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\begin{document}

\baselineskip 17 pt

\title[Completely irreducible ideals]{Commutative ideal theory without
finiteness \\ conditions: Completely Irreducible 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{irreducible ideal, completely irreducible ideal,
irredundant intersection, arithmetical ring }


\begin{abstract}   An ideal  of a ring is completely
irreducible if it is not the intersection of any set of proper overideals.  We
investigate  the structure of completely irrreducible ideals in a
commutative ring without finiteness conditions.  It is known that every ideal
of a ring is an intersection of completely irreducible ideals.
We characterize in several ways those
ideals that admit a representation as an irredundant intersection
of completely irreducible ideals, and we study the question of
uniqueness of such representations.  We characterize those
commutative rings in which every ideal is an irredundant intersection
of completely irreducible ideals.
\end{abstract}


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\maketitle


\section*{Introduction}
 
Let $R$ denote throughout a commutative ring with 1.
An ideal of $R$ is called {\it irreducible} if it is not the
intersection of two proper overideals; it  is called
{\it \ci} if it is
not the intersection of any set of proper overideals.
Our  goal in this paper is to examine the structure of
completely irreducible ideals of a commutative ring
on which there are imposed no finiteness conditions.
Other recent papers that address the structure and
ideal theory of rings without finiteness conditions
include \cite{Baz}, \cite{BS}, \cite{FoH}, \cite{FK},
  \cite{FHO}, \cite{FHO2}, \cite{FHO3}, \cite{GH},
\cite{LR}, \cite{L}.


A proper ideal $A$ of $R$ is completely irreducible if and
only if there is an element $x \in R$ such that $A$ is
maximal with respect to not containing $x$.
Indeed, the condition is clearly sufficient for $A$
to be completely irreducible since $x$ is in the intersection
of the proper overideals of $A$. On the other hand,
if $A$ is completely irreducible and $x$ is an element
that is not in $A$ but is in the
intersection of the proper overideals of $A$,
then $A$ is maximal with respect to not
containing  $x$. Evidently, maximal ideals are \ci.
If $R$ is a domain  (not a field),
then the zero ideal of $R$ is prime and irreducible, but it
is  not \ci\ (it is the intersection of all nonzero ideals).
More generally, a prime ideal is always irreducible and is
completely irreducible if and only if it is a maximal ideal.
In Theorem \ref{1.3} we characterize completely irreducible ideals
in various ways.  We deduce in Corollary \ref{1.5} that an
irreducible ideal of a Noetherian ring is completely irreducible
if and only if it is primary for a maximal ideal.


A great deal is known about the structure of the irreducible
ideals of a Noetherian ring. Indeed, the first decomposition
theorem  established by  Emmy Noether \cite{N} states that each
ideal of a Noetherian ring admits a representation as
an irredundant intersection of  finitely many
irreducible ideals; moreover,
the number of components appearing  in such a
representation is an invariant,  and  each
such representation is {\it reduced}
(i.e., no ideal in the representation can be replaced by a
strictly larger ideal to obtain the same intersection).
Another result due to Noether \cite{N}
is that a proper  irreducible ideal of a Noetherian ring is a primary ideal.
In a ring without finiteness
assumptions there may exist proper irreducible ideals
that are not primary.
Fuchs in \cite{F} introduced the concept of a
primal ideal, where a
proper ideal $A$ of $R$ is said to be {\it primal} if
the zero-divisors in
$R/A$ form an ideal. The ideal of zero-divisors is then
necessarily of the
form $P/A$ where $P$ is a prime ideal of $R$  called the {\it
adjoint prime} of
$A$. We also say that $A$ is $P$-{\it primal}.
In Fuchs \cite{F} it is  shown that proper irreducible ideals are
primal. More can be said about completely irreducible ideals.
We observe in Theorem \ref{1.3} several equivalences to
a proper ideal $C$ of $R$ being
completely irreducible among which is that  $C$ is
irreducible and $R/C$ contains a simple submodule.


A ring in which every irreducible ideal is completely
irreducible is zero-dimensional.
Corollary \ref{1.5} states that if $\dim R = 0$ and each
primary ideal of $R$ contains a power of its radical, then
every irreducible ideal of $R$
is completely irreducible.
In Section 2 we address the question:  Under what conditions is an
ideal representable  as an irredundant intersection of
completely irreducible ideals? We consider
in Section 3  the  question of uniqueness of
representation of the ideal $A$ as an irredundant intersection
of completely irreducible ideals.
In Section 4 we characterize the rings in which every ideal is
an irredundant intersection of \ci\ ideals.
We prove in Theorem \ref{5.2} that every ideal
of the ring $R$ is
an irredundant intersection of \ci\ ideals exactly if the ring is
semi-artinian, where a ring $R$ is said to be {\it semi-artinian }
if every nonzero $R$-module contains a simple $R$-module.

Proposition \ref{1.4} and Corollary \ref{1.5} characterize the completely
irreducible ideals of a Noetherian ring.  In
Section 5  we  give an explicit description of the completely
irreducible ideals of an  arithmetical ring, where  the ring
$R$ is {\it arithmetical} if for every  maximal ideal
$M$ the ideals of the localization $R_M$ are linearly ordered with respect to inclusion.
An arithmetical integral domain is a {\it Pr\"ufer domain}.
For a prime ideal $P$
of  a Pr\"ufer domain,  Fuchs and Mosteig prove in
\cite[Lemma 4.3]{FM} that the $P$-primal ideals form a semigroup under ideal
multiplication. We generalize this result in Theorem \ref{semigroup}.
For a prime ideal $P$ of an arithmetical ring,
we show  that  the regular $P$-primal ideals
form a semigroup under ideal multiplication, where an ideal is
{\it regular} if it contains a nonzerodivisor.
Theorem \ref{4.4} states that if $M$ is a maximal ideal of an arithmetical
ring $R$, then the set
$\FF$ of \ci\ regular ideals of $R$ with adjoint maximal ideal $M$ is
closed under ideal-theoretic multiplication, and $\FF$ with this
multiplication is  a totally ordered
cancellative semigroup.

A good reference for our terminology and notation is \cite{FS}.
For ideals $I,J$ of the ring $R$, the {\it 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}.  Notice  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 $P$.

Two different concepts of associated  primes of a
proper ideal $A$ of the ring $R$ are useful for us.
One of these was introduced by Krull \cite[page 742]{K1},
and following \cite{IR} we call a prime ideal $P$ of $R$ a
{\it Krull associated prime}
of $A$ if for every  $x \in P$, there exists $y \in R$ such
that $x \in A:y \subseteq P$. The prime ideal $P$ is said
to be a {\it strong Bourbaki associated prime} of $A$ if
$P = A:x$ for some $x \in R$.
%, and is said to be a
%{\it Zariski-Samuel} associated prime of $A$
%if the radical $\sqrt {A:x} = P$ for some $x \in R$.



\section{Irreducible and Completely Irreducible Ideals}

A ring  is called {\it subdirectly irreducible} if in any of its
representations as a subdirect product of rings, one of the
projections to a component is an isomorphism.
It is straightforward to see:


\begin{lemma} \label {1.1} A proper ideal $C$ of $R$ is \ci\ if and
only if the factor ring $R/C$ is subdirectly irreducible, or
equivalently if and only if the $R$-module
$R/C$ has a simple essential socle.
\qedsymbol
\end{lemma}


Thus  if $C$ is \ci, then $R/C$ contains a minimal nonzero
ideal $C^*/C$: the
intersection of all nonzero ideals of $R/C$; this is then
the essential socle of $R/C$.  We shall call
$C^*$ the {\it cover of} $C$. Clearly, $C^*/C$ is a simple
$R$-module, so $C^*/C \cong R/M$ for a maximal ideal $M$ of $R$.
(If $C$ is only assumed to be irreducible, then we can only
claim that $R/C$ is a {\it uniform} $R$-module, that is, the
intersection of any two nonzero submodules of $R/C$ is not zero.)



\begin{proposition} \label{1.2} A completely irreducible
proper  ideal $C$ of
the ring $R$  is a primal ideal whose adjoint prime is the
maximal ideal $M$ of $R$ for which $C^*/C \cong R/M$.
Furthermore,
$M$ is a strong Bourbaki associated prime of $C$.
\end{proposition}



\begin{proof} Let $x \in R$ be a representative of any  coset of $C$
generating $C^*/C \cong R/M$. Then $x \not \in C$ and $xM \subseteq
C$. Hence $M \subseteq C:x \subset R$ (proper containment);  thus
we necessarily have $M=C:x$, so  $M$ is a strong Bourbaki
associated prime of $C$. Evidently, $M$ has to be the adjoint
prime of the primal ideal $C$.
\end{proof}


Completely irreducible ideals can be characterized in various
ways as we demonstrate in Theorem \ref{1.3}.


\begin{theorem} \label {1.3} For a proper  ideal $C$ of $R$, the
following conditions are equivalent.


   {\rm (i)} $C$ is \ci;


{\rm (ii)} the factor module  $R/C$ is an essential extension of a simple module;


{\rm (iii)} $C$ is an irreducible ideal and $R/C$  contains a
simple $R$-submodule;


{\rm (iv)} $C$ is an irreducible ideal and $C$ is
properly contained in $C:M$   for
some maximal ideal $M$ of $R$;


   {\rm (v)} $C$ is irreducible with adjoint prime   a
maximal ideal $M$ of $R$,  and $M=C:x$ for some $x \in R \setminus C$;


{\rm (vi)}  $C = C_{(M)}$ for some maximal ideal $M$ of $R$ and $CR_M$ is a
completely irreducible ideal of $R_M$.
\end{theorem}


\begin{proof} The equivalence of (i) and (ii) and of (iii) and (iv)
is obvious, and so is
the implication (i) $\Rightarrow$ (iv). To prove
(iii) $\Rightarrow$ (i), observe that a simple submodule of
a uniform module is an
essential socle of the module.


By Proposition \ref{1.2},  (i) $\Rightarrow$ (v) is clear.
On the other hand, Condition (v) implies that $x+C$
generates a simple $R$-submodule in
$R/C$. By the uniformity of $R/C$ this simple submodule
is an essential socle of $R/C$, and (i) holds.

The equivalence of (v) and (vi) follows from the following two
observations:  (a)  if $M$ is a
maximal ideal containing $C$, then  $(C:x)R_M = CR_M:_{R_M}x$ for any $x \in
R$;  and (b) $C_{(M)}$ is irreducible if and only if $CR_M$ is an irreducible ideal
of $R_M$ (see Remark 1.6 of \cite{FHO}).  \end{proof}

A ring is {\it Laskerian} if every ideal has  a finite primary decomposition.

\begin{proposition} \label{1.4}
If every proper ideal of the ring $R$ is an intersection of primary
ideals (possibly an infinite intersection),
then every  completely irreducible proper ideal
of $R$ is   primary for a maximal ideal.
In particular, if $R$ is a
Laskerian ring (or a Noetherian ring), then every
completely irreducible proper ideal of $R$ is primary for a
maximal ideal.
\end{proposition}


\begin{proof} Let $C$ be a completely irreducible ideal
of $R$ with adjoint prime $M$ and
let $x \in R$ be a representative of any  coset of $C$
generating $C^*/C \cong R/M$. Then $x \not \in C$ and $xM \subseteq
C$. Since $C$ is an intersection of primary ideals, there
exists a primary ideal $Q$ of $R$ such that $C \subseteq Q$
and $x \not\in Q$. Since $C$ is completely irreducible,
it follows that $C = Q$. Therefore $C$ is a primary ideal
that is $M$-primal, so $C$ is $M$-primary.
\end{proof}



\begin{corollary} \label{1.5}  Let $M$ be a maximal
ideal of the ring $R$ and let $C$ be an
irreducible $M$-primary ideal. If $M^n \subseteq C$
for some positive integer $n$, then $C$ is
completely irreducible.  Thus if $M$ is
finitely generated, then
every irreducible $M$-primary ideal is \ci.
Therefore  an irreducible proper ideal of a Noetherian ring
is completely irreducible if and only if it
is primary for a maximal ideal.
\end{corollary}


\begin{proof} If  $M^n \subseteq C$ for some
positive integer $n$, then $C \subset  (C:M)$.
Hence by Theorem \ref{1.3}(iv), $C$ is completely irreducible.
The last statement now follows from Proposition \ref{1.4}.
\end{proof}


\begin{remark}\label{1.6}
We are interested in describing the rings in which
every irreducible ideal is completely irreducible.
Since prime ideals are irreducible and
a prime ideal is completely irreducible if
and only if it is maximal, the condition that
every irreducible ideal in $R$ is
completely irreducible implies that  all  prime ideals of $R$
are  maximal and $\dim R = 0$. If $R$ is a reduced
zero-dimensional ring, then $R_P$ is a field for
each $P \in \Spec R$ and every primal ideal of $R$ is
maximal. Therefore every irreducible ideal of a
reduced zero-dimensional ring is completely irreducible.

Recall that a ring $R$ is {\it semi-artinian} if every nonzero
$R$-module contains a simple $R$-module.
If $R$ is semi-artinian, then all
irreducible ideals of $R$ are completely irreducible
(see also Lemma 2.4 in Dilworth-Crawley \cite{DC}).
The existence of a reduced
zero-dimensional ring $R$ having no principal maximal
ideals shows that a ring in which all irreducible
ideals are completely irreducible need not be
semi-artinian. In Example \ref{1.7} we give two specific
ways to obtain an  example
of such a ring $R$.
\end{remark}


\begin{example} \label{1.7}

\begin{enumerate}
\item
Let $L$ denote the algebraic closure
of the field $\Q$ of rational numbers and let $D$ be the
integral closure of the ring $\Z$ of integers in the field $L$.
Fix a prime integer $p$ and let $R = D/\sqrt{pD}$. Then $R$ is a
reduced zero-dimensional ring having the property that no
maximal ideal of $R$ is finitely generated.
\item
Let $\Q^{\omega}$ denote the product of countably infinitely
many copies of the field $\Q$ of rational numbers and let
$I$ denote the direct sum ideal of $\Q^{\omega}$. Then
$R = \Q^{\omega}/I$  is a
reduced zero-dimensional ring in which  no
maximal ideal is finitely generated.
\end{enumerate}
\end{example}



\begin{remark}\label{1.8}
 Let $M$ be a
maximal ideal of the ring $R$. By Theorem \ref{1.3},
an irreducible $M$-primary ideal $A$ is
completely irreducible  if and
only if $A \subset  (A:M)$.  If $(R,M)$ is a rank-one
nondiscrete valuation domain and $A = xR$ is a principal
$M$-primary ideal, then $A$ is irreducible, but not
completely irreducible. (Since the value group $G$ of $R$ is
a dense subset of $\mathbb R$, the value of $x$ is the limit
of smaller elements of $G$.)
On the other hand,
if  $\dim R = 0$ and if each primary ideal of
$R$ contains a power of its radical, then
Corollary \ref{1.5} implies that all irreducible
ideals of $R$ are \ci.
\end{remark}








Let $A \subseteq C$ be ideals of the ring $R$. If  $C$ is
completely irreducible,  we
call $C$ a {\it relevant \ci\ divisor} of $A$ if $A$ has
a decomposition as the intersection of \ci\ ideals in which $C$ is
relevant (i.e. it cannot be omitted). If $A$ admits an
irredundant decomposition with \ci\ ideals, then all the ideals in
this decomposition
are relevant. On the other hand, a prime ideal that is not a maximal
ideal has no  relevant \ci\ divisors.  In Proposition \ref{1.10} we
characterize the  relevant \ci\ divisors
of an ideal.


\begin{proposition} \label{1.10}
A \ci\ ideal $C$ containing the
ideal $A$ is  a  relevant \ci\ divisor of $A$ if
and only if the submodule $C/A$ of $R/A$ is not
essential.
\end{proposition}

\begin{proof} First suppose $C$ is a  relevant \ci\ divisor of $A$,
and $A = C \cap \bigcap_{i \in I} C_i$ is a decomposition with
\ci\ ideals  $C_i$, where $C$ cannot be omitted. Then
there exist an element  $x \in \bigcap_{i \in I} C_i \setminus C$
and an $r \in R$ such that $rx \in C^* \setminus C$.
Let $u = rx$ and let $\overline u$ denote the image of $u$
in $R/A$. Then $\overline u \not\in C/A$, while if
$M$ denotes the maximal ideal of $R$ such that $C^*/C \cong R/M$,
then $Mu  \subseteq C \cap \bigcap_{i \in I} C_i =A$
implies $R \overline u$ is a simple submodule of $R/A$.
Hence the submodule $C/A$ of $R/A$ is not essential.

Conversely, assume $C/A$ is not essential in $R/A$, i.e. there exists
$x \in R \setminus A$ such that $(Rx+A) \cap C = A$. Write the
ideal $(Rx+A) $ as an intersection of \ci\ ideals $C_i \ (i \in I)$.
Then $A = C \cap \bigcap_{i \in I} C_i$, where $C$ is relevant.
\end {proof}


\begin{corollary} \label{1.11} For every \ci\ ideal $C$
containing the ideal $A$, the cover $C^\star$ satisfies:
$C^\star/A$ is an essential submodule in $R/A$.
\end{corollary}

\begin{proof}
If $C$ is not a relevant
completely irreducible divisor of $A$,
then $C/A$ is an essential submodule of $R/A$, so $C^*/A$ is
essential in this case. On the other hand,  if
$C$ is relevant, then for any $x \in R \setminus C$, there
exists $r \in R$ such that $rx \in C^* \setminus C$.
\end{proof}


\begin{corollary} \label{1.12}
Let $A$ be a proper ideal of the ring $R$. There
exists a relevant completely irreducible divisor of $A$
if and only if the socle  $\Soc (R/A) \ne 0$.
\end{corollary}

\begin{proof} Assume that $C$ is a relevant completely
irreducible divisor of $A$. If $A = C$, then
$\Soc (R/A) = \Soc (R/C) = C^*/C \ne 0$. If $A \subset C$,
then $A = C \cap B$, where $A \subset B$. Let $b \in B \setminus A$.
Then $b \not\in C$, so there exists $r \in R$ such that
$rb \in C^* \setminus C$. It follows that $Mrb \subseteq C \cap B =A$,
where $M$ is the adjoint prime of $C$. Therefore
$(A + rbR)/A \subseteq \Soc (R/A)$ and $\Soc (R/A) \ne 0$.

Conversely, assume that $\Soc (R/A) \ne 0$. Then there
exists $x \in R \setminus A$ such that $xM \subseteq A$
for some maximal ideal $M$ of $R$ with $A \subseteq M$.
Let $B = A + xR$. Then $B/A \cong R/M$.
Let $C$ be an ideal of $R$ containing $A$ and
maximal with respect to $x \not\in C$. Then
$C$ is completely irreducible and $A = C \cap B$.
Therefore $C$ is a relevant completely irreducible
divisor of $A$.
\end{proof}


\begin{remark} \label{1.13}
The ring $R$ of Example \ref{1.7} is a zero-dimensional
reduced ring for which $\Soc R = 0$. Thus in this
ring, the ideal $(0)$  has no
relevant completely irreducible divisors.
\end{remark}

In Proposition \ref{1.99},
we observe a connection between relevant completely irreducible
divisors of an ideal and
maximal ideals that are strong
Bourbaki associated primes of the ideal.

\begin{proposition} \label{1.99}
Let $A$ be a proper ideal of the ring $R$.
A maximal ideal $M$ of $R$ is a strong Bourbaki
associated prime of $A$ if and only if there
exists a relevant completely irreducible divisor of $A$
that is $M$-primal.
\end{proposition}

\begin{proof}
Suppose $C$ is a relevant completely irreducible divisor of
$A$ that is $M$-primal. The proof of Proposition \ref{1.10}
establishes the existence of an element $u \in R \setminus A$
such that $Mu \subseteq A$. Hence $M = A:u$ is a strong
Bourbaki associated prime of $A$. Conversely, if
$M = A:x$, then $x \not\in A$. Let $C$ be an ideal of
$R$ that contains $A$ and is maximal without $x$.
Then $M = C:x$ and $C$ is $M$-primal. Since
every ideal is an intersection of completely irreducible ideals,
there exist completely irreducible ideals $C_i$
such that $A + Rx = \cap_{i \in I}C_i$. Then  $(A + Rx)/A$ being
simple implies that $A = C \cap (\cap_{i \in I}C_i)$,
where $C$ is clearly a relevant component.
\end{proof}




\section{Irredundant Intersections}


We  consider  under what conditions an
ideal may be represented as an irredundant intersection of \ci\ ideals,
where, as usual, {\it irredundant} means that none of the components may be
omitted without changing the intersection.


If $\{B_i\}_{i \in I}$ is a family of $R$-modules, then by
an {\it interdirect product} of this family we mean an
$R$-submodule of the direct product $\prod_{i \in I}B_i$
that contains the direct sum $\bigoplus_{i \in I}B_i$.
Similarly, if $\{R_i\}_{i \in I}$ is a family of rings, by an
{\it interdirect product} of this family of rings we mean a subring
of the direct product $\prod_{i \in I}R_i$ that contains
the direct sum $\bigoplus_{i \in I}R_i$ and
the identity of the direct product.



In Lemma \ref{2.1} and later in this section, we use the following notation.
Let $\{C_i\}_{i \in I}$ be a family of completely
irreducible ideals of the ring $R$ and let
$A = \bigcap_{i \in I}C_i$.  For each $j \in I$ let
$\overline C_j = \bigcap _{i \in I, i \ne j} C_i $.
We frequently use  the  following basic fact.


\begin{lemma} \label{2.1} Let $A = \bigcap _{i \in I} C_i $
be an irredundant representation  of the proper ideal $A$ with \ci\
ideals $C_i$.


{\rm (i)} There are elements $u_i \in R \setminus C_i \ (i \in I) $ such that
$C^*_i= C_i+ Ru_i$ for $i \in I$ and $u_i \in C_j$ for all
$j \ne i$.


{\rm (ii)} For each $i \in I$, the $R$-module $\overline C_i/A$
has an essential socle generated by $u_i+A$.


{\rm (iii)} In the representation $A = \bigcap_{i \in I}C_i$,
no  $C_i$ can be replaced by a
larger ideal and still have the  intersection be equal to $A$.


{\rm (iv)} With $u_i$ as in (i), let $U_i = (Ru_i + A)/A$.
Then $U_i$ is a simple $R$-module and the socle of $R/A$
is an interdirect product   of the $U_i$.
\end{lemma}


\begin{proof} (i) Since the intersection of the $C_i \ (i \in I)$ is
irredundant, for each $i \in I$ there exists an
element $x_i \in \overline{C}_i$ such that
$x_i \notin C_i$. By Lemma \ref{1.1},
$R/C_i$ is subdirectly
irreducible, so some multiple $rx_i \notin C_i$ has the property
that its image in $R/C_i$ is in $\Soc~(R/C_i)$.
If we choose $u_i = rx_i$ for each $i \in I$, then we
obtain elements with the desired properties.


(ii) From (i) it is clear that $\langle u_i + A \rangle$ is in
the socle of $\overline C_i/A$. By way of contradiction, suppose that
there is a cyclic $R$-module $\langle v + A \rangle$ contained in
$\overline C_i/A$ and independent of  $\langle u_i + A \rangle$.
As  $C_i$ is \ci, there is an $r \in R$ such that $rv -u_i \in C_i$.
Then $rv - u_i \in C_i \cap \overline C_i = A$, contradicting the
independence of $\langle v + A \rangle$ and $\langle u_i + A \rangle$.



(iii) If we replace $C_i$ by a larger ideal, then the intersection
will contain $u_i$, so it will no longer represent $A$.


(iv) As $M_iu_i \subseteq A$ for a maximal ideal $M_i$, it is
clear that the
$U_i = \langle u_i +A \rangle$ are simple submodules of  $R/A$.
  From (i) it follows that they are independent, so Soc~$(R/A)$
contains their direct sum $D = \oplus_{i \in I}U_i$.
Suppose there is a $v \in R$ such
   that $v + A \in $ Soc~$(R/A)$, but not in $D$, say $Mv \subseteq A$
for a maximal ideal $M$ of $R$. Thus $v \notin C_j$
for some $j \in I$. If $v \notin C_j+Ru_j$, then $Mv \subseteq C_j$
implies $C_j = (C_j+ Ru_j) \cap (C_j+ Rv)$,
                                   contradicting the
irreducibility of $C_j$. Hence $v \in C_j+Ru_j$ for all $j$
with $v \notin C_j$. This means that the canonical injection
$R/A \to \prod_{j \in I}R/C_j$ maps $v$ to an element
of $\prod_{j \in I}(Ru_j + C_j)/C_j \cong \prod_{j \in I}U_j$.
\end{proof}


We remark that part (iii) of Lemma \ref{2.1}
is also a consequence of the stronger statement: if
$A = B \cap C$, where $C$ is irreducible and relevant for the decomposition
of $A$, then $C$ cannot be replaced by any proper overideal (see
Noether \cite[Hilfssatz II]{N}). In fact, if $C$ is properly
contained in $C'$, then
$A= B \cap C'$ would lead to $C = C + (B \cap C') = (C + B) \cap C'$,
   contradicting the irreducibility of $C$.


\begin{remark}\label{2.2} It will be useful to keep in mind that
  (iv) of Lemma \ref{2.1} implies that
  $$  \bigoplus_{i \in I} \Soc (\overline C_i/A) \subseteq \Soc~(R/A)
\subseteq  \prod_{i \in I} \Soc (\overline C_i/A). \leqno (1)$$
Notice also that  $\Soc (\overline C_i/A) \cong \Soc (C_i
+\overline C_i)/C_i= \Soc~(R/C_i) = C_i^*/C_i$. \end{remark}


Next we exhibit an example where $\Soc (R/A) $ is the direct product
of the $\Soc (\overline C_i/A).$


\begin{example}  \label{2.3} For each nonnegative integer $n$,
let $Z_n = \langle x_n \rangle$ be a cyclic group of
order $p$, where $p$ is a fixed prime integer. Consider
the product $\prod_{n < \w} Z_n$ with trivial multiplication
and let $R$ be  the ring
obtained by adjoining an identity to $\prod_{n < \w} Z_n$
in such a way that as an additive group $R = \Z \oplus
\prod_{n < \w}Z_n$ and multiplication is defined as
$(a,b)\times (c,d) = (ac, ad + cb)$, where
$a, c \in \Z$ and $b, d \in \prod_{n < \w}Z_n$.


For each nonnegative integer $m$, let
$C_m \cong  \prod_{n < \w, n \ne m} Z_n$ denote the
subgroup of $\prod_{n < \w} Z_n$ of elements having zero coordinate
in component $Z_m$. Then $C_m$ is an ideal of $R$ that is
maximal with respect to not  containing the element that is
$x_m$ in component $Z_m$ and zero in all other components.
Thus $C_m$ is \ci. We have
$\bigcap_{n < \w} C_n = 0$ is an irredundant intersection, and in
this example, the first containment relation in (1) is strict,
while the second becomes equality.
\end {example}


For an irredundant intersection $A = \bigcap_{i \in I}C_i$
with \ci\ ideals $C_i$, we demonstrate in Example \ref{2.4}
that  $\Soc~(R/A)$ may fail to be an essential
submodule of $R/A$.


\begin{example}  \label{2.4} There is a ring $R$ in which the zero
  ideal is an irredundant intersection of \ci\ ideals and $\Soc R$ is
  not an essential submodule of
$R$. Let $\{p_n \ | \ n < \w\}$ be the set of prime numbers, and let
$Z_n$ denote the ring $\vZ/p_n^2\vZ \ (n < \w).$ In the direct product
$P = \prod_{n < \w} Z_n$ consider the subring $R$ generated by
$1=(1,1, \dots, 1, \dots), e = (p_0, p_1, \dots, p_n, \dots)$, and
$r_n= (0, \dots, 0,p_n, 0, \dots)$ for each $n < \w$. Then
$e^2=0, er_n =0$ and $r_n r_m=0$ for all $n,m < \w$. Consequently,
a typical element of $R$ has the form $a = m \cdot 1+ se + n_0r_0 +
\dots + n_kr_k$ for some $k \geq 0$, where $m,s, n_i \in \vZ$. The ideal
$Re$ is just the infinite cyclic subgroup generated by $e$.
Define $C_n$ as an ideal containing all $r_i \ (i< \w, i \ne n)$,
  $r_n -  p_0^n p_1^n \cdots p_n^n e$,   and being maximal with respect to
$r_n \notin C_n$.   Clearly, the intersection $\bigcap_{n < \w} C_n$
  contains  no nonzero element of the socle
$\Soc R = \oplus_{n < \w}Rr_n$, and
is evidently irredundant.  To see that it cannot  contain any  element
  outside the socle either, assume by way of contradiction that $a$
as above is contained in every $C_n$. Then either a multiple of $se$ or $ae=
me$ is contained in every $C_n$, which is  impossible, since
$p_0^n p_1^n \cdots p_n^n e \notin C_n.$  However, the socle of $R$ is not
essential in $R$, as the ideal $Re$ intersects $\Soc R$ trivially.
\end {example}


Let $A$ be a proper ideal of the ring $R$.  In Theorem \ref{2.5} we
relate representations of $A$ as an intersection of bigger ideals
with representations of $R/A$ as a subdirect product.





\begin{theorem} \label{2.5} For a proper  ideal $A$ of the ring $R$
the following are equivalent:


{\rm (i)} $A$ is the irredundant intersection of  \ci\ ideals;


{\rm (ii)}  $R/A$ is an irredundant subdirect product  of subdirectly
irreducible $R$-algebras  $R_i$.
\end{theorem}


\begin{proof} (i) $\Rightarrow$ (ii) Let $\{C_i\}_{i \in I}$ be
a family of completely irreducible ideals of $R$ such that
$A = \bigcap_{i \in I}C_i$ is an irredundant intersection.
Then  $R/A$ can be identified via
the canonical map  $\phi: R \to \prod _{i \in I} R/C_i$
with the subring $\phi (R)$ of the direct product
$\prod _{i \in I} R/C_i$, where each component $R/C_i$ contains
   a simple $R$-module $C_i^*/C_i$ as an essential socle, so
each component is
subdirectly irreducible. If we drop a component $R/C_j$, then
  $ \f$ will no longer induce an embedding of $R/A$ in
  the product
$\prod_{i \in I, i \ne j}R/C_i$, so the subdirect representation
is irredundant.



(ii) $\Rightarrow$ (i)  Assume that  $R/A $ is an irredundant
subdirect product  of the subdirectly irreducible $R$-algebras
$R_i \ (i \in I)$. Let
$T = \prod_{i \in I}R_i$, and let $\pi_i: R/A \to R_i$ be the $i$th canonical projection.
The  kernel of $\pi_i$  is $C_i/A$,  where  $C_i$ is a completely irreducible ideal of $R$
properly containing $A$, and  $R_i$ is isomorphic to $R/C_i$. Clearly,
$A = \bigcap_{i \in I}C_i$ and this representation is irredundant.
\end{proof}


\begin{example} \label{2.6}  Let $F_p$ denote the prime
field of characteristic
$p$, for a prime  $p$. Define $R$ as the direct product $\prod_p
F_p$ with $p$ running over an infinite set of primes. Identify the
product $C_q$ of the $F_p$ with $p
\ne q$ ($q$ a fixed prime)  as a subset  of $R$ consisting
of all tuples having a zero in position $q$. Under this
identification, $C_q$  is a \ci\ ideal of $R$, and $0 = \bigcap _{q }
C_q $ is an irredundant intersection.
(In this example, $\Soc R$ is an essential submodule of $R$.) \end{example}



In the preceding example, we could have used equally well several subrings
(containing $ \oplus_p F_p$) of $R$, e.g. the von Neumann
regular ring over which all  von Neumann  regular rings  are
unital algebras; see  Fuchs-Halperin \cite{FH}.
\medskip



Another criterion for the existence of an irredundant intersection
representation of an ideal $A$ with \ci\ ideals can be given in
terms of the injective hull of $R/A$.


\begin{theorem}\label{2.7} The proper ideal $A$ of the ring $R$
admits an irredundant representation as an intersection of \ci\
ideals $C_i \ (i \in I)$ if and only if the injective hull $E(R/A)$
of $R/A$ is an interdirect product  of injective hulls of simple $R$-modules.
\end{theorem}


\begin{proof} The `only if' part of the theorem is an immediate
consequence of Theorem \ref{2.5}, since the components $R/C_i$
in the subdirect product  are
$R$-modules each with a simple essential submodule.


Conversely, assume $E(R/A)$ is an interdirect product  of injective
hulls $E(S_i)$ of simple $R$-modules $S_i$.
Define $R_i = \pi_i(R/A)$, where $\pi_i$ denotes the
$i$th coordinate projection of the product of the $E(S_i)$. Then
$R/A$ is evidently a subdirect product of the $R_i$, and the
$R_i$ are subdirectly irreducible. For each $i$,
the submodule $E(S_i)$ of $E(R/A)$ has nonzero intersection with
$R/A$, and this intersection is contained in the kernel of $\pi_j$
for each $j \ne i$. This shows that the representation is
irredundant.
By  Theorem \ref{2.5}, $A$ admits an irredundant
representation as an intersection of  \ci\ ideals.
\end{proof}



For a proper ideal $A$ of $R$, let  $S(A)$ denote the
set of elements $r \in R$ that are not prime to $A$, i.e.
that satisfy $A  \subset A:r$.



\begin{proposition} \label{proposition 2.8} Let $A$ be a
proper ideal of a ring $R$.
If $A$ is an irredundant intersection of \ci\ ideals $C_i$, then
   $S(A)$ is the union of maximal ideals of $R$, each  of
   which is a strong Bourbaki associated prime of $A$.
\end{proposition}


\begin{proof}
Given $r \in R$, we have $r \in S(A) \iff A \subset A:r$. Since
$A:r = \bigcap_{i \in I}(C_i:r)$, we have, by Lemma~\ref{2.1}(iii),
$r \in S(A) \iff C_i \subset C_i:r$ for some $i$. Letting
$M_i = \{t \in R \,|\, C_i \subset C_i:t \}$, the adjoint prime
of $C_i$, we see that $S(A) = \bigcup_{i \in I}M_i$.

The elements $u_i \in R$ introduced in
   Lemma~\ref{2.1} satisfy $M_i = C_i:u_i$ and
$u_i \in C_j$ for $j \ne i$. Thus $A: u_i = M_i$, showing that
$M_i$ is a strong Bourbaki associated prime of the ideal $A$.
\end{proof}


We can now prove:


\begin{corollary} \label{2.9} Let $A= \bigcap_{i \in I}\ C_i$ be an
irredundant intersection, where the $C_i$ are  \ci\ ideals.
Then $A$ is an $M$-primal
ideal if and only if all the $C_i$ are $M$-primal.
\end{corollary}


\begin{proof} If all the \ci\ ideals $C_i$ are $M$-primal, then by
Proposition \ref{proposition 2.8} $S(A)= M$, so $A$ is $M$-primal.
On the other hand, if
$A$ is $M$-primal, i.e. if  $S(A)=M$, then again by Proposition \ref{proposition 2.8},
all the adjoint primes of the $C_i$ must be equal to $M$.
\end{proof}


This leads to a coarser intersection decomposition of ideals admitting
irredundant intersections with \ci\ ideals.

\begin{corollary}\label{2.10}
Assume that the   proper  ideal $A$ of $R$ has an irredundant
representation as an intersection of \ci\ ideals.  Then $A$  is an
irredundant intersection of primal ideals with distinct adjoint maximal ideals
such that each primal component is an irredundant intersection of \ci\ ideals.
\end{corollary}
\begin{proof}
Let $A= \bigcap_{i \in I}\ C_i$ be an
irredundant intersection, where the $C_i$ are  \ci\ ideals.
For each maximal ideal $M$ that is the adjoint prime
of some $C_i$, , let $A^{[M]}$ denote the intersection
of the $C_i$ that are $M$-primal. By Corollary \ref{2.9},
$A^{[M]}$ is $M$-primal.
Thus $A =\bigcap_{M \in
\Max R}\ A^{[M]}$ is a decomposition of $A$ into
an irredundant intersection of primal ideals with
distinct adjoint primes $M$, and each $A^{[M]}$ is an irredundant
intersection of \ci\ ideals.
\end{proof}

  Since the isolated $M$-component $A_{(M)}$ of an ideal $A$ is the
intersection of all $M$-primal overideals of $A$, it is clear that
$A_{(M)} \subseteq A^{[M]}$ for each  $M$.

It would be interesting to
know if the converse of Corollary \ref{2.10} is also true.


\begin{example} \label{2.11} It is possible that an ideal $A$ is an irredundant
intersection of \ci\ ideals $\{C_i\}_{i \in I}$
and  also the intersection of relevant \ci\
divisors $\{C'_i\}_{i \in I'}$, where no subset of $\{C'_i\}_{i \in I'}$
gives an
irredundant representation of $A$.  Let $R$ be the ring defined in Example
\ref{2.4}. It has been proved there that its zero ideal admits a representation
as  an irredundant intersection of \ci\ ideals. Define  $A_n \ ( n>0) $ to be  an
ideal containing all $r_i \ (i< \w, i \ne n)$, containing $ e$, and maximal with respect
to not containing
$r_n$. Furthermore, for each $ n < \w$ let  $B_n \ ( n < \w)$ be an ideal that
contains all $r_i$ with $i > 0$, contains the element $r_0 - p_0^n p_1^n
\cdots p_n^n e$,
  and is maximal with respect to not containing  $r_0$. In view of Proposition \ref{1.10}, all of $A_n,
B_n$ are relevant \ci\ divisors of $A$. Evidently, $A = \bigcap_{n >0}\ A_n
\cap \bigcap _{n < \w} B_n$. Here each of the $A_n$, but none of
$B_n$, is relevant for the intersection. Indeed, infinitely many $B_n$ can be
deleted from this intersection as long as infinitely many $B_n$ remain. In view
of this, this intersection cannot be made irredundant by canceling components.
\end{example}


\section{The Question of Uniqueness  }

 

In her seminal paper \cite{N} E. Noether proved that in a Noetherian ring
every proper ideal $A$ is the finite irredundant intersection of irreducible
ideals, and this intersection has the following uniqueness
properties: (1) the number of irreducible components is unique, (2)
the components satisfy  the  Replacement  Property as defined in (\ref{3.2}), and
(3) the components are primary ideals and  their  set of prime
radicals (along with their multiplicities) is uniquely determined by $A$.


Our next goal is to prove, if possible, analogous results for ideals $A$ that
admit a representation as an irredundant intersection of \ci\ ideals.
Of course,
the role of primary ideals will be taken over by primal ideals, and the set
of prime radicals will be replaced by the set of adjoint maximal ideals.


  In our attempt to  generalize  the Noetherian situation
the first problem we face is that the
cardinality of the set of \ci\ components in an irredundant intersection need
not be unique.  This is illustrated by the following example.


\begin{example} \label{3.1}  Let $\k$ be an infinite cardinal and $\l$ any
cardinal such that $\k < \l \le 2^\k$. There exists a ring $R$ that contains an
ideal which is the irredundant intersection of $\k$ \ci\ ideals and also
the irredundant intersection of $\l$ \ci\ ideals.
Fix a prime integer $p$ and let
$P = \prod_{\a < \k} Z_\a$ be the direct product of
of $\k$ cyclic groups $Z_\a$
of order $p$.
Give $P$ the trivial multiplication and make $R = \Z \oplus P$ into a ring
as in Example \ref{2.3}.
Evidently $P = \Soc R$ and $|P| = 2^\k$.
For $\b <\k$,  the product $P_\b =  \prod_{\a < \k, \a \ne \b} Z_\a$
may be viewed
  as an ideal in $R$; it is \ci, since it is maximal with respect to
intersecting  $Z_\b$ in zero.
Clearly, $0 = \bigcap_{\b < \k} P_\b$ is an irredundant intersection
with $\k$ components. Pick a subgroup $P'$ of $P$ that contains the
direct sum $\oplus_{\a < \k} Z_\a$ and has index $\l$ in $P$. Now
$P = P' \oplus \bigoplus_{\g <\l} W_\g$ for cyclic groups $W_\g$
of order $p$. Consider \ci\ ideals $C_\a$ that contain $\bigoplus_{\g <\l}
W_\g$ and are maximal with respect to intersecting $Z_\a$ in zero,
and consider as well  \ci\ ideals $C_\g'$
that contain $P'$ and are maximal with respect to intersecting  $W_\g$ in zero. Then the
intersection $0 = \bigcap_{\a < \k} C_\a \cap \bigcap_{\g < \l} C_\g'$
is irredundant with $\l$ components.
\end{example}


As far as uniqueness of irredundant decompositions is concerned,
we have the following:


\begin{theorem}\label{3.2} Let
$$A= \bigcap _{i \in I} C_i =  \bigcap _{j \in J} B_j$$
be two irredundant intersection representations of the proper
ideal $A$ of the ring $R$ with \ci\ ideals $C_i, B_j$. Then:


{\rm (i)} Each $M \in \Max R$ that occurs as the adjoint prime of some
primal ideal  $C_i$ also occurs as the adjoint prime of some
  $B_j$. If it occurs a finite number of times in one intersection,
  then it occurs the same number of times in the other intersection.


{\rm (ii)} The intersections have the Replacement Property: for each
$C_k$ there is a $B_j$ such that replacing $C_k$
  by $B_j$ in the first intersection, we obtain an
irredundant representation  of $A$.
\end{theorem}


\begin{proof} We start the proof with (ii). (Note that in the
first paragraph  of its proof irredundancy is irrelevant.)


(ii)  Write $\overline C_k = \bigcap_{i \ne k} C_i$ and
$A_{kj} = \overline  C_k \cap B_j$. The $R$-module
$(C_k + \overline  C_k)/C_k \cong \overline C_k/(C_k \cap \overline C_k)
\cong \overline  C_k / A$ has an
essential simple socle. Therefore, $A = \bigcap_{j \in J}
A_{kj} $ with $A_{kj} \subseteq \overline C_k$ implies that one
of $A_{kj} \ (j \in J)$ must be equal to $A$, say, $\overline  C_k
\cap B_j =A$. Thus $C_k$ can be replaced by $B_j$.


To show that if $A = \bigcap _{i \in I} C_i $ is irredundant, then
so is the new intersection, assume by way of contradiction that
$C_\ell\ (\ell \ne k)$ can be omitted from the new intersection, i.e.
   $A= \overline  C_k  \cap \overline  C_\ell \cap B_j$ (where $C_k$ and
$C_\ell$ are missing). What
has been proved is applied to this decomposition and the
original one to conclude that here
$B_j$ can be replaced by one of the $C_i$. After we do this, at least
one of $C_k$ and $C_\ell$   would be missing from the intersection,
contradicting the irredundancy of the decomposition
$A = \bigcap _{i \in I} C_i$.


(i) Both $C_k$ and its replacement $B_j$ satisfy $\overline C_k \cap
C_k = A = \overline C_k \cap B_j$. Hence the covers satisfy
$C_k^*/C_k \cong \Soc (\overline C_k / A) \cong B_j^*/B_j$. It follows
that both $C_k$ and $B_j$ are $M$-primal for the maximal ideal $M$
that satisfies $\Soc (\overline C_k / A) \cong R/M$. Thus an $M$-primal
\ci\ ideal is always replaced by another $M$-primal ideal in an
irredundant intersection.


Now if the first intersection contains $n$ \ci\ ideals that are $M$-primal,
then successively replacing them by (necessarily distinct) \ci\
ideals from the second intersection, it follows that the second
intersection contains at least $n$ $M$-primal ideals. Hence the claim
is evident.
\end{proof}


Dilworth-Crawley \cite[Theorem 4.2]{DC} prove an analogue of (ii)
for compactly generated modular lattices under the hypothesis that
the lattice is atomic (in our setting, this corresponds to the case
where $\Soc~(R/A)$ is essential in $R/A$). They also characterize
atomic lattices where the components in irredundant intersections are
unique. Since we do not have atomicity in general (see Example \ref{2.4}),
in our search for uniqueness an independent approach is needed.

\begin{remark}  In \cite{HO1} the rings are characterized for which every
ideal can be represented uniquely as an irredundant intersection of completely
irreducible ideals.  These rings are necessarily arithmetical.  Indeed,
it is shown in  \cite{HO1} that
 if $\A$ is the set of ideals of
a ring $R$ that are finite intersections of completely irreducible ideals, then
$R$ is arithmetical if and only if for each $A \in \A$ the representation of
$A$ as an irredundant intersection of completely irreducible ideals is unique.
\end{remark}




\section{When Every Ideal is an Irredundant Intersection }


In this section we characterize the rings in which every ideal is
an irredundant intersection of \ci\ ideals.
We start with the following
observation.


\begin{lemma} \label{5.1}
Consider the following statements about a
proper ideal $A$ of the ring $R$.
\begin{itemize}
\item[(i)]     $A$ is an irredundant intersection of completely
irreducible ideals.
\item[(ii)]  Soc $(R/A)$ is an essential submodule of $R/A$.
\end{itemize}
Then (ii) implies (i). On the other hand,
if (i) holds for every
proper ideal of $R$, then also (ii) holds for
every proper ideal of $R$.
\end{lemma}


\begin{proof}
Assume that (ii) holds and  let
$E(R/A)$ denote
the injective hull of $R/A$. Express  $\Soc~(R/A)$ as the direct sum
$\oplus_{i \in I}S_i$ of simple $R$-modules $S_i$, and  let $E(S_i)$
be a maximal essential extension of $S_i$ in $E(R/A)$. Then
$E(S_i)$ is an injective hull of $S_i$ and $\oplus_{i \in I}E(S_i)$
is an essential submodule of $E(R/A)$. The canonical map
of the direct sum $\oplus_{i \in I}E(S_i)$ into the direct
product $\prod_{i \in I}E(S_i)$ extends to an injection of
$E(R/A)$ into $\prod_{i \in I}E(S_i)$. Thus $E(R/A)$
is an interdirect sum of the $E(S_i)$. By Theorem \ref{2.7},
$A$ is an irredundant intersection of completely irreducible ideals.



Assume that (i) holds for every proper ideal of $R$ and let
$A$ be a proper ideal of $R$.
By Lemma \ref{2.1}, $\Soc~(R/A)$
cannot be trivial. Let $B/A$ be a maximal essential extension of
$\Soc~(R/A)$ in $R/A$. Then $R/B$ has trivial socle, so $B$ cannot
admit an irredundant representation as intersection of \ci\ ideals.
It follows that $B=R$. This means that $\Soc~(R/A)$
   is an essential submodule in $R/A$.
\end{proof}


Recall that  a ring $R$ is semi-artinian if every nonzero
$R$-module contains a simple $R$-module.
 We will call
a domain {\it almost semi-artinian} if every proper factor ring is
semi-artinian.



\begin{theorem} \label{5.2} Every ideal of a ring  is
an irredundant intersection of \ci\ ideals exactly if the ring is
semi-artinian.
Every nonzero
ideal  of  an integral  domain is  an
irredundant intersection of \ci\ ideals if and only if the
integral domain  is  almost semi-artinian.
\end{theorem}


\begin{proof} Lemma \ref{5.1}   implies that  all the
ideals of a ring $R$ have the  stated property if and only if
$\Soc~(R/A)$ is essential in $R/A$ for all proper ideals $A$ of
$R$. This is the case exactly if $R$ is semi-artinian.


The claim concerning domains is an immediate consequence of
the first part.
\end{proof}


\begin{example} A ring that is perfect in the sense of Bass
\cite{Ba} is semi-artinian. Perfect rings can be defined in several
equivalent ways; for example, the ring $R$ is
perfect is equivalent to each of the following statements:
(i)  every flat $R$-module is projective, (ii)
$R$ satisfies the minimum condition for principal ideals
(see e.g. Anderson-Fuller \cite[p. 315 ]{AF}).
\end {example}


\begin{example} Consider the B\'ezout domain $R$ of  dimension
   1 constructed by
Heinzer-Ohm \cite[Example 2.2]{HO}. For this ring $R$, $\Spec R$ is
not Noetherian,   but  all of its localizations
at maximal ideals are rank-one discrete valuation domains.
Moreover, all but one of its maximal ideals are  principal.
It can be seen that this example is almost semi-artinian.  Thus
an almost semi-artinian domain need not have Noetherian spectrum.
\end {example}


More interesting examples are those almost semi-artinian
domains that are also $h$-local, where an integral domain
is {\it h-local} if each nonzero element is contained
in only finitely many maximal ideals and each nonzero
prime ideal is contained in a unique maximal ideal.


\begin{example}  The {\it almost perfect domains} introduced by
Bazzoni-Salce \cite{BS}  are defined as domains such that
every proper factor ring is a perfect ring.
   They prove that an almost perfect domain can be
characterized as an $h$-local domain $R$ such that for every
nonzero proper ideal $A$,   the factor ring $R/A$ contains a simple
   $R$-module.
We are interested in obtaining  more information about the ideal structure of
almost perfect domains.
An almost perfect domain is a Matlis domain of dimension $\le 1$
with Noetherian prime spectra,
where an integral domain $R$ is a {\it Matlis domain} if its
field of fractions has projective dimension at most 1 as an
$R$-module. The inclusion here is strict since a rank-one
nondiscrete valuation domain is a Matlis domain of dimension 1 that
fails to be almost perfect.
\end{example}


   In an almost perfect domain,  irreducible ideals are primary.
     Indeed, by \cite{F1} a
necessary and sufficient condition for an irreducible ideal
$A$ of a ring $R$  to be primary is that
any strictly ascending chain of the form
$$A \subset  A:b_{1} \subset  A: b_{1}b_{2} \subset \dots
\subset A: b_{1}b_{2}\cdots b_{n} \subset  \dots $$
   for any sequence $b_{1}, b_{2},
\dots, b_{n}, \dots $ in $R$ terminates. This is evidently true
if the descending chain condition holds for principal ideals mod $A$.


Since an almost perfect domain $R$ is $h$-local, every nonzero ideal
$A$ of $R$ is contained in at most a finite number of maximal ideals,
$M_{1}, \dots, M_{n}$. Consequently, we can write
$$A = A_{(M_{1})} \cap \dots \cap A_{(M_{n})}.$$
The localization at a maximal ideal of a domain of  dimension one is
again of dimension one,  and if $R$
is a one-dimensional quasilocal domain with maximal ideal $M$,
then every nonzero proper
ideal of $R$ is $M$-primary.  Consequently,
the isolated $M_i$-component $A_{(M_{i})}$ must be $M_i$-primary
for each $i = 1, \ldots, n$.

We record as Theorem \ref{5.6} an immediate consequence
of \cite[Theorem 8]{BH}.

\begin{theorem} \label {5.6} In an almost perfect domain $R$, every
nonzero ideal $A$ is the product of pairwise comaximal primary
ideals:
$$ A = B_{1} B_{2}\cdots B_{n},$$
where the $B_{i}$ are primary ideals with distinct maximal
ideals $M_{i}$ as radicals.
This product representation is unique up to the order of the factors.
Since an almost perfect domain is almost semi-artinian,
 the $B_i$ are irredundant
intersections of completely irreducible $M_i$-primary ideals.
\qedsymbol
\end{theorem}

   In particular, it follows that an almost perfect domain is Laskerian.
We recall that a ring is said to be {\it strongly Laskerian}
if it is Laskerian and every primary ideal contains a power of its
radical. If $A$ is a nonzero ideal of a one-dimensional
strongly Laskerian domain $R$, then
$R/A$ satisfies the minimum condition for principal ideals, for
$R/A \cong \prod_{i=1}^nR_i$, where each $R_i$ is a zero-dimensional
ring with nilpotent maximal ideal $M_i$. If $r_i$ is  a positive
integer such that $M_i^{r_i} = 0$, then a strictly descending chain of
principal ideals in $R_i$ has length at most $r_i$. It follows that
a strictly descending chain of principal ideals in $R/A$ has length
at most $r_1 + \cdots + r_n$.
Thus $R$ is almost perfect.

\begin{remark} \label{5.7} The representations  of the ideals $B_i$ of
Theorem \ref{5.6} as irredundant intersections of completely irreducible
$M_i$-primary ideals may be infinite intersections. For example,
if $R$ with maximal ideal $M$ is a one-dimensional local strongly Laskerian
domain that is not Noetherian, then $M/M^2$ is infinite dimensional as a
vector space over $R/M$. Since $R$ is almost semi-artinian, every
nonzero ideal of $R$ is an irredundant intersection of completely
irreducible ideals. However, every representation of $M^2$ as an
intersection of completely irreducible ideals is an infinite intersection
(see for example part (iv) of Lemma \ref{2.1}).
\end{remark}


\section{Arithmetical Rings}


Proposition \ref{1.4} and Corollary \ref{1.5} characterize the completely
irreducible ideals of a Noetherian ring.  In this section we give an
explicit description of the completely irreducible ideals in a much
different setting, that of an arithmetical ring.  These rings arise naturally
in the consideration of irreducible ideals.  In Theorem 1.8 and Remark 1.6 of
\cite{FHO} it is shown that the following are equivalent:
(i) the  ring $R$ is arithmetical, (ii)
every primal ideal of $R$ is irreducible, (iii) the ideal
$A_{(P)}$ is irreducible for every
ideal $A$ and prime ideal $P$ of $R$ with $A \subseteq P$.

We first consider the special case of a valuation ring,  where
by a {\it valuation ring} we mean a ring in which the ideals are
linearly ordered with respect to inclusion.




   \begin{lemma} \label{4.1} Let $M$ be a maximal ideal of
the ring $R$. The following statements are equivalent.
\begin{itemize}
\item[(i)]  $R$ is a valuation ring with maximal ideal $M$.
\item[(ii)]  The set of proper completely irreducible ideals of $R$ is precisely the
set of ideals of the form $rM$, where $r$ is a nonzero
element of $R$.
\item[(iii)] For every nonzero $r \in R$, the ideal $rM$ is irreducible.
\end{itemize}
\end{lemma}



\begin{proof} (i) $\Rightarrow$ (ii)  Let $C$ be a completely
irreducible ideal of $R$. Since $C^*/C$ is a simple
$R$-module and $R$ is a valuation ring, $C^*$ must be a
principal ideal of $R$ and $C = MC^*$.   On the other hand,
every  ideal of
the form $rM$, $r \in R$ and $r$ nonzero, is completely
irreducible by Theorem~\ref{1.3}(v) and the fact that every
ideal of a valuation ring is irreducible.


(ii) $\Rightarrow$ (iii) This is clear.


(iii) $\Rightarrow$ (i) We first observe that (iii) implies $R$
is quasilocal. For if $r$
is not in $M$ but is  in some other maximal ideal of $R$,
then $rM = M \cap rR$ is an irredundant intersection,
contradicting the fact that $rM$ is irreducible.

Let $x$ and $y$ be nonzero
elements of $R$.  We show that the ideals $xR$ and $yR$ are
comparable. Notice that  $M = xM:x$ and $M = yM:y$.
The ideals $xM$ and $yM$ are irreducible, so by Theorem \ref{1.3}(v)
$xM$ and $yM$ are completely
irreducible. Thus it follows that $xR$ and $yR$ are the respective covers (unique minimal
overideals) of $xM$ and $yM$.
Let $I = (x,y)R$  and let    $A$
be  an ideal of $R$ with $MI = Mx + My \subset A$. Since $xR$ is
the cover of $xM$ and $yR$ is the cover of $yM$,   we must
have $x \in A$ and $y \in A$. Also   $MI \subset I$.
Therefore $I$ is the unique minimal overideal of $MI$, and
$MI$ is completely irreducible with cover $I$.
In particular, $MI$ is irreducible, so $I/MI$ has dimension one as
an $R/M$-vector space. Thus $I = xR + MI$ or $I = yR + MI$.
Since $R$ is quasilocal,  it follows that $I = xR$ or $I =
yR$, proving that $xR$ and $yR$ are comparable.    \end{proof}


\begin{lemma} \label{factor}   Let $A$ be an ideal of a ring $R$.  If $M$ is a
maximal ideal of $R$, then $MA_{(M)} = (MA)_{(M)}$.   \end{lemma}

\begin{proof}  Note first that $MA_{(M)} \subseteq (MA)_{(M)}$.  For if $m \in
M$ and $x \in A_{(M)}$, then there exists $y \in R \setminus M$ such that $xy
\in A$, so that $mxy \in MA$ and $mx \in (MA)_{(M)}$.
To complete the proof we verify locally that
$(MA)_{(M)}=  MA_{(M)}$.  Clearly $(MA)_{(M)}R_M = MAR_M = MA_{(M)}R_M$, so
suppose $N$ is a maximal ideal of $R$ distinct from $M$.  Then $MA_{(M)}
\subseteq (MA)_{(M)}$ implies that $A_{(M)}R_N = MA_{(M)}R_N \subseteq
(MA)_{(M)}R_N \subseteq A_{(M)}R_N$; hence $MA_{(M)}R_N = (MA)_{(M)}R_N$.
\end{proof}



\begin{theorem} \label{4.2} A ring $R$ is arithmetical if and only if the
completely irreducible proper ideals of $R$ are precisely the ideals of the form
$MB_{(M)}$,  where $M$ is a maximal ideal and $B$ is a principal ideal
having the property that $BR_{M} \ne 0$.
\end{theorem}


\begin{proof}
Suppose that $R$ is an arithmetical ring.  Let $C$ be a
completely irreducible proper
ideal of $R$  and let $M$ be the adjoint (maximal) ideal of $C$.  By
Theorem~\ref{1.3},
$CR_M$ is completely irreducible, and by Lemma~\ref{4.1},
$CR_M =  (r/1)MR_{M}$, where $r \in R$ is such that  its image $r/1$ in
$R_M$ is nonzero.  Define $B = rR$.  Since for every ideal $A$ of $R$,
the ideal $A_{(M)}$ is the preimage of $AR_M$ under the canonical mapping $R \rightarrow
R_M$, and  since $C = C_{(M)}$, we have  $C = (MB)_{(M)}$.  Therefore by
Lemma~\ref{factor} $C = MB_{(M)}$, and every completely irreducible ideal
of $R$ has the stated form. By Lemmas~\ref{4.1} and~\ref{factor} and Theorem
\ref{1.3} (vi), every ideal of the stated form is completely irreducible.


Conversely, assume that the completely irreducible proper ideals of $R$
are precisely the ideals of the form
$MB_{(M)}$,  where $M$ is a maximal ideal and $B$ is a principal ideal
of $R$ having the property that $BR_{M} \ne 0$. It suffices to prove that
$R_M$ is a valuation ring for every maximal ideal  $M$
of $R$.  By Lemma~\ref{4.1} it suffices
to show that $(r/1)MR_M$ is irreducible for all elements $r \in R$
such that $r/1 \in R_M$ is nonzero. By assumption,  $M(rR)_{(M)}$ is an
irreducible ideal of $R$, and this implies that  $(r/1)MR_M$ is an irreducible
ideal of $R_M$ (Remark 1.6 of \cite{FHO}). \end{proof}


An ideal $A$ of a domain $R$ is said to be {\it archimedean}
if its only endomorphisms are multiplications by elements
of $R$, i.e. if its endomorphism ring,  $\End(A)$, is equal to $R$ \cite[page~71]{FS}.
We observe in Remark \ref{4.3} that completely irreducible ideals
of a  Pr\"ufer domain need not be archimedean.

\begin{remark} \label{4.3}
Let $R = \mathbb Z + x\mathbb Q[[x]]$ be the ring of power series in $x$ with
rational coefficients whose constant term is an integer. It is known
that $R$ is a Pr\"ufer domain, $\dim R = 2$ and $P = x\mathbb Q[[x]]$ is 
the unique nonmaximal nonzero prime ideal of $R$. Moreover $P$ is
contained in each maximal
ideal of $R$, and for $p \in \mathbb Z$ a prime
integer, $pR$ is a maximal ideal of $R$. Let $V$ denote
the valuation overring $R_{2R} = \mathbb Z_{2\mathbb Z} + P$, 
and consider the principal ideal $A = xV$ of $V$. Since $x \in P$,
$A$ is also an ideal of $R$ and is completely irreducible as an
ideal of $R$ with adjoint maximal ideal $2R$ and cover $(x/2)V$.
Observe that $\End(A)$ is the proper valuation overring $V$ of $R$. 
Therefore $A$ is not archimedean as an ideal of $R$. 
Since $A$ is completely irreducible, it is not
possible to represent $A$ as an intersection of archimedean 
ideals of $R$.
\end{remark}



For a prime ideal $P$
of  a Pr\"ufer domain,  Fuchs and Mosteig prove in
\cite[Lemma 4.3]{FM} that the $P$-primal ideals form a semigroup under ideal
multiplication.  We generalize this result in
Theorem~\ref{semigroup}. In  the proof of Theorem~\ref{semigroup},
we use the following lemma.


\begin{lemma} \label{M-primal} Let $M$ be a maximal ideal of an
arithmetical ring $R$.  If $A$ is a finitely generated ideal of
$R$ contained in $M$ such that $AR_{M} \ne 0$, then $A_{(M)}$
is an $M$-primal ideal of $R$. \end{lemma}

\begin{proof} Observe that  $A_{(M)}$
is an $M$-primal ideal of $R$ if and only if $M$ is the unique
maximal Krull associated prime of $A_{(M)}$.   If $P$ is a Krull
associated prime of $A_{(M)}$, then necessarily $P \subseteq M$,
so to show that $A_{(M)}$ is an $M$-primal ideal it suffices to
show that $M$ is a Krull associated prime of $A_{(M)}$.  Now by
Lemma 2.4 of \cite{FHO}, $M$ is a Krull associated prime of
$A_{(M)}$ if and only if $MR_M$ is a Krull associated prime of
$AR_M$.  Also, a consequence of Lemma 1.5 in \cite{FHO2} is that
every nonzero principal ideal of a valuation ring $V$ has adjoint
prime the maximal ideal of $V$.  Thus $AR_M$ is $MR_M$-primal,
and it follows that $A_{(M)}$ is an $M$-primal ideal of $R$.
\end{proof}


\begin{theorem}\label{semigroup}
Let $P$ be a prime ideal of an arithmetical ring $R$.  If $A$ and $B$ are
 $P$-primal ideals such that for all maximal ideals $M$ of $R$, $ABR_M$ is a
nonzero ideal of $R_M$, then $AB$ is a $P$-primal ideal of $R$.
 In particular the set of regular
$P$-primal ideals of $R$ is a semigroup under ideal multiplication.
\end{theorem}



\begin{proof}
  We first make two observations:


(a) {\it If $C$ is an ideal of $R$, then $P$ is a Krull
associated prime of $C$ if and only if $PR_M$ is a Krull
associated prime of $CR_M$ for some (equivalently,
all) maximal ideal(s) $M$ containing $P$.}  This is proved in
Lemma 2.4 of \cite{FHO}.


(b)  {\it If $C$ is a primal ideal of $R$, then $P$ is the
adjoint prime of $C$ if and only if for each maximal ideal $M$
of $R$ containing $P$, $CR_M$ is a $PR_M$-primal ideal of
$R_M$.} This is an application of (a) and the fact that an ideal
$C$ is $P$-primal if and only if $P$ is the unique maximal member
of the set of Krull associated primes of $A$.



We now prove the theorem.
Suppose first that $R$ is a valuation ring.  Then $AB$ is
irreducible, hence primal, with adjoint prime, say $Q$.  We claim
that $Q \subseteq P$.  If this is not the case then $P \subset
Q$, and since $AB$ is $Q$-primal, there exists $q \in Q
\setminus P$ such that $qx \in AB$ for some $x \in Q \setminus
AB$.    Since $B$ is $P$-primal, $B:q = B$.  Also, $B \subseteq
qR$, so $B = qC$ for some ideal $C$.  Necessarily $B  \subseteq C
\subseteq B:q = B$, so this forces $B= C$.  Thus  $qB = B$ and
$qAB = AB$. Since $R$ is a valuation ring and $x \not \in
AB$, we have $AB \subseteq xR$. Thus $xqR \subseteq AB = qAB
\subseteq xqR$, and we conclude that $xqR = AB$. It follows that
$xq^2R = qAB = AB = xqR$, so that $xq = rxq^2$ for some $r \in
R$.  But this implies that $(1-rq)xq = 0$, and since $1-rq$ is a
unit in $R$, $xq = 0$, contradicting the assumption that $xqR =
AB \ne 0$.  Thus  $Q \subseteq P$.


We show that in fact $Q = P$.  If this is not the case, then $Q \subset  P$ and
there exists  $p \in P \setminus Q$. Since $AB$ is $Q$-primal,
$AB:p^2 = AB$. Since $A$ and $B$ are $P$-primal, there exist $x \in P
\setminus A$ and $y \in P \setminus B$ such that $xp \in A$ and
$yp \in B$.  Hence $xy \in AB :p^2 = AB$.  Since $R$ is a valuation
ring, $A \subset  xR$ and $B \subset  yR$.  We conclude that $xyR
\subseteq AB \subseteq xyR$.  Thus by Lemma~\ref{M-primal} and
the fact that $AB \ne 0$, we have that $AB$ is an $M$-primal
ideal of $R$.  This forces $Q = P = M$, a contradiction to the
assumption that $Q \subset P$.  Hence $P = Q$ and $AB$ is
$P$-primal.


We consider now the general case in which $R$ is not necessarily a valuation
ring.  As noted above in (b), to prove that $AB$ is $P$-primal,
it suffices to show that $P$ is the unique maximal member of the
set of Krull associated primes of $AB$.  By (b) $AR_M$ and $BR_M$
are $PR_M$-primal ideals of $R$, and since $R_M$ is a valuation
ring we have by the case considered above that $ABR_M$ is a
$PR_M$-primal ideal.  By (a) $P$ is a Krull associated prime of
$AB$.  To complete the proof it remains to show that $P$ is the
unique maximal member of the set of Krull associated primes of
$AB$.  Suppose that $Q$ is a Krull associated prime of $AB$,
and let $M$ be a maximal ideal of $R$ containing $Q$.  By (a)
$QR_M$ is  a Krull associated prime of $ABR_M$. Since $ABR_M$ is
$PR_M$-primal,  $QR_M \subseteq PR_M$, so that $Q \subseteq P$.
This completes the proof that  every Krull associated prime of
$AB$ is contained in $P$.    \end{proof}
 
 
\begin{remark}\label{4.33} Simple examples show that even
in a valuation ring $R$ with maximal ideal $M$ it is
possible to have a proper  ideal $A$ that is completely
irreducible (and therefore $M$-primal) such that
$A^2$ is not $M$-primal.
Let $V$ be a valuation domain whose value group is $\G=\Z \oplus \Z$ with
lexicographic ordering.  Then the maximal ideal of $V$ is principal
generated by an element $x$ where the value $v(x) = (0,1)$
is the smallest positive
element in $\G$.  Let $y \in V$ have value $(1,0)$, and let
$Q$ be the prime ideal of $V$ of height one. Thus $Q$ is
generated by elements $z$ having value  $v(z) = (1,b)$, where $b \in \Z$
may be negative.   Notice that $Q^2 \subset yV \subset Q$
and that $Q^2$ is $Q$-primary. Define
$R= V/Q^2$. Then $R$ is  a valuation ring with maximal ideal $M= xR$.
The ideal $A= yR$ is completely irreducible, so $M$-primal.
However,  $A^2=(0)$ is not $M$-primal.  For a specific
realization of this example,  let $x, y$ be
indeterminates over a field $k$, let $W = k(x)[y]_{(y)}$,
let $Q = yW$, and set $V = k[x]_{(x)} + Q$. Then $V$
is a rank-two valuation domain with principal maximal
ideal $xV$ and height-one prime $Q$. The
valuation ring $R = V/Q^2$ has maximal ideal
$M = xR$. Since $Q^2$ is $Q$-primary, the ideal (0) of $R$
is not $M$-primal. The principal ideal  $A = yR$ is a completely
irreducible ideal and therefore is $M$-primal.
However, $A^2 = (0)$ is not $M$-primal.
\end{remark}


If we consider only those \ci\ regular ideals whose adjoint prime
is a fixed maximal ideal $M$, then we can verify:



\begin{theorem} \label{4.4}
Let $R$ be an arithmetical ring  and let $M$ be a maximal ideal
of $R$.  The set $\FF$ of \ci\ regular ideals of $R$
with adjoint maximal ideal $M$ is
closed under ideal-theoretic multiplication, and $\FF$ with this
multiplication is  a totally ordered
cancellative semigroup. \end{theorem}


\begin{proof}
We show first that: ($\star$) {\it If $I$ and $J$ are
principal  ideals of $R$ and $I_{(M)}$ and $J_{(M)}$ are
regular ideals of $R$, then $(IJ)_{(M)} = I_{(M)}J_{(M)}$.}  Suppose
that $I$ or $J$, say $I$, is not contained in $M$.   Then $IJR_M
= JR_M$, so that $(IJ)_{(M)} = J_{(M)}$, and since $I \not
\subseteq M$, $I_{(M)} = R$.  Thus if $I$ or $J$ is not
contained in $M$, then ($\star$) holds.  Otherwise, both $I$ and
$J$ are contained in $M$, and by Lemma~\ref{M-primal} $I_{(M)}$
and $J_{(M)}$ are $M$-primal ideals of $R$.
 As a product of regular $M$-primal
ideals, $I_{(M)}J_{(M)}$ is $M$-primal by
Theorem~\ref{semigroup}.  Thus $(I_{(M)}J_{(M)})_{(M)} =
I_{(M)}J_{(M)}$, so since $IJ \subseteq I_{(M)}J_{(M)}$ we have
that $(IJ)_{(M)} \subseteq I_{(M)}J_{(M)}$, and it remains to
verify the reverse inclusion.  Suppose that $x \in I:b$ and $y
\in J:c$ for some $b,c \in R\setminus M$.  Then $xy \in IJ:bc
\subseteq (IJ)_{(M)}$ since $bc \not \in M$.  It follows then
that $I_{(M)}J_{(M)} \subseteq (IJ)_{(M)}$, and this proves
($\star$).


Now let $A$ and $B$ be in $\FF$. Then $AB$ is an
$M$-primal ideal by Theorem~\ref{semigroup}.
%In particular, $AB$ is an
%irreducible ideal of the arithmetical ring $R$.
By Theorem~\ref{4.2}, $A = MJ_{(M)}$ and $B = MK_{(M)}$ for some principal  ideals
$J$ and $K$ of $R$.  Since $A$ and $B$ are regular ideals, so are
$J_{(M)}$ and $K_{(M)}$.  Now if $M \ne M^2$, then since  by
Theorem~\ref{semigroup},  $M^2$ is primal, hence irreducible,
it follows that $M/M^2$ is a one-dimensional $R/M$-vector space.
 Hence if $M \ne M^2$, $M = I_{(M)}$ for some principal
ideal $I \subseteq M$.  In this case we
have by ($\star$) (and the fact that $M$ is regular) that $AB =
I_{(M)}J_{(M)}K_{(M)}M = (IJK)_{(M)}M$,  so $AB \in \FF$ by
Theorem~\ref{4.2}.  Otherwise, if $M = M^2$, then
again by ($\star$) $AB = (IJ)_{(M)}M$ and  by
Theorem~\ref{4.2},   $AB \in \FF$.


Now assume $AC = BC$, where $A,B,C$ are in $\FF$.
Using Theorem~\ref{4.2}, write $A =
(MJ)_{(M)}$, $B = (MK)_{(M)}$ and $C= (ML)_{(M)}$ for principal ideals
$J,K,$ and $L$ of $R$.  Then $ACR_M = BCR_M$ implies that
$M^2JLR_M = M^2KLR_M$.  Moreover  $LR_M$ is a regular principal
ideal of $R_M$ since $C$ is regular and $C \subseteq L_{(M)}$
implies $CR_M \subseteq LR_M$.
  Thus $LR_M$ is a cancellation ideal of $R_M$ and  $JM^2R_M =
KM^2R_M$.  If $MR_M = M^2R_M$, then clearly $AR_M = CR_M$.
Otherwise, if $MR_M \ne M^2R_M$, then the irreduciblity of
$M^2R_M$ implies (as above) that $MR_M$ is principal and
regular, in which case also $AR_M = CR_M$.  Therefore $A =
A_{(M)} =  C_{(M)} = C$.


Since $R$ is an arithmetical ring, the ideals of $R_M$
are linearly ordered with respect to inclusion  and thus
so are  their preimages under the canonical map
$R \to R_M$.  Therefore for ideals $A$ and $B$ of $R$,
the ideals $A_{(M)}$ and $B_{(M)}$ of $R$ are comparable.  It
follows that the ideals in $\FF$ are totally ordered under
inclusion. \end{proof}

\medskip
\noindent
{\bf Acknowledgement.} We thank the referee for numerous helpful
suggestions about this paper.


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\end{document}

A representation of $A$ as an intersection of bigger ideals,
$A = \bigcap_{i \in I}C_i$, gives a representation of
$R/A$ as a subdirect product $R/A \hookrightarrow \prod_{i \in I}R/C_i$.
Conversely, a representation of $R/A$ as a subdirect product of
$\prod_{i \in I}R_i$ gives a representation of $A$ as an intersection
of bigger ideals $C_i$, where $C_i$ is the kernel of the
composition $R \to R/A \to \prod_{j \in I}R_j \to R_i$ and
this last map is the canonical projection of the product to its
$i$th  component. Since, in a subdirect product representation,
the maps $R/A \to R_i$ are required to be surjective, these are,
up to isomorphism, one-to-one correspondences. Thus a representation
of $A$ as an intersection of bigger ideals corresponds to a subdirect
representation of the ring $R/A$. The requirement that each $C_i$ contain
$A$ properly corresponds to the requirement that none of the maps
$R/A \to R_i$ is an isomorphism.  The ideals $C_i$ are completely
irreducible if and only if the rings $R_i = R/C_i$ are subdirectly
irreducible. Moreover, the representation of $A$ as an intersection
$A = \bigcap_{i \in I}C_i$
is irredundant if and only if  the subdirect representation
$R/A = \prod_{i \in I}R/C_i$ is irredundant  (that is, the
kernel becomes nontrivial if one deletes any component).   We
summarize this discussion in Theorem \ref{2.5}.







\begin{remark}\label{1.8}
 Let $M$ be a
maximal ideal of the ring $R$. By Theorem \ref{1.3},
an irreducible $M$-primary ideal $A$ is
completely irreducible  if and
only if $A \subset  (A:M)$.  If $(R,M)$ is a rank-one
nondiscrete valuation domain and $A = xR$ is a principal
$M$-primary ideal, then $A$ is irreducible, but not
completely irreducible. On the other hand,
if  $\dim R = 0$ and if each primary ideal of
$R$ contains a power of its radical, then
Corollary \ref{1.5} implies that all irreducible
ideals of $R$ are \ci.
\end{remark}






\begin{question} \label{1.9}
Let $(R,M)$ be a zero-dimensional quasilocal ring.
What are necessary and sufficient conditions in order
that every irreducible ideal of $R$ is completely
irreducible?
\end{question}

\begin{corollary} \label{4.3} In a Pr\"ufer domain,  \ci\ ideals are
archimedean. \end{corollary}


\begin{proof}
Let $C$ be a completely irreducible proper  ideal in the \Pd\ $R$, and let
$M$ denote its adjoint maximal ideal. As in \cite{FHPR},
we have
$$
\End C = \bigcap_{N \in \Max R} \End C_N.
$$
By Theorem~\ref{4.2}, we
can write $C = (rR_M \cap R)M$ for some $r \in R$.
Localizing at a maximal ideal $N \ne M$, we have $M_N = R_N$;
   thus
$$
C_N = (rR_MR_N \cap R_N) M_N = rR_MR_N \cap R_N =
rR_N,
$$
so $\End C_N = R_N$ whenever $N \ne M$. If we
localize at $M$, then we get  $C_M=(rR_M  \cap R_M)M_M
= rM_M$, whence $\End C_M = R_M$ follows since
maximal ideals of valuation domains are archimedean. Hence
$\End C = \bigcap_{N \in \Max R} R_N = R$.
\end{proof}




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