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\title[\ci\ ideals]{Unique Irredundant
Intersections  of \\ Completely Irreducible Ideals }





\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} \keywords{arithmetical ring,
valuation ring, chain ring, Pr\"ufer domain, almost Dedekind domain,
irreducible ideal, irredundant intersection}



\begin{abstract}
An ideal of a commutative ring  is completely irreducible if it is
not the intersection of any set of proper overideals. It is known
that every ideal is an intersection of completely irreducible
ideals. We characterize the rings for which every ideal can be
represented uniquely as an irredundant intersection of completely
irreducible ideals as precisely the rings in which every proper
ideal is an irredundant intersection of powers of maximal ideals.
We prove that every nonzero ideal of an integral domain $R$ has a
unique representation as an intersection of completely irreducible
ideals if and only if $R$ is an almost Dedekind domain with the
property that for each proper ideal $A$ the ring $R/A$ has at least
one finitely generated maximal ideal.  We characterize the rings for
which every proper ideal is an irredundant intersection of powers of
prime ideals as precisely the rings $R$ for which (i) $R_M$ is a
Noetherian valuation ring for each maximal ideal $M$,  and (ii)
every ideal of $R$ is an irredundant intersection of irreducible
ideals.
\end{abstract}



\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. In this paper we
characterize the rings for which every ideal can be represented
uniquely as an irredundant intersection of completely irreducible
ideals.  We prove in  Theorem~\ref{3.5} that such rings are
necessarily arithmetical, and in Theorem~\ref{main} that the
completely irreducible ideals of such a ring are precisely the
powers of maximal ideals.



We recall that a ring is said to be {\it arithmetical} if its
localization at each maximal ideal is a valuation ring, where by a
{\it valuation ring} we mean a ring in which the ideals are linearly
ordered with respect to inclusion, i.e., the ideals  form a chain.
An arithmetical integral domain is a {\it Pr\"ufer domain}.  An
integral domain $R$ is said to be {\it almost Dedekind} if $R_M$ is
a Noetherian valuation domain for each maximal ideal $M$. The
completely irreducible ideals of an arithmetical ring are explicitly
described in \cite{FHO4} (see Remark~\ref{4.1} below).  Thus the
ideals  of a ring in  which every ideal is
uniquely represented as an irredundant intersection of completely irreducible
ideals can be decomposed into ``simpler'' ideals belonging to a
well-understood class.



Let $\A$ be the set of ideals of the ring $R$ that are finite
intersections of completely irreducible ideals. We prove in Theorem
\ref{3.5} that every ideal $A \in \A$ is a unique irredundant
intersection of completely irreducible ideals if and only if $R$ is
arithmetical.  Moreover, if $R$ is arithmetical, then the components
are unique in every irredundant intersection of irreducible ideals,
even in the case of an infinite intersection.



In  Corollary~\ref{arithmetical case} we prove that in a
zero-dimensional ring $R$ every ideal is a unique irredundant
intersection of irreducible ideals if and only if $\Spec(R)$ is a
scattered topological space.  Combining this with
Theorem~\ref{irredundant irreducibles}, we obtain that for a
zero-dimensional ring $R$ every ideal is a unique irredundant
intersection of irreducible ideals if and only if $R$ is an
arithmetical ring such that for every radical ideal $J$ of $R$,
$R/J$ has a principal maximal ideal.
%every radical
%ideal is an irredundant intersection of maximal ideals. Thus the condition
%in a zero-dimensional ring $R$ in order that every ideal be a unique
%irredundant intersection of irreducible ideals is an invariant of the
%topological space $\Spec R$.  We show in fact that
As we record in Question~\ref{irreducible
question}, the classification of rings of positive dimension  for
which every ideal is a unique irredundant intersection of
irreducible (not necessarily {\it completely} irreducible) ideals
remains an open problem.


In Theorem~\ref{prime power} we prove that every proper irreducible
ideal of a ring $R$ is a power of a maximal ideal if and only if
$R_M$ is a Noetherian valuation ring for every maximal ideal  $M$  of
$R$.  We then characterize in Theorem~\ref{main} the rings for which
every ideal can be represented uniquely as an irredundant
intersection of completely irreducible ideals. We  show that this
class of rings coincides with the class of rings for which every
proper ideal is an irredundant intersection of powers of maximal
ideals.


Theorem~\ref{main} motivates our consideration in Section 4 of the
class of rings $R$ in which every proper ideal  is an irredundant
intersection of powers of prime ideals. We observe that this class
of rings properly includes the ZPI rings of classical interest,
i.e., those rings for which every proper ideal is a product of prime
ideals. We prove in Theorem~\ref{ZPI generalization} that the
following two conditions are equivalent in a ring $R$: (i) every
ideal of $R$ can be represented as an irredundant intersection of
powers of prime ideals, and (ii) $R_M$ is a Noetherian valuation
ring for each maximal ideal $M$ of $R$ and every ideal of $R$ can be
represented as an irredundant intersection of irreducible ideals.




\begin{notation}
For ideals $I,J$ of the ring $R$, 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}.
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 $P$.
\end{notation}



Two different notions 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 Zariski-Samuel} associated
prime of $A$ if there exists $x \in R$ such that $\sqrt {A:x} = P$.
We denote by $\Ass(A)$ the set of Krull associated primes of $A$ and
by $\ZZ(A)$ the set of Zariski-Samuel associated primes of $A$. It
is true in general that $\ZZ(A) \subseteq \Ass(A)$.




\begin{remark}
In \cite{F} Fuchs defines  a {\it primal ideal} of a ring $R$  as an
ideal $A$ having the property that the zero divisors in $R/A$ form
an ideal.  This ideal is necessarily prime and hence of the form
$P/A$ for some prime ideal $P$ of $R$. The ideal $P$ is called the
{\it adjoint prime} of $A$.  If $A$ is a $P$-primal ideal (that is,
$A$ is a primal ideal with adjoint prime $P$) then $A = A_{(P)}$
\cite[Theorem 3.4]{FHO}.  Moreover a prime ideal $P$ of a ring $R$
is a Krull associated prime of an ideal $A$ if and only if $A_{(P)}$
is a $P$-primal ideal of $R$ \cite[Theorem 3.4]{FHO}.  Every
irreducible ideal is primal \cite{F}; hence every completely
irreducible ideal is primal.  In addition,  the adjoint prime of a
completely irreducible ideal is a maximal ideal \cite[Proposition
1.2]{FHO4}.
\end{remark}


{\it Acknowledgement.} Our work in this paper is an outgrowth of our
collaboration with Laszlo Fuchs in the articles \cite{FHO},
\cite{FHO2}, \cite{FHO4} and  \cite{FHO5}. We would like to thank
Laszlo for generously sharing with us his  keen mathematical insight
as we have collaborated with him  on these topics.


\section{Unique irredundant intersections of irreducible ideals}



In this section we consider irredundant intersections of irreducible
ideals. This allows us to develop several technical
characterizations needed in Section~3 and Section~4. We give
sufficient conditions for a ring to have the property that every
ideal is a unique irredundant intersection of irreducible ideals,
and we characterize this class of rings in the zero-dimensional
case.  We observe in Theorem~\ref{3.5} that the requirement of
uniqueness places this problem in the setting of arithmetical rings,
that is: {\it If every ideal of a ring $R$ can be represented
uniquely as an irredundant intersection of irreducible ideals,  then
$R$ is arithmetical.}



\begin{lemma} \label{3.4}
Let $(R,M)$ be a quasilocal ring. If $R$ is not a valuation ring,
then there exists an ideal of $R$ that has two different
representations as an irredundant intersection of two completely
irreducible ideals. More precisely, there exist an ideal $A$ of $R$
and distinct completely irreducible ideals $C_1, C_2, C_3$ of $R$
such that $A = C_1 \cap C_2 = C_1 \cap C_3$ are two different
irredundant representations of $A$.
\end{lemma}




\begin{proof} Assume $R$ is not a valuation ring. Since every ideal
in $R$ is an intersection of completely irreducible ideals, there
exist incomparable completely irreducible ideals $C_1$ and $C_2$ of
$R$. Let $A = C_1 \cap C_2$. Let $C^*_1$ and $C^*_2$ denote the
unique minimal overideals to the completely irreducible ideals $C_1$
and $C_2$, respectively.  There exist elements $x \in (C_1^* \cap
C_2) \setminus A$ and $y \in (C_1 \cap C_2^*) \setminus A$. Since $x
\in C_2$ and $y \in C_1$, we have $\Soc R/A = (A + (x,y)R)/A$ is a
$2$-dimensional vector space over $R/M$ and $x+y \not\in C_1 \cup
C_2$. Let $C_3$ be an ideal containing $A + (x+y)R$ that is maximal
with respect to $x \not\in R$. Then $C_3$ is completely irreducible,
distinct from $C_1$ and $C_2$,  and $A = C_1 \cap C_3$.
\end{proof}




\begin{theorem}\label{3.5}
Let $\A$ be the set of ideals of the ring $R$ that are finite
intersections of completely irreducible ideals. Every ideal $A \in
\A$ has a unique representation as an irredundant intersection of
completely irreducible ideals if and only if $R$ is arithmetical.
Moreover, if $R$ is arithmetical, then the components are unique in
every irredundant intersection of irreducible ideals, even in the
case of an infinite intersection.
\end{theorem}




\begin{proof} By  Corollary 5.6 of \cite{FHO} in an
arithmetical ring the components in any irredundant intersection of
irreducible ideals are unique.     Assume, conversely, that for each
$A \in \A$, the representation of $A$ as an irredundant intersection
of completely irreducible ideals is unique.  This property then also
holds in $R_M$ for each maximal ideal $M$ of $R$.  For if $M$ is a
maximal ideal of $R$ and $C$ is a completely irreducible ideal of
$R_M$, then the preimage $B$ of $C$ in $R$ under  the mapping $R
\rightarrow R_M$ is completely irreducible and $C = BR_M$ (see
\cite[Theorem 1.3]{FHO4}).     By Lemma \ref{3.4}, $R_M$ is a
valuation ring. Therefore $R$ is arithmetical. \end{proof}




\begin{lemma} \label{every zs} Let $R$ be a ring in which every proper ideal
has a Zariski-Samuel associated prime ideal.  If $A$ is a proper
ideal of $R$, then $A = \bigcap_{P \in \ZZ(A)}A_{(P)}$.
\end{lemma}



\begin{proof}
To show that  $A  = \bigcap_{P \in \ZZ(A)}A_{(P)}$, it suffices to
verify the inclusion $\supseteq$, so suppose $x \in R \setminus A$.
Then  the proper ideal $A:x$ has a Zariski-Samuel associated prime
ideal; that is, there exist a prime ideal $P$ and $y \in R\setminus
(A:x)$ such that $P = \sqrt{(A:x):y}$.  Since $(A:x):y = A:xy$, we
have $P \in \ZZ(A)$. Moreover, $A:xy \subseteq P$ implies that  $xy
\not \in A_{(P)}$. Thus $x \not \in A_{(P)}$.  It follows that
$\bigcap_{P \in \ZZ(A)}A_{(P)} \subseteq A$, and the proof is
complete.
\end{proof}




\begin{lemma} \label{zs} (i) Let $R$ be a ring in which  every radical ideal
$J$  has a minimal prime divisor $P$ such that $P/J$ is the radical
of a finitely generated ideal of $R/J$. If   $A$ is  a proper ideal
of $R$, then $A = \bigcap_{P \in \ZZ(A)}A_{(P)}$.


(ii) Assume the ring  $R$  satisfies the ascending chain condition
on prime ideals. For a proper ideal $A$ of $R$, let $\ZZ^*(A)$
denote the maximal elements of $\ZZ(A)$.  If $A = \bigcap_{P \in
\ZZ(A)}A_{(P)}$, then $A = \bigcap_{P \in \ZZ^*(A)}A_{(P)}$ and this
second intersection is irredundant. If also $R$ is arithmetical,
then this second intersection is the unique representation of $A$ as
an irredundant intersection of irreducible ideals.
\end{lemma}



\begin{proof} (i)  To prove the first claim it suffices by Lemma~\ref{every
zs} to show that every proper ideal $B$  of $R$ has a Zariski-Samuel
associated prime ideal.   Set $J = \sqrt{B}$.  By assumption there
exists a minimal prime divisor $P$ of $J$  such that $P = \sqrt{J +
C}$ for some finitely generated ideal $C$ of $R$.  Since $P$ is
minimal over $J$, we have $JR_P = PR_P$.  Thus $J_{(P)} = P$, so $C
\subseteq J_{(P)}$ and since $C$ is finitely generated, there exists
$x \in R \setminus P$ such that $xC \subseteq J$.  Hence $J + C
\subseteq J:x$, and since $x \not \in P$ we have  $J:x \subseteq P$.
Thus $P = \sqrt{J:x}$. It follows that  $P =  \bigcup_{n>0}
\sqrt{B:x^n}$. For  $x \not \in P$ and $B \subseteq P$ implies
$\sqrt{B:x^n} \subseteq P$ for all $n>0$,   and if $a \in
P=\sqrt{J:x}$, then there exists $k>0$ such that $a^kx \in J$, so
$a^{kn}x^n \in B$ for some $n>0$; hence $a \in \sqrt{B:x^n}$. Since
$C$ is finitely generated and contained in $P$, we have  $C
\subseteq \sqrt{B:x^n}$ for some $n>0$.  Also, $J =\sqrt{B}
\subseteq \sqrt{B:x^n}$, so we have $P = \sqrt{C+J} \subseteq
\sqrt{B:x^n} \subseteq P$, proving that $P$ is a Zariski-Samuel
associated prime of $B$.



(ii) Now assume that $A$ is a proper ideal of a ring $R$ that
satisfies the ascending chain condition on prime ideals. Then for
each $Q \in \ZZ(A)$ there exists $P \in \ZZ^*(A)$ such that $Q
\subseteq P$.  It follows that  $A_{(P)} \subseteq A_{(Q)}$.
Therefore $A = \bigcap_{P \in \ZZ(A)}A_{(P)}$ implies that $A =
\bigcap_{P \in \ZZ^*(A)}A_{(P)}$.  Moreover,  this second
intersection is irredundant. For if $P \in \ZZ^*(A)$ and $x \in
R\setminus A$ is such that $P = \sqrt{A:x}$, then $x \in \bigcap_{Q
\in \ZZ^*(A)\setminus\{P\}}A_{(Q)} \setminus A_{(P)}$. If we also
assume that $R$ is arithmetical, then each $A_{(P)}$  is irreducible
(cf. Remark 1.6 of \cite{FHO}), and as we have observed above in
Theorem~\ref{3.5}, in an arithmetical ring any representation of an
ideal as an irredundant intersection of irreducible ideals is
unique, so the proof of Lemma~\ref{zs} is complete.
\end{proof}



\begin{remark} \label{key example} The prime spectrum of a  ring $R$ is
Noetherian as a topological space if and only if every radical ideal
of $R$ is the radical of a finitely generated ideal \cite{OP}.  Thus
if $R$ is an arithmetical ring with Noetherian prime spectrum, then
by Lemma~\ref{zs} every ideal of $R$ can be represented uniquely as
an irredundant intersection of irreducible ideals. However,  even
for zero-dimensional arithmetical rings, there exist examples of
rings without Noetherian prime spectrum for which every ideal is a
unique irredundant intersection of irreducible ideals. We postpone a
discussion of such examples until Remark~\ref{Boolean}.
\end{remark}




\begin{lemma} \label{split} Let  $A$ be a proper ideal of the ring $R$
and assume that    $J = \sqrt{A}$ is an irredundant intersection $J
= M \cap B$, where $M$ is a maximal ideal of $R$ and $B$ is a proper
ideal not contained in $M$.
   Then  $M$ is a minimal prime divisor of $A$ and $M/A$ is the
radical of a principal ideal of $R/A$. Moreover $M/J$ is a principal
maximal ideal of $R/J$. \end{lemma}




\begin{proof}   Since $J = M \cap B$ is irredundant and  $M$ is maximal
we have $R = M + B$, so $J = MB$. Thus there exist $x \in M
\setminus B$ and $y \in B \setminus M$ such that $1 = x + y$.  It
follows  that $M = xR + J$, for if $w \in M$, then $w = wx + wy \in
xR + MB = xR + J$.  Hence $M/J$ is a principal maximal ideal of
$R/J$.  To  show that  $M$ is minimal over $A$ is equivalent to
showing that $M$ is minimal over $J$,  and for this it is enough to
show that $R_{M}/JR_{M}$ is a field.  Since the maximal ideal of
$R_{M}/JR_{M}$ is generated by $x/1 + JR_{M}$, it suffices to show
$x/1$ is in the ideal $JR_{M}$.  This is indeed the case since $yx
\in J$ and $y \not \in M$.




Since the prime ideals containing $A$ are precisely those that
contain $J$, and $M$ is the only prime ideal that contains both $J$
and $x$, we have $M = \sqrt{xR + A}$. It follows that $M/A$ is the
radical of a principal ideal of $R/A$.
\end{proof}




\begin{lemma} \label{PM} Let $A$ be an ideal
of the  ring $R$.  If $M$ and $N$ are distinct maximal ideals of $R$
with $A \subseteq M \cap N$ and if $A_{(M)} \subseteq N$, then there
is a prime ideal $P$ of $R$ such that $A_{(M)} \subseteq P
\subsetneq  M \cap N$. \end{lemma}




\begin{proof}  Define $S$ to be the
multiplicatively closed set $\{xy:x \in R \setminus M, y \in
R\setminus N\}$. We observe that $A_{(M)} \subseteq N$ implies
$A_{(M)} \cap S = \emptyset$. For suppose there exists an element $r
\in A_{(M)} \cap S$. Then $r = xy$, with $x \in R \setminus M$ and
$y \in R \setminus N$. Also $r \in A_{(M)}$ implies there exists $x'
\in R \setminus M$ such that $x'r = a \in A$, and $a = x'xy$ implies
$y \in A_{(M)}$. But $y \not\in N$. Thus $A_{(M)} \subseteq N$
implies $A_{(M)} \cap S = \emptyset$. Hence  there is a prime ideal
$P$ of $R$ containing $A_{(M)}$ such that $P \cap S = \emptyset$. It
follows that $A_{(M)} \subseteq P \subsetneq M \cap N$.
\end{proof}



A topological space $X$ is {\it scattered} if every nonempty subset
of $X$ contains a point that is isolated in the relative topology.
If $R$ is a zero-dimensional ring, then $\Spec(R)$ is scattered if
and only if for each nonempty family  $\{M_i:i \in I\}$ of maximal
ideals of $R$, there exists $i \in I$ such that $\bigcap_{j \ne
i}M_j \not \subseteq M_i$.



\begin{theorem}\label{irredundant irreducibles}
Let $R$ be a zero-dimensional ring. The following statements are
equivalent.
\begin{itemize}



\item[(i)]  $\Spec(R)$ is a scattered space.


\item[(ii)]  For every  radical ideal $J$ of $R$, there is a maximal
ideal $M$ containing $J$  such that $M/J$ is a principal ideal of
$R/J$.



\item[(iii)]  For every radical ideal $J$ of $R$, there is a maximal ideal
$M$ containing $J$  such that $M/J$ is the radical of a finitely
generated ideal of $R/J$.



\item[(iv)] For every proper  ideal $A$ of $R$, the set
${\mathcal{Z}}(A)$ of Zariski-Samuel associated primes of $A$ is
nonempty.



\item[(v)]  For every  proper ideal $A$ of $R$, $A  = \bigcap_{P \in
{\mathcal{Z}}(A)}A_{(P)}$.


\item[(vi)]   Every  radical ideal of $R$ is
an irredundant intersection of maximal ideals.




\end{itemize}
{\noindent}Moreover if $R$ satisfies (i)-(vi) and $A$ is a proper
ideal of $R$, then $A = \bigcap_{P \in {\mathcal{Z}}(A)}A_{(P)}$ is
an irredundant intersection. \end{theorem}


\begin{proof}
First observe that if $A$ is a proper ideal of $R$, then each member
of ${\mathcal{Z}(A)}$ is a maximal ideal of $R$   since $R$ is
zero-dimensional. Therefore by Lemma~\ref{zs}, if $A =   \bigcap_{P
\in {\mathcal{Z}}(A)} A_{(P)}$, then this intersection is
irredundant.



(i) $\Rightarrow$ (ii)  If $J = \bigcap_{i \in I}M_i$ is an
intersection of maximal ideals $M_i$ of $R$, then by (i)  there
exists $i \in I$ such that $\bigcap_{j \ne i}M_j \not \subseteq
M_i$.  By Lemma~\ref{split}, $M_i/J$ is a principal ideal of $R/J$.




%(i) $\Rightarrow$ (ii)  Apply Lemma~\ref{split}.


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


(iii) $\Rightarrow$ (iv) Apply Lemma~\ref{zs}.


(iv) $\Rightarrow$ (v) Apply Lemma~\ref{every zs}.


(v) $\Rightarrow$ (vi) Let $J$ be a radical ideal of $R$.  By (v) $J
= \bigcap_{P \in {\mathcal{Z}}(J)} J_{(P)}$, and by the remark at
the beginning of the proof this intersection is irredundant, so to
complete the proof it suffices to observe that each $J_{(P)}$, $P
\in {\mathcal{Z}}(J)$, is a maximal ideal of $R$.      Indeed, since
$J$ is a radical ideal, so is $J:x$ for every $x \in R$.  It follows
that $J_{(P)}$, as a set union of radical ideals, is a radical
ideal.  Since $R$ is zero-dimensional, $J_{(P)}$ is contained only
in $P$ (Lemma~\ref{PM}). Thus $J_{(P)} = P$, and it follows that
$J$ is an irredundant intersection of maximal ideals of $R$.



(vi) $\Rightarrow$ (i)  Let $\{M_i:i \in I_1\}$ be a collection of
maximal ideals of $R$, and let $J = \bigcap_{i \in I_1}M_i$.  By
(vi) there is a collection $\{N_i:i \in I_2\}$ of maximal ideals
such that $J = \bigcap_{i \in I_2}N_i$ is an irredundant
intersection.
%We prove that
%$\{N_i : i \in I_2\} \subseteq \{M_i : i \in I_1\}$.
Fix $i_2 \in I_2$, and let $J_2 = \bigcap_{i \ne i_2}N_i$.  The
intersection $J  = N_{i_2} \cap J_{2}$ is irredundant, and $J
\subseteq M_i$ implies either $N_{i_2} = M_i$ or $J_{2} \subseteq
M_i$. If $N_{i_2} \ne  M_i$ for all $i \in I_1$, then $J_{2}
\subseteq \bigcap_{i \in I_1}M_i = J$, a contradiction. Hence
$N_{i_2} = M_{i_1}$ for some $i_1 \in I_1$.  Set $J_1 = \bigcap_{i
\ne i_1}M_i$.  Since $N_{i_2} \cap J_2 = J \subseteq J_1$ and
$N_{i_2} \not \subseteq M_i$ for any $i \in I_1$ with $i \ne i_1$,
it follows that $J_2 \subseteq J_1$.  Thus $J = M_{i_1} \cap J_1$
and $J_1 \not \subseteq M_{i_1}$ since otherwise $J_2 \subseteq J_1
\subseteq M_{i_1} = N_{i_2}$, a contradiction. This proves
$\Spec(R)$ is scattered. \end{proof}





\begin{corollary}\label{arithmetical case}
Let $R$ be a zero-dimensional ring. The following statements are
equivalent.
\begin{itemize}
\item[(i)]  Every ideal of $R$ can be represented
uniquely as an irredundant intersection of irreducible ideals.
\item[(ii)] $R$ is an arithmetical ring with scattered prime
spectrum. \end{itemize} \end{corollary}



\begin{proof} (i) $\Rightarrow$ (ii) By Theorem~\ref{3.5},  $R$ is 
arithmetical.
If an ideal $A$ of an arithmetical ring $R$ is an irredundant intersection $A
= \bigcap_i A_i$ of irreducible ideals $A_i$, then each $A_i$ is an
isolated component of $A$, that is, $A_i = A_{(P_i)}$ for some prime
ideal $P_i$ of $R$ (cf. Lemma 5.5 of \cite{FHO}).  If $A$ is a
radical ideal, then $A_{(P_i)}$ is a prime ideal by Lemma~\ref{PM},
so it follows that $A$ is an irredundant intersection of prime
ideals.  Since $R$ is zero-dimensional, $R$ satisfies the equivalent
conditions of Theorem~\ref{irredundant irreducibles}.



(ii) $\Rightarrow$ (i)  Apply Lemma~\ref{zs} and
Theorem~\ref{irredundant irreducibles}. \end{proof}





In Corollary~\ref{arithmetical case}  we have characterized  the
zero-dimensional rings for which every ideal can be represented
uniquely as an irredundant intersection of irreducible ideals. We
record the following characterization for one-dimensional integral
domains.



\begin{corollary}\label{one-dim case}
Let $R$ be a one-dimensional integral domain. The following
statements are equivalent.
\begin{itemize}
\item[(i)]  Every ideal of $R$ can be represented
uniquely as an irredundant intersection of irreducible ideals.
\item[(ii)] $R$ is a Pr\"ufer domain and for each nonzero
proper ideal $A$, the ring $R/A$ has a scattered spectrum.
\item[(iii)] $R$ is a Pr\"ufer domain and for each proper ideal $A$,
the ring $R/A$ contains at least one maximal ideal that is the
radical of a finitely generated ideal.
\end{itemize}
\end{corollary}







\begin{remark}  \label{Boolean}
(i) An almost Dedekind domain $R$ with at most finitely many maximal
ideals that are not finitely generated satisfies (i)-(iii) of
Corollary~\ref{one-dim case}.  Thus if $A$ is any proper nonzero
ideal of $R$, then $R/A$ is satisfies the equivalent conditions of
Corollary~\ref{arithmetical case}.
  One reference for such an example that is non-Noetherian is
\cite[Example~2.2]{HO}.


(ii) In \cite{O}, the example of (i) is generalized to show that if
$X$ is a compact totally disconnected topological space, then there
is an almost Dedekind domain $R$ such that $\Max(R)$ is homeomorphic
to $X$. Since a compact scattered space is totally disconnected,
such a space can be realized as the maximal spectrum of an almost
Dedekind domain.  Several examples are discussed in \cite{O}; we
mention one here. Let $(X,\leq)$ be a well-ordered set. Then $X$ is
a compact scattered space with respect to the order topology on $X$,
and the isolated points of $X$ are precisely the smallest element of
$X$ and the immediate successors of elements in $X$ \cite[Example
17.3, p. 272]{Ko}.  Hence by suitable choices of the space $X$, one
obtains examples of scattered spaces having infinitely many
non-isolated points.  If $R$ is an almost Dedekind domain with
$\Max(R)$ homeomorphic to $X$, then $R$ has nonzero Jacobson radical
and the isolated points of $X$ correspond to finitely generated
maximal ideals of $R$ \cite{O}.  The ring $R$ satisfies the
equivalent conditions of Corollary~\ref{one-dim case} and every
proper homomorphic image of $R$ satisfies the conditions of
Corollary~\ref{arithmetical case}.
   \end{remark}





It would be interesting to discover how to extend the
characterizations of this section to resolve the following:




\begin{question}\label{irreducible question} What rings $R$ have the property
that every ideal of $R$ can be  represented uniquely as an
irredundant intersection of irreducible ideals? \end{question}




Necessarily such rings are arithmetical.  Moreover this question is
equivalent to Question 5.17(1) of \cite{FHO}, which asks for a
classification of the arithmetical rings $R$ having the property
that  every proper ideal $A$ can be written as an irredundant
intersection $A = \bigcap_{P \in C}A_{(P)}$ for some set $C$ of
Krull associated primes of $A$. (Apply Lemma 5.5 and Corollary 5.6
of \cite{FHO}.)





\section{irredundant intersections of completely
irreducible ideals}




In this section we characterize the rings in  which every ideal is a
unique irredundant intersection of completely irreducible ideals.



\begin{remark}\label{4.1} In \cite[Theorem 4.3]{FHO4} the
completely irreducible ideals of an arithmetical ring are explicitly
described: {\it           A ring $R$ is arithmetical if and only if
the proper completely irreducible 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{remark}



\begin{lemma}\label{proper principal}  If $R$ is a valuation ring having a
proper nonzero principal ideal $rR$ that is completely irreducible,
then the maximal ideal $M$ of $R$ is principal and every $M$-primary
ideal of $R$ is a power of $M$. \end{lemma}




\begin{proof}
By Remark \ref{4.1}, $rR = sM$ for some $s \in R$. If $M = M^2$,
then $rM = sM^2 = sM = rR$, but this implies $r \in rM$ which
contradicts the fact that $r \ne 0$.  Hence $M \ne M^2$, and since
$M^2$ is irreducible, it follows that the $R/M$-vector space $M/M^2$
has dimension one.
Consequently, $M$ is a principal ideal of $R$.  It follows that  every
$M$-primary ideal of $R$ is a power of $M$.
  \end{proof}




We recall that a principal ideal ring $R$ is called a {\em special }
PIR if $R$ is a local ring with maximal ideal $M$ and every ideal of
$R$ is a power of $M$ \cite[page 245]{ZS}.





\begin{lemma}\label{special PIR}  Every ideal of
a ring $R$ is completely irreducible if and only if  $R$ is a
special PIR.
\end{lemma}




\begin{proof} Suppose that every ideal of $R$ is
completely irreducible.  Then any two ideals of $R$ are comparable,
so $R$ is a valuation ring and hence is quasilocal. In Remark 1.6 of
\cite{FHO4} it is noted that if every irreducible ideal of a ring is
completely irreducible, then the ring is zero-dimensional.  Thus
$R$ is zero-dimensional, and the claim follows from
Lemma~\ref{proper principal}.




Conversely, suppose that $R$ is local with maximal ideal $M$ and
every ideal of $R$ is a power of $M$.  Then $R$ is a valuation ring
and either $R$ is a field or $M \ne M^2$, so as in the proof of the
preceding lemma, $M$ is a principal ideal of $R$.  The claim now
follows from Remark~\ref{4.1}. \end{proof}




\begin{proposition}  \label{locally special PIR}    Every primal ideal of a
ring $R$ is completely irreducible if and only if $R_M$ is a special
PIR for each maximal ideal $M$ of $R$.
\end{proposition}




\begin{proof} Assume every primal ideal of $R$ is completely
irreducible. By Theorem 1.8 of \cite{FHO}, $R$ is arithmetical. Let
$M$ be a maximal ideal of $R$. If $B$ is an ideal of $R_M$ and $A$
is the preimage of $B$ in $R$ under the canonical mapping $R
\rightarrow R_M$, then $AR_M = B$ and $A = A_{(M)}$ is irreducible
and hence primal and therefore completely irreducible. (The
irreduciblity of $A = A_{(M)}$ follows because $R$ is arithmetical,
so $AR_M$ is irreducible; see Remark 1.6 of \cite{FHO}.) It is shown
in Theorem 1.3 of \cite{FHO4} that an ideal $C$ of (any ring) $R$ is
completely irreducible if and only if for some maximal ideal $N$ of
$R$, $C = C_{(N)}$ and $CR_N$ is a completely irreducible ideal of
$R_N$.   Thus we conclude that $B$ is a completely irreducible ideal
of $R_M$. Consequently, every ideal of $R_M$ is completely
irreducible. By  Lemma~\ref{special PIR}, $R_M$ is a special PIR.


Assume, conversely, that $R_M$ is a special PIR for every maximal
ideal $M$ of $R$. Then $R$ is arithmetical and zero-dimensional. If
$A$ is a primal ideal of $R$, then $A$ is $M$-primal for some
maximal ideal $M$.  By Lemma~\ref{special PIR},  every ideal of
$R_M$ is completely irreducible.  In particular,  $AR_M$ is
completely irreducible,    and it follows  (as above)  from Theorem
1.3 of \cite{FHO4}  that $A = A_{(M)}$ is completely irreducible.
\end{proof}



\begin{theorem} \label{prime power}  The following statements are equivalent
for a ring $R$.
\begin{itemize}
\item[(i)]
Every proper irreducible ideal of $R$ is a power of a prime ideal.
\item[(ii)] For every  maximal
ideal $M$ of $R$, $R_M$ is a Noetherian valuation ring.
\end{itemize}
\end{theorem}




\begin{proof}(i) $\Rightarrow$ (ii)
We first establish the following three claims.



Claim (1) {\it If $R$ is a quasilocal ring with maximal ideal $M \ne
M^2$ and every proper irreducible ideal of $R$ is a power of a prime
ideal, then $R$ is a Noetherian valuation ring.} We show first that
$M$ is a principal ideal. Let $m \in M \setminus M^2$.  Since every
ideal is an intersection of completely irreducible ideals, it
follows from our assumption that $mR = \bigcap_{i \in I}P_i^{e_i}$
for some prime ideals $P_i$ of $R$ such that for each $i$,
$P_i^{e_i}$ is a completely irreducible ideal of $R$. Since the only
completely irreducible prime ideal of $R$ is $M$, it follows that
for each $i$ with $P_i \ne M$, it must be that $e_i>1$.  In fact, if
$e_i>1$, then $m \in P_i^{e_i} \subseteq P_i^{2} \subseteq M^2$,
contrary to assumption. This forces $\{P_i\}_{i \in I} = \{M\}$ and
$mR = M$, proving that $M$ is a principal ideal of $R$.


Under the  assumptions of claim (1),  we show that  $R$ is a
zero-dimensional ring or a one-dimensional domain.  If $P$ is a
prime ideal of $R$ properly contained in $M =mR$, then $P \subsetneq
mR$, so that $P = mA$ for some proper ideal $A$ of $R$.  Since $P$
is a prime ideal of $R$ and $m \not \in P$, it follows that $A = P$.
Thus  for all prime ideals $P$ properly contained in $M$, $P = mP$
and $P \subseteq \bigcap_{k = 1}^{\infty}M^k$.  Now suppose that
there exists a nonmaximal prime ideal $P$ of $R$.  We claim that $P
= (0)$.   Let $y \in P$.   Since every ideal of $R$ is an
intersection of irreducible ideals, $yR = \bigcap_{i \in
I}P_i^{e_i}$ for some prime ideals $P_i$ of $R$.  If $\{P_i\}_{i \in
I} = \{M\}$, then $\bigcap_{k=1}^{\infty}M^k \subseteq yR \subseteq
P$.  As already noted, $P \subseteq \bigcap_{k=1}^{\infty}M^k$, so
we have in this case that $yR = P$.  Thus $P = mP$ implies $y = ymr$
for some $r \in R$ and $y(1-mr) = 0$. Since $R$ is quasilocal,
$1-mr$ is a unit and  $y = 0$. Assume  $\{P_i\}_{i \in I} \ne
\{M\}$. Since $P_i \ne M$ implies $P_i \subseteq M^k$ for all $k>0$,
we may assume that $P_i \ne M$ for each $i$.  In particular for each
$i$, $P_i = mP_i$ and we have $yR = \bigcap_{i \in I}P_i^{e_i} =
m(\bigcap_{i \in I}P_i^{e_i}) = ymR$. From the fact that $R$ is
quasilocal, we conclude $y = 0$.  This shows that if there exists a
nonmaximal prime ideal $P$ of $R$, then $P = 0$. We conclude that
$R$ is either a zero-dimensional ring or a one-dimensional domain.
In the case that $R$ is zero-dimensional, by assumption every
irreducible ideal of $R$ is a power of $M$; in the case that $R$ is
one-dimensional, every nonzero irreducible ideal of $R$ is a power
of $M$.     Since every ideal of a ring is an intersection of
irreducible ideals, we conclude that $R$ is a Noetherian valuation
ring.


Claim (2) {\it If $R$ is a ring such that every proper irreducible
ideal is a power of a prime ideal, then for each prime ideal $Q$ of
$R$, every proper irreducible ideal of $R_Q$ is a power of a prime
ideal.} Indeed, if $A$ is a proper  irreducible ideal of $R_Q$, let
$B$ denote the preimage of $A$ under the mapping $R \rightarrow
R_Q$.  Then $B$ is an irreducible ideal of $R$, and
by assumption $B = P^{e}$ for some prime ideal $P$ of $R$ and $e >0$.
Thus $A = BR_Q =
P^{e}R_Q$, so that every proper irreducible ideal of $R_Q$ is a
power of a prime ideal of $R_Q$.



Claim (3)  {\it If $R$ is a quasilocal domain in which every proper
irreducible ideal is a power of a prime ideal, then $R$ is either a
field or a discrete rank-one valuation domain (DVR)}.  Suppose that
$R$ is not a field and let $M$  be the maximal ideal of $R$. Suppose
that there exists a prime ideal $P$ of $R$ properly contained in
$M$.  We claim that $P = (0)$.  Let $y \in M \setminus P$, and let
$Q$ be a prime ideal of $R$ minimal over $P + Ry$. Since $QR_Q$ is
the radical of $PR_Q+yR_Q$, it is also the radical of $A:=PR_Q +
y^2R_Q$. Furthermore $A \ne QR_Q$, so by claim (2) $A = Q^eR_Q$ for
some $e > 1$ since $A$ is an intersection of irreducible ideals and
the only prime ideal containing $A$ is $QR_Q$. In particular $QR_Q
\ne Q^2R_Q$.  Thus $R_Q$ is a quasilocal ring such that every proper
irreducible ideal of $R_Q$ is a power of a prime ideal and $QR_Q \ne
Q^2R_Q$.  This places us in the setting of claim (1), so we conclude
that $R_Q$ is a DVR.  In particular, $PR_Q = 0$. Thus (since $R$ is
a domain) $P = 0$ and $R$ is a one-dimensional domain.  Now if $B$
is a nonzero irreducible ideal of $R$ properly contained in $M$,
then $B = M^k$ for some $k>1$. Hence $M \ne M^2$ and by claim (1),
$R$ is a DVR.



We now prove that (i) $\Rightarrow$ (ii) in full generality.  Let
$R$ be a ring satisfying (i) and let $M$ be a maximal ideal of $R$.
By claim (2), every proper irreducible ideal of $R_M$ is a power of
a prime ideal. Thus to prove that $R_M$ is a Noetherian valuation
ring, it suffices by claim (1) to show that $R_M$ is a field or
$MR_M \ne M^2R_M$. Suppose there exists a prime ideal $P$ of $R$
properly contained in $M$. Then every proper irreducible ideal of
$R_M/P_M$ is a power of a prime ideal, so by claim (3) $R_M/P_M$ is
a DVR.  In particular, $MR_M \ne M^2R_M$, as claimed.  Otherwise, if
$R_M$ is zero-dimensional and if $R_M$ is not a field, then there is
an irreducible ideal $B$ of $R_M$ properly contained in $MR_M$.  By
assumption, $B = M^kR_M$ for some $k>1$.  Hence $MR_M \ne M^2R_M$ in
this case also.  Thus by claim (1) $R_M$ is a Noetherian valuation
ring.



(ii) $\Rightarrow$ (i) Let $A$ be a proper  irreducible ideal of $R$
with adjoint prime ideal $P$, and let $M$ be a maximal ideal of $R$
containing $P$.  Then $A = A_{(M)}$ since $A$ is $P$-primal.  Now
$AR_M$ is the zero ideal of $R_M$ or a power of $MR_M$.  If $AR_M =
M^kR_M$ for some $k$, then the preimages $A_{(M)}$ and $(M^k)_{(M)}$
of $AR_M$ and $M^kR_M$, respectively, under the mapping $R
\rightarrow R_M$ are equal; that is, $A = A_{(M)} = (M^k)_{(M)}$.
Since $M$ is a maximal ideal of $R$, then $(M^k)_{(M)} = M^k$.  For
it is enough to verify this equality locally and if $N$ is a maximal
ideal of $R$, then $(M^k)_{(M)}R_N = M^kR_N$. Thus $A = M^k$ in case
$AR_M$ is a power of $MR_M$.
Otherwise, $AR_M$ is not a power of $MR_M$, so by (ii) $AR_M = (0)R_M$, where
$(0)R_M$ is a prime ideal of $R_M$; in particular, $AR_M = QR_M$ for
some prime ideal $Q$ of $R$.  Now $A = A_{(M)} = Q_{(M)} = Q$, so
that $A$ is a prime ideal of $R$.  This proves that every
irreducible ideal of $R$ is a power of a prime ideal of $R$.
\end{proof}



\begin{corollary} \label{decompose} If $A$ is a proper ideal of an
arithmetical ring $R$ such that every primal ideal of $R/A$ is
completely irreducible, then $A$ is an intersection $A =
\bigcap_{i}M_i^{e_i}$ of powers of maximal ideals $M_i$ of $R$.
\end{corollary}


\begin{proof}   By
Proposition~\ref{locally special PIR} $R_M/A_M$ is a special PIR for
all maximal ideals $M$ of $R$ containing $A$.  Since every ideal of
a ring is an intersection of completely irreducible ideals, we may
write $A = \bigcap_{i}A_i$, where each $A_i$ is completely
irreducible.   Since $A_i$ is completely irreducible, $A_i$ is
primal; hence $A_i = (A_i)_{(M_i)}$ for some maximal ideal $M_i$ of
$R$.  Since $R/A_i$ is a zero-dimensional ring,
$M_i$ is by Lemma~\ref{PM} the unique maximal ideal of $R$ containing
$A_i$.  Moreover by Theorem~\ref{prime power} for each $i$, $A_i =
M_i^{e_i}+A$ for some  $e_i>0$.    Hence $A_i = (M_i^{e_i})_{(M_i)}
+ A_{(M_i)}$. Since $R_{M_i}$ is a valuation ring, the ideals
$A_{(M_i)}$ and $(M_i^{e_i})_{(M_i)}$ are comparable, and since
$A_{(M_i)} \subseteq A_i$, we conclude $A_i = (M_i^{e_i})_{(M_i)}$.
As noted in the proof of (ii) $\Rightarrow$ (i) of
Theorem~\ref{prime power}, $M_i^{e_i} = (M_i^{e_i})_{(M_i)}$.  Thus
$A_i = M_i^{e_i}$, proving that $A$ is an intersection of powers of
maximal ideals of $R$. \end{proof}




\begin{theorem}\label{main}  The following statements are equivalent for a
ring $R$.
\begin{itemize}
\item[(i)]  Every ideal of $R$ can be represented uniquely as an irredundant
intersection of completely irreducible ideals.


\item[(ii)] $R$ has a scattered prime spectrum and  $R_M$ is a special PIR
for each maximal ideal $M$ of $R$.


\item[(iii)] For every proper ideal $A$ of $R$, $R/A$ has a finitely
generated maximal ideal, and for every maximal ideal $M$ of $R$,
$R_M$ is a special PIR.


\item[(iv)] Every proper ideal $A$ of
$R$ is an irredundant intersection $A = \bigcap_i M_i^{e_i}$ of
powers of maximal ideals $M_i$. \end{itemize} \end{theorem}




\begin{proof} (i) $\Rightarrow$ (ii) Since an irreducible ideal cannot
be expressed as an irredundant intersection of two distinct proper
overideals,  (i) implies that every irreducible ideal of $R$ is
completely irreducible, so by Proposition~\ref{locally special PIR}
$R_M$ is a special PIR for every maximal ideal $M$ of $R$.  Thus $R$
is zero-dimensional and by Corollary~\ref{arithmetical case}
$\Spec(R)$ is a scattered space.



(ii) $\Rightarrow$ (iii) Let $A$ be a proper ideal of $R$.    By
Theorem~\ref{irredundant irreducibles} there exists $y \in R$ and a
maximal ideal $M$ of $R$ such that $M = yR+ \sqrt{A}$.  Since $R_M$
is a special PIR, there exists $z \in M$ such that $MR_M = zR_M$.
Define $B = (y,z)R+A$.  We claim $M = B$.  Clearly $MR_M = BR_M$ and
by construction $B$ has radical $M$. Therefore $MR_N = BR_N$ for
each maximal ideal $N$ of $R$, so $B = M$.  Hence $M/A$ is a
finitely generated ideal of $R/A$.


(iii) $\Rightarrow$ (iv)  By (iii) $R$ is an arithmetical ring since
each localization at a maximal ideal is a valuation ring.  Let $A$
be a proper ideal of $R$.    By Theorem~\ref{irredundant
irreducibles} and Corollary~\ref{arithmetical case} $A$ is an
irredundant intersection of irreducible ideals, and by
Proposition~\ref{locally special PIR} every primal (hence
irreducible) ideal is completely irreducible. Also, it follows from
Corollary~\ref{decompose} that every completely irreducible proper
ideal of $R$ is a power of a maximal ideal of $R$.  Hence $A$ is an
irredundant intersection of powers of maximal ideals of $R$.




(iv) $\Rightarrow$ (i) From (iv) it follows that every proper
irreducible ideal of $R$ is a power of a maximal ideal of $R$, so
$R$ is  zero-dimensional. Thus,
by Theorem~\ref{prime power}, $R_M$ is a special PIR  for each maximal 
ideal $M$
of $R$.  Now to show that (i) holds, it suffices by (iv) and
Theorem~\ref{3.5} to show that $M^e$ is completely irreducible for
each maximal ideal $M$ of $R$ and integer $e>0$. As noted in the
proof of Theorem~\ref{prime power} (ii) $\Rightarrow$ (i), $M^e =
(M^e)_{(M)}$.  Since $R_M$ is a valuation ring, the ideal $M^eR_M$
of $R_M$ is irreducible, so $M^e$, as the preimage of $M^eR_M$ under
the mapping $R \rightarrow R_M$, is an irreducible ideal of $R$.
Thus every power of a maximal ideal of $R$ is irreducible.  Since
for each maximal ideal $M$ of $R$ the ring $R_M$ is as special PIR,
we have by Proposition~\ref{locally special PIR} that every
irreducible ideal, hence every power of a maximal ideal, is
completely irreducible.
\end{proof}



\begin{remark}\label{semilocal}
(i)  A Noetherian ring  satisfying the equivalent conditions of
Theorem~\ref{main} is necessarily semilocal since it is
zero-dimensional. Conversely, any semilocal ring satisfying the
equivalent conditions of Theorem~\ref{main} is Noetherian.    It
follows that such a ring  is a finite product of special PIRs (see
for example Corollary~\ref{ZPI corollary}).


(ii) Rings that are proper homomorphic images of the almost Dedekind
domains described in Remark~\ref{Boolean} (ii) satisfy the
equivalent conditions of Theorem~\ref{main}.  In particular, if $X$
is a compact scattered space having both isolated and non-isolated
points, then there is an almost Dedekind domain $R$ such that
$\Max(R)$ is homeomorphic to $X$.  Since $\Max(R)$ has an isolated
point, the Jacobson radical $J$ of $R$ is nonzero.  The ring $R/J$
has a scattered prime spectrum and $R/J$ satisfies the equivalent
conditions of Theorem~\ref{main}.  Since $\Spec(R/J)$ has a
non-isolated point, $R/J$ is not semilocal; hence by (i) $R/J$ is
not Noetherian.
   \end{remark}




A ring satisfying the equivalent conditions of Theorem~\ref{main} is
locally Noetherian at each maximal ideal, but, as noted in
Remark~\ref{semilocal}, the ring itself need not be Noetherian.
However from Theorem~\ref{main} (iii) it follows that every non-unit
in such a ring is contained in a finitely generated maximal ideal.





\begin{corollary}\label{almost Dedekind}  The following statements are
equivalent for a domain $R$.
\begin{itemize}
\item[(i)]  Every  nonzero proper
ideal of $R$ has a unique representation as an irredundant
intersection of powers of maximal ideals.


\item[(ii)]  $R$ is almost Dedekind and for every proper ideal $A$,
the ring $R/A$ has at least one finitely generated maximal ideal.



%\item[(iii)] $R$ is almost Dedekind and for each nonzero
%radical ideal $J$  there is a maximal ideal $M$
%and a finitely generated ideal $B$
%such that $M = \sqrt{J+B}$.


\item[(iii)] $R$ is an almost Dedekind domain and for each proper nonzero
ideal $A$ of $R$, $\Spec(R/A)$ is a scattered space.


\item[(iv)] Every nonzero proper ideal
of $R$ can be represented uniquely
as an irredundant intersection of completely irreducible ideals.


\end{itemize}
\end{corollary}


\begin{proof}
(i) $\Rightarrow$ (ii) Statement (i) implies that $R/A$ is
zero-dimensional for every nonzero proper ideal $A$ of $R$.
Therefore $\dim R \le 1$. If $A$ and $B$ are nonzero ideals
contained in the maximal ideal $M$, then by Theorem~\ref{main},
$R_M/(A \cap B)R_M$ is a special PIR.  Thus $AR_M$ and $BR_M$ are
comparable ideals of $R_M$. It follows that $R_M$ is a valuation
domain. Indeed,  since $R_M/AR_M$ is a special PIR for any nonzero
ideal $A \subseteq M$, it follows that $R_M$ is a Noetherian
valuation domain. This proves that $R$ is almost Dedekind.   Now let
$A$ be  a proper ideal of $R$.  If $A$ is nonzero,  then by (i)
every proper ideal of $R/A$ can be represented as an irredundant
intersection of powers of maximal ideals of $R/A$.  Hence by
Theorem~\ref{main} $R/A$ has a finitely generated maximal ideal. On
the other hand, if $\dim R = 1$ and $A = 0$, let $a$ be a nonzero
nonunit of $R$.   Then $R/aR$ has a finitely generated maximal
ideal, say $M/aR$, where $M$ is a maximal ideal of $R$.  Thus $M$ is
a finitely generated ideal of $R$.


(ii) $\Rightarrow$ (iii) Apply Theorem~\ref{irredundant
irreducibles}.



(iii) $\Rightarrow$ (iv)  If $A$ is a nonzero proper ideal of $R$,
then by (iii),  $R/A$ satisfies the equivalent conditions of
Theorem~\ref{main}.  Hence  $A$ is an irredundant intersection of
completely irreducible ideals.




(iv) $\Rightarrow$ (i)  Suppose $R$ is not a field.  For any nonzero
ideal $B$ of $R$, $R/B$ is by Theorem~\ref{main} a zero-dimensional
ring.  Hence $R$ is a one-dimensional domain.  Let $A$ be a proper
nonzero ideal of $R$.
  Then $A$ is an irredundant intersection $A =
\bigcap_{i \in I} A_i$ of completely irreducible ideals $A_i$ of
$R$. Let $i \in I$.  For each maximal ideal $M$ of $R$  containing
$A_i$, $R_M/(A_i)_M$ is a special PIR, so by
Proposition~\ref{locally special PIR} every primal ideal of $R/A_i$
is completely irreducible.  Since $A_i$ is completely irreducible,
we have by Corollary~\ref{decompose} that $A_i$ is a power of a
maximal ideal of $R$.    Thus (i) follows.
      \end{proof}


\begin{remark} As mentioned in Remark~\ref{Boolean}, it is
shown in \cite{O} that every compact scattered space can be realized
as $\Max(R)$ for an almost Dedekind domain $R$. If $R$ is such a
domain, then for every nonzero ideal $A$, $\Spec(R/A)$ is a
scattered space and $R$ satisfies the equivalent conditions of
Corollary~\ref{almost Dedekind}.
\end{remark}







\begin{corollary} \label{main corollary}
The following statements are equivalent for a ring $R$.
\begin{itemize}
\item[(i)]  Every proper ideal of $R$ can be represented uniquely as a finite
irredundant intersection of completely irreducible ideals.
\item[(ii)] Every proper ideal
of $R$ is a finite product of  maximal ideals.
\item[(iii)]  $R$ is isomorphic to a
finite direct product of special PIRs. \end{itemize}
\end{corollary}




\begin{proof} (i) $\Rightarrow$ (ii) By Theorem~\ref{main} $R$ is a
zero-dimensional ring.  Now $(0) = A_1 \cap \cdots \cap A_n$ for
some completely irreducible ideals $A_i$ of $R$.  For each $i$,
$A_i$ is primal, so $A_i = (A_i)_{(M_i)}$ for some maximal ideal
$M_i$ of $R$. By Lemma~\ref{PM} $M_i$ is the unique maximal ideal of
$R$ containing $A_i$. It follows that the only  maximal ideals of
$R$ are $M_1,M_2,\ldots,M_n$. Now apply Theorem~\ref{main} to obtain
(ii).




(ii) $\Rightarrow$ (iii)  By (ii) the zero ideal $(0)$
can be represented uniquely as a finite intersection of powers
$M_i^{e_i}$, $i =1,\ldots,n$, of distinct  maximal ideals $M_i$ of
$R$.  Thus $R \cong \prod_{i=1}^n R/M_i^{e_i}$. By
Theorem~\ref{main}, each $R/M_i^{e_i}$ is a special PIR.




(iii) $\Rightarrow$ (i) Since $R$ is Noetherian, every ideal of $R$
is an intersection of finitely many irreducible ideals.  Also, by
Proposition~\ref{locally special PIR} every irreducible ideal of $R$
is completely irreducible, so  (i) follows from Theorem~\ref{3.5}.
\end{proof}



In view of  the fact that every ideal of a ring is an intersection
of irreducible ideals, if a ring has the property that every
irreducible ideal is a power of a prime ideal, then every proper
ideal is an intersection of powers of prime ideals. This motivates
us to ask:



\begin{question} If every proper ideal $A$ of a ring $R$ is an intersection $A
= \bigcap_{i} P_i^{e_i}$ of powers of prime ideals, does $R$ satisfy
the equivalent conditions of Theorem~\ref{prime power}?
\end{question}




Since every  ideal of a ring  is an intersection of completely
irreducible ideals, this question is equivalent to:  {\it If 
every completely irreducible proper ideal of $R$ is a
power of a prime ideal, is every irreducible proper ideal of $R$ a
power of a prime ideal?}



\section{A generalization of ZPI rings}



A ring is a ZPI (Zerlegung in Primideale) ring if every proper ideal
in the ring is a product of prime ideals.  It is well known that the
ZPI domains are precisely the Dedekind domains.  The general case is
also well-understood: a ring is a ZPI ring if and only if it is
isomorphic to a finite product of Dedekind domains and special PIRs
(see Chapter IX, Section 2, of \cite{LM}).  As an application of our
previous results, we  consider in this section  a related class of
rings, those for which every proper ideal is an irredundant
intersection of powers of prime ideals.  We observe in
Remark~\ref{example} and Corollary~\ref{ZPI corollary} that this
class of rings properly  contains  the class of ZPI rings.




\begin{theorem}\label{ZPI generalization}
The following statements are equivalent for a ring $R$.
\begin{itemize}
\item[(i)]  Every proper ideal $A$ of $R$
can be represented uniquely as an irredundant intersection $A =
\bigcap_{i}P_i^{e_i}$ of powers  of prime ideals $P_i$.
\item[(ii)]
Every proper ideal $A$ of $R$ is an irredundant intersection $A =
\bigcap_{i} P_i^{e_i}$ of powers of prime ideals $P_i$.
\item[(iii)]  Every ideal of $R$ can be represented
as an irredundant intersection of irreducible ideals and
$R_M$ is  a Noetherian valuation ring for each maximal ideal $M$ of
$R$.
%\item[(iv)]  Every radical ideal of $R$ has a Zariski-Samuel associated prime
%and for each maximal ideal $M$ of $R$, $R_M$ is a Noetherian valuation ring.
                      \end{itemize} \end{theorem}




\begin{proof} Before proving Theorem~\ref{ZPI generalization},
we  establish ($\star$):  {\it If $R$ is a ring in which every
proper irreducible ideal is a power of a prime ideal, then every
power of a prime ideal of $R$ is an irreducible ideal.}  By
Theorem~\ref{prime power}, $R_M$ is a Noetherian valuation ring for
each maximal ideal $M$ of $R$.   Let $P$ be a prime ideal of $R$ and
let $k>0$. Let  $M$ be a maximal ideal of $R$ containing $P$.  It
suffices  to prove that $(P^k)_{(M)} = P^k$, since this implies that
$P^k$, as a preimage of an irreducible ideal under the mapping $R
\rightarrow R_M$, is itself irreducible. If $P$ is a maximal ideal
of $R$ (i.e. $P = M$), then as in the proof of (ii) $\Rightarrow$
(i) of Theorem~\ref{prime power}, $(P^k)_{(M)} = P^k$. Otherwise if
$P$ is not a maximal ideal of $R$, then $P$ is  a minimal prime
ideal and $PR_M = (0)R_M$. Hence $P^kR_M = PR_M$, so that
$(P^k)_{(M)} = P_{(M)} = P$.  Thus for any maximal ideal $M$
containing $P$, $(P^k)_{(M)} = P$. Now $P^k = \bigcap_{N \supseteq
P}(P^k)_{(N)}$, where $N$ ranges over the set of maximal ideals of
$R$ containing $P$. Therefore we conclude $P^k = P$, so that $P^k$
is clearly an irreducible ideal of $R$. This shows that every power
of a prime ideal is an irreducible ideal.




We now verify the theorem.





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




(ii) $\Rightarrow$ (iii) Since an irreducible ideal cannot be
expressed as an irredundant intersection of distinct overideals,
(ii) implies that  every proper irreducible ideal is a power of a
prime ideal. By Theorem~\ref{prime power}, for each maximal ideal
$M$, $R_M$ is a Noetherian valuation ring. By our above observation
($\star$),  every power of a prime ideal is irreducible so the
assertion that every proper ideal of $R$ can be represented  as an
irredundant intersection of irreducible ideals is a consequence of
(ii).





(iii) $\Rightarrow$ (i) By Theorem~\ref{prime power},  every proper irreducible
ideal of $R$ is a power of a prime ideal of $R$.  If $A$ is a proper
ideal of $R$, then by (iii) there is an irredundant intersection $A
= \bigcap_i P_i^{e_i}$ of powers of prime ideals $P_i$ of $R$.
By ($\star$), the set of proper irreducible
ideals of $R$ is precisely the set of powers of prime ideals of $R$.
Thus from Theorem~\ref{3.5} it follows that the representation $A =
\bigcap_i P_i^{e_i}$ is unique among irredundant intersections of
powers of prime ideals.                \end{proof}


\begin{corollary}\label{ZPI domain}  A domain $R$ satisfies the equivalent
conditions of Theorem~\ref{ZPI generalization} if and only if $R$ is
an almost Dedekind domain such that for each proper ideal $A$ of
$R$, the ring $R/A$ has at least one finitely generated maximal ideal.
\end{corollary}


\begin{proof}  Apply Theorem~\ref{ZPI generalization} and Corollary~\ref{almost
Dedekind}.
\end{proof}



\begin{remark} \label{represent}
If $P$ is a nonmaximal prime ideal of a ring $R$ satisfying (i) -
(iii) of Theorem~\ref{ZPI generalization}, then as noted in the
proof of ($\star$) of this theorem, $P^k = P$ for all $k>0$.  Thus
by (i) every proper ideal $A$ of $R$ is an irredundant intersection:
$A = (\bigcap_{i \in I}M_i^{e_i}) \cap (\bigcap_{j \in J} P_j)$,
where each $M_i$ is a maximal ideal of $R$ and each $P_j$ is a
one-dimensional prime ideal of $R$.  (We admit here the possibility
that $I$ or $J$ is empty.)  Moreover,  since $R$ is arithmetical,
the set $\{M_i\} \cup \{P_j\}$ consists of pairwise comaximal prime
ideals. \end{remark}




\begin{remark}\label{example} (i)  By Theorem~\ref{main},  if $R$ is an
almost Dedekind domain with nonzero Jacobson radical and $R$  has a
scattered prime spectrum and at least one non-isolated point, then
$R$ satisfies the conditions of Corollary~\ref{ZPI domain} but is
not a ZPI ring.  See Remark~\ref{Boolean} for such an example.


(ii)  If $R = R_1 \times R_2 \times \cdots \times R_n$ is a finite
product of rings satisfying the equivalent conditions of
Theorem~\ref{ZPI generalization}, then $R$ also satisfies the
conditions of the theorem.  To prove this we may assume $n =2$. Let
$e_1$ and $e_2$ be idempotent elements of $R$ such that $Re_i =
R_i$, $i = 1, 2$.  A maximal ideal $M$ of $R$ contains either $e_1$
or $e_2$, but not both. If $e_1 \in M$, then $M = R_1 \times M_2$,
where $M_2$ is a maximal ideal of $R_2$ and $R_M \cong (R_2)_{M_2}$.
Similarly, if $e_2 \in M$, then $M = M_1 \times R_2$, where $M_1$ is
a maximal ideal of $R_1$ and $R_M \cong (R_1)_{M_1}$.  Thus each
localization of $R$ at a maximal ideal is a Noetherian valuation
ring. If $B = B_1 \times B_2$ is an arbitrary ideal of $R$, we
obtain an irredundant representation of $B$ as an intersection of
irreducible ideals as follows: there exist irreducible ideals
$A_{1i}$ of $R_1$ that intersect irredundantly in $B_1$ and
irreducible ideals $A_{2i}$ of $R_2$ that intersect irredundantly in
$B_2$. Each of the ideals $A_{1i} \times R_2$ and $R_1 \times
A_{2i}$ is irreducible in $R$ and $B = B_1 \times B_2$ is
represented irredundantly as the intersection of this collection of
ideals. Therefore $R = R_1 \times R_2$ also satisfies the conditions
of Theorem~\ref{ZPI generalization}.


(iii)  If $R$ is a finite product of domains satisfying the
conditions of Corollary~\ref{ZPI domain} and rings satisfying the
conditions of Theorem~\ref{main}, then by (ii) $R$ is a
one-dimensional ring with zero-divisors having the property that
every proper ideal is an irredundant intersection of powers of prime
ideals.        \end{remark}





\begin{corollary} \label{ZPI corollary} The following statements are equivalent
for a ring $R$.
\begin{itemize}
\item[(i)]  $R$ is a ZPI ring.
\item[(ii)]
There is an isomorphism of rings, $R \cong R_1 \times R_2 \times
\cdots \times R_n$, where each $R_i$ is a Dedekind domain or a
special PIR.
\item[(iii)] Every proper ideal $A$ of $R$ is a finite intersection
$A = \bigcap_{i=1}^{n}P_i^{e_i}$ of powers of prime ideals $P_i$ of
$R$.
    \item[(iv)]  Every proper ideal $A$ of $R$ is a finite
irredundant intersection $A = \bigcap_{i=1}^{n}P_i^{e_i}$ of powers
of prime ideals $P_i$ of $R$, and this representation of $A$ is
unique among irredundant representations of $A$ as an intersection
of powers of prime ideals.
\end{itemize}
\end{corollary}




\begin{proof}
(i) $\Leftrightarrow$ (ii)   This equivalence can be found in
Theorem IX.9.10 of \cite{LM}.


(ii) $\Rightarrow$ (iii) Write $R = Re_1 \oplus \cdots \oplus Re_n$,
where the $e_i$ are orthogonal idempotents and for each $i$, $R_i
\cong Re_i$.  For each $i$ define $B_i = \sum_{j \ne i}Re_j$.    If
$A$ is an ideal of $R$, then $A = \bigcap_{i=1}^n (A+ B_i)$, and
since each $R_i$ is a Dedekind domain or a special PIR, each ideal
$A+B_i$ is a finite intersection of powers of prime ideals.  Hence
(iii) follows.



(iii) $\Rightarrow$ (iv)  By (iii) $R$ satisfies the equivalent
conditions of Theorem~\ref{ZPI generalization}, so that by property
($\star$) in the proof of this theorem, every power of a prime ideal
of $R$ is irreducible. In an arithmetical ring, an irredundant
intersection of irreducible ideals of $R$ is unique
(Theorem~\ref{3.5}).  Thus (iv) follows.



(iv) $\Rightarrow$ (i)
Let $A$ be a proper ideal of $R$.   By Theorem~\ref{ZPI
generalization} and Remark~\ref{represent},  $A$ is an irredundant
intersection of comaximal powers of prime ideals.  By the uniqueness
assertion in (iv) this  intersection must be finite, so $A$ is a
product of these same powers of prime ideals.
\end{proof}




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\end{document}

