# Ring Structure of Integer-Valued Rational Functions

Baian Liu

###### Abstract

Integer-valued rational functions are a natural generalization of integer-
valued polynomials. Given a domain $D$, the collection of all integer-valued
rational functions over $D$ forms a ring extension
$\operatorname{Int{}^{\text{R}}}(D)$ of $D$. For a valuation domain $V$, we
characterize when $\operatorname{Int{}^{\text{R}}}(V)$ is a Prüfer domain and
when $\operatorname{Int{}^{\text{R}}}(V)$ is a Bézout domain. We also extend
the classification of when $\operatorname{Int{}^{\text{R}}}(D)$ is a Prüfer
domain.

##  1 Introduction

Integer-valued polynomials appear in many areas of mathematics, including
Hilbert polynomials of polynomial rings and interpolation formulas. A natural
generalization of integer-valued polynomials is integer-valued rational
functions. One of the first analyses of integer-valued rational functions was
done by Kochen [Koc69]. Kochen uses rings of integer-valued rational functions
to help determine when Diophantine equations have an integral solution over a
$p$-adically closed field.

We are interested in studying sets of integer-valued rational functions over a
ring as a ring itself. We will be investigating the ring-theoretic properties
of these rings of integer-valued rational functions on their own and in
relation to the ring over which they are defined.

We start with a domain $D$. Then we want to define the ring of integer-valued
rational functions $\operatorname{Int{}^{\text{R}}}(D)$ over $D$, as well as
some notions to help us investigate $\operatorname{Int{}^{\text{R}}}(D)$.

######  Definition 1.1.

Let $D$ be a domain and $K$ its field of fractions. We define the ring of
integer-valued rational functions over $D$ to be

$\operatorname{Int{}^{\text{R}}}(D)=\\{\varphi\in K(x)\mid\varphi(D)\subseteq
D\\}.$

######  Remark 1.2.

The set $\operatorname{Int{}^{\text{R}}}(D)$ is in fact a ring which is also
closed under composition.

Requiring $\varphi(D)\subseteq D$ means that $\varphi$ cannot have any poles
in $D$. Since a rational function only has finitely many poles, we many choose
to ignore these finitely many elements. This turns out to not change the set
of rational functions we are considering, since if $D$ is not a field, a
rational function $\varphi\in K(x)$ such that $\varphi(d)\in D$ for all but
finitely many $d\in D$ is in $\operatorname{Int{}^{\text{R}}}(D)$ [CC97, p.
260].

We can also obtain a ring by taking the collection of all rational functions
that are integer-valued on some subset of the field of fractions. These rings
can help give a more nuanced description of what properties of $D$ lead to
certain properties of $\operatorname{Int{}^{\text{R}}}(D)$.

######  Definition 1.3.

Take $D$ to be a domain and $K$ its field of fractions. Let $E$ be some subset
of $K$. We can more generally define the ring of integer-valued rational
functions on $E$ over $D$ to be

$\operatorname{Int{}^{\text{R}}}(E,D)=\\{\varphi\in
K(x)\mid\varphi(E)\subseteq D\\}.$

In particular, for the ring $\operatorname{Int{}^{\text{R}}}(K,D)$, we may
also choose to ignore finitely many values in determining if a rational
function is integer-valued on $K$.

######  Proposition 1.4.

Let $D$ be a domain that is not a field and $K$ its field of fractions. If
$\varphi\in K(x)$ is such that $\varphi(r)\in D$ for almost all $r\in K$, then
$\varphi\in\operatorname{Int{}^{\text{R}}}(K,D)$.

###### Proof.

Let $b\in D$ such that $b\neq 0$. Define
$\psi_{b}(x)=\varphi\mathopen{}\mathclose{{}\left(\frac{x}{b}}\right)$. By
assumption, $\psi_{b}$ is almost integer-valued on $D$, so $\psi_{b}$ is
integer-valued on $D$. Now let $d\in K$. We can write $d=\frac{a}{b}$ for some
$a,b\in D$ with $b\neq 0$. Then $\varphi(d)=\psi_{b}(a)\in D$. Thus,
$\varphi\in\operatorname{Int{}^{\text{R}}}(K,D)$. ∎

The definition of rings of integer-valued rational functions appears to be
very similar to that of rings of integer-valued polynomials. Despite this, we
will see that the behavior of rings of integer-valued rational functions can
be vastly different from that of rings of integer-valued polynomials. For
example, rings of integer-valued rational functions are not as sensitive to
infinite residue fields. In the case of integer-valued polynomials, if $D$ is
a domain with infinite residue fields, then $\operatorname{Int}(D)=D[x]$
[CC97, p. 10].

One notion that marks a difference between integer-valued rational functions
and polynomials is the notion of unit-valued polynomials. Take $D$ to be a
domain. Because the inverse of a nonconstant polynomial is not a polynomial,
unit-valued polynomials are not units in $\operatorname{Int}(D)$. However, the
inverse of a polynomial is a rational function, so unit-valued polynomials are
units in $\operatorname{Int{}^{\text{R}}}(D)$.

######  Definition 1.5.

Let $D$ be a domain and let $K$ be the field of fractions of $D$. Then a
polynomial $f\in D[x]$ is unit-valued over $D$ if $f(D)\subseteq D^{\times}$.
We will denote the set of all unit-valued polynomials by

$T=\\{f\in D[x]\mid f(D)\subseteq D^{\times}\\}.$

######  Remark 1.6.

Let $\operatorname{Int}(D)$ denote the ring of unit-valued polynomials of $D$.
We can see that $T$ is a multiplicative subset of $\operatorname{Int}(D)$ and
we have the containment

$T^{-1}\operatorname{Int}(D)\subseteq\operatorname{Int{}^{\text{R}}}(D).$

The containment
$T^{-1}\operatorname{Int}(D)\subseteq\operatorname{Int{}^{\text{R}}}(D)$ can
be strict (see Example 1.13), so unit-valued polynomials do not explain all of
the differences between $\operatorname{Int}(D)$ and
$\operatorname{Int{}^{\text{R}}}(D)$. Nevertheless, unit-valued polynomials
are a useful tool for describing some of the structure of
$\operatorname{Int{}^{\text{R}}}(D)$. In addition, unit-valued polynomials
over a domain $D$ are closely linked to the residue fields of $D$, so residue
fields of $D$ can help describe the structure of
$\operatorname{Int{}^{\text{R}}}(D)$.

######  Proposition 1.7.

Let $D$ be a local domain with maximal ideal $\mathfrak{m}$. Then $f\in D[x]$
is unit-valued if and only if $f\mod\mathfrak{m}$ has no roots in
$D/\mathfrak{m}$.

###### Proof.

We have that $f\in D[x]$ is unit-valued if and only if
$f(d)\notin\mathfrak{m}$ for any $d\in D$, which happens if and only if
$f\mod\mathfrak{m}$ has no roots in $D/\mathfrak{m}$.

∎

######  Corollary 1.8.

Let $D$ be a domain. Then $f\in D[x]$ is unit-valued if and only if for every
maximal ideal $\mathfrak{m}$ of $D$, the polynomial $f\mod\mathfrak{m}$ has no
roots in $D/\mathfrak{m}$.

###### Proof.

If $f$ is unit-valued over $D$, then $f$ is valued in $D\setminus\mathfrak{m}$
for each maximal ideal $\mathfrak{m}$, so $f\mod\mathfrak{m}$ has no roots in
$D/\mathfrak{m}$. Conversely, if $f$ is unit-valued over $D_{\mathfrak{m}}$
for every maximal ideal $\mathfrak{m}$ of $D$, then $f(a),\frac{1}{f(a)}\in D$
for each $a\in D$, which means $f$ is unit-valued over $D$. ∎

There exist domains $D$ such that
$\operatorname{Int{}^{\text{R}}}(D)=\operatorname{Int}(D)$. In other words,
all of the integer-valued rational functions over $D$ are polynomials. This
means analyzing $\operatorname{Int{}^{\text{R}}}(D)$ is no different than
analyzing $\operatorname{Int}(D)$, so we are only interested in domains $D$
such that $\operatorname{Int{}^{\text{R}}}(D)\neq\operatorname{Int}(D)$. One
observation we make is that if $D$ is a domain such that there exists a
nonconstant unit-valued polynomial $f$, then
$\frac{1}{f}\in\operatorname{Int{}^{\text{R}}}(D)\setminus\operatorname{Int}(D)$.
A ring with such a property is called a non-D-ring.

######  Definition 1.9.

Let $D$ be a domain. We call $D$ a non-D-ring if there is a nonconstant unit-
valued polynomial $f\in D[x]$.

######  Example 1.10.

Let $D$ be a domain with a nonzero Jacobson radical, such as a semi-local
domain or a local domain. Then $D$ is a non-D-ring. To see this, take $d$ to
be a nonzero element in the Jacobson radical. Then $dx+1$ is a nonconstant
unit-valued polynomial.

The Jacobson radical of $D$ need not to be nonzero for $D$ to be a non-D-ring,
as seen in the following example.

######  Example 1.11.

[Lop88, Example 1.11] The domain
$D={\mathbb{Z}}\mathopen{}\mathclose{{}\left[\frac{1}{p}\,\middle|\,p\equiv
1,2\pmod{4},\text{$p$ is a prime}}\right]$ has unit-valued polynomial
$x^{2}+1$ and is therefore a non-D-ring.

The property of being a non-D-ring turns out to be exactly the one we want to
consider in order to study rings of integer-valued rational functions.

######  Proposition 1.12.

[GM76, Proposition 1] Let $D$ be a domain. Then $D$ is a non-D-ring if and
only if $\operatorname{Int{}^{\text{R}}}(D)\neq\operatorname{Int}(D)$.

In general, the ring $\operatorname{Int{}^{\text{R}}}(D)$ is not the
localization of $\operatorname{Int}(D)$ by unit-valued polynomials. The
following is an example of a domain $D$ such that
$\operatorname{Int{}^{\text{R}}}(D)\neq T^{-1}\operatorname{Int}(D)$.

######  Example 1.13.

Let $V$ be a valuation domain with an infinite residue field and a principal
maximal ideal, generated by some $t\in V$. Then we claim that
$\varphi(x)\coloneqq\frac{t}{x^{2}+t}\in\operatorname{Int{}^{\text{R}}}(V)\setminus
T^{-1}\operatorname{Int}(V)$.

Let $v$ be the valuation associated with $V$. Take $d\in V$. If $v(d)=0$, then
$v(\varphi(d))=v(t)\geq 0$. If $v(d)>0$, then $v(\varphi(d))=0$. Thus,
$\varphi\in\operatorname{Int{}^{\text{R}}}(V)$.

Now suppose that $\varphi=\frac{f}{g}$, where $f\in\operatorname{Int}(V)$ and
$g\in T$. Since $V$ is local with infinite residue field, we have
$\operatorname{Int}(V)=V[x]$. Then we obtain

$g=\frac{x^{2}+t}{t}\cdot f.$

Let $d\in V$ such that $v(d)=0$. Evaluate at $x=d$ to get
$g(d)=\frac{d^{2}+t}{t}\cdot f(d)$. We see that $v(f(d))=v(t)$, so
$f(d)=0\mod(t)$. Since $V/(t)$ is infinite, we must have that $f(x)\mod(t)$ is
the zero polynomial. However, evaluating $x=0$, we get that $g(0)=f(0)$, so
$f(0)$ is a unit and $f(x)\mod(t)$ cannot be the zero polynomial, a
contradiction. Thus, $\varphi\notin T^{-1}\operatorname{Int}(V)$.

We can study the structure of $\operatorname{Int{}^{\text{R}}}(E,D)$ via its
ideals. In particular, we can analyze its prime and maximal ideals. Since
$\operatorname{Int{}^{\text{R}}}(E,D)$ consists of functions valued in $D$, we
can define some of the ideals of $\operatorname{Int{}^{\text{R}}}(E,D)$ via
ideals of $D$. We call these pointed ideals.

######  Definition 1.14.

Let $D$ be a domain with field of fractions $K$. Take $E$ to be a subset of
$K$. Also let $I$ be an ideal of $D$ and $a\in E$. Then define

$\mathfrak{I}_{I,a}=\\{\varphi\in\operatorname{Int{}^{\text{R}}}(E,D)\mid\varphi(a)\in
I\\}.$

Ideals of $\operatorname{Int{}^{\text{R}}}(E,D)$ this form are called pointed
ideals. If $I$ is a prime ideal $\mathfrak{p}$ of $D$, we use the notation
$\mathfrak{P}_{\mathfrak{p},a}=\mathfrak{I}_{\mathfrak{p},a}$ and call these
pointed prime ideal.

If $\mathfrak{m}$ is a maximal ideal of $D$, then we use the notation
$\mathfrak{M}_{\mathfrak{m},a}$ for $\mathfrak{P}_{\mathfrak{m},a}$. We call
ideals of $\operatorname{Int{}^{\text{R}}}(E,D)$ of this form pointed maximal
ideals.

######  Remark 1.15.

The notation $\mathfrak{P}_{I,a}$ does not indicate the ring $D$ and subset
$E$, so $D$ and $E$ are understood from context.

Note that if $\mathfrak{p}$ is a prime ideal of $D$, then
$\mathfrak{P}_{\mathfrak{p},a}$ is a prime ideal of
$\operatorname{Int{}^{\text{R}}}(E,D)$, so it is justified to call
$\mathfrak{P}_{\mathfrak{p},a}$ a pointed prime ideal. Moreover, the pointed
maximal ideal $\mathfrak{M}_{\mathfrak{m},a}$ is indeed a maximal ideal of
$\operatorname{Int{}^{\text{R}}}(E,D)$.

######  Proposition 1.16.

Let $D$ be a domain and $E$ a subset of the field of fractions. If
$\mathfrak{p}$ is a prime ideal of $D$, then for any $a\in E$, we have
$\operatorname{Int{}^{\text{R}}}(E,D)/\mathfrak{P}_{\mathfrak{p},\alpha}\cong
D/\mathfrak{p}$.

###### Proof.

Consider the map $\operatorname{Int{}^{\text{R}}}(E,D)\to D/\mathfrak{p}$
given by $\varphi\mapsto\varphi(a)\mod\mathfrak{p}$. This map is surjective
since the constant functions are in $\operatorname{Int{}^{\text{R}}}(E,D)$.
Furthermore, the kernel of this map are rational functions in
$\operatorname{Int{}^{\text{R}}}(E,D)$ such that their evaluation at $a$
modulo $\mathfrak{p}$ is 0, so the kernel is exactly
$\mathfrak{P}_{\mathfrak{p},\alpha}$. Thus,
$\operatorname{Int{}^{\text{R}}}(E,D)/\mathfrak{P}_{\mathfrak{p},\alpha}\cong
D/\mathfrak{p}$. ∎

######  Remark 1.17.

In particular, for any maximal ideal $\mathfrak{m}$ of $D$, we have
$\operatorname{Int{}^{\text{R}}}(E,D)/\mathfrak{M}_{\mathfrak{m},a}\cong
D/\mathfrak{m}$, implying that $\mathfrak{M}_{\mathfrak{m},a}$ is a maximal
ideal in $\operatorname{Int{}^{\text{R}}}(E,D)$.

In general, the pointed ideals of $\operatorname{Int{}^{\text{R}}}(E,D)$ are
not sufficient to describe all of the ideals of
$\operatorname{Int{}^{\text{R}}}(E,D)$. Also, in general, the pointed prime
ideals do not describe all of the prime ideals either, and even the pointed
maximal ideals do not describe all of the maximal ideals in general. However,
we can use ultrafilters to describe more of the ideals. We first give the
definition of an ultrafilter.

######  Definition 1.18.

Let $S$ be a set. An ultrafilter $\mathcal{U}$ on $S$ is a set of subsets of
$S$ such that

  * •

$\emptyset\notin\mathcal{U}$,

  * •

if $A\subseteq B\subseteq S$ and $A\in\mathcal{U}$, then $B\in\mathcal{U}$,

  * •

if $A,B\in\mathcal{U}$, then $A\cap B\in\mathcal{U}$,

  * •

if $A\subseteq S$, then $A\in\mathcal{U}$ or $S\setminus A\in\mathcal{U}$.

Fix an element $s\in S$. The collection of all subsets of $S$ containing $s$
is an ultrafilter. We call ultrafilters of this form principal and non-
principal otherwise.

######  Remark 1.19.

If $S$ is an infinite set, there exist non-principal ultrafilters on $S$ by
Zorn’s Lemma.

Now we use ultrafilters to take ultrafilter limits of a set of prime ideals.

We can define a prime ideal from a collection of prime ideals using
ultrafilters in a more general setting. The general construction does not
limit us to rings of integer-valued rational functions.

######  Definition 1.20.

Let $R$ be a commutative ring and
$\\{\mathfrak{p}_{\lambda}\\}_{\lambda\in\Lambda}$ a subset of
$\operatorname{Spec}(R)$. Consider an element $a\in R$. The characteristic set
of $a$ with respect to $\\{\mathfrak{p}_{\lambda}\\}$ is defined as

$\Phi_{a}\coloneqq\\{\mathfrak{p}_{\lambda}\mid
a\in\mathfrak{p}_{\lambda}\\}.$

Take $\mathcal{U}$ to be an ultrafilter of $\\{\mathfrak{p}_{\lambda}\\}$. We
define the ultrafilter limit of $\\{\mathfrak{p}_{\lambda}\\}$ with respect to
$\mathcal{U}$ as

$\lim\limits_{\mathcal{U}}\mathfrak{p}_{\lambda}\coloneqq\\{a\in
R\mid\Phi_{a}\in\mathcal{U}\\}.$

######  Remark 1.21.

Since any ultrafilter on $\\{\mathfrak{p}_{\lambda}\\}$ can be extended
uniquely to an ultrafilter on $\operatorname{Spec}(R)$ containing
$\\{\mathfrak{p}_{\lambda}\\}$ and any ultrafilter on $\operatorname{Spec}(R)$
containing $\\{\mathfrak{p}_{\lambda}\\}$ can be obtained this way, we may
take $\mathcal{U}$ to be an ultrafilter of $\\{\mathfrak{p}_{\lambda}\\}$ or
an ultrafilter of $\operatorname{Spec}(R)$ containing
$\\{\mathfrak{p}_{\lambda}\\}$.

Using the definition of an ultrafilter, we may confirm that
$\lim\limits_{\mathcal{U}}\mathfrak{p}_{\lambda}$ is a prime ideal. In
particular, if we take $\\{\mathfrak{p}_{\lambda}\\}$ to be a set of pointed
prime ideals of $\operatorname{Int{}^{\text{R}}}(E,D)$, then the ultrafilter
limit of $\\{\mathfrak{p}_{\lambda}\\}$ with respect a non-principal
ultrafilter can yield a prime ideal of $\operatorname{Int{}^{\text{R}}}(E,D)$
that is not a pointed prime ideal.

For rings of integer-valued polynomials, we have
$S^{-1}\operatorname{Int}(D)\subseteq\operatorname{Int{}^{\text{R}}}(S^{-1}D)$
for any multiplicative subset $S$ of a domain $D$ [CC97, Proposition I.2.2].
However, for rings of integer-valued rational functions, we don’t necessarily
have inclusion of $S^{-1}\operatorname{Int{}^{\text{R}}}(D)$ in
$\operatorname{Int{}^{\text{R}}}(S^{-1}D)$ for $S$ a multiplicative subset of
$D$.

######  Example 1.22.

Let $k$ be a field and let $K=k(s,t)$ with a valuation
$v:K\to{\mathbb{Z}}\oplus{\mathbb{Z}}\cup\\{\infty\\}$ given by

$v\mathopen{}\mathclose{{}\left(\sum_{i,j}a_{ij}s^{i}t^{j}}\right)=\min\limits_{i,j}\\{(i,j)\\}$

for each nonzero $\sum_{i,j}a_{ij}s^{i}t^{j}\in k[s,t]$, where each $a_{ij}\in
k\setminus\\{0\\}$, and extended uniquely to $K$. The value group is ordered
lexicographically. Let $D$ be the associated valuation domain. Its prime
spectrum is $(0)\subsetneq(s,s/t,s/t^{2},s/t^{3},\dots)\subsetneq(t)$. If
$S=D\setminus(s,s/t,s/t^{2},s/t^{3},\dots)$, then
$\frac{1/t}{x-1/t}\in\operatorname{Int{}^{\text{R}}}(D)$ but
$\frac{1/t}{x-1/t}\notin\operatorname{Int{}^{\text{R}}}(S^{-1}D)$.

In Section 2, we discuss rings of integer-valued rational functions over
valuation domains. For a valuation domain $V$, we completely determine when
$\operatorname{Int{}^{\text{R}}}(V)$ is a Prüfer domain. We also completely
determine when $\operatorname{Int{}^{\text{R}}}(V)$ is a Bézout domain. When
$\operatorname{Int{}^{\text{R}}}(V)$ is not a Prüfer domain, we determine
prime ideals that are not essential.

In Section 3, we consider integer-valued rational functions over a Prüfer
domain $D$. We give some conditions when $\operatorname{Int{}^{\text{R}}}(D)$
is not Prüfer and a family of Prüfer domains such that
$\operatorname{Int{}^{\text{R}}}(D)$ is Prüfer for each domain $D$ in this
family.

##  2 Integer-valued rational functions over valuation domains

In this section, we let $V$ be a valuation domain, $K$ its field of fractions,
$\mathfrak{m}$ the maximal ideal of $V$, $v$ the associated valuation, and
$\Gamma$ the value group. We investigate whether
$\operatorname{Int{}^{\text{R}}}(V)$ is a Prüfer domain or not. Prüfer domains
are of particular interest since they possess many nice properties, such as
satisfying a generalized version of the Chinese Remainder Theorem or being a
generalized notion of a Dedekind domain. Prüfer domains can also be seen as a
global version of a valuation domain.

Not only are valuations powerful tools that assist in analyzing the ring of
integer-valued rational function, but valuations also induce a topology that
interacts well with integer-valued rational functions. The following proof is
a modification of [CC97, Proposition X.2.1].

######  Proposition 2.1.

Let $D$ be a domain with field of fractions $K$ and $E$ a subset of $K$. Let
$v:K^{\times}\to\Gamma$ be a valuation such that the induced valuation ring
$V$ contains $D$. Then each element of $\operatorname{Int{}^{\text{R}}}(E,D)$
is a continuous function from $E$ to $D$ with respect to the topology induced
by the valuation.

###### Proof.

Let $\varphi\in\operatorname{Int{}^{\text{R}}}(E,D)$. We can write
$\varphi=\frac{f}{g}$ for $f,g\in K[x]$ relatively prime polynomials. Fix
$a\in E$, and let $b\in E$ and $\varepsilon\in\Gamma$. We calculate

$\varphi(b)-\varphi(a)=\frac{f(b)}{g(b)}-\frac{f(a)}{g(a)}=\frac{f(b)-f(a)}{g(a)}-\frac{f(b)}{g(b)}\cdot\frac{g(b)-g(a)}{g(a)}.$

Since $g(a)\neq 0$, we can say that $v(g(a))=\gamma$ for some
$\gamma\in\Gamma$. Let $\delta\in\Gamma$. Since $f$ and $g$ are continuous
with respect to the topology induced by the valuation, there is some
$\delta\in\Gamma$ such that $v(a-b)>\delta$ implies
$v(f(a)-f(b)),v(g(a)-g(b))>\varepsilon+\gamma$.

Now, we see that if $v(a-b)>\delta$, then we have
$v(\varphi(b)-\varphi(a))\geq\min\mathopen{}\mathclose{{}\left\\{v\mathopen{}\mathclose{{}\left(\frac{f(b)-f(a)}{g(a)}}\right),v\mathopen{}\mathclose{{}\left(\frac{f(b)}{g(b)}\cdot\frac{g(b)-g(a)}{g(a)}}\right)}\right\\}$.
We know that
$v\mathopen{}\mathclose{{}\left(\frac{f(b)-f(a)}{g(a)}}\right)>\varepsilon+\gamma-\gamma=\varepsilon$
and
$v\mathopen{}\mathclose{{}\left(\frac{f(b)}{g(b)}\cdot\frac{g(b)-g(a)}{g(a)}}\right)=v\mathopen{}\mathclose{{}\left(\frac{f(b)}{g(b)}}\right)+v\mathopen{}\mathclose{{}\left(\frac{g(b)-g(a)}{g(a)}}\right)>0+\varepsilon+\gamma-\gamma=\varepsilon$
since $\frac{f(b)}{g(b)}\in D\subseteq V$. Thus,
$v(\varphi(b)-\varphi(a))>\varepsilon$, showing that $\varphi$ is continuous
at $a$, which means that $\varphi$ is continuous since $a\in E$ was
arbitrarily chosen. ∎

For studying rings of integer-valued rational functions, there are certain
Prüfer domains that are of particular interest to us. The following
definitions are from [CL98]:

######  Definition 2.2.

A Prüfer domain $D$ is monic if there is a monic unit-valued polynomial in
$D$.

######  Definition 2.3.

Let $D$ be a Prüfer domain. The Prüfer domain $D$ is singular if there exists
a family $\Lambda$ of maximal ideals of $D$ such that

  * •

$D=\bigcap\limits_{\mathfrak{m}\in\Lambda}D_{\mathfrak{m}}$,

  * •

for each $\mathfrak{m}\in\Lambda$, the maximal ideal of $D_{\mathfrak{m}}$ is
a principal ideal, generated by some $t_{\mathfrak{m}}\in D_{\mathfrak{m}}$,
and

  * •

there is an element $t\in D$ and an integer $n$ such that, for each
$\mathfrak{m}\in\Lambda$,
$0<v_{\mathfrak{m}}(t)<nv_{\mathfrak{m}}(t_{\mathfrak{m}})$, where
$v_{\mathfrak{m}}$ is the valuation associated with $D_{\mathfrak{m}}$.

An important invariant of a Prüfer domain $D$ is the Picard group of $D$,
denoted as $\operatorname{Pic}(D)$. The Picard group is defined to be the set
of finitely generated ideals of $D$ modulo the principal ideals of $D$. The
group operation is ideal multiplication.

The following result is stated only for $\operatorname{Int{}^{\text{R}}}(D)$
in [Lop94], but the same proof can be used to get the same statement about
$\operatorname{Int{}^{\text{R}}}(E,D)$.

######  Theorem 2.4.

[Lop94] Let $D$ be a monic Prüfer domain with $E$ a subset of $K$, the
quotient field of $D$. Then $\operatorname{Int{}^{\text{R}}}(E,D)$ is a Prüfer
domain with torsion Picard group.

For a Prüfer domain $D,$ the Picard group can be seen as a way to measure how
far $D$ is from being a Bézout domain, since a Bézout domain has trivial
Picard group. This next result shows that rings of integer-valued rational
functions over singular Prüfer domains are Bézout domains, which implies that
they are Prüfer domains as well.

######  Theorem 2.5.

[CL98] Let $D$ be a singular Prüfer domain with $E$ a subset of $K$, the
quotient field of $D$. Then $\operatorname{Int{}^{\text{R}}}(E,D)$ is a Bézout
domain.

We take the results of [Lop94, CL98] and restrict ourselves to looking at
rings of integer-vauled rational functions over valuation rings.

######  Remark 2.6.

Note that a valuation domain is monic if and only if its residue field is not
algebraically closed, and a valuation domain is singular if and only if its
maximal ideal is principal.

######  Corollary 2.7.

If $V$ is a valuation domain with a residue field that is not algebraically
closed or a principal maximal ideal, then $\operatorname{Int{}^{\text{R}}}(V)$
is a Prüfer domain.

In this section, we will explore the converse. To that end, we need a few
definitions.

######  Definition 2.8.

Let $\Gamma$ be an abelian group with a total order on its elements. We say
that $\Gamma$ is a totally ordered abelian group if for
$\alpha,\beta,\gamma,\delta\in\Gamma$, $\alpha\leq\beta$ and
$\gamma\leq\delta$ imply that $\alpha+\gamma\leq\beta+\delta$.

######  Definition 2.9.

A totally ordered abelian group $\Gamma$ is divisible if for all
$\gamma\in\Gamma$ and nonzero $n\in{\mathbb{Z}}$, there exists
$\delta\in\Gamma$ such that $n\delta=\gamma$.

######  Definition 2.10.

Let $\Gamma$ be a totally ordered abelian group. We can define its divisible
hull

${\mathbb{Q}}\Gamma=\mathopen{}\mathclose{{}\left\\{\frac{\gamma}{n}\mid\gamma\in\Gamma,n\in{\mathbb{Z}}_{>0}}\right\\}/\sim,$

where $\frac{\gamma}{n}\sim\frac{\gamma^{\prime}}{m}$ if
$m\gamma=n\gamma^{\prime}$ in $\Gamma$. Then define the group operation to be
$\frac{\gamma}{n}+\frac{\gamma^{\prime}}{m}=\frac{m\gamma+n\gamma^{\prime}}{nm}$
and the ordering to be $\frac{\gamma}{n}\leq\frac{\gamma^{\prime}}{m}$ if and
only if $m\gamma\leq n\gamma^{\prime}$. Equivalently, we may define
${\mathbb{Q}}\Gamma=\Gamma\otimes_{\mathbb{Z}}{\mathbb{Q}}$.

Furthermore, if $v:K\to\Gamma\cup\\{\infty\\}$ is a valuation, we can extend
the valuation to
$\tilde{v}:K(t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma)\to{\mathbb{Q}}\Gamma\cup\\{\infty\\}$
defined as the monomial extension mapping $t_{\gamma}$ to $\gamma$ for every
$\gamma\in{\mathbb{Q}}\Gamma$. More explicitly, for nonzero elements of
$K[t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma]$, we define $\tilde{v}$ to be

$\tilde{v}\mathopen{}\mathclose{{}\left(\sum\limits_{i_{1},\dots,i_{n}}a_{i_{1}\cdots
i_{n}}t_{\gamma_{1}}^{c_{1,i_{1}\cdots i_{n}}}\cdots
t_{\gamma_{n}}^{c_{n,i_{1}\cdots
i_{n}}}}\right)=\min\mathopen{}\mathclose{{}\left\\{v(a_{i_{1}\dots
i_{n}})+\sum_{k=1}^{n}c_{k,i_{1}\cdots i_{n}}\gamma_{k}}\right\\},$

where each $a_{i_{1}\cdots i_{n}}\in K$. Then this extends uniquely to
$K(t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma)$.

######  Remark 2.11.

The divisible hull ${\mathbb{Q}}\Gamma$ of $\Gamma$ is divisible and extends
the ordering on $\Gamma$.

######  Remark 2.12.

If $\mathfrak{m}$ is principal, then the value group $\Gamma$ is not
divisible. Say $\mathfrak{m}=(\varpi)$ for some $\varpi\in V$. There does not
exist an element $a\in V$ such that $2v(a)=v(\varpi)$.

In order to consider many valuations at once, we introduce the notions of
minimum valuation functions and local polynomials. The minimal valuation
function is closely related to monomial valuations.

######  Definition 2.13.

Take a nonzero polynomial $f\in K[x]$ and write it as
$f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ for $a_{0},a_{1},\dots,a_{n}\in K$. We
define the minimum valuation function of $f$ as
$\operatorname{minval}_{f,v}:\Gamma\to\Gamma$ by

$\gamma\mapsto\min\\{v(a_{0}),v(a_{1})+\gamma,v(a_{2})+2\gamma,\dots,v(a_{n})+n\gamma\\}$

for each $\gamma\in\Gamma$. We will denote $\operatorname{minval}_{f,v}$ as
$\operatorname{minval}_{f}$ if the valuation $v$ is clear from context. It is
oftentimes helpful to think of $\operatorname{minval}_{f}$ as a function from
${\mathbb{Q}}{\Gamma}$ to ${\mathbb{Q}}{\Gamma}$ defined as
$\gamma\mapsto\min\\{v(a_{0}),v(a_{1})+\gamma,v(a_{2})+2\gamma,\dots,v(a_{n})+n\gamma\\}$
for each $\gamma\in{\mathbb{Q}}{\Gamma}$.

In the same setup, taking $t\in K$, we can define the local polynomial of $f$
at $t$ to be

$\operatorname{loc}_{f,v,t}(x)=\frac{f(tx)}{a_{d}t^{d}}\mod\mathfrak{m},$

where $d=\max\\{i\in\\{0,1,\dots,n\\}\mid
v(a_{i})+iv(t)=\operatorname{minval}_{f}(v(t))\\}$. Again, we may omit the
valuation $v$ in $\operatorname{loc}_{f,v,t}(x)$ and write
$\operatorname{loc}_{f,t}(x)$ if the valuation is clear from the context. A
priori, we do not know if the coefficients of $\frac{f(tx)}{a_{d}t^{d}}$ are
in $V$. We need to confirm this so that the local polynomial is well-defined.

######  Remark 2.14.

Using the notation above, we compare the minimum valuation function to
monomial valuations. In Definition 3.3 of [Per18], the monomial valuation
(centered at 0) $v_{0,\gamma}$ is defined to be
$v_{0,\gamma}(f)=\min\\{v(a_{i})+i\gamma\mid i=0,\dots,n\\}$. This shows that
$\operatorname{minval}_{f}(\gamma)=v_{0,\gamma}(f)$, so we can think of the
minimum valuation function as ranging over various monomial valuations.

######  Proposition 2.15.

Let $f\in K[x]$ be some nonzero polynomial and $t\in K$. Write
$f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ for $a_{0},a_{1},\dots,a_{n}\in K$. Then
the local polynomial of $f$ at $t$ is a well-defined monic polynomial in
$V/\mathfrak{m}[x]$ of degree $d$, where

$d\coloneqq\max\\{i\in\\{0,1,\dots,n\\}\mid
v(a_{i}t^{i})=v(a_{i})+iv(t)=\operatorname{minval}_{f}(v(t))\\}.$

Moreover, if $i_{1}<\cdots<i_{s}$ are the indices $i$ such that
$\operatorname{minval}_{f}(v(t))=v(a_{i})+iv(t)$, then

$\operatorname{loc}_{f,t}(x)=r_{i_{1}}x^{i_{1}}+\cdots+r_{i_{s-1}}x^{i_{s-1}}+x^{i_{s}},$

for some nonzero elements $r_{i_{1}},\dots,r_{i_{s-1}}\in V/\mathfrak{m}$.

###### Proof.

We have that $f(tx)=a_{0}+a_{1}tx+a_{1}t^{2}x^{2}+\cdots+a_{n}t^{n}x^{n}$.
Then consider each coefficient of $\frac{f(tx)}{a_{d}t^{d}}$. We calculate
that $v\mathopen{}\mathclose{{}\left(\frac{a_{i}t^{i}}{a_{d}t^{d}}}\right)\geq
0$ for all $i$ since $\operatorname{minval}_{f}(v(t))\leq v(a_{i}t^{i})$ for
all $i$. This shows that $\frac{f(tx)}{a_{d}t^{d}}\in V[x]$. More
specifically,
$v\mathopen{}\mathclose{{}\left(\frac{a_{i}t^{i}}{a_{d}t^{d}}}\right)=0$ if
and only if $v(a_{i}t^{i})=\operatorname{minval}_{f}(v(t))$. Therefore, for
$i>d$, we have
$v\mathopen{}\mathclose{{}\left(\frac{a_{i}t^{i}}{a_{d}t^{d}}}\right)>0$.
Moreover, the coefficient of the degree $d$ term of $\frac{f(tx)}{a_{d}t^{d}}$
is 1, so $\frac{f(tx)}{a_{d}t^{d}}\mod\mathfrak{m}$ has degree $d$ with
leading coefficient 1.

Additionally, since
$v\mathopen{}\mathclose{{}\left(\frac{a_{i}t^{i}}{a_{d}t^{d}}}\right)=0$ if
and only if $v(a_{i}t^{i})=\operatorname{minval}_{f}(v(t))$, we know that
$x^{i}$ has a nonzero coefficient in $\operatorname{loc}_{f,t}(x)$ exactly
when $\operatorname{minval}_{f}(v(t))=v(a_{i})+iv(t)$. ∎

Now we establish the minimal valuation function as a piecewise linear
function. Furthermore, we deduce that the slopes of the minimal valuation
function can be obtained from the highest and lowest degree terms in certain
local polynomials.

######  Proposition 2.16.

For a nonzero $f\in K[x]$, the function $\operatorname{minval}_{f}$ has the
following form evaluated at $\gamma\in{\mathbb{Q}}\Gamma$

$\operatorname{minval}_{f}(\gamma)=\begin{cases}c_{1}\gamma+\beta_{1},&\gamma\leq\delta_{1},\\\
c_{2}\gamma+\beta_{2},&\delta_{1}\leq\gamma\leq\delta_{2},\\\ \vdots\\\
c_{k-1}\gamma+\beta_{k-1},&\delta_{k-2}\leq\gamma\leq\delta_{k-1},\\\
c_{k}\gamma+\beta_{k},&\delta_{k-1}\leq\gamma,\end{cases}$

where $c_{1},\dots,c_{k}\in{\mathbb{N}}$ such that $c_{1}>\cdots>c_{k}$;
$\beta_{1},\dots,\beta_{k}\in\Gamma$; and
$\delta_{1},\dots,\delta_{k-1}\in{\mathbb{Q}}{\Gamma}$ such that
$\delta_{1}<\cdots<\delta_{k-1}$.

Moreover, suppose that $t\in K$ is such that $v(t)=\delta_{i}$ for some
$i\in\\{1,\dots,k-1\\}$. Write
$\operatorname{loc}_{f,t}(x)=r_{j_{i,t,1}}x^{j_{i,1}}+\cdots+r_{j_{i,t,s_{i}-1}}x^{j_{i,s_{i}-1}}+x^{j_{i,s_{i}}}$
with $j_{i,1}<\cdots<j_{i,s_{i}}$, where
$r_{j_{i,t,1}},\dots,r_{j_{i,t,s_{i}-1}}\in V/\mathfrak{m}$ and are all
nonzero. Then $c_{i+1}=j_{i,1}$ and $c_{i}=j_{i,s_{i}}$.

###### Proof.

Write $f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $a_{0},a_{1},\dots,a_{n}\in
K$. We know that
$\operatorname{minval}_{f}(\gamma)=\min\\{j\gamma+v(a_{j})\mid
j\in\\{0,\dots,n\\}\\}$ for all $\gamma\in{\mathbb{Q}}\Gamma$. Then let
$\delta_{1}<\cdots<\delta_{k-1}$ be the elements of ${\mathbb{Q}}{\Gamma}$
such that $\operatorname{minval}_{f}(\delta_{i})=j\delta_{i}+v(a_{j})$ for at
least two indices $j$. Since
$j\gamma+v(a_{j})=j^{\prime}\gamma+v(a_{j^{\prime}})$ for $j\neq j^{\prime}$
if and only if $\gamma=\frac{v(a_{j^{\prime}})-v(a_{j})}{j-j^{\prime}}$, we
know that $\\{\delta_{1},\dots,\delta_{k-1}\\}$ is a subset of the finite set
$\mathopen{}\mathclose{{}\left\\{\frac{v(a_{j^{\prime}})-v(a_{j})}{j-j^{\prime}}\,\middle|\,j,j^{\prime}\in\\{0,\dots,n\\},j\neq
j^{\prime}}\right\\}$. For convenience, set $\delta_{0}=-\infty$ and
$\delta_{k}=\infty$.

For $i\in\\{1,\dots,k-1\\}$, we set

$c_{i}\coloneqq\max\\{j\in\\{0,1,\dots,n\\}\mid\operatorname{minval}_{f}(\delta_{i})=j\delta_{i}+v(a_{j})\\}.$

Also set

$c_{k}\coloneqq\min\\{j\in\\{0,1,\dots,n\\}\mid\operatorname{minval}_{f}(\delta_{k-1})=j\delta_{k-1}+v(a_{j})\\}.$

Next, set $\beta_{i}\coloneqq v(a_{c_{i}})$ for $i\in\\{1,\dots,k\\}$.

Fix $i$ in $\\{1,\dots,k-1\\}$. We want to show
$\operatorname{minval}_{f}(\gamma)=c_{i}\gamma+\beta_{i}$ for all
$\gamma\in{\mathbb{Q}}\Gamma$ such that
$\delta_{i-1}\leq\gamma\leq\delta_{i}$. Suppose not. Then there exists
$\gamma\in{\mathbb{Q}}\Gamma$ such that $\delta_{i-1}\leq\gamma<\delta_{i}$
and $\operatorname{minval}_{f}(\gamma)<c_{i}\gamma+\beta_{i}$. We now have
that
$\operatorname{minval}_{f}(\gamma)=j\gamma+v(a_{j})<c_{i}\gamma+\beta_{i}$ for
some $j\in\\{0,\dots,n\\}$ different from $c_{i}$. Thus,
$(j-c_{i})\gamma<\beta_{i}-v(a_{j})$. Also note that $j\delta_{i}+v(a_{j})\geq
c_{i}\delta_{i}+\beta_{i}$, so $(j-c_{i})\delta_{i}\geq\beta_{i}-v(a_{j})$.
These inequalities imply that
$(j-c_{i})\gamma<\beta_{i}-v(a_{j})\leq(j-c_{i})\delta_{i}$. Because
$\gamma<\delta_{i}$, we can deduce that $j>c_{i}$. Using the inequalities
again shows that $\gamma<\frac{\beta_{i}-v(a_{j})}{j-c_{i}}\leq\delta_{i}$. We
want both inequalities to be strict. If
$\frac{\beta_{i}-v(a_{j})}{j-c_{i}}=\delta_{i}$, then
$j\delta_{i}+v(a_{j})=c_{i}\delta_{i}+\beta_{i}$. The fact that $j>c_{i}$
contradicts the maximality of $c_{i}$. Thus,
$\gamma<\frac{\beta_{i}-v(a_{j})}{j-c_{i}}<\delta_{i}$. Due to the fact that
$j\delta+v(a_{j})=c_{i}\delta+\beta_{i}$ for
$\delta=\frac{\beta_{i}-v(a_{j})}{j-c_{i}}$ and the way
$\delta_{1},\dots,\delta_{k-1}$ are picked out, we know that
$\operatorname{minval}_{f}\mathopen{}\mathclose{{}\left(\frac{\beta_{i}-v(a_{j})}{j-c_{i}}}\right)<c_{i}\cdot\frac{\beta_{i}-v(a_{j})}{j-c_{i}}+\beta_{i}$.
We replace $\gamma$ with $\frac{\beta_{i}-v(a_{j})}{j-c_{i}}$ and repeat the
argument. Since
$\delta_{i-1}\leq\gamma<\frac{\beta_{i}-v(a_{j})}{j-c_{i}}<\delta_{i}$ and
there are only finitely many elements of the form
$\frac{\beta_{i}-v(a_{j})}{j-c_{i}}$ for $j\neq c_{i}$, this argument cannot
be repeated infinitely. Therefore, a contradiction is reached eventually after
a finite number of repetitions. This shows that
$\operatorname{minval}_{f}(\gamma)=c_{i}\gamma+\beta_{i}$ for all
$\gamma\in{\mathbb{Q}}\Gamma$ such that
$\delta_{i-1}\leq\gamma\leq\delta_{i}$. A similar argument will show that
$\operatorname{minval}_{f}(\gamma)=c_{k}\gamma+\beta_{k}$ for all
$\gamma\in{\mathbb{Q}}\Gamma$ such that $\gamma\geq\delta_{k-1}$.

Again, fix $i\in\\{1,\dots,k-1\\}$. Suppose that $t\in K$ is any element such
that $v(t)=\delta_{i}$. Write
$\operatorname{loc}_{f,t}(x)=r_{j_{i,t,1}}x^{j_{i,1}}+\cdots+r_{j_{i,t,s_{i}-1}}x^{j_{i,s_{i}-1}}+x^{j_{i,s_{i}}}$
with $j_{i,1}<\cdots<j_{i,s_{i}}$, where
$r_{j_{i,t,1}},\dots,r_{j_{i,t,s_{i}-1}}\in V/\mathfrak{m}$ and are all
nonzero and each $s_{i}\geq 2$. The local polynomial has this form due to
Proposition 2.15. We also furthermore know that $j_{i,1}<\cdots<j_{i,s_{i}}$
are all the indices $j$ such that
$\operatorname{minval}_{f}(\delta_{i})=j\delta_{i}+v(a_{j})$. We then have
$c_{i}=j_{i,s_{i}}$ since $c_{i}$ is the maximum of all such indices. If
$i=k-1$, we have $c_{k}=j_{i,1}$ by the definition of $c_{k}$. Now suppose
that $i<k-1$ and $c_{i+1}>j_{i,1}$ for a contradiction. The fact that
$j_{i,1}\delta_{i}+\beta_{j_{i,1}}=c_{i+1}\delta_{i}+\beta_{i+1}$ implies

$\displaystyle j_{i,1}\delta_{i+1}+\beta_{j_{i,1}}$
$\displaystyle=j_{i,1}\delta_{i}+\beta_{j_{i,1}}+j_{i,1}(\delta_{i+1}-\delta_{i})$
$\displaystyle=c_{i+1}\delta_{i}+\beta_{i+1}+j_{i,1}(\delta_{i+1}-\delta_{i})$
$\displaystyle<c_{i+1}\delta_{i}+\beta_{i+1}+c_{i+1}(\delta_{i+1}-\delta_{i})$
$\displaystyle=c_{i+1}\delta_{i+1}+\beta_{i+1}$
$\displaystyle=\operatorname{minval}_{f}(\delta_{i+1}),$

contradicting the definition of $\operatorname{minval}_{f}(\delta_{i+1})$.
This shows that $c_{i+1}=j_{i,1}$. Note also that $s_{i}>1$ by the definition
of $\delta_{i}$. Thus, $c_{i+1}=j_{i,1}<j_{i,s_{i}}=c_{i}$, which implies that
$c_{1}>\cdots>c_{k}$. ∎

######  Remark 2.17.

We see that if the value group $\Gamma$ is divisible, then
$\delta_{1},\dots,\delta_{k}$ given above are always in $\Gamma$.

Now we give some results about how information about the valuation of the
polynomial evaluations can be extracted from the minimum valuation polynomial
and the local polynomials.

######  Lemma 2.18.

Let $f\in K[x]$ be a nonzero polynomial. For all nonzero $t\in K$, we have
$\operatorname{minval}_{f}(v(t))\leq v(f(t))$.

###### Proof.

Write $f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $a_{0},\dots,a_{n}\in K$.
Then

$v(f(t))=v\mathopen{}\mathclose{{}\left(\sum_{i=0}^{n}a_{i}t^{i}}\right)\geq\min\\{v(a_{i}t^{i})\mid
i\in\\{0,\dots,n\\}\\}=\operatorname{minval}_{f}(v(t)).$

∎

In other words, the minimum valuation function serves as a lower bound for the
valuation of polynomial evaluations. We now characterize when this lower bound
is strict or not using a local polynomial.

######  Proposition 2.19.

Let $f\in K[x]$ be nonzero and $t\in K$. Then there exists an $s\in K$ with
$v(s)=v(t)$ such that $v(f(s))>\operatorname{minval}_{f}(v(t))$ if and only if
$\operatorname{loc}_{f,t}(x)$ has a nonzero root. More specifically, for $u\in
V$ such that $v(u)=0$, we have that $v(f(tu))>\operatorname{minval}_{f}(v(t))$
if $\operatorname{loc}_{f,t}(u+\mathfrak{m})=0$ and
$v(f(tu))=\operatorname{minval}_{f}(v(t))$ if
$\operatorname{loc}_{f,t}(u+\mathfrak{m})\neq 0$.

###### Proof.

Write $f(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}$ with each $a_{i}\in K$. Let
$i_{1}<\cdots<i_{r}$ be all the indices $i$ such that
$\operatorname{minval}_{f}(v(t))=iv(t)+v(a_{i})$. Then

$\frac{f(tx)}{a_{i_{r}}t^{i_{r}}}=\frac{a_{0}}{a_{i_{r}}t^{i_{r}}}+\frac{a_{1}t}{a_{i_{r}}t^{i_{r}}}x+\frac{a_{2}t^{2}}{a_{i_{r}}t^{i_{r}}}x^{2}+\cdots\frac{a_{n}t^{n}}{a_{i_{r}}t^{i_{r}}}x^{n}.$

Taking this modulo $\mathfrak{m}$, we get

$\operatorname{loc}_{f,t}(x)=\frac{a_{i_{1}}t^{i_{1}}}{a_{i_{r}}t^{i_{r}}}x^{i_{1}}+\cdots+\frac{a_{i_{r-1}}t^{i_{r-1}}}{a_{i_{r}}t^{i_{r}}}x^{i_{r-1}}+x^{i_{r}}\mod\mathfrak{m}.$

Take $s\in K$ with $v(s)=v(t)$. Set $u=\frac{s}{t}$. Note that $u\neq
0\mod\mathfrak{m}$. Then
$v(f(s))>\operatorname{minval}_{f}(v(t))=v(a_{i_{r}}t^{i_{r}})$ if and only if

$\frac{f(tu)}{a_{i_{r}}t^{i_{r}}}=\frac{a_{0}}{a_{i_{r}}t^{i_{r}}}+\frac{a_{1}t}{a_{i_{r}}t^{i_{r}}}u+\frac{a_{2}t^{2}}{a_{i_{r}}t^{i_{r}}}u^{2}+\cdots\frac{a_{n}t^{n}}{a_{i_{r}}t^{i_{r}}}u^{n}$

has valuation under $v$ that is strictly greater than 0, which can happen if
and only if

$\operatorname{loc}_{f,t}(u+\mathfrak{m})=\frac{a_{i_{1}}t^{i_{1}}}{a_{i_{r}}t^{i_{r}}}u^{i_{1}}+\cdots+\frac{a_{i_{r-1}}t^{i_{r-1}}}{a_{i_{r}}t^{i_{r}}}u^{i_{r-1}}+u^{i_{r}}\mod\mathfrak{m}=0.$

∎

We observe that most of the time, the local polynomial is a monomial, which
does not have nonzero roots. Thus, the minimum valuation function gives the
valuation of the polynomial evaluations most of the time.

######  Corollary 2.20.

Take $f\in K[x]$ to be a nonzero polynomial. Then using the notation of
Proposition 2.16, if $t\in K$ is such that $v(t)\neq\delta_{i}$ for all
$i\in\\{1,\dots,k-1\\}$, then $v(f(t))=\operatorname{minval}_{f}(v(t))$.

###### Proof.

Write $f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$, where $a_{0},\dots,a_{n}\in K$.
Since $v(t)\neq\delta_{i}$ for any $i$, we have that
$\operatorname{minval}_{f}(v(t))=jv(t)+v(a_{j})$ for a unique $j$ in
$\\{0,\dots,n\\}$ by the definition of $\delta_{1},\dots,\delta_{k-1}$ in
Proposition 2.16. Thus, $\operatorname{loc}_{f,t}(x)=x^{j}$, which has no
nonzero roots, so $v(f(t))=\operatorname{minval}_{f}(v(t))$ by Proposition
2.19. ∎

If we further assume that $V$ is a valuation domain with infinite residue
field, we can say something about the finitely many exceptions
$\delta_{1},\dots,\delta_{k-1}$ of the value group. In this case, every
element of the value group has some element of that value whose evaluation has
valuation determined by the minimum valuation function.

######  Corollary 2.21.

Suppose $V$ has infinite residue field. If $f_{1},\dots,f_{m}\in K[x]$ are a
finite number of nonzero polynomials, then for any $\gamma\in\Gamma$, there
exists an $a\in K$ with $v(a)=\gamma$ such that
$v(f_{i}(a))=\operatorname{minval}_{f_{i}}(\gamma)$ for each $i=1,\dots,m$.

###### Proof.

Fix $\gamma\in\Gamma$ and $t\in K$ with $v(t)=\gamma$. Let $u\in K$ with
$v(u)=0$. We know for any nonzero polynomial $f\in K[x]$ that
$v(f(tu))>\operatorname{minval}_{f}(v(t))$ if and only if
$\operatorname{loc}_{f,t}(u+\mathfrak{m})=0$. However,
$\operatorname{loc}_{f_{1},t},\dots,\operatorname{loc}_{f_{m},t}$ can
collectively only have a finite number of nonzero roots in $V/\mathfrak{m}$.
Thus, there exists $u\in V^{\times}$ such that
$\operatorname{loc}_{f_{i},t}(u+\mathfrak{m})\neq 0$ for all $i$ and hence
$v(f_{i}(tu))=\operatorname{minval}_{f_{i}}(\gamma)$ for each $i=1,\dots,m$.

∎

######  Proposition 2.22.

Let $f,g\in K[x]$ be nonzero polynomials. Then

$\operatorname{minval}_{fg}=\operatorname{minval}_{f}+\operatorname{minval}_{g}.$

###### Proof.

We will view $f$ and $g$ as being in
$K(t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma)[x]$ and take the monomial
valuation $v$ that extends $V$ mapping $t_{\gamma}$ to $\gamma$ for each
$\gamma\in{\mathbb{Q}}\Gamma$.

By Corollary 2.20, we know that for all but finitely many
$\gamma\in{\mathbb{Q}}\Gamma$, we have that
$v(f(a))=\operatorname{minval}_{f}(\gamma)$,
$v(g(a))=\operatorname{minval}_{g}(\gamma)$, and
$v((fg)(a))=\operatorname{minval}_{fg}(\gamma)$ for all $a\in
K(t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma)$ such that $v(a)=\gamma$. For
such values of $\gamma$, we take $a_{\gamma}\in
K(t_{\gamma}\mid\gamma\in{\mathbb{Q}}\Gamma)$ such that
$v(a_{\gamma})=\gamma$. This means that

$\operatorname{minval}_{fg}(\gamma)=v((fg)(a_{\gamma}))=v(f(a_{\gamma}))+v(g(a_{\gamma}))=\operatorname{minval}_{f}(\gamma)+\operatorname{minval}_{g}(\gamma)$

holds for all but finitely many $\gamma\in{\mathbb{Q}}\Gamma$. Since
$\operatorname{minval}_{fg},\operatorname{minval}_{f}$, and
$\operatorname{minval}_{g}$ are all piecewise linear functions from
${\mathbb{Q}}\Gamma$ to ${\mathbb{Q}}\Gamma$ by Proposition 2.16, we have
equality for all $\gamma\in{\mathbb{Q}}\Gamma$ and in particular for all
$\gamma\in\Gamma$. ∎

The previous proposition also follows using the fact that
$\operatorname{minval}_{fg}(\gamma)=v_{0,\gamma}(fg)=v_{0,\gamma}(f)+v_{0,\gamma}(g)=\operatorname{minval}_{f}(\gamma)+\operatorname{minval}_{g}(\gamma)$
for every $\gamma\in{\mathbb{Q}}\Gamma$.

We now use this fact to define the minimum valuation function of a nonzero
rational function.

######  Definition 2.23.

Let $\varphi\in K(x)$ be a nonzero rational function. Write
$\varphi=\frac{f}{g}$ for some $f,g\in K[x]$. For $\gamma\in\Gamma$, we define
$\operatorname{minval}_{\varphi}(\gamma)=\operatorname{minval}_{f}(\gamma)-\operatorname{minval}_{g}(\gamma)$,
the minimum valuation function of $\varphi$.

This is well defined. If $\frac{f}{g}=\frac{F}{G}$ for some $F,G\in K[x]$,
then $fG=gF$ and then
$\operatorname{minval}_{f}+\operatorname{minval}_{G}=\operatorname{minval}_{g}+\operatorname{minval}_{F}$,
which means that
$\operatorname{minval}_{f}-\operatorname{minval}_{g}=\operatorname{minval}_{F}-\operatorname{minval}_{G}$.

There is an analog for Proposition 2.16 giving the form of the minimum
valuation function of a rational function. Note that the ordering on the
coefficients of $\gamma$ is lost and these coefficients can be negative.

######  Proposition 2.24.

For a nonzero $\varphi\in K(x)$, the function
$\operatorname{minval}_{\varphi}$ has the following form evaluated at
$\gamma\in{\mathbb{Q}}\Gamma$

$\operatorname{minval}_{\varphi}(\gamma)=\begin{cases}c_{1}\gamma+\beta_{1},&\gamma\leq\delta_{1},\\\
c_{2}\gamma+\beta_{2},&\delta_{1}\leq\gamma\leq\delta_{2},\\\ \vdots\\\
c_{k-1}\gamma+\beta_{k-1},&\delta_{k-2}\leq\gamma\leq\delta_{k-1},\\\
c_{k}\gamma+\beta_{k},&\delta_{k-1}\leq\gamma,\end{cases}$

where $c_{1},\dots,c_{k}\in{\mathbb{Z}}$;
$\beta_{1},\dots,\beta_{k}\in\Gamma$; and
$\delta_{1},\dots,\delta_{k-1}\in{\mathbb{Q}}{\Gamma}$ such that
$\delta_{1}<\cdots<\delta_{k-1}$.

###### Proof.

Write $\varphi=\frac{f}{g}$ for some polynomials $f,g\in K[x]$. By Proposition
2.16, we know there are
$\delta_{1},\dots,\delta_{k-1},\delta_{1}^{\prime},\dots,\delta_{k^{\prime}-1}^{\prime}$
such that
$-\infty=\delta_{0}<\delta_{1}<\cdots<\delta_{k-1}<\delta_{k}=\infty$ and
$-\infty=\delta_{0}^{\prime}<\delta_{1}^{\prime}<\cdots<\delta_{k^{\prime}-1}^{\prime}<\delta_{k^{\prime}}=\infty$
and that for all $\gamma\in{\mathbb{Q}}\Gamma$ such that $\gamma$ is between
$\delta_{i}$ and $\delta_{i+1}$, we have some $c_{i}\in{\mathbb{N}}$ and
$\beta_{i}\in\Gamma$ such that

$\operatorname{minval}_{f}(\gamma)=c_{i}\gamma+\beta_{i}$

and for all $\gamma\in{\mathbb{Q}}\Gamma$ such that $\gamma$ is between
$\delta_{i}^{\prime}$ and $\delta_{i+1}^{\prime}$, we have some
$c_{i}^{\prime}\in{\mathbb{N}}$ and $\beta_{i}^{\prime}\in\Gamma$ such that

$\operatorname{minval}_{g}(\gamma)=c_{i}^{\prime}\gamma+\beta_{i}^{\prime}.$

Now order the elements of the set
$\\{\delta_{0},\dots,\delta_{k},\delta_{0}^{\prime},\dots,\delta_{k}^{\prime}\\}$
and rename the elements
$\delta_{0}^{\prime\prime},\dots,\delta_{k^{\prime\prime}}^{\prime\prime}$ so
that
$-\infty=\delta_{0}^{\prime\prime}<\delta_{1}^{\prime\prime}<\cdots<\delta_{k^{\prime\prime}-1}^{\prime\prime}<\delta_{k^{\prime\prime}}^{\prime\prime}=\infty$.
Let $r$ be such that $r\in\\{0,\dots,k^{\prime\prime}-1\\}$. We know that the
interval between $\delta_{r}^{\prime\prime}$ and $\delta_{r+1}^{\prime\prime}$
is contained in the interval between $\delta_{i}$ and $\delta_{i+1}$ and also
contained in the interval between $\delta_{j}^{\prime}$ and
$\delta_{j+1}^{\prime}$ for some $i$ and $j$. Thus, for all
$\gamma\in{\mathbb{Q}}\Gamma$ such that
$\delta_{r}^{\prime\prime}\leq\gamma\leq\delta_{r+1}^{\prime\prime}$, we have

$\operatorname{minval}_{\varphi}(\gamma)=\operatorname{minval}_{f}(\gamma)-\operatorname{minval}_{g}(\gamma)=(c_{i}-c_{j}^{\prime})\gamma+(\beta_{i}-\beta_{j}^{\prime}),$

giving us the desired form for $\operatorname{minval}_{\varphi}$. ∎

Even though the ordering on the coefficients of $\gamma$ in
$\operatorname{minval}_{\varphi}$ is lost, the coefficients can still give
information about the powers that appear in the local polynomials.

######  Lemma 2.25.

Take $\varphi\in K(x)$ to be nonzero and $\alpha\in\Gamma$. There exist
$\varepsilon\in{\mathbb{Q}}\Gamma$ with $\varepsilon>0$ small enough,
$c,c^{\prime}\in{\mathbb{Z}}$, and $\beta,\beta^{\prime}\in\Gamma$ such that

$\operatorname{minval}_{\varphi}(\gamma)=\begin{cases}c\gamma+\beta,&\text{if
$\alpha-\varepsilon<\gamma<\alpha$},\\\
c^{\prime}\gamma+\beta^{\prime},&\text{if
$\alpha<\gamma<\alpha+\varepsilon$}.\end{cases}$

Write $\varphi=\frac{f}{g}$ for some $f,g\in K[x]$. Take $t\in K$ such that
$v(t)=\gamma$. We can write
$\operatorname{loc}_{f,t}=a_{i_{1}}x^{i_{1}}+\cdots+a_{i_{r}}x^{i_{r}}$ and
$\operatorname{loc}_{g,t}=b_{j_{1}}x^{j_{1}}+\cdots+b_{j_{s}}x^{j_{s}}$ for
some nonzero $a_{i_{1}},\dots,a_{i_{r}},b_{j_{1}},\dots,b_{j_{s}}\in
V/\mathfrak{m}$. Then

$c=i_{r}-j_{s}\quad\text{and}\quad c^{\prime}=i_{1}-j_{1}.$

###### Proof.

Due to Proposition 2.16, we can make $\varepsilon$ small enough so that there
exist $\zeta,\zeta^{\prime},\eta,\eta^{\prime}\in\Gamma$ such that

$\operatorname{minval}_{f}(\gamma)=\begin{cases}i_{r}\gamma+\zeta,&\text{if
$\alpha-\varepsilon<\gamma<\alpha$},\\\ i_{1}\gamma+\zeta^{\prime},&\text{if
$\alpha<\gamma<\alpha+\varepsilon$}\end{cases}$

and

$\operatorname{minval}_{g}(\gamma)=\begin{cases}j_{s}\gamma+\eta,&\text{if
$\alpha-\varepsilon<\gamma<\alpha$},\\\ j_{1}\gamma+\eta^{\prime},&\text{if
$\alpha<\gamma<\alpha+\varepsilon$}.\end{cases}$

Since
$\operatorname{minval}_{\varphi}=\operatorname{minval}_{f}-\operatorname{minval}_{g}$,
we obtain $c=i_{r}-j_{s}$ and $c^{\prime}=i_{1}-j_{1}$. ∎

However, for rational functions, there is no analog of Lemma 2.18, since it is
possible that $\operatorname{minval}_{\varphi}(v(t))$ is greater than, equal
to, or less than $v(\varphi(t))$. To calculate how
$\operatorname{minval}_{\varphi}(v(t))$ compares with $v(\varphi(t))$, we can
try to apply Proposition 2.19 to the local polynomials of the numerator and
the denominator of $\varphi$. This does not give a definite answer in the case
when the local polynomials have a common nonzero root, so further calculations
are needed in this case. Nevertheless, there are analogs of Corollary 2.20 and
Corollary 2.21 that say $\operatorname{minval}_{\varphi}(v(t))$ and
$v(\varphi(t))$ are equal most of the time.

######  Lemma 2.26.

Take $\varphi\in K(x)$ to be a nonzero rational function. For all but finitely
many $\gamma\in\Gamma$, we have that
$v(\varphi(t))=\operatorname{minval}_{\varphi}(v(t))$ for all $t\in K$ such
that $v(t)=\gamma$.

###### Proof.

We write $\varphi=\frac{f}{g}$ for some $f,g\in K[x]$. Corollary 2.20 tells us
that for all but finitely many $\gamma\in\Gamma$, we have
$v(f(t))=\operatorname{minval}_{f}(v(t))$ for all $t\in K$ such that
$v(t)=\gamma$. Similarly, for all but finitely many $\gamma\in\Gamma$, we have
$v(g(t))=\operatorname{minval}_{g}(v(t))$ for all $t\in K$ such that
$v(t)=\gamma$. There are still only finitely many values of $\gamma\in\Gamma$
we need to exclude. Thus, for all but finitely many $\gamma\in\Gamma$, we have
that

$v(\varphi(t))=v(f(t))-v(g(t))=\operatorname{minval}_{f}(v(t))-\operatorname{minval}_{g}(v(t))=\operatorname{minval}_{\varphi}(v(t))$

for all $t\in K$ such that $v(t)=\gamma$. ∎

######  Proposition 2.27.

Suppose the residue field of $V$ is infinite. If we have nonzero
$\varphi_{1},\dots,\varphi_{m}\in K(x)$, then for any $\gamma\in\Gamma$, there
exists $a\in K$ with $v(a)=\gamma$ such that
$\operatorname{minval}_{\varphi_{i}}(\gamma)=v(\varphi_{i}(a))$ for all $i$.

###### Proof.

Write $\varphi_{i}=\frac{f_{i}}{g_{i}}$ with $f_{i},g_{i}\in K[x]$ for all
$i$. By applying Corollary 2.21 to $f_{1},\dots,f_{m},g_{1},\dots,g_{m}$, we
see that for any $\gamma\in\Gamma$, there exists $a\in K$ with $v(a)=\gamma$
such that $\operatorname{minval}_{f_{i}}(\gamma)=v(f_{i}(a))$ and
$\operatorname{minval}_{g_{i}}(\gamma)=v(g_{i}(a))$ for all $i$. Putting these
together yields

$\operatorname{minval}_{\varphi_{i}}(\gamma)=\operatorname{minval}_{f_{i}}(\gamma)-\operatorname{minval}_{g_{i}}(\gamma)=v(f_{i}(a))-v(g_{i}(a))=v(\varphi_{i}(a))$

for all $i$. ∎

The following lemma will be useful in describing the case of a valuation
domain with algebraically closed residue field and maximal ideal that is not
principal.

######  Lemma 2.28.

Suppose that $V/\mathfrak{m}$ is algebraically closed and $\mathfrak{m}$ is
not a principal ideal of $V$. Let $\varphi\in K(x)$ be a nonzero rational
function such that there exist $\alpha,\varepsilon\in\Gamma$ with
$\varepsilon>0$ so that

$\operatorname{minval}_{\varphi}(\gamma)=\begin{cases}c_{1}\gamma+\beta_{1},&\text{
if $\alpha-\varepsilon\leq\gamma\leq\alpha$,}\\\ c_{2}\gamma+\beta_{2},&\text{
if $\alpha\leq\gamma\leq\alpha+\varepsilon$,}\end{cases}$

for some $c_{1},c_{2}\in{\mathbb{Z}}$ and $\beta_{1},\beta_{2}\in\Gamma$.

If $c_{1}>c_{2}$, then there exists $a\in K$ with $v(a)=\alpha$ and
$v(\varphi(a))>\operatorname{minval}_{\varphi}(\alpha)$.

If $c_{1}<c_{2}$, then there exists $a\in K$ with $v(a)=\alpha$ and
$v(\varphi(a))<\operatorname{minval}_{\varphi}(\alpha)$.

###### Proof.

Write $\varphi=\frac{f}{g}$ for some $f,g\in K[x]$. Let $b$ be an element of
$K$ such that $v(b)=\alpha$. Then we can completely factor
$\operatorname{loc}_{f,b}(x)$ and $\operatorname{loc}_{g,b}(x)$ as
$V/\mathfrak{m}$ is algebraically closed, so those polynomials have the forms

$\operatorname{loc}_{f,b}(x)=x^{i}(x-\xi_{1})^{e_{1}}\cdots(x-\xi_{n})^{e_{n}}\quad\text{and}\quad\operatorname{loc}_{g,b}(x)=x^{j}(x-\xi_{1})^{e_{1}^{\prime}}\cdots(x-\xi_{n})^{e_{n}^{\prime}},$

where each $\xi_{k}\in V/\mathfrak{m}$ are nonzero and
$i,j,e_{k},e_{k}^{\prime}\in{\mathbb{N}}$ for $k=1,2,\dots,n$.

Now by Proposition 2.16, we note that there exists some
$\varepsilon^{\prime}\in\Gamma$ with $0<\varepsilon^{\prime}\leq\varepsilon$
such that

$\operatorname{minval}_{f}(\gamma)=\begin{cases}d_{1}\gamma+\delta_{1},&\text{
if $\alpha-\varepsilon^{\prime}\leq\gamma\leq\alpha$,}\\\
d_{2}\gamma+\delta_{2},&\text{ if
$\alpha\leq\gamma\leq\alpha+\varepsilon^{\prime}$}\end{cases}$

and

$\operatorname{minval}_{g}(\gamma)=\begin{cases}d_{1}^{\prime}\gamma+\delta_{1}^{\prime},&\text{
if $\alpha-\varepsilon^{\prime}\leq\gamma\leq\alpha$,}\\\
d_{2}^{\prime}\gamma+\delta_{2}^{\prime},&\text{ if
$\alpha\leq\gamma\leq\alpha+\varepsilon^{\prime}$,}\end{cases}$

for some $d_{1},d_{2},d_{1}^{\prime},d_{2}^{\prime}\in{\mathbb{N}}$ and
$\delta_{1},\delta_{2},\delta_{1}^{\prime},\delta_{2}^{\prime}\in\Gamma$. From
this, we get that

$d_{1}-d_{1}^{\prime}=c_{1}\quad\text{and}\quad d_{2}-d_{2}^{\prime}=c_{2}$

by Lemma 2.25. We also have

$(i+e_{1}+\cdots+e_{n})-i=d_{1}-d_{2}\quad\text{and}\quad(j+e_{1}^{\prime}+\cdots+e_{n}^{\prime})-j=d_{1}^{\prime}-d_{2}^{\prime}$

by Proposition 2.16. Putting it all together, we get that

$(e_{1}+\cdots+e_{n})-(e_{1}^{\prime}+\cdots+e_{n}^{\prime})=c_{1}-c_{2}>0.$

This means that there exists $\ell\in\\{1,\dots,n\\}$ such that
$e_{\ell}>e_{\ell}^{\prime}$.

Let $u\in V$ be such that $u+\mathfrak{m}=\xi_{\ell}$. We will lift
$\operatorname{loc}_{f,b}(x)$ and $\operatorname{loc}_{g,b}(x)$ back to
$V[x]$. We now have

$\frac{f(bx)}{t}=(x-u)^{e_{\ell}}f_{1}(x)+f_{2}(x)\quad\text{and}\quad\frac{g(bx)}{t^{\prime}}=(x-u)^{e_{\ell}^{\prime}}g_{1}(x)+g_{2}(x),$

where $t,t^{\prime}\in K$ such that $v(t)=\operatorname{minval}_{f}(v(b))$ and
$v(t^{\prime})=\operatorname{minval}_{g}(v(b))$, $f_{1},g_{1}\in V[x]$ such
that $f_{1}(u),g_{1}(u)\notin\mathfrak{m}$, and
$f_{2},g_{2}\in\mathfrak{m}[x]$.

Let $h\in\mathfrak{m}$ such that
$v(h)<\frac{1}{e_{\ell}}\min\\{\operatorname{minval}_{f_{2}}(0),\operatorname{minval}_{g_{2}}(0)\\}$.
This is possible since $\mathfrak{m}$ is not principal. Set $a\coloneqq
b(u+h)$. Note that $v(a)=v(b)=\alpha$. We then calculate

$\varphi(a)=\frac{t(h^{e_{\ell}}f_{1}(u+h)+f_{2}(u+h))}{t^{\prime}(h^{e_{\ell}^{\prime}}g_{1}(u+h)+g_{2}(u+h))}.$

Then,
$v(\varphi(a))=\operatorname{minval}_{\varphi}(\alpha)+(e_{\ell}-e_{\ell}^{\prime})v(h)>\operatorname{minval}_{\varphi}(\alpha)$,
as desired.

If $c_{1}<c_{2}$, then apply the $c_{1}>c_{2}$ case to $\frac{1}{\varphi}$ to
get the desired result. ∎

We use the previous lemma to determine that
$\operatorname{Int{}^{\text{R}}}(V)$ is not a Prüfer domain for $V$ a
valuation domain with algebraically closed residue field and maximal ideal
that is not principal.

######  Theorem 2.29.

Suppose that $V/\mathfrak{m}$ is algebraically closed and $\mathfrak{m}$ is
not a principal ideal of $V$. Then $\operatorname{Int{}^{\text{R}}}(V)$ is not
Prüfer.

###### Proof.

Aiming for a contradiction, we assume that
$\operatorname{Int{}^{\text{R}}}(V)$ is Prüfer.

Let $d\in\mathfrak{m}$. Since $\operatorname{Int{}^{\text{R}}}(V)$ is Prüfer,
the finitely-generated ideal $(x,d)$ is invertible. This means that there are
$\varphi,\psi\in(x,d)^{-1}$ such that $x\varphi+d\psi=1$. By Proposition 2.27,
there exists some $b\in V$ such that $v(b)=v(d)$ and
$v(\varphi(b))=\operatorname{minval}_{\varphi}(v(d))$ and
$v(\psi(b))=\operatorname{minval}_{\psi}(v(d))$. Evaluating $x\varphi+d\psi=1$
at $x=b$, we obtain

$b\varphi(b)+d\psi(b)=1.$

We have $x\varphi,d\psi\in\operatorname{Int{}^{\text{R}}}(V)$, so
$b\varphi(b),d\psi(b)\in V$. Thus, we have $v(b\varphi(b))=0$ or
$v(d\psi(b))=0$. Then we get
$\operatorname{minval}_{\varphi}(v(d))=v(\varphi(b))=-v(b)=-v(d)$ or
$\operatorname{minval}_{\psi}(v(d))=v(\psi(b))=-v(d)$.

Either way, we have some function $\rho\in(x,d)^{-1}$ such that
$v(b\rho(b))=0$ and $v(\rho(b))=\operatorname{minval}_{\rho}(v(d))=-v(d)$. By
Proposition 2.24, there exists some $\varepsilon\in{\mathbb{Q}}\Gamma$ with
$\varepsilon>0$ such that there exist some $c,c^{\prime}\in{\mathbb{Z}}$ and
$\beta,\beta^{\prime}\in\Gamma$ so that

$\operatorname{minval}_{\rho}(\gamma)=\begin{cases}c\gamma+\beta,&v(d)-\varepsilon\leq\gamma\leq
v(d),\\\ c^{\prime}\gamma+\beta^{\prime},&v(d)\leq\gamma\leq
v(d)+\varepsilon.\end{cases}$

Since $\mathfrak{m}$ is not principal, due to Lemma 2.26, there exists
$b^{\prime}\in V$ with the property that $v(d)-\varepsilon<v(b^{\prime})<v(d)$
and $v(\rho(b^{\prime}))=cv(b^{\prime})+\beta$. Since
$x\rho\in\operatorname{Int{}^{\text{R}}}(V)$, we have

$v(b^{\prime}\rho(b^{\prime}))=v(b^{\prime})+cv(b^{\prime})+\beta=(c+1)v(b^{\prime})+\beta\geq
0=v(b\rho(b))=(c+1)v(b)+\beta.$

This implies $(c+1)v(b^{\prime})\geq(c+1)v(b)$ and thus
$0\geq(c+1)(v(b)-v(b^{\prime}))$. We know that $v(b)>v(b^{\prime})$ so we must
have $c+1\leq 0$. In other words, $c\leq-1$.

Now there exists $b^{\prime\prime}\in V$ so that
$v(d)<v(b^{\prime\prime})<v(d)+\varepsilon$ and
$v(\rho(b^{\prime\prime}))=c^{\prime}v(b^{\prime\prime})+\beta^{\prime}$.
Because $d\rho\in\operatorname{Int{}^{\text{R}}}(V)$, we get that

$v(d\rho(b^{\prime\prime})))=v(d)+c^{\prime}v(b^{\prime\prime})+\beta^{\prime}\geq
0=v(b\rho(b))=v(d)+c^{\prime}v(b)+\beta^{\prime}.$

Thus, $c^{\prime}v(b^{\prime\prime})\geq c^{\prime}v(b)$. This implies that
$c^{\prime}\geq 0$ since $v(b^{\prime\prime})>v(b)$.

Since $c<c^{\prime}$, there exists $a\in V$ such that
$v(\rho(a))<\operatorname{minval}_{\rho}(v(d))=-v(d)$ and $v(a)=v(d)$
according to Lemma 2.28. This implies that $v(a\rho(a))<0$, contradicting the
fact that $x\rho\in\operatorname{Int{}^{\text{R}}}(V)$. We can conclude that
$(x,d)$ is not invertible, so $\operatorname{Int{}^{\text{R}}}(V)$ cannot be
Prüfer. ∎

This result combined with the results of [Lop94, CL98] completely classifies
the case of when $\operatorname{Int{}^{\text{R}}}(V)$ is a Prüfer domain given
that $V$ is a valuation domain.

######  Corollary 2.30.

Let $V$ be a valuation domain. Then $\operatorname{Int{}^{\text{R}}}(V)$ is a
Prüfer domain if and only if $V/\mathfrak{m}$ is not algebraically closed or
$\mathfrak{m}$ is a principal ideal of $V$.

In fact, for a valuation domain $V$, most of the time when
$\operatorname{Int{}^{\text{R}}}(V)$ is Prüfer, the ring
$\operatorname{Int{}^{\text{R}}}(V)$ is also Bézout. We know that
$\operatorname{Int{}^{\text{R}}}(V)$ is Bézout when $V$ has a principal
maximal ideal or there exist two nonconstant, monic, unit-valued polynomials
over $V$ of coprime degrees [CL98, Theorem 3.5 and Corollary 3.3]. Note that
the latter condition is equivalent to saying that there exist two nonconstant
polynomials of coprime degrees over $V/\mathfrak{m}$ with no roots in
$V/\mathfrak{m}$, where $\mathfrak{m}$ is the maximal ideal of $V$. We will
now completely characterize when the ring integer-valued rational functions
over a valuation domain is a Bézout domain. We first require a lemma about the
minimum valuation functions of generators of finitely-generated ideals in
$\operatorname{Int{}^{\text{R}}}(V)$.

######  Lemma 2.31.

Suppose $V/\mathfrak{m}$ is infinite. Let
$\varphi_{1},\dots,\varphi_{n},\psi_{1},\dots,\psi_{m}\in\operatorname{Int{}^{\text{R}}}(V)$
be nonzero integer-valued rational functions such that

$(\varphi_{1},\dots,\varphi_{n})=(\psi_{1},\dots,\psi_{m})$

as ideals of $\operatorname{Int{}^{\text{R}}}(V)$. Then

$\min\\{\operatorname{minval}_{\varphi_{1}}(\gamma),\dots,\operatorname{minval}_{\varphi_{n}}(\gamma)\\}=\min\\{\operatorname{minval}_{\psi_{1}}(\gamma),\dots,\operatorname{minval}_{\psi_{m}}(\gamma)\\}$

for all $\gamma\in\Gamma$ such that $\gamma\geq 0$.

###### Proof.

Let $\gamma\in\Gamma$ with $\gamma\geq 0$. Since $V$ has an infinite residue
field, there exists $d\in V$ such that $v(d)=\gamma$ and
$\operatorname{minval}_{\varphi_{i}}(\gamma)=v(\varphi_{i}(d)),\operatorname{minval}_{\psi_{j}}(\gamma)=v(\psi(d))$
for all $i$ and $j$ by Proposition 2.27. Because
$(\varphi_{1}(d),\dots,\varphi_{n}(d))=(\psi_{1}(d),\dots,\psi_{m}(d))$, it
follows that

$\min\\{v(\varphi_{1}(d)),\dots,v(\varphi_{n}(d))\\}=\min\\{v(\psi_{1}(d)),\dots,v(\psi_{m}(d))\\}.$

Therefore, we obtain

$\min\\{\operatorname{minval}_{\varphi_{1}}(\gamma),\dots,\operatorname{minval}_{\varphi_{n}}(\gamma)\\}=\min\\{\operatorname{minval}_{\psi_{1}}(\gamma),\dots,\operatorname{minval}_{\psi_{m}}(\gamma)\\}.$

∎

Now we characterize when $\operatorname{Int{}^{\text{R}}}(V)$ is a Bézout
domain.

######  Proposition 2.32.

Suppose that $\mathfrak{m}$ is not principal and there does not exist two
nonconstant polynomials of coprime degrees over $V/\mathfrak{m}$ with no roots
in $V/\mathfrak{m}$. Then $\operatorname{Int{}^{\text{R}}}(V)$ is not Bézout.

###### Proof.

Let $t\in\mathfrak{m}$. We want to show that the finitely-generated ideal
$(x,t)$ of $\operatorname{Int{}^{\text{R}}}(V)$ is not principal. Suppose on
the contrary that $(x,t)=(\varphi)$ for some
$\varphi\in\operatorname{Int{}^{\text{R}}}(V)$.

Note that $V/\mathfrak{m}$ is necessarily an infinite field, so by Lemma 2.31,

$\operatorname{minval}_{\varphi}(\gamma)=\min\\{\operatorname{minval}_{x}(\gamma),\operatorname{minval}_{t}(\gamma)\\}=\begin{cases}\gamma,&\text{if
$0\leq\gamma\leq v(t)$},\\\ v(t),&\text{if $\gamma\geq v(t)$},\end{cases}$

for each $\gamma\in\Gamma$ such that $\gamma\geq 0$. Therefore, if we write
$\varphi=\frac{f}{g}$ for some $f,g\in V[x]$, then
$\deg(\operatorname{loc}_{f,t})=\deg(\operatorname{loc}_{g,t})+1$ and the
degree of the lowest degree monomial of $\operatorname{loc}_{f,t}$ and
$\operatorname{loc}_{g,t}$ are the same by Lemma 2.25. Plus, we claim it is
impossible for every nonzero element of $V/\mathfrak{m}$ to be a root of
$\operatorname{loc}_{f,t}$ and $\operatorname{loc}_{g,t}$ of the same
multiplicity. Here, we allow for the possibility that an element is a root of
multiplicity $0$, meaning it is not a root. Suppose $\xi_{1},\dots,\xi_{n}$
are the nonzero roots of $\operatorname{loc}_{f,t}$ and
$\operatorname{loc}_{g,t}$ with each $\xi_{i}$ appearing in
$\operatorname{loc}_{f,t}$ and $\operatorname{loc}_{g,t}$ with multiplicity
$e_{i}$. Then the polynomials
$F(x)=\frac{\operatorname{loc}_{f,t}(x)}{x^{m}(x-\xi_{1})^{e_{1}}\cdots(x-\xi_{n})^{e_{n}}}$
and
$G(x)=\frac{\operatorname{loc}_{g,t}(x)}{x^{m}(x-\xi_{1})^{e_{1}}\cdots(x-\xi_{n})^{e_{n}}}$
both have no roots over $V/\mathfrak{m}$. Moreover, since
$\deg(F)=\deg(G)+1\geq 3$, we know that $\gcd(\deg(F),\deg(G))=1$. Thus, the
assumption about $V/\mathfrak{m}$ is contradicted.

Thus, there exists some nonzero element $\xi\in V/\mathfrak{m}$ such that
$\xi$ is a root of $\operatorname{loc}_{f,t}$ of multiplicity $c_{1}$ and
$\xi$ is a root of $\operatorname{loc}_{g,t}$ of multiplicity $c_{2}$ with
$c_{1}\neq c_{2}$. Let $u\in V$ be a lift of $\xi$. Then by lifting
$\operatorname{loc}_{f,t}$, we obtain

$\frac{f(tx)}{b}=(x-u)^{c_{1}}f_{1}(x)+f_{2}(x),$

where $b\in V$ is some element such that
$v(b)=\operatorname{minval}_{f}(v(t))$, $f_{1}(x)\in V[x]$ is such that
$f_{1}(u)\notin\mathfrak{m}$, and $f_{2}(x)\in\mathfrak{m}[x]$. We similarly
obtain $\frac{g(tx)}{b^{\prime}}=(x-u)^{c_{2}}g_{1}(x)+g_{2}(x)$, where
$b^{\prime}\in V$ is such that
$v(b^{\prime})=\operatorname{minval}_{g}(v(t))$, $g_{1}(x)\in V[x]$ is such
that $g_{1}(u)\notin\mathfrak{m}$, and $g_{2}(x)\in\mathfrak{m}[x]$. Then
there exists an element $h\in\mathfrak{m}$ such that
$v(h)<\frac{1}{\max\\{c_{1},c_{2}\\}}\min\\{\operatorname{minval}_{f_{2}}(0),\operatorname{minval}_{g_{2}}(0)\\}$
and

$v\mathopen{}\mathclose{{}\left(\frac{f(t(u+h))}{b}}\right)=v(h^{c_{1}}f_{1}(u+h)+f_{2}(u+h))=v(h^{c_{1}}f_{1}(u+h))=c_{1}v(h).$

Therefore, $v(f(t(u+h)))=c_{1}v(h)+\operatorname{minval}_{f}(v(t))$. A similar
calculation yields $v(g(t(u+h)))=c_{2}v(h)+\operatorname{minval}_{g}(v(t))$.
Now, we have

$v(\varphi(t(u+h)))=(c_{1}-c_{2})v(h)+\operatorname{minval}_{\varphi}(v(t))=(c_{1}-c_{2})v(h)+v(t).$

We must have $v(\varphi(t(u+h)))=\min\\{v(t(u+h)),v(t)\\}=v(t)$ because
$\varphi$ generates $(x,t)$. This implies that $c_{1}=c_{2}$, a contradiction.
Thus, the existence of $\varphi$ is impossible, meaning that
$\operatorname{Int{}^{\text{R}}}(V)$ cannot be Bézout. ∎

This, along with [CL98, Theorem 3.5 and Corollary 3.3], gives us a complete
characterization of when $\operatorname{Int{}^{\text{R}}}(V)$ is a Bézout
domain.

######  Corollary 2.33.

The ring $\operatorname{Int{}^{\text{R}}}(V)$ is a Bézout domain if and only
if $\mathfrak{m}$ is principal or there exist two nonconstant polynomials of
coprime degrees over $V/\mathfrak{m}$ with no roots in $V/\mathfrak{m}$.

###  2.1 When $\operatorname{Int{}^{\text{R}}}(V)$ is not a Prüfer domain

We have completely classified the conditions on $V$ that make
$\operatorname{Int{}^{\text{R}}}(V)$ Prüfer. We consider the case when
$\operatorname{Int{}^{\text{R}}}(V)$ is not Prüfer and try to understand in
what ways $\operatorname{Int{}^{\text{R}}}(V)$ fails to be Prüfer.

If $\operatorname{Int{}^{\text{R}}}(V)$ is not Prüfer, then we know that $V$
has algebraically closed residue field and maximal ideal that is not
principal. We first consider the case when we additionally assume that the
value group that is not divisible. Since $\operatorname{Int{}^{\text{R}}}(V)$
is not Prüfer, there must be some prime ideal $\mathfrak{P}$ of
$\operatorname{Int{}^{\text{R}}}(V)$ such that
$\operatorname{Int{}^{\text{R}}}(V)_{\mathfrak{P}}$ is not a valuation domain.
We will show that such a prime ideal $\mathfrak{P}$ cannot be a pointed
maximal ideal. First, we will need a lemma.

######  Lemma 2.34.

Let $D$ be a domain and $\mathfrak{p}$ be a prime ideal. Let $a,b$ be two
elements of $K$, the field of fractions of $D$, such that $b\neq 0$. Then
$\frac{a}{b}\in D_{\mathfrak{p}}$ if and only if there exists $c\in K$ such
that $\frac{a}{c}\in D$ and $\frac{b}{c}\in D\setminus\mathfrak{p}$.

###### Proof.

Suppose $\frac{a}{b}\in D_{\mathfrak{p}}$. This implies that
$\frac{a}{b}=\frac{r}{s}$ for some $r\in D$ and $s\in D\setminus\mathfrak{p}$.
Now set $c\coloneqq\frac{b}{s}$. We see that $\frac{a}{b/s}=r\in D$ and
$\frac{b}{b/s}=s\in D\setminus\mathfrak{p}$.

On the other hand, if there exists $c\in K$ such that $\frac{a}{c}\in D$ and
$\frac{b}{c}\in D\setminus\mathfrak{p}$, then $\frac{a}{b}=\frac{a/c}{b/c}\in
D_{\mathfrak{p}}$. ∎

We will show that for a valuation domain $V$ with algebraically closed residue
field, maximal ideal that is not principal, and value group that is not
divisible that $\operatorname{Int{}^{\text{R}}}(V)$ localized at a pointed
maximal ideal is a valuation domain. This is true in a more general setting so
we give the result with weaker assumptions.

######  Proposition 2.35.

Suppose that $\Gamma$ is not divisible. Let $E$ be a subset of $K$ and take
$a\in E$. Then

$\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},a}}=\\{\varphi\in
K(x)\mid\varphi(a)\in V\\},$

a valuation domain.

###### Proof.

It suffices to assume without loss of generality that $0\in E$ and show that
$\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},0}}=\\{\varphi\in
K(x)\mid\varphi(0)\in V\\}$ since
$\operatorname{Int{}^{\text{R}}}(E,V)\cong\operatorname{Int{}^{\text{R}}}(E-a,V)$
for all $a\in E$.

We see that
$\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},0}}\subseteq\\{\varphi\in
K(x)\mid\varphi(0)\in V\\}$, so we want to show the reverse inclusion. Let
$\varphi\in K(x)$ be a nonzero rational function such that $\varphi(0)\in V$.
We can write $\varphi=\frac{f}{g}$ such that $f,g\in K[x]$ and $v(g(0))=0$.

Since $v(g(0))=0$, we know that $v(g(a))=0$ for all $a\in K$ such that $v(a)$
is sufficiently large due to the fact that the valuation of each monomial in
$g(a)$ except the constant can be arbitrarily large depending on $v(a)$. We
also know that by Corollary 2.20 that
$v(f(a))=\operatorname{minval}_{f}(v(a))$ for all $a\in K$ such that $v(a)$ is
sufficiently large. Since $v(f(0))\geq 0$, we can ensure that $v(f(a))\geq 0$
for all $a\in K$ such that $v(a)$ is sufficiently large. The reasoning for
this is similar to that for $g(x)$, except $f(x)$ might not have a constant
term. From this, we deduce that there exists $\delta\in\Gamma$ with
$\delta\geq 0$ such that $v(f(d))\geq v(g(d))=0$ for all $d\in K$ with
$v(d)>\delta$.

Since $\Gamma$ is not divisible, there exists
$\eta\in{\mathbb{Q}}\Gamma\setminus\Gamma$ such that $\eta>\delta$. There also
exists an $n\in{\mathbb{N}}$ with $n>0$ such that $n\eta\in\Gamma$. Let
$m\coloneqq\max\\{\deg f,\deg g\\}$. Set $h(x)\coloneqq\frac{1}{t}x^{mn}+1$,
where $t\in V$ is such that $v(t)=mn\eta$. We want to show that
$\frac{f}{h},\frac{g}{h}\in\operatorname{Int{}^{\text{R}}}(E,V)$ and
$\frac{g}{h}\notin\mathfrak{M}_{\mathfrak{m},0}$.

Now let $d\in E$. Then

$v(h(d))=\begin{cases}0,&\text{if $v(d)>\eta$},\\\ mn(v(d)-\eta),&\text{if
$v(d)<\eta$}.\end{cases}.$

From this, we can gather that for all $d\in E$ with $v(d)>\eta$, we have
$v\mathopen{}\mathclose{{}\left(\frac{f}{h}(d)}\right)\geq 0$ and
$v\mathopen{}\mathclose{{}\left(\frac{g}{h}(d)}\right)\geq 0$. Furthermore,
$\frac{g}{h}(0)=g(0)\notin\mathfrak{m}$.

Now fix $d\in E$ such that $v(d)<\eta$. We know
$v(f(d))\geq\operatorname{minval}_{f}(v(d))=cv(d)+\beta$ for some
$c\in{\mathbb{Z}}$ and $\beta\in\Gamma$. Notice that
$c\eta+\beta\geq\operatorname{minval}_{f}(\eta)$ by the definition of
$\operatorname{minval}_{f}$ and thus
$c\eta+\beta\geq\operatorname{minval}_{f}(\eta)\geq 0$. This implies
$\beta\geq-c\eta$. Now we obtain

$v(f(d))\geq cv(d)+\beta\geq c(v(d)-\eta)>mn(v(d)-\eta)=v(h(d))$

because $c\leq m<mn$ and $v(d)-\eta<0$. This shows that
$\frac{f}{h}(d)\in\mathfrak{m}$. We can show that
$\frac{g}{h}(d)\in\mathfrak{m}$ the same way. Thus,
$\frac{f}{h},\frac{g}{h}\in\operatorname{Int{}^{\text{R}}}(E,V)$ and
$\frac{g}{h}\notin\mathfrak{M}_{\mathfrak{m},0}$, which implies that
$\frac{f}{g}=\varphi\in\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},a}}$.

This shows that
$\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},a}}=\\{\varphi\in
K(x)\mid\varphi(a)\in V\\}$, so
$\operatorname{Int{}^{\text{R}}}(E,V)_{\mathfrak{M}_{\mathfrak{m},a}}$ is a
valuation domain.

∎

Suppose we have a valuation domain $V$ such that $\mathfrak{m}$ is not a
principal ideal of $V$ and $V/\mathfrak{m}$ is algebraically closed. We know
that $\operatorname{Int{}^{\text{R}}}(V)$ is not Prüfer, so there must be some
prime ideal $\mathfrak{P}$ of $\operatorname{Int{}^{\text{R}}}(V)$ such that
$\operatorname{Int{}^{\text{R}}}(V)_{\mathfrak{P}}$ is not a valuation domain.
We have just seen that this ideal cannot be a pointed maximal ideal if
$\Gamma$ is not divisible. We will give an example of such a prime in the form
of an ultrafilter limit of pointed maximal ideals. We will also use the idea
of a pseudo-divergent sequence [Per18, Definition 2.1]. Note that we do not
need to assume that the value group of $V$ is not divisible.

######  Proposition 2.36.

Suppose that $\mathfrak{m}$ is not a principal ideal of $V$ and
$V/\mathfrak{m}$ is algebraically closed. Fix some $d\in V$ with $v(d)>0$. Let
$\\{d_{i}\\}_{i=1}^{\infty}\subseteq V$ such that $v(d_{i})>v(d)$ for each $i$
and for each $\varepsilon\in\Gamma$ with $\varepsilon>0$, there exists an $i$
such that $v(d_{i})<v(d)+\varepsilon$. Then let $\mathcal{U}$ be a non-
principal ultrafilter of
$\\{\mathfrak{M}_{\mathfrak{m},d_{i}}\\}_{i=1}^{\infty}$. Then
$\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}$ is a prime ideal
of $\operatorname{Int{}^{\text{R}}}(V)$ such that
$\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$
is not valuation domain.

###### Proof.

First note that the existence of $\\{d_{i}\\}_{i=1}^{\infty}\subseteq V$ such
that $v(d_{i})>v(d)$ for each $i$ and for each $\varepsilon\in\Gamma$ with
$\varepsilon>0$, there exists an $i$ such that $v(d_{i})<v(d)+\varepsilon$
depends on the maximal ideal not being principal.

We will show that
$\frac{x+d}{d}\notin\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$.
A similar argument will show that
$\frac{d}{x+d}\notin\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$.
From this, we will see that
$\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$
is not valuation domain.

Suppose on the contrary that
$\frac{x+d}{d}\in\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$.
Then we may write $\frac{x+d}{d}=\frac{\varphi}{\psi}$, where
$\varphi,\psi\in\operatorname{Int{}^{\text{R}}}(V)$ and
$\psi\notin\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}$. Set
$\rho=\frac{d}{\psi}$. Then $\frac{x+d}{\rho}=\varphi$ and
$\frac{d}{\rho}=\psi$.

We then get that $\\{\mathfrak{M}_{\mathfrak{m},d_{i}}\mid
i\in{\mathbb{N}},\psi\notin\mathfrak{M}_{\mathfrak{m},d_{i}}\\}\in\mathcal{U}$.
Because $\mathcal{U}$ is not principal,
$\\{\mathfrak{M}_{\mathfrak{m},d_{i}}\mid
i\in{\mathbb{N}},\psi\notin\mathfrak{M}_{\mathfrak{m},d_{i}}\\}$ is an
infinite set. Therefore, there exists $i\in{\mathbb{N}}$ such that $v(d_{i})$
is arbitrarily close to $v(d)$ and
$\psi\notin\mathfrak{M}_{\mathfrak{m},d_{i}}$, or equivalently,
$v(\psi(d_{i}))=0$. This shows that there exists some $\varepsilon\in\Gamma$
with $\varepsilon>0$ such that $\operatorname{minval}_{\psi}(\gamma)=0$ for
$\gamma\in\Gamma$ such that $v(d)\leq\gamma\leq v(d)+\varepsilon$. Since
$\operatorname{minval}_{\rho}=\operatorname{minval}_{d}-\operatorname{minval}_{\psi}$,
we obtain that $\operatorname{minval}_{\rho}(\gamma)=v(d)$ for $\gamma$ such
that $v(d)\leq\gamma\leq v(d)+\varepsilon$.

On the other hand, we can make $\varepsilon$ small enough so that
$\operatorname{minval}_{\rho}(\gamma)=c\gamma+\beta$ for $\gamma$ such that
$v(d)-\varepsilon\leq\gamma\leq v(d)$ for some $c\in{\mathbb{Z}}$ and
$\beta\in\Gamma$. We know that $\operatorname{minval}_{\rho}(v(d))=v(d)$, so
$\beta=(1-c)v(d)$. Next, since $\varphi\in\operatorname{Int{}^{\text{R}}}(V)$,
for $\gamma\in\Gamma$ with $\gamma\geq 0$, we have that
$\operatorname{minval}_{x+d}(\gamma)-\operatorname{minval}_{\rho}(\gamma)=\operatorname{minval}_{\varphi}(\gamma)\geq
0$. Lemma 2.26 implies this inequality for almost all such values of $\gamma$,
and by the form of the minimum valuation function given in Proposition 2.24,
the inequality holds for all $\gamma\geq 0$. Thus, for $\gamma$ such that
$v(d)-\varepsilon<\gamma\leq v(d)$, we get

$\gamma=\operatorname{minval}_{x+d}(\gamma)\geq\operatorname{minval}_{\rho}(\gamma)=c\gamma+(1-c)v(d).$

This implies that $(1-c)\gamma\geq(1-c)v(d)$, but we have $\gamma\leq v(d)$,
so it must be the case that $1-c\leq 0$, or equivalently, $c\geq 1$.

By Lemma 2.28, there exists $a\in V$ with $v(a)=v(d)$ such that
$v(\rho(a))>v(d)$, but then
$v(\psi(a))=v\mathopen{}\mathclose{{}\left(\frac{d}{\rho(a)}}\right)<0$,
contradicting the fact that $\psi\in\operatorname{Int{}^{\text{R}}}(V)$. Thus,
$\frac{x+d}{d}\notin\operatorname{Int{}^{\text{R}}}(V)_{\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}}$.

∎

######  Remark 2.37.

Suppose that $\mathfrak{m}$ is not a principal ideal of $V$, that
$V/\mathfrak{m}$ is algebraically closed, and that $\Gamma$ is not divisible.
Then $\operatorname{Int{}^{\text{R}}}(V)$ is an example of an essential
domain, a domain that can be written as the intersection of some family of
essential valuation overrings, that is not a P$v$MD. Another example of an
essential domain that is not a P$v$MD can be found in [HO73].

We can write

$\operatorname{Int{}^{\text{R}}}(V)=\bigcap_{a\in
V}\operatorname{Int{}^{\text{R}}}(V)_{\mathfrak{M}_{\mathfrak{m},a}}.$

For every $a\in V$, we know that $\mathfrak{M}_{\mathfrak{m},a}$ is essential
by Proposition 2.35. This means that $\operatorname{Int{}^{\text{R}}}(V)$ is
an essential domain. Furthermore, for every $a\in V$, the ideal
$\mathfrak{M}_{\mathfrak{m},a}$ being essential implies that
$\mathfrak{M}_{\mathfrak{m},a}$ is a $t$-ideal [Kan89, Lemma 3.17]. Using the
notation of the previous proposition, we know that
$\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}$ is a t-ideal
since the ultrafilter limit of t-ideals is a t-ideal [CLT00, Proposition 2.5].
However, the ideal
$\lim\limits_{\mathcal{U}}\mathfrak{M}_{\mathfrak{m},d_{i}}$ is a $t$-maximal
ideal of $\operatorname{Int{}^{\text{R}}}(V)$ that is not essential. Thus,
$\operatorname{Int{}^{\text{R}}}(V)$ is not a P$v$MD.

We will now consider the case where $V/\mathfrak{m}$ is algebraically closed
and the value group is divisible (which implies that $\mathfrak{m}$ is not
principal). We can actually detect that $\operatorname{Int{}^{\text{R}}}(V)$
is not Prüfer by localizing at a pointed maximal ideal.

######  Proposition 2.38.

Suppose that $V/\mathfrak{m}$ is algebraically closed and $\Gamma$ is
divisible. Then the localization of $\operatorname{Int{}^{\text{R}}}(V)$ at
any pointed maximal ideal is not a valuation ring.

###### Proof.

Let $a\in V$. Mapping $x\mapsto x-a$ and fixing $V$ determines an automorphism
for $\operatorname{Int{}^{\text{R}}}(V)$ for any $a\in V$, we can study the
behavior of localizing $\operatorname{Int{}^{\text{R}}}(V)$ at
$\mathfrak{M}_{\mathfrak{m},a}$ by only considering the localization at
$\mathfrak{M}_{\mathfrak{m},0}$.

Suppose for a contradiction that
$W\coloneqq\operatorname{Int{}^{\text{R}}}(V)_{\mathfrak{M}_{\mathfrak{m},0}}$
is a valuation domain. Fix a nonzero $d\in\mathfrak{m}$. We have
$\frac{d}{x+d}\in W$ or $\frac{x+d}{d}\in W$. Thus, by Lemma 2.34, we have
some $\varphi\in K(x)$ such that
$\frac{d}{\varphi}\in\operatorname{Int{}^{\text{R}}}(V)$ and
$\frac{x+d}{\varphi}\in\operatorname{Int{}^{\text{R}}}(V)$, and additionally,
$\frac{d}{\varphi}\notin\mathfrak{M}_{\mathfrak{m},0}$ or
$\frac{x+d}{\varphi}\notin\mathfrak{M}_{\mathfrak{m},0}$.

We know that $\operatorname{minval}_{\varphi}$ has the form

$\operatorname{minval}_{\varphi}(\gamma)=\begin{cases}c_{1}\gamma+\beta_{1},&\gamma\leq\delta_{1},\\\
c_{2}\gamma+\beta_{2},&\delta_{1}\leq\gamma\leq\delta_{2},\\\ \vdots\\\
c_{n-1}\gamma+\beta_{n-1},&\delta_{n-2}\leq\gamma\leq\delta_{n-1},\\\
c_{n}\gamma+\beta_{n},&\delta_{n-1}\leq\gamma,\end{cases}$

for some $c_{i}\in{\mathbb{Z}}$, $\beta_{i}\in\Gamma$,
$\delta_{i}\in{\mathbb{Q}}\Gamma$ by Proposition 2.24. We may choose
$\delta_{n-1}$ such that $c_{n-1}\neq c_{n}$. We claim that we must have
$c_{n}=0$ and $\beta_{n}=v(d)$. If
$\frac{x+d}{\varphi}\notin\mathfrak{M}_{\mathfrak{m},0}$, then
$v\mathopen{}\mathclose{{}\left(\frac{0+d}{\varphi(0)}}\right)=0$, so
$v(\varphi(0))=v(d)$. If
$\frac{d}{\varphi}\notin\mathfrak{M}_{\mathfrak{m},0}$, we similarly have
$v(\varphi(0))=v(d)$. Either way, we have $v(\varphi(0))=v(d)$. Since $d\neq
0$, we have that $\varphi(0)\neq 0$. Then $v(\varphi(a))=v(\varphi(0))$ for
$a\in K$ such that $v(a)$ is sufficiently large. This implies that
$\operatorname{minval}_{\varphi}(\gamma)$ is constant and equal to
$v(\varphi(0))=v(d)$ for $\gamma\in\Gamma$ sufficiently large by Lemma 2.26.
Therefore, $c_{n}=0$ and $\beta_{n}=v(d)$.

We also claim that $\delta_{n-1}\geq v(d)$. If not, by Lemma 2.26, there
exists $b\in V$ such that $\max\\{\delta_{n-1},0\\}<v(b)<v(d)$ and
$v(\varphi(b))=\operatorname{minval}_{\varphi}(v(b))=v(d)$. However, this
would imply that

$v\mathopen{}\mathclose{{}\left(\frac{b+d}{\varphi(b)}}\right)=v(b)-v(d)<0,$

contradicting the fact that
$\frac{x+d}{\varphi}\in\operatorname{Int{}^{\text{R}}}(V)$.

Next, we claim that $c_{n-1}>c_{n}=0$. If $c_{n-1}<0$, we can find $b\in V$
such that $\max\\{\delta_{n-2},0\\}<v(b)<\delta_{n-1}$ and
$v(\varphi(b))=\operatorname{minval}_{\varphi}(v(b))=c_{n-1}v(b)+\beta_{n-1}$.
Since $c_{n-1}\delta_{n-1}+\beta_{n-1}=v(d)$, we have that
$v(\varphi(b))-v(d)=c_{n-1}(v(b)-\delta_{n-1})>0$ as both $c_{n-1}$ and
$v(b)-\delta_{n-1}$ are less than 0. This contradicts the fact that
$\frac{d}{\varphi}\in\operatorname{Int{}^{\text{R}}}(V)$.

Because $\Gamma$ is divisible, we know that $\delta_{n-1}\in\Gamma$.
Furthermore, since $c_{n-1}>c_{n}$, by Lemma 2.28, there exists some element
$a\in V$ so that we have $v(a)=\delta_{n-1}$ and
$v(\varphi(a))>\operatorname{minval}_{\varphi}(\delta_{n-1})=v(d)$. This
contradicts $\frac{d}{\varphi}\in\operatorname{Int{}^{\text{R}}}(V)$. Thus,
the assumption that $W$ is a valuation domain is false. ∎

We end this section by noting that for a valuation domain $V$ such that
$V/\mathfrak{m}$ is algebraically closed and $\mathfrak{m}$ is not a principal
ideal of $V$, even though $\operatorname{Int{}^{\text{R}}}(V)$ is not a Prüfer
domain, there are subsets $E$ of the field of fractions $K$ of $V$ such that
$\operatorname{Int{}^{\text{R}}}(E,V)$ is a Prüfer domain. For example, if $E$
is a singleton, then $\operatorname{Int{}^{\text{R}}}(E,V)$ is a valuation
domain and therefore a Prüfer domain. Likewise, there are other subsets $E$
such that $\operatorname{Int{}^{\text{R}}}(E,V)$ is not a Prüfer domain. We
can obtain from the proofs of Theorem 2.29 and Lemma 2.28 some conditions $E$
such that $\operatorname{Int{}^{\text{R}}}(E,V)$ is not Prüfer. One case is
indicated in the following result.

######  Proposition 2.39.

Suppose $V/\mathfrak{m}$ is algebraically closed and $\mathfrak{m}$ is not a
principal ideal of $V$. Also suppose that $V$ is an essential valuation
overring of a domain $D$ centered on a maximal ideal of $D$. Then
$\operatorname{Int{}^{\text{R}}}(D,V)$ is not Prüfer.

###### Proof.

Let $v$ be the valuation associated with $V$ and $\Gamma$ the value group.
This follows from the proofs of Theorem 2.29 and Lemma 2.28 and the facts that
$V/\mathfrak{m}\cong D/(D\cap\mathfrak{m})$ and for every $\gamma\in\Gamma$,
there exists $d\in D$ such that $v(d)=\gamma$. ∎

##  3 Integer-valued rational functions over Prüfer domains

Let $D$ be a domain. We want to know what conditions on $D$ makes
$\operatorname{Int{}^{\text{R}}}(D)$ a Prüfer domain. The case for rings of
integer-valued polynomials has been answered [Lop98, CCF00]. If
$\operatorname{Int{}^{\text{R}}}(D)$ is a Prüfer domain, then $D$ is a Prüfer
domain since homomorphic images of Prüfer domains are Prüfer domains. In
[CL98], we see that if $D$ is a monic or singular Prüfer domain, then
$\operatorname{Int{}^{\text{R}}}(D)$ is Prüfer. We want to first investigate a
few cases when $D$ is a Prüfer domain but $\operatorname{Int{}^{\text{R}}}(D)$
is not a Prüfer domain.

One possible obstruction to $\operatorname{Int{}^{\text{R}}}(E,D)$ being a
Prüfer domain is an valuation overring $V$ of $D$ that yields a domain
$\operatorname{Int{}^{\text{R}}}(E,V)$ that is not Prüfer.

######  Proposition 3.1.

Let $D$ be a domain with field of fractions $K$ and $E\subseteq K$ a subset.
Also let $V$ be a valuation overring of $D$ such that
$\operatorname{Int{}^{\text{R}}}(E,D)$ and
$\operatorname{Int{}^{\text{R}}}(E,V)$ have the same field of fractions. If
$\operatorname{Int{}^{\text{R}}}(E,V)$ is not Prüfer, then
$\operatorname{Int{}^{\text{R}}}(E,D)$ is not Prüfer either.

###### Proof.

Since $D\subseteq V$, we have
$\operatorname{Int{}^{\text{R}}}(E,D)\subseteq\operatorname{Int{}^{\text{R}}}(E,V)$,
so $\operatorname{Int{}^{\text{R}}}(E,V)$ is an overring of
$\operatorname{Int{}^{\text{R}}}(E,D)$ that is not Prüfer. Thus,
$\operatorname{Int{}^{\text{R}}}(E,D)$ is not Prüfer as every overring of a
Prüfer domain is a Prüfer domain. ∎

######  Corollary 3.2.

Let $D$ be a domain with field of fractions $K$. Suppose there exists a
valuation overring $V$ of $D$ centered on a maximal ideal of $D$ such that the
residue field of $V$ is algebraically closed and the maximal ideal of $V$ is
not principal. Then $\operatorname{Int{}^{\text{R}}}(D)$ is not Prüfer.

###### Proof.

Since $D\subseteq\operatorname{Int{}^{\text{R}}}(D)$ and
$x\in\operatorname{Int{}^{\text{R}}}(D)$, we have that the field of fractions
of $\operatorname{Int{}^{\text{R}}}(D)$ is $K(x)$. Thus,
$\operatorname{Int{}^{\text{R}}}(D,V)$ is an overring of
$\operatorname{Int{}^{\text{R}}}(D)$. Moreover,
$\operatorname{Int{}^{\text{R}}}(D,V)$ is not Prüfer by Proposition 2.39, so
$\operatorname{Int{}^{\text{R}}}(D)$ is not Prüfer either. ∎

Given a Prüfer domain $D$, obstructions to
$\operatorname{Int{}^{\text{R}}}(D)$ being a Prüfer domain do not necessarily
come locally from a single valuation overring. The obstruction can come from a
collection of valuation overrings that behave collectively like a valuation
overring with algebraically closed residue field and maximal ideal that is not
principal, but individually, each valuation overring does not have both
algebraically closed residue field and maximal ideal that is not principal.
The construction of such domains is done via sequential domains, a
generalization of sequence domains in [Lop97].

######  Definition 3.3.

Let $D$ be a domain with field of fractions $K$. We say that $D$ is a
sequential domain if there exist for each $i\in{\mathbb{N}}\setminus\\{0\\}$,
valuations $v_{i}:K\to\Gamma_{i}\cup\\{\infty\\}$, where $\Gamma_{i}$ is a
totally ordered abelian group, such that

  * •

each associated valuation domain $V_{i}$ is an essential overring of $D$ such
that $D=\bigcap_{i=1}^{\infty}V_{i}$,

  * •

there is a common totally ordered abelian group $\Gamma$ with embeddings
$\Gamma_{i}\hookrightarrow\Gamma$ such that for each $d\in D$,
$\\{v_{i}(d)\\}_{i=1}^{\infty}$ is eventually constant viewed as a sequence in
$\Gamma$, and

  * •

there exists $\varpi\in D$ such that $v_{i}(\varpi)$ is not eventually 0.

Since $v_{i}(d)$ is eventually constant for all $d\in D$, there is valuation
$v_{\infty}$ defined by $v_{\infty}(d)=\lim\limits_{i\to\infty}v_{i}(d)$ with
associated valuation domain $V_{\infty}$ and maximal ideal
$\mathfrak{m}_{\infty}$. We also have an embedding of the value group
$\Gamma_{\infty}\hookrightarrow\Gamma$.

We say a sequential domain $D$ has the unbounded ramification property if for
all $\gamma\in{\mathbb{Q}}\Gamma_{\infty}$ with $\gamma>0$ and
$N\in{\mathbb{N}}$, there is some $\gamma^{\prime}\in\Gamma_{n}$ with
$\gamma^{\prime}>0$ for some $n\geq N$ such that $\gamma^{\prime}<\gamma$
considered as elements of ${\mathbb{Q}}\Gamma$.

From Theorem 2.29, we know that if $V$ is a valuation domain with
algebraically closed residue field and maximal ideal that is not principal,
then $\operatorname{Int{}^{\text{R}}}(V)$ is not a Prüfer domain. A sequential
domain can spread out the obstructions that make the ring of integer-valued
rational functions not a Prüfer domain. The unbounded ramification property
mimics the property of having a maximal ideal that is not principal, even if
all of the valuation overrings have a principal maximal ideal. There is an
example following the proposition illustrating this phenomenon.

The following proposition uses the notation from the definition of a
sequential domain.

######  Proposition 3.4.

Let $D$ be a sequential domain with the unbounded ramification property. Also
suppose that $V_{\infty}/\mathfrak{m}_{\infty}$ is algebraically closed. Then
$\operatorname{Int{}^{\text{R}}}(D)$ is not a Prüfer domain.

###### Proof.

We can assume that $D$ is Prüfer since if $D$ is not Prüfer, then
$\operatorname{Int{}^{\text{R}}}(D)$ is not Prüfer as well. We will suppose
that $(x,\varpi)\subseteq\operatorname{Int{}^{\text{R}}}(D)$ is invertible for
a contradiction. Then there exist $\varphi,\psi\in(x,\varpi)^{-1}$ such that
$\varphi\cdot x+\psi\cdot\varpi=1$. Let $\alpha\coloneqq v_{\infty}(\varpi)$.
There exists $a\in D$ with $v_{\infty}(a)=\alpha$ such that
$\operatorname{minval}_{\varphi,v_{\infty}}(\alpha)=v_{\infty}(\varphi(a))$
and $\operatorname{minval}_{\psi,v_{\infty}}(\alpha)=v_{\infty}(\psi(a))$.
Considering $\varphi(x)x+\psi(x)\varpi=1$, we deduce that

$0\geq\min\\{v_{\infty}(\varphi(a))+\alpha,v_{\infty}(\psi(a))+\alpha\\}.$

Because $\varphi,\psi\in(x,\varpi)^{-1}$, we have
$v_{\infty}(\varphi(a))+\alpha,v_{\infty}(\psi(a))+\alpha\geq 0$. This means
that $v_{\infty}(\varphi(a))+\alpha=0$ or $v_{\infty}(\psi(a))+\alpha=0$.
Either way, we have some $\rho\in(x,\varpi)^{-1}$ such that
$\operatorname{minval}_{\rho,v_{\infty}}(\alpha)=-\alpha$.

Now for some $\varepsilon\in{\mathbb{Q}}\Gamma_{\infty}$ with $\varepsilon>0$,
we have

$\operatorname{minval}_{\rho,v_{\infty}}(\gamma)=\begin{cases}c\gamma+\beta,&\alpha-\varepsilon<\gamma\leq\alpha,\\\
c^{\prime}\gamma+\beta^{\prime},&\alpha\leq\gamma<\alpha+\varepsilon,\\\
\end{cases}$

where $c,c^{\prime}\in{\mathbb{Z}}$ and $\beta,\beta^{\prime}\in\Gamma$. Since
there are a finite number of coefficients which appear in $\rho$ and their
values under $v_{i}$ is eventually constant and equal to their values under
$v_{\infty}$, there exists $N\in{\mathbb{N}}$ such that for all $i\geq N$, we
have
$\operatorname{minval}_{\rho,v_{\infty}}=\operatorname{minval}_{\rho,v_{i}}$
and additionally $\alpha\in\Gamma_{i}$. Because $D$ has the unbounded
ramification property, there exists some $n\geq N$ such that there exists
$\delta\in\Gamma_{n}$ such that $0<\delta<\varepsilon$. Then $\alpha-\delta$
and $\alpha+\delta$ are both in $\Gamma_{n}$. Since $\rho\in(x,\varpi)^{-1}$
as a fractional ideal of $\operatorname{Int{}^{\text{R}}}(D)$, we have
$\operatorname{minval}_{\rho,v_{n}}\geq-\min\\{\operatorname{minval}_{x,v_{n}},\operatorname{minval}_{\varpi,v_{n}}\\}$
on $(\Gamma_{n})_{\geq 0}$. Combining this with the fact that
$\alpha-\varepsilon<\alpha-\delta<\alpha+\delta<\alpha+\varepsilon$, we force
$c\leq-1$ and $c^{\prime}\geq 0$. Therefore, $c^{\prime}-c>0$. Also note that
$\operatorname{minval}_{\rho,v_{i}}(\alpha)=-\alpha$ for all $i\geq N$.

Write $\rho=\frac{f}{g}$ for some $f,g\in D[x]$. Since
$V_{\infty}/\mathfrak{m}_{\infty}$ is algebraically closed,
$\operatorname{loc}_{f,\varpi,v_{\infty}}$ and
$\operatorname{loc}_{g,\varpi,v_{\infty}}$ factor completely modulo
$\mathfrak{m}_{\infty}$. This means we can write

$\displaystyle\frac{f(\varpi x)}{t}$
$\displaystyle=x^{m}(x-u_{1})^{e_{1}}\cdots(x-u_{r})^{e_{r}}+h_{1}(x),$
$\displaystyle\frac{g(\varpi x)}{t^{\prime}}$
$\displaystyle=x^{m^{\prime}}(x-u_{1}^{\prime})^{e_{1}^{\prime}}\cdots(x-u_{r})^{e_{r}^{\prime}}+h_{2}(x),$

for some $t,t^{\prime}\in K$ with
$v_{\infty}(t)=\operatorname{minval}_{f,v_{\infty}}(\alpha)$,
$v_{\infty}(t^{\prime})=\operatorname{minval}_{g,v_{\infty}}(\alpha)$,
$u_{1},\dots,u_{r}\in D$ representing distinct nonzero residues modulo
$D\cap\mathfrak{m}_{\infty}$, and $h_{1},h_{2}\in\mathfrak{m}_{\infty}[x]$. We
by Lemma 2.25 have that

$(e_{1}^{\prime}-e_{1})+\cdots+(e_{r}^{\prime}-e_{r})=c^{\prime}-c>0,$

so without loss of generality by permuting the indices, we can assume that
$e_{1}^{\prime}>e_{1}$. Write $h_{1}(x)=\sum b_{j}x^{j}$ and $h_{2}(x)=\sum
b_{j}^{\prime}x^{j}$ with each $b_{j}$ and $b_{j}^{\prime}$ being in
$\mathfrak{m}_{\infty}$. Furthermore, we have some $M\geq N$ such that for all
$i\geq M$, we get $v_{i}(u_{\ell})=0$ and
$v_{i}(u_{\ell}-u_{\ell^{\prime}})=0$ for $\ell,\ell^{\prime}$ distinct,
$v_{i}(b_{j})=v_{\infty}(b_{j})>0$, and
$v_{i}(b_{j}^{\prime})=v_{\infty}(b_{j}^{\prime})>0$. Because $D$ has the
unbounded ramification property, there exists $M^{\prime}\geq M$ such that
there exists $d\in D$ with
$0<v_{M^{\prime}}(d)<\frac{1}{e_{1}^{\prime}}\min\limits_{j}\\{v_{\infty}(b_{j}),v_{\infty}(b_{j}^{\prime})\\}$.

Now we evaluate $\rho$ at $x=\varpi(d+u_{1})$. We get

$\displaystyle\rho(\varpi(d+u_{1}))$
$\displaystyle=\frac{f(\varpi(d+u_{1}))}{g(\varpi(d+u_{1}))}$
$\displaystyle=\frac{t}{t^{\prime}}\cdot\frac{d^{e_{1}}(d+u_{1})^{m}(d+u_{1}-u_{2})^{e_{2}}\cdots(d+u_{1}-u_{r})^{e_{r}}+h_{1}(d+u_{1})}{d^{e_{1}^{\prime}}(d+u_{1})^{m^{\prime}}(d+u_{1}-u_{2})^{e_{2}^{\prime}}\cdots(d+u_{1}-u_{r})^{e_{r}^{\prime}}+h_{2}(d+u_{1})}$

We have that
$v_{M^{\prime}}(d^{e_{1}}(d+u_{1})^{m}(d+u_{1}-u_{2})^{e_{2}}\cdots(d+u_{1}-u_{r})^{e_{r}})=e_{1}v_{M^{\prime}}(d)$
and $v_{M^{\prime}}(h_{1}(d+u_{1}))>e_{1}v_{M^{\prime}}(d)$ since each
coefficient of the polynomial $h_{1}$ has coefficients with $v_{M^{\prime}}$
valuation strictly greater than $e_{1}v_{M^{\prime}}(d)$. Thus,

$v_{M^{\prime}}(d^{e_{1}}(d+u_{1})^{m}(d+u_{1}-u_{2})^{e_{2}}\cdots(d+u_{1}-u_{r})^{e_{r}}+h_{1}(d+u_{1}))=e_{1}v_{M^{\prime}}(d).$

Similarly,

$v_{M^{\prime}}(d^{e_{1}^{\prime}}(d+u_{1})^{m^{\prime}}(d+u_{1}-u_{2})^{e_{2}^{\prime}}\cdots(d+u_{1}-u_{r})^{e_{r}^{\prime}}+h_{2}(d+u_{1}))=e_{1}^{\prime}v_{M^{\prime}}(d).$

This means that

$\displaystyle v_{M^{\prime}}(\rho(\varpi(d+u_{1})))$
$\displaystyle=v_{M^{\prime}}\mathopen{}\mathclose{{}\left(\frac{t}{t^{\prime}}}\right)+e_{1}v_{M^{\prime}}(d)-e_{1}^{\prime}v_{M^{\prime}}(d)$
$\displaystyle=\operatorname{minval}_{\rho,v_{M^{\prime}}}(\alpha)+(e_{1}-e_{1}^{\prime})v_{M^{\prime}}(d)$
$\displaystyle=-\alpha+(e_{1}-e_{1}^{\prime})v_{M^{\prime}}(d)$
$\displaystyle<-\alpha.$

This is a contradiction since
$\rho\cdot\varpi\in\operatorname{Int{}^{\text{R}}}(D)$ but
$v_{M^{\prime}}(\rho(\varpi(d+u_{1}))\varpi)<0$. ∎

######  Example 3.5.

Let $k$ be an uncountable algebraically closed field and form the field
$K=k(s,t_{1},t_{2},\dots)$. Take
$\\{\alpha_{1},\alpha_{2},\dots\\}\subseteq{\mathbb{R}}_{>0}$ to be a
${\mathbb{Q}}$-linearly independent subset. For each $i\in{\mathbb{N}}$,
define $v_{i}:K^{\times}\to\mathopen{}\mathclose{{}\left(\sum\limits_{j\neq
i}{\mathbb{Z}}\alpha_{j}}\right)\oplus{\mathbb{Z}}\oplus\frac{1}{i}{\mathbb{Z}}$,
ordered lexicographically, as the valuation such that

$v_{i}(k^{\times})=\\{(0,0,0)\\},v_{i}(t_{i})=\mathopen{}\mathclose{{}\left(0,0,\frac{1}{i}}\right),v_{i}(s)=(0,1,0),v_{i}(t_{j})=(\alpha_{j},0,0),$

for $j\neq i$ and $v_{i}(f)$ for $f\in k[s,t_{1},t_{2},\dots]$ is the minimum
of the $v_{i}$ values of each monomial of $f$. Then each $v_{i}$ extends
uniquely to $K$. Let $V_{i}$ be the associated valuation domain of $v_{i}$.
Define the domain $D=\bigcap\limits_{i=1}^{\infty}V_{i}$. We see that $D$ is a
sequential domain with field of fractions $K$ (since $k\subseteq D$ and
$s,t_{1},t_{2},\dots\in D$) where each $V_{i}$ has an algebraically closed
residue field and a principal maximal ideal, generated by $t_{i}$.
Additionally, each value group embeds naturally into
$\mathopen{}\mathclose{{}\left(\sum\limits_{i=1}^{\infty}{\mathbb{Z}}\alpha_{j}}\right)\oplus{\mathbb{Z}}\oplus{\mathbb{Q}}$,
and $v_{\infty}$ gives rise to the associated valuation overring $V_{\infty}$,
which also has algebraically closed residue field and principal maximal ideal,
generated by $s$. This domain $D$ also has the unbounded ramification
property, since $v_{i}(t_{i})$ can be arbitrarily small. We also know that $D$
is a Bézout domain and therefore a Prüfer domain by [OR06, Theorem 6.6] since
$D$ is the intersection of countably many valuation domains containing $k$, an
uncountable subfield. However, by the previous proposition,
$\operatorname{Int{}^{\text{R}}}(D)$ is not a Prüfer domain.

Notice that each $V_{i}$ for $i\in{\mathbb{N}}\setminus\\{0\\}$ has a
principal maximal ideal. The same holds true for $V_{\infty}$. Thus, Corollary
3.2 cannot be used immediately to show that
$\operatorname{Int{}^{\text{R}}}(D)$ is not a Prüfer domain.

###  3.1 Intersection of monic and singular Prüfer domains

We know that if $D$ is a Prüfer domain that is monic or singular, then
$\operatorname{Int{}^{\text{R}}}(D)$ is also a Prüfer domain. We want to see
if there are other instances of Prüfer domains whose ring of integer-valued
rational functions is also a Prüfer domain. For this reason, we consider
Prüfer domains $D$ that are neither monic nor singular. Some domains of this
form are formed by the intersection of a monic Prüfer domain and a singular
Prüfer domain.

We first consider a family of Prüfer domains obtained from intersecting a
monic Prüfer domain and a singular Prüfer domain with some extra conditions.
We then give some corollaries with conditions that are easier to verify. This
includes the case of a finite intersection of valuation domains. Examples will
follow to showcase the different conditions.

######  Theorem 3.6.

Let $D$ be a Prüfer domain with $K$ as the field of fractions. Suppose that
$D$ can be written as $D=D_{1}\cap D_{2}$, where $D_{1}$ is a monic Prüfer
overring of $D$ and $D_{2}$ is a singular Prüfer overring of $D$. Suppose
there exist $n\in{\mathbb{N}}$, a collection $\\{V_{\lambda}\\}$ of valuation
overrings of $D_{1}$, and a collection $\\{W_{\mu}\\}$ of valuation overrings
of $D_{2}$ such that

  * •

$D_{1}=\bigcap\limits_{\lambda}V_{\lambda}$ and
$D_{2}=\bigcap\limits_{\mu}W_{\mu}$,

  * •

the maximal ideal of $W_{\mu}$ is generated by some $\varpi_{\mu}\in W_{\mu}$,

  * •

there exists some $d\in D_{2}$ such that $0<w_{\mu}(d)<nw_{\mu}(\varpi_{\mu})$
for all $\mu$, where $w_{\mu}$ is the valuation corresponding to $W_{\mu}$,

  * •

$v_{\lambda}(d)=0$ for all $\lambda$, where $v_{\lambda}$ is the valuation
corresponding to $V_{\lambda}$, and

  * •

there exist polynomials $f,g,h_{1},h_{2}\in D[x]$ such that

    * –

$f$ is monic of degree $n$ such that $f(D_{1})\subseteq D_{1}^{\times}$,

    * –

$g(D_{1})\subseteq D_{1}^{\times}$, $g$ has leading coefficient $d$, $\deg
g=n$, and $w_{\mu}(g(a))=w_{\mu}(da^{n})$ for all $\mu$ and for all $a\in K$
such that $w_{\mu}(a)<0$,

    * –

$h_{1}$ is monic and $\deg h_{1}=n$, $\deg h_{2}<n$, and

    * –

$dfh_{1}+gh_{2}$ is unit-valued for each $V_{\lambda}$ and $W_{\mu}$.

Then $\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer domain with torsion
Picard group for any subset $E\subseteq K$.

###### Proof.

Fix a subset $E\subseteq K$.

Set $\theta(x)=\frac{df(x)}{g(x)}$. Fix $\mu$. Let $a\in K$. If
$w_{\mu}(a)\geq 0$, then $w_{\mu}((dfh_{1}+gh_{2})(a))=0$, which implies
$w_{\mu}((gh_{2})(a))=0$ and thus $w_{\mu}(g(a))=0$. Then we calculate

$w_{\mu}(\theta(a))=\begin{cases}w_{\mu}(df(a))\geq 0,&\text{if
$w_{\mu}(a)\geq 0$},\\\ w_{\mu}(da^{n})-w_{\mu}(da^{n})=0,&\text{if
$w_{\mu}(a)<0$}.\end{cases}$

Similarly, if we fixed $\lambda$, then since $g(D_{1})\subseteq
D_{1}^{\times}$ and the leading coefficient of $g$ is a unit of $D_{1}$, we
know that $g$ is unit-valued for $V_{\lambda}$ [Lop94, Proposition 2.1]. Let
$a\in K$, we have that

$v_{\lambda}(\theta(a))=\begin{cases}v_{\lambda}(df(a))=0,&\text{if
$v_{\lambda}(a)\geq 0$},\\\
v_{\lambda}(da^{n})-v_{\lambda}(da^{n})=0,&\text{if
$v_{\lambda}(a)<0$}.\end{cases}$

This shows that $\theta\in\operatorname{Int{}^{\text{R}}}(K,D)$.

Now let $a,b\in K$ with $b\neq 0$. We define

$a\diamond
b\coloneqq\theta\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)b^{n}h_{1}\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)+b^{n}h_{2}\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right).$

We claim that $v(a\diamond b)=\min\\{v(a^{n}),v(b^{n})\\}$ for any valuation
$v=v_{\lambda}$ or $v=w_{\mu}$.

Fix a valuation $v=v_{\lambda}$ or $v=w_{\mu}$. Let $a,b\in K$ with $b\neq 0$.
We have that

$\frac{a\diamond
b}{b^{n}}=\theta\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)h_{1}\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)+h_{2}\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)=\mathopen{}\mathclose{{}\left(\frac{dfh_{1}+gh_{2}}{g}}\right)\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right),$

so if $v(a)\geq v(b)$, then $v(a\diamond b)=v(b^{n})$ since both
$dfh_{1}+gh_{2}$ and $g$ are unit-valued over the valuation ring associated
with $v$.

Now suppose that $v(a)<v(b)$. Write
$h_{1}(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}+x^{n}$ and
$h_{2}(x)=b_{0}+b_{1}x+\cdots+b_{r}x^{r}$ with $a_{i},b_{i}\in D$ and some
$r<n$. Then we have

$a\diamond
b\coloneqq\theta\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)(a_{0}b^{n}+a_{1}ab^{n-1}+\cdots+a_{n-1}a^{n-1}b+a^{n})+(b_{0}b^{n}+b_{1}ab^{n-1}+\cdots+b_{r}a^{r}b^{n-r}).$

If $v(a)<v(b)$, then
$v\mathopen{}\mathclose{{}\left(\theta\mathopen{}\mathclose{{}\left(\frac{a}{b}}\right)}\right)=0$,
so $v(a\diamond b)=v(a^{n})$, as desired.

Now take $\varphi,\psi\in\operatorname{Int{}^{\text{R}}}(E,D)$ with $\psi\neq
0$. We claim that $(\varphi,\psi)^{n}$ is generated by

$\rho\coloneqq\theta\mathopen{}\mathclose{{}\left(\frac{\varphi}{\psi}}\right)(a_{0}\psi^{n}+a_{1}\varphi\psi^{n-1}+\cdots+a_{n-1}\varphi^{n-1}\psi+\varphi^{n})+(b_{0}\psi^{n}+b_{1}\varphi\psi^{n-1}+\cdots+b_{r}\varphi^{r}\psi^{n-r}).$

We see that $\rho\in(\varphi,\psi)^{n}$ since
$\theta\in\operatorname{Int{}^{\text{R}}}(K,D)$ and $a_{i},b_{i}\in D$.
Furthermore, let $j,k\in{\mathbb{N}}$ such that $j+k=n$. Then for each $a\in
K$ such that $\psi(a)\neq 0$, we have

$v(\rho(a))=v(\varphi(a)\diamond\psi(a))=\min\\{v(\varphi(a))^{n},v(\psi(a))^{n}\\}\leq
v(\varphi(a)^{j}\psi(a)^{k})$

for all valuations $v=v_{\lambda}$ or $w_{\mu}$. This implies that $\rho$
divides $\varphi^{j}\psi^{k}$ in $\operatorname{Int{}^{\text{R}}}(K,D)$ and
therefore also in $\operatorname{Int{}^{\text{R}}}(E,D)$. Thus,
$(\varphi,\psi)^{n}\subseteq(\rho)$. We then get that
$(\varphi,\psi)^{n}=(\rho)$.

Since a power of an ideal of $\operatorname{Int{}^{\text{R}}}(E,D)$ generated
by two elements is principal, we also know that this ideal is invertible.
Thus, $\operatorname{Int{}^{\text{R}}}(E,D)$ is a Prüfer domain.

Now let $(\varphi_{1},\dots,\varphi_{m})$ be a finitely-generated, and thus
invertible, ideal of $\operatorname{Int{}^{\text{R}}}(E,D)$. We can ensure
that $\varphi_{1},\dots,\varphi_{m}$ are all nonzero. Then as before, for each
$a\in K$ except for the finitely many values such that $a$ is a pole for some
$\varphi_{i}$, we see that

$v((((\varphi_{1}(a)\diamond\varphi_{2}(a))\diamond\varphi_{3}(a)^{n})\diamond\varphi_{4}(a)^{n^{2}})\diamond\cdots\diamond\varphi_{m}(a)^{n^{m-2}})=\min_{i}\\{n^{m-1}v(\varphi_{i}(a))\\}$

for all valuations $v=v_{\lambda}$ or $v=w_{\mu}$. Using the same arguments as
before, it follows that $(\varphi_{1},\dots,\varphi_{m})^{n^{m-1}}$ is
principal. Thus, the Picard group of $\operatorname{Int{}^{\text{R}}}(E,D)$ is
torsion. ∎

######  Corollary 3.7.

Let $D$ be a Prüfer domain with $K$ as the field of fractions. Suppose that
$D$ can be written as $D=D_{1}\cap D_{2}$, where $D_{1}$ is a monic Prüfer
overring of $D$ and $D_{2}$ is a singular Prüfer overring of $D$. Suppose
there exist $n\in{\mathbb{N}}$, a collection $\\{V_{\lambda}\\}$ of valuation
overrings of $D_{1}$, and a collection $\\{W_{\mu}\\}$ of valuation overrings
of $D_{2}$ such that

  * •

$D_{1}=\bigcap\limits_{\lambda}V_{\lambda}$ and
$D_{2}=\bigcap\limits_{\mu}W_{\mu}$,

  * •

the maximal ideal of $W_{\mu}$ is generated by some $\varpi_{\mu}\in W_{\mu}$,

  * •

there exists some $d\in D_{2}$ such that $0<w_{\mu}(d)<nw_{\mu}(\varpi_{\mu})$
for all $\mu$, where $w_{\mu}$ is the valuation corresponding to $W_{\mu}$,

  * •

$v_{\lambda}(d-1)>0$ for all $\lambda$, where $v_{\lambda}$ is the valuation
corresponding to $V_{\lambda}$, and

  * •

there exist a monic polynomial $f\in D[x]$ of degree $n$ such that
$f(D_{1})\subseteq D_{1}^{\times}$ and $f(0)\in D^{\times}$.

Then $\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer domain with torsion
Picard group for any subset $E\subseteq K$.

###### Proof.

We verify the conditions of Theorem 3.6.

We see that $v_{\lambda}(d)=0$ for all $\lambda$ from the fact that
$v_{\lambda}(d-1)>0$. Furthermore, $g(x)\coloneqq d(f(x)-f(0))+f(0)$ is unit-
valued over $D_{1}$. This is because for any $a\in D_{1}$ and any $\lambda$,
we have that $v_{\lambda}(g(a)-f(a))=v_{\lambda}((d-1)(f(a)-f(0)))>0$, so
$v_{\lambda}(g(a))=v_{\lambda}(f(a))=0$. Observe that the leading coefficient
of $g$ is $d$. Also, fix a $\mu$ and let $a\in K$ such that $w_{\mu}(a)<0$.
Then $w_{\mu}(g(a))=w_{\mu}(d(f(a)-f(0))+f(0))=w_{\mu}(da^{n})$ since
$w_{\mu}(f(a)-f(0))=w_{\mu}(a^{n})$.

Lastly, we set $h_{1}(x)\coloneqq f(x)-f(0)$ and $h_{2}(x)\coloneqq f(0)$. We
have $n=\deg h_{1}>\deg h_{2}$ and $h_{1}$ is monic. Next, we must check that
$dfh_{1}+gh_{2}$ is unit-valued for all $V_{\lambda}$ and all $W_{\mu}$. We
have

$\displaystyle(dfh_{1}+gh_{2})(x)$
$\displaystyle=df(x)(f(x)-f(0))+(d(f(x)-f(0))+f(0))f(0)$
$\displaystyle=df(x)^{2}-df(x)f(0)+df(x)f(0)-df(0)^{2}+f(0)^{2}$
$\displaystyle=df(x)^{2}-df(0)^{2}+f(0)^{2}.$

For any $a\in D$, we have $w_{\mu}((dfh_{1}+gh_{2})(a))=w_{\mu}(f(0)^{2})=0$
for all $\mu$ and

$v_{\lambda}((dfh_{1}+gh_{2})(a)-f(a)^{2})=v_{\lambda}((d-1)(f(a)^{2}-f(0)))>0$

so $v_{\lambda}((dfh_{1}+gh_{2})(a))=0$ for all $\lambda$. The domain $D$
satisfies all of the hypotheses of Theorem 3.6, so
$\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer domain with torsion Picard
group for any subset $E\subseteq K$. ∎

The intersection of finitely many valuation domains with the same field of
fractions is a Prüfer domain [Gil92, Theorem 22.8]. If these valuation domains
are pairwise independent, we many use the approximation theorem for
independent valuations [Gil92, Theorem 22.9] to obtain the following result.

######  Corollary 3.8.

Let $V_{1},\dots,V_{r}$ be pairwise independent valuation rings on the field
$K$. Suppose that each $V_{i}$ has a principal maximal ideal or a residue
field that is not algebraically closed and set $D=V_{1}\cap\cdots\cap V_{n}$.
Then $\operatorname{Int{}^{\text{R}}}(E,D)$ is a Prüfer domain with torsion
Picard group for every subset $E$ of $K$.

###### Proof.

We can order the indices so that $V_{1},\dots,V_{s}$ for some $s\leq r$ are
such that for each $i=1,\dots,s$, the residue field of $V_{i}$ is not
algebraically closed and the residue fields for $V_{s+1},\dots,V_{r}$ are
algebraically closed. For $i\leq s$, this means there is some monic
nonconstant unit valued polynomial $f_{i}$ for $V_{i}$. Taking appropriate
powers of each of the $f_{i}$, we can assume that all of the $f_{i}$ have the
same degree $n$. Now for each $i=1,\dots,s$, write

$f_{i}(x)=x^{n}+\sum_{j=0}^{n-1}a_{ij}x^{j},$

where each $a_{ij}\in V_{i}$. Then by the approximation theorem for pairwise
independent valuations, we have that for each $j=0,1,\dots,n-1$ an element
$A_{j}\in K$ such that

$v_{i}(A_{j}-a_{ij})>0\text{ for all $i=1,\dots,s$ and }v_{i}(A_{j})=0\text{
for all $i>s$},$

where $v_{i}$ denotes the valuation corresponding to $V_{i}$. Form

$F(x)\coloneqq x^{n}+\sum_{j=0}^{n-1}A_{j}x^{j}.$

We claim that $F\in D[x]$ is unit-valued for $D_{1}\coloneqq
V_{1}\cap\cdots\cap V_{s}$. Let $a\in D_{1}$ and let $i\in\\{1,\dots,s\\}$.
Then

$v_{i}(F(a)-f_{i}(a))=v_{i}\mathopen{}\mathclose{{}\left(\sum_{j=0}^{n-1}(A_{j}-a_{ij})a^{j}}\right)>0$

and $v_{i}(f_{i}(a))=0$ imply that $v_{i}(F(a))=0$. Therefore, $F(a)\in
D_{1}^{\times}$. Furthermore, $F(0)=A_{0}$ and $v_{i}(A_{0})=0$ for all
$i=1,\dots,r$, so $F(0)\in D^{\times}$.

Now let $D_{2}=V_{s+1}\cap\cdots\cap V_{r}$. For $i=s+1,s+2,\dots,r$, the
residue field of $V_{i}$ is algebraically closed so the maximal ideal of
$V_{i}$ must be principal by hypothesis. Say the maximal ideal of $V_{i}$ is
generated by some $\varpi_{i}\in V_{i}$. Then by the approximation theorem
again, we have some $d\in K$ such that

$v_{i}(d-1)>0\text{ for $i\leq s$ and }v_{i}(d)=v_{i}(\varpi_{i})\text{ for
$i>s$}.$

Then we verify that $d\in D_{2}$ and
$0<v_{i}(d)=v_{i}(\varpi_{i})<nv_{i}(\varpi)$ for all $i>s$. This means $D$
satisfies all of the hypotheses of the previous corollary and thus
$\operatorname{Int{}^{\text{R}}}(E,D)$ is a Prüfer domain with torsion Picard
group for any subset $E\subseteq K$. ∎

Here is an example of a Prüfer domain $D$ with a finite number of maximal
ideals that is neither singular nor monic. The previous corollary allows us to
determine that $\operatorname{Int{}^{\text{R}}}(E,D)$ is a Prüfer domain for
any subset $E\subseteq K$.

######  Example 3.9.

Let $K=k(s,t)$, where $k$ is any real closed field. Also, let
$\alpha\in{\mathbb{R}}$ with $\alpha>0$ be irrational. Define a valuation
$v_{1}:K\to{\mathbb{Z}}[\alpha]\cup\\{\infty\\}$ as follows. We first define
$v_{1}$ on $k[s,t]\setminus\\{0\\}$ by

$v_{1}\mathopen{}\mathclose{{}\left(\sum
a_{i_{1}i_{2}}s^{i_{1}}t^{i_{2}}}\right)=\min\\{i_{1}+i_{2}\alpha\mid
a_{i_{1}i_{2}}\neq 0\\},$

where $\sum a_{i_{1}i_{2}}s^{i_{1}}t^{i_{2}}\in k[s,t]\setminus\\{0\\}$. Then
this function uniquely extends to a valuation on $K$.

Now we define a valuation
$v_{2}:K\to{\mathbb{Z}}\oplus{\mathbb{Z}}\cup\\{\infty\\}$, ordered
lexicographically. We first define another valuation
$w_{2}:k(s)\to{\mathbb{Z}}\cup\\{\infty\\}$ as the valuation corresponding to
the valuation ring $k[s]_{(s)}$ so that $w_{2}(s)=1$. Let $\Delta\coloneqq
k[s]_{(s)}/sk[s]_{(s)}$, the residue field of $k[s]_{(s)}$. Then define
$w_{2}^{\prime}:\Delta(t)\to{\mathbb{Z}}\cup\\{\infty\\}$ as the valuation
corresponding to $\Delta[t]_{(t^{2}+1)}$ so that $w_{2}^{\prime}(t^{2}+1)=1$.
Now we can define $v_{2}$ on $k[s]_{(s)}[t]\setminus\\{0\\}$. Let $f(t)\in
k[s]_{(s)}[t]$. We write $f(t)=dg(t)$ for some $d\in k[s]_{(s)}$ and $g(t)\in
k[s]_{(s)}[t]\setminus sk[s]_{(s)}[t]$. Now, we have

$v_{2}(f)=(w_{2}(d),w_{2}^{\prime}(g\mod sk[s]_{(s)})),$

which extends uniquely to a valuation on $K$.

Let $V_{1}$ and $V_{2}$ be the valuation rings corresponding to $v_{1}$ and
$v_{2}$, respectively. Then $V_{1}$ has residue field that isomorphic to $k$,
which is not algebraically closed, and the maximal ideal is not principal. As
for $V_{2}$, the ring $V_{2}$ has residue field isomorphic to
$k[t]/(t^{2}+1)$, which is algebraically closed, and the maximal ideal is
principal. Therefore, $D=V_{1}\cap V_{2}$ is a Prüfer domain that is neither
monic nor singular, but $V_{1}$ and $V_{2}$ are independent valuations on $K$,
so $\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer for any subset $E$ of $K$
by the previous corollary.

The next example is a Prüfer domain $D$ that is not the intersection of
finitely many valuation domains, but we can use Corollary 3.7 to determine
that $\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer. In the next two
examples, we make use of the fact that the intersection of a countable number
of valuation domains with the same field of fractions all containing a common
uncountable field is a Bézout domain and thus a Prüfer domain [OR06, Theorem
6.6].

######  Example 3.10.

Let $K=k(t_{1},t_{2},\dots)$, where $k$ is an uncountable algebraically closed
field. Also let $\\{\alpha_{1},\alpha_{2},\dots\\}\subseteq{\mathbb{R}}_{>0}$
be a ${\mathbb{Q}}$-linearly independent subset of ${\mathbb{R}}$. For
$i\in{\mathbb{N}}\setminus\\{0\\}$, we define

$\displaystyle v_{i}\mathopen{}\mathclose{{}\left(\sum
a_{e_{1}e_{2}\cdots}t_{1}^{e_{1}}t_{2}^{e_{2}}\cdots}\right)$
$\displaystyle=\min\mathopen{}\mathclose{{}\left\\{\sum_{j=2}^{\infty}e_{j}\alpha_{i+j}\,\middle|\,a_{e_{1}e_{2}\cdots}\neq
0}\right\\},$ $\displaystyle w_{i}\mathopen{}\mathclose{{}\left(\sum
a_{e_{1}e_{2}\cdots}a_{e_{1}e_{2}\cdots}t_{1}^{e_{1}}(t_{2}+1)^{e_{2}}t_{3}^{e_{3}}t_{4}^{e_{4}}\cdots}\right)$
$\displaystyle=\min\mathopen{}\mathclose{{}\left\\{\mathopen{}\mathclose{{}\left(\sum_{j\neq
2}e_{j}\alpha_{i+j},e_{2}}\right)\,\middle|\,a_{e_{1}e_{2}\cdots}\neq
0}\right\\},$

where both sums on the left range over
$(e_{1},e_{2},\dots)\in\bigoplus\limits_{\ell=1}^{\infty}{\mathbb{N}}$ and
each $a_{e_{1}e_{2}\cdots}\in k$ with all but finitely many
$a_{e_{1}e_{2}\cdots}$ are $0$. The value group of each $w_{i}$ is
$\mathopen{}\mathclose{{}\left(\sum\limits_{j=2}{\mathbb{Z}}\alpha_{i+j}}\right)\oplus{\mathbb{Z}}$
endowed with left-to-right lexicographic ordering. These functions extend
uniquely to valuations on $K$, since $K$ is the field of fractions of
$k[t_{1},t_{2},\dots]=k[t_{1},t_{2}+1,t_{3},t_{4},\dots]$. Let $V_{i}$ and
$W_{i}$ be the valuation domains corresponding to $v_{i}$ and $w_{i}$,
respectively.

For the valuation
$v_{i}:K\to\sum\limits_{j=2}^{\infty}{\mathbb{Z}}\alpha_{i+j}\cup\\{\infty\\}$,
$V_{i}$ has residue field isomorphic to $k(t)$, which is not algebraically
closed, and the maximal ideal of $V_{i}$ is not principal.

As for $w_{i}:K\to\mathopen{}\mathclose{{}\left(\sum\limits_{j\neq
2}{\mathbb{Z}}\alpha_{i+j}}\right)\oplus{\mathbb{Z}}\cup\\{\infty\\}$, we have
$W_{i}$ having residue field isomorphic to $k$, which is algebraically closed
and the maximal ideal of $W_{i}$ is principal, generated by $t_{2}+1$.

Set $D_{1}\coloneqq\bigcap\limits_{i=1}^{\infty}V_{i}$. We know that $D_{1}$
is a Prüfer domain since $D_{1}$ is the intersection of countably many
valuation domains with a common uncountable subfield $k$. We have that
$x^{2}-(t_{1}+1)$ is unit valued for $D_{1}$, so $D_{1}$ is monic.
Furthermore, $K$ is the field of fractions of $D_{1}$ since $k\subseteq D_{1}$
and $t_{1},t_{2},\dots\in D_{1}$.

Set $D_{2}\coloneqq\bigcap\limits_{i=1}^{\infty}W_{i}$. As with $D_{1}$, we
can verify that $D_{2}$ is a Prüfer domain with field of fractions $K$. We
also have $0<w_{i}(t_{2}+1)<2w_{i}(t_{2}+1)$, so $D_{2}$ is singular.

Now form $D\coloneqq D_{1}\cap D_{2}$. Since $k\subseteq D$ and
$t_{1},t_{2},\dots\in D$, we have that $K$ is the field of fractions of $D$ as
well. Note that $D$ is also the intersection of countably many valuation
domains with a common uncountable subfield $k$, so $D$ is Prüfer.

Now we verify the remaining conditions for $D$ for Corollary 3.7. Here, we’ll
use $n=2$. We have that $f(x)\coloneqq x^{2}-(t_{1}+1)\in D[x]$. Also,
$v_{i}(-(t_{1}+1))=0$ and $w_{i}(-(t_{1}+1))=0$ for all $i$, so $f(0)\in
D^{\times}$. Plus, $v_{i}((t_{2}+1)-1)=v_{i}(t_{2})=\alpha_{i+2}>0$ for all
$i$. Therefore, by the corollary, we have that
$\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer for any subset $E$ of $K$.

The full power of Theorem 3.6 is used to determine that
$\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer in the following example.

######  Example 3.11.

We let $K=k(t_{1},t_{2},\dots)$, where $k$ is an uncountable algebraically
closed field. We also take
$\\{\alpha_{1},\alpha_{2},\dots\\}\subseteq{\mathbb{R}}_{>0}$ to be a
${\mathbb{Q}}$-linearly independent subset of ${\mathbb{R}}$. For
$i\in{\mathbb{N}}\setminus\\{0\\}$, we define

$\displaystyle v_{i}\mathopen{}\mathclose{{}\left(\sum
a_{e_{1}e_{2}\cdots}t_{1}^{e_{1}}t_{2}^{e_{2}}\cdots}\right)$
$\displaystyle=\min\mathopen{}\mathclose{{}\left\\{\sum_{j=i+1}^{\infty}e_{j}\alpha_{i+j}\,\middle|\,a_{e_{1}e_{2}\cdots}\neq
0}\right\\},$ $\displaystyle w_{i}\mathopen{}\mathclose{{}\left(\sum
a_{e_{1}e_{2}\cdots}t_{1}^{e_{1}}t_{2}^{e_{2}}\cdots}\right)$
$\displaystyle=\min\mathopen{}\mathclose{{}\left\\{\mathopen{}\mathclose{{}\left(\sum_{j=2}^{\infty}e_{j}\alpha_{i+j},e_{1}}\right)\,\middle|\,a_{e_{1}e_{2}\cdots}\neq
0}\right\\},$

where both sums on the left range over
$(e_{1},e_{2},\dots)\in\bigoplus\limits_{\ell=1}^{\infty}{\mathbb{N}}$ and
each $a_{e_{1}e_{2}\cdots}\in k$ with all but finitely many
$a_{e_{1}e_{2}\cdots}$ are $0$. The value group of each $w_{i}$ is
$\mathopen{}\mathclose{{}\left(\sum\limits_{j=2}{\mathbb{Z}}\alpha_{i+j}}\right)\oplus{\mathbb{Z}}$
endowed with left-to-right lexicographic ordering. These functions extend
uniquely to valuations on $K$. Let $V_{i}$ and $W_{i}$ be the valuation rings
corresponding to $v_{i}$ and $w_{i}$, respectively.

For the valuation
$v_{i}:K\to\sum\limits_{j=i+1}^{\infty}{\mathbb{Z}}\alpha_{i+j}\cup\\{\infty\\}$,
$V_{i}$ has residue field isomorphic to $k(t_{1},\dots,t_{i})$, which is not
algebraically closed, and the maximal ideal of $V_{i}$ is not principal.

As for
$w_{i}:K\to\mathopen{}\mathclose{{}\left(\sum\limits_{j=2}{\mathbb{Z}}\alpha_{i+j}}\right)\oplus{\mathbb{Z}}\cup\\{\infty\\}$,
we have $W_{i}$ having residue field isomorphic to $k$, which is algebraically
closed and the maximal ideal of $W_{i}$ is principal, generated by $t_{1}$.

Set $D_{1}\coloneqq\bigcap\limits_{i=1}^{\infty}V_{i}$,
$D_{2}\coloneqq\bigcap\limits_{i=1}^{\infty}W_{i}$, and $D\coloneqq D_{1}\cap
D_{2}$. Since $D_{1},D_{2}$, and $D$ are all countable intersections of
valuation domains all containing $k$, an uncountable field, we know that
$D_{1},D_{2}$, and $D$ are all Prüfer domains. Furthermore,
$t_{1},t_{2},\dots,\in D$ and $k\subseteq D$, so $K$ is the field of fractions
$D_{1},D_{2},$ and $D$. We see that $x^{2}-t_{1}$ is unit-valued for $D_{1}$,
so $D_{1}$ is monic. We also have $0<w_{i}(t_{1})<2w_{i}(t_{1})$, so $D_{2}$
is singular.

There does not exist $d\in D_{2}$ and $n\in{\mathbb{N}}$ such that
$0<w_{i}(d)<nw_{i}(t_{1})$ for all $i$ and $v_{i}(d-1)>0$ for all $i$. This is
because there exists $i$ large enough such that $v_{i}(d-1)=0$ for any choice
of $d$. Therefore, the conditions in Corollary 3.7 are not satisfied.

Nevertheless, we can use Theorem 3.6. Set $n=2,d=t_{1},f(x)=x^{2}-t_{1}$,
$g(x)=t_{1}x^{2}-1$, $h_{1}(x)=x^{2}$ and $h_{2}(x)=1$. We know that
$v_{i}(t_{1})=0$ for all $i$. We verify that $x^{2}-t_{1}$ is unit-valued over
$D_{1}$ since $x^{2}-t_{1}$ has no roots over the residue field
$k(t_{1},t_{2},\dots,t_{i})$ of $V_{i}$ for any $i$. Similarly, $g(x)$ is
unit-valued over $D_{1}$ for the same reason. Also, $\deg g=2$ and for all
$a\in K$ and $i$ such that $w_{i}(g(a)))<0$, we have $w_{i}(t_{1}a^{2})$.

Lastly, we check that

$dfh_{1}+gh_{2}=t_{1}x^{4}+(t_{1}-t_{1}^{2})x^{2}-1$

is unit-valued over each $V_{i}$ and each $W_{i}$. Fix an $i$. Let
$\mathfrak{m}_{i}$ be the maximal ideal of $V_{i}$. Since
$V_{i}/\mathfrak{m}_{i}$ is isomorphic to $k(t_{1},\dots,t_{i})$, we let $v$
be the valuation corresponding to $k(t_{2},\dots,t_{i})[t_{1}]_{(t_{1})}$. If
$t_{1}x^{4}+(t_{1}-t_{1}^{2})x^{2}-1\mod\mathfrak{m}_{i}$ has a root $\xi$,
then $v(\xi)=-\frac{v(t_{1})}{4}$, which is impossible. Lastly, notice that
$dfh_{1}+gh_{2}$ is equivalent to $-1$ modulo the maximal ideal of any
$W_{i}$, which makes $dfh_{1}+gh_{2}$ unit-valued for each $W_{i}$. Thus,
$\operatorname{Int{}^{\text{R}}}(E,D)$ is Prüfer for any subset $E$ of $K$ by
Theorem 3.6.

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