# Non-integrally closed Kronecker function rings and integral domains with a
unique minimal overring

Lorenzo Guerrieri  Jagiellonian University, Instytut Matematyki, 30-348 Kraków
Email address<EMAIL_ADDRESS>K. Alan Loper  Department of
Mathematics, Ohio State University – Newark, Newark, OH 43055 Email address:
<EMAIL_ADDRESS>
###### Abstract

It is well-known that an integrally closed domain $D$ can be express as the
intersection of its valuation overrings but, if $D$ is not a Prüfer domain,
the most of valuation overrings of $D$ cannot be seen as localizations of $D$.
The Kronecker function ring of $D$ is a classical construction of a Prüfer
domain which is an overring of $D[t]$, and its localizations at prime ideals
are of the form $V(t)$ where $V$ runs through the valuation overrings of $D$.
This fact can be generalized to arbitrary integral domains by expressing them
as intersections of overrings which admit a unique minimal overring. In this
article we first continue the study of rings admitting a unique minimal
overring extending known results obtained in the 70’s and constructing
examples where the integral closure is very far from being a valuation domain.
Then we extend the definition of Kronecker function ring to the non-integrally
closed setting by studying intersections of Nagata rings of the form $A(t)$
for $A$ an integral domain admitting a unique minimal overring.

MSC: 13A15, 13A18, 13B02, 13B21, 13B30, 13F05.
Keywords: Kronecker function ring, Nagata ring, intersection of integral
domains, integral closure.

##  1 Introduction

Let $D$ be a local, integrally closed integral domain with maximal ideal
$\mathfrak{m}$ and quotient field $K$. Let $\alpha\in K$ be such that $\alpha$
and $1/\alpha$ are not in $D$. A theorem of Seidenberg [31, Theorem 7]
indicates that $\mathfrak{m}$ extends to a nonmaximal prime ideal in the ring
$D[\alpha]$ and gives details concerning the structure of the maximal ideals
of $D[\alpha]$ which contain $\mathfrak{m}$. Here, we only note that there are
infinitely many. Clearly, these same results hold for $D[1/\alpha]$. The fact
that $D$ is integrally closed insures that $D[\alpha]\cap D[1/\alpha]=D$. We
can think of the introduction of $\alpha$ and $1/\alpha$ as being like
striking a large crystal (the ring $D$) with a hammer and shattering it into
many smaller pieces.

Of course, the scenario described above does not quite always work. In
particular, it is impossible to choose the element $\alpha$ when $D$ is a
valuation domain. In accordance with the depiction above, a classical theorem
by Krull states that every integrally closed domain is the intersection of all
its valuation overrings. For this reason, valuation rings are the indivisible
atoms in a canonical decomposition of an integrally closed domain. This makes
it seem that valuation rings would be a powerful tool in studying the
structure of rings. This is often true, but there are several things that can
make it difficult.

First, note that it is hard to describe in a simple way all the valuation
overrings of a given ring. For instance, let $F$ be any field. The ring
$D=F[x,y]$ of polynomials in two variables has uncountably many valuation
overrings, but only few of them are easy to see when looking at the
localizations of $D$. Krull provided an answer to this difficulty. The
Kronecker function ring $Kr(D)$ of $D$ is an overring of the polynomial ring
$D[t]$ with the property that that there is an easy correspondence between the
valuation overrings of $D$ and the localizations of $Kr(D)$ at prime ideals
[25], [26], [12]. A second difficulty is harder to overcome. If one is
inclined to analyze a ring by the means of studying its valuation overrings,
it can be disconcerting to be faced with an integral domain that is not
integrally closed. Of course, one can turn to tools of an entirely different
nature. The aim of this paper is to generalize the notions of valuation ring
and Kronecker function ring to a non-integrally closed setting.

When semistar operations were first introduced one of the major selling points
was the possibility of using semistar operations to build a Kronecker function
ring for a domain which is not integrally closed [9]. However, in this case
the Kronecker function ring that is produced is simply a conventional
Kronecker function ring of an integrally closed overring of $D$. Our real
purpose is to produce a non-integrally closed Kronecker function ring which
mirrors the non-integrally closed character of $D$. The first step in such a
process is to generalize the notion of a valuation domain. We need two
properties to hold for such a generalization.

  * •

We need a generalized valuation ring to actually be a valuation ring if it is
integrally closed.

  * •

We need for a domain which is not integrally closed to be equal to the
intersection of the generalized valuation rings that contain it.

Suppose now $D$ to be not integrally closed. Let $\alpha$ be an element of the
quotient field $K$ which is not in $D$. Let $A_{\alpha}$ be an overring of $D$
which is maximal with respect to the property of not containing the element
$\alpha$. Such a domain exists by application of Zorn’s Lemma and it is
necessarily local. We call $A_{\alpha}$ a maximal excluding domain. It has
been observed that maximal excluding domains are exactly the integral domains
admitting a unique minimal overring in the sense of [13] (see also [16]). This
is our proposed notion for a generalization of valuation rings.

The structure of this paper is as follows. Section 2 is devoted to review the
already known facts concerning Kronecker function rings and maximal excluding
domains. We provide in this section all the needed references.

Section 3 is devoted to constructing examples of maximal excluding domains.
Almost all valuation rings are maximal excluding (specifically all those with
branched maximal ideal). One might think that this fact, combined with the
fact that we are specifically working with rings that are not integrally
closed, would mean that the integral closure of a maximal excluding domain is
a valuation ring, or at least a Prüfer domain. This is known to not be true
[15]. Here we show how to construct maximal excluding domains using
generalized power series rings. Our construction is interesting since the
maximal excluding domains constructed can be farther from being Prüfer than
examples that are already in the literature. We can indeed construct examples
that have infinitely many incomparable prime ideals of the same height. We
also construct maximal excluding domains of a more ordinary character using
pullbacks.

In section 4 we work in earnest on the theory of non-integrally closed
Kronecker function rings. In the classical setting we begin with an integrally
closed domain $D$. In this case, the key feature of $Kr(D)$ is that if we
localize at any prime ideal we get a ring of the form $V(t)$ where $V$ is a
valuation overring of $D$ and $V(t)$ is what is known as the Nagata ring of
$V$. Suppose that $D$ is not integrally closed and is expressed as an
intersection of maximal excluding domains. We can define a new generalized
version of Kronecker function ring to be the intersection of the Nagata rings
of the maximal excluding domains. We should not expect in this setting to
always recover all the maximal excluding domains back by localizing this
generalized Kronecker function ring. For example, if the integral closure of
$D$ is a Prüfer domain then perhaps we should expect any of these generalized
Kronecker function rings to be just equal to the Nagata ring $D(t)$. We
explore various settings where this happens. In particular we prove that the
only Kronecker function ring of $D$ is $D(t)$ if the integral closure of $D$
is Prüfer, semilocal and has sufficiently large residue fields. Along the way
we study general properties about whether the operation of Nagata rings
extension commutes with intersection of integral domains.

Finally, in Section 5 we explore various settings where: we start with a
domain $D$, which is not integrally closed, we express $D$ as an intersection
of maximal excluding domains, and we intersect the rings $A(t)$ where $A$ runs
through the maximal excluding rings in our collection. We first demonstrate
that a localization of this intersection at a maximal ideal has the form
$C(t)$ where $C$ is an overring of $D$ (providing also several example where
this ring $C$ is maximal excluding). Then, we demonstrate that the integral
closure of this intersection is a classical Kronecker function ring of the
integral closure of $D$. We analyze also a simpler construction of the form
$R=Kr(\overline{D})\cap A(t)$ where $\overline{D}$ is the integral closure of
a domain $D$ and $D=\overline{D}\cap A$ for $A$ a semilocal overring. We can
study the properties of this ring $R$ more in general without requiring $A$ to
be maximal excluding. To give examples and applications we involve the maximal
excluding rings coming from the constructions in Section 3. Furthermore, along
all the paper, we left several open questions for further research.

##  2 Preliminaries

We fix our notation for this article and we recall all the definitions and
results that we will need about Nagata rings, Kronecker function rings, and
integral domains maximal with respect to excluding a given element of their
quotient field.

All the rings we consider will be integral domains, having the same unit
element $1$. For an integral domain $D$, we denote its quotient field by
$\mathcal{Q}(D)$. If $D$ is local, we denote its unique maximal ideal by
$\mathfrak{m}_{D}$. An overring of $D$ is an integral domain $A$ such that
$D\subseteq A\subseteq\mathcal{Q}(D)$. If $A$ is an overring and $A$ and $D$
are both local, we say that $A$ dominates $D$ if
$\mathfrak{m}_{A}\supseteq\mathfrak{m}_{D}$. The set of all valuation
overrings dominating $D$ is standardly called Zar$(D)$ (the name comes from
Zariski’s definition). Given a valuation overring $V$ we denote by $v$ the
associated valuation and by $G_{V}$ its value group. The integral closure of
an integral domain $D$ in a field $F\supseteq\mathcal{Q}(D)$ is denoted by
$\overline{D}^{F}$. If $F=\mathcal{Q}(D)$, we simply use the notation
$\overline{D}$.

###  2.1 Nagata rings and Kronecker function rings

The main reference we consider for Nagata rings and Kronecker function rings
is Gilmer’s book [12]. For an historical introduction we refer to the paper
[7] and to its bibliography. These subjects have been widely studied in the
last 60 years. There are many other relevant references including [25], [26],
[27], [3], [18], [8], [9].

Given an integral domain $D$ and an indeterminate $t$ over the quotient field
of $D$, consider a polynomial $f(t)\in D[t]$. The content of $f$ is the ideal
$c(f)\subseteq D$ generated by the coefficients of $f$. The Nagata ring of $D$
is defined as

$D(t):=\left\\{\frac{f}{g}\,:\,f,g\in D[t],\,c(g)=D\right\\}.$

This ring is the localization of the ring $D[t]$ at the multiplicatively
closed set generated by the polynomials whose content is equal to the unit
ideal.

We will use several known facts about the Nagata ring. When $\mathfrak{m}$ is
a maximal ideal of $D$, the extension $\mathfrak{m}D(t)$ is a maximal ideal of
$D(t)$, and $D(t)_{\mathfrak{m}D(t)}=D_{\mathfrak{m}}(t)$ (cf. [22,
Proposition 2.1]). Therefore, since $D=\bigcap_{\mathfrak{m}\subseteq
D}D_{\mathfrak{m}}$, we get $D(t)=\bigcap_{\mathfrak{m}\subseteq
D}D_{\mathfrak{m}}(t).$ We can thus say that the operation of Nagata ring
extension commutes with the intersection of localizations at the maximal
ideals. By [14, Theorem 3], Nagata ring extension commutes also with integral
closure. We have that the integral closure of $D(t)$ in its quotient field is
$\overline{D}(t)$.

If $V$ is a valuation domain, the ring $V(t)$ is also a valuation domain,
called the trivial extension of $v$ to the field $\mathcal{Q}(V)(t)$. The
value of a polynomial $f=\sum_{k=0}^{n}a_{k}t^{k}\in\mathcal{Q}(V)[t]$ with
respect to this valuation is equal to $\mbox{\rm
min}_{k=0,\ldots,n}\\{v(a_{k})\\}$.

Let $D$ be an integrally closed domain. By a classical theorem of Krull,
$D=\bigcap_{V\in\mbox{ \footnotesize Zar}(D)}V$. The Kronecker function ring
of $D$ can be defined as the intersection

$Kr(D)=\bigcap_{V\in\mbox{\footnotesize Zar}(D)}V(t).$

In the literature, the definition of Kronecker function ring is commonly given
using the e.a.b star operations of the ring $D$, but for the purpose of this
paper, where we study intersections of Nagata rings, we give this as an
equivalent definition. It is well-known that the Kronecker function ring of
$D$ is always a Bezout domain (indipendently of the properties of $D$) and its
localizations at the prime ideals are all the trivial extensions of the
valuation overrings of $D$. Moreover, $Kr(D)\supseteq D(t)$ and they coincide
if and only if $D$ is a Prüfer domain. Given a subset
$\mathcal{F}\subseteq\mbox{\rm Zar }(D)$ such that
$D=\bigcap_{V\in\mathcal{F}}V$, one can define another ring
$Kr^{\mathcal{F}}(D)=\bigcap_{V\in\mathcal{F}}V(t).$ This last ring is an
overring of the Kronecker function ring $Kr(D)$. Also these overrings
$Kr^{\mathcal{F}}(D)$ are commonly called Kronecker function rings of $D$.

###  2.2 Integral domains maximal with respect to excluding an element of
their quotient field

In this paper we are interested in studying intersections of Nagata rings of
overrings of an integral domain. For this reason we need to understand the
properties of those integral domains which cannot be expressed as intersection
of proper overrings. These rings have been already considered in the
literature. Gilmer and Heinzer in [13] consider integral domains admitting a
unique minimal overring, in the sense that an integral domain $D$ has an
overring $A$ such that for any other overring $B$ of $D$ there are inclusions
$D\subseteq A\subseteq B$. Several properties of these domains have been
studied in [13], [16] and other papers, but generally these rings are still
quite mysterious and difficult to identify (in the case they are not
integrally closed). It is observed that an integral domain $D$ admits a unique
minimal overring in the sense of Gilmer and Heinzer if and only if it is
maximal with respect to the property of not containing some element
$\alpha\in\mathcal{Q}(D)$ (obviously if $D$ is maximal with respect to
excluding $\alpha$, then $D[\alpha]$ is the unique minimal overring of $D$).
Clearly these properties are also equivalent to the fact that $D$ cannot be
written as intersection of proper overrings. In particular any such a ring $D$
has to be local and its maximal ideal a $t$-ideal (see [12] for information
about the star operation $t$). It is also easy to observe that any integral
domain can be written as the intersection of its overrings that are maximal
with respect to excluding some element of the quotient field.

In [28], Papick consider a slightly weaker notion of unique minimal overring.
To avoid confusion with this notion we call an integral domain maximal with
respect to excluding an element of its quotient field, a maximal excluding
domain. For a survey about this topic and about the more general concept of
minimal ring extensions we refer to [29].

If not otherwise specified, the proofs of all the results that we mention in
the following paragraphs can be found in [13], [16]. The first important known
fact is that an integrally closed maximal excluding domain is necessarily a
valuation domain with branched maximal ideal (in particular any valuation
domain of finite dimension is maximal excluding). Another relevant class of
maximal excluding domains, not necessarily integrally closed, is the class of
local domains such that every ideal is divisorial, for instance local
Gorenstein noetherian domains of dimension one (cf. [20], [10], [4]).

If a domain $D$, maximal with respect to excluding an element $\alpha$, is not
integrally closed, then the unique minimal overring $D[\alpha]$ is an integral
extension of $D$ and is semilocal with at most two maximal ideals.
Furthermore, $\mathfrak{m}_{D}D[\alpha]\subseteq D$.

If $D[\alpha]$ has exactly two maximal ideals, then the integral closure of
$D$ is a Prüfer domain obtained as the intersection of two valuation rings. In
this case, if $D[\alpha]=\overline{D}$, the structure of $D$ has been
described in [21, Theorem 14] with the use of pullback diagrams. Also if the
maximal ideal $\mathfrak{m}_{D}$ of $D$ coincides with the maximal ideal of
$D[\alpha]$, then the integral closure of $D$ is a valuation overring $V$ of
$D$ such that $\mathfrak{m}_{V}=\mathfrak{m}_{D}$. In the case $D[\alpha]$ is
local and its maximal ideal is strictly larger than $\mathfrak{m}_{D}$, the
integral closure of $D$ may not be a Prüfer domain. The most known example of
this situation is the subject of the paper [15]. In this case $D$ is one-
dimensional and its integral closure is a PVD but not a valuation domain (a
PVD is a local domain sharing its maximal ideal with a valuation overring
[19]). A similar example appears here as Example 3.8.

Many questions about non-integrally closed maximal excluding domains are still
open. Already in [21], one can find examples of maximal excluding domains
whose prime ideals are not linearly ordered (but the integral closure is a
Prüfer domain). In Section 3 of this paper we show that a maximal excluding
domain of dimension at least 2 can have infinitely many prime ideals of the
same height and its integral closure may not even be a PVD (see Theorem 3.13,
Example 3.14).

The complete integral closure of a domain $D$ is the ring of the elements
$x\in\mathcal{Q}(D)$ for which there exists $d\in D$ such that $dx^{n}\in D$
for every $n\geq 1$. The complete integral closure of the example described by
Gilmer and Hoffmann in [15] is a valuation domain. The same happens for the
classes of maximal excluding domains that we consider here in Section 3. We
leave the following open question.

######  Question 2.1.

Let $D$ be a maximal excluding domain. Is the complete integral closure of $D$
always a Prüfer domain? If $D$ is one-dimensional and $\overline{D}$ is local,
is $\overline{D}$ a PVD?

##  3 Constructions of maximal excluding domains

In this section we characterize maximal excluding domains that can be obtained
as pullbacks and as generalized power series rings over a field, defined by
submonoids of the positive part of a totally ordered abelian group. In both
cases, as a consequence of Theorems 3.1 and 3.10, we get that the complete
integral closures of the rings in these families are valuation rings.

###  3.1 Maximal excluding domains obtained as pullbacks

Let $T$ be a local domain with maximal ideal $\mathfrak{m}$ and let $B$ be an
integral domain having quotient field $\kappa:=\frac{T}{\mathfrak{m}}.$ Let
$\phi:T\to\kappa$ be the canonical surjective map. Define the ring
$D:=\phi^{-1}(B)$ as in the pullback diagram:

${D}$${B}$${T}$${\kappa}$

For exhaustive information about the properties of rings of this form the
reader may consult [12], [11].

######  Theorem 3.1.

The integral domain $D$ is maximal excluding if and only if $B$ is maximal
excluding and $T$ is a valution domain.

###### Proof.

First suppose that $B$ is not maximal excluding. Thus, there are two proper
overrings $B_{1},B_{2}$ of $B$ such that $B=B_{1}\cap B_{2}$. It follows that
$D=\phi^{-1}(B_{1})\cap\phi^{-1}(B_{2})$ is an intersection of two proper
overrings, hence is not maximal excluding.

Thus, suppose that $B$ is maximal excluding with unique minimal overring
$B[\alpha^{\prime}]$. Let $\alpha$ be an element of $T$ such that
$\phi(\alpha)=\alpha^{\prime}.$ Clearly $\alpha$ is a unit in $T$.

Assume that $T$ is a valuation ring and pick $z\in\mathcal{Q}(D)\setminus D$.
In particular $z\not\in\mathfrak{m}$. We need to show that $\alpha\in D[z]$.
If $z\not\in T$ then, $\frac{1}{z}\in\mathfrak{m}\subseteq D$. It follows that
$\frac{\alpha}{z}\in\mathfrak{m}\subseteq D$ and $\alpha=z\frac{\alpha}{z}\in
D[z]$. Next assume $z$ to be a unit in $T$. Therefore $\phi(z)\in\kappa$.
Since $B$ is maximal excluding, we get $\alpha^{\prime}\in B[\phi(z)]$. Since
$\mathfrak{m}\subseteq D$, we get $\alpha\in\phi^{-1}(B[\phi(z)])=D[z].$

Finally consider the case where $T$ is not a valuation domain. Let $V$ be a
valuation overring of $T$ such that $\mathfrak{m}\subseteq\mathfrak{m}_{V}$.
By way of contradiction suppose that $D$ is maximal with respect to excluding
an element $\beta$. We must have $\beta\in D[\alpha]\subseteq T$. If there
exists $z\in\mathfrak{m}_{V}\setminus\mathfrak{m}$, then we find the
contradiction $\beta\not\in D[z]$. Indeed, if $\beta\in D[z]$, then there
exists $d_{0}\in D$ such that $\beta-
d_{0}=z(d_{1}+zd_{2}+\ldots+z^{n-1}d_{n})\in\mathfrak{m}_{V}\cap
T=\mathfrak{m}\subseteq D$. This is a contradiction since $\beta\not\in D$. If
instead $\mathfrak{m}_{V}=\mathfrak{m}$, let $\phi^{\prime}$ be the canonical
quotient map $V\to V/\mathfrak{m}_{V}$. Observe that $\phi^{\prime}|_{T}=\phi$
and the image of $\phi^{\prime}$ is a proper field extension of $\kappa$. We
show that there exists an element $z\in V\setminus T$ such that
$\kappa(\phi^{\prime}(z))\supsetneq\kappa$ and $\beta\not\in D[z]$. Setting
$\theta:=\phi^{\prime}(z)$ for some $z\in V\setminus T$, this is equivalent to
showing that $\alpha^{\prime}\not\in B[\theta]$. If $\theta$ is transcendental
over $\kappa$, this is obvious. If it is algebraic, let us assume it to be of
degree $n$. Let $f$ be the minimal polynomial of $\theta$ over $F$. Using that
$\kappa$ is the quotient field of $B$, we can find $b\in B$ such that the
constant term of the minimal polynomial of $b\theta$, which is equal to
$b^{n}f(0)$, is in $B$. Replace $\theta$ by $b\theta$ to have $f(0)\in B$.
Assume there exists a relation
$\alpha^{\prime}=b_{0}+b_{1}\theta+b_{2}\theta^{2}+\ldots+b_{t}\theta^{t}$.
For $k\geq n,$ using the relation given by $f(\theta)=0$, the term
$\theta^{k}$ can be replaced by a linear combination of
$\theta,\ldots,\theta^{n-1}$ with coefficient in $\kappa$ and constant term in
$B$. This yields another equation of algebraic dependence of $\theta$ over
$\kappa$ of degree at most $n-1$ but with constant term not in $B$. This is a
contradiction. ∎

######  Corollary 3.2.

With the notation of the previous theorem assume that $B$ is a field and
$\kappa$ a field extension. Then $D$ is maximal excluding if and only if $T$
is a valuation domain and the extension $\kappa/B$ is a minimal field
extension.

###### Proof.

The condition that $\kappa/B$ is a minimal field extension is clearly
necessary. The other conditions can be proved exactly as in Theorem 3.1
choosing $\alpha^{\prime}$ to be any element of $\kappa\setminus B$. ∎

###  3.2 Maximal excluding domains in generalized power series rings

In the following $G$ will denote a totally ordered abelian group (with
additive notation and order relation $\leq$), $\boldsymbol{0}$ will denote its
unit element and $G_{\geq 0}$ the subsemigroup of elements of $G$ larger than
or equal to $\boldsymbol{0}$. We consider a family of generalized power series
rings, defined according to [5]. The literature about generalized power series
rings is very extensive, an interested reader can consult also [30], [23],
[24] and several other papers by P. Ribemboim and other authors.

Let $S$ be a subsemigroup of $G_{\geq 0}$ containing $\boldsymbol{0}$. A
subset $A$ of $S$ is Artinian if does not contain any infinite descending
sequence of elements with respect to $\leq$ (equivalently, if $A$ admits a
minimal element). Let $K$ be any field. The generalized power series ring
$[[K^{S}]]$ is defined formally as the set of all the maps $S\to K$ such that
the set $\mbox{supp}(f):=\\{s\in S\,|\,f(s)\neq 0\\}$ is Artinian. The
operations on this ring are pointwise addition and the convolution product
defined as in [5, Section 2]. The ring $[[K^{S}]]$ is a commutative integral
domain with unit element equal to the map $e$ such that $e(\boldsymbol{0})=1$
and $e(s)=0$ for $s\neq\boldsymbol{0}$. Notice that in [5] and in the
subsequent papers, the authors consider a more general case of this
construction where the group $G$ is not necessarily totally ordered and $K$ is
replaced by an arbitrary commutative ring $R$. The classical semigroup ring
$K[X^{s},s\in S]$ can be embedded canonically in $[[K^{S}]]$ sending the
homogeneous element $X^{s}$ to the map $f$ such that $f(s)=1$ and $f(g)=0$ for
every $g\neq s$. For simplicity of notation we also denote the image of this
element in $[[K^{S}]]$ by $X^{s}$ and we say that these elements are monomial
elements. A general element of $[[K^{S}]]$ can be now expressed as a possibly
infinite sum $f=\sum_{s\in S}u_{s}X^{s}$ with $u_{s}\in K$ and such that
$\mbox{supp}(f)$ is Artinian. The units of $[[K^{S}]]$ can be described using
[5, Proposition 5] (which is based on a result proved in [17]). It turns out
that $f$ is a unit if and only if $f(\boldsymbol{0})$ is a nonzero element of
$K$. This fact makes $[[K^{S}]]$ a local domain. It is easy to observe that,
if $S=G_{\geq 0}$, the ring $[[K^{G_{\geq 0}}]]$ is a valuation domain with
value group $G$ and the value of an element $f\in[[K^{G_{\geq 0}}]]$ is
exactly the minimal element of $\mbox{supp}(f)$. Let us set $V:=[[K^{G_{\geq
0}}]]$.

We want to describe the maximal excluding domains of the form $[[K^{S}]]$ with
$S\subseteq G_{\geq 0}$. In the following we always assume that $[[K^{S}]]$
and $V$ have the same quotient field. The first result we prove is inspired by
the case $G=\mathbb{Z}$ which has been considered in [10]. If $G=\mathbb{Z}$,
$V$ is isomorphic to a standard power series ring in one variable over a the
field $K$. The set $S$ is a numerical semigroup and the subring $[[K^{S}]]$ is
maximal excluding if and only if $S$ is symmetric or pseudo-symmetric.

######  Theorem 3.3.

Let $S$ be a proper submonoid of $G_{\geq 0}$ and let $D:=[[K^{S}]]\subsetneq
V$. Assume that $D$ and $V$ have the same quotient field. Take $a\in G_{\geq
0}\setminus S$. The following conditions are equivalent:

  1. (1)

$D$ is maximal with respect to excluding the element $X^{a}$.

  2. (2)

For every $g\in G$, $g\neq\frac{a}{2}$, $X^{g}\in D$ if and only if
$X^{a-g}\not\in D$.

###### Proof.

Let $v$ be the valuation associated to $V$ and call $\mathfrak{m}_{V}$ the
maximal ideal of $V$. Assume $D$ to be maximal with respect to excluding the
element $y=X^{a}$. Let us first show that $y\mathfrak{m}_{V}\subseteq D$. If
there exists $z\in y\mathfrak{m}_{V}\setminus D$, then by definition of
maximal excluding domain $y=d_{0}+d_{1}z+\ldots+d_{n}z^{n}$ with
$d_{0},\ldots,d_{n}\in D$. Hence $v(y-d_{0})\geq v(z)>v(y)$, implying that
$d_{0}$ is an element of $D$ of value $v(y)=a$. This is a contradiction. It
follows that if $g>a$, then $X^{g}\in D$ and if $g<\boldsymbol{0}$, then
$X^{a-g}\in D$.

Consider now $g\in G$ such that $\boldsymbol{0}<g<a$ and $2g\neq a$. If both
$X^{g},X^{a-g}\in D$, then $y=X^{g}X^{a-g}\in D$, contradicting the
assumption. Suppose both $X^{g},X^{a-g}\not\in D$. In this case we show that
one among $D[X^{g}]$ and $D[X^{a-g}]$ is an overring of $D$ not containing
$y$. Suppose $y\in D[X^{g}]$. Hence

$X^{a}=d_{0}+d_{1}X^{g}+d_{2}X^{2g}+\ldots+d_{n}X^{ng},$

with $d_{0},\ldots,d_{n}\in D$. Using that $X^{g}\in D$ for every $g>a$, if
$2g>a$ the equation reduces to $X^{a}=d_{0}+d_{1}X^{g}$. But this yields a
contradiction since no term of $d_{1}$ can contain $X^{a-g}$ and no term of
$d_{0}$ can contain $X^{a}$. If instead $2g<a$, we obtain that $y\not\in
D[X^{a-g}]$. For this write

$X^{a}=c_{0}+c_{1}X^{a-g}+c_{2}X^{2a-2g}+\ldots+c_{n}X^{na-ng},$

with $c_{0},\ldots,c_{n}\in D$. Now observe that $2a-2g>a$ and conclude in the
same way as before.

Assume now condition $(2)$ and prove that $D$ is maximal with respect to
excluding $y$. Using that $D$ is a subring of $V$, we know that $D$ cannot
contain $X^{g}$ with $g<0$. Hence $X^{g}\in D$ for every $g>a$. In particular
$D$ contains all the elements of $V$ of value larger than $a$, thus
$y\mathfrak{m}_{V}\subseteq D$.

Let $z$ be an element of the quotient field of $D$. We want to show that $y\in
D[z]$. In the case $z\not\in V$, we get $\frac{y}{z}\in
y\mathfrak{m}_{V}\subseteq D$. Thus $y=z\frac{y}{z}\in D[z]$. Otherwise, pick
$z\in V\setminus D$. By subtracting elements of $D$, we reduce to the case
where $v(z)=\mbox{min supp}(z)\in G_{\geq 0}\setminus S.$ Set $b:=v(z)$. Then
$z=uX^{b}+h$ with $u\in K$ and $v(h)>b$. By hypothesis $X^{a-b}\in D$, hence
$X^{a-b}z=uy+h^{\prime}\in D[z]$. But $v(h^{\prime})>(a-b)+b=a$, implying
$h^{\prime}\in D$. It follows that $y\in D[z]$. ∎

######  Example 3.4.

Let $G$ be a submonoid of $(\mathbb{R},+)$. Let $a$ be a positive real number.
Set

$S=\\{\boldsymbol{0}\\}\cup\left\\{g\in
G\,|\,\frac{a}{2}<g<a\right\\}\cup\\{g\in G\,|\,g>a\\}.$

Clearly $S$ is a monoid contained in $G_{\geq 0}$ and by Theorem 3.3, the ring
$D:=[[K^{S}]]$ is maximal excluding with unique minimal overring $D[X^{a}]$.
It is easy to check that the integral closure of $D$ is $V:=[[K^{G_{\geq
0}}]]$.

Before presenting more examples we describe how to compute the integral
closure of certain domains of the form $[[K^{S}]]$. This result is similar to
the corresponding case for semigroup rings, where the integral closure
coincides with the root closure, see [2], [1]. For this, we need to introduce
some new notation and from now on we assume $\dim(V)=n<\infty$. Write the
group $G\cong G_{1}\oplus\ldots\oplus G_{n}$. For $i=1,\ldots,n$ define the
semigroup

$\widehat{G_{i}}=\\{(g_{1},\ldots,g_{n})\in G\,|\,g_{k}=0\mbox{ for }k<i\mbox{
and }g_{i}>0\\}\subseteq G_{\geq 0}.$

Set also $\widehat{G_{0}}=\\{\boldsymbol{0}\\}$. Observe that $G_{\geq
0}=\bigcup_{i=0}^{n}\widehat{G_{i}}$ and
$\widehat{G_{i}}+\widehat{G_{j}}\subseteq\widehat{G}_{\footnotesize\mbox{min}(i,j)}$
for any $i,j$. Denote by $\mathfrak{q}_{i}$ the prime ideal of $V$ of height
$i$. Then $g\in\widehat{G_{i}}$ if and only if
$X^{g}\in\mathfrak{q}_{i}\setminus\mathfrak{q}_{i-1}.$ Given any submonoid $S$
of $G_{\geq 0}$, we define $S_{i}:=S\cap\widehat{G_{i}}$. Clearly
$S=\bigcup_{i=0}^{n}S_{i}$. Define also

$\overline{S}:=\bigcup_{n\in\mathbb{N}}\\{g\in G\,|\,ng\in S\\}.$

Also $\overline{S}$ is a monoid contained in $G_{\geq 0}$. It is
straightforward to see that for every $i=1,\ldots,n$,
$\overline{S_{i}}=\overline{S}\cap\widehat{G_{i}}$.

######  Proposition 3.5.

Let $S$ be a proper submonoid of $G_{\geq 0}$ such that $G=\langle S\rangle$.
Let $D:=[[K^{S}]].$ Suppose that for every $i=1,\ldots,n$, there exists
$a_{i}\in S_{i}$ such that $\\{g\in\overline{S_{i}}\,|\,g\geq
a_{i}\\}\subseteq S_{i}.$ Suppose also that the ring $R=[[K^{\overline{S}}]]$
is integrally closed. Then $R$ is the integral closure of $D$.

###### Proof.

For $f\in R$, we show that $f$ is integral over $D$. We can write
$f=f_{1}+\ldots+f_{n}$ in such a way that the support of $f_{i}$ is in
$\overline{S_{i}}\cup\\{\boldsymbol{0}\\}$. It is sufficient to show that any
$f_{i}$ is integral over $D$. But for this, if $g_{i}$ is the minimum of the
support of $f_{i}$, by the Archimedean property of real numbers one can find a
positive integer $e_{i}$ such that $e_{i}g_{i}\geq a_{i}$. In this way by
assumption we get $f_{i}^{e_{i}}\in D$. ∎

We can immediately describe the integral closure in the case when $V$ is one-
dimensional.

######  Corollary 3.6.

Let $S$ be a submonoid of $G_{\geq 0}$ and let $D:=[[K^{S}]]\subseteq
V=[[K^{G_{\geq 0}}]]$. Suppose that $\dim(V)=1$, $D$ is maximal excluding and
$D$ and $V$ have the same quotient field. Then $V$ is the integral closure of
$D$.

###### Proof.

In this case the group $G$ is a submonoid of $(\mathbb{R},+)$. By Theorem 3.3,
there exists $a\in G_{\geq 0}$ such that $g\in S$, for every $g>a$. Given
$s\in G_{\geq 0}$, there exists some $n\in\mathbb{N}$ such that $ns>a$. Hence,
$\overline{S}=G_{\geq 0}$ and $V$ is the integral closure of $D$ by
Proposition 3.5. ∎

In the next two examples the valuation ring $V$ has dimension 2.

######  Example 3.7.

Let $G=\mathbb{Z}\times\mathbb{Z}$, ordered by lexicographic order, and let
$V=[[K^{G_{\geq 0}}]].$ The ring $V$ is a discrete valuation ring of rank 2.
Call $X$ the monomial element of $V$ of value $(0,1)$ and $Y$ the monomial
element of $V$ of value $(1,0)$. Consider the submonoid

$S=\left\\{(0,n)\,|\,n\geq 0\right\\}\cup\left\\{(1,n)\,|\,n\geq
1\right\\}\cup\left\\{(m,n)\,|\,m\geq 2,n\in\mathbb{Z}\right\\}\subseteq
G_{\geq 0}.$

By Theorem 3.3 the ring $D=[[K^{S}]]=K[[Y\mathfrak{m}_{V},X]]$ is maximal
excluding with unique minimal overring $D[Y]$. The integral closure is clearly
$V$ by Proposition 3.5, since $2(1,n)\in S$ for every $n\in\mathbb{Z}$. By an
analogous argument also $D[Y]$ is maximal excluding with unique minimal
overring $D[\frac{Y}{X}]$. Iterating, we observe that there exists an infinite
ascending chain

$D\subseteq D[Y]\subseteq D\left[\frac{Y}{X}\right]\subseteq
D\left[\frac{Y}{X^{2}}\right]\subseteq\ldots\subseteq\bigcup_{n=0}^{\infty}D\left[\frac{Y}{X^{n}}\right]=V,$

such that each ring is maximal excluding, the next one is the unique minimal
overring, and the integral closure of all these ring is $V$. In this example,
the ring $D$ has dimension 2 and two nonzero prime ideals. The maximal ideal
$\mathfrak{m}$ is generated by $X$ and $YX$, the height one prime is
$P=\mathfrak{p}\cap D$ where $\mathfrak{p}$ is the height one prime of $V$.
The quotient $D/P$ is a DVR with maximal ideal generated by the image of $X$.
The localization $D_{P}$ coincides with the localization $V_{\mathfrak{p}}$.
The quotient ring $D/(X)$ is an Artinian ring having dimension 2 as $K$-vector
space.

######  Example 3.8.

This example is a sort of dual of the previous one. Again let
$G=\mathbb{Z}\times\mathbb{Z}$, ordered by lexicographic order, let
$V=[[K^{G_{\geq 0}}]]$ and define $X,Y$ as in Example 3.7. Set

$S=\left\\{(0,0)\right\\}\cup\left\\{(1,n)\,|\,n\in\mathbb{Z}\setminus\\{0\\}\right\\}\cup\left\\{(m,n)\,|\,m\geq
2,n\in\mathbb{Z}\right\\}\subseteq G_{\geq 0}.$

It is easy to check that $S$ is a monoid. The ring
$D=[[K^{S}]]=K[[Y\mathfrak{m}_{V},\frac{Y}{X},\frac{Y}{X^{2}},\ldots]]$ is
maximal excluding by Theorem 3.3 and the unique minimal overring is $D[Y]$. We
notice that this ring $D[Y]$ is equal to $K+YK((X))[[Y]]$, thus it is an
integrally closed PVD but not a valuation domain ($X,X^{-1}\not\in D[Y]$). By
Proposition 3.5, we observe that $D[Y]$ is the integral closure of $D$.
Indeed, $Y$ is clearly integral over $D$ while all the powers of $X$ are not
integral since for every $m,n\in\mathbb{N}$, $m(0,n)=(0,mn)\not\in S$. Let $W$
be the rank one valuation overring of $V$. The maximal ideal of $W$ coincides
with the height one prime $\mathfrak{p}$ of $V$ and is also equal to the
maximal ideal $\mathfrak{m}$ of $D[Y]$. Hence, in this example $D$ is one-
dimensional and its complete integral closure $W$ is a valuation domain.

######  Remark 3.9.

Consider the same notation used above and assume $V$ countably many prime
ideals. In the next part of this section we can always reduce to assume that
if $D$ is maximal excluding with unique minimal overring $D[Y]$, then $Y$ is
an element of the height one prime of $V$. Indeed, given any non-maximal prime
ideal $\mathfrak{q}$ of $V$, it is well-known that $V$ is the pullback of the
valuation ring $V/\mathfrak{q}$ with respect to the quotient map from
$V_{\mathfrak{q}}$ to its residue field. The ring $V/\mathfrak{q}$ is maximal
excluding, hence if the ideal $YV$ has height $i>1$, we can choose
$\mathfrak{q}$ to be the prime ideal of height $i-1$ and apply Theorem 3.1,
reducing to the case where the height of $YV$ is $1$.

In all the above examples, the complete integral closure of $D$ is the rank
one valuation overring of $V$. We prove now that this happens in general for
all the maximal excluding rings in this family, provided that they share the
same quotient field with $V$ and $V$ has countably many prime ideals.

######  Theorem 3.10.

Let $S$ be a submonoid of $G_{\geq 0}$ and let $D:=[[K^{S}]]\subseteq
V=[[K^{G_{\geq 0}}]]$. Suppose that $D$ is maximal excluding, $D$ and $V$ have
the same quotient field, and $V$ has countably many prime ideals. Then the
rank one valuation overring $W$ of $V$ is the complete integral closure of
$D$.

###### Proof.

The ring $W$ is completely integrally closed. Hence, it is sufficient to show
that every element of $W$ is almost integral over $D$. If $\dim(V)=1$, then
$V=W$ and the result follows by Corollary 3.6. Assume $\dim(V)\geq 2$ and let
$\mathfrak{p}$ be the height one prime ideal of $V$, which coincides with the
maximal ideal of $W$. We can write $G\cong G_{W}\oplus G_{V/\mathfrak{p}}$
where $G_{W}$ is the rank one value group associated to $W$ and
$G_{V/\mathfrak{p}}$ is the value group of the valuation ring
$V/\mathfrak{p}$.

Say that the unique minimal overring of $D$ is $D[y]$ with
$y\in\mathfrak{m}_{V}$. Let $v(y)=(a_{1},a_{2})$ with $a_{1}\geq 0$. Choose
$f\in\mathfrak{p}$ such that $v(f)=(s_{1},s_{2})$ with $s_{1}>a_{1}$. In
particular $v(f)>v(y)$. Since by Theorem 3.3, $y\mathfrak{m}_{V}\subseteq D$
we get $f\in D$. Pick $h\in W$. If $h\in V$, then for every $n\geq 1$,
$v(h^{n}f)\geq v(f)>v(y)$. If $h\not\in V$, then $v(h)=(0,-g)$, for $g$ a
positive element of $G_{V/\mathfrak{p}}$. Thus
$v(h^{n}f)=(s_{1},-ng+s_{2})>(a_{1},a_{2})=v(y)$. In both cases this implies
$h^{n}f\in D$. ∎

We now analyze further the integral closure of $[[K^{S}]]$ to show that the
integral closure of a maximal excluding domain can be very far from being
Prüfer. We show that it does not even need to be a PVD and can have infinitely
many incomparable prime ideals. For this, we assume $1<\dim(V)=n<\infty$. In
the following, we let the semigroups $\widehat{G_{i}}$ and $S_{i}$ for
$i=1,\ldots,n$ be defined as above in this section. We show how given
arbitrary $S_{2},\ldots,S_{n}$ we can construct a monoid $S$ such that
$D=[[K^{S}]]$ is maximal excluding. By Remark 3.9 we can restrict to consider
only the case where the excluded element is in the height one prime of $V$.

######  Lemma 3.11.

Let $G\cong G_{1}\oplus\ldots\oplus G_{n}$ be a totally ordered abelian group
of rank $n>1$. For each $i=2,\ldots,n$, fix a (possibly empty) semigroup
$S_{i}\subseteq\widehat{G_{i}}$. Suppose that $S_{i}+S_{j}\subseteq
S_{\footnotesize\mbox{\rm min}(i,j)}$ for any $i,j=2,\ldots,n$. Also set
$S_{0}:=\widehat{G_{0}}=\\{\boldsymbol{0}\\}$. Then, it is possible to
construct $S_{1}\subseteq\widehat{G_{1}}$ such that
$S:=\bigcup_{i=0}^{n}S_{i}$ is a monoid and the ring $D=[[K^{S}]]$ is maximal
excluding. Moreover, $D$ has the same quotient field of $V=[[K^{G_{\geq
0}}]]$.

###### Proof.

Fix $a=(a_{1},\ldots,a_{n})\in\widehat{G_{1}}$ (in particular $a_{1}>0$).
Define $S_{1}=S_{1}^{\prime}\cup S_{1}^{\prime\prime}\cup S_{1}^{*}$ where

$S_{1}^{\prime}=\\{g\in\widehat{G_{1}}\,|\,g>a\\},\quad
S_{1}^{\prime\prime}=\\{g\in\widehat{G_{1}}\,|\,g<a,\,g_{1}=a_{1},\mbox{ and
}a-g\not\in S_{2}\cup\ldots\cup S_{n}\\},$

and $S_{1}^{*}\subseteq\\{g\in\widehat{G_{1}}\,|\,g_{1}<a_{1}\\}$ is empty if
$\\{g\in\widehat{G_{1}}\,|\,g_{1}<a_{1}\\}=\emptyset$ or otherwise is
constructed in such a way that:

  * •

$g\in S_{1}^{*}$ if and only if $a-g\not\in S_{1}^{*}$.

  * •

$S_{1}^{*}+S_{1}^{*}\subseteq S_{1}$.

  * •

$S_{j}+S_{1}^{*}\subseteq S_{1}^{*}$ for every $j=2,\ldots,n$.

To get some ideas on how one can concretely construct such $S_{1}^{*}$ see
Remark 3.12. We prove now that $S$ is a monoid.

We have to show that for every $j=1,\ldots,n$, the set $S_{j}+S_{1}\subseteq
S_{1}$. Clearly $S_{j}+S_{1}^{\prime}\subseteq S_{1}^{\prime}$ for every $j$.
Also
$(S_{1}^{\prime\prime}+S_{1}^{\prime\prime})\cup(S_{1}^{\prime\prime}+S_{1}^{*})\subseteq
S_{1}^{\prime}$. By the hypothesis on $S_{1}^{*}$, to conclude we only need to
check the inclusion $S_{j}+S_{1}^{\prime\prime}\subseteq S_{1}$ for $j\geq 2$.
Pick $g=(0,\ldots,0,g_{j},\ldots,g_{n})\in S_{j}$ and
$h=(a_{1},h_{2},\ldots,h_{n})\in S_{1}^{\prime\prime}$. If $g+h>a$, then
$g+h\in S_{1}^{\prime}\subseteq S_{1}$. We cannot have $g+h=a$ since this
would contradict the definition of $S_{1}^{\prime\prime}$. Hence, suppose
$g+h<a$ and let us show that $g+h\in S_{1}^{\prime\prime}$. If by way of
contradiction $a-(g+h)\in S_{2}\cup\ldots\cup S_{n}$, we would have

$a-h=g+(a-g-h)\in S_{j}+(S_{2}\cup\ldots\cup
S_{n})\subseteq(S_{2}\cup\ldots\cup S_{n}).$

Also this contradicts the definition of $S_{1}^{\prime\prime}$ and shows that
$S$ is a monoid.

Set $D:=[[K^{S}]]$. By construction $X^{a^{2}},X^{a^{3}}$ are in $D$, hence
$X^{-a}$ is in the quotient field of $D$. For every $g\in G_{\geq 0}$,
$X^{g}=X^{a+g}X^{-a}$. This implies that $D$ has the same quotient field as
$V$. Using Theorem 3.3, it is straightforward to check that $D$ is maximal
with respect to excluding the element $X^{a}$. ∎

######  Remark 3.12.

For the purpose of constructing examples, the conditions defining the set
$S_{1}^{*}$ in the above proof are not very explicit. A good way to satisfy
the third condition is assuming that, if $g_{1}>0$, for every
$g_{2},\ldots,g_{n}$, the element $g=(g_{1},g_{2},\ldots,g_{n})\in S_{1}^{*}$
if and only if $(g_{1},0,\ldots,0)\in S_{1}^{*}$. For the first two
conditions, one can choose a set such that the projection on the first
component behaves as the set described in Example 3.4. An easier assumption,
which can still produce many nice examples, is the following: set
$G_{1}=\mathbb{Z}$ and $a=(1,0,\ldots,0)$. In this way the set
$S_{1}^{*}=\emptyset$ and the proof of Lemma 3.11 can be simplified. However,
there are other possible choices to construct a set $S_{1}^{*}$ satisfying the
required conditions.

We now prove the main theorem concerning the rings which occur as integral
closure of a maximal excluding domain of the form $D=[[K^{S}]]$.

The notation for the next theorem is slightly different from the one used
until now. Let $n>1$ and let $G^{\prime}\cong G_{2}\oplus\ldots\oplus G_{n}$
be a totally ordered abelian group. Set $V^{\prime}:=[[K^{G^{\prime}_{\geq
0}}]]$. Let $H$ be any submonoid of $G^{\prime}_{\geq 0}$ such that
$\overline{H}=H$ and suppose that $A=[[K^{H}]]\subseteq V^{\prime}$ is
integrally closed. We do not require $A$ and $V^{\prime}$ to have the same
quotient field. Given another totally ordered abelian group $G_{1}$ of rank
one, set $G:=G_{1}\oplus G^{\prime}.$ Define $V:=[[K^{G_{\geq 0}}]]$. Let $W$
be the rank one valuation overring of $V$ and $\kappa$ be the residue field of
$W$.

######  Theorem 3.13.

With the notation and the assumptions stated above, there exists a submonoid
$S$ of $G_{\geq 0}$ such that:

  1. (i)

The ring $D=[[K^{S}]]\subseteq V$ is maximal excluding and has the same
quotient field as $V$.

  2. (ii)

The integral closure $\overline{D}$ of $D$ occurs as the pullback in the
following diagram:

${\overline{D}}$${A}$${W}$${\kappa}$

Conversely, if $D=[[K^{S^{\prime}}]]\subseteq V$ is a maximal excluding domain
having the same quotient field as $V$, then the integral closure of $D$ is the
pullback of some integrally closed local domain $A^{\prime}$ with respect to
the quotient map $W\to\kappa$.

###### Proof.

First notice that, by quotienting $G$ with respect to $G_{1}$, the quotient
field of $V^{\prime}$ can be identified with $\kappa$ and $V$ is the pullback
of $V^{\prime}$ with respect to the map $W\to\kappa$. Hence $A$ can be
embedded in $\kappa$.

For $i=0,\ldots,n$, define $\widehat{G_{i}}$ and $\widehat{G^{\prime}_{i}}$ as
above in this section. Decompose $H$ as the union
$H=H_{0}\cup\bigcup_{i=2}^{n}H_{i}$, where
$H_{i}=H\cap\widehat{G^{\prime}_{i}}$. Since for $i\geq 2$ there exists a
bijective map between $\widehat{G_{i}}$ and $\widehat{G^{\prime}_{i}}$, we can
identify $H_{i}$ with a subsemigroup of $\widehat{G_{i}}$.

To construct $S$, set $S_{0}=\\{\boldsymbol{0}\\}$ and for $i=2,\ldots,n$,
define $S_{i}=H_{i}$. Since $H$ is a monoid, the semigroups
$S_{2},\ldots,S_{n}$ defined in this way obviously satisfy the hypothesis of
Lemma 3.11. Use Lemma 3.11 to produce $S_{1}\subseteq\widehat{G_{1}}$ such
that $S:=\bigcup_{i=0}^{n}S_{i}$ is a monoid and the ring $D=[[K^{S}]]$
satisfies condition $(i)$.

By construction, the ring $D$ is maximal with respect to excluding an element
$a\in\widehat{G_{1}}$ and the set $\\{g\in\widehat{G_{1}}\,|\,g>a\\}\subseteq
S$. Given $g\in\widehat{G_{1}}$, since $G_{1}$ has rank one, there exists
$n\geq 1$ such that $ng>a$. Thus $\overline{S_{1}}=\widehat{G_{1}}$. Clearly,
$\overline{S}=\widehat{G_{1}}\cup H$ and $[[K^{\overline{S}}]]$ is integrally
closed by standard properties of pullback diagrams. By Proposition 3.5,
$\overline{D}=[[K^{\overline{S}}]]$. Let $\pi:W\to\kappa$ be the canonical
quotient map. We show that $\overline{D}=\pi^{-1}(A)$. Observe that, given
$g\in G$, $\pi(X^{g})=0$ if and only if $g_{1}>0$ and if $g_{1}=0$,
$\pi(X^{g})=X^{g^{\prime}}$ where $g^{\prime}$ is the component of $g$ in
$G^{\prime}$. It easily follows that $\pi(\overline{D})=A$. Since
$\mbox{ker}(\pi)\subseteq\overline{D}$, we obtain $\overline{D}=\pi^{-1}(A)$
and finally prove $(ii)$.

Conversely, if $D=[[K^{S^{\prime}}]]\subseteq V$ is a maximal excluding domain
having the same quotient field as $V$, by Remark 3.9, we can assume $D$
maximal with respect to excluding an element $X^{a}$ with
$a\in\widehat{G_{1}}$. Using Theorem 3.3 and the same argument as above we get
$\overline{S_{1}^{\prime}}=\widehat{G_{1}}$. The thesis now follows setting
$H^{\prime}:=\bigcup_{i=2}^{n}\overline{S_{i}}\cup\\{\boldsymbol{0}\\}$ and
letting $A^{\prime}$ be the integral closure of $[[K^{H^{\prime}}]]$. ∎

The next example shows that both $D$ and its integral closure can have
infinitely many incomparable prime ideals.

######  Example 3.14.

Using the notation of Theorem 3.13, let
$G^{\prime}=G_{2}=\mathbb{Z}[\sqrt{2}]$ and $G_{1}=\mathbb{Z}$. Choose

$H=\\{a+b\sqrt{2},\,|\,a,b\geq 0\\}\subseteq G^{\prime}_{\geq 0}.$

We notice that $H=\overline{H}$. Indeed, if for some $n\geq 1$ and
$a,b,c,d\in\mathbb{Z}$, $n(c+d\sqrt{2})=a+\sqrt{2}b\in H$, using the
irrationality of $\sqrt{2}$, we must have $nc=a$ and $nd=b$, forcing $c,d\geq
0$.

Define $X$ to be the monomial element of $V$ of value $(0,1)$ and $Z$ to be
the monomial element of $V$ of value $(0,\sqrt{2})$. The ring
$A=[[K^{H}]]\cong K[[X,Z]]$ is a regular local ring of dimension 2. The
quotient field of $A$ is properly contained in the quotient field of
$V^{\prime}=[[K^{G^{\prime}}]]$.

Now construct $S$ as in Theorem 3.13. Looking at the notation of Lemma 3.11
and at Remark 3.12, we notice that, if we choose $a=(1,0)$, we get
$S_{1}^{*}=\emptyset$. Therefore Lemma 3.11 gives a precise way to construct
$S$ that does not require further choices. Let $Y$ be the monomial element of
$V$ of value $(1,0)$. The ring $D=[[K^{S}]]$ is maximal with respect to
excluding $Y$. The integral closure $\overline{D}$ is the pullback of $A$ with
respect to the map $W\to\kappa$. Hence, $\overline{D}$ is local and has
infinitely many incomparable prime ideals (of height 2). Therefore it can be
neither a PVD nor Prüfer. In this example we can identify $W$ with
$K((X,Z))[[Y]]$. Then, if $\mathfrak{m}_{W}$ is the maximal ideal of $W$, we
have $\overline{D}\cong K[[X,Z]]+\mathfrak{m}_{V}$.

The prime ideals of height 2 of $\overline{D}$ are principal, generated by
elements that are also elements of $D$ as a consequence of the way $S$ is
defined in Theorem 3.13. For this reason, they all contract to distinct prime
ideals of $D$. It follows that also $D$ has infinitely many incomparable prime
ideals.

##  4 Intersections of Nagata extensions of overrings

In this section we consider intersections of Nagata rings in order to extend
the concept of Kronecker function rings to non-integrally closed domains. For
simplicity we suppose all the rings in this section to have finite Krull
dimension. In particular every integrally closed maximal excluding domain will
be a valuation domain.

Let $D$ be an integral domain. We say that a collection of overrings
$\mathcal{F}=\\{D_{i}\\}_{i\in\Lambda}$ is a defining family for $D$ if
$\bigcap_{D_{i}\in\mathcal{F}}D_{i}=D.$ From what observed in Section 2, every
integral domain always admits a defining family formed by maximal excluding
overrings. Moreover we can always consider defining families of $D$ where all
the non-integrally closed rings do not contain the integral closure of $D$.
Indeed $D$ can be always express as

$D=\overline{D}\cap\bigcap_{\alpha\in\mathcal{Q}(D)\setminus\overline{D}}A_{\alpha}$

where $A_{\alpha}$ is a non-integrally closed overring of $D$ maximal with
respect to excluding the element $\alpha$.

######  Definition 4.1.

Let $D$ be an integral domain and let $\mathcal{F}$ be a defining family of
$D$. We say that the ring

$Kr(D)^{\mathcal{F}}=\bigcap_{D_{i}\in\mathcal{F}}D_{i}(t)$

is the Kronecker function ring of $D$ with respect to $\mathcal{F}$.

When $D$ is integrally closed and the family $\mathcal{F}$ is contained in
Zar$(D)$, the ring $Kr(D)^{\mathcal{F}}$ is a classical Kronecker function
ring. We already recalled in Section 2, that an integrally closed domain is
Prüfer if and only if its Kronecker function ring $Kr(D)$ is equal to the
Nagata ring $D(t)$ (in this case the Nagata ring is the unique Kronecker
function ring). Also if $D$ is a maximal excluding domain, since the only
defining family of $D$ is $\\{D\\}$, we get that the only Kronecker function
ring, according to Definition 4.1 is $D(t)$. However, in general Prüfer
domains and maximal excluding domains may behave differently with respect to
intersecting Nagata rings of overrings. For a Prüfer domain $D$, the operation
of Nagata ring extension, mapping an overring $A$ to its Nagata ring $A(t)$,
commutes with arbitrary intersection of overrings. Conversely, one may observe
that if the integral closure of an arbitrary integral domain is not Prüfer,
then the operation of Nagata ring extension does not commute with intersection
for some collections of overrings. This is due to the fact that the Kronecker
function ring of an integrally closed domain that is not Prüfer is always
strictly bigger than the Nagata ring.

We dedicate this section to investigating whether an intersection of Nagata
rings is a Nagata ring. The next section is dedicated to constructing relevant
families of non-integrally closed Kronecker function rings that behave
similarly to the classical integrally closed ones. The following question
arises naturally:

######  Question 4.2.

Let $D$ be an integral domain and suppose that $\overline{D}$ is a Prüfer
domain. If $\mathcal{F}$ is an arbitrary collection of overrings of $D$, then
is $\bigcap_{A\in\mathcal{F}}A(t)=(\bigcap_{A\in\mathcal{F}}A)(t)$? In
particular, is $\overline{D}$ a Prüfer domain if and only if Nagata ring
extension commutes with intersection for every collections of overrings of
$D$?

A positive answer to the above question would imply that all the Kronecker
function rings of a domain $D$ such that $\overline{D}$ is Prüfer coincide
with $D(t)$. We are able to give a positive answer to this question in Theorem
4.6 in the case $\overline{D}$ is a semilocal Prüfer domain of finite
dimension and the residue fields of $D$ are sufficiently large. Our result
implies a positive answer in the case $D$ has finite dimension and the
integral closure of $D$ is a valuation ring or the intersection of two
valuation rings, without other assumptions on the residue fields. We start by
proving a lemma, which points out a condition for an intersection of local
domains to be local. Given two (non-necessarily local) integral domains
$A\subseteq B$, we say that $B$ dominates $A$ if every maximal ideal of $A$ is
contained in some maximal ideal of $B$ and every maximal ideal of $B$
contracts to some maximal ideal of $A$. If $A$ and $B$ are local this means
simply that $\mathfrak{m}_{A}\subseteq\mathfrak{m}_{B}$.

######  Lemma 4.3.

Let $D$ be an integral domain. Let $\\{D_{i}\\}_{i\in\Lambda}$ be a collection
of local overrings of $D$ such that $D=\bigcap_{i}D_{i}$. Suppose that there
exists a valuation overring $V$ that contains all the rings $D_{i}$ and
dominates all of them, except at most one. Then $D$ is local.

###### Proof.

Call $D_{1}$ the ring in the family $\mathcal{F}$ possibly not dominated by
$V$. Pick two non-units $x,y\in D$. Then
$x,y\in\bigcup_{i\in\Lambda}(\mathfrak{m}_{D_{i}}\cap
D)\subseteq(\mathfrak{m}_{D_{1}}\cap D)\cup(\mathfrak{m}_{V}\cap
D)=\mathfrak{m}_{D_{1}}\cap D.$ It follows that
$x+y\in\mathfrak{m}_{D_{1}}\cap D$ is not a unit in $D$. Notice also that if
$D_{1}$ is dominated by $V$, then also $D$ is dominated by $V$. ∎

The main assumption that we need to prove Theorem 4.6 is related to the
cardinalities of the residue fields of an integral domain $D$. Notice that $D$
contains $s$ units $u_{1},\ldots,u_{s}$ such that $u_{i}-u_{j}$ is still a
unit for every $i,j$ if and only if all the residue fields of $D$ have
cardinality at least $s+1$. For domains containing an infinite field or
containing $\mathbb{Z}(t)$, Theorem 4.6 can be proved without further
assumptions.

######  Lemma 4.4.

Let $D$ be a semilocal domain with $s$ maximal ideals. Suppose that all the
residue fields of $D$ have cardinality at least $s$. Fix a set of units
$U\subseteq D$, of cardinality at least $s-1$ and such that $u_{i}-u_{j}$ is a
unit for every distinct $u_{i},u_{j}\in U$. Let $x_{1},\ldots,x_{n}\in D$ be
elements such that $(x_{1},\ldots,x_{n})=D$. Then there exists a unit of $D$
of the form $c=\sum_{k=1}^{n}\delta_{k}x_{k}$ with $\delta_{k}\in
U\cup\\{0\\}$ for every $k$.

###### Proof.

We prove it by induction on $n$. In the case $n=1$, we clearly must have that
$x_{1}$ is a unit, and thus $\delta_{1}x_{1}$ is a unit for every
$\delta_{1}\in U$. Thus assume the thesis true for every set of at most $n-1$
elements. Therefore if after reordering the generators
$(x_{1},\ldots,x_{n-1})=D$, we conclude by applying the inductive hypothesis
and setting $\delta_{n}=0$.

Call $\mathfrak{m}_{1},\ldots,\mathfrak{m}_{s}$ the maximal ideals of $D$.
Possibly relabeling, suppose that $x_{1},\ldots,x_{n-1}\in\mathfrak{m}_{s}$,
and there exist $1\leq s^{\prime}<s$ such that
$x_{n}\in(\mathfrak{m}_{1}\cap\ldots\cap\mathfrak{m}_{s^{\prime}})\setminus(\mathfrak{m}_{s^{\prime}+1}\cup\ldots\cup\mathfrak{m}_{s})$.
The ideal generated by $x_{1},\ldots,x_{n-1}$ cannot be contained in any of
the maximal ideals $\mathfrak{m}_{1},\ldots,\mathfrak{m}_{s^{\prime}}$.
Consider the ring $A=S^{-1}D$ where
$S=D\setminus(\mathfrak{m}_{1}\cup\ldots\cup\mathfrak{m}_{s^{\prime}})$. The
ring $A$ is semilocal and contains $U$. Moreover $(x_{1},\ldots,x_{n-1})A=A$.
By inductive hypothesis there exists a unit of $A$ of the form
$c^{\prime}=\sum_{k=1}^{n-1}\delta_{k}x_{k}$ with $\delta_{k}\in
U\cup\\{0\\}$. Observing that $c^{\prime}\in D$, we obtain
$c^{\prime}\not\in\mathfrak{m}_{1}\cup\ldots\cup\mathfrak{m}_{s^{\prime}}$.

Take now two elements of the form $d_{1}=c^{\prime}+ux_{n}$,
$d_{2}=c^{\prime}+vx_{n}$ with $u,v\in U$, $u\neq v$. We claim that
$d_{1},d_{2}$ cannot be contained in a common maximal ideal. Indeed if
$d_{1},d_{2}\in\mathfrak{m}_{i}$, then we would have
$d_{1}-d_{2}=(u-v)x_{n}\in\mathfrak{m}_{i}$ and
$vd_{1}-ud_{2}=(v-u)c^{\prime}\in\mathfrak{m}_{i}$, implying that
$c^{\prime},x_{n}\in\mathfrak{m}_{i}$, since $u-v$ is a unit. This is a
contradiction with the choice of $c^{\prime}$. Moreover any element of the
form $c^{\prime}+ux_{n}$ cannot be in a maximal ideal containing either
$c^{\prime}$ or $x_{n}$. Using now that $U$ contains at least $s-1$ elements
and $D$ has only $s$ maximal ideals we can find an element of the form
$c^{\prime}+\delta_{n}x_{n}$ which is a unit in $D$. ∎

######  Proposition 4.5.

Let $D$ be an integral domain with quotient field $K$. Let
$\mathcal{F}=\\{D_{i}\\}_{i\in\Lambda}$ be a defining family of $D$. Suppose
that all the rings in $\mathcal{F}$ are contained in a common overring $T$ and
one ring $D_{1}\in\mathcal{F}$ is semilocal with $s$ maximal ideals. Also
suppose that all the residue fields of $D$ have cardinality at least $s$ and
$T$ dominates $D_{i}$ for every $i$, except possibly $D_{1}$. Then
$D(t)=\bigcap_{i}D_{i}(t).$

###### Proof.

As in Lemma 4.4, we can fix a set of units $U\subseteq D$, of cardinality at
least $s-1$ such that $q_{i}-q_{j}$ is a unit for every distinct
$q_{i},q_{j}\in U$. Clearly the set $U$ is also contained in $D_{1}$.

We always have the inclusion $D(t)\subseteq\bigcap_{i}D_{i}(t).$ To prove the
opposite inclusion pick $\phi\in\bigcap_{i}D_{i}(t)$. Such an element $\phi$
can be always written as $\frac{f}{g}$ with $f,g\in D[t]$. Say that
$f=\sum_{k=0}^{m}a_{k}t^{k}$ and $g=\sum_{k=0}^{n}b_{k}t^{k}$.

Since $\phi\in\bigcap_{i}D_{i}(t)$, for every $i$ we can find $x_{i}\in K$
such that $\frac{a_{k}}{x_{i}},\frac{b_{k}}{x_{i}}\in D_{i}$ for every $k$ and
the ideal $\frac{1}{x_{i}}(b_{0},\ldots,b_{n})D_{i}$ is the unit ideal. Hence
we can choose an element $c_{i}\in(b_{0},\ldots,b_{n})D_{i}$ such that
$c_{i}=x_{i}u_{i}$ with $u_{i}$ a unit in $D_{i}$. In particular $x_{i}\in
D_{i}$. For $j\neq i$, since $\frac{b_{k}}{x_{j}}\in D_{j}$ for every $k$, we
get $\frac{c_{i}}{x_{j}}=\frac{x_{i}}{x_{j}}u_{i}\in T$. Similarly,
$\frac{c_{j}}{x_{i}}=\frac{x_{j}}{x_{i}}u_{j}\in T$ and, since $u_{i},u_{j}$
are units in $T$, we obtain that also $\frac{x_{i}}{x_{j}}$ is a unit in $T$
for every $i,j$.

Let us now consider the semilocal domain $D_{1}$. The ideal
$\frac{1}{x_{1}}(b_{0},\ldots,b_{n})D_{1}=D_{1}$. By Lemma 4.4 we can choose
the element $c_{1}=\sum_{l=0}^{n}\delta_{l}b_{l}$ with all $\delta_{l}\in
U\cup\\{0\\}\subseteq D$. Thus $c_{1}\in D$. For every $i$,
$\frac{c_{1}}{x_{i}}\in D_{i}$ and from what said above
$\frac{c_{1}}{x_{i}}=\frac{x_{1}}{x_{i}}u_{1}$ is a unit of $T$. Since, for
$i\neq 1$, the overring $T$ dominates $D_{i}$, then $\frac{c_{1}}{x_{i}}$ has
to be a unit of $D_{i}$. It follows that for every $i$ we can write
$c_{1}=x_{i}w_{i}$ with $x_{i},w_{i}\in D_{i}$ and $w_{i}$ a unit in $D_{i}$
(for $i=1$, set $w_{1}=u_{1}$). Hence,
$\frac{a_{k}}{c_{1}},\frac{b_{k}}{c_{1}}\in\bigcap_{i}D_{i}=D$. This implies
that $\frac{f}{c_{1}},\frac{g}{c_{1}}\in D[t]$ and, since
$c_{1}\in(b_{0},\ldots,b_{n})D$ we get that the ideal of $D$ generated by the
coefficients of $\frac{g}{c_{1}}$ coincides with $D$. This implies
$\phi=\frac{f}{g}\frac{c_{1}}{c_{1}}\in D(t)$. ∎

######  Theorem 4.6.

Let $D$ be an integral domain such that $\overline{D}$ is a semilocal Prüfer
domain of finite dimension with $s$ maximal ideals. Suppose that all the
residue fields of $D$ have cardinality at least $s$. Let $\mathcal{F}$ be any
collection of local overrings of $D$ such that $\bigcap_{A\in\mathcal{F}}A=D$.
Then $\bigcap_{A\in\mathcal{F}}A(t)=D(t)$.

###### Proof.

The fact that $\overline{D}$ is a semilocal Prüfer domain of finite dimension
is equivalent to saying that $D$ has only finitely many valuation overrings,
that we call $V_{1},\ldots,V_{n}$. For $i=1,\ldots,n$, set

$\mathcal{F}_{i}:=\\{A\in\mathcal{F}\,:\,A\mbox{ is dominated by }V_{i}\\}.$

Observe that every ring in $\mathcal{F}$ has to belong to some
$\mathcal{F}_{i}$. Let $A_{i}:=\bigcap_{A\in\mathcal{F}_{i}}A$. Notice that by
Lemma 4.3, $A_{i}$ is local and dominated by $V_{i}$. By Proposition 4.5,
$\bigcap_{A\in\mathcal{F}_{i}}A(t)=A_{i}(t)$.

Since $\overline{D}$ is a Prüfer domain, its Nagata ring coincides with the
Kronecker function ring and therefore $\overline{D}(t)=V_{1}(t)\cap\ldots\cap
V_{n}(t)$. Hence we have the equality

$\bigcap_{A\in\mathcal{F}}A(t)=\bigcap_{i=1}^{n}A_{i}(t)=\left(\bigcap_{i=1}^{n}A_{i}(t)\right)\cap\overline{D}(t)=\bigcap_{i=1}^{n}(A_{i}(t)\cap\overline{D}(t)).$

Set $D_{i}:=A_{i}\cap\overline{D}$. There are exactly $s$ minimal valuation
overrings of $\overline{D}$ and $A_{i}$ is dominated by an overring of one of
them, hence $D_{i}$ is a proper intersection of $s$ local rings, therefore is
semilocal with at most $s$ maximal ideals. Moreover,
$D=\bigcap_{i=1}^{n}D_{i}$ and all the $D_{i}$ are dominated by $\overline{D}$
(clearly $\overline{D}$ is the integral closure of all of them). Again by
Proposition 4.5, $D(t)=\bigcap_{i=1}^{n}D_{i}(t)$. To conclude we only need to
prove that $D_{i}(t)=(A_{i}\cap\overline{D})(t)=A_{i}(t)\cap\overline{D}(t)$.
This follows by Proposition 4.5 since both $A_{i}$ and $\overline{D}$ are
contained in $V_{i}$, $A_{i}$ is dominated by $V_{i}$, and $\overline{D}$ is
semilocal. ∎

Notice that the above theorem implies that if $\overline{D}$ is a Prüfer
domain of finite dimension with at most 2 maximal ideals, then for every
defining family $\mathcal{F}$, we have $Kr^{\mathcal{F}}(D)=D(t)$. We ask
whether the same result of Theorem 4.6 holds in general removing the
assumptions on the residue fields of $D$.

We pass now to study the case of an intersection of a semilocal domain $A$ and
a valuation domain $V$, having the same quotient field. This case is relevant
for the results in Section 5. Again, in the case $A$ is local, the thesis of
the next theorem holds without assumptions on the residue fields.

######  Theorem 4.7.

Let $V$ be a valuation domain of finite dimension and let $A$ be a semilocal
domain such that $\mathcal{Q}(A)=\mathcal{Q}(V)=\mathcal{Q}(V\cap A)$. Assume
that $A$ has $s$ maximal ideals $\mathfrak{m}_{1},\ldots,\mathfrak{m}_{s}$ and
all the residue fields of $A\cap V$ have cardinality at least $s+1$. The
following assertions hold:

  1. (i)

$A\cap V$ is local if and only if $A$ is local and is dominated by an overring
of $V$. In this case $\mathfrak{m}_{A}$ is a common prime ideal of $A$ and
$A\cap V$, $A=(A\cap V)_{\mathfrak{m}_{A}}$, and $\frac{A\cap
V}{\mathfrak{m}_{A}}$ is a valuation domain with quotient field
$\frac{A}{\mathfrak{m}_{A}}$.

  2. (ii)

If $A\cap V$ is not local, then $A_{\mathfrak{m}_{i}}=(A\cap
V)_{\mathfrak{m}_{i}\cap V}$ for every $i=1,\ldots,s$.

  3. (iii)

If $A$ and $V$ have no common proper overrings, then $V=(A\cap
V)_{\mathfrak{m}_{V}\cap A}.$

  4. (iv)

If $A\cap V$ is not local and $A\subseteq V_{Q}$ for some prime ideal $Q$ of
$V$, maximal with respect to this property, then $(A\cap
V)_{\mathfrak{m}_{V}\cap A}=A_{Q\cap A}\cap V.$

  5. (v)

$(A\cap V)(t)=A(t)\cap V(t)$.

###### Proof.

We can assume $A\nsubseteq V$, otherwise all the results are obvious. The
assumption on the residue fields of $A\cap V$ implies that if
$z\in\mathfrak{m}_{i}$ for some $i$, then there exists a unit $q\in A\cap V$
such that $z+q$ is a unit in $A$.
(i) If $A$ is local and dominated by an overring of $V$, the implication
follows by Lemma 4.3. Conversely if $A\cap V$ is local we claim that
$\mathfrak{m}_{i}\subseteq V$ for every $i$. Indeed, if there exists
$z\in\mathfrak{m}_{i}\setminus V$, we get $z^{-1}\in\mathfrak{m}_{V}\setminus
A$. Choose a unit $q\in A\cap V$ such that $z+q$ is a unit in $A$. Observing
that

$\frac{z}{q+z}=\frac{1}{qz^{-1}+1}\in\mathfrak{m}_{i}\cap(V\setminus\mathfrak{m}_{V}),\quad\frac{qz^{-1}}{1+qz^{-1}}=\frac{q}{z+q}\in\mathfrak{m}_{V}\cap(A\setminus\bigcup_{j=1}^{s}\mathfrak{m}_{j}),$

we find that $\frac{z}{q+z}$ and $\frac{q}{z+q}$ are two non-units of $A\cap
V$ whose sum is a unit. This is a contradiction. Hence we have
$\mathfrak{m}_{i}\subseteq V$ for every $i$. Pick an element $s\in A\setminus
V$. For every $n\geq 1$ and every $x\in\bigcup_{j=1}^{s}\mathfrak{m}_{j}$ we
get $s^{n}x\in V$, showing that the radical of $xV$ is properly contained in
the radical of $s^{-1}V$. Therefore, there exists a non-maximal prime ideal
$Q$ of $V$ such that $\bigcup_{j=1}^{s}\mathfrak{m}_{j}\subseteq Q=QV_{Q}$ and
$A\subseteq V_{Q}$. In particular the unique maximal ideal of $A$ is
$\mathfrak{m}_{A}=Q\cap A$ and $A$ is dominated by $V_{Q}$. Using that
$\mathfrak{m}_{A}\subseteq V$, we obtain that $\mathfrak{m}_{A}$ is a prime
ideal of $A\cap V$. For $a\in A\setminus V$, we have $a^{-1}\in(A\cap
V)\setminus\mathfrak{m}_{A}$, thus $A=(A\cap V)_{\mathfrak{m}_{A}}$. If
$x,y\in(A\cap V)\setminus\mathfrak{m}_{A}$, then the fractions
$\frac{x}{y},\frac{y}{x}$ are in $A$ and at least one of them is in $V$,
implying that $\frac{A\cap V}{\mathfrak{m}_{A}}$ is a valuation domain.
(ii) It is sufficient to show that $A\subseteq(A\cap V)_{\mathfrak{m}_{i}\cap
V}$. For $z\in A\setminus V$, if $z^{-1}\in A$ then
$z=\frac{1}{z^{-1}}\in(A\cap V)_{\mathfrak{m}_{i}\cap V}.$ If $z$ is not a
unit, pick again a unit $q\in A\cap V$ such that $q+z$ is a unit in $A$. Set
$u=\frac{1}{q+z}$ and observe that $u$ is a unit of $A$ such that
$uz=\frac{1}{qz^{-1}+1}\in V$ and $u\in V$. Hence $z=\frac{zu}{u}\in(A\cap
V)_{\mathfrak{m}_{i}\cap V}$.
(iii) For $x\in V\setminus A$, if $x^{-1}\in A$, we conclude that $x\in(A\cap
V)_{\mathfrak{m}_{V}\cap A}$ as in item (ii) by choosing $u=\frac{1}{1+qx}$
with $q$ a unit in $A\cap V$ such that $q+x^{-1}$ is a unit in $A$. Suppose
that $x,x^{-1}\not\in A$ and write $x=\frac{a}{b}$ with $a,b$ both non-units
in $A$ and $v(a)\geq v(b)$. We want to find a unit $u$ of $A$ such that
$v(u)=-v(b)$. In this way we can write $x=\frac{au}{bu}\in(A\cap
V)_{\mathfrak{m}_{V}\cap A}$.

If there exists a unit $t$ of $A$ such that $v(t)>v(b)$, it is sufficient to
set $u=\frac{1}{t+qb}$ where $q\in A\cap V$ is a unit such that $t+qb$ is a
unit in $A$. Suppose by way of contradiction that there are no units of $A$
with value larger than or equal $v(b)$. This implies that $v(b)>0$ (because
$1$ is a unit of both $A$ and $V$). By assumption on the residue fields, every
element of $A$ can be expressed as the sum of at most two units, thus, since
$A\nsubseteq V$, there exists some unit $s$ of $A$ such that $v(s)<0$. For
every such $s$, we get that the radical of $s^{-1}V$ properly contains the
radical of $bV$. In particular all the units of $A$ are contained in a proper
overring $V_{Q}$ of $V$. This forces $A\subseteq V_{Q}$. This is a
contradiction.
(iv) Clearly $V_{Q}$ dominates $A_{Q\cap A}$. By Lemma 4.3, $A_{Q\cap A}\cap
V$ is local and contains $A\cap V$. In particular it contains the localization
$(A\cap V)_{\mathfrak{m}_{V}\cap A}.$ For $x\in A_{Q\cap A}\cap V$ we argue as
we did for item (iii) to show that $x=\frac{au}{bu}\in(A\cap
V)_{\mathfrak{m}_{V}\cap A}$. Now we can choose $a,b$ non-units in $A$ such
that $b\not\in Q\cap A$. By the choice of $Q$, there exists some unit $s$ of
$A$ such that the radical of $sV$ is the unique prime ideal $Q^{\prime}$ of
$V$ such that $\mbox{\rm ht }(Q^{\prime})=\mbox{\rm ht }(Q)+1$. Hence, there
exists a unit $t$ of $A$ such that $v(t)>v(b)$. We can conclude using the same
proof as for item (iii).
(v) We consider different cases. If $A\cap V$ is local, by item (i), $A$ is
dominated by a valuation overring of $V$. The thesis follows now by
Proposition 4.5. If $A\cap V$ is not local and $A$ and $V$ have no common
overrings, by items (ii)-(iii), all the rings $A_{\mathfrak{m}_{i}}$ and $V$
are precisely the localizations of $A\cap V$ at the maximal ideals. The thesis
follows since Nagata ring extension commutes with localizations at the maximal
ideals. If instead we are in the situation described by item (iv), we use item
(ii) and localization to say that $(A\cap V)(t)=A(t)\cap(A_{Q\cap A}\cap
V)(t)$. Using that $A_{Q\cap A}\cap V$ is local we can split further
$(A_{Q\cap A}\cap V)(t)=A_{Q\cap A}(t)\cap V(t)$ and conclude since
$A\subseteq A_{Q\cap A}$. ∎

Also about Theorem 4.7, we ask whether the same results hold removing the
assumption on the cardinality of the residue fields. We conclude this section
by showing that under some mild conditions, Nagata ring extension does not
commute with intersection also in the non-integrally closed case. Various
constructions of this kind will be analyzed in the next section (see
Constructions 5.3 and 5.13).

######  Proposition 4.8.

Let $D$ be a local integral domain and let
$\mathcal{F}=\\{D_{i}\\}_{i\in\Lambda}$ be a defining family of $D$. If there
exists $z\in\mathcal{Q}(D)$ such that $z,z^{-1}\not\in D$ and for every $i$,
$z\in D_{i}$ or $z^{-1}\in D_{i}$, then
$D(t)\neq\bigcap_{i\in\Lambda}D_{i}(t).$ In particular if $D=T\cap A$ with $T$
integrally closed, and there exists $z\in A$ such that $z,z^{-1}\not\in D$.
Then $D(t)\neq\bigcap_{V\supseteq T}V(t)\cap A(t)$.

###### Proof.

Consider

$\phi=\frac{1}{t+z}=\frac{z^{-1}}{z^{-1}t+1}.$

Clearly $\phi\in\bigcap_{i\in\Lambda}D_{i}(t)$. Let us show that $\phi\not\in
D(t)$. We can multiply by a common factor to have both numerator and
denominator of $\phi$ inside $D$. Hence pick any $d\in D$ such that $dz\in D$
and write $\phi=\frac{d}{dt+dz}$. If $\phi\in D(t)$, since $D$ is local we
would have that either $d$ or $dz$ is a unit in $D$. But $d$ cannot be a unit,
since $dz\in D$ and $z\not\in D$. If $dz$ were a unit, we would have
$z^{-1}=d(dz)^{-1}\in D$. In any case this leads to a contradiction. ∎

######  Remark 4.9.

For a local domain $D$ that is not a valuation ring and an overring $A$, the
existence of an element $z\in A$ such that $z,z^{-1}\not\in D$ is usually
satisfied. Indeed, such an element $z$ does not exist if and only if $A=D_{Q}$
for a prime ideal $Q$ of $D$ such that $Q=QD_{Q}$, and $\frac{D}{Q}$ is a
valuation ring. For this pick $z\in A$ not a unit and observe that
$z\in\mathfrak{m}_{D}$. Thus for $x,y$ not units in $A$, if $x+y$ was a unit
in $A$, we would get $1=\frac{x}{x+y}+\frac{y}{x+y}\in\mathfrak{m}_{D}$, which
is impossible and implies that $A$ is local. The maximal ideal of $A$ is equal
to a prime ideal $Q$ of $D$ which forces $Q=QD_{Q}$ and $A=D_{Q}$. Finally, if
$u,v$ are element of $D\setminus Q$, then $\frac{u}{v},\frac{v}{u}$ are units
in $A$ and at least one of them has to be in $D$, implying that $\frac{D}{Q}$
is a valuation domain. Notice that by Theorem 4.7-(i), a domain $D$ of this
form may arise as intersection $D=A\cap V$ for some valuation ring $V$ which
has an overring dominating $A$.

The above remark suggests an example of local domain $D$ not maximal
excluding, such that $\overline{D}$ is not Prüfer, but still we have
$Kr^{\mathcal{F}}(D)=D(t)$ for some defining family $\mathcal{F}$. However,
Nagata ring extension clearly does not commute with intersection for all the
overrings of $D$.

######  Example 4.10.

Let $T$ be the maximal excluding generalized power series ring of Example 3.8,
defined over the field $K=\mathbb{Q}$. Set
$D=\mathbb{Z}_{(p)}+\mathfrak{m}_{T}$ for a prime number $p$. The integral
closure of $D$ is $\overline{D}=\mathbb{Z}_{(p)}+\mathfrak{m}_{\overline{T}}$
and it is not Prüfer since $\overline{T}$ is not. Also $D$ is not maximal
excluding by Theorem 3.1 because $T$ is not a valuation ring. We can write
$D=\overline{D}\cap T$ and consider the defining family $\mathcal{F}=\mbox{\rm
Zar }(D)\cup\\{T\\}$, setting $Kr^{\mathcal{F}}(D)=Kr(\overline{D})\cap T(t)$.
Let $V$ be a valuation overring of $D$ such that $\mathfrak{m}_{V}=pV$.
Observing that there are no rings properly contained between $D$ and $T$ and
$p^{-1}\in T\setminus V$, we get $V\cap T=D$ (of course $T$ is dominated by an
overring of $V$, as expected by Theorem 4.7-(i)). Now, Theorem 5.15-(i) from
the next section applies to this setting to show that the localization of
$Kr^{\mathcal{F}}(D)$ at the center of $V(t)$ is equal to $D(t)$. Thus
$Kr^{\mathcal{F}}(D)=D(t)$.

We leave another open question.

######  Question 4.11.

Is it possible to find an integral domain $D$ not maximal excluding, such that
$\overline{D}$ is not Prüfer but $Kr^{\mathcal{F}}(D)=D(t)$ for every defining
family $\mathcal{F}$ of $D$? Is this true for the ring $D$ defined in Example
4.10?

##  5 Constructions of non-integrally closed Kronecker function rings

In this section we construct non-integrally closed rings of the form
$Kr^{\mathcal{F}}(D)$, according to the notation of Definition 4.1. The two
main questions that we investigate for such rings are: understanding what the
integral closure is and studying if they behave locally like classical
Kronecker function rings, in the sense that the localizations at maximal
ideals are Nagata ring extensions of some overring of the base ring $D$. We
immediately observe that in general the integral closure of
$Kr^{\mathcal{F}}(D)$ may not coincide with the Kronecker function ring
$Kr(\overline{D})$. Indeed we have:

######  Remark 5.1.

Let $D$ be an integral domain, $\mathcal{F}$ a defining family for $D$ and set
$R=Kr^{\mathcal{F}}(D)$. Suppose that $\overline{R}=Kr(\overline{D}).$ Then
the integral closure of every $A\in\mathcal{F}$ is a Prüfer domain. For this
simply recall that $Kr(\overline{D})$ is Prüfer and integral closure commutes
with Nagata ring extension.

In the following we first consider cases where the integral closure of
$Kr^{\mathcal{F}}(D)$ is equal to $Kr(\overline{D})$. Then, recalling that the
integral closure of a maximal excluding domain may not be a Prüfer domain, we
give in Theorem 5.15 and Example 5.12, examples such that the integral closure
is a proper non-Prüfer subring of $Kr(\overline{D})$.

Regarding the local behavior of $Kr^{\mathcal{F}}(D)$, we find that in all our
constructions, the localizations at the maximal ideals are Nagata ring
extensions of overrings of $D$. For this reason we leave the following general
question for further research:

######  Question 5.2.

Let $D$ be an integral domain and let $\mathcal{F}$ be any defining family of
$D$. Let $\mathfrak{p}$ be a maximal ideal of the ring
$R=\bigcap_{A\in\mathcal{F}}A(t)$. Is $R_{\mathfrak{p}}=C(t)$ for some
overring $C$ of $D$?

Our first construction is based on integral domains whose integral closure is
obtained by adding the generators of a finite algebraic (Galois) field
extension. We restrict to work with $D$ a local domain, since one can always
reduce from the global case to the local one by localizing at each maximal
ideal.

######  Construction 5.3.

Let $D$ be a local domain with quotient field $F$. Let $F^{\prime}$ be a
subfield of $F$ such that $\mathbb{Q}\subseteq F^{\prime}$ and $F$ is a finite
Galois extension of $F^{\prime}$, generated as $F^{\prime}$-vector space by
$\theta_{1}=1,\theta_{2},\ldots,\theta_{n}$. Call $K$ the field generated over
$\mathbb{Q}$ by $\theta_{1},\theta_{2},\ldots,\theta_{n}$ and all their
conjugates with respect to the action of Gal$(F/F^{\prime})$. We assume any
intermediate field extension $E/E^{\prime}$ such that $F^{\prime}\subseteq
E^{\prime}\subseteq E\subseteq F$ to be a Galois extension generated by
$\vartheta_{1},\ldots,\vartheta_{m}$ and satisfying one of the following
condition (see Examples 5.4 and 5.5 for examples):

  * ($\ast_{1}$)

The field $K$ is contained in $\overline{D}$.

  * ($\ast_{2}$)

$\\{\vartheta_{1},\ldots,\vartheta_{m}\\}=\\{1,\vartheta,\vartheta^{2},\ldots,\vartheta^{m-1}\\}$
where $\vartheta$ is a simple root of degree $m$ over $E^{\prime}$ and
$\overline{D}$ contains $\xi\vartheta^{m}$ for every $m$-th root of unity
$\xi$.

Suppose there exists a local integrally closed domain $D^{\prime}$ containing
$\mathbb{Q}$, having quotient field $F^{\prime}$ such that
$\mathfrak{m}_{D^{\prime}}\subseteq\mathfrak{m}_{D}$,
$D=D^{\prime}[\mathfrak{m}_{D}]_{\mathfrak{m}_{D}}$, and
$\overline{D^{\prime}}^{F}=\overline{D}=D[\theta_{1},\ldots,\theta_{n}]$.

Given a valuation overring $V$ of $D^{\prime}$ such that
$\mathfrak{m}_{V}\supseteq\mathfrak{m}_{D^{\prime}}$, define
$A_{V}:=V[\mathfrak{m}_{D}]_{(\mathfrak{m}_{V},\mathfrak{m}_{D})}$. Clearly
$D\subseteq\bigcap_{V\supseteq D^{\prime}}A_{V}\subseteq\overline{D}$. Suppose
that $\bigcap_{V\supseteq D^{\prime}}A_{V}=D$, and in that case define
$\mathcal{F}$ to be the defining family of $D$ containing all the maximal
excluding overrings $A$ such that $A\supseteq A_{V}$ for some $V$ valuation
overring of $D^{\prime}$. Set $R:=Kr^{\mathcal{F}}(D)$.

The main examples of rings $D$ that we can obtain from Construction 5.3 are
certain kinds of $K+\mathfrak{m}$ constructions and of algebras over fields,
as shown in the next examples.

######  Example 5.4.

Let $D^{\prime}=K^{\prime}+\mathfrak{n}$ be an integrally closed domain with
quotient field $F^{\prime}$, such that $\mathbb{Q}\subseteq K^{\prime}$. Let
$K$ be a finite Galois extension of $K^{\prime}$ generated as
$K^{\prime}$-vector space by $\theta_{1},\ldots,\theta_{n}$ and such that
$K\cap F^{\prime}=K^{\prime}$. Call $F$ the finite Galois extension of
$F^{\prime}$ generated by the same elements. Then define
$D=K^{\prime}+\mathfrak{m}$ with
$\mathfrak{m}=\sum_{i=1}^{n}\theta_{i}\mathfrak{n},$ and observe that
$\overline{D}=\overline{D^{\prime}}^{F}=K+\mathfrak{m}$. Clearly the field
extension $F/F^{\prime}$ satisfies condition ($\ast_{1}$). By construction the
rings $A_{V}$ do not contain any element of $K\setminus K^{\prime}$. Therefore
$\bigcap_{V\supseteq D^{\prime}}A_{V}=D$.

######  Example 5.5.

Consider a $K$-algebra of the form
$D=\sum_{i=1}^{c}K[[x_{i}^{e_{1i}},\ldots,x_{i}^{e_{l_{i}i}}]]$ (or the
corresponding localized polynomial ring version), where $x_{1},\ldots,x_{c}$
are indeterminates over $K$, the exponents $1\leq e_{1i}<\ldots<e_{l_{i}i}$
generate a numerical semigroup, and $K$ contains the $e_{1i}$-th cyclotomic
field for all $i=1,\ldots,c$. In this case
$\overline{D}=K[[x_{1},\ldots,x_{c}]]$ and
$D^{\prime}=K[[x_{1}^{e_{11}},\ldots,x_{c}^{e_{1c}}]].$ The field extension
$F/F^{\prime}$ satisfies condition ($\ast_{2}$). To see that
$\bigcap_{V\supseteq D^{\prime}}A_{V}=D$, observe that if $V$ is a valuation
overring of $D^{\prime}$ (with
$\mathfrak{m}_{V}\supseteq\mathfrak{m}_{D^{\prime}}$) such that
$x_{1}^{e_{11}}$ has infinitely larger value than all the other
$x_{i}^{e_{1i}}$, then all the elements of the form
$x_{1}^{s}f(x_{2},\ldots,x_{c})$ are not in $A_{V}$ if $s$ is not in the
semigroup generated by $e_{11},\ldots,e_{l_{1}1}$. Similarly one can show that
each element of $\overline{D}\setminus D$ does not belong to some ring
$A_{V}$. Simple examples that one may consider in this family are rings such
as $K[[x^{2},x^{3},y]]$ and $K[[x^{2},x^{3},y^{2},y^{3}]]$.

######  Lemma 5.6.

Take the assumptions and notations of Construction 5.3. Let $V$ be a valuation
overring of $D^{\prime}$. Then any element $\sum_{j=1}^{n}a_{j}\theta_{j}\in
F$ with $a_{j}\in F^{\prime}$ is integral over $V$ if and only if each summand
$a_{j}\theta_{j}$ is integral over $V$.

###### Proof.

Let $\sigma_{1},\ldots,\sigma_{n}$ be the elements of Gal$(F/F^{\prime})$. By
restricting each time to the automorphisms with fixed field $E^{\prime}$, we
can reduce to prove this result only for the main extension $F/F^{\prime}$,
assuming that it satisfies either condition $(\ast_{1})$ or condition
$(\ast_{2})$.

Set $B_{V}:=\overline{V}^{F}.$ Let $\alpha=\sum_{j=1}^{n}a_{j}\theta_{j}\in
B_{V}$ with $a_{j}\in F^{\prime}$. Since $F/F^{\prime}$ is a Galois extension,
it is well-known that the integral closure of $V$ in $F$ is the intersection
of all the extensions of $V$ to the field $F$, and such extensions are all
conjugates by the elements of Gal$(F/F^{\prime})$ (see for instance [6,
Section 3.2]). Therefore, also all the conjugates of $\alpha,$
$\sigma_{i}(\alpha)$ are elements of $B_{V}.$ It follows that any linear
combination of $\sigma_{1}(\alpha),\ldots,\sigma_{n}(\alpha)$ with
coefficients in $B_{V}$ is in $B_{V}.$ Now, set up a linear square system to
express $a_{j}\theta_{j}$ as a linear combination of
$\sigma_{1}(\alpha),\ldots,\sigma_{n}(\alpha)$. The equations we obtain are of
the form $a_{j}\theta_{j}=a_{j}\sum_{k=1}^{n}b_{k}\sigma_{k}(\theta_{j})$ and
$0=a_{i}\sum_{k=1}^{n}b_{k}\sigma_{k}(\theta_{i})$ for $i\neq j$. These can be
solved independently of $a_{1},\ldots,a_{n}$ for a choice of coefficients
$b_{1},\ldots,b_{n}$ in the field $K$, which is generated over $\mathbb{Q}$ by
the elements $\sigma_{i}(\theta_{k})$. Assuming condition $(\ast_{1})$, we
have that $K\subseteq\overline{D}\subseteq B_{V}$. The square matrix defining
the linear system is

$\begin{bmatrix}1&1&\ldots&1\\\
\theta_{2}&\sigma_{2}(\theta_{2})&\ldots&\sigma_{n}(\theta_{2})\\\
\vdots&\vdots&\ldots&\vdots\\\
\theta_{n}&\sigma_{2}(\theta_{n})&\ldots&\sigma_{n}(\theta_{n})\end{bmatrix}.$

The determinant of this matrix lives in $K$ and is equal to the square root of
the discriminant of the field extension $F/F^{\prime}$. This extension is
separable, hence the discriminant is nonzero. Since $K\subseteq B_{V}$, the
system can be solved over $B_{V}$ by Cramer’s rule yielding
$a_{j}\theta_{j}\in B_{V}$ for every $j$. In case condition $(\ast_{2})$ is
satisfied, the field $F$ is generated over $F^{\prime}$ by simple roots, and
by hypothesis we have $\theta_{j}=\theta_{2}^{j-1}$ for $j=1,\ldots,n$.
Moreover, the $n$-th cyclotomic field $\mathbb{Q}(\xi)$ is contained in
$\overline{D}$, hence in $B_{V}$. To solve the same linear system as above
with respect to $a_{j}\theta_{j}$, we reduce to solve a linear square system
in $\mathbb{Q}(\xi)$ whose matrix is a Vandermonde matrix in the entries
$1,\xi,\ldots,\xi^{n-1}$. Since $\mathbb{Q}(\xi)\subseteq B_{V}$ we again
obtain $a_{j}\theta_{j}\in B_{V}$ for every $j$. ∎

The next lemma gives us a sufficient condition to have a localization of an
overring of $D(t)$ equal to the Nagata ring extension of an overring of $D$.

######  Lemma 5.7.

Let $D$ be an integral domain and let $A$ be a local overring. Suppose $R$ to
be an integral domain such that $D(t)\subseteq R\subseteq A(t)$. Let
$\mathfrak{p}=\mathfrak{m}_{A(t)}\cap R$ and suppose $A\subseteq
R_{\mathfrak{p}}$. Then $R_{\mathfrak{p}}=A(t)$.

###### Proof.

By assumption $t\in R$, thus there are inclusions $A[t]\subseteq
R_{\mathfrak{p}}\subseteq A(t)$. The ring $A(t)$ is the localization of $A[t]$
at the prime ideal $\mathfrak{m}_{A}[t]=\mathfrak{m}_{A(t)}\cap A[t]$. From
the fact that $\mathfrak{p}R_{\mathfrak{p}}\subseteq\mathfrak{m}_{A(t)}$, it
follows that $\mathfrak{q}:=\mathfrak{p}R_{\mathfrak{p}}\cap
A[t]\subseteq\mathfrak{m}_{A}[t]$. This implies that
$A(t)\subseteq(A[t])_{\mathfrak{q}}\subseteq R_{\mathfrak{p}}$. The thesis
follows. ∎

######  Theorem 5.8.

Let $R$ be defined as in Construction 5.3. Then
$\overline{R}=Kr(\overline{D})$ and all the localizations of $R$ at the
maximal ideals are of the form $R_{\mathfrak{p}}=A_{V}(t)$ where
$\mathfrak{p}=\mathfrak{m}_{A_{V}(t)}\cap R$.

###### Proof.

By construction, for every valuation overring $V$ dominating $D^{\prime}$, the
integral closure of $A_{V}$ is $B_{V}:=\overline{V}^{F}$, which is the
intersection of all the extensions of $V$ to $F$, of which there are finitely
many. Hence $\overline{A_{V}}$ is a semilocal Prüfer domain. By Theorem 4.6,
Nagata ring extension commutes with intersection for all the overrings of
$A_{V}$ (for this we use the assumption that $\mathbb{Q}\subseteq
D^{\prime}\subseteq A_{V}$). It follows that $R=\bigcap_{V\supseteq
D^{\prime}}A_{V}(t)$ and therefore

$Kr(D^{\prime})=\bigcap_{\footnotesize
V\in\mbox{Zar}(D^{\prime})}V(t)\subseteq R\subseteq
Kr(\overline{D})=\bigcap_{\footnotesize V\in\mbox{Zar}(D^{\prime})}B_{V}(t).$

Pick now an element $\Phi\in Kr(\overline{D})$. The field extension
$F(t)/F^{\prime}(t)$ is clearly a finite Galois extension generated by
$\theta_{1},\ldots,\theta_{n}$ and satisfies one of the conditions
$(\ast_{1})$, $(\ast_{2})$, since so does the field extension $F/F^{\prime}.$
Hence, there exist $\phi_{1}(t),\ldots,\phi_{n}(t)\in F^{\prime}(t)$ such that

$\Phi=\sum_{j=1}^{n}\phi_{j}(t)\theta_{j}\in
Kr(\overline{D})=\bigcap_{\footnotesize V\in\mbox{Zar}(D^{\prime})}B_{V}(t).$

By Proposition 5.6, this yields $\phi_{j}(t)\theta_{j}\in Kr(\overline{D})$
for every $j=1,\ldots,n$. If now all $\theta_{j}$ are units in $\overline{D}$,
this gives

$\phi_{j}(t)\in\bigcap_{\footnotesize V\in\mbox{Zar}(D^{\prime})}B_{V}(t)\cap
F^{\prime}(t)=\bigcap_{\footnotesize
V\in\mbox{Zar}(D^{\prime})}V(t)=Kr(D^{\prime})\subseteq R$

and shows that $\Phi$ is integral over $R$ since so are all $\theta_{j}$. If
instead all $\theta_{j}$ are simple roots, saying that $\theta_{j}^{e_{j}}\in
D^{\prime}$, we obtain $(\phi_{j}(t)\theta_{j})^{e_{j}}\in
Kr(\overline{D})\cap F^{\prime}(t)\subseteq R,$ again showing that $\Phi$ is
integral over $R$, being a sum of integral elements.

Now, using that $\overline{R}=Kr(\overline{D})$, we obtain that all the
maximal ideals of $R$ are centers of trivial extensions of valuation overrings
of $D$. In particular, since $R\subseteq
A_{V}(t)\subseteq\overline{A_{V}}(t)=B_{V}(t)$, every maximal ideal of $R$ is
of the form $\mathfrak{p}=\mathfrak{m}_{A_{V}(t)}\cap R$. This proves the
inclusions $R_{\mathfrak{p}}\subseteq A_{V}(t)$ and
$\mathfrak{p}R_{\mathfrak{p}}\subseteq\mathfrak{m}_{A_{V}(t)}$. We show now
that $A_{V}\subseteq R_{\mathfrak{p}}$. For this observe first that
$\mathfrak{m}_{D}\subseteq\mathfrak{m}_{A_{V}}\cap R\subseteq\mathfrak{p}$.
Then pick $\alpha\in V$. The fraction $\frac{1}{t+\alpha}\in
Kr(D^{\prime})\subseteq R$ but $v(\frac{1}{t+\alpha})=0$, hence
$\frac{1}{t+\alpha}\not\in\mathfrak{m}_{V(t)}=\mathfrak{m}_{A_{V}(t)}\cap
V(t)$. This implies $\frac{1}{t+\alpha}\not\in\mathfrak{p}$, thus $\alpha\in
R_{\mathfrak{p}}$. It is clear that
$\mathfrak{m}_{V(t)}\subseteq\mathfrak{p}$. It follows that $A_{V}\subseteq
R_{\mathfrak{p}}$. The thesis now follows by Lemma 5.7. ∎

The above theorem shows that the localizations of $R$ at prime ideals coincide
with the localizations of some Nagata ring extensions of overrings of $D$.
Essentially by the same proof we can obtain the same result also for some
integral extensions of $R$.

######  Corollary 5.9.

Take the assumptions and notations of Construction 5.3. For each ring $A_{V}$,
let $C_{V}$ be a local integral extension. Let $T=\bigcap_{V\supseteq
D^{\prime}}C_{V}$ and suppose $T\neq\overline{D}$. Let
$R^{*}=\bigcap_{V\supseteq D^{\prime}}C_{V}(t)$. Then the ring
$R^{*}=Kr^{\mathcal{F}^{*}}(T)$ for some defining family $\mathcal{F}^{*}$ of
$T$ and the integral closure of $R^{*}$ is $Kr(\overline{D})$. Moreover, let
$\mathfrak{p}=\mathfrak{m}_{C_{V}(t)}\cap R$. If $C_{V}\subseteq
R^{*}_{\mathfrak{p}}$, then $R^{*}_{\mathfrak{p}}=C_{V}(t)$.

###### Proof.

The integral closure of any of the rings $C_{V}$ is semilocal and Prüfer.
Hence, as in the proof of Theorem 5.8, the family $\mathcal{F}^{*}$ can be
chosen to be the family of all the maximal excluding overrings of the rings
$C_{V}$. By definition we have inclusions $R\subseteq R^{*}\subseteq
R\subseteq Kr(\overline{D})$, showing that
$\overline{R^{*}}=\overline{R}=Kr(\overline{D})$. The last part about
localizations follows by Lemma 5.7. ∎

Corollary 5.9 can be applied to the following situation.

######  Proposition 5.10.

Let $D=K^{\prime}+\mathfrak{m}$ be constructed as in Example 5.4. Let
$C_{V}=V[\sum_{i=1}^{n}\theta_{i}\mathfrak{m}_{V}]$ and let
$\mathfrak{p}=\mathfrak{m}_{C_{V}(t)}\cap R$. Then $C_{V}$ is local and
$A_{V}\subseteq C_{V}\subseteq R^{*}_{\mathfrak{p}}$.

###### Proof.

From the inclusions
$\mathfrak{m}_{D}\subseteq\sum_{i=1}^{n}\theta_{i}\mathfrak{m}_{D^{\prime}}\subseteq\sum_{i=1}^{n}\theta_{i}\mathfrak{m}_{V}$,
it follows that $A_{V}\subseteq C_{V}$ and
$\overline{C_{V}}=\overline{V}^{F}$. Let $\sigma_{1},\ldots,\sigma_{n}$ be the
elements of Gal$(F/F^{\prime})$ with $\sigma_{1}$ equal to the identity. To
see that $C_{V}$ is local, pick $\alpha=a+\sum_{i=1}^{n}\theta_{i}b_{i}$ with
$b_{i}\in\mathfrak{m}_{V}$ and $a$ a unit in $V$. Observe that
$\sigma_{j}(\theta_{i})b_{i}\in\mathfrak{m}_{V}C_{V}\subseteq\mbox{Jac}(\overline{V}^{F})$
for every $i,j$. In particular the product
$\prod_{i=1}^{n}\sigma_{i}(\alpha)=a^{n}+z$, with $z\in\mathfrak{m}_{V}C_{V}$,
is a unit in $C_{V}$. It follows that

$\frac{1}{\alpha}=\frac{\prod_{i=2}^{n}\sigma_{i}(\alpha)}{\prod_{i=1}^{n}\sigma_{i}(\alpha)}\in
C_{V}.$

Now let us prove that $C_{V}\subseteq R^{*}_{\mathfrak{p}}$. By the same
argument used in the proof of Theorem 5.8, we obtain $V\subseteq
R^{*}_{\mathfrak{p}}$. We need to show that $z\theta_{i}\in
R^{*}_{\mathfrak{p}}$ for $z\in\mathfrak{m}_{V}$. For this consider the
fraction $\phi=\frac{1}{t+z\theta_{i}}$. For all the valuation overrings
$V^{\prime}$ of $D^{\prime}$ containing $z$, we have $\phi\in C_{V}(t)$. For
the valuation overrings $V^{\prime}$ not containing $z$, we have
$z^{-1}\in\mathfrak{m}_{V^{\prime}}$, $z^{-1}\theta_{i}^{-1}\in C_{V}$, and
$\phi=\frac{z^{-1}\theta_{i}^{-1}}{z^{-1}\theta_{i}^{-1}t+1}\in C_{V}(t)$.
Thus $\phi\in\bigcap_{V\supseteq D^{\prime}}C_{V}(t)=R$. In particular, since
$\phi$ is a unit in $C_{V}$, we obtain $\phi^{-1}\in R^{*}_{\mathfrak{p}}$ and
therefore $z\theta_{i}\in R^{*}_{\mathfrak{p}}$. ∎

In general the rings $A_{V}$ may be too small for being maximal excluding. In
particular if $\overline{V}^{F}$ has more than two maximal ideals we do not
expect $A_{V}$ to be maximal excluding. In some cases it can be shown that
$A_{V}$ is maximal excluding. For $D=K[[x^{2},x^{3},y]]$ and
$D^{\prime}=K[[x^{2},y]]$, we have $A_{V}=V[x^{3}]_{\mathfrak{m}_{V},x^{3}}$.
For $V=K[[x^{2},y,\frac{y}{x^{2}},\frac{y}{x^{4}},\ldots]]$, the ring
$A_{V}=K[[x^{2},x^{3},y,\frac{y}{x},\frac{y}{x^{2}},\ldots]]$ is maximal
excluding since it can be expressed as pullback as in Theorem 3.1. Also, if
$V=K[[y,x^{2},\frac{x^{2}}{y},\frac{x^{2}}{y^{2}},\ldots]]$, the ring
$A_{V}=K[[y,x^{3},x^{2},\frac{x^{2}}{y},\frac{x^{2}}{y^{2}},\ldots]]$ is a
generalized power series ring and is maximal with respect to excluding the
element $\frac{x^{3}}{y}$ by Theorem 3.3. In the next examples we construct
rings of the form $Kr^{\mathcal{F}}(D)$ such that the rings $A_{V}$ or $C_{V}$
are all maximal excluding.

######  Example 5.11.

Adopting the same notation of Example 5.4 and Proposition 5.10, let
$D^{\prime}=\mathbb{Q}+\mathfrak{n}$,
$D=\mathbb{Q}+\mathfrak{m}=\mathbb{Q}+\mathfrak{n}+i\mathfrak{n}$ and
$\overline{D}=\mathbb{Q}(i)+\mathfrak{m}$. For a valuation overring $V$ of
$D^{\prime}$, set $C_{V}=V[i\mathfrak{m}_{V}]$. We show that $C_{V}$ is
maximal with respect to excluding the element $i$. It is clear that
$C_{V}\subseteq V(i)$ is a minimal ring extension. We have two possible cases.
In the first $a^{2}+1\not\in\mathfrak{m}_{V}$ for every $a\in V$. It is easy
to check that in this case there is only one valuation $W$ of $F$ extending
$V$, and $w(a+ib)=\mbox{\rm min}(v(a),v(b))$. Hence, $W=V(i)$ and its residue
field is isomorphic to $\frac{V}{\mathfrak{m}_{V}}(i)$. Now [13, Theorem 2.4],
or alternatively our Corollary 3.2, shows that $C_{V}$ is maximal excluding.
In the other case we have two distinct conjugated valuations $W_{1}$ and
$W_{2}$ extending $V$ to $F$. We observe that $V(i)=W_{1}\cap W_{2}$, indeed
for $a+ib\in W_{1}\cap W_{2}$ we cannot have $v(a)=v(b)<0$, otherwise the
conjugate $a-ib$ would not be in the same ring. Again [13, Theorem 2.4] shows
that $C_{V}$ is maximal excluding. The structure of $C_{V}$ in this case is
described by [21, Theorem 14].

In the next example we replace the intersection of all the valuation overrings
of $D^{\prime}$ by the intersection of a smaller collection, which still form
a defining family for $D^{\prime}$. The Kronecker function ring associated to
this family is an overring of the canonical Kronecker function ring.

######  Example 5.12.

Let $\kappa$ be a field, $x$ an indetreminate over $\kappa$, and $T$ a local
integrally closed domain of the form $\kappa(x)+\mathfrak{m}$. Set
$W=\kappa[[x]]$ and $B=\kappa[[x^{2},x^{3}]]$ (one can also take any other
maximal excluding domain having $W$ as integral closure). Define $D$ to be the
pullback $B+\mathfrak{m}$. If we assume that $T$ is not a valuation ring, the
ring $D$ is not maximal excluding by Theorem 3.1. For $F=\mathcal{Q}(D)$, let
$F^{\prime}$ be the largest subfield of $F$ containing $x^{2}$ but not $x$.
The ring $D^{\prime}$ in this case is set to be
$\kappa[[x^{2}]]+(\mathfrak{m}\cap F^{\prime})$ and we set also
$T^{\prime}=D^{\prime}_{\mathfrak{m}\cap
F^{\prime}}=\kappa(x^{2})+(\mathfrak{m}\cap F^{\prime})$. The extension
$F/F^{\prime}$ satisfies the assumption of Construction 5.3. We can now
express $D^{\prime}$ as the intersection of the valuation overrings $V$ of the
form $V=\pi_{V}^{-1}(\kappa[[x^{2}]])$ where $\pi_{V}$ is the quotient map
from a valuation overring $V^{\prime}$ of $T^{\prime}$ to its residue field.
For any of such $V$, we can define as previously
$A_{V}=V[\mathfrak{m}_{D}]_{\mathfrak{m}_{V},\mathfrak{m}_{D}}=V[x^{3},\mathfrak{m}]$.
In particular $A_{V}$ is the pullback of $B$ with respect to the quotient map
from the valuation overring $V^{\prime}(x)$ of $T$ to its residue field.
Therefore $A_{V}$ is maximal excluding by Theorem 3.1. Let $\mathcal{F}$ be
the family of all the rings $A_{V}$. The same proof used for Theorem 5.8 shows
that, if $R=Kr^{\mathcal{F}}(D)$, then $\overline{R}$ is the Kronecker
function ring of $\overline{D}$ obtained by intersecting all the valuations of
the form $V^{\prime}(x)(t)$. Furthermore, all the localizations of $R$ at the
maximal ideals are of the form $A_{V}(t)$.

We now move on to discuss a second construction which produces rings that are
very close the the Kronecker function ring of $\overline{D}$, in the sense
that they locally coincide for almost all the maximal ideals. This
construction allows also to describe cases where the integral closure of $D$
is not finitely generated over $D$ or where the integral closure of some
overring in the defining family of $D$ is not Prüfer. The rings we define here
are not necessarily of the form $Kr^{\mathcal{F}}(D)$, if the integral domain
$A$ is neither maximal excluding nor has Prüfer integral closure. However,
Theorem 4.7 give us the tools to study a more general situation.

######  Construction 5.13.

Let $D$ be an integral domain and suppose that there exists a semilocal
overring $A$ such that $D=\overline{D}\cap A$. Suppose that also
$\overline{A}$ is semilocal with $s$ maximal ideals and all the residue fields
of $D$ have cardinality at least $s+1$. Let $R^{\prime}=Kr(\overline{D})\cap
A(t)$.

Notice that if $D$ arises as in Construction 5.3 and $A$ contains an
intersection of rings of the form $A_{V}$, then the above ring $R^{\prime}$ is
an overring of the ring $R$, contained in $Kr(\overline{D})$. By Theorem 5.8,
one immediately gets $\overline{R^{\prime}}=Kr(\overline{D})$. Easy examples
that illustrate this construction are

$D=K[[x^{2},x^{3},y]]=K[[x,y]]\cap K((y))[[x^{2},x^{3}]],$

and the ring $D$ defined in Example 4.10.

Before proving the main theorem about this construction, we need to
investigate the integral closure of local rings of the form $A\cap V$
considered in Theorem 4.7-(i). Obviously there is an inclusion
$\overline{A\cap V}\subseteq\overline{A}\cap V$. However, the containment may
be strict. For instance take $A=\mathbb{Q}+X\mathbb{Q}(i)[[X]]$ and
$V=\mathbb{Z}_{(5)}[\frac{2+i}{5}]+X\mathbb{Q}(i)[[X]]$. These rings have the
same quotient field and $V$ is a valuation domain. We have $A\cap
V=\mathbb{Z}_{(5)}+X\mathbb{Q}(i)[[X]]$, $\overline{A}=\mathbb{Q}(i)[[X]]$,
and

$\overline{A\cap
V}=\mathbb{Z}_{(5)}(i)+X\mathbb{Q}(i)[[X]]\subsetneq\overline{A}\cap V=V.$

Next proposition analyzes the relation between $\overline{A\cap V}$ and
$\overline{A}\cap V$.

######  Proposition 5.14.

Let $A$ be a local domain, $V$ a valuation ring such that
$\mathcal{Q}(A)=\mathcal{Q}(V)=\mathcal{Q}(A\cap V)$ and suppose that $A$ is
dominated by a localization $V_{Q}$ of $V$. Let
$\mathfrak{q}=Q\cap\overline{A}$ be the center of $V_{Q}$ in $\overline{A}$
and set $B=\overline{A\cap V}$ and $C=\overline{A}_{\mathfrak{q}}\cap V$. Then
the following assertions hold:

  1. (1)

$B^{\prime}:=\frac{B}{Q\cap B}$ and $C^{\prime}:=\frac{C}{Q\cap C}$ are Prüfer
domains with the same quotient field.

  2. (2)

$C^{\prime}$ is the localization of $B^{\prime}$ at the maximal ideal defined
as the center of the valuation ring $\frac{V}{Q}$.

###### Proof.

Since $V_{Q}$ dominates $A$, then $\mathfrak{q}$ is a maximal ideal of
$\overline{A}$ and $V_{Q}$ dominates $\overline{A}_{\mathfrak{q}}$. Thus $C$
is local and its structure is described in Theorem 4.7-(i). In particular
$Q\cap C=\mathfrak{q}\overline{A}_{\mathfrak{q}}$ and $C^{\prime}$ is a
valuation domain with quotient field
$\kappa=\frac{\overline{A}_{\mathfrak{q}}}{\mathfrak{q}\overline{A}_{\mathfrak{q}}}=\frac{\overline{A}}{\mathfrak{q}}.$
For the same reason, again by Theorem 4.7-(i), we have that
$A^{\prime}:=\frac{A\cap V}{\mathfrak{m}_{A}}$ is a valuation domain with
quotient field $\kappa^{\prime}=\frac{A}{\mathfrak{m}_{A}}$. By standard
results on quotient rings of integral extensions, since $Q\cap(A\cap
V)=\mathfrak{m}_{A}$, we get that $B^{\prime}$ is an integral extension of
$A^{\prime}$ and $\kappa$ is an algebraic field extension of
$\kappa^{\prime}$. The integral closure of a valuation domain in an algebraic
field extension of the quotient field is a Prüfer domain. The ring
$B^{\prime}$ is clearly integrally closed, thus is a Prüfer domain. It is
clear that $B^{\prime}\subseteq C^{\prime}$. To prove (1) it remains to show
that they have the same quotient field. For this we prove that
$\overline{A}\cap V$ is contained in a ring of fractions $S^{-1}B$ for $S$ a
multiplicatively closed set of $B$ with $S\cap Q=\emptyset$. Define $S=(A\cap
V)\setminus\mathfrak{m}_{A}$. This set $S$ is multiplicatively closed since
$\mathfrak{m}_{A}$ is a prime ideal of $A\cap V$ and $S\cap
Q=S\cap\mathfrak{m}_{A}=\emptyset$. Pick $x\in\overline{A}\cap V$. Then
$x^{n}+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}=0$ for $a_{0},\ldots,a_{n-1}\in A$.
If $a_{0},\ldots,a_{n-1}\in V$ then $x\in B$. Thus suppose that at least one
of them is not in $V$ and set $a=\prod_{a_{k}\not\in V}a_{k}$. Since
$\mathfrak{m}_{A}\subseteq Q\subseteq V$, we have that $a$ is a unit in $A$
and $a^{-1}\in V$. Multiplying the above equation of integral dependence of
$x$ by $a^{-n}$ yields an equation of integral dependence of $a^{-1}x$ over
$A\cap V$. It follows that $a^{-1}x\in B$ and $x\in S^{-1}B$ because
$a^{-1}\in S$. From this, going modulo the contraction of $Q$ and observing
that $\frac{\overline{A}\cap V}{\mathfrak{q}}$ and $C^{\prime}$ have the same
quotient field, we obtain that $B^{\prime}$ and $C^{\prime}$ have the same
quotient field. Finally to prove (2), we simply notice that $V$ dominates both
$A\cap V$ and $C$, therefore it dominates the localization of $B$ at the
maximal ideal $\mathfrak{m}_{V}\cap B=\mathfrak{m}_{C}\cap B$. It follows that
$C^{\prime}$ dominates the localization of $B^{\prime}$ at the center of
$\frac{V}{Q}$. Since $B^{\prime}$ and $C^{\prime}$ are Prüfer domains with the
same quotient field, $C^{\prime}$ must be equal to that localization of
$B^{\prime}$. ∎

######  Theorem 5.15.

Let $D$, $A$ and $R^{\prime}$ be defined as in Construction 5.13. The
following assertions hold:

  1. (i)

For a valuation overring $V$ of $D$, let $\mathfrak{p}=\mathfrak{m}_{V(t)}\cap
R^{\prime}$. Then $R^{\prime}_{\mathfrak{p}}=(A\cap V)_{\mathfrak{m}_{V}\cap
A}(t)$. Moreover, if $A$ and $V$ have no common proper overrings then
$R^{\prime}_{\mathfrak{p}}=V(t)$.

  2. (ii)

For every valuation overring $\mathcal{V}$ of $R^{\prime}$, there exists a
valuation overring $V$ of $D$ such that $\mathcal{V}\supseteq(A\cap V)(t)$.

  3. (iii)

The localizations of $R^{\prime}$ at its maximal ideals are Nagata ring
extensions of overrings of $R$.

  4. (iv)

The integral closure of $R^{\prime}$ is $Kr(\overline{D})\cap\overline{A}(t)$.

###### Proof.

It is well-known that the number of maximal ideals of $A$ is less or equal
than the number of maximal ideals of $\overline{A}$. The assumption on the
residue fields of $D$ implies that also the residue fields of $A\cap V$ and
$\overline{A}\cap V$ have cardinality at least $s$ for every valuation
overring $V$ of $D$. Therefore the intersections $A\cap V$ and
$\overline{A}\cap V$ satisfy the hypothesis of Theorem 4.7.
(i) By Theorem 4.7-(iv) we immediately get $(A\cap V)(t)=A(t)\cap V(t)$.
Hence, $R^{\prime}\subseteq(A\cap V)(t)$ and $\mathfrak{p}$ is equal also to
the contraction in $R^{\prime}$ of the maximal ideal of $(A\cap
V)_{\mathfrak{m}_{V}\cap A}$. This implies
$R^{\prime}_{\mathfrak{p}}\subseteq(A\cap V)_{\mathfrak{m}_{V}\cap A}(t)$. We
only need to show $(A\cap V)_{\mathfrak{m}_{V}\cap A}\subseteq R^{\prime}$ and
apply Lemma 5.7. Let $a,s\in A\cap V$ with $s\not\in\mathfrak{m}_{V}$.
Consider

$F=\frac{1}{t+a},\quad G=\frac{1}{t+s^{-1}}=\frac{s}{st+1}.$

Clearly $F,G\in Kr(\overline{D})\cap A(t)=R^{\prime}$. Moreover, $v(a)\geq 0$,
$v(s)=0$, and $F,G$ are units in $V(t)$. Thus $F^{-1}=t+a$ and
$G^{-1}=t+s^{-1}$ are in $R^{\prime}_{\mathfrak{p}}$. This proves
$\frac{a}{s}\in R^{\prime}_{\mathfrak{p}},$ because $t\in R^{\prime}$. If $A$
and $V$ have no common proper overrings we get
$R^{\prime}_{\mathfrak{p}}=V(t)$ by Theorem 4.7-(iii).
(ii) Set $V=\mathcal{V}\cap\mathcal{Q}(D)$ (of course we allow also the case
$V=\mathcal{Q}(D)$). Clearly $V\supseteq R^{\prime}\cap\mathcal{Q}(D)\supseteq
D(t)\cap\mathcal{Q}(D)=D$. Since $t\in R^{\prime}$ we get $(A\cap
V)[t]\subseteq V[t]\subseteq\mathcal{V}$. Pick a polynomial
$f=\sum_{k=0}^{m}a_{k}t^{k}\in(A\cap V)[t]$ such that $c(f)=A\cap V$. Let
$a_{k}$ be a coefficient of $f$. In every valuation overring of
$Kr(\overline{D})$, the value of $a_{k}$ is larger than or equal to the value
of $f$. Hence $a_{k}f^{-1}\in Kr(\overline{D})\cap
A(t)=R^{\prime}\subseteq\mathcal{V}.$ Since $c(f)=A\cap V$ and $A$ is
semilocal, by Lemma 4.4 there exist $u_{0},\ldots,u_{m}\in D$, that are either
units or zeros, such that $a=\sum_{k=0}^{m}a_{k}u_{k}$ is a unit in $A\cap V$.
It follows that $af^{-1}\in R^{\prime}\subseteq\mathcal{V}$. Now $a$ is a unit
in $A\cap V$, hence a unit in $\mathcal{V}$. This implies
$f^{-1}\in\mathcal{V}$ and shows $(A\cap V)(t)\subseteq\mathcal{V}$.
(iii) Every prime ideal of $R^{\prime}$ is the center of some valuation
overring of $R^{\prime}$. By item (ii), every valuation overring $\mathcal{V}$
of $R^{\prime}$ contains an intersection of the form $(A\cap V)(t)=A(t)\cap
V(t)$ for some valuation overring $V$ of $D$. By item (i), $(A\cap
V)_{\mathfrak{m}_{V}\cap A}(t)$ is the localization of $R^{\prime}$ at the
center of the valuation overring $V(t)$. In particular, given a maximal ideal
$\mathfrak{m}$ of $A$, the ring $A_{\mathfrak{m}}(t)$ is the localization of
$R^{\prime}$ at the center of any valuation overring $V(t)$ such that $V$
dominates $A_{\mathfrak{m}}$. Let $\mathfrak{p}=\mathfrak{m}_{\mathcal{V}}\cap
R^{\prime}$. By Theorem 4.7, since localization at the maximal ideals commutes
with Nagata ring extension, $\mathcal{V}$ contains a ring $C(t)$ which is
equal either to $A_{\mathfrak{m}}(t)$ for $\mathfrak{m}$ a maximal ideal of
$A$ or to $(A\cap V)_{\mathfrak{m}_{V}\cap A}(t)$. For what said above, this
ring $C(t)$ is a localization of $R^{\prime}$ at a prime ideal and, if
$\mathfrak{q}=\mathfrak{m}_{\mathcal{V}}\cap C(t)$, then $\mathfrak{q}\cap
R^{\prime}=\mathfrak{p}$. Hence $R^{\prime}_{\mathfrak{p}}$ is the
localization of $C(t)$ at $\mathfrak{q}$. In particular, since we already know
that the rings $C(t)$ are localizations of $R^{\prime}$ at some prime ideal,
it follows that the localizations of $R^{\prime}$ at its maximal ideals are
all of the form $C(t)$ for some overring $C$ of $D$.
(iv) We argue showing that every valuation overring of $R^{\prime}$ contains
$\overline{A}(t)\cap V(t)=(\overline{A}\cap V)(t)$ for some valuation overring
$V$ of $D$. Let $\mathcal{V}$ be a valuation overring of $R^{\prime}$ and let
$V=\mathcal{V}\cap\mathcal{Q}(D)$. By item (ii) we already know that
$\mathcal{V}\supseteq(A\cap V)(t)$. If $\mathcal{V}\supseteq A(t)$, we are
done since then $\mathcal{V}\supseteq\overline{A(t)}=\overline{A}(t)$.
Otherwise by Theorem 4.7, localizing at the maximal ideals of $A\cap V$ we
find that $\mathcal{V}\supseteq(A\cap V)_{\mathfrak{m}_{V}\cap A}(t).$ If
$(A\cap V)_{\mathfrak{m}_{V}\cap A}=V$ we can conclude. If not, again by
Theorem 4.7, we reduce to the case where there exists a prime ideal $Q$ of $V$
such that $V_{Q}$ contains $A$ and $(A\cap V)_{\mathfrak{m}_{V}\cap
A}=A_{Q\cap A}\cap V.$ Without loss of generality, we simplify the notation
possibly replacing $A$ by $A_{Q\cap A}$, and reduce to the case where $A$ and
$A\cap V$ are local, $V_{Q}$ dominates $A$, and $\mathcal{V}\supseteq(A\cap
V)(t)$. Obviously, using that Nagata ring extension commutes with integral
closure, we get $\mathcal{V}\supseteq(\overline{A\cap V})(t)$. Now, if
$\overline{A\cap V}=\overline{A}\cap V$ we are clearly done. If not we use the
result of Proposition 5.14. Let $\mathfrak{q}=Q\cap\overline{A}$ and set
$B=\overline{A\cap V}$ and $C=\overline{A}_{\mathfrak{q}}\cap V$. It is
sufficient to prove that $\mathcal{V}\supseteq C(t)$. Since $\mathcal{V}$
contains $V$ and $t$, then $\mathcal{V}\supseteq C[t]$. Notice also that the
maximal ideal $\mathfrak{m}_{V}\cap B$ of $B$ is contained in
$\mathfrak{m}_{\mathcal{V}}$. Thus, by localization we find that
$\mathcal{V}\supseteq B_{\mathfrak{m}_{V}\cap B}(t)$. Pick $f\in C[t]$ such
that $c(f)=C$. We want to prove that $f^{-1}\in\mathcal{V}$. For $h\in C[t]$,
if $h^{-1}\in\mathcal{V}$ and
$g\in\mathfrak{m}_{V}[t]\subseteq\mathfrak{m}_{\mathcal{V}}$ we get
$(h+g)^{-1}\in\mathcal{V}$. Hence, we want to find $g\in\mathfrak{m}_{V}[t]$
such that $f+g\in B_{\mathfrak{m}_{V}\cap B}[t]$ and
$c(f+g)=B_{\mathfrak{m}_{V}\cap B}$. But Proposition 5.14 implies that
$B_{\mathfrak{m}_{V}\cap B}$ and $C$ coincide after going modulo the
contractions of $Q$ on both rings. In particular, if $x\in C$, then there
exists $z\in Q\cap C\subseteq Q\subseteq\mathfrak{m}_{V}$ such that $x+z\in
B_{\mathfrak{m}_{V}\cap B}$. Furthermore, if $x$ is a unit in $C$ then $x+z$
is a unit in $B_{\mathfrak{m}_{V}\cap B}$. Applying this procedure to all the
coefficients of $f$ that are not in $B_{\mathfrak{m}_{V}\cap B}$, one can
construct a polynomial $g\in Q[t]\subseteq\mathfrak{m}_{V}[t]$ such that
$f+g\in B_{\mathfrak{m}_{V}\cap B}[t]$ and $c(f+g)=B_{\mathfrak{m}_{V}\cap
B}$. ∎

######  Remark 5.16.

With the setting of the above theorem, if $\overline{A}$ is Prüfer, we
automatically have $\overline{R^{\prime}}=Kr(\overline{D})$. In this case,
since we assume all the rings to have finite dimension, $\overline{A}$ has
only finitely many valuation rings, hence all but finitely many localizations
of $R^{\prime}$ at prime ideals are valuation domains of the form $V(t)$. The
other localizations are localizations of $A(t)$ or of domains of the form
$(A\cap V)(t)$.

For instance, let $D=K[[x^{2},x^{3},y]]$, $A=K((y))[[x^{2},x^{3}]]$, and
$R^{\prime}=Kr(\overline{D})\cap A(t)$. Then
$\overline{R^{\prime}}=Kr(\overline{D})$ because $\overline{A}$ is Prüfer. In
this case all but two localizations of $R^{\prime}$ at its prime ideals are of
the form $V(t)$. The two remaining localizations are $A(t)$ and $(A\cap W)(t)$
where $W$ is the rank two valuation overring of $D$ having $\overline{A}$ as
unique minimal overring.

We construct now examples of Kronecker function ring of $D$ of the form
$Kr(\overline{D})\cap A(t)$ such that $A$ is maximal excluding, but
$\overline{A}$ is not a valuation. Therefore
$\overline{R^{\prime}}=Kr(\overline{D})\cap\overline{A}(t)\subsetneq
Kr(\overline{D})$.

######  Example 5.17.

Let $K$ be any field and let $D^{\prime}=[[K^{S}]]$ be a maximal excluding
generalized power series ring. Let $z$ be an indeterminate over
$\mathcal{Q}(D)$ and set $D=D^{\prime}[[z]]$. Let $A$ be the generalized power
series ring over $K$ obtained by adding to $D$ all the elements of the form
$\frac{z}{y}$ for $y\in\mathfrak{m}_{D^{\prime}}$ a monomial element. This
ring $A$ can be represented as pullback of $D^{\prime}$ with respect to the
quotient map from $\mathcal{Q}(D^{\prime})[[z]]$ to its residue field. Hence,
by Theorem 3.1, $A$ is maximal excluding. By standard properties of pullback
diagrams (see [11]), $\overline{A}$ is the pullback of $\overline{D^{\prime}}$
with respect to the same quotient map. Moreover, $D=\overline{D}\cap A$. Let
$R^{\prime}=Kr(\overline{D})\cap A(t)$.

Notice that $A$ and $\overline{A}$ are both local, therefore $D$, $A$ and
$R^{\prime}$ satisfy the hypothesis of Construction 5.13 without restrictions
on the field $K$. The generalized power series rings of Example 3.8 and
Example 3.14 provide examples where $\overline{A}$ is not a Prüfer domain.
Those from Examples 3.4 and 3.7 provide examples where $\overline{D}$ is not
finitely generated as $D$-module.

## Acknowledgements

The first author is supported by the grants MAESTRO NCN -
UMO-2019/34/A/ST1/00263 - Research in Commutative Algebra and Representation
Theory and NAWA POWROTY - PPN/PPO/2018/1/00013/U/00001 - Applications of Lie
algebras to Commutative Algebra. The first author would like to thank Ohio
State University for supporting his research visit in Columbus in May 2022.
Both authors would like to thank Marco D’Anna and Carmelo A. Finocchiaro for
giving them the possibility to visit University of Catania in June 2022 and
for interesting conversations about the content of this paper.

## References

  * [1] D. D. Anderson, D. F. Anderson, Divisibility properties of graded domains, Can. J. Math., Vol. XXXIV, No. 1, 196-215 (1982)
  * [2] D. F. Anderson, Root Closure in Integral Domains, J. of Algebra 79, 51-59 (1982)
  * [3] J. Arnold, On the ideal theory of the Kronecker function ring and the domain $D(X)$, Canad. J. Math. 21 (1969), 558–563.
  * [4] S. Bazzoni, L. Salce. Warfield domains, J. Algebra 185, 836-868, (1996).
  * [5] G. Elliott, P. Ribemboim, Fields of generalized power series, Arch. Math., Vol. 54, 365-371 (1990)
  * [6] A. J. Engler, A. Prestel, Valued fields, Springer, Berlin, 2005, https://doi.org/10.1007/3-540-30035-X
  * [7] M. Fontana, K. A. Loper, An historical overview of Kronecker function rings, Nagata rings, and related star and semistar operations, Multiplicative ideal theory in commutative algebra, 169–187, Springer, New York, 2006.
  * [8] M. Fontana, K. A. Loper, Kronecker function rings: a general approach, in “Ideal Theoretic Methods in Commutative Algebra” (D.D. Anderson and I.J. Papick, Editors), M. Dekker Lecture Notes Pure Appl. Math. 220 (2001), 189–205.
  * [9] M. Fontana, K. A. Loper, Nagata rings, Kronecker function rings and related semistar operations, Comm. Algebra 31 (2003), 4775–4805.
  * [10] R. Fröberg, C. Gottlieb, R. Häggkvist, Gorenstein rings as maximal subrings of $k[[X]]$ with fixed conductor, Comm. Algebra, 16:1621–1625, 1988.
  * [11] S. Gabelli, E. Houston, Ideal theory in pullbacks, Non-Noetherian Commutative Ring Theory, edited by S. Chapman and S. Glaz, Kluwer, Dordrecht, 2000.
  * [12] R. Gilmer, Multiplicative ideal theory, Marcel Dekker, New York, 1972.
  * [13] R. Gilmer, W. Heinzer. Intersections of quotient rings of an integral domain, J. Math. Kyoto Univ., 7 (1967), 133-150.
  * [14] R. Gilmer, J. Hoffmann, A characterization of Prüfer domains in terms of polynomials, Pacific Journal of Mathematics Vol. 60, No. 1, 1975
  * [15] R. Gilmer, J. Hoffmann, The integral closure need not be a Prüfer domain, Mathematika 21(1974), 233-238
  * [16] R. Gilmer, J. Huckaba, Maximal overrings of an integral domain not containing a given element, Comm. in Algebra 2(1974), 377-401.
  * [17] H. Hahn, Über die nichtarchimedischen Grö$\beta$ensysteme. Sitz. Akad. Wiss. Wien 116, 601-655 (1907).
  * [18] F. Halter-Koch, Kronecker function rings and generalized integral closures, Comm. Algebra 31 (2003), 45–59.
  * [19] J.R. Hedstrom, E.G. Houston, Pseudo-valuation domains, Pacific J. Math. 75 (1) (1978) 137–147.
  * [20] W. Heinzer, Integral domains in which each non-zero ideal is divisorial, Mathematika, 15(2), 164-170 (1968)
  * [21] A. Jaballah, Maximal Non-Prüfer and Maximal Non-Integrally Closed Subrings of a Field, Journal of Algebra and its Applications, Vol. 11, No. 5 (2012)
  * [22] B. G. Kang, Prüfer $v$-multiplication domains and the ring $R[X]_{N_{v}}$, J.Algebra 123 (1989) 151–170.
  * [23] H. Kim, Y. S. Park, Krull domains of generalized power series, J. Algebra, 237(2001), 292-301.
  * [24] H. Kim, E. O. Kwon, T. I. Kwon, Weak Normality and Strong t-closedness of Generalized Power Series Rings, Kyungpook Math. J. 48(2008), 443-455
  * [25] W. Krull, Beiträge zur Arithmetik kommutativer Integritätsbereiche, I - II. Math. Z. 41 (1936), 545–577; 665–679.
  * [26] W. Krull, Beiträge zur Arithmetik kommutativer Integritätsbereiche, VIII. Math. Z. 48 (1942/43), 533–552.
  * [27] M. Nagata, Local rings, Interscience, New York, 1962.
  * [28] I. Papick, Local minimal overrings, Can. J. Math. Vol. XXVIII, No. 4, 788-792 (1976)
  * [29] G. Picavet, M. Picavet-L’Hermitte, About minimal morphisms, in: Multiplicative Ideal Theory in Commutative Algebra, Springer-Verlag, New York (2006), 369–386.
  * [30] P. Ribenboim, Special properties of generalized power series, J. Algebra, 173(1995), 566-586.
  * [31] A. Seidenberg, A note on the dimension theory of rings, Pacific J. Math. 3 (1953), 505-512.