\mid i=1,\ldots,n\\}$.

In particular,
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is a PID if
and only if $E_{p}$ contains at most one element which is transcendental over
$\mathbb{Q}$ and unramified over $\mathbb{Q}_{p}$.

###### Proof.

Let $R_{p}=\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$.
Note that, if $E_{p}$ is the empty set, then $R_{p}=\mathbb{Q}[X]$. We assume
henceforth that $E_{p}\not=\emptyset$.

Suppose $R_{p}$ is a Dedekind domain with finite residue fields. Let $V$ be a
unitary valuation overring of $R_{p}$. By Theorem 1.1, there exists
$\alpha_{0}\in\overline{\mathbb{Z}_{p}}$ such that $V=V_{p,\alpha_{0}}$. Let
$M=\mathfrak{M}_{p,\alpha_{0}}$ be the center of $V_{p,\alpha_{0}}$ on
$R_{p}$. Since $M$ is finitely generated and $R_{p}$ is Prüfer, by [12,
Proposition 1.4] $M\subseteq\mathfrak{M}_{p,\alpha_{p}}$ for some
$\alpha_{p}\in E_{p}$ (we may exclude the non-unitary prime ideals of $R_{p}$
because they do not contain $p$, hence neither $M$). Since $M$ is maximal, it
follows that $M=\mathfrak{M}_{p,\alpha_{p}}$, which means that $\alpha_{0}$
and $\alpha_{p}$ are conjugate over $\mathbb{Q}_{p}$ by [19, Theorem 3.2].
Hence, without loss of generality, we may suppose that $\alpha_{0}\in E_{p}$.

Let now
$\widehat{R}_{p}=\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})$
and
$\mathcal{R}_{p}=\textnormal{Int}_{\overline{\mathbb{Q}_{p}}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}(E_{p},\overline{\mathbb{Z}_{p}})$.
By the previous paragraph and by Lemma 2.9, every unitary valuation overring
of $\widehat{R}_{p}$ is equal to $\widehat{V}_{p,\alpha_{p}}$ for some
$\alpha_{p}\in E_{p}$. In particular, since
$\mathbb{Q}_{p}(X)\subset\overline{\mathbb{Q}_{p}}(X)$ is a Galois extension,
each unitary valuation overring of $\mathcal{R}_{p}$ is equal to
$\overline{\widehat{V}}_{p,\alpha_{p}^{\prime}}=\\{\psi\in\overline{\mathbb{Q}_{p}}(X)\mid\psi(\alpha_{p}^{\prime})\in\overline{\mathbb{Z}_{p}}\\}$,
for some $\alpha_{p}^{\prime}\in\overline{\mathbb{Z}_{p}}$ which is conjugate
over $\mathbb{Q}_{p}$ of some element $\alpha_{p}\in E_{p}$. In fact,
$\overline{\widehat{V}}_{p,\alpha_{p}}$ is an extension of
$\widehat{V}_{p,\alpha_{p}}$ to $\overline{\mathbb{Q}_{p}}(X)$ and every other
extension of $\widehat{V}_{p,\alpha}$ is equal to
$\overline{\widehat{V}}_{p,\alpha_{p}^{\prime}}$, since the absolute Galois
group acts transitively over such extensions ([1, Chapt. VI, §8, 6.,
Corollaire 1]).

We claim that each pseudo-monotone sequence $E=\\{s_{n}\\}_{n\in\mathbb{N}}$
contained in $E_{p}$ is Cauchy. For such a sequence $E$, the valuation domain
$\overline{\mathbb{Z}_{p}}_{E}=\\{\psi\in\overline{\mathbb{Q}_{p}}(X)\mid\psi(s_{n})\in\overline{\mathbb{Z}_{p}},\text{
for almost all }n\in\mathbb{N}\\}$ of $\overline{\mathbb{Q}_{p}}(X)$ (see [20,
Theorem 3.4]) is clearly an overring of $\mathcal{R}_{p}$. The above argument
shows that
$\overline{\mathbb{Z}_{p}}_{E}=\overline{\widehat{V}}_{p,\alpha_{p}^{\prime}}$
for some $\alpha_{p}^{\prime}\in\overline{\mathbb{Z}_{p}}$ conjugate over
$\mathbb{Q}_{p}$ to some element of $E_{p}$. By [20, Proposition 3.7 (c) &
Proposition 5.1], $E$ is neither pseudo-stationary, nor strictly pseudo-
monotone with non-zero breadth ideal.

In particular, $E_{p}$ is a compact subset of $\overline{\mathbb{Z}_{p}}$
([3]).

Finally, we show that since $R_{p}$ is Noetherian, it follows that every
element of $E_{p}$ is isolated, hence $E_{p}$ is finite. Suppose otherwise,
and let $\\{\alpha_{n}\\}_{n\in\mathbb{N}}\subset E_{p}$ be a Cauchy sequence
converging to some $\alpha_{0}\in E_{p}$. By assumption
$\mathfrak{M}_{p,\alpha_{0}}$ is finitely generated, say by some polynomials
$f_{1},\ldots,f_{n}$; hence, for each $i=1,\ldots,n$,
$v_{p}(f_{i}(\alpha_{0}))>0$ implies $v_{p}(f_{i}(\alpha_{n}))>0$ for all $n$
sufficiently large, hence
$\mathfrak{M}_{p,\alpha_{0}}\subseteq\mathfrak{M}_{p,\alpha_{n}}$, so these
ideals are equal for all $n$ sufficiently large. In particular, by (2.7),
$V_{p,\alpha_{0}}=V_{p,\alpha_{n}}$ for all such $n$’s. This is a
contradiction with the fact that $V_{p,\alpha}=V_{p,\beta}$ if and only if
$\alpha,\beta\in\overline{\mathbb{Q}_{p}}$ are conjugate over $\mathbb{Q}_{p}$
([19, Theorem 3.2]).

Suppose now that $E_{p}\subset\overline{\mathbb{Z}_{p}}$ is a finite subset of
transcendental elements over $\mathbb{Q}$. In particular, $E_{p}$ has bounded
degree, so, by Theorem 2.4, $R_{p}$ is Prüfer. By (2.3), $R_{p}$ is equal to
an intersection of DVRs which are essential over it. Moreover, each non-zero
$f\in R_{p}$ belongs to finitely many maximal ideals, since $E_{p}$ is finite
and $f$ has finitely many irreducible factors in $\mathbb{Q}[X]$. Hence,
$R_{p}$ is a Krull domain, so, by [11, Theorem 43.16], $R_{p}$ is a Dedekind
domain. Finally, $R_{p}$ has finite residue fields, because each of the
valuation overring of $R_{p}$ (namely, $V_{p,\alpha_{p}}$, $\alpha_{p}\in
E_{p}$ and $\mathbb{Q}[X]_{(q)}$, $q\in\mathcal{P}^{\text{irr}}$) has finite
residue field.

The claim regarding the class group follows easily from [8, Theorem], taking
into account the representation (2.3). If
$E_{p}=\\{\alpha_{1},\ldots,\alpha_{n}\\}$, let
$\mathbb{e}=(e_{\alpha_{1}},\ldots,e_{\alpha_{n}})\in\mathbb{Z}^{n}$ and
$e=GCD(e_{\alpha_{1}},\ldots,e_{\alpha_{n}})$. Then, the class group of
$R_{p}$ is isomorphic to

$\mathbb{Z}^{n}/<\mathbb{e}>\cong\mathbb{Z}/e\mathbb{Z}\oplus\mathbb{Z}^{n-1}.$

The last claim follows at once from the description of the class group. ∎

###  2.2 Global case

In general, if, for each $p\in\mathbb{P}$,
$E_{p}\subset\overline{\mathbb{Z}_{p}}$ is a finite subset of transcendental
elements over $\mathbb{Q}$ and $\underline{E}=\prod_{p}E_{p}$, then, by [6,
Corollary 2.6],
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is an almost Dedekind domain. However, in general, it is not Noetherian, that
is, a Dedekind domain. See for example the construction of [6, Theorem 3.1],
in which the polynomial $X$ is divisible by infinitely many primes
$p\in\mathbb{P}$. In general, an almost Dedekind domain $R$ is Dedekind if and
only if it has finite character, that is, each non-zero $f\in R$ belongs to
finitely many maximal ideals of $R$ ([11, Theorem 37.2]), or, equivalently,
$v(f)\not=0$ only for finitely many valuation overrings $V$ of $R$ (which are
only DVRs). We aim to characterize the subsets $\underline{E}=\prod_{p}E_{p}$
of $\overline{\widehat{\mathbb{Z}}}$ for which
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is Dedekind.

We introduce the following definition.

######  Definition 2.11.

We say that $\underline{E}$ is _polynomially factorizable_ if, for each
$g\in\mathbb{Z}[X]$ and $\alpha=(\alpha_{p})\in\underline{E}$, there exist
$n,d\in\mathbb{Z}$, $n,d\geq 1$ such that $\frac{g(\alpha)^{n}}{d}$ is a unit
of $\overline{\widehat{\mathbb{Z}}}$, that is,
$v_{p}(\frac{g(\alpha_{p})^{n}}{d})=0$, $\forall p\in\mathbb{P}$.

Note that
$g(\alpha)^{n}=(g(\alpha_{p})^{n})\in\overline{\widehat{\mathbb{Z}}}$. Loosely
speaking, a subset $\underline{E}$ of $\overline{\widehat{\mathbb{Z}}}$ is
polynomially factorizable if, for every $g\in\mathbb{Z}[X]$ and
$\alpha\in\underline{E}$, $g(\alpha)\in\overline{\widehat{\mathbb{Z}}}$ is
divisible only by finitely many primes $p\in\mathbb{P}$ (up to some exponent
$n\geq 1$), or, equivalently, all but finitely many components of $g(\alpha)$
are units. Note that, if the above condition of the definition holds, then
$g(\alpha)^{n}$ and $d$ generate the same principal ideal of
$\overline{\widehat{\mathbb{Z}}}$.

The next lemma gives a simple characterization of polynomially factorizable
subsets $\underline{E}$ of $\overline{\widehat{\mathbb{Z}}}$ in terms of the
finiteness of some sets of primes associated to every polynomial in
$\mathbb{Z}[X]$. For every $g\in\mathbb{Z}[X]$ and subset
$\underline{E}=\prod_{p}E_{p}\subseteq\overline{\widehat{\mathbb{Z}}}$, we set

$\mathbb{P}_{g,\underline{E}}=\\{p\in\mathbb{P}\mid\exists\alpha_{p}\in
E_{p}\text{ such that }v_{p}(g(\alpha_{p}))>0\\}$

######  Lemma 2.12.

Let $g\in\mathbb{Z}[X]$ and
$\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$, where
each $E_{p}\subset\overline{\mathbb{Z}_{p}}$ is a closed set of transcendental
elements over $\mathbb{Q}$. Then the following conditions are equivalent:

  * i)

the set $\mathbb{P}_{g,\underline{E}}$ is finite.

  * ii)

for each $\alpha\in\underline{E}$, there exist $n\geq 1$ and $d\in\mathbb{Z}$
such that $\frac{g(\alpha)^{n}}{d}$ is a unit of
$\overline{\widehat{\mathbb{Z}}}$.

###### Proof.

We use the following easy remark: for
$\alpha=(\alpha_{p})\in\widehat{\mathbb{Z}}=\prod_{p}\mathbb{Z}_{p}$, the set
$\\{p\in\mathbb{P}\mid v_{p}(\alpha_{p})>0\\}$ is finite if and only if there
exists $d\in\mathbb{Z}$ such that
$\alpha\widehat{\mathbb{Z}}=d\widehat{\mathbb{Z}}$.

Suppose i) holds and let $\alpha=(\alpha_{p})\in\underline{E}$. By assumption,
there are only finitely many $p\in\mathbb{P}$ such that
$v_{p}(g(\alpha_{p}))>0$, say, $p_{1},\ldots,p_{k}$. In particular, there
exists $n\in\mathbb{N}$ such that $nv_{p}(g(\alpha_{p}))=a_{p}\in\mathbb{Z}$
for each prime $p$ (where $a_{p}=0$ for all
$p\notin\\{p_{1},\ldots,p_{k}\\}$). Hence, if we let
$d=\prod_{i=1}^{k}p_{i}^{a_{p_{i}}}$ we get
$v_{p}(g(\alpha_{p})^{n})=v_{p}(d)$ for all $p\in\mathbb{P}$, hence ii) holds.

Assume now that ii) holds and suppose that $\mathbb{P}_{g,\underline{E}}$ is
infinite. For each $p\in\mathbb{P}_{g,\underline{E}}$, let $\alpha_{p}\in
E_{p}$ be such that $v_{p}(g(\alpha_{p}))>0$ and consider the element
$\alpha=(\alpha_{p})\in\underline{E}$, where $\alpha_{p}$ is any element of
$E_{p}$ for $p\notin\mathbb{P}_{g,\underline{E}}$. If there is no $n\geq 1$
such that $nv_{p}(g(\alpha_{p}))=a_{p}\in\mathbb{Z}$ for all $p\in\mathbb{P}$
we immediately get a contradiction. Suppose instead that such an $n$ exists.
Since $a_{p}$ is non-zero for infinitely many $p\in\mathbb{P}$, there is no
$d\in\mathbb{Z}$ such that $v_{p}(\frac{g(\alpha_{p})^{n}}{d})=0$ for each
$p\in\mathbb{P}$, which again is a contradiction.

∎

######  Remark 2.13.

By Lemma 2.12, it follows easily that a subset
$\underline{E}\subseteq\overline{\widehat{\mathbb{Z}}}$ is polynomially
factorizable if and only if $\mathbb{P}_{g,\underline{E}}$ is finite for each
irreducible $g\in\mathbb{Z}[X]$. In fact, if $g=\prod_{i}g_{i}$, where
$g_{i}\in\mathbb{Z}[X]$ are irreducible, then
$\mathbb{P}_{g,\underline{E}}=\bigcup_{i}\mathbb{P}_{g_{i},\underline{E}}$.

It is well-known that, given $q\in\mathbb{Z}[X]$, there exist infinitely many
$p\in\mathbb{P}$ for which there exists $n\in\mathbb{Z}$ such that $q(n)$ is
divisible by $p$ (see for example the proof of [2, Prop. V.2.8]). In
particular, $\widehat{\mathbb{Z}}$ is not polynomially factorizable by Lemma
2.12.

The next lemma describes the Picard group of
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
in terms of the Picard groups of the localizations
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$,
$p\in\mathbb{P}$ (see Lemma 2.5).

######  Lemma 2.14.

Let $\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ be a
subset. Then

$\text{Pic}(\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}}))\cong\bigoplus_{p\in\mathbb{P}}\text{Pic}(\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}}))$

###### Proof.

Let
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
and $R_{p}=(\mathbb{Z}\setminus p\mathbb{Z})^{-1}R$, for $p\in\mathbb{P}$; by
Lemma 2.5,
$R_{p}=\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$. The
proof follows by the same arguments as in [13, Theorem 1], so we sketch the
proof and refer to the proof of the cited result for the details. By a
classical argument (see for example [17, Lemma 1]), every finitely generated
ideal $J$ of $R$ (in particular, every invertible ideal of $R$) is isomorphic
to a finitely generated unitary ideal $I$, that is,
$I\cap\mathbb{Z}=d\mathbb{Z}\not=(0)$. For such an ideal,
$(I\cap\mathbb{Z})_{(p)}=\mathbb{Z}_{(p)}$ for all $p\in\mathbb{P}$ not
dividing $d$, so $IR_{p}=R_{p}$. This argument shows that we have a well-
defined map from $\text{Pic}(R)$ to
$\bigoplus_{p\in\mathbb{P}}\text{Pic}(R_{p})$.

If $I$ is a unitary ideal of $R$, say $I\cap\mathbb{Z}=d\mathbb{Z}$, is such
that $IR_{p}$ is principal, it is generated by $d$. Hence, $I$ and $dR$ have
the same localizations at each prime $p\in\mathbb{P}$, so they are equal. This
shows that the previous map is injective.

For the surjectivity, it is sufficient to show that, if $J_{p}$ is an
invertible unitary ideal of $R_{p}$, for some $p\in\mathbb{P}$, then there
exists an invertible ideal $J$ of $R$ such that $JR_{p}=J_{p}$ and
$JR_{q}=R_{q}$ for each $q\in\mathbb{P}\setminus\\{p\\}$. The ideal
$J=J_{p}\cap R$ has the required properties. ∎

We may now characterize when a generalized ring of integer-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is Dedekind and describe its class group.

######  Theorem 2.15.

Let $\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ be a
subset. Then
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a Dedekind domain with finite residue fields if and only if $E_{p}$ is a
finite set of transcendental elements over $\mathbb{Q}$ for each prime $p$ and
$\underline{E}$ is polynomially factorizable.

If this condition holds, then the class group of
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\widehat{\mathbb{Z}})$ is equal
to a direct sum of a countable family of finitely generated abelian groups.

###### Proof.

Let
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
and suppose the conditions for $\underline{E}$ in the statement are satisfied.
In particular, $\underline{E}$ is locally bounded so, by Theorem 2.4, $R$ is
Prüfer. For $R$ to be Dedekind, it is sufficient to show that it is a Krull
domain ([11, Theorem 43.16]). By assumption, each of the valuation overrings
of $R$ in the representation (2.2) is a DVR with finite residue field. We have
to show that $R$ has finite character, that is, for each non-zero
$f=\frac{g}{n}\in R$, $g\in\mathbb{Z}[X]$ and
$n\in\mathbb{Z}\setminus\\{0\\}$, $f$ is contained in only finitely many
maximal ideals of $R$. As in the proof of Proposition 2.10, $f$ is contained
in only finitely many non-unitary prime ideals of $R$. We now check the
maximal unitary ideals of $R$, described in the Proposition 2.6, which contain
$f$. Since the denominator $n$ of $f$ is divisible by only finitely many
$p\in\mathbb{P}$, $f$ is contained in only finitely many maximal unitary
ideals if and only if the same condition holds for $g$. Since $E_{p}$ is
finite for each $p\in\mathbb{P}$, this is equivalent to the finiteness of the
set $\mathbb{P}_{g,\underline{E}}$. Since $\underline{E}$ is polynomially
factorizable, by Lemma 2.12, $\mathbb{P}_{g,\underline{E}}$ is finite.

Conversely, if
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a Dedekind domain with finite residue fields, then, for each prime $p$, the
overring $\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is a
Dedekind domain with finite residue fields ([11, Theorem 40.1]). By
Proposition 2.10, $E_{p}$ is a finite subset of $\overline{\mathbb{Z}_{p}}$
formed by transcendental elements over $\mathbb{Q}$ (so, in particular,
$\underline{E}$ is locally bounded). If there exists some $g\in\mathbb{Z}[X]$
such that the set $\mathbb{P}_{g,\underline{E}}$ is infinite, then $g(X)$
would be contained in infinitely many unitary prime ideals of
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$,
a contradiction with [11, Theorem 37.2]. Therefore, $\underline{E}$ is
polynomially factorizable by Lemma 2.12.

The final claim regarding the class group follows from Lemma 2.14 and
Proposition 2.10. ∎

The next corollary is a generalization of [14, Lemma 3.3]: it characterizes
for which $\alpha\in\overline{\widehat{\mathbb{Z}}}$ the ring
$\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\overline{\widehat{\mathbb{Z}}})$
is a PID.

######  Corollary 2.16.

Let $\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ be a
subset. Then
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a PID with finite residue fields if and only if, for each prime $p$,
$E_{p}$ contains at most one element of $\overline{\mathbb{Z}_{p}}$,
unramified over $\mathbb{Q}_{p}$ and transcendental over $\mathbb{Q}$ and
$\underline{E}$ is polynomially factorizable.

Note that if the conditions of the corollary occur, namely,
$E_{p}=\\{\alpha_{p}\\}$ for each $p\in\mathbb{P}$, then $\underline{E}$ is
the singleton $\\{\alpha\\}$, where
$\alpha=(\alpha_{p})\in\overline{\widehat{\mathbb{Z}}}$. The condition that
$\underline{E}$ is polynomially factorizable appears in another equivalent
forms in [14, Lemma 3.3] and [15, Proposition 1.1], in the case
$\alpha\in\widehat{\mathbb{Z}}$.

###### Proof.

The proof follows from Theorem 2.15, Lemma 2.14 and Proposition 2.10. ∎

An argument similar to the one in the proof of [8, Theorem] shows that a PID
$\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\overline{\widehat{\mathbb{Z}}})$
as in the statement of Corollary 2.16 is never an Euclidean domain.

We now show that each Dedekind domain with finite residue fields between
$\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ is indeed a generalized ring of integer-
valued polynomials.

######  Theorem 2.17.

Let $R$ be a Dedekind domain with finite residue fields such that
$\mathbb{Z}[X]\subset R\subseteq\mathbb{Q}[X]$. Then $R$ is equal to
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$,
for some subset
$\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ such that
$E_{p}$ is a finite set of transcendental elements over $\mathbb{Q}$ for each
prime $p$ and $\underline{E}$ is polynomially factorizable.

In particular, the class group of $R$ is isomorphic to a direct sum of a
countable family of finitely generated abelian groups.

###### Proof.

Let $\mathbb{P}_{R}=\\{p\in\mathbb{P}\mid\exists P\in\text{Spec}(R)\text{ such
that }P\cap\mathbb{Z}=p\mathbb{Z}\\}$. Clearly, $\mathbb{P}_{R}$ is empty if
and only if $R=\mathbb{Q}[X]$; in this case for $\underline{E}$ equal to the
empty set we have the claim. Suppose $\mathbb{P}_{R}$ is not empty. For each
$p\in\mathbb{P}_{R}$, we denote by $\mathbb{P}_{R,p}$ the set of unitary prime
ideals of $R$ lying above $p$. By assumption, for each $P\in\mathbb{P}_{R,p}$,
$p\in\mathbb{P}$, $R_{P}$ is a DVR of $\mathbb{Q}(X)$ with finite residue
field extending $\mathbb{Z}_{(p)}$. In particular, by Theorem 1.1, there
exists $\alpha_{p}\in\overline{\mathbb{Q}_{p}}$, transcendental over
$\mathbb{Q}$, such that $R_{P}=V_{p,\alpha_{p}}$. Note that, since $X\in
R\subset R_{P}$, necessarily $\alpha_{p}\in\overline{\mathbb{Z}_{p}}$. Let
$E_{p}$ be the subset of $\overline{\mathbb{Z}_{p}}$ formed by such
$\alpha_{p}$’s, for each $P\in\mathbb{P}_{R,p}$. Since $R$ is Dedekind and by
(2.2) and (2.3), we have the following equalities:

$R=\bigcap_{p\in\mathbb{P}_{R}}\bigcap_{P\in\mathbb{P}_{R,p}}R_{P}\cap\mathbb{Q}[X]=\bigcap_{p\in\mathbb{P}_{R}}\bigcap_{\alpha_{p}\in
E_{p}}V_{p,\alpha_{p}}\cap\mathbb{Q}[X]=\bigcap_{p\in\mathbb{P}_{R}}\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$

where
$\underline{E}=\prod_{p\in\mathbb{P}_{R}}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$.
By Theorem 2.15, for each $p\in\mathbb{P}$, $E_{p}$ is a finite subset of
$\overline{\mathbb{Z}_{p}}$ of transcendental elements over $\mathbb{Q}$,
$\underline{E}$ is polynomially factorizable and the class group of $R$ is
isomorphic to a direct sum of a countable family of finitely generated abelian
groups. ∎

It has been showed in [14, Proposition 3.4] that the cardinality of the set of
$\alpha\in\widehat{\mathbb{Z}}$ such that
$\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\widehat{\mathbb{Z}})$ is a PID is
$2^{\aleph_{0}}$. The next corollary describes all the PIDs between
$\mathbb{Z}[X]$ and $\mathbb{Q}[X]$.

######  Corollary 2.18.

Let $R$ be a PID with finite residue fields such that $\mathbb{Z}[X]\subset
R\subset\mathbb{Q}[X]$. Then $R$ is equal to
$\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\overline{\widehat{\mathbb{Z}}})$,
for some $\alpha=(\alpha_{p})\in\overline{\widehat{\mathbb{Z}}}$ such that,
for each $p\in\mathbb{P}$, $\alpha_{p}$ is transcendental over $\mathbb{Q}$,
$\alpha_{p}$ is unramified over $\mathbb{Q}_{p}$ and $\\{\alpha\\}$ is
polynomially factorizable.

###### Proof.

The proof follows from Theorem 2.17 and Corollary 2.16. ∎

###  2.3 Equality of generalized rings of integer-valued polynomials

Given two locally bounded closed subsets $\underline{E},\underline{F}$ of
$\overline{\widehat{\mathbb{Z}}}$, we characterize when the associated
generalized ring of integer-valued polynomials
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$,
$\textnormal{Int}_{\mathbb{Q}}(\underline{F},\overline{\widehat{\mathbb{Z}}})$
are the same.

The following is a general result about integral extensions of ring of
integer-valued polynomials. For an integral domain $D$ with quotient field
$K$, let $\overline{K}$ and $\overline{D}$ be the algebraic closure of $K$ and
the absolute integral closure of $D$, respectively. We let
$G_{K}=\text{Gal}(\overline{K}/K)$ be the absolute Galois group of $K$. For a
subset $\Omega$ of $\overline{K}$ we set
$G_{K}(\Omega)=\\{\sigma(a)\mid\sigma\in
G_{K},a\in\Omega\\}=\bigcup_{\sigma\in G_{K}}\sigma(\Omega)$. We say that
$\Omega$ is _$G_{K}$ -invariant_ if $G_{K}(\Omega)=\Omega$. Note that in
general we have

$\textnormal{Int}_{K}(\Omega,\overline{D})=\textnormal{Int}_{K}(G_{K}(\Omega),\overline{D})$
(2.19)

because if $f(\alpha)\in\overline{D}$ for some $f\in K[X]$ and
$\alpha\in\overline{D}$, then, for every $\sigma\in G_{K}$, we have
$\sigma(f(\alpha))=f(\sigma(\alpha))\in\overline{D}$.

######  Lemma 2.20.

let $D$ be an integrally closed domain with quotient field $K$. Let
$\Omega\subset\overline{D}$ be $G_{K}$-invariant. Let $F$ be an algebraic
extension of $K$ containing $\Omega$. Then
$\textnormal{Int}_{F}(\Omega,\overline{D})$ is the integral closure of
$\textnormal{Int}_{K}(\Omega,\overline{D})$ in $F(X)$.

###### Proof.

We just need to show that
$\textnormal{Int}_{K}(\Omega,\overline{D})\subseteq\textnormal{Int}_{F}(\Omega,\overline{D})$
is an integral ring extension, since it is well-known that
$\textnormal{Int}_{F}(\Omega,\overline{D})$ is integrally closed ([2,
Proposition IV.4.1]).

Without loss of generality, we may enlarge $F$ and suppose that $F$ is normal
over $K$ (e.g., we may take $F=\overline{K}$). Let
$f\in\textnormal{Int}_{F}(\Omega,\overline{D})\subset F[X]$. In particular,
$f$ is integral over $K[X]$, i.e., it satisfies a monic equation of the form:

$f^{n}+g_{n-1}f^{n-1}+\ldots+g_{1}f+g_{0}=0$

where $g_{i}\in K[X]$, for $i=0,\ldots,n-1$. We claim that
$g_{i}\in\textnormal{Int}_{K}(\Omega,\overline{D})$, for $i=0,\ldots,n-1$,
which will prove the claim. In fact, let
$\Phi(T)=T^{n}+g_{n-1}T^{n-1}+\ldots+g_{0}\in K[X][T]$. The roots of $\Phi(T)$
are the conjugates of $f$ under the action of the Galois group
$\text{Gal}(F(X)/K(X))\cong{\rm Gal}(F/K)$, which acts on the coefficients of
the polynomial $f$. If $\sigma\in\text{Gal}(F/K)$, then
$\sigma(f)\in\textnormal{Int}_{F}(\Omega,\overline{D})$. In fact, for each
$\alpha\in\Omega$, since $\Omega$ is ${\rm Gal}(F/K)$-invariant, we have
$\alpha=\sigma(\alpha^{\prime})$ for some $\alpha^{\prime}\in\Omega$,
therefore $\sigma(f)(\alpha)=\sigma(f(\alpha^{\prime}))$ which still is an
element of $\overline{D}$ (which likewise is left invariant under the action
of ${\rm Gal}(F/K)$). Now, since each coefficient $g_{i}$ of $\Phi(T)$ is an
elementary symmetric function of the elements
$\sigma(f),\sigma\in\text{Gal}(F/K)$, we have $g_{i}(\alpha)\in\overline{D}$,
for each $\alpha\in\Omega$, thus
$g_{i}\in\textnormal{Int}_{K}(\Omega,\overline{D})$, as claimed. ∎

In order to ease the notation, for each prime $p$, we set
$G_{p}=G_{\mathbb{Q}_{p}}$, the absolute Galois group of $\mathbb{Q}_{p}$.

######  Theorem 2.21.

Let $\underline{E}=\prod_{p}E_{p},\underline{F}=\prod_{p}F_{p}$ be locally
bounded closed subsets of $\overline{\widehat{\mathbb{Z}}}$. Then
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\textnormal{Int}_{\mathbb{Q}}(\underline{F},\overline{\widehat{\mathbb{Z}}})$
if and only if $G_{p}(E_{p})=G_{p}(F_{p})$, for each $p\in\mathbb{P}$.

###### Proof.

Clearly,
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\textnormal{Int}_{\mathbb{Q}}(\underline{F},\overline{\widehat{\mathbb{Z}}})$
if and only if the two rings have the same localization at each
$p\in\mathbb{P}$, that is, by Lemma 2.5,
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}}(F_{p},\overline{\mathbb{Z}_{p}})$.
Such a condition is equivalent to
$\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}_{p}}(F_{p},\overline{\mathbb{Z}_{p}})$.
In fact, one implication is obvious because
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})$ is the
contraction to $\mathbb{Q}[X]$ of
$\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})$.
Conversely, suppose that
$\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}}(F_{p},\overline{\mathbb{Z}_{p}})$
and let
$f\in\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})$, say
$f(X)=\sum_{i}\alpha_{i}X^{i}$. We can choose $g\in\mathbb{Q}[X]$ sufficiently
$v_{p}$-adically close to $f(X)$, that is, $g(X)=\sum_{i}a_{i}X^{i}$, where
$v_{p}(\alpha_{i}-a_{i})\geq n$ for each $i\geq 0$, where $n\in\mathbb{N}$ is
arbitrary large. In particular, $h=f-g\in p^{n}\mathbb{Z}_{p}[X]$, so, if
$\alpha_{p}\in E_{p}$, it follows that
$g(\alpha_{p})=f(\alpha_{p})+h(\alpha_{p})\in\overline{\mathbb{Z}_{p}}$.
Hence,
$g\in\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{\mathbb{Q}}(F_{p},\overline{\mathbb{Z}_{p}})$.
If now $\beta_{p}\in F_{p}$, we have
$f(\beta_{p})=g(\beta_{p})+h(\beta_{p})\in\overline{\mathbb{Z}_{p}}$, which
proves that
$f\in\textnormal{Int}_{\mathbb{Q}_{p}}(F_{p},\overline{\mathbb{Z}_{p}})$. The
other containment
$\textnormal{Int}_{\mathbb{Q}_{p}}(F_{p},\overline{\mathbb{Z}_{p}})\subseteq\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})$
follows in the same way.

Let $p\in\mathbb{P}$ be a fixed prime and set
$\widehat{R}_{p,E_{p}}=\textnormal{Int}_{\mathbb{Q}_{p}}(E_{p},\overline{\mathbb{Z}_{p}})$
and
$\widehat{R}_{p,F_{p}}=\textnormal{Int}_{\mathbb{Q}_{p}}(F_{p},\overline{\mathbb{Z}_{p}})$.
Since $E_{p},F_{p}$ are subsets of $\overline{\mathbb{Z}_{p}}$ of bounded
degree, there exists a finite Galois extension $K$ of $\mathbb{Q}_{p}$
containing both of them. By (2.19),
$\widehat{R}_{p,E_{p}}=\textnormal{Int}_{\mathbb{Q}_{p}}(G_{p}(E_{p}),\overline{\mathbb{Z}_{p}})$
and
$\widehat{R}_{p,F_{p}}=\textnormal{Int}_{\mathbb{Q}_{p}}(G_{p}(F_{p}),\overline{\mathbb{Z}_{p}})$.
Clearly, $\widehat{R}_{p,E_{p}}$ and $\widehat{R}_{p,F_{p}}$ are equal if and
only if they have the same integral closure in $K(X)$. By Lemma 2.20, this
amounts to say that

$\textnormal{Int}_{K}(G_{p}(E_{p}),\overline{\mathbb{Z}_{p}})=\textnormal{Int}_{K}(G_{p}(F_{p}),\overline{\mathbb{Z}_{p}}).$
(2.22)

Note that the rings of the equality (2.22) are equal to
$\textnormal{Int}_{K}(G_{p}(E_{p}),O_{K})$,
$\textnormal{Int}_{K}(G_{p}(F_{p}),O_{K})$, respectively, where $O_{K}$ is the
ring of integers of $K$. Moreover, $G_{p}(E_{p})$ is a closed subset of
$O_{K}$, being a finite union of closed sets $\sigma(E_{p})$,
$\sigma\in\text{Gal}(K/\mathbb{Q}_{p})$. Similarly, $G_{p}(F_{p})$ is closed.

Finally, by [18, Lemma 2], equation (2.22) holds if and only if
$G_{p}(E_{p})=G_{p}(F_{p})$.∎

In particular, Theorem 2.21 implies that the rings
$\textnormal{Int}_{\mathbb{Q}}(\\{\alpha\\},\widehat{\mathbb{Z}})$,
$\alpha\in\widehat{\mathbb{Z}}$, are in one-to-one correspondence with the
elements of $\widehat{\mathbb{Z}}$.

##  3 Construction of a Dedekind domain with prescribed class group

We review Chang’s construction in [6] that we mentioned in the Introduction
and modify it in order to show that, given a group $G$ which is the direct sum
of a countable family of finitely generated abelian groups, there exists a
Dedekind domain $R$, $\mathbb{Z}[X]\subset R\subseteq\mathbb{Q}[X]$ with class
group $G$. As in Eakin and Heinzer’s result of [8], we show first that the
ring constructed by Chang can also be represented as a generalized ring of
integer-valued polynomials. In [6, Lemma 3.4] it is proved that for each
$n\in\mathbb{N}$ and $p\in\mathbb{P}$, there exists a DVR of $\mathbb{Q}(X)$
which is a residually algebraic extension of $\mathbb{Z}_{(p)}$ with
ramification index equal to $n$; by means of Theorem 1.1, we can give an
explicit representation of such an extension in terms of a valuation domain
$V_{p,\alpha}$ associated to some $\alpha\in\overline{\mathbb{Z}_{p}}$ which
generates a totally ramified extension of $\mathbb{Q}_{p}$ of degree $n$.

Let $I$ be a countable set and $G=\bigoplus_{i\in I}G_{i}$ be a direct sum of
finitely generated abelian groups $G_{i}$. Suppose that for each $i\in I$ we
have

$G_{i}\cong\mathbb{Z}^{m_{i}}\oplus\mathbb{Z}/n_{i,1}\mathbb{Z}\oplus\ldots\oplus\mathbb{Z}/n_{i,k_{i}}\mathbb{Z}$

for some uniquely determined non-negative integers
$m_{i},n_{i,1},\ldots,n_{i,k_{i}}$ with $n_{i,j}\mid n_{i,j+1}$. We partition
$\mathbb{P}$ into a family of finite subsets $\\{\mathbb{P}_{i}\\}_{i\in I}$
each of which contains arbitrary chosen $1+k_{i}$ primes, namely
$\mathbb{P}_{i}=\\{p_{i},q_{i,1},\ldots,q_{i,k_{i}}\\}$ and correspondingly
for each $i\in I$ we fix the following $1+k_{i}$ sets:

  * i)

$E_{p_{i}}$ is a finite subset of $\mathbb{Z}_{p_{i}}$ of $m_{i}+1$ elements
$\\{\alpha_{p_{i},1},\ldots,\alpha_{p_{i},m_{i}+1}\\}$ which are
transcendental over $\mathbb{Q}$.

  * ii)

for $j=1,\ldots,k_{i}$, $E_{q_{i,j}}=\\{\alpha_{q_{i,j}}\\}$ a singleton of
$\overline{\mathbb{Z}_{q_{i,j}}}$ such that $\alpha_{q_{i,j}}$ is
transcendental over $\mathbb{Q}$ and $n_{i,j}=e_{\alpha_{q_{i,j}}}$, the
ramification index of $\mathbb{Q}_{p}(\alpha_{q_{i,j}})$ over
$\mathbb{Q}_{p}$.

We set $\underline{E}_{i}=E_{p_{i}}\times\prod_{j=1}^{k_{i}}E_{q_{i,j}}$ and
also

$R_{i}=\textnormal{Int}_{\mathbb{Q}}(E_{p_{i}},\mathbb{Z}_{p_{i}})\cap\bigcap_{j=1}^{k_{i}}\textnormal{Int}_{\mathbb{Q}}(E_{q_{i,j}},\overline{\mathbb{Z}}_{q_{i,j}})=\textnormal{Int}_{\mathbb{Q}}(\underline{E}_{i},\overline{\widehat{\mathbb{Z}}}).$

Since each of the unitary valuation overrings of $R_{i}$, namely
$V_{p,\alpha_{p}}$, $p\in\mathbb{P}_{i}$ and $\alpha_{p}\in E_{p}$, is a DVR
which is residually algebraic over $\mathbb{F}_{p}$ ([19, Proposition 2.2]),
by [8, Theorem & Corollary] $R_{i}$ is a Dedekind domain with class group
isomorphic to $G_{i}$.

We also set

$R=\bigcap_{i\in
I}R_{i}=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$

where $\underline{E}=\prod_{i}\underline{E}_{i}$. By [6, Corollary 2.6], $R$
is an almost Dedekind domain with class group isomorphic to $G$.

As we already mentioned at the beginning of §2.2, the ring
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is not Dedekind in general. By Theorem 2.15, this happens precisely when
$\underline{E}$ is polynomially factorizable. By a suitable modification of
the above construction, we are going to show that there exists a polynomially
factorizable subset $\underline{E}$ of $\overline{\widehat{\mathbb{Z}}}$ such
that $R$ is Dedekind with class group isomorphic to $G$, thus giving a
positive answer to [6, Question 3.7].

######  Theorem 3.1.

Let $G$ be a direct sum of a countable family $\\{G_{i}\\}_{i\in I}$ of
finitely generated abelian groups. Then there exists a Dedekind domain $R$
between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ with class group isomorphic to
$G$. Moreover, for each $i\in I$, there exists a multiplicative subset $S_{i}$
of $\mathbb{Z}$ such that $S_{i}^{-1}R$ is a Dedekind domain with class group
$G_{i}$.

###### Proof.

We keep the notation used in the above construction. Let
$\mathbb{P}_{r}=\bigcup_{i\in I}(\mathbb{P}_{i}\setminus\\{p_{i}\\})$. For
each $q=q_{i,j}\in\mathbb{P}_{r}$, for some $i\in I$ and
$j\in\\{1,\ldots,k_{i}\\}$, we set $n_{q}=n_{i,j}$. We choose a uniformizer
$\tilde{q}$ of $\mathbb{Z}_{q}$ which is transcendental over $\mathbb{Q}$. Let
$\tilde{\alpha}_{q}\in\overline{\mathbb{Z}_{q}}$ be a root of the Eisenstein
polynomial $X^{n_{q}}-\tilde{q}$. Clearly, $\tilde{\alpha}_{q}$ is still
transcendental over $\mathbb{Q}$ and it is well-known that
$\mathbb{Q}_{q}(\tilde{\alpha}_{q})$ is a totally ramified extension of
$\mathbb{Q}_{q}$ of degree $n_{q}$. We let now
$\alpha_{q}=\tilde{\alpha}_{q}+\lfloor\log q\rfloor$: this is another
generator of $\mathbb{Q}_{q}(\tilde{\alpha}_{q})$ over $\mathbb{Q}_{q}$ which
still is transcendental over $\mathbb{Q}$ and has $v_{q}$-adic valuation zero.
We then set $E_{q}=\\{\alpha_{q}\\}$ in the above construction.

Similarly, for each $p=p_{i}\in\mathbb{P}\setminus\mathbb{P}_{r}$, for some
$i\in I$, let $m_{p}=m_{p_{i}}$. We choose distinct elements
$\alpha_{p,i}\in\lfloor\log p\rfloor+p\mathbb{Z}_{p}$, for
$i=1,\ldots,m_{p}+1$, which are transcendental over $\mathbb{Q}$ and set
$E_{p}=\\{\alpha_{p,1},\ldots,\alpha_{p,m_{p}}\\}$.

We show now that with these choices the subset
$\underline{E}=\prod_{p}E_{p}\subset\overline{\widehat{\mathbb{Z}}}$ is
polynomially factorizable, and therefore the corresponding domain
$R=\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
is a Dedekind domain by Theorem 2.15. By Lemma 2.12, we need to show that for
each $g\in\mathbb{Z}[X]$, $\mathbb{P}_{g,\underline{E}}$ is finite. Let
$g\in\mathbb{Z}[X]$ be a fixed polynomial, and let
$\alpha=(\alpha_{p})\in\underline{E}$.

  * -

if $p\in\mathbb{P}\setminus\mathbb{P}_{r}$, then $\alpha_{p}=pa+\lfloor\log
p\rfloor$, for some $a\in\mathbb{Z}_{p}$.

  * -

if $p\in\mathbb{P}_{r}$, then $\alpha_{p}=\tilde{\alpha}_{p}+\lfloor\log
p\rfloor$, where $\tilde{\alpha}_{p}$ is a root of an Eisenstein polynomial,
so, in particular, $v_{p}(\tilde{\alpha}_{p})>0$.

In either case, for each $p\in\mathbb{P}$, let $\pi_{p}$ be a uniformizer of
$\mathbb{Q}_{p}(\alpha_{p})$ (which is just $p$ if $p\notin\mathbb{P}_{r}$).
We then have

$g(\alpha_{p})\equiv g(\lfloor\log p\rfloor)\pmod{\pi_{p}}.$

Now, for all $p$ sufficiently large, $g(\lfloor\log p\rfloor)$ is not
divisible by $p$, since $\lim_{x\to\infty}\frac{g(\log x)}{x}=0$. Hence,
$\mathbb{P}_{g,\underline{E}}$ is finite.

The fact that
$\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})$
has class group equal to $G$ follows either by [6, Corollary 2.6] or by
applying Lemma 2.14 and Proposition 2.10, by noting that
$\text{Pic}(\textnormal{Int}_{\mathbb{Q}}(E_{p},\overline{\mathbb{Z}_{p}}))=\mathbb{Z}^{m_{p}}$
for each $p\in\mathbb{P}\setminus\mathbb{P}_{r}$ and
$\text{Pic}(\textnormal{Int}_{\mathbb{Q}}(E_{q},\overline{\mathbb{Z}_{q}}))=\mathbb{Z}/n_{q}\mathbb{Z}$
for each $q\in\mathbb{P}_{r}$.

For the last claim, if $i\in I$, we let $S_{i}$ be the multiplicative subset
of $\mathbb{Z}$ generated by $\mathbb{P}\setminus\mathbb{P}_{i}$. Then, by
Lemma 2.5,
$S_{i}^{-1}\textnormal{Int}_{\mathbb{Q}}(\underline{E},\overline{\widehat{\mathbb{Z}}})=\textnormal{Int}_{\mathbb{Q}}(\underline{E}_{i},\overline{\widehat{\mathbb{Z}}})$
which has class group isomorphic to $G_{i}$ by Lemma 2.14 and Proposition
2.10. ∎

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