# Closures and co-closures attached to FCP ring extensions

Gabriel Picavet and Martine Picavet-L’Hermitte  Mathématiques
8 Rue du Forez, 63670 - Le Cendre
France  picavet.mathu (at) orange.fr

###### Abstract.

The paper deals with ring extensions $R\subseteq S$ and the poset $[R,S]$ of
their subextensions, with a special look at FCP extensions (extensions such
that $[R,S]$ is Artinian and Noetherian). When the extension has FCP, we show
that there exists a co-integral closure, that is a least element
$\underline{R}$ in $[R,S]$ such that $\underline{R}\subseteq S$ is integral.
Replacing the integral property by the integrally closed property, we are able
to prove a similar result for an FCP extension. The radicial closure of $R$ in
$S$ is well known. We are able to exhibit a suitable separable closure of $R$
in $S$ in case the extension has FCP, and then results are similar to those of
field theory. The FCP property being always guaranteed, we discuss when an
extension has a co-subintegral or a co-infra-integral closure. Our theory is
made easier by using anodal extensions. These (co)-closures exist for example
when the extension is catenarian, an interesting special case for the study of
distributive extensions to appear in a forthcoming paper.

###### Key words and phrases:

FCP extension, minimal extension, integral extension, Prüfer extension,
closure, co-closure

###### 2010 Mathematics Subject Classification:

Primary: 13B02, 13B22; Secondary: 13B40

##  1\. Introduction and Notation

We consider the category of commutative and unital rings, whose epimorphisms
will be involved. If $R\subseteq S$ is a (ring) extension, we denote by
$[R,S]$ the set of all $R$-subalgebras of $S$. We set
$]R,S[:=[R,S]\setminus\\{R,S\\}$ (with a similar definition for $[R,S[$ or
$]R,S]$). Then, $[R,S]$ endowed with the partial order $\subseteq$ is a
lattice, called the lattice of the extension.

Now, $(R:S)$ is the conductor of the extension $R\subseteq S$. The integral
closure of $R$ in $S$ is denoted by $\overline{R}^{S}$ (or by $\overline{R}$
if no confusion can occur).

The aim of this paper is to introduce or revisit various relative closures or
hulls associated to ring extensions. But we will mainly work in the context of
FCP or FIP extensions, whose meanings are defined below. We emphasize that
these restricted contexts produce results that may not be valid in an
arbitrary context. Actually, some of them are established because we need them
in forthcoming papers, for example when we characterize extensions, whose
lattices are distributive. Note also that many of them have their own
interest.

The extension $R\subseteq S$ is said to have FIP (for the “finitely many
intermediate algebras property”) or is an FIP extension if $[R,S]$ is finite.
A chain of $R$-subalgebras of $S$ is a set of elements of $[R,S]$ that are
pairwise comparable with respect to inclusion. We will say that $R\subseteq S$
is chained if $[R,S]$ is a chain. We also say that the extension $R\subseteq
S$ has FCP (or is an FCP extension) if each chain in $[R,S]$ is finite, or
equivalently, its lattice is Artinian and Noetherian. Clearly, each extension
that satisfies FIP must also satisfy FCP. Dobbs and the authors characterized
FCP and FIP extensions [4].

Our main tool are the minimal (ring) extensions, a concept that was introduced
by Ferrand-Olivier [8]. Recall that an extension $R\subset S$ is called
minimal if $[R,S]=\\{R,S\\}$. An extension $R\subseteq S$ is called simple if
$S=R[t]$ for some $t\in S$. A minimal extension is simple. The key connection
between the above ideas is that if $R\subseteq S$ has FCP, then any maximal
(necessarily finite) chain $\mathcal{C}$ of $R$-subalgebras of $S$,
$R=R_{0}\subset R_{1}\subset\cdots\subset R_{n-1}\subset R_{n}=S$, with length
$\ell(\mathcal{C}):=n<\infty$, results from juxtaposing $n$ minimal extensions
$R_{i}\subset R_{i+1},\ 0\leq i\leq n-1$. An FCP extension is finitely
generated, and (module) finite if integral. For any extension $R\subseteq S$,
the length $\ell[R,S]$ of $[R,S]$ is the supremum of the lengths of chains of
$R$-subalgebras of $S$. Notice that if $R\subseteq S$ has FCP, then there does
exist some maximal chain of $R$-subalgebras of $S$ with length $\ell[R,S]$ [5,
Theorem 4.11]. An FCP extension $R\subset S$ is called catenarian if all the
maximal chains of $[R,S]$ have the same length. We study such extensions in
[25].

Recall that an extension $R\subseteq S$ is called Prüfer if $R\subseteq T$ is
a flat epimorphism for each $T\in[R,S]$ (or equivalently, if $R\subseteq S$ is
a normal pair) [11, Theorem 5.2, p. 47]. It follows that a Prüfer integral
extension is an isomorphism. The Prüfer hull of an extension $R\subseteq S$ is
the greatest Prüfer subextension $\widetilde{R}$ of $[R,S]$ [19]. In [21], we
defined an extension $R\subseteq S$ to be quasi-Prüfer if it can be factored
$R\subseteq R^{\prime}\subseteq S$, where $R\subseteq R^{\prime}$ is integral
and $R^{\prime}\subseteq S$ is Prüfer. An FCP extension is quasi-Prüfer [21,
Corollary 3.4].

An extension $R\subseteq S$ is called almost-Prüfer if $\widetilde{R}\subseteq
S$ is integral, or equivalently, when $R\subseteq S$ is FIP, if
$S=\widetilde{R}\overline{R}$ [21, Theorem 4.6]. An almost-Prüfer extension is
quasi-Prüfer.

###  1.1. A summary of the main results

Any undefined material is explained in the next sections and at the end of
this section, where the reader may find standard results on minimal
extensions.

Ring extensions admit numerous closures and we add in this paper some more.
One of our aims is to define dual closures, for example like the co-integral
closure of a ring extension. These dual closures exist for FCP extensions but
the reader is warned that they do not for arbitrary extensions. We think that
the main reason is that FCP extensions have Artinian and Noetherian lattices
although proofs most of time rely on induction over tower of minimal
extensions.

For a ring extension $R\subseteq S$, Section 2 recalls some known facts about
the seminormalization and the $t$-closure of $R$ in $S$. We also recall less
known facts about the $u$-closure of $R$ in $S$, that can be found in a
comprehensive study of the first author [15]. Also some technical results
about these closures, that are needed in the sequel, are established in the
FCP context. We have postponed deeper results on the $u$-closure in Section 5,
because it is interwoven with another closure. Theorem 5.8 shows that the
t-closure of an integral FCP extension is the composite of its u-closure and
its seminormalization.

We show that an FCP extension $R\subseteq S$ has a co-integral closure
$\underline{R}$ containing $\widetilde{R}$, with equality if and only if the
extension is almost-Prüfer (Theorems 3.5 and 3.7). We give an FIP example of
an almost-Prüfer extension (Example 3.9).

In Section 3, we consider the elements $T\in[R,S]$ such that $R\subseteq T$ is
integrally closed. Evidently the Prüfer hull of $R$ in $S$ is such an element.
We are interested in extensions that have a greatest integrally closed
subextension, for example if the extension is chained. Another example is a
quasi-Prüfer extension (for example, an FCP extension). In some cases, this
greatest element $T$ verifies $T\subseteq S$ is integral, and could have been
considered as a co-integral closure, but it is not always the case. Actually
the right co-integral closure is defined in Section 3.

We consider in Section 4 some closures linked to the properties of their
residual field extensions. Kubota introduced the radicial closure of a ring
extension [12], reconsidered by Manaresi under another name [14]. The second
author considered a weaker closure [27]. Another type of ring extensions, that
are linked to residual extensions, are the unramified extensions of Algebraic
Geometry. In order to unify these concepts, if $\mathcal{P}$ is a property of
field extensions, we say that an extension is a $\kappa$-$\mathcal{P}$
extension if all its residual extensions verify $\mathcal{P}$. For example a
radicial extension is a $\kappa$-radicial $i$-extension. As a first result we
show that an unramified radicial FCP integral extension is trivial
(Proposition 4.2). We give some properties of unramified extensions in the FCP
context. We introduce a new concept: the $\kappa$-separable extensions, within
the framework of integral extensions. This concept is more tractable than the
unramified property. They coincide in some cases: an integral $t$-closed FCP
extension is unramified if and only if it is $\kappa$-separable (Corollary
4.4). An important result is that an FCP integral extension $R\subseteq S$
admits a $\kappa$-separable closure ${}^{\square}_{S}R$ such that
${}^{\square}_{S}R\subseteq S$ is radicial (Theorem 4.12 and Corollary 4.13).

In Section 5, we give properties of $u$-integral extensions and anodal
($u$-closed) extensions, valid in the FCP context. For example, an FCP
integral extension is $u$-integral if and only if it is locally an
epimorphism, in which case the extension is unramified (Theorem 5.7).
Moreover, an integral FCP extension is anodal if and only if it is an
$i$-extension (Proposition 5.2). We also examine the links between the
$u$-closure and the other closures or properties previously considered. We
then introduce the unramified closure ${}^{\omega}_{S}R$ of a ring extension
$R\subseteq S$ and show that ${}^{\omega}_{S}R\subseteq S$ is $u$-closed in
case the extension is integral and has FCP (Proposition 5.13). Results of
Section 5 about the radicial property may not hold.

In Section 6, we consider dual closures of the seminormalizations and
$t$-closures of ring extensions. This may seem weird, but in a forthcoming
paper on distributive FCP extensions, these new closures are an essential
tool. The reader is warned that these closures may not exist but exist in the
context we have to consider. A co-subintegral closure of a ring extension
$R\subseteq S$ is the least $T\in[R,S]$ such that $T\subseteq S$ is
subintegral. The co-infra-integral closure is similarly defined. We give some
generalities about these closures and their properties. For example, an infra-
integral FCP extension admits a co-subintegral closure. We then compare all
the closures we have introduced with these dual closures. Now considering an
integral FCP extension $R\subseteq S$ and setting $T=\cap[U\in[R,S]\mid
U\subseteq S$ subintegral$]$, we get that the co-subintegral closure exists if
and only if $T\subseteq S$ is catenarian (resp.; subintegral), in which case
$T$ is the co-subintegral closure (Theorem 6.15). As a corollary, we have that
if in addition the extension is catenarian, the co-subintegral closure exists.
We have similar results for the co-infra-integral closure. But the proofs are
much more involved. The rest of the paper exhibits examples where these dual
closures exist or not.

Section 7 explores the cardinality of lattices of integral FIP extensions.
Some results are linked to co-subintegral closures and complements. A striking
result is Theorem 7.11 which says that if $R\subset S$ is an FIP integral
extension over the local ring $R$, then $|[R,S]|$ is the sum of the number of
the complements of the t-closure ${}_{S}^{t}R$ in all extensions
$R^{\prime}\subseteq S^{\prime}$, for each
$(R^{\prime},S^{\prime})\in[R,{}_{S}^{t}R]\times[{}_{S}^{t}R,S]$.

###  1.2. Definitions and notation

A local ring is here what is called elsewhere a quasi-local ring. As usual,
Spec$(R)$ and Max$(R)$ are the set of prime and maximal ideals of a ring $R$.
For an extension $R\subseteq S$ and an ideal $I$ of $R$, we write
$\mathrm{V}_{S}(I):=\\{P\in\mathrm{Spec}(S)\mid I\subseteq P\\}$ and
$\mathrm{D}_{S}(I)$ for its complement. The support of an $R$-module $E$ is
$\mathrm{Supp}_{R}(E):=\\{P\in\mathrm{Spec}(R)\mid E_{P}\neq 0\\}$, and
$\mathrm{MSupp}_{R}(E):=\mathrm{Supp}_{R}(E)\cap\mathrm{Max}(R)$. We denote by
${\mathrm{L}}_{R}(E)$ the length of $E$ as an $R$-module.

When $R\subseteq S$ is an extension, we will set
$\mathrm{Supp}(T/R):=\mathrm{Supp}_{R}(T/R)$ and
$\mathrm{Supp}(S/T):=\mathrm{Supp}_{R}(S/T)$ for each $T\in[R,S]$, unless
otherwise specified.

Let $R\subset S$ be a ring extension. For $T\in[R,S]$, an element $U\in[R,S]$
is called a complement of $T$ if $T\cap U=R$ and $TU=S$. If $T$ has a unique
complement, this complement is denoted by $T^{o}$.

If $R\subset S$ is an extension and $T\in]R,S[$, we say that the extension
splits at $T$, if $\mathrm{MSupp}(S/T)\cap\mathrm{MSupp}(T/R)=\emptyset$. In
[26, Theorem 4.8], we proved that if an FCP extension splits at $T$, then $T$
has a unique complement $T^{o}$.

Let $R\subset S$ be an FCP extension with
$\mathrm{MSupp}(S/R):=\\{M_{i}\\}_{i=1}^{n}$. In [23, the paragraph before
Proposition 2.21], we say that $R\subset S$ is a $\mathcal{B}$-extension if
the map $\varphi:[R,S]\to\prod_{i=1}^{n}[R_{M_{i}},S_{M_{i}}]$ defined by
$\varphi(T):=(T_{M_{i}})_{i=1}^{n}$ is bijective. In this case, $\varphi$ is
an order isomorphism. An integral FCP extension is a $\mathcal{B}$-extension
[4, Theorem 3.6]. If, in addition, $R\subset S$ has FIP, then
$|[R,S]|=\prod_{i=1}^{n}|[R_{M_{i}},S_{M_{i}}]|$.

If $R\subseteq S$ is a ring extension and $P\in\mathrm{Spec}(R)$, then $S_{P}$
is both the localization $S_{R\setminus P}$ as a ring and the localization at
$P$ of the $R$-module $S$. We denote by $\kappa_{R}(P)$ the residual field
$R_{P}/PR_{P}$ at $P$. If $R\subseteq S$ is a ring extension and
$Q\in\mathrm{Spec}(S)$, there exists a residual field extension
$\kappa_{R}(Q\cap R)\to\kappa_{S}(Q)$. Moreover, for $P\in\mathrm{Spec}(R)$,
the ring morphism $\kappa(P)\to\kappa(P)\otimes_{R}S=S_{P}/PS_{P}$ is called
the fiber morphism at $P$, whose target is the fiber of the extension at $P$.

A ring extension $R\subseteq S$ is called an i-extension if the natural map
$\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is injective.

Finally, $|X|$ is the cardinality of a set $X$, $\subset$ denotes proper
inclusion and for a positive integer $n$, we set
$\mathbb{N}_{n}:=\\{1,\ldots,n\\}$. The characteristic of an integral domain
$k$ is denoted by $\mathrm{c}(k)$.

###  1.3. Results on minimal extensions

######  Definition 1.1.

[26, Definition 2.1] An extension $R\subset S$ is called $M$-crucial if
$\mathrm{MSupp}(S/R)=\\{M\\}$. Such a maximal ideal $M$ is called the crucial
(maximal) ideal $\mathcal{C}(R,S)$ of $R\subset S$.

######  Theorem 1.2.

[8, Théorème 2.2] A minimal extension is simple, crucial and is either
integral (finite) or a flat epimorphism.

In [21], a minimal flat epimorphism is called a Prüfer minimal extension. An
FCP Prüfer extension has FIP and is a tower of finitely many Prüfer minimal
extensions [21, Proposition 1.3].

Three types of minimal integral extensions exist, characterized in the next
theorem, (a consequence of the fundamental lemma of Ferrand-Olivier), whence
four types of minimal extensions exists and are mutually exclusive.

######  Theorem 1.3.

[4, Theorems 2.2 and 2.3] Let $R\subset T$ be an extension and $M:=(R:T)$.
Then $R\subset T$ is minimal and finite if and only if $M\in\mathrm{Max}(R)$
and one of the following three conditions holds:

(a) inert case: $M\in\mathrm{Max}(T)$ and $R/M\to T/M$ is a minimal field
extension.

(b) decomposed case: There exist $M_{1},M_{2}\in\mathrm{Max}(T)$ such that
$M=M_{1}\cap M_{2}$ and the natural maps $R/M\to T/M_{1}$ and $R/M\to T/M_{2}$
are both isomorphisms; or, equivalently, there exists $q\in T\setminus R$ such
that $T=R[q]$ and $q^{2}-q\in M$.

(c) ramified case: There exists $M^{\prime}\in\mathrm{Max}(T)$ such that
${M^{\prime}}^{2}\subseteq M\subset M^{\prime},\ [T/M:R/M]=2$, and the natural
map $R/M\to T/M^{\prime}$ is an isomorphism; or, equivalently, there exists
$q\in T\setminus R$ such that $T=R[q]$ and $q^{2}\in M$.

In each of the above cases, $M=\mathcal{C}(R,T)$.

##  2\. Recall on closures involved in ring extensions

###  2.1. Definitions

The following definitions are needed for our study. We first consider more or
less classical concepts.

######  Definition 2.1.

An integral extension $R\subseteq S$ is called infra-integral [18] (resp.;
subintegral [30]) if all its residual extensions are isomorphisms (resp.; and
is an i-extension).

We next consider some other concepts that will allow us to give an elementwise
interpretation of the above definitions. Two of them are classical and the
third was developed by the first author and does not seem well known, but is
essential in the sequel.

######  Definition 2.2.

According to [15, Definitions 1.1, 1.2 and 1.5, Proposition 1.6], [18] and
[30], and setting $p_{r}(X):=X^{2}-rX\in R[X]$, where $R\subseteq S$ is a ring
extension, $R\subseteq S$ is called:

  * •

s-elementary (resp.; t-elementary, u-elementary ) if $S=R[b]$, where
$p_{0}(b),bp_{0}(b)\in R$ (resp.; $p_{r}(b),bp_{r}(b)\in R$ for some $r\in R,\
p_{1}(b),bp_{1}(b)\in R$).

In the following, the letter x denotes s, t or u.

  * •

cx-elementary if $R\subseteq S$ is a tower of finitely many x-elementary
extensions.

  * •

x-integral if there exists a directed set $\\{S_{i}\\}_{i\in I}\subseteq[R,S]$
such that $R\subseteq S_{i}$ is cx-elementary and $S=\cup_{i\in I}S_{i}$.

In particular, an s-elementary (resp.; u-elementary) extension is a
t-elementary extension. In the same way, a cs-elementary (resp.; cu-
elementary) extension is a ct-elementary extension and an s-integral (resp.;
u-integral) extension is a t-integral extension.

######  Proposition 2.3.

An s-integral (resp.; t-integral) extension is nothing but a subintegral [30]
(resp.; infra-integral [18]) extension.

######  Definition 2.4.

An extension $R\subseteq S$ is called s-closed (or seminormal) (resp.;
t-closed, u-closed (or anodal)) if an element $b\in S$ is in $R$ whenever
$p_{0}(b),bp_{0}(b)\in R$ (resp.; $p_{r}(b),bp_{r}(b)\in R$ for some $r\in R,\
p_{1}(b),bp_{1}(b)\in R$).

Let x$\in\\{$s,t,u$\\}$. The x-closure ${}_{S}^{x}R$ of $R$ in $S$ is the
smallest element $B\in[R,S]$ such that $B\subseteq S$ is x-closed and the
greatest element $B^{\prime}\in[R,S]$ such that $R\subseteq B^{\prime}$ is
x-integral. It follows that ${}_{S}^{u}R\subseteq{}_{S}^{t}R$. Note that the
$s$-closure is actually the seminormalization ${}_{S}^{+}R$ of $R$ in $S$ and
is the greatest subintegral extension of $R$ in $S$. Note also that the
$t$-closure ${}_{S}^{t}R$ is the greatest infra-integral extension of $R$ in
$S$.

In some next subsections, other closures will be defined.

The canonical decomposition of an arbitrary ring extension $R\subset S$ is
$R\subseteq{}_{S}^{+}R\subseteq{}_{S}^{t}R\subseteq\overline{R}\subseteq S$.
One of our aims is to investigate when a dual decomposition exists.

######  Proposition 2.5.

An FCP Prüfer extension is $t$-closed.

###### Proof.

Let $R\subset S$ be an FCP Prüfer extension. Since an FCP Prüfer extension is
a tower of finitely many Prüfer minimal extensions, let
$R_{0}:=R\subset\ldots\subset R_{i}\subset\ldots\subset R_{n}:=S$ be a maximal
chain; so that, each $R_{i}\subset R_{i+1}$ is minimal Prüfer for
$i\in\\{0,\ldots,n-1\\}$. Let $b\in S$ such that $b^{2}-rb,b^{3}-rb^{2}\in
R\subseteq R_{n-1}$ for some $r\in R\subseteq R_{n-1}$. Then, either $b$ or
$b-r$ is in $R_{n-1}$ by [8, Proposition 3.1]; so that, $b\in R_{n-1}$. An
easy induction shows that $b\in R$ and $R\subset S$ is t-closed. ∎

###  2.2.  Technical results needed in the sequel

The next proposition describes the link between the elements of the canonical
decomposition and minimal extensions.

######  Proposition 2.6.

[22, Proposition 4.5] Let there be an integral extension $R\subset S$ and a
maximal chain $R=R_{0}\subset\cdots\subset R_{i}\subset\cdots\subset R_{n}=S$
of subextensions, each $R_{i}\subset R_{i+1}$ being minimal. The following
statements hold:

  1. (1)

$R\subset S$ is subintegral if and only if each $R_{i}\subset R_{i+1}$ is
ramified.

  2. (2)

$R\subset S$ is seminormal and infra-integral if and only if each
$R_{i}\subset R_{i+1}$ is decomposed.

  3. (3)

$R\subset S$ is infra-integral if and only if each $R_{i}\subset R_{i+1}$ is
either decomposed or ramified.

  4. (4)

$R\subset S$ is t-closed if and only if each $R_{i}\subset R_{i+1}$ is inert.

  5. (5)

$R\subset S$ is seminormal if and only if each $R_{i}\subset R_{i+1}$ is
either decomposed or inert.

If either (1) or (4) holds, then $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is
bijective.

######  Lemma 2.7.

[4, Corollary 3.2] If there exists a maximal chain
$R=R_{0}\subset\cdots\subset R_{i}\subset\cdots\subset R_{n}=S$ of extensions,
such that each $R_{i}\subset R_{i+1}$ is minimal, then
$\mathrm{Supp}(S/R)=\\{\mathcal{C}(R_{i},R_{i+1})\cap R\mid
i=0,\ldots,n-1\\}$.

We emphasize on the following useful result.

######  Proposition 2.8.

[26, Proposition 2.4] Let $R\subset S$ be an integral FCP extension. Then,
$R\subset S$ is seminormal if and only if $(R:S)$ is an intersection of
finitely many maximal ideals of $S$.

If this condition holds, the intersection giving $(R:S)$ is irredundant in $R$
and $S$.

######  Proposition 2.9.

Let $R\subseteq S$ be an FCP integral extension with conductor $I$. Then
$\mathrm{V}_{R}(I)$ has finitely many elements and is contained in
$\mathrm{Max}(R)$; so that,
$\mathrm{V}_{R}(I)=\mathrm{Supp}(S/R)=\mathrm{MSupp}(S/R)$. Moreover, for each
$Q\in\mathrm{Spec}(S)$ such that $R\cap Q\in\mathrm{D}_{R}(I)$, the residual
extension $\kappa(Q\cap R)\to\kappa(Q)$ is an isomorphism.

###### Proof.

Since $R\subset S$ is an integral FCP extension, $R/I$ is an Artinian ring [4,
Theorem 4.2] and the extension is finite; so that,
$\mathrm{V}_{R}(I)=\mathrm{Supp}(S/R)=\mathrm{MSupp}(S/R)$, where the last
equation is valid because $R/I$ is Artinian.

Let $Q\in\mathrm{Spec}(S)$ such that $P:=R\cap Q\in\mathrm{D}(I)$. Since
$P\not\in\mathrm{Supp}(S/R)$, we have that $R_{P}=S_{P}=S_{Q}$, whence
$\kappa(Q\cap R)\cong\kappa(Q)$. ∎

######  Lemma 2.10.

Let $R\subseteq S$ be an FCP integral t-closed extension. Then, for each
$M\in\mathrm{MSupp}(S/R)$, there exists a unique $N\in\mathrm{Max}(S)$ lying
above $M$. Moreover, $(R_{M}:S_{M})=MS_{N}=NS_{N}$ and
$\kappa(N)=S_{N}/(R_{M}:S_{M})$. We denote by (${\dagger}$) this last
statement for a later use.

###### Proof.

Let $M\in\mathrm{MSupp}(S/R)$. Since $R\subset S$ is an FCP t-closed
extension, so is $R_{M}\subset S_{M}$ and
$(R_{M}:S_{M})=MR_{M}\in\mathrm{Max}(S_{M})$ by [4, Proposition 4.10]; so
that, $MR_{M}=MS_{M}$. Moreover, there exists a unique $N\in\mathrm{Max}(S)$
lying above $M$ by Proposition 2.6 and $S_{M}=S_{N}$ according to [2,
Proposition 2, page 40]. These facts combine to yield that
$I:=(R_{M}:S_{M})=MS_{N}=NS_{N}$. In particular, $\kappa(M)=R_{M}/I$ and
$\kappa(N)=S_{N}/NS_{N}=S_{N}/I$. ∎

##  3\. Co-integral closures

Let $R\subset S$ be an almost-Prüfer FCP extension. Since $\widetilde{R}$ is
the least $T\in[R,S]$ such that $T\subseteq S$ is integral ([21, Proposition
4.17]), we can ask if, for any FCP extension $R\subset S$, there exists a
least $T\in[R,S]$ such that $T\subseteq S$ is integral. The answer is yes, as
we are going to show. But this statement may not hold for any arbitrary ring
extension. To see this we will use for instance [7, Example 2.1] where the
following situation holds. Let $p$ be a prime number, $k$ a field with
$\mathrm{c}(k)=p$, and $n$ a prime number such that $n>p$. Let $X$ be an
indeterminate over $k$, and set $K:=k(X),\ F:=k(X^{n})$ and
$L:=(X^{n}+X^{p})$. Then, we get that $k=F\cap L\subset K=FL$ is not
algebraic, and then not integral, while $F\subset K$ and $L\subset K$ are
minimal field extensions, and then integral. Actually intersecting two minimal
integral extension whose targets are $S$ is a first test for the validity of
the result we have in view.

######  Proposition 3.1.

Let $R\subset S$ be an FCP extension and $T,U\in[R,S]$ be such that $T\subset
S$ and $U\subset S$ are minimal integral extensions with $T\neq U$. Then,
$T\cap U\subset S$ is integral with $T\cap U\neq T,U$.

###### Proof.

Set $V:=T\cap U,\ M:=(T:S)$ and $N:=(U:S)$.

Assume first that $M\neq N$. Then, $V\subset S$ is integral according to [7,
Proposition 6.6], since so are $V\subset T$ and $V\subset U$.

Assume now that $M=N$. It follows by [7, Proposition 5.7] that
$M\in\mathrm{Max}(V)$. But we also have $M\in\mathrm{Max}(T)$ and
$M\in\mathrm{Max}(U)$ with $M=(V:S)$. It follows that $V/M\subset T/M$ and
$V/M\subset U/M$ are FCP field extensions, and then integral by [4, Theorem
4.2], since any minimal subextension of $V/M\subset T/M$ and $V/M\subset U/M$
is inert by Theorem 1.3. Hence, $V/M\subset S/M$ is an integral extension, and
so is $V\subset S$. Obviously, $T\cap U\neq T,U$ since $T\neq U$. ∎

To get the general result, the following lemmata are needed.

######  Lemma 3.2.

Let $R\subset S$ be an FCP extension and $T,U\in[R,S]$, such that $T\subset S$
is integral and $U\subset S$ is minimal integral with $U\not\in[T,S]$. Then,
there exists $W\in[T\cap U,T]$ such that $W\subset T$ is minimal integral.

###### Proof.

Let $\\{T_{i}\\}_{i=0}^{n}\subseteq[R,S]$ be a maximal chain such that
$T_{0}:=S$ and $T_{n}:=T$; so that, each $T_{i+1}\subset T_{i}$ is minimal
integral. For each $i\in\mathbb{N}_{n}$, set $V_{i}:=T_{i}\cap U$, giving
$V:=V_{n}=T\cap U$. Obviously, we have $V_{i+1}=T_{i+1}\cap V_{i}$. We intend
to build by induction on $i\in\mathbb{N}_{n}$ a set
$\\{W_{i}\\}_{i=1}^{n}\subseteq[V,S]$ such that $W_{i}\in[V_{i},T_{i}]$ with
$W_{i}\subset T_{i}$ minimal integral for each $i\in\mathbb{N}_{n}$. Our
induction hypothesis is the following: for $i\in\mathbb{N}_{n}$, there exists
$W_{i}\in[V_{i},T_{i}]$ with $W_{i}\subset T_{i}$ minimal integral and either
$W_{i}\neq T_{i+1}$ or $W_{i}=T_{i+1}\neq V_{i}$ when $i<n$.

For $i=1$, since $T_{1}\subset S$ and $U\subset S$ are minimal integral, we
know by Proposition 3.1 that $V_{1}\subset T_{1}$ is an integral FCP
extension. Of course, $V_{1}\neq T_{1}$ because $T_{1}\neq U$. In particular,
there exists $W_{1}\in[V_{1},T_{1}]$ with $W_{1}\subset T_{1}$ minimal
integral. Assume that $W_{1}=T_{2}$ leading to $V_{2}=T_{2}\cap U=W_{1}\cap
U\supseteq V_{1}\cap U=V_{1}$; so that, $V_{2}=V_{1}$. If $T_{2}=V_{1}$, then
$T\subseteq T_{2}\subseteq U$ leads to a contradiction. It follows that
$T_{2}\neq V_{1}$.

Assume that the induction hypothesis holds for some $i\in\mathbb{N}_{n-1}$; so
that, $W_{i}\subset T_{i}$ is minimal integral for some
$W_{i}\in[V_{i},T_{i}]$ such that either $W_{i}\neq T_{i+1}$ or
$W_{i}=T_{i+1}\neq V_{i}$.

First, assume that $W_{i}\neq T_{i+1}$ and set $W^{\prime}_{i+1}:=T_{i+1}\cap
W_{i}$. We claim that $W^{\prime}_{i+1}\neq T_{i+1}$. Otherwise,
$T_{i+1}\subseteq W_{i}$, which leads to $T_{i+1}=W_{i}$, a contradiction.
Then, $W^{\prime}_{i+1}\subset T_{i+1}$ is integral by Proposition 3.1 and
there exists $W_{i+1}\in[W^{\prime}_{i+1},T_{i+1}]$ such that $W_{i+1}\subset
T_{i+1}$ is minimal integral. If $i=n-1$, we are done. If $i<n-1$, assume that
$W_{i+1}=T_{i+2}$. Then, $V_{i+2}=T_{i+2}\cap U=W_{i+1}\cap U\supseteq
V_{i+1}\cap U=V_{i+1}$; so that, $V_{i+2}=V_{i+1}$. If $T_{i+2}=V_{i+1}$, then
$T\subseteq T_{i+2}\subseteq U$ leads to a contradiction. It follows that
$T_{i+2}\neq V_{i+1}$. Then, the induction hypothesis holds. We therefore have
the following commutative diagram:

$\begin{matrix}V_{i}&\to&W_{i}&{}&\to&{}&T_{i}\\\
\uparrow&{}&\uparrow&{}&{}&{}&\uparrow\\\
V_{i+1}&\to&W^{\prime}_{i+1}&\to&W_{i+1}&\to&T_{i+1}\end{matrix}$

Now, assume that $W_{i}=T_{i+1}\supset V_{i}$. It follows that $V_{i}\subseteq
W_{i}\cap U=T_{i+1}\cap U=V_{i+1}\subseteq V_{i}$, which gives $V_{i+1}=V_{i}$
and $V_{i+1}\subset T_{i+1}$. Then, there exists $W_{i+1}\in[V_{i+1},T_{i+1}]$
such that $W_{i+1}\subset T_{i+1}$ is minimal integral. If $i=n-1$, we are
done. If $i<n-2$, assume that $W_{i+1}=T_{i+2}=V_{i+1}$, then $T\subseteq
T_{i+2}\subseteq U$ leads to a contradiction. It follows that $T_{i+2}\neq
V_{i+1}$. Then, the induction hypothesis holds for $i+1$ and then for any
$i\in\mathbb{N}_{n}$. In particular, there exists
$W_{n}\in[V_{n},T_{n}]\setminus\\{T_{i}\\}_{i=0}^{n}$ such that $W_{n}\subset
T_{n}$ is minimal integral, that is there exists $W\in[T\cap U,T]$ such that
$W\subset T$ is minimal integral. ∎

######  Lemma 3.3.

Let $R\subset S$ be an FCP extension and $T,U\in[R,S]$, such that $T\subset S$
is integral and $U\subset S$ is minimal integral, with $U\not\in[T,S]$. Then,
$T\cap U\subset S$ is an integral extension.

###### Proof.

Set $V:=T\cap U$. We are going to show by induction that there exists a
maximal finite chain $\\{W_{j}\\}_{j=0}^{m}$ such that $W_{0}:=T$ and
$W_{m}:=V$, where $W_{j+1}\subset W_{j}$ is minimal integral. Our induction
hypothesis is the following: for $k\in\\{0,\ldots,m-1\\}$, where $m$ is some
positive integer such that $m\leq\ell[T\cap U,T]$, there exists a maximal
finite chain $\\{W_{j}\\}_{j=0}^{k}\subseteq[V,T]$ such that $W_{0}:=T$ and
$W_{j+1}\subset W_{j}$ is minimal integral for any $j\in\\{0,\ldots,k\\}$. The
induction hypothesis holds for $k=0$ by Lemma 3.2. Assume that the induction
hypothesis holds for some $k\in\\{0,\ldots,m-1\\}$. Set
$T^{\prime}:=W_{k}\in[V,T]$; so that, $T^{\prime}\subset S$ is an integral
extension. Moreover, $V\subseteq T^{\prime}\cap U\subseteq T\cap U=V$; so
that, $T^{\prime}\cap U=V$. If $U\in[T^{\prime},S]$, then $T^{\prime}\cap
U=T^{\prime}=V$, and $V\subset S$ is integral. Assume now that
$U\not\in[T^{\prime},S]$. We apply Lemma 3.2 to the extensions
$T^{\prime}\subset S$ which is integral and $U\subset S$ which is minimal
integral. Then, there exists $W_{k+1}\in[V,T^{\prime}]\subseteq[V,T]$ such
that $W_{k+1}\subset T^{\prime}=W_{k}$ is minimal integral. It follows that
the induction hypothesis holds for $k+1$, and we have a decreasing maximal
chain in $[V,T]$ which stops at some $W_{m}=V$ since $R\subset S$ has FCP. In
particular, $V\subset T$ is an integral extension, and so is $V\subset S$. ∎

######  Proposition 3.4.

Let $R\subset S$ be an FCP extension and $T,U\in[R,S]$ such that $T\subset S$
and $U\subset S$ are integral. Then, $T\cap U\subset S$ is integral.

###### Proof.

Set $V:=T\cap U$. If $V\in\\{T,U\\}$, we have the result. Now, assume that
$V\neq T,U$. Let $\\{U_{i}\\}_{i=0}^{n}$ be a maximal chain such that
$U_{0}:=S$ and $U_{n}:=U$. Then, $U_{i+1}\subset U_{i}$ is minimal integral
for each $i\in\\{0,\ldots,n\\}$. Set $V_{i}:=T\cap U_{i}$ for each
$i\in\\{0,\ldots,n\\}$. We are going to prove by induction on $i$, that
$V_{i}\subseteq S$ is integral for any $i\in\\{0,\ldots,n\\}$. The induction
hypothesis holds obviously for $i=0$ and is Lemma 3.3 for $i=1$ if
$U_{1}\not\in[T,S]$. If $U_{1}\in[T,S]$, then $V_{1}=T$; so that,
$V_{1}\subset S$ is integral. Assume that the induction hypothesis holds for
some $i\in\\{0,\ldots,n-1\\}$, that is $V_{i}\subseteq S$ is integral. We have
$V_{i+1}=T\cap U_{i+1}=T\cap U_{i}\cap U_{i+1}=V_{i}\cap U_{i+1}$. Since
$V_{i}\subseteq S$ is integral, so is $V_{i}\subseteq U_{i}$. Moreover,
$U_{i+1}\subset U_{i}$ is minimal integral. Using again Lemma 3.3 for the
extensions $V_{i}\subseteq U_{i}$ and $U_{i+1}\subset U_{i}$ if
$U_{i+1}\not\in[V_{i},U_{i}]$, we get that $V_{i+1}\subset V_{i}$ is integral,
and so is $V_{i+1}\subset S$. If $U_{i+1}\in[V_{i},U_{i}]$, then
$V_{i+1}=U_{i+1}\cap V_{i}=V_{i}$, and then the result also holds. The
induction hypothesis holds for $i+1$, and then for any $i\in\\{0,\ldots,n\\}$,
and in particular for $i=n$; so that, $V\subset S$ is integral. ∎

The previous results combine to yield that, for an FCP extension $R\subset S$,
we can introduce a new closure, namely the co-integral closure as it is shown
in the following Theorem.

######  Theorem 3.5.

Let $R\subset S$ be an FCP extension. There exists a least $T\in[R,S]$ such
that $T\subseteq S$ is integral. We call such $T$ the co-integral closure of
$R\subset S$ and denote it by $\underline{R}$. Moreover,
$\underline{R}=\cap[U\in[R,S]\mid U\subseteq S$ integral$\\}$ and
$\widetilde{R}\subseteq\underline{R}$.

###### Proof.

Set $\mathcal{E}:=\\{T\in[R,S]\mid T\subseteq S$ integral$\\}$. Then
$S\in\mathcal{E}$, and, since $[R,S]$ is an Artinian lattice, $\mathcal{E}$
admits a minimal element. We claim that this minimal element is unique.
Indeed, assume that there exist $T,U$ two distinct minimal elements of
$\mathcal{E}$; so that, $T\neq U$ where $T\subset S$ and $U\subset S$ are
integral extensions. Then, $T\cap U\subset S$ is integral according to
Proposition 3.4, a contradiction with the minimality of $T$. It follows that
the minimal element of $\mathcal{E}$, that we denote $\underline{R}$, is the
least $T\in[R,S]$ such that $T\subseteq S$ is integral. In particular,
$\underline{R}=\cap[U,\in[R,S]\mid U\subseteq S$ integral$\\}$ and any
$T\in[\underline{R},S]$ is such that $T\subseteq S$ is integral.

Now, we show that $\widetilde{R}\subseteq\underline{R}$. Let $V$ be the
integral closure of $R$ in $\underline{R}$ and set $W:=\widetilde{R}V$. We
have the following commutative diagram:

$\begin{matrix}\widetilde{R}&\to&W=\widetilde{R}V&\to&S\\\
\uparrow&{}&\uparrow&{}&\uparrow\\\ R&\to&V&\to&\underline{R}\end{matrix}$

Since $R\subseteq V$ is integral, so is $\widetilde{R}\subseteq W$ and
$R\subseteq W$ is almost-Prüfer. According to [21, Proposition 4.16],
$R_{P}\subseteq W_{P}$ is either integral $(*)$ or Prüfer $(**)$ for each
$P\in\mathrm{Spec}(R)$. In case $(*)$, we get $R_{P}=(\widetilde{R})_{P}$; so
that, $V_{P}=W_{P}$. In case $(**)$, we get $R_{P}=V_{P}$; so that,
$(\widetilde{R})_{P}=W_{P}$, leading to $V_{P}\subseteq W_{P}$ is Prüfer.
Then, $V\subseteq W$ is Prüfer, as is $V\subseteq\underline{R}$, and so is
$V\subseteq W\underline{R}$ by [11, Corollary 5.11, page 53]. Since
$V\subseteq\underline{R}\subseteq W\underline{R}$ holds, we deduce that
$\underline{R}\subseteq W\underline{R}$ is Prüfer. But,
$W\underline{R}\in[\underline{R},S]$ implies that $\underline{R}\subseteq
W\underline{R}$ is integral, giving $\underline{R}=W\underline{R}$. To end,
$\widetilde{R}\subseteq W\subseteq W\underline{R}=\underline{R}$. ∎

######  Corollary 3.6.

Let $R\subset S$ be an FCP extension. Let $T$ be the co-integral closure of
$R\subset S$ and $U$ be the integral closure of $R$ in $T$. Then, $U\subseteq
S$ is almost-Prüfer, and $T$ is the Prüfer hull of $U\subseteq S$.

###### Proof.

Since $U\subseteq T$ is Prüfer and $T\subseteq S$ is integral, we get that
$U\subseteq S$ is almost-Prüfer, and $T$ is the Prüfer hull of $U\subseteq S$.
∎

######  Theorem 3.7.

Let $R\subset S$ be an FCP extension. Then, $R\subseteq S$ is almost-Prüfer if
and only if $\widetilde{R}=\underline{R}$.

###### Proof.

Assume that $\widetilde{R}=\underline{R}$. Then, $R\subseteq\widetilde{R}$ is
Prüfer and $\widetilde{R}=\underline{R}\subseteq S$ is integral; so that,
$R\subseteq S$ is almost-Prüfer.

Conversely, assume that $R\subseteq S$ is almost-Prüfer. According to [21,
Proposition 4.17], $\widetilde{R}$ is the least $T\in[R,S]$ such that
$T\subseteq S$ is integral, which implies that $\widetilde{R}=\underline{R}$.
∎

The previous Theorem shows that for an FCP extension $R\subseteq S$ which is
not almost-Prüfer, then $\widetilde{R}\subset\underline{R}$. In particular,
$R\subset\underline{R}$ is not Prüfer, and then not integrally closed, because
an integrally closed FCP extension is Prüfer. This situation holds in Example
3.9.

If $R\subseteq S$ is an almost-Prüfer extension, the closures $\overline{R}$
and $\underline{R}$ are linked as we can see in the following Proposition
which improves [26, Theorem 6.13], since $\widetilde{R}=\underline{R}$.

######  Proposition 3.8.

If $R\subset S$ is an almost-Prüfer FCP extension, there are order-
isomorphisms:

  1. (1)

$\varphi:[R,\overline{R}]\to[\underline{R},S]$ defined by
$\varphi(T):=T\underline{R}$ for any $T\in[R,\overline{R}]$, and whose inverse
$\psi:[\underline{R},S]\to[R,\overline{R}]$ is defined by
$\psi(U):=U\cap\overline{R}$ for any $U\in[\underline{R},S]$.

  2. (2)

$\theta:[R,\widetilde{R}]\to[\overline{R},S]$ defined by
$\theta(T):=T\overline{R}$ for any $T\in[R,\widetilde{R}]$, and whose inverse
$\rho:[\overline{R},S]\to[R,\widetilde{R}]$ is defined by
$\rho(U):=U\cap\widetilde{R}$ for any $U\in[\overline{R},S]$.

  3. (3)

$\gamma:[R,S]\to[R,\overline{R}]\times[R,\widetilde{R}]$ defined by
$\gamma(W):=(W\cap\overline{R},W\cap\widetilde{R})$ for any $W\in[R,S]$, and
whose inverse $\delta:[R,\overline{R}]\times[R,\widetilde{R}]\to[R,S]$ is
defined by $\delta((W,W^{\prime})):=WW^{\prime}$ for any
$(W,W^{\prime})\in[R,\overline{R}]\times[R,\widetilde{R}]$.

###### Proof.

Since $R\subset S$ is an almost-Prüfer FCP extension, [26, Proposition 4.3]
says that $R\subset S$ splits at $\widetilde{R}$ and $\overline{R}$ with
$\widetilde{R}=(\overline{R})^{o}$ and Theorem 3.7 asserts that
$\underline{R}=\widetilde{R}$.

(1) According to [26, Theorem 6.13 (4)], the map
$\varphi:[R,\overline{R}]\to[\underline{R},S]$ is an order-isomorphism.
Moreover, $\psi:[\underline{R},S]\to[R,\overline{R}]$ defined by
$\psi(U):=U\cap\overline{R}$ clearly exists and preserves order. Let
$T\in[R,\overline{R}]$ and $U\in[\underline{R},S]$. Then,
$T\cap\widetilde{R}\subseteq\overline{R}\cap\widetilde{R}=R$ and
$S=\widetilde{R}\overline{R}\subseteq U\overline{R}$. It follows that
$T=T\underline{R}\cap\overline{R}=\psi[\varphi(T)]$ and
$U=(U\cap\overline{R})\underline{R}=\varphi[\psi(U)]$, a consequence of [21,
Proposition 4.21]. Then $\psi$ is an order-isomorphism such that
$\psi={\varphi}^{-1}$.

(2) Similarly $\theta:[R,\widetilde{R}]\to[\overline{R},S]$ is an order-
isomorphism by [26, Theorem 6.13 (3) ] and
$\rho:[\overline{R},S]\to[R,\widetilde{R}]$ defined by
$\rho(U):=U\cap\widetilde{R}$ exists and preserves order. Let
$T\in[R,\widetilde{R}]$ and $U\in[\overline{R},S]$. Then,
$T\cap\overline{R}\subseteq\overline{R}\cap\widetilde{R}=R$ and
$S=\widetilde{R}\overline{R}\subseteq U\widetilde{R}$. It follows that
$U=(U\cap\widetilde{R})\overline{R}=\theta[\rho(U)]$, a consequence of [21,
Proposition 4.21]. Now, [21, Corollary 4.3] implies that
$T=T\overline{R}\cap\widetilde{R}=\rho[\theta(T)]$. Then $\rho$ is an order-
isomorphism such that $\rho={\theta}^{-1}$.

(3) According to [26, Theorem 6.13 (5)],
$\delta^{\prime}:[R,\widetilde{R}]\times[R,\overline{R}]\to[R,S]$ defined by
$\delta^{\prime}((W,W^{\prime})):=WW^{\prime}$ for any
$(W,W^{\prime})\in[R,\widetilde{R}]\times[R,\overline{R}]$ is an order-
isomorphism, and so is obviously
$\delta:[R,\overline{R}]\times[R,\widetilde{R}]\to[R,S]$ defined by
$\delta((W,W^{\prime})):=WW^{\prime}$ for any
$(W,W^{\prime})\in[R,\overline{R}]\times[R,\widetilde{R}]$. Moreover, the map
$\gamma:[R,S]\to[R,\overline{R}]\times[R,\widetilde{R}]$ defined by
$\gamma(W):=(W\cap\overline{R},W\cap\widetilde{R})$ for any $W\in[R,S]$ exists
and preserves order.

Let $W\in[R,S]$. Then,
$U:=\delta[\gamma(W)]=(W\cap\overline{R})(W\cap\widetilde{R})\ (*)$. Taking
into account [21, Proposition 4.18] and localizing $(*)$ at any
$M\in\mathrm{MSupp}(S/R)=\mathrm{MSupp}(\overline{R}/R)\cup\mathrm{MSupp}(\widetilde{R}/R)$,
with
$\mathrm{MSupp}(\overline{R}/R)\cap\mathrm{MSupp}(\widetilde{R}/R)=\emptyset$,
we get the following:

If $M\in\mathrm{MSupp}(\overline{R}/R)$, then $R_{M}=(\widetilde{R})_{M}$ and
$S_{M}=(\overline{R})_{M}$; so that, $U_{M}=W_{M}$. If
$M\in\mathrm{MSupp}(\widetilde{R}/R)$, then $R_{M}=(\overline{R})_{M}$ and
$S_{M}=(\widetilde{R})_{M}$; so that, $U_{M}=W_{M}$. It follows that $U=W$ and
$\delta[\gamma(W)]=W$ for any $W\in[R,S]$. Then, $\delta\circ\gamma$ is the
identity on $[R,S]$. Since $\delta$ is an order-isomorphism, so is
$\gamma=\delta^{-1}$. ∎

In the context of algebraic orders, we built an FIP extension $R\subset K$
which is not almost-Prüfer [26, Example 6.15], but where we exhibit its
almost-Prüfer closure. As the co-integral closure was not still defined, we
recall this example here and show that $K$ is the co-integral closure of the
extension.

######  Example 3.9.

[26, Example 6.15] Let $K$ be an algebraic number field, $T^{\prime}$ be its
ring of algebraic integers, and $R^{\prime}$ be an algebraic order of $K$ such
that $R^{\prime}\subset T^{\prime}$ is a minimal inert extension. Set
$M^{\prime}:=(R^{\prime}:T^{\prime})\in\mathrm{Max}(R^{\prime})$; so that,
$M^{\prime}\in\mathrm{Max}(T^{\prime})$. Let
$N^{\prime}\in\mathrm{Max}(R^{\prime}),\ M^{\prime}\neq N^{\prime}$. Set
$\Sigma:=R^{\prime}\setminus(M^{\prime}\cup N^{\prime})$, which is a
multiplicatively closed subset. Then, $R:=R^{\prime}_{\Sigma}$ is a semilocal
ring with two maximal ideals $M:=M^{\prime}_{\Sigma}$ and
$N:=N^{\prime}_{\Sigma}$. Moreover $T:=T^{\prime}_{\Sigma}$ is also a
semilocal ring with two maximal ideals, and such that $R\subset T$ is a
minimal inert extension with $M:=M^{\prime}_{\Sigma}=(R:T)\in\mathrm{Max}(T)$.
The other maximal ideal of $T$ is $P:=NT$. In particular, $T$ is a Prüfer
domain with quotient field $K$, $T\subset T_{M}$ is minimal Prüfer with
$P=\mathcal{C}(T,T_{M})$ and $\\{N\\}=\mathrm{MSupp}_{R}(T_{M}/T)$. Moreover,
$T$ is the integral closure of $R\subset K$, and $R\subset K$ is an FCP
extension. There is a commutative diagram with
$[R,K]=\\{R,T,R_{M},T_{M},T_{P},K\\}$:

$\begin{matrix}{}&{}&R_{M}&{}&{}&{}&{}\\\ {}&\nearrow&{}&\searrow&{}&{}&{}\\\
R&{}&{}&{}&T_{M}&\to&K\\\ {}&\searrow&{}&\nearrow&{}&\nearrow&{}\\\
{}&{}&T&\to&T_{P}&{}&{}\end{matrix}$

where $R\subset T_{M}$ is almost-Prüfer with $R_{M}\subset T_{M}$ minimal
inert as $R\subset T$, and $R\subset R_{M}$ is Prüfer minimal as $T\subset
T_{M},\ T_{M}\subset K,\ T\subset T_{P}$ and $T_{P}\subset K$.

When considering the extension $R\subset K$, obviously, there does not exist
some $V\in[R,K[$ such that $V\subset K$ is integral, which implies that $K$ is
the co-integral closure of $R\subset K$. And $R_{M}$ is the Prüfer hull of
$R\subset K$, and also of $R\subset T_{M}$. Then, as it is said after Theorem
3.7, the co-integral closure and the Prüfer hull do not always coincide. But,
in the present example, as $R\subset T_{M}$ is almost-Prüfer, $R_{M}$ is also
the co-integral closure of $R\subset T_{M}$.

######  Example 3.10.

Let $R\subseteq S$ be an extension. Each $T\in[R,S]$ such that $T\subseteq S$
is integrally closed contains $\overline{R}$. Dually, things are not so
simple. For an extension $R\subseteq S$, we set $[R,S]_{IC}:=\\{T\in[R,S]\mid
T\cap\overline{R}=R\\}$, i.e. $T\in[R,S]_{IC}$ if and only if $R\subseteq T$
is integrally closed. In particular, $\widetilde{R}\in[R,S]_{IC}$. Clearly $R$
belongs to $[R,S]_{IC}$ and this set is $\cup$-inductive in the poset
$([R,S];\subseteq)$. By Zorn Lemma, it contains maximal elements. It also
contains $T\in[R,S]$ such that $T\setminus R$ is a mcs by [29, Théorème 1(d)].

Let $\mathcal{M}$ be a maximal element of $[R,S]_{IC}$ and $s\in
S\setminus\mathcal{M}$. Then $\mathcal{M}\subset\mathcal{M}[s]$ cannot be
integrally closed because otherwise $R\subseteq\mathcal{M}[s]$ would be
integrally closed. It follows that there is some
$p(s)\in\mathcal{M}[s]\setminus\mathcal{M}$ which is integral over
$\mathcal{M}$, and then $s$ is algebraic over $\mathcal{M}$. Therefore, $S$ is
algebraic over $\mathcal{M}$. If we choose some maximal $\mathcal{M}$,
containing $\widetilde{R}$, we get a tower
$R\subseteq\widetilde{R}\subseteq\mathcal{M}\subseteq S$, where the last
extension is algebraic. In particular, for each $s\in S$, there is some
$m\in\mathcal{M}$ such that $ms$ is integral over $\mathcal{M}$.

There exist some extensions $R\subseteq S$ such that $[R,S]_{IC}$ has a
greatest element, for example if the extension is chained. Other examples are
quasi-Prüfer extensions like FCP extensions, because such extensions are
integrally closed if and only if they are Prüfer and subextensions of quasi-
Prüfer extensions are quasi-Prüfer [21, Corollary 3.3]. In this case the
greatest element of $[R,S]_{IC}$ is $\widetilde{R}$. Actually,
$\widetilde{R}\subseteq S$, is quasi-Prüfer, whence a $\mathcal{P}$-extension;
that is, each element $s\in S$ is a zero of a polynomial $p(X)$ over
$\widetilde{R}$, whose content is $\widetilde{R}$. In particular $S$ is
algebraic over $\widetilde{R}$.

Another example is given by “completely” distributive extensions $R\subseteq
S$ (extensions such that for each $V\in[R,S]$ and for any family
$\\{W_{i}\\}_{i\in I}$ of elements of $[R,S]$, we have $(\prod_{i\in
I}W_{i})\cap V=\prod_{i\in I}(W_{i}\cap V)$). The greatest element of
$[R,S]_{IC}$ is the product of all elements of $[R,S]_{IC}$. Denote by
$\\{T_{i}|i\in I\\}$ the elements of $[R,S]_{IC}$, by $U$ their product and
let $T$ be one of them. Then it is enough to show that $U$ is integrally
closed over $R$, because $T\subseteq U$. But we have
$U\cap\overline{R}=\prod_{i\in I}(T_{i}\cap\overline{R})=R$, which proves that
$U$ is integrally closed over $R$.

Note that the same proof holds if $R\subseteq S$ is distributive and has FCP
(whence has FIP by [23, Lemma 2.10]).

If $[R,S]_{IC}=\\{T_{1},\ldots,T_{n}\\}$, we thus get that
$\widetilde{R}=T_{1}\cdots T_{n}$.

We can also consider for an extension $R\subseteq S$, the set of all
$T\in[R,S]$ such that $R\subseteq T$ is seminormal (resp.; t-closed,
u-closed). An element $T\in[R,S]$ is such that $R\subseteq T$ is seminormal
(resp.; t-closed, u-closed) if and only if ${}^{+}_{S}R\cap T=R$ (resp.;
${}^{t}_{S}R\cap T;{}^{u}_{S}R\cap T=R$). Then most of the preceding
considerations hold.

##  4\. Radicial and separable ring extensions

We first introduce some notation. If $\Pi$ is a property of field extensions,
we say that a ring extension is a $\kappa$-$\Pi$-extension if all its residual
field extensions verify $\Pi$. In the sequel of the section, we will consider
the separable and radicial properties, where in this paper a purely
inseparable field extension is called radicial.

######  Example 4.1.

An FCP t-closed extension $R\subset S$ is $\kappa$-catenarian if and only if
$R\subset S$ is catenarian. Indeed, according to [25, Proposition 4.3],
$R\subset S$ is catenarian if and only if $R_{M}\subset S_{M}$ is catenarian
for each $M\in\mathrm{MSupp}(S/R)$. Since $R\subset S$ is an FCP t-closed
extension, we infer from property (${\dagger}$) of Lemma 2.10 that for each
$M\in\mathrm{MSupp}(S/R)$, there exists a unique $N\in\mathrm{Max}(S)$ lying
above $M$. Moreover, $(R_{M}:S_{M})=MR_{M}=NS_{N}=NS_{M}$ and
$\kappa(N)=S_{N}/(R_{M}:S_{M})$. Set $I:=(R_{M}:S_{M})$. In particular,
$\kappa(M)=R_{M}/I$ and $\kappa(N)=S_{N}/I$. Then, [25, Corollary 3.7] shows
that $R_{M}\subset S_{M}$ is catenarian for each
$M\in\mathrm{MSupp}(S/R)\Leftrightarrow R_{M}/I\subset S_{M}/I$ is catenarian
for each $M\in\mathrm{MSupp}(S/R)\Leftrightarrow\kappa(M)\subset\kappa(N)$ is
catenarian for each $M\in\mathrm{MSupp}(S/R)\Leftrightarrow\kappa(N\cap
R)\subseteq\kappa(N)$ is catenarian for each $N\in\mathrm{Max}(S)$ (because of
the bijection $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$) $\Leftrightarrow R\subset
S$ is $\kappa$-catenarian.

Kubota introduced the radicial closure of a ring extension, focussing his
results on integral ring extensions [12]. Later on Manaresi reintroduced this
notion and named it weak normalization [14]. We keep the notation of Manaresi
(but not her terminology): for a ring extension $R\subseteq S$, that we will
suppose integral for the rest of the Section, we denote by ${}^{*}_{S}R$ the
radicial closure of $R$ in $S$. We recall from [14] or [12], that
${}^{*}_{S}R:=\\{x\in S\mid x\otimes 1-1\otimes
x\in\mathrm{Nil}(S\otimes_{R}S)\\}$. Moreover, ${}^{*}_{S}R$ is the greatest
subextension $R\subseteq T$ of $R\subseteq S$, such that $R\subseteq T$ is a
radicial extension; that is, the ring morphism $R\to T$ is universally an
$i$-morphism, or equivalently, an $i$-morphism which is $\kappa$-radicial [9,
Definition 3.7.2, p. 248] and [12, Proposition 1]. For instance, a subintegral
extension is radicial.

Yanagihara proved that a ring extension $R\subseteq S$ is radicial if and only
if $S$ is the filtered union of all $U\in[R,S]$ such that $R\subseteq U$ is a
finite tower of finitely many subextensions of the type $T\subseteq T[x]$,
with $x\in S$ and $T\in[R,S]$, such that either $x^{p}\in T$ and $px\in T$ for
some prime integer $p$ or $x^{2},x^{3}\in T$ [31, Theorem 2].

Let $R\subset S$ be an integral ring extension. In [27], the second author
says that $R\subset S$ is weakly infra-integral if it is $\kappa$-radicial.
She defined the strong t-closure, that is here termed the $\kappa$-radicial
closure, of a ring extension $R\subseteq S$ as the greatest subalgebra
$V:={}^{\circ}_{S}R\in[R,S]$ such that $R\subseteq V$ is $\kappa$-radicial
[27, Definition 3.1]. In particular, ${}^{*}_{R}S\subseteq{}^{\circ}_{S}R$.

Let $f:R\to S$ be a ring morphism and its co-diagonal ring morphism
$\bigtriangledown_{f}:S\otimes_{R}S\to S$. Denote by $I$ the kernel of the
surjective ring morphism $\bigtriangledown_{f}$. If $f$ is of finite type as
an algebra, then $I$ is an ideal of finite type and $I=I^{2}$ if and only if
$I$ is generated by an idempotent. We recall that a ring morphism $f:R\to S$
is called formally unramified if $\Omega_{R}(S):=I/I^{2}$, the $S$-module of
differentials of $f$, is zero and unramified if it is formally unramified and
of finite type. We see that a ring morphism of finite type $f:R\to S$ is
unramified if and only if $I$ is generated by an idempotent. In this case, $f$
is called separable by some authors because $\bigtriangledown_{f}$ defines $S$
as an $S\otimes_{R}S$-projective module, [28, Proposition 9, p. 32, Chapitre
III]. We will not use this terminology. But in this paper we will say that an
integral extension $R\subset S$ is $\kappa$-separable if its residual
extensions are separable, in which case $R\subseteq U$ and $U\subseteq S$ are
$\kappa$-separable for any $U\in[R,S]$. To check that an integral FCP
extension $R\subseteq S$ with conductor $I$ is $\kappa$-separable it is enough
to verify that the extensions $\kappa(Q\cap R)\to\kappa(Q)$ are separable for
each maximal ideal $Q\in\mathrm{V}_{S}(I)$ by Proposition 2.9. Note that
$R\subseteq U$ is $\kappa$-separable for any $U\in[R,{}^{t}_{S}R]$. Note also
that a $\kappa$-radicial and $\kappa$-separable extension is infra-integral.

The following Proposition belongs to the algebraic geometry folklore, but we
can not give any reference. So we give an outline of its proof.

######  Proposition 4.2.

A finite type extension $R\subseteq S$ is a ring epimorphism if and only if it
is radicial and unramified. Therefore, an integral FCP extension is trivial
when it is radicial and unramified.

###### Proof.

One implication is clear because an epimorphism is an $i$-extension and the
residual extensions are isomorphisms [13, Proposition 1.5, page 109]. Now if
$R\subseteq S$ is radicial and unramified, then the kernel of
$\bigtriangledown:S\otimes_{R}S\to S$ is generated by an idempotent which is
nilpotent. Therefore $\bigtriangledown$ is an isomorphism and the extension is
an epimorphism [13, Proposition 1.5, p.109]. The last assertion is a
consequence of the following facts: each subextension of an integral FCP
extension is (module) finite and a module finite ring epimorphism is
surjective [13, Proposition 1.7, p.111]. ∎

We have the following results that may not hold if the extension is not FCP:

######  Proposition 4.3.

Let $R\subseteq S$ be an integral FCP extension. The following statements are
equivalent:

  1. (1)

The extension is unramified;

  2. (2)

For all $Q\in\mathrm{Spec}(S)$, the ring morphism $R_{R\cap Q}\to S_{Q}$ is
formally unramified;

  3. (3)

$QS_{Q}=PS_{Q}$ and $\kappa(P)\to\kappa(Q)$ is a finite separable field
extension for each $Q\in\mathrm{Spec}(S)$ and $P:=Q\cap R$;

  4. (4)

The extension is $\kappa$-separable and for each $Q\in\mathrm{Spec}(S)$ and
$P:=Q\cap R$, the ideal $PS_{P}$ of $S_{P}$ is semiprime;

  5. (5)

$\kappa(P)\to\kappa(P)\otimes_{R}S$ is unramified for each
$P\in\mathrm{Spec}(R)$.

It follows that an unramified extension is universally $\kappa$-separable.

###### Proof.

(1) $\Leftrightarrow$ (2) is [10, Exercise 1.14 (vi)]. To prove that (1)
$\Leftrightarrow$ (4), we use [28, Exercise (1), p.38] which states that a
ring morphism of finite type $R\to S$ is unramified if and only if
$QS_{Q}=PS_{Q}$ and $\kappa(P)\to\kappa(Q)$ is a finite separable field
extension for each $Q\in\mathrm{Spec}(S)$ and $P:=Q\cap R$, which proves (1)
$\Leftrightarrow$ (3). We next prove that (3) $\Leftrightarrow$ (4). Suppose
that the extension is unramified, we can assume that $R$ is local with maximal
ideal $P$. Then $R/P\subseteq S/PS$ is étale over the field $R/P$. It follows
that $S/PS$ is reduced, and $PS$ is semiprime by [28, Proposition 1, page 74].
To prove the converse, if $PS$ is semiprime, we observe that $PS$ is the
intersection of the (finitely many) maximal ideals of $S$ lying over $P$,
because $R\subseteq S$ is finite. It is now easy to show that $PS_{Q}=QS_{Q}$.

Now (1) $\Leftrightarrow$ (5) because of [10, Property 2.2.vi, page 8] and the
fact that the unramified property is stable under any base change. ∎

######  Corollary 4.4.

Let $R\subseteq S$ be an integral FCP t-closed extension. Then, $R\subseteq S$
is unramified if and only if $R\subseteq S$ is $\kappa$-separable.

###### Proof.

According to Proposition 4.3, $R\subseteq S$ is unramified if and only if
$QS_{Q}=PS_{Q}$ and $\kappa(P)\to\kappa(Q)$ is a finite separable field
extension for each $Q\in\mathrm{Spec}(S)$ and $P:=Q\cap R$. But the condition
(${\dagger}$) of Lemma 2.10 shows that $QS_{Q}=PS_{Q}$ for each
$Q\in\mathrm{Spec}(S)$ and $P:=Q\cap R$. Then, $R\subseteq S$ is unramified
$\Leftrightarrow\kappa(P)\to\kappa(Q)$ is a finite separable field extension
for each $Q\in\mathrm{Spec}(S)$ and $P:=Q\cap R\Leftrightarrow R\subseteq S$
is $\kappa$-separable. ∎

The following facts are known. If $R\subset S\subset T$ is an unramified tower
of extension, then $T\subset S$ is unramified and a tower of unramified
extensions is unramified. But $R\subset S$ does not need to be unramified,
except if $R$ is a field [10, Lemme 3.4, p.11]. However, we can provide the
following result that can be used on a tower of finitely many minimal
extensions, the last one being pure. Recall that a ring morphism $f:R\to S$ is
called pure if $R^{\prime}\to R^{\prime}\otimes_{R}S$ is injective for each
ring morphism $R\to R^{\prime}$ (or equivalently, $M\to M\otimes_{R}S$ is
injective for each $R$-module $M$). If $f$ is pure, then $f^{-1}(IS)=I$ for
each ideal $I$ of $R$; so that, $f$ is injective. Moreover, a pure ring
morphism $R\to S$ verifies the condition (O), i.e. an $R$-module $M$ is zero
if $M\otimes_{R}S=0$.

######  Proposition 4.5.

Let $R\subset S\subset T$ be an unramified FCP extension, such that $S\subset
T$ is pure. Then $R\subset S$ is unramified.

###### Proof.

Consider a prime ideal $P$ of $R$ and the fiber morphism
$\kappa(P)\to\kappa(P)\otimes_{R}S\to\kappa(P)\otimes_{R}T$. The last morphism
is injective because deduced from the pure morphism $S\subset T$ by the base
change $S\to\kappa(P)\otimes_{R}S$. Now it is enough to use Proposition
4.3(5), because over a field a subextension of an unramified extension is
unramified [10, Lemma 3.4, p.11]. ∎

The next result gives a necessary and sufficient condition in order that the
above result holds.

######  Proposition 4.6.

Let $R\subseteq S$ be an integral seminormal FCP extension and $T\in[R,S]$.
Then $R\subseteq S$ is unramified if and only if $R\subseteq T$ and
$T\subseteq S$ are unramified.

###### Proof.

We need only to prove that if $R\subseteq S$ is unramified then so is
$R\subseteq T$. Indeed the unramified property is stable under composition and
stable under left division, without any other hypothesis by [10, Properties
2.2.ii and 2.2.v, p. 8]. Let $Q\in\mathrm{Spec}(T),\ P:=Q\cap R$ and let
$M\in\mathrm{Spec}(S)$ be lying over $Q$; so that, $P=R\cap M$. Since
$\kappa(P)\to\kappa(M)$ is a separable field extension, so is
$\kappa(P)\to\kappa(Q)$. Then, according to Proposition 4.3, to show that
$R\subseteq T$ is unramified, it is enough to show that $PT_{P}$ is semiprime
in $T_{P}$, that is an intersection of prime ideals of $T_{P}$. But
$R_{P}\subseteq T_{P}$ is still a seminormal FCP integral extension with
$R_{P}$ a local ring. It follows that we may assume that $(R,P)$ is a local
ring, and we have to show that $PT$ is semiprime in $T$. Set $C:=(R:T)$, which
is an intersection of maximal ideals of $T$ and $R$, according to Proposition
2.8; so that, $C=P$. To conclude, $R\subseteq T$ is unramified. ∎

By [16, Lemma 2.2], if $I$ is the conductor ideal of an extension $R\subseteq
S$, then $R\subseteq S$ is unramified if and only if $R/I\subseteq S//I$ is
unramified. It follows that an inert minimal extension $R\subset S$ with
conductor $M$ is unramified if and only if the field extension $R/M\subset
S/M$ is separable, in which case the extension $R\subseteq S$ is called
separable inert.

######  Remark 4.7.

(1) In the previous Proposition, the condition that $R\subset S$ is seminormal
is necessary. Indeed, assume that $R\subset S$ is an integral FCP extension
which is not seminormal. Then, $R\neq{}^{+}_{S}R$, and there exists
$T\in[R,{}^{+}_{S}R]$ such that $R\subset T$ is minimal ramified. It follows
that $P:=(R:T)\in\mathrm{Max}(R)$. Let $Q\in\mathrm{Max}(T)$ be the (unique
because of Theorem 1.3) maximal ideal of $T$ lying above $P$ and let
$M\in\mathrm{Max}(S)$ be lying above $P$, and then above $Q$. Even if
$R\subset S$ is unramified; so that, $\kappa(P)\to\kappa(M)$ is separable by
Proposition 4.3, and then $\kappa(P)\to\kappa(Q)$ is separable, $R\subset T$
is not unramified. Indeed, $R_{P}\subset T_{P}$ is still minimal ramified with
$PR_{P}=PT_{P}=PT_{Q}\subset QT_{P}=QT_{Q}$ because $T_{P}=T_{Q}$ by [2,
Proposition 2, page 40].

(2) The situation of (1) occurs in the following example: Let $(R,M)$ be a
SPIR such that $M^{2}=0$. Set $S:=R^{2}$ and $T:={}^{+}_{S}R=R+(M\times M)$ by
[20, Proposition 2.8]. Then, $R\subset T$ is minimal ramified and $T\subset S$
is minimal decomposed because $\mathrm{Max}(S)=\\{R\times M,M\times R\\}$; so
that, $R\subset S$ is infra-integral, with residual extensions which are
isomorphisms, and then separable. To show that $R\subset S$ is unramified,
using Proposition 4.3 (4), it is enough to show that $MS$ is semiprime in $S$,
which is obvious since $MS=M\times M=(M\times R)\cap(R\times M)$. Then,
$R\subset S$ is unramified although $R\subset T$ is not.

(3) But there exist unramified extensions $R\subset U$, that cannot be
seminormal, with $T,S\in]R,U[$ such that $R\subset S,\ S\subset U,\ T\subset
U$ are unramified and $R\subset T$ is not unramified.

We keep the hypotheses of the preceding example. Let $R\subset V$ be a minimal
inert separable extension and set $M:=Rt$. It is always possible to build such
an extension. It is enough to choose the SPIR $(R,M)$ such that $k:=R/M$ has a
minimal separable field extension $K:=k[x]=k[X]/(f(X))$ and to set
$V:=R[X]/(tX,g(X)))$, where $f(X)=\bar{g}(X)\in k[X]$ is the monic polynomial
whose coefficients are the classes of coefficients of $g(X)$. Then, $R\subset
V$ is minimal inert separable. Now, set $U:=V\times R$. According to [3,
Proposition III.4], $S\subset U$ is also minimal inert separable (with
$(S:U)=M\times R$, giving the type of the minimal extension); so that,
$R\subset U$ is unramified, as the composite of two unramified extensions
$R\subset S$ and $S\subset U$. However, we still have $R\subset T$ not
unramified and $T\subset U$ unramified, since so are $T\subset S$ and
$S\subset U$.

Looking at the preceding example, we may ask under what conditions an
unramified extension is a tower of only unramified minimal extensions. We
begin by characterizing minimal unramified extensions.

######  Proposition 4.8.

Let $R\subset S$ be a minimal extension. Then, $R\subset S$ is unramified if
and only if $R\subset S$ is either decomposed or separable inert or minimal
Prüfer.

###### Proof.

We first observe that a ramified minimal extension $R\subseteq S$ with
conductor $M$ a maximal ideal is luckily not unramified, otherwise
$R/M\to(R/M)[X]/(X^{2})$ would be étale and then $(R/M)[X]/(X^{2})$ reduced.

Now a decomposed extension is modulo its conductor ideal of the form
$k\subseteq k\times k$, where $k$ is a field, whence is unramified.

A minimal inert separable extension is unramified, as we said before Remark
4.7.

At last, a minimal Prüfer extension is unramified because it is an epimorphism
by Proposition 4.2. ∎

######  Proposition 4.9.

Let $R\subset S$ be an FCP integral extension. The following statements are
equivalent:

  1. (1)

$R\subset S$ is a seminormal $\kappa$-separable extension;

  2. (2)

$R\subset S$ is a tower of finitely many decomposed minimal extensions and
inert separable extensions.

If these conditions hold, then $R\subset S$ is unramified.

###### Proof.

$R\subset S$ is a seminormal extension if and only if $R\subset S$ is a tower
of decomposed minimal extensions and inert extensions by Proposition 2.6,
which proves a part of the equivalence.

Assume that the previous conditions hold. Let $R_{0}:=R\subset\ldots\subset
R_{i}\subset\ldots\subset R_{n}:=S$ be a maximal chain where $R_{i}\subset
R_{i+1}$ is minimal for each $i\in\\{0,\ldots,n-1\\}$. Let
$Q\in\mathrm{Spec}(S)$. Set $P:=Q\cap R$ and $P_{i}:=Q\cap R_{i}$ for each
$i\in\\{0,\ldots,n-1\\}$; so that, $P_{i}=P_{i+1}\cap R_{i}$. Then,
$\kappa_{R_{i}}(P_{i})\to\kappa_{R_{i+1}}(P_{i+1})$ is either an isomorphism
or a minimal field extension. In this last case,
$\kappa_{R_{i}}(P_{i})\to\kappa_{R_{i+1}}(P_{i+1})$ is either radicial or
separable. It follows that $\kappa_{R}(P)\to\kappa_{S}(Q)$ is separable if and
only if $\kappa_{R_{i}}(P_{i})\to\kappa_{R_{i+1}}(P_{i+1})$ is either an
isomorphism or a minimal separable field extension for each
$i\in\\{0,\ldots,n-1\\}$ if and only if $R_{i}\subset R_{i+1}$ is either a
decomposed minimal extension or an inert separable minimal extension for each
$i\in\\{0,\ldots,n-1\\}$.

The last result comes from Propositions 4.8 and 4.6. ∎

We will define the unramified closure of an FCP integral extension in Section
6. We are now going to show the existence of the $\kappa$-separable closure of
an FCP integral ring extension $R\subseteq S$ as the greatest subalgebra
$V:={}^{\square}_{S}R\in[R,S]$ such that $R\subseteq V$ is $\kappa$-separable.

######  Remark 4.10.

Contrary to the unramified context, we can note that the $\kappa$-separable
and $\kappa$-radicial properties verify, due to Field Theory, and without any
additional hypotheses, the following statement. Let $R\subset S\subset T$ be a
tower of integral FCP ring extensions, then $R\subset T$ has the
$\kappa$-separable (or the $\kappa$-radicial) property if and only if so do
$R\subset S$ and $S\subset T$.

######  Lemma 4.11.

If $R\subset S$ is an integral FCP t-closed extension, there exists a greatest
subalgebra $V:={}^{\square}_{S}R\in[R,S]$ such that $R\subseteq V$ is
$\kappa$-separable.

###### Proof.

We begin to assume that $(R,M)$ is a local ring, and so is $(S,M)$ by [4,
Proposition 4.10]. Since $M=(R:S)$, there is an order preserving bijection
$[R,S]\to[R/M,S/M]$ defined by $T\mapsto T/M$ according to [3, Lemma II.3].
But $R/M\subset S/M$ is an algebraic field extension, and admits a separable
closure $W$ which is the greatest subextension of $R/M\subset S/M$ such that
$R/M\subseteq W$ is a separable algebraic field extension. Then, the previous
bijection shows that there exists a greatest subalgebra
$V:={}^{\square}_{S}R\in[R,S]$ such that $R\subseteq V$ is $\kappa$-separable.

Assume now that $R\subset S$ is an arbitrary integral FCP t-closed extension
and set $\mathrm{MSupp}(S/R):=\\{M_{1},\ldots,M_{n}\\}$. Define
$\varphi:[R,S]\to\prod_{i=1}^{n}[R_{M_{i}},S_{M_{i}}]$ by
$\varphi(T)=(T_{M_{1}},\ldots,T_{M_{n}})$. By [4, Theorem 3.6], $\varphi$ is a
bijection. Let $M_{i}\in\mathrm{MSupp}(S/R)$. Then $R_{M_{i}}\subset
S_{M_{i}}$ is an integral FCP t-closed extension, where
$(R_{M_{i}},M_{i}R_{M_{i}})$ a local ring. The first part of the proof shows
that there exists a greatest subalgebra $V_{i}\in[R_{M_{i}},S_{M_{i}}]$ such
that $R_{M_{i}}\subseteq V_{i}$ is $\kappa$-separable. Let $V$ be the unique
element of $[R,S]$ such that $V_{M_{i}}=V_{i}$ for each $i\in\mathbb{N}_{n}$.
Moreover, $R\subset S$ is an i-extension, being t-closed by Proposition 2.6.
Let $N_{i}$ be the unique maximal ideal of $S$ lying above $M_{i}$; so that,
$S_{M_{i}}\cong S_{N_{i}}$. Then, $\kappa(M_{i})=R/M_{i}\cong
R_{M_{i}}/M_{i}R_{M_{i}}$ and $\kappa(N_{i})=S/N_{i}\cong
S_{M_{i}}/M_{i}R_{M_{i}}$ because $M_{i}R_{M_{i}}=(R_{M_{i}}:S_{M_{i}})$. For
each $i\in\mathbb{N}_{n}$, set $P_{i}:=N_{i}\cap V\in\mathrm{Max}(V)$. Then,
we have the fields extensions
$\kappa(M_{i})\subseteq\kappa(P_{i})\subseteq\kappa(N_{i})$, with
$\kappa(P_{i})=V/P_{i}$ the greatest subalgebra of
$[\kappa(M_{i}),\kappa(N_{i})]$ such that
$\kappa(M_{i})\subseteq\kappa(P_{i})$ is separable. Then, there exists a
$\kappa$-separable closure $V$ of $R\subset S$. ∎

######  Theorem 4.12.

An integral FCP extension $R\subset S$ admits a greatest subalgebra
$V:={}^{\square}_{S}R\in[R,S]$ such that $R\subseteq V$ is $\kappa$-separable.
Moreover, any subextension $U$ of $S$ is $\kappa$-separable over $R$ if and
only if $U\in[R,V]$ and any $Z\in[V,S]$ is such that $Z\subseteq S$ is
$\kappa$-radicial.

###### Proof.

Let $T:={}^{t}_{S}R$ be the t-closure of $R\subset S$. Since $T\subseteq S$ is
t-closed, it admits a $\kappa$-separable closure $V$ by the previous lemma.
Let $P\in\mathrm{Max}(R),\ Q\in\mathrm{Max}(V)$ lying over $P$ and $N:=Q\cap
T$ which is also lying over $P$. Then, $\kappa(P)\cong\kappa(N)$ because
$R\subseteq T$ is infra-integral and $\kappa(N)\subseteq\kappa(Q)$ is
separable by definition of $V$. It follows that $\kappa(P)\subseteq\kappa(Q)$
is separable, and $R\subseteq V$ is $\kappa$-separable. Obviously, any
subextension of $V$ is $\kappa$-separable over $R$. Let $Z\in[V,S]$ and let
$M\in\mathrm{Max}(Z)$ be lying above $Q$, which is also lying over $P$. Then,
$\kappa(Q)\subseteq\kappa(M)$ is radicial by definition of $\kappa(Q)$; so
that, $Z\subseteq S$ is $\kappa$-radicial.

Let $U\in[R,S]$ be such that $R\subset U$ is $\kappa$-separable. If
$U\in[T,S]$, then $T\subseteq U$ is $\kappa$-separable; so that, $U\subseteq
V$. Assume that $U\not\in[T,S]$. We claim that $U\in[R,V]$. Set $W:=U\cap
T\subset T$, which is the t-closure of $R\subset U$. We have the following
commutative diagram:

$\begin{matrix}R&\to&W&\to&U&{}&{}\\\ {}&{}&\downarrow&{}&\downarrow&{}&{}\\\
{}&{}&T&\to&TU&\to&S\end{matrix}$

Since $W\subset T$ is infra-integral, there exists a directed set
$\\{W_{i}\\}_{i\in I}$ of $W$-subalgebras of $T$ such that $W\subset W_{i}$ is
ct-elementary and $T=\cup_{i\in I}W_{i}$ by [15, Definition 1.2]. It obviously
follows that there exists a directed set $\\{U_{i}\\}_{i\in I}$ of
$U$-subalgebras of $TU$ such that $U\subset U_{i}$ is ct-elementary and
$TU=\cup_{i\in I}U_{i}$. Indeed, if $A\subset B$ is a t-elementary
subextension of $[W,T]$, $UA\subset UB$ is a t-elementary subextension of
$[U,TU]$. Therefore, $U\subseteq TU$ is infra-integral, and then
$\kappa$-separable by an above remark. Hence $R\subset UT$ is
$\kappa$-separable, and so is $T\subseteq UT$; so that, $U\subseteq
UT\subseteq V$, and $U\in[R,V]$. ∎

######  Corollary 4.13.

Let $R\subset S$ be an integral FCP extension. Then,
${}^{\square}_{S}R\subseteq S$ is radicial and
${}^{t}_{S}R={}^{\square}_{S}R\cap{}^{\circ}_{S}R$.

###### Proof.

Set $T:={}^{t}_{S}R,\ V:={}^{\square}_{S}R$ and $U:={}^{\circ}_{S}R$. Let
$Q\in\mathrm{Max}(S),\ P:=Q\cap T$ and $M:=Q\cap R=P\cap R$. Since $R\subseteq
T$ is infra-integral, we have $\kappa(M)\cong\kappa(P)$. But we also have
$V={}^{\square}_{S}T$ by the proof of Theorem 4.12. Hence, $N:=Q\cap V$ is the
unique maximal ideal of $V$ lying over $P$. Moreover, $V_{P}/PT_{P}=\kappa(N)$
is the separable closure of the field extension $\kappa(P)\subseteq\kappa(Q)$
as a consequence of the construction made in Lemma 4.11; so that,
$\kappa(N)\subseteq\kappa(Q)$ is radicial. At last,
$\mathrm{Spec}(S)\to\mathrm{Spec}(V)$ is bijective since $V\subseteq S$ is
t-closed by Proposition 2.6, whence $V\subseteq S$ is radicial.

In particular, $T\subseteq U\cap V$ since the residual extensions of
$R\subseteq T$ are isomorphisms. Assume that $T\subset U\cap V$. Then, the
residual extensions of $T\subset U\cap V$ are both radicial and separable
fields extensions, that is isomorphisms; so that, $T\subset U\cap V$ is infra-
integral, a contradiction. To conclude, $T=U\cap V$. ∎

##  5\.  u-closures of integral FCP extensions

In this section, we work in the context of FCP integral extensions, where many
results of [15] about anodality will be easier to handle.

######  Proposition 5.1.

An integral FCP extension $R\subseteq S$ is u-closed if and only if
$R/I\subseteq S/I$ is u-closed for any ideal $I$ shared by $R$ and $S$.

###### Proof.

Obvious. ∎

The next result does not hold if $R\subset S$ is not FCP.

######  Proposition 5.2.

Let $R\subset S$ be an integral FCP extension. Then the following conditions
are equivalent:

  1. (1)

$R\subset S$ is u-closed;

  2. (2)

$R\subset S$ is an i-extension;

  3. (3)

${}_{S}^{+}R={}_{S}^{t}R$.

If these conditions are satisfied, any minimal subextension of $R\subset S$ is
either ramified or inert.

###### Proof.

(1) $\Rightarrow$ (2) Set $I:=(R:S)$. Proposition 5.1 shows that $R/I\subseteq
S/I$ is u-closed. Moreover, $R/I$ is an Artinian, and so a Noetherian ring by
[4, Theorem 4.2]. Then, by [15, Proposition 2.35],
$\mathrm{Spec}(S/I)\to\mathrm{Spec}(R/I)$ is bijective. It follows that
$\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is bijective.

(2) $\Rightarrow$ (1) Assume that $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is
bijective. Then [15, Theorem 2.27] asserts that $R\subset S$ is u-closed.

(2) $\Rightarrow$ (3) ${}_{S}^{+}R\neq{}_{S}^{t}R$ implies that
$\mathrm{Spec}({}_{S}^{t}R)\to\mathrm{Spec}({}_{S}^{+}R)$ is not bijective
according to Theorem 1.3 since there is a minimal decomposed subextension of
${}_{S}^{+}R\subset{}_{S}^{t}R$.

(3) $\Rightarrow$ (2) If ${}_{S}^{+}R={}_{S}^{t}R$, then
$\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is bijective since so are
$\mathrm{Spec}(S)\to\mathrm{Spec}({}_{S}^{t}R)$ and
$\mathrm{Spec}({}_{S}^{+}R)\to\mathrm{Spec}(R)$ by Proposition 2.6.

The last assertion is an easy consequence of (3). ∎

######  Corollary 5.3.

A minimal extension $R\subset S$ is u-closed if and only if $R\subset S$ is
either ramified or inert or minimal Prüfer.

###### Proof.

The equivalence in the integral case is a consequence of Proposition 5.2 since
$\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ is bijective if and only if $R\subset S$
is either ramified or inert. Assume that $R\subset S$ is a minimal Prüfer
extension. Then $R\subset S$ is u-closed since t-closed by Proposition 2.5. ∎

######  Proposition 5.4.

Let $R\subseteq S$ be a u-closed integral FCP extension. Then $T\subseteq U$
is u-closed for any $T,U\in[R,S]$ such that $T\subseteq U$.

###### Proof.

We use Proposition 5.2. Assume that $\mathrm{Spec}(U)\to\mathrm{Spec}(T)$ is
not bijective. There exist $N_{1},N_{2}\in\mathrm{Max}(U),\ N_{1}\neq N_{2}$,
such that $N_{1}\cap T=N_{2}\cap T$. In particular, $N_{1}\cap R=N_{2}\cap R$.
Since, $R\subseteq S$ is an FCP integral extension, so are $T\subset U$ and
$U\subset S$. In particular, there exist $M_{1},M_{2}\in\mathrm{Max}(S)$ such
that $N_{1}=M_{1}\cap U$ and $N_{2}=M_{2}\cap U$. It follows that $M_{1}\cap
R=M_{2}\cap R$ with $M_{1}\neq M_{2}$ since $N_{1}\neq N_{2}$, a contradiction
with $\mathrm{Spec}(S)\to\mathrm{Spec}(R)$ bijective. Then,
$\mathrm{Spec}(U)\to\mathrm{Spec}(T)$ is bijective and $T\subseteq U$ is
u-closed by Proposition 5.2. ∎

######  Proposition 5.5.

Let $R\subset S$ be an integral FCP extension.

  1. (1)

$({}_{S}^{u}R)_{M}={}_{S_{M}}^{u}R_{M}$ for any $M\in\mathrm{MSupp}(S/R)$.

  2. (2)

$({}_{S}^{u}R)/I={}_{(S/I)}^{u}(R/I)$ for any ideal $I$ shared by $R$ and $S$.

###### Proof.

Set $U:={}_{S}^{u}R$.

(1) Let $M\in\mathrm{MSupp}(S/R)$. We first show that $U_{M}\subseteq S_{M}$
is u-closed. Using Proposition 5.2, it is enough to show that
$\mathrm{Spec}(S_{M})\to\mathrm{Spec}(U_{M})$ is bijective, which is obvious
since so is $\mathrm{Spec}(S)\to\mathrm{Spec}(U)$ because $U\subseteq S$ is
u-closed. Then, ${}_{S_{M}}^{u}R_{M}\subseteq U_{M}\ (*)$.

Now, since $R\subseteq S$ is an integral FCP extension, [4, Theorem 3.6] shows
that there exists $U^{\prime}\in[R,S]$ such that
$U^{\prime}_{M}={}_{S_{M}}^{u}R_{M}$ for any $M\in\mathrm{MSupp}(S/R)$. In
particular, $U^{\prime}_{M}\subseteq S_{M}$ is u-closed for any
$M\in\mathrm{MSupp}(S/R)$. Therefore, $U^{\prime}\subseteq S$ is u-closed by
[15, Proposition 1.16]. Then, $U\subseteq U^{\prime}$, which implies
$U_{M}\subseteq U^{\prime}_{M}={}_{S_{M}}^{u}R_{M}$ for any
$M\in\mathrm{MSupp}(S/R)$; so that,
$U_{M}=({}_{S}^{u}R)_{M}={}_{S_{M}}^{u}R_{M}$ for any
$M\in\mathrm{MSupp}(S/R)$ by $(*)$.

(2) Let $I$ be an ideal shared by $R$ and $S$. Then $I$ is also an ideal of
$U$. Since $U\subseteq S$ is u-closed, so is $U/I\subseteq S/I$ by Proposition
5.1; so that, ${}_{(S/I)}^{u}(R/I)\subseteq U/I$. But
$\mathrm{Spec}(S/I)\to\mathrm{Spec}({}_{(S/I)}^{u}(R/I))$ is bijective, with
${}_{(S/I)}^{u}(R/I)\in[R/I,S/I]$. It follows that there exists
$U^{\prime\prime}\in[R,S]$ such that $U^{\prime\prime}/I={}_{(S/I)}^{u}(R/I)$,
with $U^{\prime\prime}/I\subseteq U/I$, giving $U^{\prime\prime}\subseteq U\
(**)$. Moreover, $\mathrm{Spec}(S/I)\to\mathrm{Spec}(U^{\prime\prime}/I)$ is
bijective, and so is $\mathrm{Spec}(S)\to\mathrm{Spec}(U^{\prime\prime})$. At
last, Proposition 5.2 shows that $U^{\prime\prime}\subset S$ is u-closed,
which implies $U\subseteq U^{\prime\prime}$, and, to conclude,
$U=U^{\prime\prime}$ by $(**)$, that is $U/I={}_{(S/I)}^{u}(R/I)$. ∎

######  Corollary 5.6.

Let $R\subset S$ be an integral FCP extension. The following conditions are
equivalent

  1. (1)

$R\subset S$ is u-closed (resp.; u-integral);

  2. (2)

$R_{M}\subset S_{M}$ is u-closed (resp.; u-integral) for any
$M\in\mathrm{MSupp}(S/R)$;

  3. (3)

$R/I\subset S/I$ is u-closed (resp.; u-integral) for any ideal $I$ shared by
$R$ and $S$.

###### Proof.

We use Proposition 5.5. Then $({}_{S}^{u}R)_{M}={}_{S_{M}}^{u}R_{M}$ for any
$M\in\mathrm{MSupp}(S/R)$ and $({}_{S}^{u}R)/I={}_{(S/I)}^{u}(R/I)$ for any
ideal $I$ shared by $R$ and $S$. The different equivalences are obvious since
$R\subset S$ is u-closed (resp.; u-integral) if and only if ${}_{S}^{u}R=R$
(resp.; ${}_{S}^{u}R=S$), with similar equalities for $R_{M}\subset S_{M}$ and
$R/I\subset S/I$. ∎

We recall [15, Definition 2.11] that a ring morphism $f:A\to B$ is said to be
locally an epimorphism if $A_{f^{-1}(Q)}\to B_{Q}$ is an epimorphism for every
$Q\in\mathrm{Spec}(B)$.

The next result gives a characterization of $u$-integral extensions in the FCP
context by some property of ring morphisms as it is the case for subintegral
and infra-integral extensions. In an arbitrary context this result may not
hold.

######  Theorem 5.7.

An integral FCP extension $R\subset S$ is u-integral if and only if $R\subset
S$ is locally an epimorphism, in which case $R\subset S$ is unramified.

###### Proof.

Assume first that $R\subset S$ is u-integral. Then, $R\subset S$ is locally an
epimorphism by [15, Corollary 2.23]. Conversely, assume that $R\subset S$ is
locally an epimorphism. Since ${}_{S}^{u}R\subseteq S$ is u-closed,
Proposition 5.2 asserts that $\mathrm{Spec}(S)\to\mathrm{Spec}({}_{S}^{u}R)$
is bijective. Moreover, $R\subset S$ is a finite injective locally
epimorphism. According to [15, Remark 2.34 (1)], $R\subset S$ is cu-
elementary, and, in particular, u-integral. The last statement holds by
Proposition 4.3(2) and [10, Exercise 1.14, p.7]. ∎

The seminormalization, the t-closure and the u-closure are linked as it is
shown in the following Theorem.

######  Theorem 5.8.

If $R\subset S$ is an integral FCP extension, then
$({}_{S}^{u}R)({}_{S}^{+}R)={}_{S}^{t}R$ holds. In particular, such an
extension is t-closed if and only if it is seminormal and u-closed (anodal).

###### Proof.

Since $\mathrm{Spec}(S)\to\mathrm{Spec}({}_{S}^{t}R)$ is bijective,
Proposition 5.2 shows that ${}_{S}^{t}R\subset S$ is u-closed; so that,
${}_{S}^{u}R\subseteq{}_{S}^{t}R$. We already know that
${}_{S}^{+}R\subseteq{}_{S}^{t}R$ by Definition 2.1, whence
$({}_{S}^{u}R)({}_{S}^{+}R)\subseteq{}_{S}^{t}R$ holds.

Now ${}_{S}^{u}R{}\subseteq({}_{S}^{u}R)({}_{S}^{+}R)$ and
${}_{S}^{+}R\subseteq({}_{S}^{u}R)({}_{S}^{+}R)$ combine to yield that
$({}_{S}^{u}R)({}_{S}^{+}R)\subseteq S$ is both seminormal and u-closed, and
then locally u-closed by Corollary 5.6. Hence,
$({}_{S}^{u}R)({}_{S}^{+}R)\subseteq S$ is t-closed, according to [17,
Proposition 3.17], leading to
${}_{S}^{t}R\subseteq({}_{S}^{u}R)({}_{S}^{+}R)$, and then to
$({}_{S}^{u}R)({}_{S}^{+}R)={}_{S}^{t}R$.

The last statement is [17, Proposition 3.17] since u-closedness is equivalent
to local u-closedness by Corollary 5.6. ∎

######  Theorem 5.9.

Let $R\subset S$ be an integral FCP extension. The following conditions are
equivalent:

  1. (1)

$R\subset S$ is u-integral and seminormal;

  2. (2)

$R\subset S$ is seminormal infra-integral;

  3. (3)

$|\mathrm{V}_{S}((R:S))|=|\mathrm{V}_{R}((R:S))|+\ell[R,S]$;

  4. (4)

${}_{S}^{+}R=R$ and ${}_{S}^{t}R=S$.

  5. (5)

$(R:S)$ is an irredundant intersection of $\ell[R,S]+n$ maximal ideals of $S$
and an intersection of $n$ maximal ideals of $R$ for some positive integer
$n$.

If these conditions hold, then $n=|\mathrm{V}_{R}((R:S))|$.

###### Proof.

Set $n:=|\mathrm{V}_{R}((R:S))|<\infty$ by Proposition 2.9.

(1) $\Rightarrow$ (2) Since $R\subset S$ is u-integral, then $R\subset S$ is
infra-integral by Definition 2.2 and Proposition 2.3.

(2) $\Rightarrow$ (1) Since $R\subset S$ is infra-integral, we have
$S={}_{S}^{t}R$ and since $R\subset S$ is seminormal, we have $R={}_{S}^{+}R$.
Then, Theorem 5.8 leads to
${}_{S}^{u}R=({}_{S}^{u}R)R=({}_{S}^{u}R)({}_{S}^{+}R)={}_{S}^{t}R=S$; so
that, $R\subset S$ is u-integral.

(2) $\Leftrightarrow$ (3) because of Proposition 2.6.

(2) $\Leftrightarrow$ (4) is obvious according to the definitions of
${}_{S}^{+}R$ and ${}_{S}^{t}R$.

(2) $\Rightarrow$ (5) Since $R\subset S$ is seminormal FCP,
$(R:S)=\cap_{i=1}^{m}M_{i}$, where $M_{i}\in\mathrm{Max}(S)$ for each
$i\in\mathbb{N}_{m}$ by Proposition 2.8; so that,
$m=|\mathrm{V}_{S}((R:S))|=|\mathrm{V}_{R}((R:S))|+\ell[R,S]$ by (3). But, we
also have $(R:S)=\cap_{j=1}^{n}P_{j}$, where the $P_{j}$’s are the
intersections of the $M_{i}$’s with $R$, with perhaps more than one $M_{i}$
for one $P_{j}$. Then, $m=n+\ell[R,S]$, giving (5).

(5) $\Rightarrow$ (3) Assume that $(R:S)=\cap_{i=1}^{m}M_{i}$, where
$M_{i}\in\mathrm{Max}(S)$ for each $i\in\mathbb{N}_{m}$, with $m=\ell[R,S]+n$
and that $(R:S)=\cap_{j=1}^{n}P_{j}$, where the $P_{j}\in\mathrm{Max}(R)$ for
each $j\in\mathbb{N}_{n}$. Then,
$|\mathrm{V}_{S}((R:S))|=|\mathrm{V}_{R}((R:S))|+\ell[R,S]$, giving (3). ∎

The following Proposition was proved in [15, Theorem 4.8] in a different
context.

######  Proposition 5.10.

[15, Theorem 4.8] A minimal extension $R\subset S$ is u-integral if and only
if $R\subset S$ is decomposed.

###### Proof.

Let $R\subset S$ be a minimal extension and set $M:=\mathcal{C}(R,S)$.

If $R\subset S$ is decomposed, there exists $q\in S$ such that $S=R[q],\
q^{2}-q\in M\subset R$ and $Mq\subseteq M$; so that, $q^{3}-q^{2}\in R$. Then,
$R\subset S$ is u-integral.

Conversely, assume that $R\subset S$ is u-integral. By assumption, $R\subset
S$ is a minimal integral extension. If $R\subset S$ is either minimal ramified
or minimal inert, then $R\subset S$ is an i-extension; so that, $R\subset S$
is u-closed, which implies $R=S$, a contradiction. Then, $R\subset S$ is
decomposed. ∎

######  Remark 5.11.

Let $R\subset S$ be an integral FCP extension with
${}_{S}^{+}R,{}_{S}^{t}R\neq R,S$. According to Proposition 2.6, a minimal
subextension of $R\subset{}_{S}^{+}R$ is ramified and a minimal subextension
of ${}_{S}^{+}R\subset S$ is either decomposed or inert. Similarly, any
minimal subextension of $R\subset{}_{S}^{t}R$ is either ramified or decomposed
and any minimal subextension of ${}_{S}^{t}R\subset S$ is inert. Contrary to
these cases, we have seen in Proposition 5.2 that any minimal subextension of
an u-closed extension is either ramified or inert. But a minimal ramified
extension can also be a subextension of an u-integral extension as we can see
in the following examples. We have not the same behavior with the u-closure
where we may have two minimal extensions of the same type contained and
containing the u-closure. Let $R\subset S$ be an integral FCP extension with
u-closure ${}_{S}^{u}R$. We can see in the following examples that there
exists a minimal ramified subextension of $R\subset{}_{S}^{u}R$ and a minimal
ramified subextension of ${}_{S}^{u}R\subset S$

(1) Let $R$ be an Artinian non reduced ring. Then, $R\subset R^{2}$ is an FCP
infra-integral and not seminormal extension by [20, Proposition 1.4]. It
follows that there exists $U\in[R,R^{2}]$ such that $R\subset U$ is minimal
ramified. Moreover, $R\subset R^{2}$ is u-integral by [15, Example 2.32]. This
example also shows that a subextension of an u-integral extension is not
necessarily u-integral. Indeed, $R\subset{}_{R^{2}}^{+}R$ is u-closed since
subintegral by Proposition 5.2.

(2) We continue to work with the situation of (1), with $R:=$

$(\mathbb{Z}/2\mathbb{Z})[T]/(T^{2})=(\mathbb{Z}/2\mathbb{Z})[t]$, where $t$
is the class of $T$ in $R$. Then, $R$ is an Artinian local non reduced ring.
Let $M:=Rt$ be its maximal ideal. It follows that $R\subset R^{2}$ is
u-integral. According to [20, Proposition 2.8], $R+(M\times
M)={}_{R^{2}}^{+}R$, with $R\subset R+(M\times M)$ minimal ramified. Indeed,
setting $e_{1}:=(1,0)$, we have $R+(M\times M)=R[te_{1}]$, with
$(te_{1})^{2}=0\in R$ and $Mte_{1}=Rt^{2}e_{1}=0$. And $R+(M\times M)\subset
R^{2}$ is minimal decomposed since $|\mathrm{Max}(R^{2})|=2$. Then,
$\ell[R,R^{2}]=2$. Set $S:=R\times R[x]$, where $R[x]:=R[X]/(X^{2},tX)$ and
$x$ is the class of $X$ in $R[X]/(X^{2},tX)$. Since $x^{2}=0$ and $Mx=Rtx=0$,
we get that $R\subset R[x]$ is minimal ramified, as $R^{2}\subset S$, by [3,
Proposition III.4]; so that, $\ell[R,S]=3$. We claim that $R^{2}={}_{S}^{u}R$.
Set $U:={}_{S}^{u}R$. Then, $U\neq R$ since $R\subset S$ is infra-integral and
not subintegral, because $|\mathrm{Max}(R^{2})|=2$. For a similar reason,
$U\neq S$ because $R^{2}\subset S$ is subintegral. Then, $R\subset U\subseteq
R^{2}\subset S$, which implies that $\ell[R,U]\in\\{1,2\\}$. If $\ell[R,U]=2$,
then $U=R^{2}$, and we are done. Assume that $\ell[R,U]=1$; so that, $R\subset
U$ is minimal, necessarily decomposed, since u-integral and $U\subset R^{2}$
is minimal ramified. Considering the extension $R\subset R^{2}$ and using [6,
Lemma 17], we get that $MR^{2}=M\times M$ is not a radical ideal of $R^{2}$, a
contradiction. Then, $\ell[R,U]=2$ and $U=R^{2}={}_{S}^{u}R$. This example
exhibits a subintegral minimal ramified subextension $R\subset{}_{R^{2}}^{+}R$
of $R\subset{}_{S}^{u}R=R^{2}$ as well as a subintegral minimal ramified
subextension ${}_{S}^{u}R\subset S$ of (itself) ${}_{S}^{u}R\subset S$ as we
can see in the following commutative diagram:

$\begin{matrix}{}&{}&{}&{}&{}_{S}^{+}R&{}&{}\\\
{}&{}&{}&\nearrow&{}&\searrow&{}\\\ R&\to&{}_{R^{2}}^{+}R&{}&{}&{}&S\\\
{}&{}&{}&\searrow&{}&\nearrow&{}\\\
{}&{}&{}&{}&{}_{S}^{u}R=R^{2}&{}&{}\end{matrix}$

In the preceding Remark we can observe that
$R\subset{}_{S}^{+}R\cap{}_{S}^{u}R={}_{R^{2}}^{+}R$. It may be asked under
what conditions $R={}_{S}^{+}R\cap{}_{S}^{u}R$ holds for an FCP extension
$R\subset S$. The answer is given in the following Proposition.

######  Proposition 5.12.

Let $R\subset S$ be an FCP extension. Then, $R={}_{S}^{+}R\cap{}_{S}^{u}R$ if
and only if $R\subseteq{}_{S}^{u}R$ is seminormal.

###### Proof.

Set $U:={}_{S}^{u}R$. Then, $T:={}_{S}^{+}R\cap U={}_{U}^{+}R$. It follows
that $T=R$ if and only if $R={}_{U}^{+}R$ if and only if $R\subseteq U$ is
seminormal. ∎

The $\omega$-closure of a ring extension $R\subseteq S$ is the greatest
element $U:={}^{\omega}_{S}R$ of $[R,S]$ such that $R\subseteq U$ is
unramified [15, Proposition 3.3].

######  Proposition 5.13.

Let $R\subseteq S$ be an FCP integral extension. There exists a greatest
unramified subextension $R\subseteq{}^{\omega}_{S}R$ of $R\subseteq S$. Then
${}^{\omega}_{S}R\subseteq S$ is $u$-closed; so that,
${}^{u}_{S}R\subseteq{}^{\omega}_{S}R\subseteq{}^{\square}_{S}R$.

###### Proof.

For the first part use [15, Proposition 3.3], taking into account that all
subextensions of an FCP extension are of finite type. To show that
${}^{\omega}_{S}R\subseteq S$ is $u$-closed, consider some $s\in S$, such that
$s^{2}-s,s^{3}-s^{2}\in{}^{\omega}_{S}R$, then
${}^{\omega}_{S}R\subseteq{}^{\omega}_{S}R[s]$ is $u$-integral, whence
unramified according to Theorem 5.7, and so is
$R\subseteq{}^{\omega}_{S}R[s]$. By maximality of the unramified closure, $s$
belongs to ${}^{\omega}_{S}R$. Since ${}^{\omega}_{S}R\subseteq S$ is
$u$-closed, ${}^{u}_{S}R\subseteq{}^{\omega}_{S}R$. Now
${}^{\omega}_{S}R\subseteq{}^{\square}_{S}R$ holds by Proposition 4.3 since
$R\subseteq{}^{\omega}_{S}R$ is unramified. ∎

######  Corollary 5.14.

Let $R\subset S$ be a $\kappa$-radicial FCP extension (for instance, infra-
integral). There exists some $\Sigma\in[R,S]$ such that $R\subseteq\Sigma$ is
unramified, $\Sigma\subseteq S$ is radicial and
$\Sigma={}^{\omega}_{S}R={}^{u}_{S}R$.

###### Proof.

It is enough to observe that the extension
${}^{u}_{S}R\subseteq{}^{\omega}_{S}R$ is unramified because so is
$R\subseteq{}^{\omega}_{S}R$, and radicial because $\kappa$-radicial and an
$i$-extension by Proposition 5.2. Moreover, a radicial unramified FCP
extension is trivial by Proposition 4.2. Then, set
$\Sigma:={}^{\omega}_{S}R={}^{u}_{S}R$. ∎

######  Remark 5.15.

Theorem 4.12 shows that for an integral FCP extension $R\subset S$, the
extension $R\subseteq{}^{\square}_{S}R$ is $\kappa$-separable and
${}^{\square}_{S}R\subseteq S$ is $\kappa$-radicial. The following example
shows that we cannot generalize the above Corollary to arbitrary integral
extensions, which means that we cannot replace the $\kappa$-separable property
by the unramified property. We use [22, Example 4.10] to get such a situation.

Let $K\subset L$ be a minimal separable field extension of degree 2; so that,
there exists $y\in L$ such that $L=K[y]$. Set $S:=L[X]/(X^{2})$ and let $x$ be
the class of $X$ in $S$. Set $R:=K+Kx$ and $T:=K+Lx$; so that, $R$ is a local
ring with maximal ideal $M:=Kx$ and $T$ is a local ring with maximal ideal
$M^{\prime}:=M+Rxy$. It was proved in [22, Example 4.10] that $R\subset T$ is
minimal ramified, $T\subset S$ is minimal inert with $M^{\prime}=(T:S)$ and
$T/M^{\prime}\subset S/M^{\prime}$ is a minimal separable field extension with
$|[R,S]|=3$. Moreover, $R\subset T$ is radicial because an i-extension with
residual extensions which are isomorphisms and $T\subset S$ is unramified by
Proposition 4.3. Since $|[R,S]|=3$, there does not exist $\Sigma\in[R,S]$ such
that $R\subset\Sigma$ is unramified and $\Sigma\subset S$ is radicial.

##  6\. Co-subintegral closure and co-infra-integral closure of an integral
FCP extension

Definition 2.4 recalls the definitions of several closures. Let $R\subset S$
be a ring extension. As we did in Section 3 with the co-integral closure, we
are going to characterize some closures defined as the least $T\in[R,S]$ such
that $T\subseteq S$ satisfies a given property.

Let $R\subseteq S$ be an integral FCP extension. In [6, Lemma 17], we give a
condition under which there exists $T\in[R,S]$ such that $T\subset S$ is
minimal ramified.

A more general requirement is as follows: Let $R\subseteq S$ be an FCP
extension. There exists some $U\in[R,S]$ which is the least $V\in[R,S]$ such
that $V\subseteq S$ is a subintegral extension. If such a ring $U$ exists, we
call it the co-subintegral closure of $R\subseteq S$ and denote it by
$S^{+}_{R}$ since it is a kind of dual of the seminormalization which is
nothing but the subintegral closure ${}^{+}_{S}R$ and the largest $V\in[R,S]$
such that $R\subseteq V$ is subintegral.

Similarly, we can ask if there exists some $U\in[R,S]$ which is the least
$V\in[R,S]$ such that $V\subseteq S$ is an infra-integral extension. If this
$U$ exists, we call it the co-infra-integral closure of $R\subseteq S$ and
denote it by $S^{t}_{R}$ since it is a kind of dual of the t-closure
${}^{t}_{S}R$ (the largest $V\in[R,S]$ such that $R\subseteq V$ is infra-
integral). Many of the proofs of this section are similar for the subintegral
case and the infra-integral case.

If these co-subintegral closure or co-infra-integral closures exist, we have
$\underline{R}\subseteq S^{t}_{R}\subseteq S^{+}_{R}\subseteq S$, with
$\underline{R}\subseteq S$ an integral FCP extension. That is the reason why
we limit ourselves to consider only integral FCP extensions.

######  Proposition 6.1.

Let $R\subset S$ be an integral FCP extension. The co-subintegral (resp.; co-
infra-integral) closure exists in the following cases:

  1. (1)

$R\subset S$ is subintegral (resp.; infra-integral). Then, $R=S^{+}_{R}$
(resp.; $=S^{t}_{R}$).

  2. (2)

$R\subset S$ is seminormal (resp.; t-closed). Then, $S=S^{+}_{R}$ (resp.;
$=S^{t}_{R}$) and ${}_{S}^{u}R={}_{S}^{t}R$.

###### Proof.

Obvious. ∎

######  Corollary 6.2.

Let $R\subset S$ be an integral FCP extension. Then,
$R=({}_{S}^{+}R)^{+}_{R}=({}_{S}^{+}R)^{t}_{R}=({}_{S}^{t}R)^{t}_{R}$.

###### Proof.

Since $R\subseteq{}_{S}^{+}R$ (resp.; ${}_{S}^{t}R$) is subintegral (resp.;
infra-integral), whence infra-integral, Proposition 6.1 gives the answer in
both cases. ∎

The following remark shows that such closures do not always exist.

######  Remark 6.3.

Let $k\subset K$ be a minimal radicial field extension of degree 2; so that,
$\mathrm{c}(k)=\mathrm{c}(K)=2$. Let $\alpha\in K$ be such that $K=k[\alpha]$,
with $a:=\alpha^{2}\in k$. Set $S:=K[X]/(X^{2})=K[x]=K+Kx$, where $x$ is the
class of $X$ in $S$. Then, $K\subset S$ is a minimal ramified extension (and
then subintegral and infra-integral) with $x^{2}=0$. Set $\beta:=\alpha+x\in
S$ and $L:=k[\beta]\subseteq S$. It follows that $\beta^{2}=\alpha^{2}=a\in
k$; so that, $k\subset L$ is a minimal radicial field extension with $L\cong
k[T]/(T^{2}-a)\cong K$. Of course $K\neq L$ and $S=L[x]=L+Lx$, with $L\subset
S$ a minimal ramified extension (and then subintegral and infra-integral). As
$K\cap L=k$, the extension $k\subset S$ has neither a co-subintegral closure
nor a co-infra-integral closure, since there is no $T\in[k,S]$ such that
$T\subset S$ is subintegral (resp.; infra-integral) with $T\subseteq K,L$.

######  Proposition 6.4.

Let $R\subset S$ be an integral FCP extension.

  1. (1)

If the co-subintegral (resp.; co-infra-integral) closure $T$ of $R\subset S$
exists, then $T_{M}$ is the co-subintegral (resp.; co-infra-integral) closure
of $R_{M}\subset S_{M}$ for any $M\in\mathrm{MSupp}(S/R)$.

  2. (2)

Assume that for any $M\in\mathrm{MSupp}(S/R)$, there exists a co-subintegral
(resp.; co-infra-integral) closure $T_{(M)}$ of $R_{M}\subset S_{M}$. Then,
the co-subintegral closure (resp.; co-infra-integral) $T$ of $R\subset S$
exists and satisfies $T_{M}=T_{(M)}$ for any $M\in\mathrm{MSupp}(S/R)$.

###### Proof.

(1) Assume that the co-subintegral (resp.; co-infra-integral) closure $T$ of
$R\subset S$ exists; so that, $T$ is the least $U\in[R,S]$ such that
$U\subseteq S$ is subintegral (resp.; infra-integral). Then $T_{M}\subseteq
S_{M}$ is subintegral (resp.; infra-integral) for any
$M\in\mathrm{MSupp}(S/R)$ by [30, Corollary 2.10] (resp.; [17, Proposition
1.16]). Let $V\in[R_{M},S_{M}]$ be such that $V\subseteq S_{M}$ is subintegral
(resp.; infra-integral) for some $M\in\mathrm{MSupp}(S/R)$. Since $R\subset S$
is an integral FCP extension, there exists $W\in[R,S]$ such that $W_{M}=V$ and
$W_{M^{\prime}}=S_{M^{\prime}}$ for any
$M^{\prime}\in\mathrm{MSupp}(S/R)\setminus\\{M\\}$ by [4, Theorem 3.6]. Then,
again by [30, Proposition 2.9] (resp.; [17, Proposition 3.6]), we get that
$W\subseteq S$ is subintegral (resp.; infra-integral); so that, $T\subseteq
W$, which leads to $T_{M}\subseteq W_{M}=V$. Then $T_{M}$ is the least
$U\in[R_{M},S_{M}]$ such that $U\subseteq S_{M}$ is subintegral (resp.; infra-
integral), which means that $T_{M}$ is the co-subintegral (resp.; co-infra-
integral) closure of $R_{M}\subset S_{M}$.

(2) Conversely, assume that for any $M\in\mathrm{MSupp}(S/R)$, there exists a
co-subintegral (resp.; co-infra-integral) closure $T_{(M)}$ of $R_{M}\subset
S_{M}$; so that, $T_{(M)}\subseteq S_{M}$ is subintegral (resp.; infra-
integral) for any $M\in\mathrm{MSupp}(S/R)$. We deduce from the above
reference [4, Theorem 3.6] that there exists $T\in[R,S]$ such that
$T_{M}=T_{(M)}$ for any $M\in\mathrm{MSupp}(S/R)$; so that, $T_{M}\subseteq
S_{M}$ is subintegral (resp.; infra-integral) for any
$M\in\mathrm{MSupp}(S/R)$, giving that $T\subseteq S$ is subintegral (resp.;
infra-integral) by [30, Proposition 2.9] (resp.; [17, Proposition 3.6]). Now,
let $V\in[R,S]$ be such that $V\subseteq S$ is subintegral (resp.; infra-
integral). Then, $V_{M}\subseteq S_{M}$ is subintegral (resp.; infra-integral)
for any $M\in\mathrm{MSupp}(S/R)$. It follows that $T_{M}=T_{(M)}\subseteq
V_{M}$ for any $M\in\mathrm{MSupp}(S/R)$. To end $T\subseteq V$ shows that $T$
is the co-subintegral (resp.; co-infra-integral) closure of $R\subseteq S$. ∎

Let $R\subset S$ be an FCP extension. We recall that $R$ is called unbranched
in $S$ if $\overline{R}$ is local, in which case all rings in $[R,S]$ are
local by [24, Lemma 3.29]. We also say that $R\subset S$ is unbranched. A
locally unbranched extension is an extension $R\subset S$ such that
$R_{P}\subset S_{P}$ is unbranched for each $P\in\mathrm{Spec}(R)$. A locally
unbranched integral extension is nothing but an integral $i$-extension.

######  Proposition 6.5.

If $R\subset S$ is an integral locally unbranched FCP extension such that the
co-subintegral closure exists, then $S^{+}_{R}=S^{t}_{R}$.

###### Proof.

Since $R\subset S$ is locally unbranched, so is $S^{t}_{R}\subseteq
S^{+}_{R}$; so that, $S^{t}_{R}\subseteq S$ is subintegral, and then
$S^{t}_{R}=S^{+}_{R}$. ∎

######  Theorem 6.6.

An infra-integral FCP extension $R\subset S$ has a co-subintegral closure and
${}_{S}^{u}R=S^{+}_{R}$.

###### Proof.

Since $R\subset S$ is infra-integral, ${}_{S}^{u}R\subseteq S$ is infra-
integral and u-closed, and then an i-extension, by Proposition 5.2; so that,
${}_{S}^{u}R\subseteq S$ is subintegral. In particular, ${}_{S}^{u}R$ is the
least element $T\in[R,S]$ such that, $T\subseteq S$ is subintegral since the
least element $T\in[R,S]$ such that $T\subseteq S$ is an i-extension. Then,
${}_{S}^{u}R=S^{+}_{R}$. ∎

######  Corollary 6.7.

If $R\subset S$ is integral and FCP, then $({}_{S}^{t}R)^{+}_{R}={}^{u}_{S}R$.

###### Proof.

Set $T:=({}_{S}^{t}R)^{+}_{R}$, which exists by Theorem 6.6 and
$U:={}^{u}_{S}R$. Since $T\subseteq{}_{S}^{t}R$ is subintegral, it is an
i-extension, and so is ${}_{S}^{t}R\subseteq S$, since t-closed. Then,
$T\subseteq S$ is an i-extension, and then u-closed by Proposition 5.2; so
that, $U\subseteq T\subseteq{}_{S}^{t}R$. Therefore, $U\subseteq T$ is infra-
integral and an i-extension, whence subintegral; so that, $U=T$. ∎

######  Corollary 6.8.

Let $R\subset S$ be an integral FCP extension whose co-infra-integral closure
exists. Then the co-subintegral closure of $R\subset S$ exists and
$S^{+}_{R}=S^{+}_{(S^{t}_{R})}$.

###### Proof.

Set $T:=S^{t}_{R}$. Then $T\subseteq S$ is infra-integral and admits a co-
subintegral closure $U:=S^{+}_{T}$ according to Theorem 6.6. Let $V\in[R,S]$
be such that $V\subseteq S$ is subintegral, whence infra-integral. It follows
that $U=S^{+}_{T}\subseteq V$. Then $U$ is also the co-subintegral closure of
$R\subseteq S$. ∎

######  Proposition 6.9.

An integral FCP extension $R\subset S$, which has a co-subintegral closure
distinct from $S$ is not unramified.

###### Proof.

Assume that $R\subset S$ is unramified and admits a co-subintegral closure
distinct from $S$. Then, there exists $T\in[R,S]$ such that $T\subset S$ is
minimal ramified, hence not unramified, a contradiction. ∎

######  Proposition 6.10.

Let $R\subset S$ be an integral FCP extension.

  1. (1)

Assume that the co-infra-integral closure $S^{t}_{R}$ exists. Then,
$S^{t}_{R}\subseteq T$ and $T\subseteq S$ are infra-integral for any
$T\in[S^{t}_{R},S]$.

  2. (2)

Assume that the co-subintegral $S^{+}_{R}$ closure exists. Then,
$S^{+}_{R}\subseteq T$ and $T\subseteq S$ are subintegral for any
$T\in[S^{+}_{R},S]$.

###### Proof.

We prove the two parts together. Assume that the co-infra-integral (resp.; co-
subintegral) closure exists and set $U:=S^{t}_{R}$ (resp.; $U:=S^{+}_{R}$).
According to Proposition 2.6, any maximal chain of $[U,S]$, defined by
$U=U_{0}\subset\cdots\subset U_{i}\subset\cdots\subset U_{n}=S$, where each
$U_{i}\subset U_{i+1}$ is minimal is such that each $U_{i}\subset U_{i+1}$ is
either ramified or decomposed (resp.; ramified). Let $T\in[U,S]$. Any maximal
chain of $[U,T]$ followed by a maximal chain of $[T,S]$ defines a maximal
chain of $[U,S]$. Then, using again Proposition 2.6, we get that $U\subseteq
T$ and $T\subseteq S$ are infra-integral (resp.; subintegral). ∎

The following Propositions show the links between the co-subintegral closure,
the co-infra-integral closure, the $\omega$-closure, the u-closure and the
$\kappa$-radicial closure.

######  Proposition 6.11.

Let $R\subset S$ be an integral FCP extension. Then
${}_{S}^{u}R={}^{\omega}_{S}R\cap{}^{t}_{S}R={}^{\omega}_{S}R\cap{}^{\circ}_{S}R,\
{}^{+}_{R}S={}^{*}_{R}S\cap{}^{\square}_{S}R$ and
${}^{*}_{R}S\cap{}^{\omega}_{S}R=R$.

The following statements hold:

  1. (1)

${}_{S}^{u}R={}^{\omega}_{S}R$ if and only if $R\subseteq{}^{\omega}_{S}R$ is
infra-integral;

  2. (2)

If the co-subintegral closure exists, then ${}_{S}^{u}R\subseteq
S^{+}_{R}\cap{}^{t}_{S}R$.

###### Proof.

We already know that ${}_{S}^{u}R\subseteq{}^{t}_{S}R$ by [15, Definition 1.5
and Proposition 1.6] and
${}_{S}^{u}R\subseteq{}^{\omega}_{S}R\subseteq{}^{\square}_{S}R$ by
Proposition 5.13; so that,
${}_{S}^{u}R\subseteq{}^{\omega}_{S}R\cap{}^{t}_{S}R$. Since
${}_{S}^{u}R\subseteq{}^{t}_{S}R$ is both u-closed and infra-integral, any
minimal subextension of ${}_{S}^{u}R\subseteq{}^{t}_{S}R$ is ramified by
Proposition 5.2. Since ${}_{S}^{u}R\subseteq{}^{\omega}_{S}R$ is unramified,
any minimal subextension of ${}_{S}^{u}R\subseteq{}^{\omega}_{S}R$ is either
decomposed or inert separable by Proposition 4.8. It follows that
${}_{S}^{u}R={}^{\omega}_{S}R\cap{}^{t}_{S}R$.

Since ${}^{t}_{S}R={}^{\square}_{S}R\cap{}^{\circ}_{S}R$ by Corollary 4.13, we
get that
${}_{S}^{u}R={}^{\omega}_{S}R\cap{}^{\square}_{S}R\cap{}^{\circ}_{S}R$. But,
${}^{\omega}_{S}R\subseteq{}^{\square}_{S}R$ gives
${}_{S}^{u}R={}^{\omega}_{S}R\cap{}^{\circ}_{S}R$. Now,
${}^{*}_{R}S\cap{}^{\square}_{S}R={}^{*}_{R}S\cap{}^{\circ}_{S}R\cap{}^{\square}_{S}R$
because ${}^{*}_{R}S\subseteq{}^{\circ}_{S}R$, which leads to
${}^{*}_{R}S\cap{}^{\square}_{S}R={}^{*}_{R}S\cap{}^{t}_{S}R$, since we have
${}^{t}_{S}R={}^{\circ}_{S}R\cap{}^{\square}_{S}R$. It follows that
${}^{*}_{R}S\cap{}^{\square}_{S}R={}^{+}_{R}S$ because
${}^{*}_{R}S\cap{}^{t}_{S}R={}^{+}_{R}S$ according to [27, Corollary 3.6]. To
end,
${}^{*}_{R}S\cap{}^{\omega}_{S}R={}^{*}_{R}S\cap{}^{\circ}_{S}R\cap{}^{\omega}_{S}R\cap{}^{\square}_{S}R={}^{*}_{R}S\cap{}^{\square}_{S}R\cap{}^{\omega}_{S}R\cap{}^{t}_{S}R={}_{S}^{u}R\cap{}_{S}^{+}R$
by the previous equalities. But ${}_{S}^{u}R\cap{}_{S}^{+}R=R$ according to
Proposition 4.2 because $R\subseteq{}_{S}^{u}R\cap{}_{S}^{+}R$ is both
subintegral, and then radicial, and u-integral, then also unramified by
Theorem 5.7.

In particular, ${}_{S}^{u}R={}^{\omega}_{S}R$ if and only if
${}^{\omega}_{S}R\subseteq{}^{t}_{S}R$ if and only if
$R\subseteq{}^{\omega}_{S}R$ is infra-integral.

Moreover, if the co-subintegral closure exists, $S^{+}_{R}\subseteq S$ is
subintegral, and then u-closed by Proposition 5.2. Then, ${}_{S}^{u}R\subseteq
S^{+}_{R}$, which leads to ${}_{S}^{u}R\subseteq S^{+}_{R}\cap{}^{t}_{S}R$. ∎

Let $R\subset S$ be an integral FCP extension (such that the co-infra-integral
closure exists for the first diagram). We have the following commutative
diagrams:

$\begin{matrix}{}&{}&{}_{S}^{+}R&\to&{}_{S}^{t}R&{}&{}\\\
{}&\nearrow&{}&\nearrow&{}&\searrow&{}\\\
R&\to&{}_{S}^{u}R&\to&{}^{\omega}_{S}R&\to&S\\\
{}&\searrow&{}&\searrow&{}&\nearrow&{}\\\
{}&{}&S^{t}_{R}&\to&S^{+}_{R}&{}&{}\end{matrix}$ and
$\begin{matrix}{}&{}&{}_{S}^{+}R&\to&{}_{S}^{*}R&\to&{}^{\circ}_{S}R&{}&{}\\\
{}&\nearrow&{}&\searrow&{}&\nearrow&{}&\searrow&{}\\\
R&{}&{}&{}&{}^{t}_{S}R&{}&{}&{}&S\\\
{}&\searrow&{}&\nearrow&{}&\searrow&{}&\nearrow&{}\\\
{}&{}&{}_{S}^{u}R&\to&{}^{\omega}_{S}R&\to&{}^{\square}_{S}R&{}&{}\end{matrix}$

######  Proposition 6.12.

Let $R\subset S$ be an integral FCP extension.

  1. (1)

Assume that the co-subintegral (resp.; co-infra-integral) closure exists.
Then, $S=(S^{+}_{R})({}^{+}_{S}R)$ (resp.; $=(S^{t}_{R})({}^{t}_{S}R)$).
Moreover, $S^{+}_{R}\cap{}^{+}_{S}R={}^{+}_{S^{+}_{R}}R$ (resp.;
$S^{t}_{R}\cap{}^{t}_{S}R={}^{t}_{S^{t}_{R}}R$).

  2. (2)

Assume that the co-infra-integral closure exists; so that, the co-subintegral
closure exists. Then, $({}_{S}^{u}R)(S^{t}_{R})=S^{+}_{R}$.

###### Proof.

(1) Set $U:=(S^{+}_{R})({}^{+}_{S}R)\subseteq S$. Using Proposition 6.10, we
get that $U\subseteq S$ is subintegral and by Proposition 2.6, we get that
$U\subseteq S$ is seminormal; so that, $U=S$. A similar argument shows the
equality for the infra-integral case. The two last equalities come from the
properties of the seminormalization and the t-closure.

(2) Assume that the co-infra-integral closure exists. According to Corollary
6.8, the co-subintegral closure exists and satisfies $S^{t}_{R}\subseteq
S^{+}_{R}$. We know by Proposition 6.11 that ${}_{S}^{u}R\subseteq S^{+}_{R}$.
Then, $({}_{S}^{u}R)(S^{t}_{R})\subseteq S^{+}_{R}$. Assume that
$({}_{S}^{u}R)(S^{t}_{R})\subset S^{+}_{R}$ and let
$V\in[({}_{S}^{u}R)(S^{t}_{R}),S^{+}_{R}]$ be such that $V\subset S^{+}_{R}$
is minimal. Since $S^{t}_{R}\subseteq S$ is infra-integral, $V\subset
S^{+}_{R}$ is either ramified or decomposed. Since ${}_{S}^{u}R\subseteq S$ is
u-closed, $V\subset S^{+}_{R}$ is either ramified or inert by Corollary 5.3.
Then, $V\subset S^{+}_{R}$ is ramified, and $V\subset S$ is subintegral, a
contradiction with the definition of $S^{+}_{R}$. To conclude,
$({}_{S}^{u}R)(S^{t}_{R})=S^{+}_{R}$. ∎

Thanks to Proposition 6.12, we are able to give a characterization of
extensions whose closures and co-closures play a similar role as in almost-
Prüfer extensions.

######  Corollary 6.13.

Let $R\subset S$ be an integral FCP extension.

  1. (1)

Assume that the co-subintegral closure $S^{+}_{R}$ of $R\subset S$ exists.
Then, the following conditions are equivalent:

    1. (a)

$R\subseteq S^{+}_{R}$ is seminormal,

    2. (b)

$R={}_{S}^{+}R\cap S^{+}_{R}$,

    3. (c)

$(R:S^{+}_{R})$ is an intersection of finitely many maximal ideals of
$S^{+}_{R}$.

  2. (2)

Assume that the co-infra-integral closure $S^{t}_{R}$ of $R\subset S$ exists.
Then, the following conditions are equivalent:

    1. (a)

$R\subseteq S^{t}_{R}$ is t-closed,

    2. (b)

$R={}_{S}^{t}R\cap S^{t}_{R}$,

    3. (c)

$(R:S^{t}_{R})$ is an intersection of finitely many maximal ideals of
$S^{t}_{R}$ and
$|\mathrm{V}_{S^{t}_{R}}((R:S^{t}_{R}))|=|\mathrm{V}_{R}((R:S^{t}_{R}))|$.

###### Proof.

(a) $\Leftrightarrow$ (b) in (1) and (2) by Proposition 6.12.

(1) (a) $\Leftrightarrow$ (c) by Proposition 2.8.

(2) (a) $\Leftrightarrow$ (c) by Proposition 2.8, Theorem 5.8 and Proposition
5.2.∎

######  Proposition 6.14.

Let $R\subset S$ be a locally unbranched integral FCP extension. If the co-
infra-integral closure $S^{t}_{R}$ exists, then, $S^{t}_{R}=S^{+}_{R}$.

###### Proof.

We know that $S^{+}_{R}$ exists by Corollary 6.8 and satisfies
$S^{t}_{R}\subseteq S^{+}_{R}$. In order to prove the equality of these rings,
it is enough to prove it locally thanks to Proposition 6.4. Then we may assume
that $R$ is local. But the fact that $R\subset S$ is unbranched implies that
$R\subset S$ is an $i$-extension, as $S^{t}_{R}\subseteq S$, which is then
subintegral. Then, $S^{t}_{R}=S^{+}_{R}$. ∎

In [25, Proposition 4.7], we proved that an FCP infra-integral extension is
catenarian.

######  Theorem 6.15.

Let $R\subset S$ be an integral FCP extension. Set $T:=\cap[U\in[R,S]\mid
U\subseteq S$ is subintegral$]$. Then the following conditions are equivalent:

  1. (1)

The co-subintegral closure of $R\subset S$ exists;

  2. (2)

$T\subseteq S$ is subintegral;

  3. (3)

$T\subseteq S$ is catenarian.

If these conditions hold, then $T$ is the co-subintegral closure of $R\subset
S$.

###### Proof.

Since ${}_{S}^{u}R\subseteq U$ for any $U\in[R,S]$ such that $U\subseteq S$ is
subintegral, we get that ${}_{S}^{u}R\subseteq T$; so that, $T\subseteq S$ is
u-closed, and then an i-extension by Proposition 5.2.

We prove that (1) $\Rightarrow$ (2), and more precisely, that $T$ is the co-
subintegral closure of $R\subset S$. Indeed, let $T^{\prime}$ be the co-
subintegral closure of $R\subset S$; so that, $T\subseteq T^{\prime}$. But
$T^{\prime}\subseteq U$ for any $U\in[R,S]$ such that $U\subseteq S$ is
subintegral. Then $T^{\prime}\subseteq T$ leads to $T^{\prime}=T$.

(2) $\Rightarrow$ (3) by [25, Proposition 4.7].

(3) $\Rightarrow$ (1) If $T=S$, there is no $U\in[R,S]$ such that $U\subset S$
is subintegral; so that, $S$ is the co-subintegral closure of $R\subset S$.

Assume now that $T\neq S$. Let $U\in[R,S]$ be such that $U\subset S$ is
subintegral. Then, $T\subseteq U$; so that, the co-subintegral closure of
$R\subset S$, if it exists, is the co-subintegral closure of $T\subseteq S$,
and the converse also holds. It follows that, to show the existence of the co-
subintegral closure of $R\subset S$, it is enough to show the existence of the
co-subintegral closure of $T\subset S$. Using Proposition 6.4, we are reduced
to show that for any $M\in\mathrm{MSupp}_{T}(S/T)$, there exists a co-
subintegral closure of $T_{M}\subset S_{M}$. Since $T\subset S$ is catenarian,
so is $T_{M}\subset S_{M}$ for any $M\in\mathrm{MSupp}(S/T)$ by [25,
Proposition 4.3]. Let $M\in\mathrm{MSupp}(S/T)$. Then $T_{M}\subset S_{M}$ is
catenarian and unbranched, and [25, Theorem 4.13] asserts that
$[T_{M},S_{M}]=[T_{M},{}^{t}_{S_{M}}T_{M}]\cup[{}^{t}_{S_{M}}T_{M},S_{M}]$.
Assume that there exists some $V\in[T_{M},S_{M}]$ such that $V\subset S_{M}$
is subintegral. We deduce that either $V\subset{}^{t}_{S_{M}}T_{M}\ (*)$ or
${}^{t}_{S_{M}}T_{M}\subseteq V\ (**)$. In case $(*)$, we have
${}^{t}_{S_{M}}T_{M}=S_{M}$ since $V\subset S_{M}$ is subintegral and
$T_{M}\subset S_{M}$ is subintegral; so that, $T_{M}$ is the co-subintegral
closure of $T_{M}\subset S_{M}$. But, in case $(**)$, we also have $V\subset
S_{M}$ t-closed; so that, $V=S_{M}$, a contradiction. If there does not exist
some $V\in[T_{M},S_{M}]$ such that $V\subset S_{M}$ is subintegral, $S_{M}$ is
the co-subintegral closure of $T_{M}\subset S_{M}$. In any case, there exists
a co-subintegral closure of $T_{M}\subset S_{M}$; so that, there exists a co-
subintegral closure of $T\subset S$, and also of $R\subset S$. ∎

######  Corollary 6.16.

Let $R\subset S$ be an integral FCP extension. If ${}_{S}^{u}R\subseteq S$ is
catenarian, then the co-subintegral closure of $R\subset S$ exists. In
particular, these conditions hold if $R\subset S$ is catenarian.

###### Proof.

We use the ring $T:=\cap[U\in[R,S]\mid U\subseteq S$ is subintegral$]$ defined
in Theorem 6.15. It was proved in this Theorem that ${}_{S}^{u}R\subseteq T$;
so that, $T\subseteq S$ is catenarian if so is ${}_{S}^{u}R\subseteq S$. Then,
Theorem 6.15 shows the existence of the co-subintegral closure of $R\subset
S$. ∎

This Corollary shows in another way than in Theorem 6.6 that an FCP infra-
integral extension admits a co-subintegral closure.

Although we have similar statements for the subintegral case (Theorem 6.15)
and the infra-integral case (Theorem 6.21), this last case needs some lemmata
before getting the result. To this aim, we begin by considering an FCP
extension $R\subset S$ with $T,U\in[R,S],\ T\neq U$ such that $T\subset S$ and
$U\subset S$ are both minimal non-inert and look at the behavior of $T\cap
U\subset S$.

######  Lemma 6.17.

Let $R\subset S$ be an integral FCP extension. Assume that there exist
$T,U\in[R,S],\ T\neq U$ such that $T\subset S$ and $U\subset S$ are both
minimal non-inert with the same crucial maximal ideal $M$. Set $V:=T\cap U$.
The following properties hold:

  1. (1)

$T\subset S$ and $U\subset S$ are minimal of the same type.

  2. (2)

$M\in\mathrm{Max}(V)$ and $M=(V:S)$.

  3. (3)

$V\subset T$ and $V\subset U$ are t-closed finite extensions.

  4. (4)

If $T\subset S$ and $U\subset S$ are both ramified, then $R\subset S$ admits
neither a co-subintegral closure nor a co-infra-integral closure.

###### Proof.

(1) According to [7, Proposition 5.7], $T\subset S$ and $U\subset S$ are
minimal extensions of the same type and $M\in\mathrm{Max}(V)$.

(2) Since $M$ is also an ideal of $S$, it follows that $M=(V:S)$.

(3) We get that $M\in\mathrm{Max}(V),\ \mathrm{Max}(T)$ and
$M\in\mathrm{Max}(U)$; so that, $V/M\subset T/M$ and $V/M\subset U/M$ are
finite fields extensions, and then t-closed extensions by [17, Lemme 3.10],
and so are $V\subset T$ and $V\subset U$ by [17, Théorème 3.15]. In
particular, $V\subset S$ is not infra-integral.

(4) Assume that $R\subset S$ admits a co-subintegral closure $S^{+}_{R}$. If
$T\subset S$ and $U\subset S$ are both minimal ramified, $T,U\in[S^{+}_{R},S]$
implies that $V=T\cap U\in[S^{+}_{R},S]$, a contradiction by (3) since
$S^{+}_{R}\subset S$ is subintegral. It follows that the co-infra-integral
closure also does not exists by Corollary 6.8. ∎

######  Lemma 6.18.

Let $R\subset S$ be an integral catenarian FCP extension. Let $T,U\in[R,S],\
T\neq U$ be such that $T\subset S$ and $U\subset S$ are minimal non-inert.
Then $T\cap U\subset S$ is infra-integral and $\ell[T\cap U,S]=2$.

###### Proof.

Set $V:=T\cap U,\ M:=(T:S)$ and $N:=(U:S)$. Then, $M$ and $N$ are ideals of
$S$. Assume first that $M\neq N$. According to [7, Proposition 6.6], $V\subset
T$ and $V\subset U$ are both minimal non-inert; so that, $V\subset S$ is
infra-integral.

Assume now that $M=N$. By Lemma 6.17, we get, that $V\subset T$ is t-closed.
Since $R\subset S$ has FCP, there exists $T^{\prime}\in[V,T]$ such that
$T^{\prime}\subset T$ is minimal inert by Proposition 2.6. Moreover,
$M=(T^{\prime}:T)=(T:S)$, a contradiction given by [25, Lemma 4.10], since
$R\subset S$ is catenarian.

To conclude, the only possible case is that $M\neq N$ and $T\cap U\subset S$
is infra-integral with $\ell[V,S]=2$. ∎

######  Lemma 6.19.

Let $R\subset S$ be a catenarian FCP integral extension. Let $T,U\in[R,S]$ be
such that $T\subset S$ is infra-integral and $U\subset S$ is minimal non-
inert. Then $T\cap U\subset S$ is infra-integral. If, in addition,
$T\not\subseteq U$, then $T\cap U\subset T$ is minimal non-inert.

###### Proof.

Assume first that $T\subseteq U$. Then, $T\cap U=T$; so that, $T\cap U\subset
S$ is infra-integral.

Assume now that $T\not\subseteq U$ and set $V:=T\cap U$.

We will prove that $V\subset S$ is infra-integral and $V\subset T$ is minimal
non-inert by induction on $r:=\ell[T,S]<\ell[R,S]$. For $r=1$, we have
$\ell[T,S]=1$; so that, $T\subset S$ and $U\subset S$ are minimal non-inert.
Then, Lemma 6.18 gives the result because $T\neq U$.

Assume that the result holds for some $1\leq r\leq\ell[R,S]-2$. Let
$T,U\in[R,S],\ T\not\subseteq U$ be such that $T\subset S$ is infra-integral
and $U\subset S$ is minimal non-inert, with $\ell[T,S]=r+1$. Since $r\geq 1$,
$T\subset S$ is not minimal; so that, there exists $T^{\prime}\in[T,S]$ such
that $T\subset T^{\prime}$ is minimal non-inert. Since $R\subset S$ is
catenarian, we have $\ell[T,S]=\ell[T,T^{\prime}]+\ell[T^{\prime},S]$, which
implies that $\ell[T^{\prime},S]=r$, and the induction hypothesis holds for
$T^{\prime}\subset S$ and $U\subset S$ because $T^{\prime}\not\subseteq U$.
Setting $V^{\prime}:=T^{\prime}\cap U$, we get that $V^{\prime}\subset S$ is
infra-integral, with $V^{\prime}\subset T^{\prime}$ minimal (non-inert). We
have the following commutative diagram:

$\begin{matrix}T&\to&T^{\prime}&\to&S\\\ \uparrow&{}&\uparrow&{}&\uparrow\\\
V&\to&V^{\prime}&\to&U\end{matrix}$

Applying Lemma 6.18 to the extensions $T\subset T^{\prime}$ and
$V^{\prime}\subset T^{\prime}$, it follows that $T\cap V^{\prime}=T\cap
T^{\prime}\cap U=T^{\prime}\cap V=V\subset T^{\prime}$ is infra-integral, and
so is $V\subset S$. In fact, $V^{\prime}\neq T$. Otherwise, $V=T$ leads to
$T\subseteq U$, a contradiction. Moreover, $V\subset T$ is minimal (non-
inert). Then, the induction hypothesis holds for $r+1$, and then for any $r$.
∎

######  Lemma 6.20.

Let $R\subset S$ be an integral catenarian FCP extension. If $T,U\in[R,S]$ are
such that $T\subset S$ and $U\subset S$ are infra-integral, then $T\cap
U\subset S$ is infra-integral.

###### Proof.

Set $n:=\ell[U,S]$. There exists a maximal chain $\\{U_{i}\\}_{i=0}^{n}$ such
that $U_{0}:=S$ and $U_{n}:=U$, with $U_{i+1}\subset U_{i}$ minimal infra-
integral. For each $i\in\mathbb{N}_{n}$, set $V_{i}:=T\cap U_{i}$. We prove by
induction on $i\in\mathbb{N}_{n}$ that $V_{i}\subset S$ is infra-integral for
each $i\in\mathbb{N}_{n}$.

For $i=1$, Lemma 6.19 gives the result.

Assume that the induction hypothesis holds for some $i\in\mathbb{N}_{n-1}$; so
that, $V_{i}\subset S$ is infra-integral. In particular, $V_{i}\subseteq
U_{i}$ is infra-integral and $U_{i+1}\subset U_{i}$ is minimal infra-integral.
As $V_{i+1}=T\cap U_{i+1}$ and $U_{i+1}\subset U_{i}$, we get that
$V_{i+1}=T\cap U_{i+1}\cap U_{i}=V_{i}\cap U_{i+1}$. Applying Lemma 6.19 to
the extensions $V_{i}\subseteq U_{i}$ and $U_{i+1}\subset U_{i}$, we get that
$V_{i+1}\subseteq U_{i}$ is infra-integral, and so is $V_{i+1}\subset S$. To
conclude, the induction hypothesis holds for any $i\in\mathbb{N}_{n}$. In
particular, for $i=n$, it follows that $V_{n}=T\cap U\subset S$ is infra-
integral. ∎

######  Theorem 6.21.

Let $R\subset S$ be an integral FCP extension. Set $T:=\cap[U\in[R,S]\mid
U\subseteq S$ is infra-integral $]$. Then the following conditions are
equivalent:

  1. (1)

The co-infra-integral closure of $R\subset S$ exists;

  2. (2)

$T\subseteq S$ is infra-integral;

  3. (3)

$T\subseteq S$ is catenarian.

If these conditions hold, then $T$ is the co-infra-integral closure of
$R\subset S$.

###### Proof.

(1) $\Rightarrow$ (2). We prove first that $T$ is the co-infra-integral
closure of $R\subset S$. Indeed, let $T^{\prime}$ be the co-infra-integral
closure of $R\subset S$; so that, $T\subseteq T^{\prime}$. But
$T^{\prime}\subseteq U$ for any $U\in[R,S]$ such that $U\subseteq S$ is infra-
integral. Hence, $T^{\prime}\subseteq T$ leads to $T^{\prime}=T$ and then (1)
$\Rightarrow$ (2).

(2) $\Rightarrow$ (1) Since any $U\in[R,S]$ such that $U\subseteq S$ is infra-
integral verifies $T\subseteq U$, it follows that $T$ is the co-infra-integral
closure of $R\subset S$.

(2) $\Rightarrow$ (3) by [25, Proposition 4.7].

(3) $\Rightarrow$ (2) Since $R\subset S$ has FCP, set $m:=\ell[T,S]$. If
$T=S$, there is no $U\in[R,S]$ such that $U\subset S$ is infra-integral; so
that, $S$ is the co-infra-integral closure of $R\subset S$. Assume that $m\geq
1$. Then, $T\subset S$. We show by induction on $k\in\mathbb{N}_{m}$ that
there exists $U_{k}\in[T,S]$ such that $U_{k}\subset S$ is infra-integral with
$\ell[U_{k},S]=k$. Since $m\geq 1$, there is some $U_{1}\in[T,S[$ such that
$U_{1}\subset S$ is minimal infra-integral; so that, the induction hypothesis
holds for $k=1$. Assume that the induction hypothesis holds for some
$k\in\mathbb{N}_{m-1}$; so that, there exists $U_{k}\in[T,S]$ such that
$U_{k}\subset S$ is infra-integral with $\ell[U_{k},S]=k$. In particular,
$T\subset U_{k}$ since $k<m$. Assume that for any $V\in[R,S]$ such that
$V\subset S$ is infra-integral we have $U_{k}\subseteq V$. Then, $T=U_{k}$, a
contradiction. This implies that there exists some $V\in[R,S]$ such that
$V\subset S$ is infra-integral, with $U_{k}\not\subseteq V$, giving
$T\subseteq V\cap U_{k}\subset U_{k}$. By Lemma 6.20, $V\cap U_{k}\subset
U_{k}$ is infra-integral, and there exists $U_{k+1}\in[V\cap U_{k},U_{k}]$
such that $U_{k+1}\subset U_{k}$ is minimal infra-integral; so that,
$\ell[U_{k+1},S]=k+1$. Then, the induction hypothesis holds for any
$k\in\mathbb{N}_{m}$. Of course, for $k=m$, we get that $U_{m}=T$, and
$T\subset S$ is infra-integral. ∎

######  Corollary 6.22.

Let $R\subset S$ be an integral catenarian FCP extension. Then the co-infra-
integral closure of $R\subset S$ exists.

###### Proof.

We use the ring $T:=\cap[U\in[R,S]\mid U\subseteq S$ is infra-integral$]$
defined in Theorem 6.21. Then $T\subseteq S$ is catenarian and Theorem 6.21
shows the existence of the co-infra-integral closure of $R\subset S$. ∎

######  Proposition 6.23.

Let $R\subset S$ be an integral FCP extension with co-subintegral (resp.; co-
infra-integral) closure $T$. For any $U\in[R,S],\ TU$ is the co-subintegral
(resp.; co-infra-integral) closure of $U\subseteq S$, that is
$(S^{+}_{R})U=S^{+}_{U}$ (resp.; $(S^{t}_{R})U=S^{t}_{U}$).

###### Proof.

Since $TU\in[T,S]$, we get that $TU\subseteq S$ is subintegral (resp.; infra-
integral), with $TU\in[U,S]$. Let $V\in[U,S]$ be such that $V\subseteq S$ is
subintegral (resp.; infra-integral). It follows that $T\subseteq V$; so that,
$TU\subseteq V$, and $TU$ is the least $V\in[U,S]$ such that $V\subseteq S$ is
subintegral (resp.; infra-integral). Then, $TU$ is the co-subintegral (resp.;
co-infra-integral) closure of $U\subseteq S$. ∎

######  Remark 6.24.

Let $R\subset S$ be an integral FCP extension with co-subintegral closure $T$
and let $U\in[R,S]$. Contrary to other closures like integral closure,
seminormalization or t-closure, there is no reason for $U\cap T$ to be the co-
subintegral closure of $R\subseteq U$ because $U\cap T$ is not necessarily the
least $V\in[R,T]$ such that $U\cap T\subseteq U$ is subintegral. The example
given in Remark 5.11 and Theorem 6.6 show that $R^{2}$ is the co-subintegral
closure of $R\subset S$. But $R^{2}\cap{}^{+}_{S}R={}^{+}_{R^{2}}R$ is not the
co-subintegral closure of $R\subset{}^{+}_{S}R$ since $R\subset{}^{+}_{S}R$ is
subintegral. In this example, since $R\subset S$ is infra-integral, $R$ is
both the co-infra-integral closure of $R\subset S$ and $R\subset{}^{+}_{S}R$.

We recall a result from [26].

######  Proposition 6.25.

[26, Proposition 5.21] An integral FCP extension $R\subset S$, that is split
at ${}^{+}_{S}R$ (resp.; ${}^{t}_{S}R$), has a co-subintegral (resp.; co-
infra-integral) closure $S^{+}_{R}$ (resp.; $S^{t}_{R}$), which is its
complement.

######  Remark 6.26.

(1) The last result of Proposition 6.25 could also be gotten by Proposition
6.12.

(2) Let $R\subset S$ be an integral FCP extension such that the co-subintegral
closure $S^{+}_{R}$ exists and is a complement of ${}^{+}_{S}R$. Then
$R\subset S$ does not need to split at ${}^{+}_{S}R$. See the example of
Remarks 6.24 and 5.11, and take $R_{1}:=R+(M\times M)$. Then,
${}^{+}_{S}R_{1}={}^{+}_{S}R$ and $S^{+}_{R_{1}}=S^{+}_{R}=R^{2}$. We get
${}^{+}_{S}R_{1}\cap S^{+}_{R_{1}}=R_{1}$ and
$({}^{+}_{S}R_{1})(S^{+}_{R_{1}})=S$. But $R_{1}$ is a local ring; so that,
$R_{1}\subset S$ does not split at ${}^{+}_{S}R_{1}$.

We end this section by giving some examples where we can check if the co-
subintegral closure exists or not.

The following Lemmata are special cases of Lemma 6.17 and explain more
precisely the situation of Remark 6.3.

######  Lemma 6.27.

Let $K\subset S$ and $L\subset S$ be two minimal ramified extensions where $K$
and $L$ are distinct fields. Set $k:=K\cap L$. The following properties hold:

  1. (1)

There exists $x\in S,\ x\neq 0$ such that $x^{2}=0$ with $S=K+Kx=L+Lx$, which
is a local ring. Moreover, $M:=Kx=Lx$ is the maximal ideal of $S$.

  2. (2)

There exists a $k$-isomorphism of $k$-algebras $\varphi:K\to L$ defined, for
any $a\in K$, by $\varphi(a)=b$, where $b$ is the unique element of $L$ such
that $ax=bx$.

  3. (3)

$k$ is a field and the extension $k\subset S$ does not admit a co-subintegral
closure.

###### Proof.

(1) Since $K\subset S$ is ramified, there exists $x\in S,\ x\neq 0$ such that
$x^{2}=0$ with $S=K+Kx$, which is a local ring with maximal ideal $M:=Kx$ by
Theorem 1.3. Since $x^{2}=0$, we have $x\not\in L$ because $L$ is a field; so
that, $S=L[x]$. Then, $S=L+Lx$ and $M=Lx$. In particular, $Kx=Lx\ (*)$.

(2) Let $a\in K$. There exists $b\in L$ such that $ax=bx$ by $(*)$. We claim
that this $b$ is unique for such $a$. Let $b^{\prime}\in L$ such that
$ax=b^{\prime}x$; so that, $ax=bx=b^{\prime}x$, which leads to
$(b-b^{\prime})x=0$. In particular, $b-b^{\prime}\in M$. If not,
$b-b^{\prime}$ is a unit of $S$ since $S$ is a local ring, which leads to
$x=0$, a contradiction with (1). Then, $b-b^{\prime}\in M\cap L=\\{0\\}$; so
that, $b=b^{\prime}$. Define $\varphi:K\to L$ by $\varphi(a)=b$, for any $a\in
K$, where $b$ is the unique element of $L$ such that $ax=bx$. If $a\in k=K\cap
L$, the equality $ax=ax$ shows that $\varphi(a)=a$, giving that $\varphi$ is
the identity on $k$. For $a,a^{\prime}\in K$, we obviously have
$\varphi(a+a^{\prime})=\varphi(a)+\varphi(a^{\prime})$. Set $b:=\varphi(a)$
and $b^{\prime}:=\varphi(a^{\prime})$; so that, $ax=bx$ and
$a^{\prime}x=b^{\prime}x$. Then,
$aa^{\prime}x=ab^{\prime}x=b^{\prime}ax=b^{\prime}bx$ leads to
$\varphi(aa^{\prime})=\varphi(a)\varphi(a^{\prime})$. In particular, this
relation shows that $\varphi$ is a $k$-algebra injective morphism, because
between two fields, and is obviously surjective. To conclude, $\varphi$ is a
$k$-algebra isomorphism.

(3) $k$ is obviously a field, and $k=K\cap L\subset S$ is not subintegral
since there is a residual extension $k\subset S/M\cong K$, which is not an
isomorphism. Anyway, when $k\subset S$ has FCP, Lemma 6.17(4) shows that
$k\subset S$ does not admit a co-subintegral closure since $(K:S)=(L:S)=0$,
the maximal ideal of both $K$ and $L$. ∎

######  Lemma 6.28.

Let $K\subset S$ and $L\subset S$ be two minimal decomposed extensions where
$K$ and $L$ are distinct fields. Set $k:=K\cap L$. The following properties
hold:

  1. (1)

There exists $x\in S,\ x\neq 0$ such that $x^{2}=x$ with $S=K+Kx=L+Lx$.
Moreover, $M_{1}:=Kx=Lx$ and $M_{2}:=K(1-x)=L(1-x)$ are the maximal ideals of
$S$.

  2. (2)

There exists a $k$-isomorphism of $k$-algebras $\varphi:K\to L$ defined, for
any $a\in K$, by $\varphi(a)=b$, where $b$ is the unique element of $L$ such
that $ax=bx$.

  3. (3)

$k$ is a field and the extension $k\subset S$ does not admit a co-infra-
integral closure.

###### Proof.

(1) Since $K\subset S$ is decomposed, there exists $x\in S,\ x\neq 0,1$ such
that $x^{2}=x$ with $S=K+Kx$, which has maximal ideals $M_{1}:=Kx$ and
$M_{2}:=K(1-x)$ by Theorem 1.3. Since $x^{2}-x=0$, we have $x\not\in L$ since
$L$ is a field; so that, $S=L[x]$. Then, $S=L+Lx$ and $M_{1}=Lx$ because $x\in
M_{1}$. In a similar way, $M_{2}:=L(1-x)$. In particular, $Kx=Lx\ (*)$.

(2) Let $a\in K$. There exists $b\in L$ such that $ax=bx$ by $(*)$. As in
Lemma 6.27, we claim that this $b$ is unique for such $a$. Let $b^{\prime}\in
L$ such that $ax=b^{\prime}x$; so that, $ax=bx=b^{\prime}x$, which would lead
to $(b-b^{\prime})x=0$. In particular, $b-b^{\prime}$ is not a unit of $S$,
which leads to $x=0$, a contradiction with (1). Then,
$b-b^{\prime}\in(M_{1}\cup M_{2})\cap L=\\{0\\}$; so that, $b=b^{\prime}$.
Define $\varphi:K\to L$ by $\varphi(a)=b$, for any $a\in K$, where $b$ is the
unique element of $L$ such that $ax=bx$. If $a\in k=K\cap L$, the equality
$ax=ax$ shows that $\varphi(a)=a$, giving that $\varphi$ is the identity on
$k$. For $a,a^{\prime}\in K$, we obviously have
$\varphi(a+a^{\prime})=\varphi(a)+\varphi(a^{\prime})$. Set $b:=\varphi(a)$
and $b^{\prime}:=\varphi(a^{\prime})$; so that, $ax=bx$ and
$a^{\prime}x=b^{\prime}x$. Then,
$aa^{\prime}x=ab^{\prime}x=b^{\prime}ax=b^{\prime}bx$ leads to
$\varphi(aa^{\prime})=\varphi(a)\varphi(a^{\prime})$. In particular, this
relation shows that $\varphi$ is a $k$-algebra injective morphism, because
between two fields, and is obviously surjective. To conclude, $\varphi$ is a
$k$-algebra isomorphism.

(3) $k$ is obviously a field, and $k=K\cap L\subset S$ is not infra-integral
since there are two residual extensions $k\subset S/M_{i}\cong K,\ i=1,2$,
which are not an isomorphism. Anyway, if $k\subset S$ has FCP, since
$(K:S)=(L:S)=0$, the maximal ideal of both $K$ and $L$, a proof similar as the
one of Lemma 6.17 (4) shows that $k\subset S$ does not admit a co-infra-
integral closure. ∎

We remark that there do not exist $K\subset S$ and $L\subset S$ which are two
minimal infra-integral extensions of different types, where $K$ and $L$ are
distinct fields. Indeed, if $K\subset S$ is minimal ramified, then $S$ is
local, while it has two maximal ideals if $L\subset S$ is minimal decomposed,
a contradiction. See also [7, Proposition 5.7].

The situation of Lemma 6.28 is satisfied in the following example.

######  Example 6.29.

Let $K:=\mathbb{Q}[j]$, where $j=(-1+i\sqrt{3})/2$, and
$S:=K[X]/(X^{2}-X)=K[x]$, where $x$ is the class of $X$ in $S$ and satisfies
$x^{2}=x$. Then, $K\subset S$ is a minimal decomposed extension, hence infra-
integral, and $\mathbb{Q}\subset K$ is a minimal separable field extension of
degree 2, and then t-closed. Set $\alpha:=-j+(1+2j)x\in S\setminus K$ because
$x\not\in K$. A short calculation shows that $\alpha$ is a zero of the
irreducible polynomial $X^{2}-X+1\in\mathbb{Q}[X]$; so that,
$L:=\mathbb{Q}[\alpha]$ is a field, with $L\in[\mathbb{Q},S],\ L\neq K$,
because $\alpha\not\in K$, and $\mathbb{Q}\subset L$ a minimal separable field
extension of degree 2. Moreover, $x\not\in L$ since $L$ is a field and
$x^{2}-x=0$. Since $(2x-1)^{2}=1$, this shows that $2x-1$ is a unit of $L[x]$.
Then, $j=(\alpha-x)(2x-1)^{-1}\in L[x]$; so that, $S=L[x]$. It follows that
$L\subset S$ is a minimal decomposed extension, and then infra-integral. As
$\mathbb{Q}=K\cap L$, Lemma 6.28 shows that $\mathbb{Q}\subset S$ does not
admit a co-infra-integral closure.

By Lemma 6.17, we get that an FCP extension of the form $T\cap U\subset S$,
where $T\subset S$ and $U\subset S$ are minimal ramified with the same crucial
maximal ideal does not admit a co-subintegral closure. If $(T:S)\neq(U:S)$,
then [7, Proposition 6.6] asserts that $T\cap U\subset S$ is subintegral, and
then $T\cap U$ is the co-subintegral closure of $T\cap U\subset S$. The
following Proposition exhibits cases where situation of Lemma 6.17 does not
occur.

######  Proposition 6.30.

Let $R\subset S$ be an unbranched integral FCP extension, where $(R,P)$ and
$(S,N)$ are local rings. Assume that $R/P\subset S/N$ is a finite separable
field extension. Then, there does not exist $T,U\in[R,S],\ T\neq U$ such that
$T\subset S$ and $U\subset S$ are minimal ramified with the same crucial
maximal ideal.

###### Proof.

Assume that there exist $T,U\in[R,S],\ T\neq U$ such that $T\subset S$ and
$U\subset S$ are minimal ramified with the same crucial maximal ideal $M$. Set