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\begin{document}
\title[Commutative Ideal Theory without Finiteness Conditions]{Commutative
Ideal Theory without Finiteness Conditions: Primal Ideals}
% 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.}
% General info
\subjclass{Primary 13A15, 13F05}
\date \today
%\dedicatory{This paper is dedicated to our authors.}
\keywords{Primal ideal, associated prime, arithmetical ring, Pr\"ufer domain}
\begin{abstract}
Our goal is to establish an efficient decomposition of an
ideal $A$ of a commutative ring $R$ as an intersection of
primal ideals. We prove the existence of a
canonical primal decomposition: $A = \bigcap_{P \in \XX_A}A_{(P)}$,
where the $A_{(P)}$ are isolated
components of $A$ that are primal ideals having distinct and incomparable adjoint
primes $P$. For this purpose we define the set $\Ass(A)$ of associated primes of the
ideal $A$ to be those defined and studied by Krull.
We determine conditions for the canonical primal decomposition to
be irredundant, or residually maximal, or the unique representation of $A$ as
an irredundant intersection of isolated components of $A$. Using our canonical
primal decomposition, we obtain an affirmative answer to a question raised
by Fuchs in \cite{F2} and also prove for $P \in \Spec R$ that an
ideal $A \subseteq P$ is an intersection of $P$-primal ideals if
and only if the elements of $R \setminus P$ are prime to $A$. We
prove that the following conditions are equivalent: (i) the ring
$R$ is arithmetical, (ii) every primal ideal of $R$ is
irreducible, (iii) each proper ideal of $R$ is an intersection
of its irreducible isolated components. We classify the rings
for which the canonical primal decomposition of each proper
ideal is an irredundant decomposition of irreducible ideals as
precisely the arithmetical rings with Noetherian maximal
spectrum. In particular, the integral domains having these
equivalent properties are the Pr\"ufer domains that satisfy
(\#\#). \end{abstract}
\maketitle
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\def\fg{finitely generated}
\def\fp{finitely presented}
\def\cp{cyclically presented}
\def\fr{finite rank}
\def\tf{torsion-free}\def\vd{valuation domain}
\def\vr{valuation ring}
\def\vds{valuation domains}
\def\Dd{Dedekind domain}
\def\Dds{Dedekind domains}
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\section*{Introduction}
In her seminal paper \cite{N}, Emmy Noether proved that in a
commutative ring satisfying the ascending chain condition on
ideals every ideal is the intersection of a finite number of
irreducible ideals, and the irreducible ideals are primary
ideals. She went on to establish several intersection
decompositions, one of which featured primary ideal components
with distinct radicals.
Among rings without the ascending chain condition, the rings in which
such a decomposition holds for all ideals (called Laskerian
rings) are few and far between (see e.g. Heinzer-Lantz
\cite{HL}). One reason for this is that irreducible ideals need
not be primary. Thus, if we wish to obtain analogous
decompositions in general rings, then we have to replace primary
ideals by more general ideals. Natural candidates are the primal
ideals introduced in Fuchs \cite{F2} where it is shown that
every ideal is the intersection of (in general infinitely many)
irreducible ideals and these ideals are primal ideals. This
primal decomposition of an ideal gives components whose
structure we know, but the components are not closely tied to
the ideal and there is no indication of how many and what kind
of primal ideals are needed for the intersection. It is also
possible to represent an ideal as the intersection of its
isolated components. In this case the components are closely
tied to the ideal, but the structure of the components is not
known. In general, the isolated components of an ideal are not
primal ideals and there are almost always superfluous
components.
Our main purpose here is to establish a more efficient decomposition
of an ideal into the intersection of primal ideals.
We show in Theorem 3.5 that if we take just
those components that appear in both decompositions, then we can still
represent the ideal
even with the minimal isolated primal components.
The existence of such decompositions in the Noetherian case was
established in Fuchs \cite{F2}. In Theorem 3.5 we achieve a
canonical primal decomposition (that is often irredundant)
where the primal components belong to distinct, incomparable
primes and where these distinct primes are the maximal primes in
the set of associated primes of $A$; here we are using associated
primes in the sense defined by Krull \cite[page 742]{K1}. Also
see \cite[page 253]{Be}. With this definition of associated
primes, maximal associated primes of $A$ exist and every
associated prime is contained in a maximal associated prime.
Krull \cite[page 16]{K2} mentions that it was an undecided
problem whether or not (phrased in our terminology) the
principal component $A_{(P)}$ (for $P$ a maximal prime divisor
of $A$) is $P$-primal. Nagata in \cite{Ng1} and Gilmer in
\cite{G} gave a negative answer, but the full answer is given in
Theorem 3.4: the isolated $P$-component of $A$ is $P$-primal if
and only if $P$ is a Krull associated prime of $A$.
We establish in Corollary 3.6 an affirmative answer
to the question raised in \cite[page 3]{F2} of
whether for a maximal prime divisor $P$ of an ideal $A$
the principal $P$-component $A_{(P)}$
is equal to the intersection of all primal ideals
that contain $A$ and have their associated
adjoint primes contained in $P$. Indeed, Proposition \ref{proposition 3.8}
implies that $A_{(P)}$ is an intersection of $P$-primal ideals.
In Example \ref{example 3.9}, we present an example where the ideal
$A$ has precisely two maximal prime divisors and where the principal
component with respect to each of these maximal prime divisors is
not a primal ideal.
We prove in Theorem 4.2 that under certain additional
hypothesis the canonical primal decomposition
of Theorem 3.5 is irredundant. We establish in Proposition 4.3
under certain conditions that
a stronger form of irredundancy holds: namely,
the canonical primal decomposition of Theorem 3.5 is
residually maximal in the sense that no component can
be replaced by any of its proper residuals. For $P$
a maximal associated prime of the ideal $A$, we
give in Theorem 4.6 necessary and sufficient
conditions in order that the isolated component
$A_{(P)}$ appear in every representation of $A$
as an intersection of primal isolated components of $A$.
In Section 5 we consider irredundant decompositions
into irreducible ideals. If an ideal $A$ has a
representation as an
irredundant intersection of irreducible
isolated components, we prove in Theorem 5.1
that this intersection is residually maximal
and is the unique irredundant decomposition
of $A$ into irreducible isolated components of $A$.
Example 5.4 shows there are limitations as to any
sharpening of Theorem 5.1. In Theorem~\ref{theorem 5.9} we
prove that a ring $R$ is an arithmetical ring
if and only if each proper ideal of $R$ can
be represented as an intersection of irreducible
isolated components. Indeed, if $R$ is an arithmetical ring and $A$ can be
represented as an irredundant intersection of irreducible ideals, then this
is the unique representation of $A$ as an irredundant intersection of
irreducible ideals (Corollary~\ref{corollary 5.7}).
We also consider in Section 5 the question
of what rings $R$ have the property that the
canonical primal decomposition of each proper ideal
$A$ of $R$ is an irredundant representation of $A$
as an intersection of irreducible isolated components.
We classify in Theorem~\ref{theorem 5.15} the rings $R$ having this
property as the arithmetical rings with Noetherian
maximal spectrum.
In particular, the integral domains having this property
are precisely the Pr\"ufer domains
that satisfy condition (\#\#).
{\it Acknowledgement.} We thank Gabriel Picavet and David Rush
for helpful conversations on the topics of this paper, and for showing us
the connections of our work to the literature on Krull associated primes.
\section{Preliminaries}
Throughout $R$ will denote a commutative ring with 1.
%We say that an ideal $A$ of $R$ is {\it regular} if it
%contains a regular element, i.e., an element that is
%not a zero-divisor.
For ideals $I,J$ of $R$, the residual $I:J$ is
defined as usual by
$$
I:J = \{ x \in R \ : \ xJ \subseteq I\}.
$$
For an ideal $A$ and for a prime ideal $P$ of
$R$, we use the notation
$$
A_{(P)} = \{ x \in R \ : \ sx \in A\ {\rm for\ some}\ s \in R
\setminus P\} = \bigcup _{s \in R \setminus P}\ A:s
$$
to denote the {\it isolated $P$-component}
(isoliertes Komponentenideal) of $A$ in the sense of Krull
\cite[page 16]{K2}. It will be useful to observe that $x \in
A_{(P)}$ if and only if $A:x \not \subseteq P$. If $R$ is a
domain, then $A _{(P)} = AR_P \cap R$, where $R_P$ denotes the
localization of $R$ at $P$.
Evidently, $A _{(P)} \subseteq P$ for every prime $P$ with $A \subseteq P$. If
$P$ is a
minimal prime of $A$, then $A _{(P)}$ is known to be a $P$-primary ideal:
the {\it isolated $P$-primary component} of $A$.
\begin{lemma}
\label{lemma 1.1} For every ideal $A$, prime ideal $P$ and element $x \in R$, we
have $$(A:x)_{(P)} = A _{(P)}:x. $$
\end{lemma}
\begin{proof}
Evidently, $y \in (A:x)_{(P)} \Leftrightarrow
\exists c \notin P$ such that $cy \in A:x \Leftrightarrow
\exists c \notin P$ with $ cxy \in A \Leftrightarrow xy \in A
_{(P)} \Leftrightarrow y \in A _{(P)}:x.$
\end{proof}
An element $x \in R$ is called {\it non-prime to} $A$ if $A \subset A:x $
(proper inclusion), i.e. there exists an element $c \in R \setminus A$
such that $cx \in A$. Evidently, the set $S(A)$ of non-prime
elements to $A$ corresponds to the set of zero-divisors in the
factor ring $R/A$; it is the set union of prime ideals: the
maximal prime divisors of $A$.
An ideal $A$ is said to be {\it primal} (see \cite{F2}) if
$S(A)$ itself is an ideal, i.e. if the elements of $R$ that are
not prime to $A$ form an ideal $P$; this ideal is then a
prime ideal, called the {\it adjoint ideal} of $A$. In
this case, we also say that $A$ is a {\it $P$-primal ideal}. An
ideal $A$ is primal if and only if the sum of any two ring
elements non-prime to $A$ is again non-prime to $A$.
Irreducible ideals are in general not primary (for a necessary and sufficient
condition see Fuchs \cite{F1}). Primal ideals have the advantage
over primary ideals that the following is true without any chain
condition:
\begin{lemma}
\label{Lemma 1.2}
{\rm (Fuchs \cite{F2})}
\begin{itemize}
\item[(a)] Irreducible ideals are primal.
\item[(b)] Every ideal is the
intersection of primal ideals.
\end{itemize}
\end{lemma}
We use the following result.
\begin{lemma}
\label{Lemma 1.3}
If $A$ is a $P$-primal ideal, then
\begin{itemize}
\item[(i)] $A_{(Q)} = A$ if $Q$ is a prime containing $P$;
\item[(ii)] $A_{(Q)} \supset A$ (proper inclusion) if $Q$ is a
prime not containing $P$;
\item[(iii)] if $A_{(Q)}$ is a $Q$-primal ideal for some prime
$Q$ containing $A$, then $Q \subseteq P$.
\end{itemize}
\end{lemma}
\begin{proof}
(i) To show that $A_{(Q)} \subseteq A$, observe that $x
\in A_{(Q)}$ means $cx \in A$ for some $c \in R \setminus Q$. But
if $A$ is $P$-primal, then $c$ is prime to $A$, so $x \in A$.
(ii) There exists an $x \in P \setminus Q$. Such an $x$ is not
prime to $A$, so for some $y \notin A$ we have $xy \in A$. Then
$y \in A_{(Q)} \setminus A$.
(iii) By assumption, $Q
= \cup_{x \in R \setminus A_{(Q)}}A_{(Q)}:x$. If $Q \not
\subseteq P$, then there exists $q \in Q \setminus P$ such that
$qx \in A_{(Q)}$ for some $x \in R \setminus A_{(Q)}$. Thus
there exists $c \in R \setminus Q$ such that $cqx \in A$. Now
$q \not \in P$, so since $A = A_{(P)}$, $cx \in A$. However, $c
\not \in Q$, so $x \in A_{(Q)}$, contrary to assumption. We
conclude that $Q \subseteq P$.
\end{proof}
Primal ideals behave similarly to primary ideals, but there are notable
differences.
For instance, the intersection of two primal ideals with the same adjoint
prime need
not be primal again. A counterexample is the principal ideal $xyR$ in the
polynomial ring
$R = F[x,y]$, where $x$ and $y$ are indeterminates over a field $F$.
The ideal $xyR$ is not primal ($x,y$ are not prime to $xyR$, but
$x-y$ is), though it is the intersection of
the $(x,y)R$-primal ideals $(x^2,xy)R$ and $(xy, y^2)R$. However, if we have a
{\it reduced} intersection (i.e. none of the components may be replaced by
a larger
ideal without changing the intersection), then we can claim:
\begin{lemma}
\label{lemma 1.4}
{\rm (Fuchs \cite{F2})} A reduced intersection
$A= A_1 \cap \dots \cap A_n$ of primal ideals
$A_i$ with adjoint primes $P_i$ is again primal if and only if
there is a unique maximal member in the set $\{P_1, \dots , P_n\}$.
This prime is then the adjoint prime to $A$.
\end{lemma}
Recall that the closed sets of
the prime spectrum $\Spec R$ of the ring $R$ in the {\it Zariski topology}
are the sets $V(I)=\{P\in \Spec R \ : \ P \supseteq I\}$ with $I$ ranging over
the set of ideals of $R$. $\Spec R$ is called {\it Noetherian} if the closed
subsets in the Zariski topology satisfy the descending chain condition, or
equivalently, if the radical ideals of $R$ satisfy the ascending
chain condition. The {\it maximal spectrum} of $R$ is the set
$\Max R$ of maximal ideals of $R$ with the subspace topology from
$\Spec R$. We say that $R$ has {\it Noetherian maximal spectrum}
if the closed subsets of $\Max R$ satisfy the descending chain
condition, or equivalently, if the J-radical ideals of $R$ satisfy
the ascending chain condition, where an ideal is a {\it J-radical
ideal} if it is an intersection of maximal ideals.
Some of the following characterizations of rings with Noetherian prime
spectra will be needed in our discussions.
\begin{lemma}
\label{lemma 1.5}
{\rm (Mori \cite{M} or Ohm-Pendelton \cite{OP})} For a
commutative ring $R$ the following conditions are equivalent:
\begin{itemize}
\item[(i)] $\Spec R$ is Noetherian;
\item[(ii)] $R$ has the maximum condition on radical ideals;
\item[(iii)] the ascending chain condition holds for prime
ideals and each ideal of $R$ has only finitely many minimal
prime ideals.
\end{itemize}
\end{lemma}
We will pay special attention to the class of arithmetical rings. In
addition to providing an interesting contrast to the theory of primal
decompositions in Noetherian rings, Theorem~\ref{new theorem} and the
results of Section 5 show that this class of rings arises naturally in our
context.
Recall that a ring $R$ is a {\it chained} ring or {\it
valuation} ring if the ideals of $R$ are linearly ordered with respect to
inclusion; and $R$ is an {\it arithmetical} ring if $R_M$ is a valuation
ring for each maximal ideal $M$ of $R$. An integral domain is an
arithmetical ring if and only if it is a Pr\"ufer domain.
%There are numerous
%characterizations of arithmetical rings (and Pr\"ufer domains, in particular)
%in the literature, and we add to this list below and in Section 5.
Among several
results that are relevant to our study is the theorem of
Jensen that primary ideals of an arithmetical ring are irreducible and
that a Pr\"ufer domain $R$ has the property that each ideal of $R$ with
prime radical is irreducible if and only if each nonzero prime ideal of $R$
is contained in a unique maximal ideal \cite[Theorem 6]{J}. It is
readily seen that a ring $R$ is an arithmetical ring if and only
if for each proper ideal $A$ of $R$, $A_M$ is an irreducible
ideal of $R_M$ for every maximal ideal $M$ of $R$ containing
$A$.
\begin{remark}
\label{remark 5.5}
{\rm Let $A$ be an ideal of a ring $R$ and let $P
\in \Spec R$. The ideal
$A_P$ is irreducible in $R_P$ if and only if $A_{(P)}$ is
irreducible in $R$. This statement is clear in view of the
following facts: (i) finite intersections of ideals of $R$ behave well with
respect to the map $R \to R_P$, (ii) every ideal of $R_P$ comes
from an ideal of $R$, and (iii) $A_{(P)}$ is the preimage in
$R$ of the ideal $A_P$ of $R_P$.
From this observation we deduce the characterization: {\it A ring $R$
is arithmetical if and only if for all ideals $A$ of $R$, $A_{(M)}$ is an
irreducible ideal for all maximal ideals $M$ that contain $A$.} For if $R$
is arithmetical, then every ideal of $R_M$, $M$ a maximal ideal of $R$, is
irreducible, so if $A$ is an ideal of $R$ contained in $M$, $A_M$, and
hence $A_{(M)}$, are irreducible ideals of $R_M$ and $R$, respectively.
Conversely, as noted in the paragraph before Remark \ref{remark 5.5},
to prove $R$ is arithmetical it is
enough to show for every ideal $A$ and maximal ideal $M$ containing $A$
that $A_M$ is an irreducible ideal. This is indeed the case, since by assumption
$A_{(M)}$ is irreducible and we have from the above observation that $A_M$ is
an irreducible ideal of $R_M$.} \end{remark}
\begin{lemma}\label{product}
Let $A$ be a finitely generated ideal contained in a prime ideal $P$
of the ring $R$. If $A_P \ne 0$, then
$(AP)_{(P)}$ is a $P$-primal ideal of $R$.
\end{lemma}
\begin{proof}
Set $B = AP$. Clearly, the elements of $R$ not prime to $B_{(P)}$ are
contained in $P$, so to show $B_{(P)}$ is a $P$-primal ideal of $R$, it
suffices to prove that the elements of $P$ are not prime to $B_{(P)}$. Since
$A$ is finitely generated and $A_P \ne 0$, Nakayama's Lemma implies that
$B_P \ne A_P$, and it follows that $B_{(P)} \ne A_{(P)}$. Thus $B_{(P)}
\subset A_{(P)} \subseteq B_{(P)} : P$, so there exists $y \in R \setminus
B_{(P)}$ such that $yP \subseteq B_{(P)}$. This proves that the elements of
$P$ are not prime to $B_{(P)}$ and $B_{(P)}$ is a $P$-primal ideal.
\end{proof}
\begin{theorem}
\label{new theorem}
The following statements are equivalent for
a ring $R$.
\begin{itemize}
\item[(i)] $R$ is an arithmetical ring.
\item[(ii)] Every primal ideal of $R$ is irreducible. \end{itemize}
\end{theorem}
\begin{proof} (i) $\Rightarrow$ (ii). The implication (i) $\Rightarrow$
(ii) is proved in \cite{FM} for the case where $R$ is a domain.
Allowing zero divisors requires a simple modification of this
argument. If $A$ is a primal ideal, then $A = A_{(P)}$, where
$P$ is the adjoint prime of $A$. By (i), $A_P$, and hence
$A_{(P)}$ (Remark~\ref{remark 5.5}), are irreducible ideals, and
this proves (ii).
(ii) $\Rightarrow$ (i)
We use the
following observation: ($\dagger$) {\it Let $I = (a,b)R$ be a
finitely generated ideal contained in a maximal ideal $M$
of $R$. If $I_M \ne 0$ and $IM_{M}$ is an irreducible ideal of $R_M$, then $I_{M} =
R_{M}a$ or $I_{M} = R_{M}b$.} For the $R/M$-vector
space $I_{M}/IM_{M}$ is finite-dimensional. Since $IM_{M}$ is
irreducible, it follows that $I_{M}/IM_{M} \cong R/M$. In particular, $I_M
= aR_M + IM_M$ or $I_M = bR_M + IM_M$. By Nakayama's Lemma, $I_M = aR_M$ or
$I_M = bR_M$.
To prove (i), it suffices to show for each
maximal ideal $M$ of $R$ and $a,b \in M$ that either $R_Ma \subseteq
R_Mb$ or $R_Mb \subseteq R_Ma$, for this implies that $R_M$ is a
valuation ring. We may also assume that $I = (a, b)R$ is such
that $I_M \ne 0$. By Lemma~\ref{product},
$(IM)_{(M)}$ is an $M$-primal ideal, so by (ii) and Remark~\ref{remark 5.5},
$IMR_{M}$ is an irreducible ideal of $R_M$. Thus by the observation
($\dagger$), either $R_Ma \subseteq R_Mb$ or $R_Mb \subseteq R_Ma$.
\end{proof}
\section{Associated Primes}
We start with the study of primes associated to an ideal. There are at least
six inequivalent notions of associated primes in the literature, all of which
are equivalent in the Noetherian case. (See Heinzer-Ohm
\cite{HO} and Iroz-Rush \cite{IR}, for example.)
For our purposes, the most useful concept of
associated primes is one introduced by Krull \cite[page
742]{K1}. Let $A$ be a proper ideal of the ring $R$. Following
\cite{IR}, we define a prime ideal $P$ of a ring $R$ 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 $P$ of $R$ is called a {\it weak-Bourbaki associated
prime} to $A$ 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 $x \in R$. (This differs
slightly from the terminology of \cite{U}, where $P$ is said to
be a Zariski-Samuel associated prime of $A$ if $A:x$ is
$P$-primary for some $x \in R$. Of course if $A:x$
is $P$-primary, then $P$ is a Zariski-Samuel associated prime in
our sense.) It is clear that a Zariski-Samuel associated prime
is a weak-Bourbaki associated prime.
\begin{lemma}
\label{lemma 2.1.}
Let $A$ be a proper ideal of the ring $R$.
Every weak-Bourbaki associated prime of $A$ is a
Krull associated prime of $A$.
A prime ideal $Q$ of $R$ is a
Krull associated prime of $A$ if and only if $Q$ is a
set-theoretic union of weak-Bourbaki associated primes of $A$.
\end{lemma}
\begin{proof}
Let $P$ be a weak-Bourbaki associated prime of $A$.
Then $P$ is a minimal prime of $A:x$ for
some $x \notin A$. It follows that
the ideal $(A:x)_{(P)}$ is $P$-primary. Thus given $u \in P$,
there is a smallest integer $k \geq 1$ such that $u^k \in
(A:x)_{(P)}$. Hence $u^kv \in A:x$ for some $v \notin P$.
Evidently, $ u \in A: xu^{k-1}v$. If $A: xu^{k-1}v \subseteq P$
were not true, then we could find a $w \notin P$ with
$w \in A: xu^{k-1}v$. But then $u^{k-1}vw \in A:x$, and
$ u^{k-1} \in (A:x)_{(P)}$ is impossible. Thus
$A: xu^{k-1}v \subseteq P$, indeed. (See also
\cite[Theorem 1]{Ku} or \cite[page 346]{IR}.) It follows that a
prime ideal $Q$ of $R$ is a Krull associated prime of $A$
if it is a set-theoretic union of weak-Bourbaki associated
primes of $A$. The converse is clear, since $x \in A:y \subseteq Q$
implies that $x$ is contained in every minimal prime of $A:y$.
\end{proof}
It is obvious that every
proper ideal of the ring has weak-Bourbaki associated primes; hence every
proper ideal possesses at least one Krull associated prime. However, this is
not the case for Zariski-Samuel associated primes: Nakano
\cite{Na} gives an example of a \Pd\ in which \fg\ ideals do not
admit any Zariski-Samuel associated prime.
We are convinced that
the notion of a Krull associated prime is more useful (and perhaps more
natural) than any of the other definitions proposed in the literature,
because besides being equivalent to the classical notion of associated prime
in Noetherian rings, it has the following pleasant properties: 1) the set
union of the Krull associated primes of an ideal $A$ is the set of non-prime
elements to $A$ (so there is a sufficient supply of them); 2) every
Krull associated prime is contained in a maximal Krull associated prime; 3)
Krull associated primes are preserved under localization; and 4) they
can be recognized by a distinguished property of the corresponding isolated
components of $A$ (see Theorem 3.4).
For an ideal $A$ of $R$ we follow \cite[page 289]{B} and
define $\Ass_f(A)$ to be the set of weak-Bourbaki associated
primes of $A$. We reserve the notation $\Ass(A)$ for the set of
Krull associated primes of $A$, that is,
$$
\Ass (A)= \{P\ : \ P {\mbox{ is a
prime that is the set union of primes in }} \Ass_f(A)\}.
$$
Finally, let
$$
\XX_A = \{ P \
: P {\mbox{ is a maximal member of }}\Ass(A) \}
$$
That
there is a sufficient supply of primes in ${\mathcal{X}}_A$ (and
hence in $\Ass(A)$) is shown by the next lemma
(cf. \cite[Exercise 17(b), page 289]{B}).
\begin{lemma}
\label{lemma 2.2}
Let $A$ be a proper ideal of the ring $R$ and let $S(A)$ denote
the set of elements of $R$ that are non-prime to $A$. Then
$$S(A)=\bigcup _{P \in \Ass_f(A)} P = \bigcup _{P \in
\Ass(A)} P =\bigcup _{P \in \XX_A} P.$$
\end{lemma}
\begin{proof}
The last two equalities being obvious, it suffices to verify the
first.
Let $u \in P$ for some $P \in \Ass_f(A)$. Since $P$ is a minimal
prime of $A:x$ for some $x \in R \setminus A$, and $(A:x)_{(P)} = A _{(P)}:x $
is a $P$-primary ideal, there is an integer
$k > 0$ such that $u^k \in A_{(P)}:x$. Thus $su^kx \in A$ for some $s
\notin P$.
But $A:x \subseteq P$ implies $sx \notin A$, so $u^k \in S(A)$ and hence
$u \in S(A)$ as well.
On the other hand, if $u \in S(A)$, then there is a $v \notin A$
such that $u \in A:v$. If $P$ is a minimal prime divisor of $A:v$, then
$u \in A:v \subseteq P \in \Ass_f(A)$.
\end{proof}
The maximal prime divisors of $A$
(i.e. the ideals maximal in the set $S(A)$)
need not belong to $\Ass(A)$. This situation is illustrated by
examples given by Nagata \cite{Ng1}, by Gilmer \cite{G},
and by Example 3.8 presented below. However, we have
\begin{corollary}
\label{corollary 2.3}
If $A$ is a primal ideal of $R$, then
its adjoint prime ideal $P$ is a Krull associated prime of $A$.
\end{corollary}
\begin{proof} By Lemma~\ref{lemma 2.2}, $P = S(A)$ is as stated.
\end{proof}
If $A$ is $P$-primal, then $P$ is a Zariski-Samuel
prime of $A$ if and only if $P$ is a weak-Bourbaki prime of $A$.
For it is always
true that Zariski-Samuel implies weak-Bourbaki. On the
other hand, if $x \in R \setminus A$, then every
minimal prime of $A:x$ consists of elements non-prime
to $A$. Therefore if $A$ is $P$-primal, then every minimal prime
of $A:x$ is contained in $P$. Hence $P$ is a
minimal prime of $A:x$ implies $P = \sqrt{A:x}$
and $P$ is a Zariski-Samuel prime of $A$.
As far as the behavior of associated primes under localization is concerned,
we have the most satisfactory result
(cf. \cite[Exercise 17(d), page 289]{B}):
\begin{lemma}
\label{lemma 2.4}
Let $A$ be an ideal of the ring $R$ and let
$P$ be a prime ideal of $R$ containing $A$.
\begin{itemize}
\item[(i)] The following statements are equivalent for $P$.
\begin{itemize}
\item[(a)] $P \in \Ass_f(A)$.
\item[(b)] $P_M \in \Ass_f(A_M)$ for
every maximal ideal $M$ of $R$ containing $P$.
\item[(c)] $P_M \in \Ass_f(A_M)$ for some maximal ideal $M$ of $R$
containing $P$.
\end{itemize}
\item[(ii)] The following statements are equivalent for $P$.
\begin{itemize}
\item[(a)] $P \in \Ass(A)$.
\item[(b)] $P_M \in \Ass(A_M)$ for every maximal
ideal $M$ of $R$ containing $P$.
\item[(c)] $P_M \in \Ass(A_M)$ for some maximal ideal
$M$ of $R$ containing $P$.
\end{itemize}
\end{itemize}
\end{lemma}
\begin{proof} (i) To see that (a) implies (b), suppose
$P$ is a minimal prime over $A:r$ for some $r \in R \setminus
A$, and let $M$ be a maximal ideal of $R$ containing $P$. There
exists a prime ideal $Q$ of $R$ such that $Q_M$ is minimal over
$A_M:r$, so $A_M:r \subseteq Q_M \subseteq P_M.$ In this case,
$A:r \subseteq Q \subseteq P$, so $Q = P$. This proves (a)
implies (b). That (b) implies (c) is clear, so to prove that
(c) implies (a), suppose there is a maximal ideal $M$ of $R$
such that $P_M$ is minimal over $A_M:s$ for some $s \in R
\setminus A_M.$ Then $A:s \subseteq P$, and there exists a
minimal prime $Q$ of $A:s$ such that $A:s \subseteq Q \subseteq
P.$ This implies $A_M:s \subseteq Q_M \subseteq P_M,$ so $Q_M =
P_M$ and $Q = P$.
(ii) If $P \in \Ass(A)$, then $P$ is the union of weak-Bourbaki
primes $Q$ of $A$. Let $M$ be a maximal ideal of $R$ that
contains $P$. By (i), $Q_M$ is a weak-Bourbaki prime of $A_M,$
so $P_M$ is the union of weak-Bourbaki primes $Q_M$ of $A_M$.
This proves (a) implies (b). The assertion that (b) implies (c)
is clear. To complete the proof, we verify (c) implies (a).
Suppose $P_M \in \Ass(A_M)$ for some maximal ideal $M$
containing $P$. Then $P_M$ is the union of weak-Bourbaki primes
$Q_M$ of $A_M$. By (i), the preimage $Q$ in $R$ of
each $Q_M$ is a weak-Bourbaki prime of
$A,$ so $P$ is the union of weak-Bourbaki primes of $A$.
\end{proof}
With a hypothesis of
Noetherian spectrum we can say more
about the primes in $\Ass(A)$ and about
the set ${\mathcal{X}}_A$.
However, as Example 3.8 below illustrates, it is
possible even in a ring with Noetherian spectrum
for there to exist an ideal $A$ having a maximal prime
divisor $P$ such that $P \not \in \Ass(A)$. (This cannot
happen if the ring is Noetherian.)
\begin{lemma}
\label{lemma 2.5}
Suppose that $A$ is a proper ideal of the ring
$R$ and the spectrum of $R/A$ is Noetherian.
\begin{itemize}
\item[(i)] All the ideals in $\Ass_f(A)$ are Zariski-Samuel
associated primes.
\item[(ii)] The primes in ${\mathcal{X}}_A$ are either maximal
members in the set
\vskip .2in
{\noindent}${\rm (1)} \hskip
1.7truein { \{\sqrt {A:x} \
:\ x \in R \setminus A\}}$
\vskip .2in
\noindent
of radical
ideals or unions of incomparable
primes in $\Ass_f(A)$.
\item[(iii)] If $P = \sqrt {A:y}$ is maximal in the set
{\rm (1)}, then $A:y $ is $P $-primary.
\end{itemize}
\end{lemma}
\begin{proof} (i) By Heinzer-Ohm \cite[Lemma 3.2]{HO}, the
members of $\Ass_f(A)$ are all Zariski-Samuel associated primes, if
the radical ideals containing $A$ satisfy the ascending chain
condition (cf. Lemma 1.5). Then $\Ass(A)$ is the set union of
$\Ass_f(A)$ and unions of incomparable primes in $\Ass_f(A)$. (That the
second kind of ideals may exist is illustrated by Example 4.5
infra.)
(ii) By Lemma 1.5(ii), the set
$\{\sqrt {A:x} \ :\ x \in R \setminus A\}$ of radical
ideals contains maximal members. Let $A:y \ (y \in R \setminus A)$
be an ideal with maximal radical $\sqrt {A:y}$. Lemma
1.5(iii) guarantees that $A:y$ has at most finitely many
minimal primes, thus there are finitely many primes $P_1, \dots,
P_m$ with $\sqrt {A:y} = P_1 \cap \dots \cap P_m$, where no $P_i$
may be omitted. If $m >1$, pick a $u \in (P_1 \cap \dots \cap
P_{m-1}) \setminus P_m$ and a $v \in P_m \setminus (P_1 \cap
\dots \cap P_{m-1})$. Then $u^k \notin A:y$ for any $k >0$, so
by the maximal choice of $\sqrt {A:y}$, we must have $\sqrt
{A:yu^k} = P_1 \cap \dots \cap P_m$ for every $k>0$. There is
certainly a $k >0$ such that $(uv)^k \in A:y$. Hence we obtain
$v^k \in A:yu^k$, so $v \in \sqrt {A:yu^k} \subseteq P_1 \cap
\dots \cap P_m$, a contradiction. We have thus shown that the
radical of $A:y$ is the prime ideal $P= P_m$.
(iii) To show that $A:y$ is $P$-primary, let $r,s \in R$ such
that $rs \in A:y$, but $r \notin A:y$. Then by maximality again
$\sqrt {A:ry} = P$, whence $s \in A:ry \subseteq P$.
\end{proof}
Note that from Lemma~\ref{lemma 2.5} it follows that if $R$ is a Pr\"ufer domain and
$\Spec(R/A)$ is a Noetherian space, then every Krull associated prime of $A$
is a Zariski-Samuel associated prime of $A$.
\begin{example}
\label{example 2.6}
{\it For an ideal $A$, the set $\XX_A$ need not be finite even if
$R$ has Noetherian spectrum.} {\rm Let $F$ be a field, and $x,y$
indeterminates over $F$. Let $V$ be the discrete \vd\ obtained by localizing
the polynomial ring $F[x, y]$ at the prime ideal generated by $y$. Then $V =
F(x) + P,$ where $P$ denotes the maximal ideal of $V$.
It is easily seen that $R = F[x] + P$ is a \Pd\ with $P$ as the
unique prime ideal of height one. Evidently, the factor ring
$R/P$ is isomorphic to the polynomial ring $F[x],$
and as a result, $R$ has infinitely many maximal ideals of height 2.
Consider the ideal $A = yR.$ Manifestly, Spec $R$
is Noetherian. It is straightforward to
check that each of the height 2 maximal ideals of $R$ consists of
zero-divisors in $R/A$ and is a Krull associated prime to $A$.
Hence $\XX_A$ is an infinite set.}
\end{example}
For an ideal $A$ of a Noetherian domain $R$, it may
happen that there exist prime ideals $P \subset Q$
with $P$ and $Q$ in $\Ass(A)$ and yet there exist
infinitely many prime ideals $L$ with $P \subset L \subset Q$
and $L \not\in \Ass(A)$. Consider, for example,
$R = F[x, y, z]$, where $x, y, z$ are indeterminates over a
field $F$ and $A = (x^2, xy, xz)R$. Then $\Ass(A) = \{P, Q \}$,
where $P = xR$ and $Q = (x, y, z)R$ and there exist
infinitely many prime ideals $L$ such that
$P \subset L \subset Q$ (see, for example \cite[page 7]{Ka}.
Moreover, $P$ is the radical of $A$ and
$Q$ is the adjoint of the primal ideal $A$.
To illustrate the properties of $\Ass(A)$ in a non-Noetherian
setting, we turn our attention to arithmetical rings. In this case we can
assert very specific properties of the set $\Ass(A)$. It
is interesting to compare Proposition~\ref{proposition 2.7}
to the example of the
preceding paragraph.
\begin{proposition}
\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{proposition}
\begin{proof} By Lemma~\ref{lemma 2.4},
it is enough to prove the proposition in the case where $R$ is a
valuation ring. Furthermore, we may assume that $A = (0)$. If
$P \in \Ass(A)$, then $R_P$ is a valuation ring whose maximal
ideal consists of zero divisors. Thus if $Q \subseteq P$, then
by Lemma 1.1 of \cite{GH3}, $QR_Q$ consists of zero divisors of
$R_Q$, and it follows that $Q \in \Ass(A)$. \end{proof}
We conclude from Proposition~\ref{proposition 2.7} that if $R$ is a valuation
ring and $A \ne 0$ is a proper ideal of $R$, then $\Ass (A)$
is exactly the set of all primes between the radical of $A$ and
the adjoint prime of $A$. In this case, all the primes in
$\Ass(A)$ are Zariski-Samuel associated primes with the
possible exception of those primes that are infinite unions of
strictly smaller primes.
Let $A$ be a nonzero proper ideal of a Pr\"ufer domain $R$ and
let $M$ be a maximal ideal of $R$ with $A \subseteq M$.
Since $R_M$ is a valuation domain, $\End(AR_M) = R_P$ for
some $P \in \Spec R$ with $P \subseteq M$. Let
$$
\EE_A = \{P \in \Spec R : R_P = \End(AR_Q) {\mbox{ for some
prime ideal }} Q {\mbox{ containing }} A \}.
$$
\begin{proposition}
\label{proposition 2.8} 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{proposition}
\begin{proof}
We first show $\Ass(A) \subseteq \EE_A$.
By Proposition~\ref{proposition 2.7}, it is enough to
verify that $\XX_A \subseteq \EE_A$.
Let $P \in \XX_A$, and let $M$ be
some maximal ideal of $R$ containing $P$. Since $P \in \Ass (A)$, $P_M \in
\Ass (A_M)$ by Lemma~\ref{lemma 2.4}. If $Q_M$ is the adjoint prime of the irreducible
(hence, by Lemma 1.2, primal) ideal $A_M$, then $P_M \subseteq
Q_M$, and $Q_M \in \Ass(A_M)$ by Corollary~\ref{corollary 2.3}. Thus, by
Lemma~\ref{lemma 2.4}, $Q \in
\Ass(A)$. Since $P \subseteq Q$ and $P$ is
in $\XX_A$, it must be the case that $P=Q$. Thus $A_M$ has adjoint prime
$P_M$, and by Lemma 2.4 in \cite{FM}, End($A_M) = R_P$.
Consequently, $P \in \EE_A$.
To prove the converse, by Proposition~\ref{proposition 2.7}, it suffices
to show that every maximal element $P$ (with respect to set inclusion) of
$\EE_A$ is a member of $\Ass(A)$. Note that every member of $\EE_A$
is contained in at least one maximal element of $\EE_A$ and that such a
maximal element $P$ must satisfy $\End(A_M) = R_P$ for some maximal ideal $M$
of $R$. Let $M$ be such a maximal ideal of $R$. By Lemma 2.4
in \cite{FM}, $A_M$ is a primal ideal with adjoint prime $P_M$.
It follows that $P_M \in \Ass (A_M)$; hence $P \in \Ass (A)$ by
Lemma~\ref{lemma 2.4}. \end{proof}
The following example shows that in the setting of Pr\"ufer domains, even
with additional hypothesis such as Noetherian spectrum,
the maximal prime divisors of an ideal need
not be Krull associated primes. (Compare this to the Noetherian case, in
which maximal prime divisors are always associated primes.) In
\cite{FHO}, we consider this problem in more depth.
\begin{example}
\label{example 2.9}
{\it There exists 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
$A:M = A$ and $M \not\in \Ass(A)$.} {\rm Let $D$ be a Dedekind domain with
non-torsion class group and let $P$ be a maximal ideal of $D$ that is not the
radical of a principal ideal. Let $x$ be an indeterminate over the field of
fractions $K$ of $D$ and let $V = K[x]_{(xK[x])} = K + Q$, where $Q$ is the
maximal ideal of $V$. Define $R = D + Q$. Then $R$ is a 2-dimensional
Pr\"ufer domain with $Q$ as the unique prime ideal of $R$ of height one. The
maximal ideals of $R$ are all of height two and have the form $M' = P'R$,
where $P'$ is a maximal ideal of $D$. Thus $\Spec R$ is Noetherian. Let $E =
\bigcup_{n=1}^\infty D:_KP^n$ be the $P$-transform of $D$. Consider the ideal
$A = xER$ of $R$ and the maximal ideal $M = PR$ of $R$. Since $ED_P = K$, it
follows that $AR_M = QR_M$ and $M \not\in \Ass(A)$. However, for $P'$ a
maximal ideal of $D$ with $P' \ne P$ and $M' = P'R$, we have $AR_{M'} =
xR_{M'}$. Therefore $M' \in \Ass(A)$. Since every element of $P$ is contained
in a maximal ideal of $D$ distinct from $P$, every element of $M$ is non-prime
to $A$ and $M$ is a maximal prime divisor of $A$.} \end{example}
\section{Isolated Components and Primal Ideals}
In view of Lemma 1.2, every ideal of the ring is the
intersection of primal ideals. Also, every ideal is the
intersection of its isolated components. We would like to have
an intersection representation of an ideal with components that
are both isolated components and primal ideals with distinct
adjoint primes; in this way we will have more information about
the components and the decomposition.
In the representation of an ideal as the intersection of its isolated
components we do not need all them to obtain the
ideal: those belonging to maximal prime divisors, the
{\it principal components} (Hauptkomponenten) in the
sense of Krull \cite[page 16]{K2}, would perfectly
suffice (see the comment after Lemma 3.1). However, an objection
to using these components is that they need not be ideals
with well-understood properties, for example, they may fail to
be primal. Therefore, we focus our attention on the isolated components
belonging to the maximal Krull associated primes; they will provide
sufficiently many isolated components to represent the ideals, as is shown by
the next result.
\begin{lemma}
\label{lemma 3.1}
For every proper ideal $A$
of the ring $R$ we have $$A = \bigcap_{P \in {\mathcal{X}}_A}A_{(P)}.$$
\end{lemma}
\begin{proof}
Suppose $x \not \in A$. If $Q$ is a minimal prime
of $A:x$, then $Q$ is contained in some $P \in
{\mathcal{X}}_A$. Hence $A:x \subseteq P$, and it follows that $x
\not \in A_{(P)}$. Consequently, $x \not \in \bigcap_{P \in
{\mathcal{X}}_A}A_{(P)}$, and we conclude that $A = \bigcap_{P \in
{\mathcal{X}}_A}A_{(P)}$.
\end{proof}
Observe that every prime $P$ in $\XX_A$ is contained in a maximal
prime divisor $P'$
of $A$. The principal component of $A$ with respect to $P'$ is
obviously contained in the isolated component of $A$ with
respect to $P$. Therefore, the decomposition of Lemma 3.1
continues to hold if the intersection is taken not over
the primes in $\XX_A$ but over the maximal prime divisors of $A$.
In particular, the following result was proved by Krull
\cite[page 16]{K2}:
\begin{corollary}
\label{corollary 3.2} Every ideal $A$ is the intersection of its principal
components $A_{(P)}$, $P$ a maximal prime divisor of $A$. In particular, if
$A$ has only finitely many maximal prime
divisors, $P_1, \dots,P_n$, then $A$ is the intersection of finitely many
principal components $A_{(P_i)}$. \end{corollary}
It is natural to wonder when the isolated components, and in
particular those in the intersection presented in Lemma 3.1, are
primal ideals. Examples given by Nagata \cite{Ng1}
and by Gilmer \cite{G} show that if $P$ is a
maximal prime
divisor of the ideal $A$, then the ideal $A_{(P)}$ need
not be $P$-primal. In these examples, $A$ has infinitely many maximal prime
divisors. In Example 3.8 below, we present an example where the ideal
$A$ has precisely two maximal prime divisors and where the principal
component with respect to each of these maximal prime divisors is
not a primal ideal.
There are important special cases in which isolated components
are always primal, viz. the arithmetical rings.
\begin{lemma}
\label{lemma 3.3} If
$A$ is an ideal of an arithmetical ring, then $A_{(P)}$ is an irreducible
(hence primal) ideal for every prime ideal $P$ containing $A$. Its adjoint
prime is contained in $P$. \end{lemma}
\begin{proof}
By Remark~\ref{remark 5.5}, $A_P$ is irreducible if and only if $A_{(P)}$ is
irreducible. Since $R_P$ is a valuation ring, it follows that $A_{(P)}$ is
an irreducible ideal of $R$. By Lemma 1.3, the adjoint prime of $A_{(P)}$
is contained in $P$.
\end{proof}
Let $P$ be a prime ideal containing the ideal $A$. The
isolated $P$-component $A_{(P)}$ is a $P$-primary ideal
if and only if $P$ is a minimal prime of $A$.
The most important criterion for the primality of the isolated
components is given by the equivalence of (1) and (3) of
\cite[Proposition 1.1]{Ku} as we record in
Theorem \ref{theorem 3.4}. Since our definition of
Krull associated prime is different from that given
in \cite{Ku}, we give a short direct proof of the result.
\begin{theorem}
\label{theorem 3.4}
Let $P$ be a prime ideal containing the ideal
$A$. The isolated $P$-component $A_{(P)}$ of $A$ is $P$-primal
if and only if $P$ is a Krull associated prime of $A$.
\end{theorem}
\begin{proof}
Since the elements of $R\setminus P$ are prime to
$A_{(P)}$, $A_{(P)}$ is $P$-primal if and only if
the elements of $P$ are nonprime to $A_{(P)}$.
Equivalently, for every $x \in P$, there is a
$z \notin A_{(P)}$ such that $xz \in A_{(P)}$,
i.e., $xzu \in A$ for some $u \in R \setminus P$.
Thus $x \in A:zu$, where $zu \notin A_{(P)}$. The
last relation says that $A:zu \subseteq P$, so
$P$ is a Krull associated prime of $A$.
\end{proof}
We summarize our findings in
\begin{theorem}
\label{theorem 3.5}
Every proper ideal $A$ of a ring $R$
can be written as an intersection
\vskip .3in
{\noindent}${\rm (2)} \hskip
1.7truein {A = \bigcap_{P \in \XX_A} A_{(P )}}
$\newline
\vskip .3in
{\noindent}where the (isolated) components $A_{(P)}$ are primal
ideals with distinct, incomparable adjoint primes $P$.
\end{theorem}
We call the decomposition of Theorem 3.5, the
{\it canonical primal decomposition } of $A$. Example~\ref{example 2.6}
shows that the canonical primal decomposition
may be infinite even if Spec $R$ is Noetherian.
For a maximal prime divisor $P$ of an ideal $A$
the question was raised in \cite[page 3]{F2} whether or not
the principal $P$-component $A_{(P)}$
is equal to the intersection of all primal ideals
that contain $A$ and have their associated
adjoint prime contained in $P$.
It follows from Theorem 3.4 and observations made in \cite{F2}
that the answer to this question is affirmative
if $P \in \Ass(A)$.
We show in Corollary 3.6 that the answer
is affirmative in general.
\begin{corollary}
\label{corollary 3.6} Let $P$ be a maximal prime divisor of an
ideal $A$ of the ring $R$. Let $A_{(P)}$ denote the principal
$P$-component of $A$ and let $A^*_{(P)}$ be the intersection of
all primal ideals that contain $A$ and have their associated
adjoint prime contained in $P$. Then $A_{(P)} = A^*_{(P)}$.
\end{corollary}
\begin{proof} It is shown in \cite[page 3]{F2} that $A_{(P)} \subseteq A^*_{(P)}$
and it is easily seen that the elements of $R \setminus P$ are
prime to $A_{(P)}$. Hence if $Q \in \Ass(A_{(P)})$, then $Q \subseteq P$.
Therefore the reverse inclusion follows from
the canonical primal decomposition of $A_{(P)}$ given by Theorem 3.5.
Indeed, if $Q \in \Ass(A_{(P)})$, then by Lemma~\ref{lemma 2.4}
$Q \in \Ass(A_P)$, so that (again by Lemma~\ref{lemma 2.4}) $Q
\in \Ass(A)$. Thus by Theorem 3.5
$$
A_{(P)} = \bigcap \{A_{(Q)} \ : \ Q \in \Ass(A), Q \subseteq P
\}.
$$
\end{proof}
It is also true
that $A_{(P)}$ is an intersection of $P$-primal
ideals as we show in Proposition \ref{proposition 3.8}.
\begin{proposition}
\label{proposition 3.8}
Let $P$ be a prime ideal of
the ring $R$. An ideal $A$ contained in $P$ is an intersection
of $P$-primal ideals if and only if the elements of $R\setminus
P$ are prime to $A$. In particular, $A_{(P)}$ is an
intersection of $P$-primal ideals.
\end{proposition}
\begin{proof} If $A = \bigcap Q_i$, where each $Q_i$ is $P$-primal,
then for $s \in R\setminus P$, we have $A:s = (\bigcap Q_i) :s
= \bigcap(Q_i : s) = \bigcap Q_i = A$. Thus the elements of
$R \setminus P$ are prime to $A$.
To prove the converse,
we may pass from $R$ to $R/A$ and assume that $A = (0)$.
Then the elements of $R \setminus P$ are prime to $(0)$ and
$R$ is a subring of the localization $T = R_P$.
We observe that if $J$ is a $PT$-primal ideal of $T$,
then $I = J \cap R$ is $P$-primal in $R$. For
$T$ is a localization of $R$ implies every ideal of $T$ is the extension
of an ideal of $R$, so $IT = J$. Moreover for $r \in R$, we
have $(I:_R r)T = IT:_Tr$. It follows that $I$ is
$P$-primal in $R$. Therefore it suffices to show that if $R$
is quasilocal with maximal ideal $P$, then $(0)$ is an
intersection of $P$-primal ideals of $R$. Let $a$ be a
nonzero element of $P$ and let $Q = aP$. Then $a \not\in Q$
and $P = Q:a$, so $Q$ is $P$-primal. Thus $(0)$ is an intersection
of $P$-primal ideals. \end{proof}
\begin{example}
\label{example 3.9} {\rm {\it There exists a ring $R$
having Noetherian prime spectrum and precisely two maximal
ideals, each of which is a maximal prime divisor of $(0)$ but
not a Krull associated prime of $(0)$.} Consider the polynomial ring
$E=F[x,y,z]$ in three indeterminates over a field $F$ and the
maximal ideals $N_1 = (x,y,z)E$ and $N_2 = (x,y,z-1)E$ of $E$.
Let $D$ denote the localization of $E$ at the multiplicative set
$E \setminus (N_1 \cup N_2)$. Then $D$ is a unique factorization
domain with precisely two maximal ideals, $N_1D$ and $N_2D$.
Consider the height-two prime ideals $P_1 = (x,z)D$ and $P_2 =
(y, z-1)D$, and let $C$ denote the $D$-module $C = \bigoplus
D/p $, where $p$ varies over the height-one primes of $D$ other
than $xD$ and $yD$. Let $B = C \oplus D/P_1 \oplus D/P_2$.
Define $R$ to be the direct sum of $D$ and $B$ made into a ring via Nagata's
idealization \cite{Ng2} (see Exercises 6,7 on pp. 62-63 in
Kaplansky \cite{Ka}). It is clear that Spec $R$ is
isomorphic to Spec $D$. Thus $\Spec R$ is Noetherian. Since $D$
is an integral domain, $B$ is the nilradical of $R$ and $R$ has
precisely two maximal ideals, $M_1 = N_1D \oplus B$ and $M_2 =
N_2D \oplus B$. Moreover, every element of $M_1 \cup M_2$ is a
zero divisor, so $R$ is equal to its total quotient ring. For $i
= 1, 2$, let $A_i = \{r \in R : {\rm there \ exists \ } s \in R
\setminus M_i {\rm \ such \ that \ } rs = 0 \}$ denote the
principal component of the zero ideal of $R$ with respect to the
maximal ideal $M_i$. Then $A_1$ and $A_2$ are contained in $B$;
$A_1$ is the direct sum of $D/P_2$ with $\bigoplus D/p$, where
$p$ varies over the height-one primes of $D$ not contained in
$M_1$ and $A_2$ is the direct sum of $D/P_1$ with $\bigoplus
D/p$, where $p$ varies over the height-one primes of $D$ not
contained in $M_2$. Since $(0) : x = D/P_1$ and $(0) : y =
D/P_2$, we see that $A_1 : y = A_1$ and $A_2 : x = A_2$.
Therefore the ideal $A_i$ is not $M_i$-primal and $M_i$ is not
an associated prime of $(0)$. Indeed, the ideals $A_1$ and $A_2$
are not even primal since the set of elements non-prime to $A_i$
does not form an ideal. For example, the elements $y + z$ and
$z$ are non-prime to $A_1$, while their difference $y$ is prime
to $A_1$.} \end{example}
\section{Irredundant Primal Decompositions}
We wish to consider primal decompositions of an ideal $A$ with distinct
adjoint primes, and to examine when such decompositions are irredundant
and unique in some sense. Recall that if an ideal $A$ is
represented as an intersection of ideals $A = \bigcap_jB_j$, then a component
$B_j$ is said to be {\it relevant } in this representation if omitting
$B_j$ increases the intersection. If each component $B_j$ is relevant, then
the intersection is said to be {\it irredundant}.
In this section
we concentrate on the general case and on rings
with Noetherian spectra, while in the next section we deal with
arithmetical rings.
First, we verify the following lemma.
\begin{lemma}
\label{lemma 4.1}
Suppose
that $A$ is a proper ideal of $R$, and $P \in \XX_A$ has one of
the following properties:
\begin{itemize}
\item[(i)] $P$ is not contained in the set union of the primes
$Q \in \XX_A$ with $Q \ne P$.
\item[(ii)] $P$ is a Zariski-Samuel associated prime.
\end{itemize}
Then in the canonical primal decomposition {\rm
(2)} the isolated $P$-component $A_{(P)}$ is relevant.
\end{lemma}
\begin{proof} (i) Suppose that $P$ is not contained in the set union of the primes
$Q \in \XX_A$ with $Q \ne P$. Then there exists an element $p \in P$ that
fails to belong to any of these $Q \in \XX_A$. By Theorem 3.4,
$A_{(Q)}$ is a $Q$-primal ideal, thus $A_{(Q)} : p = A_{(Q)}$
for every $Q \ne P$. From (2) we obtain
$$A: p
=(A_{(P )}:p) \cap \bigcap_{Q \in \XX_A, Q \ne P}\ A_{(Q
)}.$$
In view of Lemma~\ref{lemma 2.2}, $A:p$ properly
contains $A$, whence it is clear that (a) $A_{(P )}\subset A_{(P
)}:p$; and (b) in (2) we cannot replace $A_{(P )}$ by the
larger $A_{(P )}:p$. Consequently, the component $A_{(P )}$ is
relevant in (2).
(ii) Suppose that $P$ is a Zariski-Samuel associated prime to $A$.
Then by definition there exists an element $y \notin A$ such that
$P = \sqrt {A:y}$. If $P \ne Q \in \XX_A$, then
$A_{(Q)}:y \subseteq Q$ is impossible, as it would imply
$P=\sqrt {A:y} \subseteq \sqrt {A_{(Q)}:y} \subseteq Q$. We
conclude that $y \in A_{(Q )}$ holds for all $Q \ne P$. In other words,
$y \in \bigcap_{Q \in \XX_A, Q \ne P}\ A_{(Q )}$.
But then $A_{(P )}:y =A:y \subseteq P$, which shows that $y \notin A_{(P )}$,
establishing the claimed relevance of $ A_{(P )}$.
\end{proof}
In conclusion, from Theorem 3.5 and Lemma 4.1 we have established:
\begin{theorem}
\label{theorem 4.2}
Let $A$ be a proper ideal of the ring $R$. If
each $P \in \XX_A$ has the property that either
{\rm (i)} $P$ is not contained in the set union of
the other primes in $\XX_A$, or
{\rm (ii)} $P$ is a Zariski-Samuel associated prime of $A$,
then the canonical primal decomposition
$A = \bigcap_{P \in \XX_A}A_{(P)}$ is irredundant.
\end{theorem}
If each $P \in \XX_A$ is a
Zariski-Samuel prime of $A$, we prove that a stronger form
of irredundancy holds. An intersection $A=\bigcap_j B_j$ is said to be
{\it residually maximal at $B_j$} if the intersection no longer equals $A$
whenever $B_j$ is replaced by a residual $B_j:c$ that
properly contains $B_j$. The intersection $A = \bigcap_j B_j$
is said to be {\it residually maximal} if it is residually
maximal at each $B_j$. Residually maximal intersections
are evidently irredundant.
\begin{proposition}
\label{proposition 4.3}
If $P \in \XX_A$ is a
Zariski-Samuel associated prime of $A$, then the canonical
primal decomposition $A = \bigcap_{Q \in \XX_A}A_{(Q)}$ is
residually maximal at $A_{(P)}$. If each $P \in \XX_A$ is a
Zariski-Samuel associated prime of $A$, then the canonical
primal decomposition is residually maximal.
\end{proposition}
\begin{proof} Supposing that the component $A_{(P)}$ can be replaced by $A_{(P)}: c$
for some $c \in R$, we show that such a residual cannot be larger than
$A_{(P)}$.
We argue as in the proof of Lemma 4.1(ii) (keeping the
same notation) that in this
case
$$
A: y =A_{(P )}:y = (A_{(P )}:c) :y =
(A_{(P)}:y) :c.
$$
Thus $c$ is prime to the $P$-primary ideal $A:y$ (see
Lemma 2.5(iii)), i.e. $c \notin P$. Therefore, $A_{(P)}: c =
A_{(P)},$ since by Theorem 3.4 $A_{(P)}$ is $P$-primal. \end{proof}
We now give several examples to illustrate our theorems.
\begin{example}
\label{example 4.4} {\it An irredundant primal decomposition of an ideal $A$ of
a Noetherian ring $R$ need not consist of isolated components.} {\rm Consider
the polynomial ring $R=F[x,y,z]$ in three indeterminates with coefficients
in a field $F$. The ideal $A= (x^2y,xy^2z^2)R$ has a primary decomposition
$$
A = xR \cap yR \cap (x^2,y^2)R
\cap (x^2,z^2)R
$$
with radicals $xR, yR, (x,y)R,
(x,z)R$, respectively. The set $\XX_A$ consists of two prime
ideals (they are Zariski-Samuel associated primes):
$$
\sqrt {A:xyz^2} = (x,y)R \quad {\rm and} \quad \sqrt
{A:xy^2z} = (x,z)R.
$$
The canonical primal
decomposition of $A$ given by (2) of Theorem 3.5 is
$$
A = A_{(x,y)R} \cap A_{(x,z)R},
$$
where
\begin{center}
$A_{(x,y)R} = (x^2y,xy^2)R$ \quad and \quad
$A_{(x,z)R} = (x^2,xz^2)R$.
\end{center}
{\noindent}Note that $A =
(x^2y,xy^2)R \cap (x^2,z^2)R$ and $A = (x^2y,y^2)R \cap
(x^2,xz^2)R$ are representations of $A$ as intersections of
primal ideals, however, the primal ideals $(x^2, z^2)R$ and
$(x^2y, y^2)R$ are not isolated components of $A$.}
\end{example}
\begin{example}
\label{example 4.5}
{\it There exists a ring $R$ with Noetherian prime spectrum such that $R$ has a
$P$-primal ideal $A$ that can be represented as an irredundant intersection of
primal isolated components whose associated
adjoint primes are different from the adjoint prime $P$
of $A$.} {\rm
Consider the localized polynomial ring $D = F[x,y]_{(x,y)F[x,y]}$, where $x,
y$ are two indeterminates over a field $F$. Visibly, $D$ is a 2-dimensional
local UFD. Let $B$ be the $D$-module which is the direct sum of the $D/pD$ as
$p$ varies over the non-associate irreducible elements of $D$.
As in Example 3.8, define $R$ to be the direct sum of $D$ and $B$
made into a ring via Nagata's idealization.
It is clear that Spec $R$ is isomorphic to
Spec $D$. Hence $\Spec R$ is Noetherian. Let
$M = (x,y)D + B$ denote the maximal ideal of $R$ and let
$A = (0)$ in $R$.
Since $M$ consists of zero-divisors, $A$ is a primal ideal of $R$
whose adjoint prime is $M$.
The set $\XX_A$ consists of the single element $M$. Moreover,
the height-one primes $P$ of $R$ have the form $P = pD + B$, where
$p$ is an irreducible element of $D$. Hence $M$ is
the union of the height one primes of $R$, each of
which is a Zariski-Samuel associated prime of $A$. For
$P = pD + B$ a height-one prime of $R$, the isolated component
$A_{(P)}$ is the submodule of $B$ that is
the direct sum of $D/qD$ as $q$ varies over the
non-associate irreducible elements of $D$ distinct from $p$.
Thus the primal ideal $A$ of $R$ can be represented with primal
components both as $A$ (one component), and also as an irredundant
intersection $A = \cap A_{(P)}$ of $P$-primal ideals,
where $P$ varies over the height-one primes
of $R$ (infinitely many components). }
\end{example}
As Examples 4.4 and 4.5 show, an ideal may have various
decompositions into the intersection of primal ideals. For
primes $P \in \Ass (A)$, the isolated $P$-component $A_{(P)}$ of
an ideal $A$ is certainly the minimal $P$-primal ideal containing
$A$, so the representation established in Lemma 3.1 is the
minimal one in the obvious sense. Since there might exist
other decompositions into primal components, the usual question
arises: to what extent is the representation of an ideal as an
intersection of primal isolated components unique? Needless to say, we
cannot expect complete uniqueness, not even for irredundant
representations with distinct adjoint primes, since this fails
already in the Noetherian case. However, the uniqueness of the
set of adjoint primes can be ascertained if we assume some sort
of maximality (in the Noetherian case, it was assumed that the
intersection was reduced; see \cite{F2}).
By a {\it primal isolated component} of an ideal $A$ we
mean a primal ideal $B$ such that $B = A_{(Q)}$ for some
prime ideal $Q$. If $P$ is the adjoint prime of $B$, then
necessarily $P \subseteq Q$. It is possible that $P$ is
properly contained in $Q$. However, regardless of whether
this containment is proper, it follows that $A_{(P)} = A_{(Q)}$.
For $P \subseteq Q$ implies $A_{(Q)} \subseteq A_{(P)}$
and by Lemma 1.3(i), $B = B_{(P)}$. Since $A \subseteq B$,
we have $B = A_{(Q)} \subseteq A_{(P)} \subseteq B$. Thus
if $B$ is a primal isolated component of $A$, then $B = A_{(P)}$
for some $P \in \Ass(A)$.
In Theorem 4.6 we pinpoint a
necessary and sufficient condition for $P \in \XX_A$ to
appear in every representation of $A$ as an intersection of
a subset of its primal isolated components.
\begin{theorem}
\label{theorem 4.6}
If $A$ is an ideal of the ring $R$ and $P
\in \XX_A$, then the following statements are equivalent for $P$:
\begin{itemize}
\item[(i)] $P$ is a Zariski-Samuel associated prime of $A$.
\item[(ii)] The isolated component $A_{(P)}$ must appear in
every representation of $A$ as an intersection of primal
isolated components of $A$.
\end{itemize}
\end{theorem}
\begin{proof} (i) $\Rightarrow$ (ii) The proof of Lemma 4.1(ii) applies,
since it requires only that $Q \in \Ass(A)$.
%$\Rightarrow$ (ii) Since $P$ is a Zariski-Samuel
%associated prime of $A$, there exists $x \in R \setminus A$ such
%that $\sqrt{A:x} = P$.
% Observe that by Theorem 3.4,
%every primal isolated component of $A$ is of the form $A_{(Q)}$
%for some $Q \in \Ass(A)$. Suppose that $\CC \subseteq \Ass(A)$
%and $$(\dagger) {\hskip .3in} A = \cap_{Q \in \CC}A_{(Q)}.$$
%We show that $A_{(P)} = A_{(Q)}$ for some $Q \in \CC$. Since $x
% \in R \setminus A$, then by ($\dagger$), $x \not \in A_{(Q)}$
%for some $Q \in \CC$ and it follows that $A:x \subseteq Q$. Thus
%$P \subseteq Q$, and because $P$ is maximal in $\Ass(A)$
%it follows that $Q = P$.
(ii) $\Rightarrow$ (i) Suppose that $P$ is not a Zariski-Samuel
associated prime of $A$. We claim that $A = \cap_{Q \ne
P}A_{(Q)}$, where $Q$ ranges over the members of $\Ass(A)$
distinct from $P$, thus contradicting (ii). To prove this, it
suffices to show that for every $y \in R \setminus A$, there
exists $Q \in \Ass(A)$ with $Q \ne P$ and $A:y \subseteq Q$. Let
$y \in R \setminus A$. By assumption $\sqrt{A:y} \ne P$, so
there exists a minimal prime $Q$ of $A:y$ with $Q \ne P$.
Necessarily, $Q \in \Ass(A)$. \end{proof}
\begin{corollary}
\label{corollary 4.7}
For an ideal $A$ of the ring $R$ the
following statements are equivalent.
\begin{itemize}
\item[(i)] For each set of prime ideals ${\mathcal{C}} \subseteq \Ass(A)$
such that $A = \bigcap_{P \in {\mathcal{C}}}A_{(P)}$, it is the case that
${\XX_A} \subseteq {\mathcal{C}}$.
\item[(ii)] Each $P \in \XX_A$ is a
Zariski-Samuel associated prime of $A$. \end{itemize}
In particular, if each $P \in \XX_A$ is a Zariski-Samuel associated prime of
$A$, then the canonical primal decomposition $A = \bigcap_{P \in
\XX_A}A_{(P)}$ is irredundant. \end{corollary}
\begin{proof} That (i) implies (ii) follows from Theorems 3.4
and~\ref{theorem 4.6}. To see that (ii) implies (i), suppose $A = \bigcap_{P
\in {\mathcal{C}}}A_{(P)}$ for some ${\mathcal{C}} \subseteq \Ass(A)$. By
Theorem~\ref{theorem 4.6}, for each $P \in \XX_A$, we have $A_{(P)} =
A_{(Q)}$, for some $Q \in {\mathcal{C}}$. Since $P,Q \in \Ass(A)$, we have
by Theorem 3.4 that $A_{(P)}$ is $P$-primal and $A_{(Q)}$ is $Q$-primal.
Since $A_{(P)} = A_{(Q)}$ we conclude $P = Q$. \end{proof}
\begin{remark}
\label{remark 4.8}
{\rm If $P \in \XX_A$ is of the form $P = \sqrt{A:x}$,
then $A:x$ is $P$-primary. For if $r, s \in R$ are
such that $rs \in A:x$ and $r \not\in A:x$,
then $A:rx$ is a
proper ideal and $P \in \XX_A$ implies that
$P = \sqrt{A:rx}$. Hence $s \in A:rx \subseteq P$,
so $A:x$ is $P$-primary.
Example 4.5 shows that the canonical primal
decomposition of $A$ given in (2) of Theorem 3.5 may be
irredundant, but not the unique representation of $A$ as an
irredundant intersection of pairwise incomparable isolated
primal components of $A$.
For the ideal $A = yR$ of Example~\ref{example 2.6}, the canonical
primal decomposition of $A$ given in (2) of Theorem 3.5 is
irredundant and yet the set $\XX_A$ is infinite. There
are also examples where $A$ is an ideal of a one-dimensional
Pr\"ufer domain and the canonical primal decomposition given
in (2) of Theorem 3.5 is redundant, but a proper subset of
$\XX_A$ gives an irredundant representation of $A$.
Indeed, it can happen that all but one of the elements of
$\XX_A$ is a
Zariski-Samuel prime of $A$ and that $A$ is the intersection
of its primal components with respect to this proper subset
of $\XX_A$ (cf. Example 2.2 in \cite{HO}).}
\end{remark}
We would like to have conditions that are both necessary
and sufficient for irredundance of the canonical
primal decomposition $A = \cap_{P \in \XX_A}A_{(P)}$.
The following interesting example shown to us by
Alan Loper and included here with his permission
shows that the sufficient conditions of
Theorem \ref{theorem 4.2} are not in general necessary conditions
for irredundance. The example is a Pr\"ufer domain $R$
having property (\#), that is, the representation
$R = \cap\{R_P : P \in \Max(R) \}$ is irredundant. The
integral domain $R$ is constructed so that it has a
nonzero principal ideal $A = yR$ that is contained in
every maximal ideal of $R$. Since $R$ is Pr\"ufer and
$A$ is principal, it follows from Proposition \ref{proposition 2.8}
that $\XX_A = \Max(R)$.
Moreover the representation $A = \cap_{P \in \XX_A}A_{(P)}$
is irredundant and yet there exists $P \in \XX_A$ such
that $P$ is contained in the union of the
prime ideals of $\XX_A$ distinct from $P$ and $P$ is
not a Zariski-Samuel associated prime of $A$.
\begin{example}
\label{example 4.9}{\rm (Alan Loper) Let $V$ be a valuation
domain on a field $K$ such that the maximal ideal $M$ of $V$
is the union of the prime
ideals properly contained in $M$. Let $E = \{a_1, a_2, a_3, a_4\}$
where $\{a_i \}_{i=1}^4$ are
nonzero elements of $M$ having the property that
$$
a_4V = a_3V \supset (a_3 - a_4)V \supset a_2V \supseteq a_1V.
$$
Thus if $v$ is a valuation on $K$ associated to the
valuation domain $V$, then
$$
v(a_4) = v(a_3) < v(a_3 - a_4) < v(a_2) \le v(a_1).
$$
Let $x$ be an indeterminate and let $D = \Int(E,V) =
\{f(x) \in K[x] : f(E) \subseteq V \}$ be the
ring of integer-valued polynomials on the set $E$.
It is well-known that $D$ is a Pr\"ufer domain \cite[Corollary 7]{Mc}.
Moreover, $D$ has precisely four maximal ideals of infinite height.
These are the maximal ideals containing $MD$. They
are $M_i$, $1 \le i \le 4$, where
$M_i = \{f(x) \in K[x] : f(a_i) \in M \}$.
The other maximal
ideals of $D$ are of height one and are the contraction
to $D$ of maximal ideals of $K[x]$. Each of the maximal
ideals of $K[x]$ other than $(x - a_i)K[x]$ intersects $D$
in a maximal ideal of $D$, while $(x-a_i)K[x] \cap D$, for
$1 \le i \le 4$, is
a height-one prime of $D$ that is contained in $M_i$.
Each of the valuation domains $D_{M_i}$ has
the property that its maximal ideal is the union of the
prime ideals it properly contains. Therefore $M_i$ in
$D$ is the union of the prime ideals it properly contains.
The following argument shows that $M_i$, for $1 \le i \le 4$,
is contained in the union
of the other maximal ideals of $D$. We have $D \subseteq K[x]$
and $M_i \cap K = M$ for each $i$ with $1 \le i \le 4$.
Therefore every element of $K$ that is in one of the $M_i$ is
in all four of the $M_i$. A
nonconstant polynomial $f(x) \in D \subseteq K[x]$ that
is in one of that $M_i$
and is not of the form $f(x) = d(x-a_j)^n$, where $d \in K$,
$n$ is a positive integer and $1 \le j \le 4$ is easily
seen to be also in
another maximal ideal of $D$. Thus assume
$f(x) = d(x-a_j)^n \in D$. By definition of $D$, $f(a_i) \in V$
for $1 \le i \le 4$. Also $f(a_j) = 0 \in M$, so $f(x) \in M_j$.
Using the fact that
$$
v(a_4) = v(a_3) < v(a_3 - a_4) < v(a_2) \le v(a_1).
$$
it follows by examining cases that at most two of the values $v(f(a_i))$
can be $0$. Therefore $f(x)$ is in at least two of the $M_j$.
Let $y$ be an
indeterminate over $K(x)$, let $Q = yK(x)[y]_{(y)}$ and let
$R = D + Q$. For each maximal ideal $P$ of $R$, we
have $P = (P \cap D)R$ and $P \cap D$ is a maximal ideal
of $D$. It follows that $M_iR$, for
$1 \le i \le 4$,
is a maximal ideal of $R$ that is contained in the
union of the other maximal ideals of $R$.
To show that $R$ satisfies property (\#), it suffices to
show that for each maximal ideal $P$ of $R$ there exists a
finitely generated ideal contained in $P$ but not contained
in any other maximal ideal of $R$.
Since $Q$ is a height-one prime ideal of $R$ that is
contained in every nonzero prime ideal of $R$ and
since $D = R/Q$, it suffices to show this property
for $D$. Each maximal ideal of $D$ is either the
contraction of a maximal ideal of $K[x]$ or else
in the set $\{M_i\}_{i=1}^4$. For each of the maximal
ideals $P$ contracted from $K[x]$ there exists an element $f$
contained in $P$ and contained in no other maximal ideal
of $D$ that is contracted from $K[x]$. By choosing an
element of $P \setminus \cup_{i=1}^4M_i$, we see that
$P$ is the radical of a finitely generated ideal of $D$.
Thus each of the maximal ideals of $D$ contracted from
$K[x]$ is the radical of a finitely generated ideal.
Since a nonzero element of $D$ is contained in only
finitely many maximal ideals of $K[x]$, we also easily
get, for $1 \le i \le 4$, the existence of
a finitely generated ideal $A_i$ contained in $M_i$
such that $A_i$ is contained in no other maximal ideal of $D$.
Therefore $D$ and $R = D + Q$ satisfy property (\#).
Let $A = yR$. Then $A$ is a nonzero principal ideal of $R$ that is
contained in each maximal ideal of $R$. Hence by
Proposition \ref{proposition 2.8}, $\XX_A = \Max(R)$.
Since $R$ satisfies property (\#), the representation of
the principal ideal $A$ as
$A = \cap_{P \in \XX_A}A_{(P)}$ is
irredundant, but $M_iR$, for $1 \le i \le 4$, is not a
Zariski-Samuel associated prime of $A$ and $M_iR$ is contained
in the union of the prime ideals of $\XX_A$ distinct from $M_iR$.
This completes the presentation of Example \ref{example 4.9}. }
\end{example}
\section{Irredundant Irreducible Decompositions}
In this section our focus is on irredundant decompositions of
ideals $A$ into irreducible isolated components $A_{(P)}$. We
prove a uniqueness statement about such decompositions, and then
characterize those rings for which the canonical
primal decomposition of every proper ideal $A$ is an
irredundant representation of $A$ as an intersection of
irreducible isolated components.
In the statement of Theorem 5.1 we use the
concept of a residually
maximal intersection as defined in Section 4.
\begin{theorem}
\label{theorem 5.1}
Let $A$ be a proper
ideal of a ring $R$. If $\CC$ is a collection of prime ideals of
$R$ and $A = \bigcap_{P
\in \CC}A_{(P)}$ is an irredundant intersection of irreducible
isolated components of $A$, then this intersection is
residually maximal and is the unique irredundant decomposition
of $A$ into irreducible isolated components of $A$.
\end{theorem}
\begin{proof}
Since the isolated components $A_{(P)}$ are
irreducible, they are primal. Without loss of generality, we
assume $P$ is the adjoint prime of $A_{(P)}$ and $\CC$
consists of the adjoint primes $P$ of the $A_{(P)}$.
Thus $\CC \subseteq \Ass(A)$, and $A =
\bigcap_{P \in \CC}A_{(P)}$ is an irredundant decomposition of $A$,
where each $P \in \CC$ is the adjoint prime of the irreducible
ideal $A_{(P)}$.
If $P$ is a prime ideal
of $R$ and $A = B \cap C$ for ideals $B$ and $C$ of $R$, then
$A_{(P)} = B_{(P)} \cap C_{(P)}$. We will use this fact
frequently in the proof.
Suppose that for some $P \in \CC$ there exists an $x \in R$ such that
$$
A = (A_{(P)}:x) \cap (\bigcap\{A_{(Q)} : Q \in \CC, Q \ne P \}.
$$
Let $C = \bigcap_{Q \ne P}A_{(Q)}$. By Lemma 1.1, we
have
$$
A_{(P)} = (A_{(P)}:x) \cap C_{(P)}.
$$
Since $A_{(P)}$ is irreducible, we have either $A_{(P)} =
A_{(P)}:x$ or $A_{(P)} = C_{(P)}$. The
latter equality, however,
contradicts the irredundancy of the decomposition of
$A$, so we have that $A_{(P)} = A_{(P)}:x$. We conclude
that any irredundant decomposition of $A$ into irreducible
isolated components is residually maximal.
It remains to show that the representation $A =
\bigcap_{P \in \CC}A_{(P)}$ is unique among irredundant
intersections of $A$ into irreducible isolated components of $A$.
Suppose $\CC' = \{Q_i\}_{i \in I} \subseteq \Ass(A)$ is such
that $A = \bigcap_{i \in I}A_{(Q_i)}$ is also an irredundant
decomposition of $A$ into irreducible components $B_i= A_{(Q_i)}$
of $A$ with adjoint primes $Q_i$. We show that for each $i' \in
I$, there exists $P_{i'} \in \CC$ such that $P_{i'} \subseteq Q_{i'}$.
Fix $j \in I$, and define $C_j = \bigcap_{i \ne j}B_i$. We
exhibit $P \in \CC$ such that $P \subseteq Q_j$.
Observe that for each $P \in \CC$,
$$
A_{(P)} = (B_j)_{(P)} \cap (C_j)_{(P)}.
$$
Since $A_{(P)}$ is irreducible, it must be the case that
$A_{(P)} = (B_j)_{(P)}$ or $A_{(P)} = (C_j)_{(P)}$.
If the
latter case holds for every $P \in \CC$, then
$$
C_j \subseteq \bigcap_{P
\in \CC}A_{(P)} = A,
$$
and it follows that $\bigcap_{i \ne j}B_i = C_j \subseteq A$.
However, this contradicts the assumption that the
intersection $A = \bigcap_{i \in I}B_i$ is irredundant, so it
must be the case that for some $P \in \CC$, $A_{(P)} =
(B_j)_{(P)}$. Since $A_{(P)}$ is $P$-primal, we
have that $(B_j)_{(P)}$ is $P$-primal. However, by
assumption, $B_j$ is $Q_j$-primal, so statement (iii) of Lemma
1.3 implies $P \subseteq Q_j$. We conclude that for each $i' \in
I$, there exists $P_{i'} \in \CC$ such that $P_{i'} \subseteq Q_{i'}$.
Applying a symmetrical argument to the decomposition $A =
\bigcap_{i \in I}B_i$, we have that for each $P \in \CC$, there
exists $i \in I$ such that $Q_i \subseteq P$. Thus $P_i \subseteq
Q_i \subseteq P$. Since the decomposition $A = \bigcap_{P \in
\CC}A_{(P)}$ is irredundant, it follows that $\CC$ consists of
incomparable primes. Thus $P_i = Q_i = P$, and $\CC \subseteq
\{Q_i:i \in I\}$. Also, the symmetry of our assumptions yields
$\{Q_i:i \in I\} \subseteq \CC$. We conclude $\CC = \{Q_i:i \in
I\}$, and it follows that
$$
\{A_{(P)}:P \in \CC\} = \{A_{(Q_i)}:i \in I\}.
$$
This proves the theorem.\end{proof}
\begin{discussion}
\label{discussion 5.2}
{\rm Assume that
$\CC$ is a subset of $\Ass(A)$ containing at least two
elements. If
$A = \bigcap_{P \in \CC}A_{(P)}$ and if the intersection is
irredundant, then
$\CC$ consists of incomparable primes. Also, for each
$P \in \CC$ there exists $Q \in \XX_A$ such that
$P \subseteq Q$, and it follows that $A_{(Q)} \subseteq A_{(P)}$.
Thus if $\CC' = \CC \setminus \{P\}$ and
$B = \bigcap_{P' \in \CC'}A_{(P')}$, then $A_{(Q)}$ is
relevant in the intersection $A = A_{(Q)} \cap B$. Since
$A_{(P)}$ is $P$-primal and $A_{(Q)}$ is $Q$-primal,
we have $A_{(P)} = A_{(Q)}$ if and only if $P = Q$.
If $A_{(Q)}$ is irreducible, we observe that $P = Q$.
For $A = A_{(P)} \cap B$ implies $A_{(Q)} = A_{(P)} \cap B_{(Q)}$,
and $P \ne Q$ and $A_{(Q)}$ irreducible imply $A_{(Q)} = B_{(Q)}$.
But this implies $B \subseteq A_{(P)}$ which contradicts the
irredundance of the intersection $A = A_{(P)} \cap B$.
Therefore, if $A_{(Q)}$ is irreducible
for each $Q \in \XX_A$, then $\CC \subseteq
\XX_A$.}
\end{discussion}
From Discussion 5.2, we deduce
the following corollary to Theorem 5.1.
\begin{corollary}
\label{corollary 5.3}
Let $A$ be an ideal of a ring $R$, and suppose $\CC
\subseteq \Ass(A)$. If $A_{(Q)}$ is irreducible for
each $Q \in \XX_A$ and $A = \bigcap_{P \in \CC}A_{(P)}$ is an
irredundant intersection, then $\CC \subseteq \XX_A$ and this is
the unique decomposition of $A$ into an intersection of
irreducible isolated components of $A$.\qed
\end{corollary}
The irreducibility and isolation of the components in Theorem 5.1
are crucial to the assertion of uniqueness in the theorem. For
$A$ as in Example 4.5 we have two distinct irredundant representations
of $A$ as an intersection of primal isolated component ideals of $A$.
Example 5.4 exhibits a situation where an ideal has two distinct
irredundant representations as an intersection of primal
isolated components, one of which consists of irreducible
isolated components.
It is interesting to contrast Theorem 5.1 with the situation for
infinite irredundant primary decompositions of ideals in general.
In this case, neither the primary ideals in
the representation nor even the primes associated to these ideals
need be unique as is shown for example by Underwood in
\cite[Example 4.18]{U}. On the other hand, it is readily seen
that in a Noetherian ring there cannot exist an infinite
irredundant intersection of ideals.
%For suppose $A = \cap_{i \in I}A_i$ is an irredundant intersection,
%where $I$ is an infinite set. We may assume $\vN \subseteq I$.
%For each $n \in \vN$, let
%$B_n = \cap \{A_i : i \in I \setminus \{1, \ldots, n\} \}$. The
%irredundance of the intersection $\cap_{i \in I}A_i$ implies
%the chain of ideals $B_1 \subseteq B_2 \subseteq \cdots $
%is strictly ascending.
\begin{example}
\label{example 5.4}
{\rm {\it An ideal may have two distinct
decompositions as an irredundant intersection of primal isolated
components, one of which consists of irreducible isolated
components.} Let $\{x_n : n \in \vN \}$ be indeterminates over
a field $F$ and let $R$ be the localization of the polynomial
ring $F[x_1, x_2, \ldots ]$ at the maximal ideal generated by
$\{x_n : n \in \vN \}$. Consider the ideal $A$ of $R$ generated
by $\{x_ix_j : i \ne j \}$. Then if $M$ is the unique
maximal ideal of $R$, $R/A$ is one-dimensional with a unique
maximal ideal $M/A$ and infinitely many minimal primes $P_n/A$,
$n \in \vN$, where $P_n = (\{x_i : i \in \vN, i \ne n \})R$.
Moreover, $x_n$ is not in $P_n$ but is in every prime of $R$
containing $A$ other than $P_n$. The ideal $A$ of $R$ is
$M$-primal and is the irredundant intersection of irreducible
isolated components $A = \bigcap_{P \in \CC}A_{(P)}$, where $\CC
= \{P_n : n \in \vN \} \subset \Ass(A)$ consists of precisely the
elements in $\Ass(A)$ other than $M$. Thus for this example, the
canonical primal decomposition of $A$ given in (2) of Theorem
3.5 is trivial (that is, $A = A_{(M)}$) and yet there exists a
representation $A = \bigcap_{P \in \CC}A_{(P)}$ that is
irredundant and consists of irreducible isolated components of
$A$ and that is, by Theorem 5.1, unique among irredundant
intersections of irreducible isolated components of $A$.
However, the representation $A = \bigcap_{P \in \CC}A_{(P)}$ is
not unique among irredundant intersections of primal isolated
components of $A$.} \end{example}
In light of Theorem 5.1, it is natural to consider rings $R$ for
which every ideal $A$ is locally irreducible in the sense that
its extension to $R_M$ is irreducible for each
maximal ideal $M$ containing $A$. As noted in Section 1, these are precisely
the arithmetical rings.
Returning to Theorem 5.1, we obtain a
stronger result for arithmetical rings because of the
following lemma which asserts that irredundant primal
decompositions in arithmetical rings must involve isolated
components.
\begin{lemma}
\label{lemma 5.6}
Let $R$ be an arithmetical
ring, and let $A$ be an ideal of $R$. Suppose $A = \bigcap_{i \in
I}A_i$ is an intersection of
ideals $A_i$ of $R$. For each $i \in I$, if $A_i$ is a primal ideal of
$R$ that is relevant to this decomposition, then $A_i = A_{(P_i)}$, where $P_i$
is the adjoint prime of $A_i$. \end{lemma}
\begin{proof} Let $i \in I$, and suppose $A_i$ is relevant to the
given decomposition. If $P_i$ is the adjoint prime of $A_i$, then $A_i =
(A_i)_{(P_i)}$ in view of Lemma 1.3. Also,
$$
A_{(P_{i})} = A_i \cap (\bigcap_{j \ne i}A_j)_{(P_i)}.
$$
By Remark 1.6, $A_{(P_i)}$ is an irreducible ideal
of $R$. Thus if $A_{(P_i)} \ne A_i$, then $A_{(P_i)} = (\bigcap_{j
\ne i}A_j)_{(P_i)}$. But this implies that $\bigcap_{j \ne
i}A_j \subseteq A_i$, contradicting the
relevancy of $A_i$ to the decomposition. We conclude that $A_{(P_i)} =
A_i$. \end{proof}
\begin{corollary}
\label{corollary 5.7} Let $R$ be an arithmetical ring, and $A$
an ideal of $R$. If $A$ has an irredundant decomposition $A
= \bigcap_{i \in I}A_i$ into irreducible ideals $A_i$, then this
intersection is residually maximal and it is the unique
irredundant decomposition of $A$ into irreducible ideals.
Moreover, if $P_i$ is the adjoint prime of $A_i$, then
$\{P_i\}_{i \in I} \subseteq \XX_A$.
\end{corollary}
\begin{proof} Apply Theorem 5.1, Corollary 5.3 and
Lemma~\ref{lemma 5.6}. \end{proof}
\begin{remark}
\label{remark 5.8}
{\rm Corollary~\ref{corollary 5.7} need not be true for rings
that are not arithmetical. In particular, the uniqueness
assertion of the corollary fails in the Noetherian case. For
example, if $x$ is an indeterminate over a field $F$ and $(R,M)$
is the local domain $F[x^2, x^3]_{(x^2, x^3)}$, then $M^2 = (x^4,
x^5)R$ is $M$-primary and has an irredundant decomposition $M^2 =
x^2R \cap x^3R$, where $x^2R$ and $x^3R$ are irreducible.
However, neither $x^2R$ nor $x^3R$ is an isolated
component of $M^2$ since the only isolated component of $M^2$
is the ideal $M^2$ itself.}
\end{remark}
The uniqueness result of Theorem 5.1 suggests the question:
What rings $R$ have the property that
for every proper ideal $A$ of $R$ the
canonical primal decomposition of $A$ given in (2) of
Theorem 3.5 is an irredundant representation of $A$ as
an intersection of irreducible isolated components?
The rest of
this section is devoted to a resolution of this question.
\begin{theorem}
\label{theorem 5.9}
The following statements are equivalent for
a ring $R$.
\begin{itemize}
\item[(i)] $R$ is an arithmetical ring.
\item[(ii)] For each proper ideal $A$ of $R$, $A =
\bigcap_{P \in \XX_A}A_{(P)}$ is a decomposition of $A$ into
irreducible ideals $A_{(P)}$.
\item[(iii)] Each proper ideal $A$ of $R$ can be represented as
an intersection of irreducible isolated components of $A$.
\end{itemize}
\end{theorem}
\begin{proof}
The implication
(i) $\Rightarrow$ (ii) follows from Theorems~\ref{new theorem} and
3.5; the implication (ii) $\Rightarrow$ (iii) is clear. To complete the
proof it suffices to show (iii) $\Rightarrow$ (i).
To show $R$ is arithmetical, we establish that $R_M$ is a
valuation ring for each maximal ideal $M$ of $R$. Let $M$ be a
maximal ideal of $R$, and let $a,b \in M$. Define $I = (a,b)R$.
Then by (iii), there is a collection $\CC$ of prime ideals of $R$
such that $IM = \bigcap_{P \in \CC} (IM)_{(P)}$, and each
$(IM)_{(P)}$ is irreducible. If $M \not \in \CC$, then
$(IM)_{(P)} = I_{(P)}$ for all $P \in \CC$. But this implies
$$
I \subseteq \bigcap_{P \in \CC}I_{(P)} = IM.
$$
Since $I$ is finitely generated, $I \not \subseteq IM$. It
follows that $M \in \CC$ and $(IM)_{(M)}$ is an irreducible ideal
of $R$. As in Remark~\ref{remark 5.5}, $IMR_M$ is an irreducible ideal of
$R_M$. Thus by the observation ($\dagger$) made in the course of proof of
Theorem~\ref{new theorem}, either $R_Ma \subseteq R_Mb$ or $R_Mb \subseteq
R_Ma$. This completes the proof.
\end{proof}
%\begin{lemma}
%\label{lemma 5.10}
%Let $A$ be a proper ideal of the ring $R$ such that $A =
%\bigcap_{P \in {\mathcal{C}}}A_{(P)}$ for some collection of prime
%ideals ${\mathcal{C}}$ of $R$. Then $A_{(P)}$, $P \in
%{\mathcal{C}}$, is relevant to this decomposition if and only if
%there exists an $x \in R \backslash A$ such that $A:x \subseteq
%P$ but $A:x \not \subseteq Q$ for any $Q \in {\mathcal{C}}$ with $Q
%\ne P$.
%\end{lemma}
%\begin{proof} The lemma follows from an observation that is an
%immediate consequence of the pertinent definitions: $\bigcap_{Q \ne
%P}A_{(Q)} \not \subseteq A_{(P)}$, where $Q$ ranges over the
%prime ideals of ${\mathcal{C}}$ distinct from $P \in {\mathcal{C}}$, if and only if
%there exists an $x \in R \backslash A$ such that $A:x \subseteq P$
%but $A:x \not \subseteq Q$ for any $Q \in {\mathcal{C}}$ with $Q \ne
%P$. \end{proof}
\begin{lemma}
\label{lemma 5.11} Let $R$ be a ring, and $A$ be an ideal of $R$
that has only finitely many minimal primes $P_1,P_2,\ldots,P_n$.
If these minimal primes are comaximal, then each $P_i$ is a
Zariski-Samuel associated prime of $A$.
\end{lemma}
\begin{proof} We may assume $n > 1$. By assumption, there exist $a
\in P_1$ and $b \in P_2\cdots P_n$ such that $a + b = 1$.
Moreover $ab$ has some power $(ab)^n$ in $A$. Then $(a^n, b^n)R
= R$ and $a^n \in A:b^n$. Hence $A:b^n$ has radical $P_1$ and
$P_1$ is a Zariski-Samuel associated prime of $A$. Similarly,
$P_2,P_3,\ldots,P_n$ are Zariski-Samuel associated primes of $A$.
\end{proof}
\begin{lemma}
\label{lemma 5.12}
Let $R$ be a ring in which incomparable
prime ideals are comaximal, so, for example an
arithmetical ring. If $A$ is a radical ideal of
$R$, then each $P \in \XX_A$ is a minimal prime of
$A$ and $A_{(P)} = P$. Thus the
following are equivalent for a radical ideal $A$ of $R$:
\begin{itemize}
\item[(i)] The representation of $A$ as the
intersection of its minimal primes is irredundant.
\item[(ii)] The canonical primal decomposition of $A$
given in (2) of Theorem 3.5 is irredundant.
\end{itemize}
Moreover, these conditions hold for every radical ideal of $R$
if and only if every ideal of $R$ has only
finitely many minimal primes.
\end{lemma}
\begin{proof} Let $A$ be a radical ideal of $R$. Since
incomparable prime ideals of $R$ are comaximal
each $P \in \XX_A$ contains a unique minimal prime
$Q$ of $A$. Since $A$ is a radical ideal,
$A_{(P)} \subseteq A_{(Q)} = Q$ are radical ideals
with $Q$ the unique minimal prime of $A_{(P)}$.
Thus $A_{(P)} = Q$ and $Q = P$. Therefore conditions
(i) and (ii) are equivalent.
It is clear that if every ideal of $R$ has only
finitely many minimal primes, then
conditions (i) and (ii) hold for each radical
ideal of $R$. To
prove the converse, let $A$ be a radical ideal of $R$ and let
$\{P_i \}_{i \in I}$ be the set of minimal primes of $A$.
By hypothesis, the representation
$A = \cap_{i \in I}P_i$ is irredundant.
To prove Lemma \ref{lemma 5.12}, it suffices to
prove the set $I$ is finite. We may assume
that $I$ has cardinality at least two. For
each $i \in I$, let $A_i = \cap P_j$, where $j \in I$
and $j \ne i$. The representation $A_i = \cap_{j \ne i}P_j$
is irredundant and each $P_j, j \ne i$, is a minimal
prime of $A_i$. Since by hypothesis the representation
of each radical ideal of $R$ as the intersection of its
minimal primes is irredundant, it follows that
$\{P_j : j \in I, j \ne i \}$ is the set of
minimal primes of $A_i$.
We observe that $A_i$ and $P_i$ are
comaximal. For if there exists a prime ideal $Q$ of $R$
with $A_i + P_i \subseteq Q$, then $Q$ must contain a
minimal prime of $A_i$. But for $j \ne i$, $P_j$ and
$P_i$ are distinct minimal primes of $A$ and thus are
incomparable and hence by hypothesis comaximal. It
follows that $\sum_{i \in I}A_i = R$.
For if $\sum_{i \in I}A_i \subseteq Q$,
where $Q$ is prime in $R$, then $A \subseteq Q$ implies
$Q$ contains some minimal prime of $A$, so $Q$ contains
$P_i$ for some $i \in I$. But $A_i$ and $P_i$ are
comaximal, a contradiction. Thus $\sum_{i \in I}A_i = R$.
Hence there exists a finite subset $J$ of $I$ such
that $\sum_{i \in J}A_i = R$. We must have $J = I$,
for by construction $\sum_{j \in J}A_j \subseteq P_i$
for $i \in I \setminus J$. \end{proof}
\begin{remark}
\label{remark 5.12} {\rm Without the hypothesis that
incomparable prime ideals of $R$ are comaximal, it can happen
that every radical ideal of $R$ is the irredundant intersection
of its minimal primes and yet there exist radical ideals of $R$
having infinitely many minimal primes. This is illustrated by
the ring $R/A$ of Example 5.4. This ring is one-dimensional reduced
with a unique maximal ideal and infinitely many minimal primes
$P_n/A$. Every radical ideal of $R/A$ is the irredundant
intersection of its minimal primes. Moreover, the zero ideal of $R/A$
is a primal radical ideal that
has infinitely many minimal primes.
We remark that even in a 2-dimensional
Pr\"ufer domain it can happen that there exists a
radical ideal $A$ with infinitely many minimal primes
such that $A$ is the irredundant intersection of its
minimal primes. Thus the hypothesis in Lemma~\ref{lemma 5.12} about
every radical ideal is necessary. This is illustrated by the
following example. }
\end{remark}
\begin{example}
\label{example 5.13}
{\rm
Let $F$ be
a field of characteristic different from 2 and let $x,y$ be
indeterminates over $F$. Let $V_{11}$ be the discrete rank-one
\vd\ obtained by localizing the polynomial ring $F[x, y]$ at the
prime ideal generated by $y$. Then $V_{11} = F(x) + P_{11}$,
where $P_{11}$ denotes the maximal ideal of $V_{11}$. Define
$R_1 = W_{11}$ to be the rank-2 valuation domain $F[x]_{xF[x]} +
P_{11}$.
Let $K_1 = F(x,y)$
and let $K_2$ be a finite algebraic extension field
of $K_1$ in which the valuation domain $V_{11}$ has
at least two extensions $V_{21}$ and $V_{22}$.
For example, one may take
$K_2 = K_1[z]$, where $z$ is a root of the polynomial
$Z^2 - y - 1$. Then
$V_{11}[z]$ is the intersection of two valuation domains
of the field $K_2$ that extend $V_{11}$ which we may label
as $V_{21}$
and $V_{22}$, where $z-1$ generates the maximal ideal of $V_{21}$
and $z+1$ generates the maximal ideal of $V_{22}$.
The valuation domains $V_{21}$ and $V_{22}$ are each
localizations of the integral closure $\overline {W_{11}}$
of $W_{11}$ in the field $K_2$, say at prime ideals
$Q_{21}$ and $Q_{22}$ of $\overline {W_{11}}$. By the
Going-Up Theorem \cite[(10.9)]{Ng2}, there exist prime ideals
$N_{21} \supset Q_{21}$ and $N_{22} \supset Q_{22}$
of $\overline {W_{11}}$ with $N_{21}$ and $N_{22}$ each
lying over the maximal ideal of $W_{11}$. Hence by
\cite[Theorem 13, page 31]{ZS}, there exist extensions $W_{21}$
and $W_{22}$ of the rank-2 valuation domain $W_{11}$ to the
field $K_2$ such that $W_{2i} \subset V_{2i}$, $i = 1, 2$.
Define $R_2 = W_{21} \cap W_{22}$. By the Theorem of
Independence of Valuations \cite[(11.11)]{Ng2}, we see that
$R_2$ is a Pr\"ufer domain with precisely 4 nonzero prime
ideals, these prime ideals being $P_{21} \subset M_{21}$
contracted from $W_{21}$ and $P_{22} \subset M_{22}$ contracted
from $W_{22}$.
Let $A_1$ be the ideal $P_{11}$ of $R_1$ and define
the ideal $A_2$ of $R_2$ as $A_2 = M_{21} \cap P_{22}$.
Observe that $A_2 \cap R_1 = A_1$ and $\Ass(A_2) =
\{M_{21}, P_{22} \}$, with $M_{21}$ and $P_{22}$ the
minimal primes of $A_2$.
There exists a finite algebraic extension field $K_3$
of $K_2$ for which the valuation domain $V_{22}$ has
at least two distinct extensions $V_{32}$ and $V_{33}$,
see, for example \cite[Lemma 1.3]{HW}.
Let $V_{31}$ be an extension of $V_{21}$ to the field
$K_3$. There exists an extension $W_{31}$ to $K_3$ of the
valuation domain $W_{21}$ such that $W_{31} \subset V_{31}$.
Also there exist extensions $W_{32}$ and $W_{33}$ of $W_{22}$
to $K_3$ such that $W_{3i} \subset V_{3i}$, $i = 2, 3$.
Define $R_3 = W_{31} \cap W_{32} \cap W_{33}$. Again by
\cite[(11.11)]{Ng2}, we see that $R_3$ is a Pr\"ufer domain with
precisely 6 nonzero prime ideals, these prime ideals being
$P_{3i} \subset M_{3i}$ contracted from $W_{3i}$, $i = 1, 2, 3$.
Define $A_3 = M_{31} \cap M_{32} \cap P_{33}$. This
representation of $A_3$ is irredundant, so $\Ass(A_3) =
\{M_{31}, M_{32}, P_{33} \}$. Moreover, $A_3 \cap R_2 = A_2$.
Continuing in this way, we obtain by induction for each
positive integer $n$ a finite algebraic extension field
$K_n$ of $F(x,y)$ and a Pr\"ufer domain
$R_n = \cap_{i =1}^n W_{ni}$, where the $W_{ni}$ are
rank-2 valuation domains on the field $K_n$
and where $R_n$ has precisely $2n$ nonzero prime ideals,
these prime ideals being $P_{ni} \subset M_{ni}$
contracted from $W_{ni}$, $i = 1, \dots, n$. Define
$A_n = (\cap_{i = 1}^{n-1} M_{ni}) \cap P_{nn}$. This
representation of $A_n$ is irredundant, so
$\Ass(A_n) = \{M_{n1}, M_{n2}, \ldots, M_{nn-1}, P_{nn} \}$.
Moreover, $A_n \cap R_{n-1} = A_{n-1}$.
Let $R = \cup_{n=1}^\infty R_n$ and $A = \cup_{n=1}^\infty A_n$.
Also define $M = \cup_{n=1}^\infty M_{nn}$ and
$P = \cup_{n=1}^\infty P_{nn}$.
Then $R$ is a 2-dimensional Pr\"ufer domain. The
maximal ideals of $R$ are $M$ and
$M_r = \cup_{n=r}^\infty M_{nr}$, $r = 1, 2, \dots$, while the
nonzero nonmaximal prime ideals of $R$ are $P$ and
$P_r = \cup_{n=r}^\infty P_{nr}$, $r = 1, 2, \dots$.
We have $A = (\cap_{r=1}^\infty M_r) \cap P$ with this
intersection irredundant. Moreover every element of $M$ is
in some $M_r$. Therefore every element of $M$ is non-prime
to $A$. Thus $M$ is a maximal prime divisor of $A$.
However, the $M$-principal component $A_{(M)}$ of $A$ is $P$,
so $M \not\in \Ass(A)$. This completes the presentation of the example.}
\end{example}
\vskip 3 pt
In the proof of Theorem~\ref{theorem 5.15}, it will be convenient
to use the following terminology from \cite{GH1} or \cite[page
62]{FHP}. An integral domain $R$ satisfies (\#) if the
intersection $R = \bigcap_{M}R_M$, where $M$ ranges over the set
of maximal ideals of $R$, is irredundant in the sense that if $M$
is any maximal ideal of $R$, then $R \ne \bigcap_{N\ne M}R_N$
when $N$ ranges over the set of all maximal ideals that are
distinct from $M$. If every overring of $R$ satisfies (\#),
then $R$ satisfies (\#\#). The relevance of condition (\#\#) in
the present context is that a Pr\"ufer domain satisfies (\#\#) if
and only if for each prime ideal $P$ of $R$, there exists a
finitely generated ideal $J$ of $R$ such that $J$ is contained
in $P$ and the only maximal ideals of $R$ that contain $J$ are
those that contain $P$ (Theorem 3 in \cite{GH1}]). See also
\cite{FHP} for a recent analysis of the conditions (\#) and
(\#\#) in Pr\"ufer domains.
\begin{lemma}
\label{lemma 5.14}
Let $R$ be an arithmetical ring such that for every
prime ideal $P$ of $R$, there exists a finitely generated ideal
$J$ contained in $P$ such that the only maximal ideals of $R$
containing $J$ are those that contain $P$. Then every ideal
of $R$ has only finitely many minimal primes. In particular,
if $R$ is a Pr\"ufer domain satisfying (\#\#), then
every ideal of $R$ has only finitely many minimal primes.
\end{lemma}
\begin{proof} Let $A$ be a proper ideal of $R$.
If $P$ is a minimal prime ideal of $A$, then
by assumption there exists a finitely generated ideal $J$ of $R$
such that $J \subseteq P$ and the only maximal ideals of $R$
containing $A$ are those that contain $P$. Hence the ideal
$P/A$ of the ring $R/A$ is the
radical of a finitely generated ideal. By Theorem
1.6 in \cite{GH2}, it follows that $R/A$ has finitely many
minimal prime ideals. Therefore $A$ has only finitely many
minimal primes in $R$. The statement about Pr\"ufer domains
satisfying (\#\#) now follows from Theorem 3 of \cite{GH1}.
\end{proof}
\begin{theorem}
\label{theorem 5.15}
The following statements are equivalent for a ring $R$.
\begin{itemize}
\item[(i)] For each proper ideal $A$ of $R$, the decomposition
$A = \bigcap_{P
\in \XX_A} A_{(P)}$ is irredundant and the components
$A_{(P)}$ are irreducible ideals.
\item[(ii)] $R$ is an arithmetical ring such that for every
prime ideal $P$ of $R$, there exists a finitely generated ideal
$J$ contained in $P$ such that the only maximal ideals of $R$
containing $J$ are those that contain $P$.
\item[(iii)] $R$ is an arithmetical ring with
Noetherian maximal spectrum.
\end{itemize}
In particular, the integral domains having these equivalent
properties are precisely the Pr\"ufer domains
that satisfy (\#\#).
\end{theorem}
\begin{proof}
(i) $\Rightarrow$ (ii) First we observe that if $B$ is a proper
ideal of $R$, then $R/B$ also satisfies (i). For if $A$ is a
proper ideal of $R$ containing $B$, then by (i) the
decomposition $A = \bigcap_{P \in \XX_A} A_{(P)}$ is irredundant
and each component $A_{(P)}$ is irreducible. It is
straightforward to verify that for each $x \in R$, $(A:x)/B =
A/B:_{R/B}x+B$, and from this it follows that $\Ass(A/B) =
\{P/B:P \in \Ass(A)\}$; hence $\XX_{A/B} =\{P/B:P \in \XX_A\}$.
Moreover for each $P \in \XX_A$, $(A/B)_{(P/B)} = A_{(P)}/B$,
so that each $(A/B)_{(P/B)}$ is irreducible. Now by Theorem
3.5,
$$
A/B = \bigcap_{P \in \XX_A}(A/B)_{(P/B)} = \bigcap_{P \in
\XX_A}A_{(P)}/B.$$
Hence the irredundancy of the canonical primal decomposition of
$A/B$ follows from the irredundancy of the canonical primal
decomposition of $A$. Thus (i) holds for $R/B$.
We now prove that (i) implies (ii). By Theorem 5.8, $R$ is an
arithmetical ring and by Lemma~\ref{lemma 5.12} each ideal of
$R$ has only finitely many minimal primes. Since a domain
homomorphic image of an arithmetical ring is a Pr\"ufer domain
and since in an arithmetical ring with finitely many minimal
primes each of the minimal primes is the radical of a finitely
generated ideal, it suffices by our above observation that (i)
passes to homomorphic images to prove that the condition in (ii)
holds in the case where $R$ is a Pr\"ufer domain. In this case,
as noted above, the condition in (ii) is equivalent to showing
$R$ satisfies (\#\#). Furthermore, to verify (\#\#), it is
enough to prove that every fractional overring $S$ of $R$ (i.e.
$S$ is an overring of $R$ such that $rS \subseteq R$ for some
nonzero $r \in R$) satisfies (\#) (see, for example, Proposition
2.5 in \cite{O}). Let $S$ be a fractional overring of $R$.
Since $R$ is a Pr\"ufer domain, each maximal ideal of $S$ is of
the form $SP$ for some prime ideal $P$ of $R$. Let $\{SP_i\}$
denote the collection of maximal ideals of $S$, where each $P_i$
is a prime ideal of $R$. Using the fact that $R_{P_i}$ is a
valuation domain, it is easy to see that $S_{SP_i} = R_{P_i}$
since $P_i$ survives in $S_{SP_i}$. We show that $S$ satisfies
(\#) by verifying that for each $j$, the quasilocal ring
$S_{SP_j}$ is relevant to the decomposition $S = \bigcap_{i}
S_{SP_i}$.
Let $SP_j$ be a
maximal ideal of $S$ and let $r$ be a nonzero element of $P_j$
such that $rS \subset P_j$. We use Proposition 2.8 to
calculate $\XX_A$, where $A = rS$. We show that
$\XX_A$ consists of the prime ideals in $\{P_i\}$ that
contain $A$. We first observe:
\smallskip
\noindent
Claim 1. If $P \in \{P_i\}$ and
$A \subseteq P$, then $P \in \Ass(A)$.
We have $\End(A_P) = \End(S_P) = R_P$. The first equality follows from
the fact that $A$ is a principal ideal of $S$, and the second
equality from the fact that $S \subseteq R_P$. By Proposition 2.8,
$P \in \Ass(A)$. We next observe:
\smallskip
\noindent
Claim 2. $\XX_A \subseteq \{P_i\}$.
Let $Q \in \XX_A$. By Proposition 2.8,
$R_Q = \End(A_M) = S_M$ for some maximal ideal $M$ of $R$. Thus
$QS \ne S$, so $QS \subseteq PS$ for some $P \in \{P_i\}$. Since
$S$ is flat over $R$, we have $Q \subseteq P$. By
Claim 1, $P \in \Ass(A)$. Since $Q \in \XX_A$ is
maximal in $\Ass(A)$, we have $Q = P$.
\smallskip
Since there are no inclusion relations among the prime
ideals in $\{P_i\}$, it follows from Claims 1 and 2
that $\XX_A = \{P \in \{P_i\} : A \subseteq P\}$.
By (i), the decomposition $A = \bigcap_{P \in \XX_A}A_{(P)}$ is
irredundant, and by an application of the pertinent definitions,
this implies there exists
$x \in R\setminus A$ such that $A: x = Ax^{-1} \cap R
\subseteq P_j$ but $Ax^{-1} \cap R \not \subseteq P_i$ for all
$i \ne j$ such that $A \subseteq P_i$. Since $R$ is a Pr\"ufer
domain and $A = Sr$, we have $Ax^{-1} \cap S = S(Ax^{-1} \cap R) \subseteq
SP_j$, and if $Ax^{-1} \cap S \subseteq SP_i$ for some $i \ne
j$, then $Ax^{-1} \cap R \subseteq SP_i \cap R = P_i$. If $A
\subseteq P_i$, this is a contradiction to the choice of $x$.
Also, since $A \subseteq A:x$, it cannot be the case that $A:x
\subseteq P_i$ for any $i$ for which $A \not \subseteq P_i$.
We conclude that
the only maximal ideal of $S$ that contains $A:_S x = rS:_S x$
is $SP_j$. It follows that $rx^{-1} \in \bigcap_{i \ne j}S_{SP_i}$
but $rx^{-1} \not\in S_{SP_j}$. Thus $S_{SP_j}$ is relevant to
the decomposition $S = \bigcap_{i}S_{SP_i}$, and this proves that
$S$ satisfies (\#). Consequently, $R$
satisfies (\#\#). This completes the proof that (i) implies (ii).
(ii) $\iff$ (iii) The equivalence of (ii) and (iii) follows
from \cite[Corollaries 1.3 and 1.5]{RW}.
(ii) $\Rightarrow$ (i)
Let $A$ be a proper ideal of $R$. By Lemma 3.1, $A = \bigcap_{P
\in {\mathcal{X}}_A} A_{(P)}$. We claim that this is an
irredundant decomposition of $A$. Let $P \in {\mathcal{X}}_A$.
By (ii), there exists a finitely generated ideal $J$ of $R$
that is contained in $P$ but in no maximal ideal of $R$ not
containing $P$.
Since $R$ is arithmetical, $P =
\bigcup_{i}P_i$ for some chain of weak-Bourbaki associated
primes of $A$. Since $J$ is finitely generated, there
exists a weak-Bourbaki associated prime $P_i$ of $A$
such that $J \subseteq P_i \subseteq P$.
It follows that $P_i \not
\subseteq Q$ for any $Q \in \XX_A$ with $Q \ne P$.
For $Q$ and $P$ distinct in $\XX_A$ are comaximal
and $J \subseteq P_i$ is contained in no maximal
ideal of $R$ not containing $P$. By
definition, $P_i$ is a minimal prime ideal of $A:x$ for some $x
\in R \setminus A$. By Lemma~\ref{lemma 5.14}, $A:x$ has only finitely
many minimal primes, so Lemma~\ref{lemma 5.11} implies that there exists $y
\in R \setminus A$ such that $P_i = \sqrt{(A:x):y} = \sqrt{A:xy}$. It
follows that $A:xy \subseteq P$ but $A:xy \not \subseteq Q$ for any $Q
\in {\mathcal{X}}_A$ with $Q \ne P$. Hence $xy \in \bigcap_{Q \ne
P}A_{(Q)}$. Since $A:xy \subseteq P$, it is case that $xy \not\in
A$, so that $xy \not \in A_{(P)}$. Therefore $\bigcap_{Q \ne
P}A_{(Q)} \not \subseteq A_{(P)}$, and we conclude that the
decomposition $A = \bigcap_{P \in {\XX_A}}A_{(P)}$ is
irredundant. \end{proof}
\begin{remark}
\label{remark 5.16} {\rm {\it A ring $R$ is reduced and satisfies
(i) and (ii) of Theorem~\ref{theorem 5.9} if and only if $R$ is a finite
direct product of Pr\"ufer domains that satisfy (\#\#).} If $R$
satisfies (i) and (ii) of Theorem~\ref{theorem 5.15}, then by
Lemma~\ref{lemma 5.14}, $R$ has finitely many minimal prime ideals. Since
$R$ is reduced, $(0)$ is the intersection of finitely many prime ideals,
and since $R$ is arithmetical, these prime ideals must be comaximal.
Consequently, $R$ is a finite direct product of Pr\"ufer domains, each
of which satisfies (\#\#). Conversely, it is not hard to verify that if $R$ is a finite
direct product of Pr\"ufer domains satisfying (\#\#), then $R$ is a reduced
ring that satisfies (ii) of Theorem~\ref{theorem 5.15}.} \end{remark}
\begin{remark}
\label{remark 5.17} {\rm {\it An arithmetical ring $R$ has
Noetherian prime spectrum if and only if $R$ satisfies condition
(ii) (or (i)) of Theorem~\ref{theorem 5.15} and $R$ has the ascending chain
condition on prime ideals.} This follows, for example, from
Theorem 1.6 in \cite{GH2} and the equivalence of (i) and (iii)
of Lemma 1.5. In particular, an arithmetical ring $R$ of finite
Krull dimension satisfies (ii) of Theorem~\ref{theorem 5.15} if
and only if $R$ has Noetherian prime spectrum. On the other
hand, it is easy to exhibit infinite-dimensional Pr\"ufer
domains satisfying (ii) of Theorem~\ref{theorem 5.15} that do
not have Noetherian prime spectrum. For example, if $V$ is a
valuation domain with value group $\oplus_{i
=1}^{\infty}{\mathbb{Z}}$ given the reverse lexicographic order,
then the maximal ideal of $V$ is the union of all the prime
ideals properly contained in it.}
\end{remark}
\begin{question}
\label{question 5.18} {\rm We close with the following questions.
\begin{itemize}
\item[(1)]
In Theorem \ref{theorem 5.15}, we have characterized the rings $R$
such that for each proper ideal $A$ of $R$ the canonical primal
decomposition of $A$ is irredundant with irreducible components.
It would be interesting to classify the rings $R$ having the property that
for every proper ideal $A$ of $R$ there exists a subset
$C$ of $\Ass(A)$ such that $A = \bigcap_{P \in C}A_{(P)}$ is
irredundant and each $A_{(P)}$ is irreducible.
In view of Theorem \ref{theorem 5.9}, a ring with this
property is an arithmetical ring, so the question is
asking to classify the arithmetical
rings $R$ for which each proper ideal $A$ can
be written as an irredundant intersection
$A = \bigcap_{P \in C}A_{(P)}$ for some subset $C$ of
$\Ass(A)$. There exist Pr\"ufer domains with this property
that do not satisfy the equivalent conditions of
Theorem \ref{theorem 5.15}. For example, if $R$ is the
almost Dedekind domain that is not Dedekind of Example 2.2
of \cite[pages 277-278]{HO}, then $R$ does not satisfy
the equivalent conditions of Theorem \ref{theorem 5.15},
but every proper ideal $A$ of $R$ is an irredundant intersection
$A = \bigcap_{P \in C}A_{(P)}$ for some subset $C$ of
$\Ass(A)$.
\item[(2)]
What Pr\"ufer domains $R$ have the property that for
each proper Zariski closed subset $V$ of $\Spec R$
there exists an ideal $A$ of $R$
such that $\Ass(A) = V$? If there exists a finitely generated
ideal $A$ such that $V = \{P \in \Spec R : A \subseteq P \}$,
then Proposition \ref{proposition 2.8} implies that
$\Ass(A) = V$.
\item[(3)] Let $R$ be an arithmetical ring. Is it possible to
characterize the posets $\Ass(A)$ of $\Spec R$ as $A$ varies
over the ideals of $R$? By Proposition \ref{proposition 2.7},
if $P \in \Ass(A)$ and $Q \in \Spec R$ with
$A \subseteq Q \subseteq P $, then $Q \in \Ass(A)$.
\item[(4)]
In \cite{FHO} we investigate for an ideal $A$ of an
arithmetical ring 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 $A$. In this
connection, it would be interesting to
characterize the
subsets $C$ of the set of maximal ideals of
an arithmetical ring (or Pr\"ufer domain) $R$
such that $C$ is the
set of maximal prime divisors of an ideal $A$ of $R$.
\end{itemize}
}
\end{question}
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