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\begin{document}
\baselineskip 16 pt
\title{Maximal prime divisors in arithmetical rings}
% Information for first author
\author{Laszlo Fuchs}
% Address of record for the research reported here
\address{Department of Mathematics, Tulane University, New
Orleans, Louisiana 70118}
\email{fuchs@tulane.edu}
% \thanks will become a 1st page footnote.
%\thanks{The first author was supported in part by NSF Grant
%\#000000.}
% Information for second author
\author{William Heinzer}
\address{Department of Mathematics, Purdue University, West
Lafayette, Indiana 47907}
\email{heinzer@math.purdue.edu}
\author{Bruce Olberding}
\address{Department of Mathematical Sciences, New Mexico State University,
Las Cruces, New Mexico 88003-8001}
\email{olberdin@emmy.nmsu.edu}
%\thanks{Support information for the second author.}
\keywords{Primal ideal, associated prime, arithmetical ring, Pr\"ufer domain}
\subjclass{Primary 13A15; Secondary 13F05}
\begin{abstract}
We investigate for an ideal $A$ of an arithmetical ring $R$ the relationship
between the set $\Max(A)$ of maximal prime divisors of $A$ and the set $\XX_A$
of
maximal members of the set of Krull associated primes of $A$. We show that
the arithmetical rings $R$ such that $\XX_A \subseteq \Max(A)$ for every
regular ideal $A$ are precisely those satisfying the ``strong'' separation
property.
%If $R$ is a Pr\"ufer domain with field of fractions $K$ and $A
%\subseteq P$ are ideals of $R$ with $P$ prime, then $\End(AR_P) = R_Q$ for
%some $Q \in \Spec R$ with $Q \subseteq P$. We prove that every element of
%the set $\EE_A = \{Q \in \Spec R : R_Q = \End(AR_P)$ for some $P \in \Spec R$
%with $A \subseteq P$ \} is contained in a maximal element of $\EE_A$ and the
%maximal elements of $\EE_A$ are precisely the maximal associated primes of $A$.
For a Pr\"ufer domain $R$, we prove that every branched prime ideal of
height greater than one is the radical of a finitely generated ideal if and
only if $\End(A)_M = \End(A_M)$ for every nonzero ideal $A$ and maximal ideal
$M$ of $R$. We use this to prove that if in addition
$R$ is a QR-domain, then every
maximal prime divisor of an ideal $A$ of $R$ is a Krull associated prime of
$A$ (i.e. $\XX_A = \Max(A)$) if and only if each branched prime ideal of $R$
of height greater than one is the radical of a finitely generated ideal.
\end{abstract}
\maketitle
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\section*{Introduction}
Let $R$ be a commutative ring with identity. Associated to a
proper ideal $A$ of $R$ is the set
$S(A) = \{x \in R: xy \in A$ for some $y \in R\setminus A \}$
of elements of $R$ that are non-prime to $A$.
The complement $R \setminus S(A)$ of elements prime to $A$
is closed under multiplication. Krull proves \cite[page 732]{K1}
that ideals maximal
with respect to not meeting a multiplicatively closed set
are prime ideals and defines the {\it maximal prime divisors}
of $A$ as the prime ideals $P$ of $R$ that contain $A$
and are maximal with respect to the
property $P \subseteq S(A)$.
Let $\Max(A)$ denote the set of
maximal prime divisors of $A$. For $P \in \Max(A)$,
Krull defines
the ideal $A_{(P)} = \{x \in R : xy \in A$ for some $y \in R\setminus P \}$
to be the {\it principal component } of $A$ with respect to $P$
and establishes the decomposition of every proper ideal $A$
of $R$ as the intersection of its principal components
$A = \bigcap_{P \in \Max(A)}A_{(P)}$ \cite[Satz 2]{K1}.
However, as we discuss in
\cite{FHO}, a drawback to this decomposition is that
we do not know what kind of ideals the $A_{(P)}$ are.
For instance, there
sometimes exist elements of $P$ that are prime to the
principal component $A_{(P)}$; indeed, the ideal $A_{(P)}$
is not in general a primal ideal, where an ideal $B$ is
said to be {\it primal} if the set $S(B)$ of elements
non-prime to $B$ is an ideal. If $B$ is primal, then
$S(B)$ is a prime ideal called the {\it adjoint prime}
of $B$.
To obtain in a commutative ring without finiteness conditions
a decomposition of the ideal $A$ that involves components
closely tied to $A$ with structure we know, in \cite{FHO}
we define the set $\XX_A$ of maximal Krull associated
primes of $A$ and obtain using the set $\XX_A$
a canonical primal decomposition of $A$.
If $R$ is a Noetherian ring and $\Ass(A)=\{P_1,P_2,\ldots,P_n\}$ is the set of
associated primes of a proper ideal $A$ of $R$, then $S(A) = \bigcup_{i=1}^n
P_i$. It follows that the maximal prime divisors of $A$ are exactly the
prime ideals of $R$ that are maximal members of $\Ass(A)$. In this sense, the
maximal prime divisors of an ideal of a Noetherian ring are well-understood.
However if $R$ is not Noetherian, the set $\Max(A)$ of maximal prime divisors
of $A$ is generally less transparent, and the primes in $\Max(A)$ need not be
``associated.'' There are several inequivalent
notions of an associated prime of an ideal of a general
commutative ring, but from our point of view, it is the {\it Krull
associated primes} that are most natural. We review this and related notions
in Section 1. Motivated by the Noetherian case, we are thus interested in the
question of when the set of maximal prime divisors of an ideal $A$ coincides
with the set of maximal members of the set of Krull associated primes of $A$.
We shall examine this question for the class of {\it arithmetical rings},
those rings $R$ such that for every maximal ideal $M$ of $R$, $R_M$ is a {\it
valuation ring}, that is, a ring for which the set of ideals is linearly
ordered by inclusion. Of particular interest is the class of
{\it Pr\"ufer domains}, namely the arithmetical integral domains.
Our purpose in the present paper is to investigate for
ideals $A$ of an arithmetical ring or Pr\"ufer domain the interrelationship
between the sets $\Max(A)$ and $\XX_A$. The connection between these two sets
of prime ideals is of special interest in the context of arithmetical
rings. Indeed, if $A$ is a proper ideal of an arithmetical ring $R$,
then $\Max(A) = \XX_A$ if and only if every prime ideal containing $A$ but not
prime to $A$ is a Krull associated prime (Corollary \ref{cor 1.3}). Also,
Krull mentions \cite[p. 16]{K2} that he does not know whether the principal
components $A_{(P)}$ of $A$, $P \in \Max(A)$, are always $P$-primal ideals.
(A primal ideal $B$ is $Q$-primal if $Q = S(B)$.) Examples show that the
answer is in the negative (see \cite{Ng1},\cite{G} and \cite[Example 3.8]{FHO});
we investigate in Section 3 arithmetical rings for which the answer
to this question is in the affirmative for all regular ideals.
(An ideal $A$ of $R$ is {\it regular} if it
contains a regular element, i.e., an element that is
not a zero-divisor.)
This is also of special interest, because
an ideal of an arithmetical ring is primal exactly if it is irreducible
(Theorem 1.8 of \cite{FHO}).
In Section 3 we examine conditions on an arithmetical ring
$R$ in order that each maximal prime divisor of a regular
ideal $A$ of $R$ be a Krull associated prime of $A$.
It follows from Theorem~\ref{theorem 6.3} that a necessary condition for
this to hold is that $R$ satisfy the strong separation
property: for regular
prime ideals $P\subset Q$ and regular element $r \in P$, there exists $s \in Q$
such that $P \subset (r, s)R \subseteq Q$. We conclude that Pr\"ufer domains
such as the ring of entire functions and the ring $\Int(\mathbb{Z})$ of
integer-valued polynomials contain ideals $A$ having the
property that there exists a
maximal prime divisor of $A$ that is not a Krull associated prime of $A$.
For a regular ideal $A$ of an arithmetical ring, we consider in
Theorem 3.7 conditions in order
that the set of maximal prime divisors of $A$ is precisely the set of prime
ideals that are maximal among the Krull associated primes of $A$. We deduce a
complete characterization of the QR-domains with this property. (By
contrast, ideals of Noetherian rings always exhibit equality between these
two sets of prime ideals.)
The results that lay the
groundwork for this theorem touch on several interesting technical aspects of
arithmetical rings and Pr\"ufer domains. In particular, there is a
connection between our problem and that of when $\End(X)$
localizes, that is, when $\End(X)_M = \End(X_M)$ for a submodule $X$ of the
quotient field of $R$ and maximal ideals $M$ of $R$.
{\it Acknowledgement.} We thank Gabriel Picavet and David Rush
for helpful conversations on the topics of this paper and \cite{FHO}, and for
showing us the connections of our work to the literature on Krull associated
primes.
\section{Krull associated primes}
In this section we briefly review the notion of a Krull associated prime of a
proper ideal of a ring. We became interested in the Krull associated primes of an
ideal because of their connection to the primal isolated components of
the ideal. For a more complete treatment of these primes and their
relation to issues involving the primal decompositions of an ideal, see
\cite{FHO}.
Let $A$ be a proper ideal of the ring $R$.
Following \cite{IR}, we
define a prime ideal $P$ of $R$ to be a {\it Krull associated prime}
of the ideal $A$ if for every element $x \in P$, there exists $y \in R$ such
that $x \in (A:y) \subseteq P$.
A prime ideal $P$ of $R$ is called a
{\it weak-Bourbaki associated prime} of $A$ if it is a minimal prime
divisor
of $(A:x)$ for some $x \in R\backslash A$. Following \cite[page 279]{HO},
we call $P$ a {\it Zariski-Samuel
associated prime} of $A$ if $P = \sqrt {(A:x)}$ for some $x \in R
\setminus A$. If $R$ is a Noetherian ring, then all three of these notions of
an associated prime coincide, but for non-Noetherian rings these notions are
in general distinct (see \cite{HO}, for example).
It is clear that a Zariski-Samuel associated
prime is a weak-Bourbaki associated prime. Moreover, it is noted in Lemma 2.1
of \cite{FHO} that $P$ is a Krull associated prime of $A$ if and only if $P$
is a set-theoretic union of weak-Bourbaki primes of
$A$. It is this characterization of Krull associated primes that we shall
use in what follows.
We denote by $\Ass(A)$ the set of Krull associated primes of $A$. Notice
that if $A$ is a proper ideal of $R$, then $\Ass(A)$ is nonempty. By
contrast, Nakano \cite{Na} gives an example of a Pr\"ufer
domain such that no finitely generated ideal has a Zariski-Samuel
associated prime.
The set $\Ass(A)$ behaves well with respect to
localization:
\begin{lemma} \label{1.1}
{\em (Lemma 2.4(ii) of \cite{FHO})} Let $A$ be an ideal of a ring $R$ and $P$
be a prime ideal of $R$ containing $A$. Then $P \in \Ass(A)$ if and only if
$P_M \in \Ass(A_M)$ for some (or
equivalently every) maximal ideal $M$ of $R$ containing $P$.
\end{lemma}
It is not hard to see that if $A$ is a proper
ideal of a ring $R$, then every member of $\Ass(A)$ is contained in a maximal
member of $\Ass(A)$. We define $\XX_A$ to be the maximal members of
$\Ass(A)$, that is, $\XX_A$ consists of the maximal Krull associated primes
of $A$. There exist examples of rings with ideals $A$ such
that $\XX_A$ is infinite. (See Example 2.6 of \cite{FHO} for example, or use
Lemma~\ref{lemma 6.2} below.) These examples are necessarily non-Noetherian.
In Lemma 2.3 of \cite{FHO}, we observe that $S(A) = \bigcup_{P \in \XX_A}P$.
In particular, $\bigcup_{P \in \XX_A}P = \bigcup_{Q \in \Max(A)}Q$. Despite
this close connection between $\XX_A$ and $\Max(A)$, one cannot conclude in
general that $\XX_A \subseteq \Max(A)$ or $\Max(A) \subseteq \XX_A$. If
$\Max(A)$ has only one member, then it is easy to see $\Max(A) = \XX_A$.
However, in Example 3.8 of \cite{FHO} we construct a ring $R$ with Noetherian
prime spectrum and an ideal $A$ of $R$ such that $\Max(A)$ has only two
elements, but neither maximal prime divisor of $A$ is in $\XX_A$. We also
construct in Example 2.9 of \cite{FHO} a 2-dimensional Pr\"ufer domain $R$ for
which $\Spec(R)$ is Noetherian and for which there exist an ideal $A$ and a
finitely generated maximal ideal $M$ such that $M$ is a maximal prime divisor
of $A$ and yet $M \not\in \Ass(A)$.
When $R$ is arithmetical, the set $\Ass(A)$ has some striking properties.
In the next sections we will need the following lemmas.
\begin{lemma} {\em (Proposition 2.7 of \cite{FHO})}
\label{proposition 2.7}
Let $R$ be an arithmetical ring and let $A$
be a proper ideal of $R$. If $P \in \Ass(A)$, then $Q \in
\Ass(A)$ for every prime ideal $Q$ of $R$ such that $A \subseteq
Q \subseteq P$.
\end{lemma}
\begin{corollary} \label{cor 1.3}
Let $A$ be a proper ideal of an arithmetical ring $R$.
Then $\XX_A = \Max(A)$ if and only if every prime ideal of $R$ that contains
$A$ and is not prime to $A$ is a Krull associated prime of $A$.
\end{corollary}
\begin{proof} Assume that $\XX_A = \Max(A)$. If $P$ contains $A$ and is
not prime to $A$, then $P$ is
contained in some $Q \in \XX_A$. Hence by Lemma 1.2, $Q \in \Ass(A)$. The
converse is clear.
\end{proof}
Let $A$ be a nonzero proper ideal of a Pr\"ufer domain $R$ and
let $Q$ be a prime ideal of $R$ with $A \subseteq Q$.
Since $R_Q$ is a valuation domain, $\End(AR_Q) = R_P$ for
some $P \in \Spec R$ with $P \subseteq Q$. Define
\[
\EE_A = \{P \in \Spec R : R_P = \End(AR_Q) {\mbox{ for some
prime ideal }} Q {\mbox{ containing }} A \}.
\]
\begin{lemma}
\label{proposition 2.8} {\em (Proposition 2.8 of \cite{FHO})} Let $A$ be a
nonzero proper ideal of a Pr\"ufer domain $R$. Then $\Ass(A) = \EE_A$. In
particular, if $P \in \XX_A$, then there exists a maximal ideal $M$ of $R$
such that $\End(A_M) = R_P$. \end{lemma}
\begin{lemma}
\label{lemma 6.2}
Let $A$ be an ideal of an arithmetical ring $R$. If $A_M$ is
a nonzero finitely generated ideal of $R_M$ for each maximal ideal $M$
of $R$, then
$\Ass(A)$ is precisely the set of prime ideals of $R$
containing $A$. \end{lemma}
\begin{proof} First assume that $R$ is a valuation ring and $A$
is a nonzero finitely generated
proper ideal of $R$. Then $A = yR$ is a principal ideal of $R$.
Let $P$ be the maximal ideal of $R$. If $A =
P$, then clearly $P \in \Ass(A)$, so suppose there exists $x \in
P \setminus A$. Since $R$ is a valuation ring, $A \subset Rx$
and $A = (A:x)xR$. If $(A:x) = A$, then $A = xA$ and $y = xya$
for some $a \in A$. But this means $y(1 - xa) = 0$. Since $1-xa$ is
a unit of $R$, this implies $y = 0$, a contradiction to
our assumption that $A \ne 0$. Therefore each $x \in P$ is
non-prime to $A$, so $A$ is a primal ideal with $S(A) = P$
and $P \in \Ass(A)$.
In the general case where $R$ is an arithmetical ring, we
conclude from Lemma 2.1 and the
preceding argument that every maximal ideal $P$ of $R$ that contains $A$ is
in $\Ass(A)$. Hence by Lemma~\ref{proposition 2.7}, every prime ideal of $R$
containing $A$ is in $\Ass(A)$. \end{proof}
\section{The case $\XX_A \subseteq \Max(A)$}
In this section we characterize when $\XX_A \subseteq \Max(A)$ for every
regular ideal $A$ of an arithmetical ring. To do this, we first recall the
notion of separated prime ideals, but we reformulate this definition to
include rings with zero-divisors in such a way that our definition agrees
with that of the separation property for domains (see pp. 91-92 of \cite{FHP}).
We define a ring $R$ to have the {\it separation property} if for each pair
of distinct comparable regular prime ideals $P \subset Q$ of $R$, there
exists a finitely generated ideal $A$ such that $P \subset A \subseteq Q$.
The ring $R$ has the {\it strong separation property} if for each pair of
comparable regular prime ideals $P \subset Q$ of $R$ and regular element $r
\in P$, there exists $s \in Q$ such that $P \subset (r,s)R \subseteq Q$.
Clearly every
one-dimensional domain has the strong separation
property. By the Altitude Theorem of Krull \cite[page 26]{Ng2}
or \cite[page 110]{Ka}, a Noetherian
domain of dimension greater than two has the separation property
but not the strong separation property. By contrast,
Pr\"ufer domains need not possess either separation property.
The Pr\"ufer domains that have the separation property have been
well-studied (see, for example, pp. 91-92 of \cite{FHP} and Lemma 2.7
of \cite{O}). Note that a B\'ezout domain $R$ has the separation
property if and only if $R$ has the strong separation
property. More generally, a Pr\"ufer domain with the QR-property
( so in particular a Pr\"ufer domain with torsion Picard group )
has the separation property if and only if it has
the strong separation property.
In the following lemma, we collect some technical characterizations of
the separation property for Pr\"ufer domains that will be needed
in the next section.
\begin{lemma}
\label{lemma 6.4}
The following statements are equivalent for a
Pr\"ufer domain $R$.
\begin{itemize}
\item[(i)] $R$ has the separation property.
\item[(ii)] For each nonzero prime ideal $P$ of $R$, $P$
is a maximal prime ideal of $\End(P$).
\item[(iii)] For each nonzero prime ideal $P$ of $R$, $\End(P_M)
=\End(P)_M$ for every maximal ideal $M$ of $R$.
\item[(iv)] For each nonzero nonmaximal prime ideal $P$ of $R$,
if $\{M_i\}$ is the
collection of maximal ideals of $R$ not containing $P$, then
$R_P \subseteq (\bigcap_{i}R_{M_i})R_M$ for each maximal ideal $M$
of $R$ containing $P$. \item[(v)] For each nonzero ideal $A$ of $R$, if
$M$ is a maximal ideal with $A \subseteq M$ and $P$ is a prime ideal such
that $\End(A)_M = R_P$, then no element of $P$ is prime to $A$.
\end{itemize} \end{lemma}
\begin{proof} (i) $\Leftrightarrow$ (ii) Lemma 4.2.38 of \cite{FHP}.
(i) $\Leftrightarrow$ (iii) Lemma 2.7 of \cite{O}.
(iii) $\Rightarrow$ (iv) By Theorem 3.2.6 of \cite{FHP} $\End(P) = R_P
\cap (\bigcap_i R_{M_i})$ and $\End(P_M) = R_P$. Thus by (iii)
$R_P = \End(P_M) = \End(P)_M = R_P \cap (\bigcap_{i}M_i)R_M$, and (iv)
follows.
(iv) $\Rightarrow$ (v) Let $A$ be a nonzero ideal of $R$ and $M$ a
maximal ideal with $A \subseteq M$. Let $P$ be a prime ideal
such that $\End(A)_M = R_P$.
It follows from Lemma~\ref{proposition 2.8} that
\[
\End(A) = (\bigcap_{Q \in \XX_A}R_Q) \cap (\bigcap_{N}
R_N),
\]
where $N$ ranges over the maximal ideals of $R$ that do
not contain $A$. Let $Q'$ be the unique member of $\XX_A$ that
is contained in $M$. By Lemma~\ref{proposition 2.8},
$\End(A_M) = R_{Q'}$. If $Q'$ is a maximal ideal of $R$, then
$Q' = M$ and $\End(A_M) = R_M$, so $R_P \subseteq R_M$ implies
that $M = P = Q' \in \Ass(A)$, and the claim is clear. It remains
to consider the case where $Q'$ is a non-maximal prime ideal of
$R$. Now
$$
R_P = \End(A)_M =(\bigcap_{Q \in \XX_A} R_{Q})R_M \cap
(\bigcap_{N}R_N)R_M.
$$
We claim that $\bigcap_{Q \in \XX_A}
R_{Q} \subseteq R_P$. If this is not the case then since $R_M$ is a
valuation domain it must be that $R_P \subsetneq
(\bigcap_{Q \in \XX_A} R_{Q})R_M$. Hence from the above representation of
$\End(A)_M$ we deduce that since $R_P$ is a valuation domain, $R_P =
(\bigcap_{N}R_N)R_M$. Thus $(\bigcap_{N}R_N)R_M \subsetneq (\bigcap_{Q
\in \XX_A} R_{Q})R_M$. By (iv), $R_{Q'} \subseteq (\bigcap_{N}R_N)R_M$
since no $N$ contains $Q'$. However $Q' \in \XX_A$, so this implies
$R_{Q'} \subsetneq R_{Q'}R_M$, but since $M$ contains $Q'$, $R_{Q'}R_M =
R_{Q'}$. This contradiction implies that $\bigcap_{Q \in \XX_A}R_Q
\subseteq R_P$, so every element $r \in P$ is contained in some $Q \in
\XX_A$. Consequently, no element of $P$ is prime to $A$.
(v) $\Rightarrow$ (iii) Suppose $P$ is a non-maximal prime ideal
of $R$, and let $M$ be a maximal ideal of $R$ containing $P$.
Let $Q$ be a prime ideal of $R$ such that $\End(P)_M = R_Q$. (Since
$\End(P)_M$ is a valuation domain such a prime $Q$ must exist). Then by (v),
the elements of $Q$ are not prime to $P$. Consequently, $P = Q$, and (iii)
follows. \end{proof}
Using the strong separation property, we can characterize when $\XX_A
\subseteq \Max(A)$ for every regular ideal $A$ of an arithmetical ring. In
fact, the strong separation property is always sufficient, regardless of
whether $R$ is arithmetical, to guarantee that $\XX_A \subseteq \Max(A)$:
\begin{lemma}
\label{lemma 6.1} If $A$ is a regular ideal of a ring $R$
having the strong separation property, then every $P \in \XX_A$
is a maximal prime divisor of $A$, i.e., $\XX_A \subseteq
\Max(A)$.
\end{lemma}
\begin{proof} Let $P \in
\XX_A$. There exists $Q \in \Max(A)$ such that
$P \subseteq Q$. We show $P = Q$. If $\XX_A$
consists only of $P$, then $A$ is $P$-primal and $P = Q$.
Suppose $\XX_A \setminus \{P\}$ is nonempty, and let
$\{P_i\} = \XX_A \setminus \{P\}$. Since $Q \in \Max(A)$,
$ Q \subseteq P \cup (\bigcup_{i}P_i)$. Assume $P$ is properly
contained in $Q$. Since $R$ has the strong
separation property and $A$ is a regular ideal, there exists $x \in Q
\setminus P$ such that $P \subset A+(x) \subseteq Q$. It
follows that $x$ is in one of the $P_i$, so $P \subset A+(x)
\subseteq P_i$. This contradicts the assumption that $P \in
\XX_A$ is maximal among the Krull associated primes of $A$. Thus $P
= Q$.\end{proof}
\begin{theorem}
\label{theorem 6.3} Let $R$ be an arithmetical ring.
The following statements are equivalent.
\begin{itemize}
\item[(i)] $R$ has the strong separation property.
\item[(ii)] For each proper regular ideal $B$ of $R$, $\XX_B
\subseteq \Max(B)$.
\end{itemize}
\end{theorem}
\begin{proof} By Lemma \ref{lemma 6.1}, (i) implies (ii). To prove
that (ii) implies (i), suppose $P \subset Q$ are regular prime
ideals of $R$ and that $a$ is a regular element
with $a \in P$. Define $B = aP$
and note that the maximal
ideals of $R$ that contain $B$ are precisely those that contain
$aR$. We first show that $\XX_B$ is the set union of $\{P\}$
and the set of maximal ideals of $R$ that contain $B$ but not
$P$.
Since $a$ is a regular element, we have $P = B:a$, so $P \in \Ass(B)$. We
claim that $P \in \XX_B$. Suppose $L \in \XX_B$ and $P \subseteq L$. Let $M$
be a maximal ideal containing $L$. By Lemma 1.7 of \cite{FHO},
$(BR_M)_{(PR_M)}$ is a $P_M$-primal ideal of $R_M$ since $BR_M$ is a finitely
generated ideal of $R_M$. We claim $BR_M = (BR_M)_{(PR_M)}$. To verify that
this is the case, we need only show that $BR_M:y = BR_M$ for all $y \in
R_M\setminus P_M$. To this end, observe first that $PR_M = yPR_M$ for all $y
\in R_M \setminus P_M$. This is because $R_M$ is a chained ring, so if $y \in
R_M \setminus P_M$, then $PR_M \subseteq yR_M$, and if $p \in PR_M$, then $p
= yu$ for some $u \in R_M$. Since $y \not \in PR_M$, it follows that $u \in
PR_M$ and $p \in yP_{M}$. Thus $PR_M = yPR_M$. Hence, with $y \in R_M
\setminus PR_M$, we have that $BR_M:y = yBR_M:y = BR_M$, since $y \not \in
PR_M$ implies that $PR_M \subseteq yR_M$ and hence that $y$ is a regular
element of $R_M$. Therefore $BR_M = (BR_M)_{PR_M}$, and we conclude that
$BR_M$ is a $PR_M$-primal ideal. However, by Lemma 1.1 $LR_M \in \Ass(BR_M)$,
so this forces $LR_M \subseteq PR_M$. Hence $L
\subseteq P$, and we may conclude that $P = L \in \XX_B$.
Now suppose $N$ is a maximal ideal of $R$ that contains $B$ but
not $P$. Then $B_N = aR_N \ne 0$, so $N$ is in
$\XX_B$ by Lemma~\ref{lemma 6.2} and Lemma 1.1. This proves that
the set $\XX_B$ contains $P$ and the
maximal ideals of $R$ that contain $B$ but not $P$. The reverse inclusion
follows from the fact that any prime ideal of $R$ that is a member of $\XX_B$
must contain $B$.
Let $\{N_i\}$ be the set
of members of $\XX_B$ distinct from $P$. As noted, each $N_i$
is a maximal ideal of $R$, and $\{N_i\}$ is precisely the set
of maximal ideals of $R$ that contain $aR$ but not $P$. Since $P
\in \XX_B$ and, by (ii), $\XX_B \subseteq \Max(B)$, we have $ Q
\not \subseteq S(B) = P \cup (\bigcup_{i}N_i),$ for otherwise $P$, as a proper
subset of $Q$, would not be in $\Max(A)$. In particular, there exists $x \in
Q$ such that $x$ is not contained in $P$ nor in any of the $N_i$. We have
$(a,x)R \subseteq Q$, and to complete the proof we show that $P \subset
(a,x)R$. It is enough by Theorem II.3.1 of \cite[page 88]{B} to show that
$P_M \subset (A,x)R_M$ for all maximal ideals $M$ of $R$ that contain
$(A,x)$. Observe that if $M$ is such a maximal ideal, then since by design
$x$ is not an element of any maximal ideal of $R$ that contains $Ra$ but not
$P$, it must be that $M$ contains $P$. Now since $R_M$ is valuation ring,
the ideals $P_{M}$ and $(a,x)R_{M}$ are comparable, so if $P_{M} \not
\subseteq (a,x)R_{M}$, then it must be that $(a,x)R_{M} \subseteq P_{M}$. But
then $x \in P_{M}$ and since $P \subseteq M$, it is the case that $P_{(M)} =
P$, and we have $x \in P$, a contradiction to the choice of $x$. Thus we
conclude that $P \subseteq (a,x)R \subseteq Q$, and $R$ has the strong
separation property. \end{proof}
It follows from Theorem \ref{theorem 6.3} that if $R$ is an
arithmetical ring that does not have the strong separation
property, then there exist a proper regular ideal $B$ of $R$ and
$P \in \XX_B$ such that $P \not\in \Max(B)$. By definition of
$\Max(B)$, this means there exists $Q \in \Max(B)$ such that
$P$ is properly contained in $Q$ and by definition of $\XX_B$,
this means $Q \not\in \Ass(B)$. Therefore the strong separation
property is a necessary condition on an arithmetical ring in
order that maximal prime divisors (of regular ideals) always be Krull primes.
\begin{discussion}
\label{discussion 6.5}
{\rm Some important examples of Pr\"ufer
domains do not have the separation property, and so by
Theorem~\ref{theorem 6.3}, these Pr\"ufer domains have ideals $B$ such that $\XX_B
\not\subseteq \Max(B)$. Thus there exist maximal prime divisors of $B$ that
are not in $\Ass(B)$. For example, neither the ring of entire functions nor
the ring of integer-valued polynomials has the separation property. Indeed,
both these rings are completely integrally closed integral domains of Krull
dimension greater than one. If $P$ is a nonzero nonmaximal prime ideal of
such a Pr\"ufer domain $R$, then $\End(P) = R$ and $P$ is not maximal in
$\End(P)$. Hence by Lemma~\ref{lemma 6.4} (ii), such domains do not have
the separation property. On the other hand, every Pr\"ufer domain
having Noetherian maximal spectrum
has the separation property (apply Theorem
4.2.39 of \cite{FHP} and \cite{RW}).} \end{discussion}
\section{The case $\XX_A = \Max(A)$}
Our main objective in this section is to investigate conditions
on an arithmetical ring $R$ in
order that $\XX_A = \Max(A)$ for each regular ideal $A$ of $R$.
We recall that an integral domain $R$ is
a {\it QR-domain} if every overring
of $R$ is a localization of $R$ with respect to some
multiplicatively closed subset of $R$. It is
well-known that QR-domains are necessarily Pr\"ufer, and that
a Pr\"ufer domain with torsion Picard group (e.g. a B\'ezout
domain) is a QR-domain. More generally, a Pr\"ufer domain $R$ is
a QR-domain if and only if the radical of every finitely
generated ideal of $R$ is the radical of a principal ideal of
$R$ \cite{P}. There exist QR-domains having non-torsion Picard group
\cite{H}. Thus the condition on a Pr\"ufer domain $R$ that the
radical of every finitely generated ideal is the radical of a
principal ideal does not imply $R$ has torsion Picard group.
However, it is clear that a QR-domain that has the separation
property also has the strong separation property.
\begin{lemma}
\label{lemma 6.6}
Let $A$ be an ideal of a Pr\"ufer domain $R$.
Suppose $Q$ is a prime ideal of $R$ that contains $A$, and $P$
is a prime ideal such that $\End(A)_Q = R_P$. If $P \in
\Ass(A)$, then $\End(A)_Q = \End(A_Q)$.
\end{lemma}
\begin{proof} Since $P \in \Ass(A)$, $A_{(P)}$ is a primal ideal with
adjoint prime $P$, and it follows that $A_P$ is a $P_P$-primal
ideal. By Lemma~\ref{proposition 2.8}, $\End(A_P) = R_P$. Thus
$\End(A_P) = \End(A)_Q$, so $A\End(A_P) = A\End(A)_Q$ implies $A_P = A_Q$.
Consequently, $ \End(A_Q) = \End(A_P) = R_P = \End(A)_Q$.\end{proof}
\begin{lemma}
\label{lemma 6.7}
Let $R$ be a Pr\"ufer domain with field of
fractions $F$, let $X$ be an $R$-submodule of $F$, and let $M$
be a maximal ideal of $R$. Then $\End(X)_M = R_P$ for some $P
\in \Spec R$ with $P \subseteq M$. Assume that $P$ is the union of prime
ideals $P_i$, where each $P_i$ is the radical of a finitely
generated ideal. Then $\End(X)_Q = \End(X_Q)$ for all prime ideals $Q$
such that $P \subseteq Q \subseteq M$. \end{lemma}
\begin{proof} Since $R_M \subseteq \End(X)_M$ and $R_M$ is a
valuation domain, $\End(X)_M = R_P$ for some prime ideal
$P \subseteq M$. If $\End(X)_M = F$, then clearly $\End(X)_M =
\End(X_M)$, so we assume $\End(X)_M \ne F$ and thus $P \ne (0)$.
Let $Q$ be a prime ideal of $R$ such that $P \subseteq Q \subseteq M$.
Since $\End(X)_M = R_P$, we have $\End(X)_Q = R_P$. Now
$R_P = \End(X)_Q \subseteq \End(X_Q) \subseteq
\End(X_P)$, so to prove Lemma \ref{lemma 6.7},
it suffices to show that $\End(X_P)
\subseteq R_P$.
Let $S = \End(X)$. Now $PS \subseteq PR_P$, so $PS \ne S$. Since $S$ is an
overring of the Pr\"ufer domain $R$, $S$ is a flat extension of $R$, so
$PS$ is a prime ideal of $S$ and $S_{PS} = R_P$. Also, $PS$ is the union
of the prime ideals $P_iS$, and each $P_iS$ is the radical of a finitely
generated ideal of $S$.
Let $L$ be a prime ideal of $S$ such that $L \subseteq PS$
and such that $L = \sqrt{I}$, where $I$ is a finitely
generated ideal of $S$. We prove there exists a nonzero $q \in F$
such that $qX_L$ is an ideal of $S_L$ that is primary for $L_L$.
The invertible ideal $I^2$ of $S$ is an intersection of principal
fractional ideals of $S$. Since $\End(X) = S$, each principal
fractional ideal of $S$ is an intersection of $S$-submodules of $F$ of the
form $qX$, $q \in F$.
Since $I^2 \subseteq L$, $I^2$ is an
intersection of ideals of $S$ of the form $L \cap qX$, where $q \in F$.
Since $I^2 \subsetneq I \subseteq L$, there exists $q \in F$ such that $I^2
\subseteq L \cap qX \subsetneq L$. Hence there exists a maximal ideal $N$
of $S$ with $I^2 \subseteq N$ such that $I^2_N \subseteq L_N \cap qX_N \subsetneq
L_N$. Since $S_N$ is a valuation domain, the $S_N$-modules $qX_N$ and $L_N$
are comparable and $I^2_N \subseteq L_N \cap qX_N \subsetneq L_N$ implies
$I^2_N \subseteq qX_N \subsetneq L_N$.
Now $\sqrt{I^2} =\sqrt{I} = L$ and $I^2 \subseteq N$ implies $L \subseteq N$.
Thus $I^2_L \subseteq qX_L
\subseteq L_L$, and we conclude that $\sqrt{qX_L} = L_L$.
We observe next that $X_P \ne F$. Since $P \ne 0$, there exists
a nonzero $L = P_iS \subseteq PS$ such that
$L = \sqrt{I}$, where $I$ is finitely generated.
As we have established in the paragraph above,
there exists a nonzero $q \in F$ such that $qX_L$ is an
ideal of $S_L$. Thus $qX_P \subseteq qX_L \subseteq S_L$, so it is not
possible that $X_P =F$.
Fix some member $L$ of the chain $\{P_iS\}$.
Since $X_P \ne F$, $L \subseteq PS$ and $R_P$ is a valuation domain,
there exists a nonzero element $s$ of $S$ such that $sX \subseteq L_L$.
Since $\End(X_P) = \End(sX_P)$ and we wish to show that $\End(X_P)
\subseteq R_P$ we may assume without loss of generality that $s = 1$; that
is, we assume for the rest of the proof that $X \subseteq L_{L}$. Define
$A = X \cap S$. Then $A$ is an ideal of $S$. Moreover $A$ is contained
in $L$ since $A_{L} \subseteq X_L \subseteq L_L$.
With the aim of applying Lemma~\ref{lemma 6.6}, we observe that $PS \in
\Ass(A)$. For each $i$ define $L_i = P_iS$. It suffices to show
each $L_i$ with $L \subseteq L_i \subseteq PS$ is in $\Ass(A)$, since this
implies that $PS = \bigcup_{L_i \supseteq L}L_i$ is a union of members of
$\Ass(A)$. Let $i$ be such that $L \subseteq L_i$. Since $L_i$ is the
radical of a finitely generated ideal of $S$, there exists (as we have
established above) a nonzero $q \in F$ such that $qX_{L_i}$ is an ideal of
$S_{L_i}$ that is primary for $(L_i)_{L_i}$. Now $A_{L_i} = X_{L_i} \cap
S_{L_i}$. Since $S_{L_i}$ is a valuation domain, $A_{L_i} = X_{L_i}$ or
$S_{L_i} \subseteq X_{L_i}$. By assumption, $X \subseteq L_L$.
Since $L \subseteq L_i$, $X_{L_i} \subseteq L_{L}$, and it is impossible that
$S_{L_i} \subseteq X_{L_i}$. Thus $A_{L_i} = X_{L_i}$. Consequently,
$qX_{L_i} = qA_{L_i}$ and
$qA_{L_i}$ is an ideal of $S_{L_i}$ that is primary for $(L_i)_{L_i}$.
Since $S_{L_i}$ is a valuation domain, it follows that $qA_{L_i}=
A_{L_i}:s$ for some $s \in S$. Thus $(L_i)_{L_i} \in \Ass(A_{L_i})$, so
by Lemma \ref{1.1}, $L_i \in \Ass(A)$. This proves $PS \in \Ass(A)$.
Since $A =X \cap S$ is an ideal of $S$, $S \subseteq \End(A)$.
For each maximal ideal $N$ of
$S$, either $A_N = X_N$ or $A_N = S_N$ It follows that $\End(A)
\subseteq \End(X) = S$, so $\End(A) = S$. Thus $\End(A)_P = S_P = R_P$, and by
Lemma~\ref{lemma 6.6}, $\End(A_P) = R_P$. (We have used here that
$S_{SP} = R_P$.) Now $A_P = X_P \cap S_P = X_P \cap R_P$. Since $R_P$ is
a valuation domain, $A_P = X_P$ or $R_P \subseteq X_P$. The latter case is
impossible since $X_P \subseteq X_L \subseteq L_L$. Thus $A_P =
X_P$. We conclude that End($X_{P}) =$ End($A_{P}) = R_P$.
\end{proof}
A prime ideal $P$ is {\it branched } if there exists a
$P$-primary ideal different from $P$. If $P$ is a nonzero
prime ideal of a Pr\"ufer domain $R$, then $P$ is branched
if and only if $P$ is not the union of the prime ideals
properly contained in $P$, and in this case if $P$ fails to
be branched, then the valuation domain $R_P$ is infinite
dimensional and there is no maximal element among the prime
ideals properly contained in $P$.
\begin{lemma}
\label{lemma 6.8}
Let $R$ be a Pr\"ufer domain having the
separation property. If there exists a finitely generated ideal
$J$ of $R$ having infinitely many minimal primes,
then there exists a submodule $X$ of the field of fractions
$F$ of $R$ and a maximal ideal $M$ containing $J$
such that $\End(X)_M \ne \End(X_M)$.
\end{lemma}
\begin{proof}
Let $P_1, P_2, P_3, \ldots$ be countably many distinct minimal
primes of $J$. Define $S = \bigcap_iR_{P_i}$. Then $S$ is a
Pr\"ufer overring of $R$ and the minimal primes of $JS$ are of
the form $QS$, where $Q \in \Spec R$ is a minimal prime of $J$.
In particular, $P_iS$ is a minimal prime of $JS$ for each $i \ge 1$.
Since $R$ has the separation property, each $P_iS $ is
a maximal ideal of $S$. This is because if $j$ is any positive
integer, then $\End(P_j) = R_{P_j} \cap (\bigcap_{N}R_N)$ by Theorem 3.2.6
of \cite{FHP}, so $\End(P_j) \subseteq S$ since the $P_i$'s are
comaximal. By Lemma~\ref{lemma 6.4}(ii) $P_j$ is a maximal ideal of
$\End(P_j)$, and since $R$ is a Pr\"ufer domain, either $P_j$ extends to a
maximal ideal $SP_j$ of $S$ or $SP_j = S$. The latter case is impossible
since $S \subseteq R_{P_j}$. Thus $SP_j$ is a maximal ideal of $S$.
Since $J$ is a nonzero finitely generated ideal, each
minimal prime of $J$ is branched.
Since $JS$ has infinitely many minimal primes, Theorem 1.6
of \cite{GH3} implies there exists
a minimal prime $QS$ of $JS$ that is not the radical of a
finitely generated ideal. If $Q = P_j$ for some $j$,
then by \cite[Theorem 2]{GH2}, we have
$S = \bigcap_{i \ne j}R_{P_i}$. Therefore, by
relabeling if necessary, we may
assume that $Q \not\in \{P_i\}_{i=1}^\infty$.
Define $A = JR_Q \cap R$. Then $AS = JR_Q \cap S$ is $QS$-primary.
In particular, $QS$ is the unique minimal prime of $AS$
and $A \not\subseteq P_iS \cap R = P_i$ for each $i \ge 1$.
Let $Q_1$ be the prime ideal of $R$ just below $Q$ (such a prime $Q_1$ exists
because $Q$ is branched). Since $R$ has the separation property, there is a
finitely generated ideal $B$ of $R$ such that $Q_1 \subset B \subseteq Q$.
Local verification shows that $Q_1 \subset B^n \subseteq Q$ for each positive
integer $n$.
For each $i \ge 1$, let
\[
x_i \in A^i \setminus (P_1 \cup \cdots \cup P_i \cup A^{i+1}).
\]
(By Prime Avoidance (see for example \cite[Lemma 3.3]{E}) such
an element $x_i$ exists.) Define
\[
J_i = B^{i+1} + x_iR.
\]
Then $Q_1 \subseteq J_i$ and $J_i \not \subseteq
(P_1 \cup \cdots \cup P_i)$. Define $X =
\sum_{i>1}J_i^{-1}S$.
Let $N = QS$ and for each $i>0$, define $N_i = P_iS$. As previously
demonstrated, $N$ and the ideals $N_i$, $i>0$, are maximal ideals of $S$.
We show first that $\End(X_{N}) \ne \End(X)_{N}$.
For each $i \ge 1$,
\[
X_{N_i} = \sum_{k=1}^{i-1}J_k^{-1}S_{N_i},
\]
since by design, $J_k^{-1}S_{N_k} = S_{N_k}$ for all $k \geq i$.
So as a finitely generated fractional ideal of
$S_{N_i}$, $X_{N_i}$ is isomorphic to $S_{N_i}$. Thus
End($X_{N_i}) = S_{N_i}$ for each $i$. We have that for each $i$, $S_{N_i} =
R_{P_i}$; hence
\[
{\rm End}(X) \subseteq \bigcap_{i}{\rm End}(X_{N_i}) = S.
\]
Therefore, End($X) = S$. In particular, $\End(X)_N = S_N$.
Thus to prove the claim that $\End(X)_N \ne \End(X_N)$, it suffices to show
that $N\End(X_N) = \End(X_N)$. To this end we verify that $X_{N} = NX_{N}$.
Since $N = SQ$, we have $S_N = R_Q$, so for each $i \ge 1$, $A^{i+1}S_N
\subset A^iS_N$. In particular, $x_{i+1}S_N \subset x_{i}S_N$ since $x_i
\in A^i \setminus A^{i+1}$ and $A^{i+1}S_N \cap R = A^{i+1}R_Q \cap R =
A^{i+1}$. It follows that $(J_{i+1})S_{N} \subset
(J_{i})S_{N}$. Since $J_i$ is a finitely generated ideal of $S$, it
follows that $(J_{i+1})S_{N} \subseteq N(J_{i})S_{N}$. Consequently,
$J_{i}^{-1} \subseteq N(J_{i+1}^{-1})S_{N}$ for each $i \ge 1$. Therefore
$X \subseteq NX_{N}$, as claimed.
For $S$-submodules $Y$ and $W$ of $F$, let $[W:Y] = \{q \in F:qY
\subseteq W\}$. Since $X_{N} = NX_{N}$ and End($N_N) =
S_N$, we have
\[
[S_N:X_N] = [N_N:NX_N] = [N_N:X_N].
\]
Moreover, $[S_N:X_N] = \bigcap_{i}J_{i}S_N$ and
\[
Q_1S_N \subseteq \bigcap_{i}J_iS_N \subseteq
\bigcap_{i}A^iS_N = Q_1S_N.
\]
The latter equality follows from the fact that since $S_N$ is a valuation
domain and $AS_N$ is a principal ideal of $S_N$, the intersection
$\bigcap_{i}A^iS_N$ is a prime ideal. Since $Q_1$ is the largest prime ideal
of $R$ properly contained in $Q$, $A \subseteq Q$ and $A \not
\subseteq Q_1$, we have $\bigcap_{i}A^iS_N = Q_1S_N.$
It follows that $[N_N:X_N] = [S_N:X_N] = (SQ_1)_N$. Since $R_N$
is a valuation domain, the maximal ideal $N_N$ of $S_N$ is
$m$-canonical (see \cite{HHP}), that is,
\[
X_N = [N_N:[N_N:X_N]] =
[N_N:(SQ_1)_N].
\]
Since $SQ_1$ is a nonmaximal prime ideal of
$S$, $[N_N:(SQ_1)_N] = S_{SQ_1}$ \cite[Theorem 4.1.21]{FHP}. Thus $X_N =
S_{SQ_1}$, and we have End($X_N) = S_{SQ_1} = R_{Q_1}$. In particular,
End($X_N) \ne R_{Q}$. Now $X_N = X_{SQ} = X_{Q}$, and if $M$ is a maximal
ideal of $R$ containing $Q$, then $S_{SQ} = S_{M}$, so $X_M = X_{Q}$. Thus
\[
R_{Q_1} = {\rm End}(X_N) = {\rm End}(X_M),
\]
while
\[
{\rm End}(X)_M = S_M = R_{Q}.
\]
Therefore End($X_M) \ne$ End($X)_M$. \end{proof}
\begin{theorem}
\label{theorem 6.9} Let $R$ be a Pr\"ufer domain with field of
fractions $F$. The following statements are equivalent.
\begin{itemize}
\item[(i)] For each finitely generated ideal $A$ of $R$,
every weak-Bourbaki associated prime of $A$ is a Zariski-Samuel
associated prime of $A$.
\item[(ii)] Every finitely generated ideal of R has only
finitely many minimal prime ideals.
\item[(iii)] Every principal ideal of R has only
finitely many minimal prime ideals.
\item[(iv)] Every
branched prime ideal of $R$ is the radical of a finitely generated
ideal.
\item[(v)] For each $R$-submodule $X$ of $F$,
$\End(X_M) = \End(X)_M$, for every maximal ideal
$M$ of $R$.
\end{itemize}
\end{theorem}
\begin{proof}
(i) $\Rightarrow$ (ii)
Let $A$ be a finitely generated ideal of $R$. Every minimal
prime $P$ of $A$ is a weak-Bourbaki prime of $A$ and therefore by
(i) a Zariski-Samuel prime of $A$ and hence
of the form $\sqrt{(A:x)}$. Since $R$ is Pr\"ufer,
$(A:x)$ is finitely generated, see, for example, \cite[Lemma 2]{GH2}.
Thus each minimal prime of $R/A$ is the radical of a
finitely generated ideal of $R/A$. By Theorem 1.6 in \cite{GH3}, $R/A$
has finitely many minimal primes. This proves (ii).
(ii) $\Rightarrow$ (iii) is obvious.
(iii) $\Rightarrow$ (iv) Let $P$ be a branched
prime of $R$. Then there exists a prime ideal $Q$ of $R$ such
$Q \subset P$ and there are no other prime ideals between $Q$ and
$P$. Let $x \in P \setminus Q$. Since the prime
ideals contained in $P$ are linearly ordered, $P$ is a minimal
prime ideal of $xR$. By (iii), $xR$ has only finitely many
minimal prime ideals. Since $R$ is Pr\"ufer,
if $B$ is the intersection of the other
minimal primes of $xR$, then $B + P = R$. Hence there
exists $y \in P$ such that $B + yR = R$. It follows that
$P$ is the radical of $(x,y)R$. This proves (iv).
(iv) $\Rightarrow$ (v) Since every
nonzero prime ideal of $R$ is the union of branched prime
ideals, this follows from Lemma~\ref{lemma 6.7}.
(v) $\Rightarrow$ (i) Let $A$ be a finitely generated ideal of $R$
and let $P$ be a weak-Bourbaki associated prime of $A$. Then
$P$ is a minimal prime of $(A:x)$ for some $x \in R$. Since
$R$ is Pr\"ufer, $(A:x)$ is finitely generated. Lemma~\ref{lemma 6.8}
implies that $(A:x)$ has only finitely many minimal primes.
Thus by Lemma 5.10 of \cite{FHO} each minimal prime of
$(A:x)$ is a Zariski-Samuel associated prime of $(A:x)$, and therefore a
Zariski-Samuel associated prime of $A$. This proves (i). \end{proof}
There is an interesting connection between the conditions of Theorem 3.4 and
``trace properties'' of Pr\"ufer domains. T. Lucas shows in \cite[Theorem
23]{L} that a Pr\"ufer domain $R$ satisfies (iv) of Theorem 3.4 if and only if
$R$ has the {\it radical trace property}, namely, for every ideal $A$ of $R$,
$AA^{-1}$ is either $R$ or a radical ideal of $R$.
\begin{corollary}
\label{corollary 6.10}
The following statements are equivalent
for a Pr\"ufer domain $R$.
\begin{itemize}
\item[(i)] Every branched
prime ideal of $R$ of height greater than one is the radical of a
finitely generated ideal of $R$.
\item[(ii)] For each nonzero ideal $A$ of $R$, $\End(A_M) =
\End(A)_M$ for all maximal ideals $M$ of $R$.
\item[(iii)] For each ideal $A$ of $R$, $\Ass(A)$ is precisely
the set of prime ideals $P$ such that $A \subseteq P$ and
$P\End(A) \ne \End(A)$.
\end{itemize}
\end{corollary}
\begin{proof} (i) $\Rightarrow$ (ii) Suppose (i) holds, and let
$A$ be a nonzero ideal of $R$. Let $M$ be a maximal ideal of
$R$. If $M$ has height one then $R_M$ is a one-dimensional
valuation domain, so End($A_M) = R_M$. Consequently, End$(A)_M
=$ End($A_M$). If $M$ has height greater than 1, then $M$ is
the union of branched prime ideals of height greater than 1, so
by (i) and Lemma~\ref{lemma 6.7}, End($A_M) =$ End($A)_M$. This proves
(ii).
(ii) $\Rightarrow$ (i) Let $P$ be a branched prime ideal of
$R$ such that there is a nonzero prime ideal $Q$ of $R$ with $Q
\subset P$. From (ii) and Lemma~\ref{lemma 6.4}, it follows that $R$ has
the separation property. Thus $Q$ is a maximal ideal of End($Q$) and
End($Q)/Q$ is the quotient field of $R/Q$. From (ii) it follows that if $X$
is an $R$-submodule of End($Q)$ with $Q \subseteq X$, then
\[
{\rm End}((X/Q)_{M/Q}) = {\rm End}(X/Q)_{M/Q}
\]
for all maximal ideals $M$ of $R$ containing $Q$. By
Theorem~\ref{theorem 6.9},
the branched prime ideal $P/Q$ of $R/Q$ is the radical of a
finitely generated ideal. Hence $P$ is the radical of an ideal
$I + Q$, where $I$ is a finitely generated ideal of $R$. Since
$R$ has the separation property, there is a finitely generated
ideal $J$ such that $Q \subset J \subseteq P$. Thus $P$ is the
radical of the finitely generated ideal $I + J$.
(ii) $\Rightarrow$ (iii) Let $A$ be an ideal of $R$, and suppose
$P \in \Ass(A)$. Then there exists $Q \in \XX_A$ such that $P
\subseteq Q$. By Lemma~\ref{proposition 2.8}, $R_Q = \End(A_M)$ for some
maximal ideal $M$ of $R$. Thus by (ii), $R_Q = \End(A)_M$, and
we have $Q\End(A) \ne \End(A)$. It follows that $P\End(A) \ne
\End(A)$.
Now suppose $P$ is a prime ideal containing $A$ such that
$P\End(A) \ne \End(A)$. Then $\End(A)_M \ne P\End(A)_M$ for some
maximal ideal $M$ of $R$, and it follows that $\End(A)_M
\subseteq R_P$. By (ii), $\End(A_M) \subseteq R_P$. If
$\End(A_M) = R_Q$ for some prime ideal $Q$, then by
Lemma~\ref{proposition 2.7} and
\ref{proposition 2.8},
it follows that $Q \in \Ass(A)$. Since $A \subseteq P \subseteq Q$,
Lemma~\ref{proposition 2.7} implies $P \in \Ass(A)$. This proves (iii).
(iii) $\Rightarrow$ (ii) Let $A$ be a proper ideal of $R$ and
$M$ be a maximal ideal of $R$ containing $A$. If $\End(A)_M =
R_P$ for some prime ideal $P$ of $R$, then $P\End(A) \ne
\End(A)$, so by (iii), $P \in \Ass(A)$. By Lemma~\ref{lemma 6.6},
$\End(A)_M = \End(A_M)$. This proves (ii). \end{proof}
\begin{remark}
\label{remark 6.11} {\rm Theorem~\ref{theorem 6.9} shows that conditions
for good behavior with respect to localization for endomorphism rings of
submodules of the fraction field involve all the primes of $R$. On the other
hand, Corollary~\ref{corollary 6.10} shows that the corresponding property
for ideals involves only conditions on the primes of height greater than one.
This phenomenon is illustrated in the article Goeters-Olberding \cite{GO},
where the statement of Corollary~\ref{corollary 6.10} that $\End(A)$ commutes
with localizations is of central importance. Our proof of Lemma~\ref{lemma
6.8} is a refinement of the argument given in Theorem 3.7 of \cite{GO}.}
\end{remark}
\begin{theorem}
\label{theorem 6.12}
Consider the following statements for an arithmetical ring
$R$.
\begin{itemize}
\item[(i)] $\XX_A = \Max(A)$ for each
proper regular ideal $A$ of $R$.
\item[(ii)] Every branched prime ideal of $R$
that properly contains a regular prime ideal is the
radical of a finitely generated ideal. \item[(iii)] Every branched
prime ideal of $R$ that properly contains a regular prime ideal is the radical
of a principal ideal. \end{itemize} Then (iii) $\Rightarrow$ (i)
$\Rightarrow$ (ii). Moreover, if $R$ is a QR-domain, then all three
statements are equivalent. \end{theorem}
\begin{proof} (i) $\Rightarrow$ (ii) We consider first the case
where $R$ is a domain. Then to prove (ii), it suffices by
Corollary~\ref{corollary 6.10} to show that for every ideal $A$ of $R$ and
maximal ideal $M$ containing $A$, $\End(A)_M = \End(A_M)$. Let
$A$ be a proper nonzero ideal of $R$ and let $M$ be a maximal
ideal such that $A \subseteq M$. Since $R_M$ is a valuation
domain, there exists a prime ideal $P$ of $R$ such that
$\End(A)_M = R_{P}$. We claim first that the elements of $P$ are not prime to
$A$. Let $\{M_\alpha\}$ denote the set of maximal ideals of $R$ that
contain $A$, and let $\{N_\beta\}$ denote the set of maximal ideals of $R$
that do not contain $A$. For each $\alpha$, there exists a prime ideal
$P_\alpha$ such that $\End(A_{M_\alpha}) = R_{P_\alpha}$. Moreover, by
Lemmas 1.2 and~\ref{proposition 2.8},
we have $\XX_A \subseteq \{P_\alpha\} \subseteq \Ass(A)$.
Since $\End(A) = (\bigcap_{\alpha}\End(A_{M_\alpha})) \cap
(\bigcap_{\beta}\End(A_{N_\beta}))$, we have
\[
R_P = \End(A)_M = (\bigcap_{\alpha}R_{P_\alpha})R_{M} \cap
(\bigcap_{\beta}R_{N_\beta})R_M.
\]
Since $R$ has the separation property, Lemma 2.3(iv) implies
that $\bigcap_{\beta}R_{N_\beta} \not \subseteq R_{P}$. Thus because $R_P$
is a valuation domain, it must be that $\bigcap_{\alpha}R_{P_\alpha}
\subseteq R_P$, and it follows that $P \subseteq \bigcup_{\alpha}P_\alpha$.
Since $S(A) = \cup_{\alpha}P_\alpha$, the elements of $P$ are
not prime to $A$, as claimed. Therefore there exists $Q \in \Max(A)$ such
that $P \subseteq Q$. By (i), $Q \in \XX_A$, so by Lemma~\ref{proposition
2.7}, $P \in \Ass(A)$ since $A \subseteq P \subseteq Q$. Thus by
Lemma~\ref{lemma 6.6}, $\End(A)_M = \End(A_M)$. This proves that (i) implies
(ii) in the case $R$ is a domain.
Now consider the general case where $R$ is not necessarily a
domain. Observe that since
statement (i) holds for $R$, statement (i) holds for $R/P$ for
all regular prime ideals $P$ of $R$. This follows from the observation that if
$A$ is an ideal of $R$ that contains $P$ then for all $x \in R$,
\[
(A/P:_{R/P}x+P) = ((A:x)+P)/P.
\]
In particular, $\Ass(A/P) = \{Q/P:Q \in \Ass(A), P\subseteq Q\}$. Also,
$\Max(A/P) = \{Q/P:Q \in \Max(A), P \subseteq Q\}$.
Thus statement (ii) holds for $R/P$ for all regular prime ideals $P$ of
$R$. Let $Q$ be a branched prime ideal of $R$ containing at least one
regular prime ideal. Then there exists
a regular prime ideal $P$ of $R$ such that $P \subset Q$. By
Theorem~\ref{theorem 6.3}, there exists a finitely generated ideal $B$
of $R$ such that $P \subseteq B \subset Q$. Moreover, since (ii) holds for
$R/P$, there exists a finitely generated ideal $C$ of $R$ such that $Q$ is
the radical of $C + P$. It follows that $Q$ is the radical of the finitely
generated ideal $C +B$.
(iii) $\Rightarrow$ (i) Let $A$ be a proper regular ideal of $R$. For each $P
\in \XX_A$ there exists $M \in \Max(A)$ such that $P \subseteq M$ and then $P
= M$ if and only if $M \in \Ass(A)$. Thus it suffices to prove that $M \in
\Max(A)$ implies $M \in \Ass(A)$. This is clear if $M/A$ has height one as
prime ideal of $R/A$. Assume that $\hgt(M/A) \ge 2$. Then $M$ is the union of a
chain of branched prime ideals $P_i$ of $R$, where $A \subseteq P_i$ and
$\hgt(P_i/A) \ge 2$. Since $A$ is a regular ideal, so is each $P_i$. Thus
by (iii), each $P_i$ is the radical of a principal ideal of $R$. Since $P_i
\subseteq M$, the elements in $P_i$ are non-prime to $A$. Since $P_i$ is the
radical of a principal ideal and $S(A) = \bigcup_{Q \in \XX_A}Q$, $P_i$ is
contained in some member of $\XX_A$. By Lemma~\ref{proposition 2.7}, each $P_i
\in \Ass(A)$. Therefore $\bigcup_iP_i = M \in \Ass(A)$.
Finally, we conclude that statements (i),(ii) and (iii) are equivalent for a
QR-domain $R$ since in a QR-domain every prime ideal that is
the radical of a finitely
generated ideal is the radical of a principal ideal. \end{proof}
\begin{corollary} If $R$ is a Pr\"ufer domain
having the property that every maximal prime
divisor of an ideal A of R is a Krull associated prime of A,
then every branched prime ideal of R of height greater
than one is the radical of a finitely generated ideal.
\end{corollary}
\begin{proof} The corollary is a consequence of Theorem~\ref{theorem 6.12} and
Corollary 1.3.
\end{proof}
Example~\ref{6.13} illustrates the fact that it is possible for a
2-dimensional Pr\"ufer domain to satisfy the hypotheses of
Theorem~\ref{theorem 6.3}, but
not Theorem~\ref{theorem 6.12}.
\begin{example}\label{6.13} {\it There exists a Pr\"ufer domain $R$ that
has the strong separation property and yet contains an ideal $A$
such that $\XX_A$ is a proper
subset of $\Max(A)$. Moreover, $R$ does not satisfy any of the conditions (i),
(ii) or (iii) of Theorem 3.7.} Let $D$ be an almost Dedekind domain (that is,
$D_M$ is a DVR for all maximal ideals $M$ of $D$) that is not Dedekind.
(See \cite[page 281]{FHP} for an example of such a domain.)
Let $F$ denote the field of
fractions of $D$, let $x$ be an indeterminate over $F$, and let $P =
xF[x]_{(x)}$. Define $R = D + P$. Then $R$ is a Pr\"ufer domain and $P$ is
the unique nonzero nonmaximal prime ideal of $R$. Moreover, if $M$ is a
maximal ideal of $R$, then $M$ contains $P$ and if $r \in M \setminus P$, then
$P \subseteq Rr \subseteq M$, so $R$ has the strong separation property. In
particular, by Theorem~\ref{theorem 6.3}, every ideal $A$ of $R$ has the
property that $\XX_A \subseteq \Max(A)$. Since $D$ is almost Dedekind, but
not Dedekind, there exists a maximal ideal $Q$ of $D$ that is not the radical
of a finitely generated ideal. Consequently, the (height 2) maximal ideal $Q
+ P$ of $R$ is not the radical of a finitely generated ideal of $R$. By
Theorem~\ref{theorem 6.12}, there exists an ideal $A$ of $R$ such that $\XX_A$
is a proper subset of $\Max(A)$. \end{example}
In Example 2.9 of \cite{FHO}, we construct a 2-dimensional
Pr\"ufer domain with Noetherian spectrum having an
ideal $A$ for which there exists a maximal ideal $M$ such
that $M$ is in $\Max(A)$ but $M$ is not a Krull associated prime
of $A$. Since this domain $R$ has Noetherian spectrum, $R$ satisfies
statement (ii) of Theorem 3.7. Hence condition (ii) of Theorem 3.7 does
not imply condition (i), even when $R$ is a 2-dimensional Pr\"ufer domain. This
leaves the question of whether in general (i) implies (iii) in Theorem 3.7:
\begin{question} Does there exist an arithmetical ring $R$ such that $R$
satisfies statement (i) but not statement (iii) of Theorem 3.7?
\end{question}
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