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\begin{document}


\title{On decomposing ideals into products \\of comaximal ideals}
\author{J. W. Brewer and W. J. Heinzer}

\maketitle
\begin{flushleft}
Department of Mathematical Sciences,
Florida Atlantic University,
Boca Raton, Florida  33431
{\em E-mail address: brewer@fau.edu}

\vspace{.15in}

Department of Mathematics, Purdue University, West Lafayette,
Indiana 47907-1395
{\em E-mail address: heinzer@math.purdue.edu}

\end{flushleft}


\abstract{
We consider integral domains $D$ for which  each
nonzero ideal $A$ (or each nonzero principal ideal $aD$)
can be written as a product
$Q_1\cdot Q_2 \cdots Q_n$, where the $Q_i$ are pairwise
comaximal and have some additional property. We determine
the structure of integral domains having this property
in the following cases: each $Q_i$ has prime radical; each
$Q_i$ is primary; each $Q_i$ is the power of a prime ideal. \hfill \ 
}

\footnotetext{
{\em 2000 Mathmatics Subject Classification Numbers:} 13A15, 13C05,
13F05, 13G05.

{\em Keywords and phrases:} comaximal ideals, prime radical,
primary ideal,  Pr\"ufer domain,  Dedekind domain.}


\markright{On decomposing ideals into products of comaximal ideals}

\baselineskip17pt

\section{Introduction}

A classical result in commutative ring theory as recorded by W. Krull in
\cite[page 24]{Krull} asserts that if $D$ is a one-dimensional Noetherian
domain, then each nonzero ideal $A$ of $D$ is a product of pairwise comaximal
primary ideals. Another result along these lines is a variant of the Chinese
Remainder Theorem: an ideal $A$ of a commutative ring $R$ is a product of
pairwise comaximal primary ideals if and only if $R/A$ is a finite direct sum
of rings $R_{i}$ where $(0)$ is a primary ideal in each $R_{i}$. In
particular, if $\dim(R/A)=0$ and $A$ is contained in only finitely many prime
ideals, the Chinese Remainder Theorem implies that $A$ is a finite product of
primary ideals. This leads us to wonder:

\medskip
\noindent
{\bf Question. }  Which integral domains $D$ have the
property that each nonzero ideal $A$ of$\ D\,$can be written as a product
$$
A=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}%
$$
where the $Q_{i}$'s are pairwise comaximal and have some additional property?

\medskip

We also consider this question for nonzero principal ideals $aD$ of D. In the
case where $aD$ is a principal ideal, a useful observation
is that the factors $Q_{i}$ are invertible ideals.
\smallskip

Obviously, any Dedekind domain has the property; in this case, each $Q_{i}$ is
a power of a maximal ideal. In this paper we settle the question in the
following cases: (i) each $Q_{i}$ has prime radical (Theorem \ref{prime
radical}); (ii) each $Q_{i}$ is primary (Theorem \ref{primary}); and  (iii) 
each
$Q_{i}$ is the power of a prime (Theorem \ref{prime powers}).

We became interested in this property from reading the paper \cite{SS} by
S\'{a}ez-Schwedt and S\'{a}nchez-Giralda. It is well known in the classical
theory of linear dynamical systems over fields that canonical forms exist for
controllable systems of any dimension. In the paper \cite{BK} of J. Brewer and
L. Klingler, it is shown by means of representation-theoretic methods, that
canonical forms are not likely to exist over arbitrary principal ideal
domains. The problem treated in \cite{SS} is to try to determine a canonical
form for two-dimensional controllable systems over principal ideal domains
(and Dedekind domains). In \cite{SS}, the property above for all nonzero
principal ideals was the property the authors needed in order to carry out
their successful program.

Questions similar to the ones considered here, but without the
comaximality, have received attention by others. We thank
the referee for suggesting the following papers and the
references listed there  \cite{AM}, \cite{L1}, \cite{L2}.
We also thank Laszlo Fuchs for several helpful comments and
for sending us the article \cite{FM} concerning ideal theory
in Pr\"ufer domains of finite character. As Professor Fuchs 
noted when we exchanged articles, our paper contains a result
identical with one of the theorems in \cite{FM}. We arrived 
at the theorem from completely different directions.

\section{Main Results}

We begin with the weakest requirement on the factors and the case of nonzero
principal ideals.

\begin{theorem}
\label{prime radical}Let $D$ be an integral domain. The following are 
equivalent:

\begin{enumerate}
\item  Each nonzero principal ideal $aD$ of $D$ can be written in the form
$aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ has prime
radical and the $Q_{i}$'s are pairwise comaximal.

\item $D$ has the following properties:

\begin{enumerate}
\item  For any maximal ideal $M$ of $D$, the set of prime ideals of $D$
contained in $M$ is linearly ordered under inclusion.

\item  Each nonzero principal ideal of $D$ has only finitely many minimal 
primes.
\end{enumerate}
\end{enumerate}
\end{theorem}

\begin{proof}
$(1)\Rightarrow(2):$ Suppose that $M$ is a maximal ideal of $D$ and that
$P_{1}$and $P_{2}$ are incomparable prime ideals of $D$ contained in $M$. Let
$a$ be in $P_{1}$ not $P_{2}$ and $b$ be in $P_{2}$ not $P_{1}$. The principal
ideal $abD$ can be written in the form $Q_{1}\cdot Q_{2}\cdot\cdots\cdot
Q_{n}$, where each $Q_{i}$ has prime radical and the $Q_{i}$'s are pairwise
comaximal. Since $abD\subseteq M$, some $Q$ is contained in $M$, say $Q_{1}$.
Moreover, since the $Q_{i}$'s are pairwise comaximal, only $Q_{1}$ is contained
in $M$. Thus, since $P_{1}$ and $P_{2}$ are contained in $M$ and contain
$abD$, $Q_{1}\subseteq P_{1}\cap P_{2}$. But, $Q_{1}$ cannot have prime
radical, for suppose that $\sqrt{Q_{1}}=$ $P$, prime. Then $P$ is contained in
$P_{1}\cap P_{2}$. Also, $P$ must contain one of $a$ or $b$. But this is a
contradiction to the choice of $a$ and $b$ and it follows that condition (a) 
holds.

Let $a$ be a nonzero element of $D$. Write $aD=Q_{1}\cdot Q_{2}\cdot
\cdots\cdot Q_{n}$, where each $Q_{i}$ has prime radical $P_{i}$ and the
$Q_{i}$'s are pairwise comaximal. If $P$ is a minimal prime of $aD$, then
$\ P$ contains some $Q_{i}$ and hence, $P=P_{i}$. It follows that
$\{P_{1},P_{2},\ldots,P_{n}\}$ is the set of minimal prime ideals of $aD$.
This proves condition (b).

$(2)\Rightarrow(1):$ Suppose that $aD$ is a nonzero principal ideal of $D$ and
that $P_{1},P_{2},\ldots,P_{n}$ are the minimal primes of $aD$. By condition
(a), the $P_{i}$'s, are pairwise comaximal. Hence, there exist $x_{i}$ in
$P_{i}$ and $y_{i}$ in $P_{j}$ for $j$ $\neq$ $i$ such that $x_{i}+y_{i}=1$.
Let $D_{i}=D[1/y_{i}]$, the localization of $D$ at the powers of the element
$y_{i}$. If $Q_{i}=$ $aD_{i}\cap D$, then $Q_{i}$ has radical $P_{i}$. Let $M$
be a maximal ideal of $D$. Then either $aD\subseteq M$ or $aD\not\subseteq M$.
If $aD\not\subseteq M$, then $aD_{M}\cap D=D$. If $aD\subseteq M$, then
$M\supset P_{i}$ for some $i$, and hence $y_{i}\notin M $; it follows that
$D_{M}$ is a term in the intersection $D[1/y_{i}]=\bigcap D_{P}$, where $P$ is
prime and $y_{i}\notin P$. Thus, $D\subseteq\bigcap\limits_{i=1}^{n}D_{i}\cap
D[1/a]\subseteq\bigcap\limits_{M\;\max\;}D_{M}=D$. Now, for every ideal $A$ of
$D$, $A=\bigcap\limits_{M\;\max\;}AD_{M}\cap D$. \ It follows that
$aD=Q_{1}\cap Q_{2}\cap\cdots\cap Q_{n}$. Moreover the $Q_{i}$'s are pairwise
comaximal, so their intersection is their product. This completes the 
proof.\medskip
\end{proof}

The proof of Theorem \ref{prime radical} can be easily modified to prove

\begin{theorem}
\label{ideal prime radical}Let $D$ be an integral domain. The following are 
equivalent:

\begin{enumerate}
\item  Each nonzero ideal $A$ of $D$ can be written in the form $A=Q_{1}\cdot
Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ has prime radical and the
$Q_{i}$'s are pairwise comaximal.

\item $D$ has the following properties:

\begin{enumerate}
\item  For any maximal ideal $M$ of $D$, the set of prime ideals of $D$
contained in $M$ is linearly ordered under inclusion.

\item  Each nonzero ideal of $D$ has only finitely many minimal primes.
\end{enumerate}
\end{enumerate}
\end{theorem}

\begin{remark}
\emph{Note that condition 2(a) above is equivalent to the following: If $P$
and $Q$ are nonzero prime ideals of $D$, then either $P$ and $Q$ are comaximal
or $P$ and $Q$ are comparable. Viewed in this way, conditions 2(a) and 2(b)
are topological conditions on the prime spectrum of $D$ with the Zariski
topology. More specifically, 2(a) says that any two proper closed irreducible
subsets of $Spec(D)$ are either comparable or disjoint; 2(b) says that each
closed subset of $Spec(D)$ has only finitely many irreducible components. The
paper \cite{HOCH} of M. Hochster establishes the existence of many examples of
integral domains having these two properties. }
\end{remark}

There are situations\ in which Theorem \ref{prime radical} and Theorem
\ref{ideal prime radical} can be combined into one result as the following
proposition illustrates.

\begin{proposition}
\label{Finite maximal} Suppose that the integral domain $D$ has the property
that each nonzero prime ideal of $D$ is contained in only finitely many
maximal ideals. The following are equivalent:

\begin{enumerate}
\item  Each nonzero ideal $A$ of $D$ can be written in the form $A=Q_{1}\cdot
Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ has prime radical and the
$Q_{i}$'s are pairwise comaximal.

\item  Each nonzero principal ideal $aD$ of $D$ can be written in the form
$aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ has prime
radical and the $Q_{i}$'s are pairwise comaximal.

\item $D$ has the following properties:

\begin{enumerate}
\item  For any maximal ideal $M$ of $D$, the set of prime ideals of $D$
contained in $M$ is linearly ordered under inclusion.

\item  Each nonzero principal ideal of $D$ has only finitely many minimal 
primes.
\end{enumerate}
\end{enumerate}
\end{proposition}

\begin{proof}
That (1) implies (2) is obvious and that (2) is equivalent to (3) is the
content of Theorem \ref{prime radical}. Thus, we have only to prove that (3)
implies (1). Since condition (2a) is the same in both theorems, we have to
verify that each nonzero ideal $A$ of $D$ has only finitely many minimal
primes. Pick a nonzero element $a\in A$ and let $P_{1},P_{2},\ldots,P_{n}$ be
the minimal primes of $aD$. By hypothesis, each $P_{i}$ is contained in only
finitely many maximal ideals, say $P_{i}$ is contained in $M_{i1}%
,\ldots,M_{ik_{i}}$. Let $P$ be a minimal prime of $A$. Then $aD\subseteq P$
and $P\supseteq P_{j}$ for some $j$. It follows that $P\subseteq M_{jl}$ for
some $l$ between $j1$ and $jk_{j}$. ($P$ might be contained in more that one such
$M_{jl}$, but that causes no difficulty.) If $Q$ is another minimal prime of
$A$ that contains $P_{j}$, then $Q\subseteq M_{ji}$ for $i\neq l$ since the
primes contained in a given maximal ideal are comparable. Therefore, there are
at most $k_{j}$ distinct minimal primes of $A$ that contain $P_{j}$. Since
each minimal prime of $A$ must contain some $P_{j}$, it follows that there are
at most $k_{1}+k_{2}+\cdots k_{n}$ minimal primes of $A$.\medskip
\end{proof}

We record the following corollary to Theorem \ref{ideal prime radical}.

\begin{corollary}
\label{Prufer}Let $D$ be a Pr\"{u}fer domain. If each nonzero element of $D$
belongs to only finitely many maximal ideals of $D$, then each nonzero ideal
of $D$ can be written in the form $Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$,
where each $Q_{i}$ has prime radical and the $Q_{i}$'s are pairwise comaximal.
\end{corollary}

\begin{proof}
If $D$ is a Pr\"{u}fer domain, then each localization of $D$ at a prime ideal
is a valuation domain. In particular, the prime ideals of $D$ contained in a
given maximal ideal of $D$ are linearly ordered under inclusion. Let $A$ be a
nonzero ideal of $D$ with $P$ a minimal prime of $A$. Since each nonzero
element of $D$ belongs to only finitely many maximal ideals, the same is true
for $A$. Let those maximal ideals be $M_{1},M_{2},\ldots,  M_{k}$. Then
$P\subseteq M_{j}$ for some $j$. Since the prime ideals of $D$ contained in a
given maximal ideal of $D$ are linearly ordered, any $M_{i}$ can contain at
most one minimal prime of $A$. Thus, $A$ has at most $k$ minimal primes and
the result follows from Theorem \ref{ideal prime radical}.\medskip
\end{proof}

\begin{remark}
\emph{Laszlo Fuchs has pointed out to us that the proof of Corollary \ref{Prufer}
also applies for a non-Pr\"ufer domain $D$:  if  each nonzero element of $D$
belongs to only finitely many maximal ideals and if the prime ideals  
contained in a
given maximal ideal of $D$ are linearly ordered under inclusion, 
then   each nonzero ideal
of $D$ can be written in the form $Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$,
where each $Q_{i}$ has prime radical and the $Q_{i}$'s are pairwise comaximal.}
\end{remark}


There exist Pr\"{u}fer domains which satisfy the hypothesis of Theorem
\ref{prime radical} and which have nonzero elements contained in infinitely
many maximal ideals. For example, if $\mathbb{Q}$ denotes the field of
rational numbers, $\mathbb{Z}$ the ring of integers,and $X$ an indeterminate,
let $M$ be the maximal ideal $X\mathbb{Q}[X]_{(X)}$ of $\mathbb{Q}[X]_{(X)}$.
Then the domain $D=\mathbb{Z}+M$ is a two-dimensional Pr\"{u}fer domain having
a unique prime ideal $M$ of height one and having infinitely many maximal
ideals corresponding to the maximal ideals of $\mathbb{Z}$. The elements of
$M$ are contained in infinitely many maximal ideals of $D$ while the elements
of $D$ not in $M$ are in only finitely many prime ideals. Thus, $D$ satisfies
the hypothesis of Theorem \ref{prime radical}

We next present an example to illustrate the fact that Theorems \ref{prime
radical} and \ref{ideal prime radical} cannot always be combined.\ Thus, for
the case when the ideals $Q_{i}$ are only assumed to have prime radical, the
condition for all principal ideals is not equivalent to the condition for all
ideals. More specifically, we construct a Pr\"{u}fer domain $D$ having the
following two properties: Each nonzero principal ideal of $D$ has only
finitely many minimal primes; $D$ has a nonzero ideal having infinitely many
minimal primes. Since a Pr\"{u}fer domain has the property that any two prime
ideals contained in a given maximal ideal are comparable, this example
effectively separates Theorem \ref{prime radical} from Theorem \ref{ideal
prime radical}. Thus, for this Pr\"{u}fer domain $D$, each nonzero principal
ideal $aD$ can be written in the form $aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot
Q_{n}$, where each $Q_{i}$ has prime radical and the $Q_{i}$'s are pairwise
comaximal, but there exists a nonzero ideal $A$ of $D$ which cannot be so 
written.\medskip

\begin{example}
\emph{We are going to construct a Pr\"{u}fer domain $D$ having for each
positive integer $n$ precisely two primes of height $n$, $M_{n}$  which is
maximal, and $P_{n}$ which is nonmaximal. This gives rise to the diagram below
describing the prime ideal lattice of $D$ (the union of the $P_{n}$ is a
unique maximal ideal of infinite height). }

\emph{%
\[%
\begin{tabular}
[c]{ccc}%
$\vdots$ &  & $\vdots$\\
$M_{4}$ &  & $P_{4}$\\
& $\nwarrow$ & $\uparrow$\\
$M_{3}$ &  & $P_{3}$\\
& $\nwarrow$ & $\uparrow$\\
$M_{2}$ &  & $P_{2}$\\
& $\nwarrow$ & $\uparrow$\\
$M_{1}$ &  & $P_{1}$\\
& $\nwarrow\nearrow$ & \\
& $(0)$ &
\end{tabular}
\]
}

\emph{We construct $D$ to be the union of a chain of Pr\"{u}fer subdomains
$D_{n}$, where $D_{n}$ is $n$-dimensional and $D_{n}$ has two maximal ideals 
of
height $n$ and has for each positive integer $m<n$ two primes of height $m$,
exactly one of which is maximal. The construction is such that there exists an
inclusion map of $D_{n}$ into $D_{n+1}$ so that the two maximal ideals of
$D_{n+1}$ of height $n+1$ both lie over the same maximal ideal of $D_{n}$.
Moreover there is also a nonmaximal prime of $D_{n+1}$ of height $n$ lying
over this same maximal ideal of $D_{n}$. For each $m<n$, the maximal ideal of
$D_{n+1}$ of height $m$ lies over the maximal ideal of $D_{n}$ of height $m$
and the nonmaximal prime of $D_{n+1}$ of height $m$ lies over the nonmaximal
prime of $D_{n}$ of height $m$. }

\emph{Let $A := \cap_{n=1}^{\infty}M_{n}$. We show that each of the $M_{n}$'s
is a minimal prime of $A$, so that $A$ has infinitely many minimal primes. This
follows from the fact that $A_{n}:=A\cap D_{n}$ is the Jacobson radical of
$D_{n}$ and $D_{n}$ has precisely $n+1$ maximal ideals, one of height $i$ for
each $i$ between $1$ and $n-1$ and two of height $n$. We have $A_{n}D_{n+1}$
properly contained in $A_{n+1}$ and $A=\cup_{n=1}^{\infty}A_{n}$. As we go up
from $D_{n}$ to $D_{n+1}$, our ideal $A$ is picking up more minimal primes and
in the union has infinitely many. Now each principal ideal of $D$ has the form
$aD$, where $a$ is in $D_{n}$ for some $n$. Then $aD_{n}$ has at most $n+1$
minimal primes and by the construction, $aD$ has at most $n+1$ minimal primes.
The point is that in passing from $D_{n}$ to $D_{n+1}$, there are three  prime
ideals of $D_{n+1}$ lying over the same maximal ideal of height $n$ of $D_{n}%
$. If $a$ is contained in one of them, then it is in all
three  and has precisely
one of them as a minimal prime. }

\emph{We now spell out a way to construct such a Pr\"{u}fer domain $D$. (We
will be intersecting finite families of valuation domains, and using the
theorem on independence of valuations as given by Nagata in \cite[(11.11),
page 38]{NAG}.) For $k$ a field and $x_{1},x_{2},\ldots$ indeterminates over
$k$, we construct a chain $D_{n}$ of the type discussed above, where $D_{n}$
has fraction field $k(x_{1},\ldots,x_{n})$ : Let $D_{1}$ be the intersection
of the DVRs $(k[x_{1}])_{(x_{_{1}})} = V_{1}$ and $(k[x_{1}])_{(x_{1}%
-1)}=W_{1}$. Let
\[
M_{11} = (x_{1} - 1)D_{1} \mbox{ and  } P_{11} = x_{1}D_{1}.
\]
denote the maximal ideals of $D_{1}$. We construct $D_{2}$ on the field
$k(x_{1},x_{2})$ as an intersection of
three  valuation domains. We denote by
$W_{1}(x_{2})$ the valuation domain $(k(x_{2})[x_{1}])_{(x_{1}-1)}$, i.e.,
$W_{1}[x_{2}]$ localized at the extension of the maximal ideal of $W_{1}$ to
this polynomial ring. This valuation domain $W_{1}(x_{2})$ is sometimes called
the Gaussian or trivial extension of $W_{1}$ to the simple transcendental
field extension generated by $x_{2}$. We define valuation domains $W_{2}$ and
$V_{2}$ of rank 2 on $k(x_{2},y_{2})$ both of which extend $V_{1}$. Let
$N_{2}$ denote the maximal ideal of $V_{1}(x_{2})$ and let
\[
V_{2}=(k[x_{2}])_{(x_{2})}+N_{2} \mbox{ and } W_{2}=(k[x_{2}])_{(x_{2}%
-1)}+N_{2}.
\]
Define $D_{2} := W_{1}(x_{2}) \cap W_{2} \cap V_{2}$. Using \cite[(11.11)]%
{NAG}, we see that $D_{2}$ has two maximal ideals of height $2$,
\[
M_{22} := (x_{2}-1)D_{2} \mbox{  and  } P_{22} := x_{2}D_{2},
\]
and two prime ideals of height one, $M_{21} := (x_{1}-1)D_{2}$ which is
maximal, and $P_{21} := N_{2} \cap D_{2}$ which contains $x_{1}$ and is not
maximal. Observe that $M_{22}, P_{22}$ and $P_{21}$ all have the property that
their intersection with $D_{1}$ is $P_{11}$. }

\emph{We construct $D_{3}$ on $k(x_{1},x_{2},x_{3})$ as an intersection of
four valuation domains. Let $N_{3}$ denote the maximal ideal of 
$V_{2}(x_{3})$.
Define
\[
V_{3} :=(k[x_{3}])_{(x_{3})}+N_{3} \mbox{   and  } W_{3}=(k[x_{3}%
])_{(x_{3}-1)}+N_{3}.
\]
Define $D_{3} := W_{1}(x_{2},x_{3}) \cap W_{2}(x_{3}) \cap W_{3} \cap V_{3}$.
Using \cite[(11.11)]{NAG}, we see that $D_{3}$ has two maximal ideals of
height $3$,
\[
M_{33} := (x_{3}-1)D_{3} \mbox{  and  } P_{33} := x_{3}D_{3},
\]
two prime ideals of height two, $M_{32} := (x_{2}-1)D_{3}$ which is maximal,
and $P_{32} := N_{3} \cap D_{3}$ which contains $x_{2}$ and is not maximal,
and two prime ideals of height one, $M_{31} := (x_{1}-1)D_{3}$ which is
maximal and $P_{31}$ which is the contraction to $D_{3}$ of the height-one
prime of $V_{3}$. }

\emph{Let $t \ge3$ be a fixed positive integer and assume for each positive
integer $n \le t$ we have constructed $n+1$ valuation domains
\[
W_{1}(x_{2},\dots,x_{n}),W_{2}(x_{3},\dots,x_{n}),\dots,W_{i}(x_{i+1},
\dots,x_{n}), \dots, W_{n} \mbox{ and } V_{n}
\]
on the field $k(x_{1},\dots,x_{n})$ such that $W_{m}$ has rank $m$ for each
$m$ with $1\leq m\leq n$, $V_{n}$ has rank $n$, and
\[
D_{n} := W_{1}(x_{2},\dots,x_{n}) \cap W_{2}(x_{3},\dots,x_{n}) \cap\dots\cap W_{n} \cap V_{n}
\]
has the following properties: }

\begin{enumerate}
\item \emph{$D_{n}$ has two maximal ideals of height $n$, $M_{nn} =
(x_{n}-1)D_{n}$ and $P_{nn} = (x_{n})D_{n}$. }

\item \emph{For each positive integer $m < n$, $D_{n}$ has two prime ideal of
height $m$, $M_{nm} = (x_{m} - 1)D_{n}$, which is maximal, and $P_{nm}$ which
is nonmaximal. }

\item \emph{For positive integers $m \le n \le s \le t$, the prime ideal
$P_{sm}$ of $D_{s}$ intersects $D_{n}$ in $P_{nm}$. }
\end{enumerate}

\emph{It is clear that we can continue the construction by defining $t+2$
valuation domains on $k(x_{1}, \dots, x_{t+1})$ as follows. Let $N_{t+1}$
denote the maximal ideal of $V_{t}(x_{t+1})$ and define $V_{t+1}%
=k[x_{t+1}]_{(x_{t+1})}+N_{t+1}$ and $W_{t+1}=k[x_{t+1}]_{(x_{t+1}-1)}%
+N_{t+1}$. For $1 \le i \le t$, define the Gaussian extension $W_{i}(x_{i+1},
\dots, x_{t+1})$. Then $D_{t+1}$ defined as the intersection of these $t+2$
valuation domains on $k(x_{1},\dots,x_{t+1})$ gives the properties listed
above for positive integers $m \le n \le s \le t+1$. Therefore, by induction,
we have the stated properties for all positive integers $m \le n \le s$. }

\emph{We define $D=\cup_{n=1}^{\infty}D_{n}$. For each positive integer $m$,
$D$ has two prime ideals of height $m$, $M_{m} := \cup_{n=m}^{\infty}M_{nm}$,
which is maximal, and $P_{m} := \cup_{n=m}^{\infty}P_{nm}$, which is
nonmaximal. $D$ also has a unique maximal ideal $M_{\infty}:= \cup
_{n=1}^{\infty}P_{nn}$ having infinite height. This completes the construction
of the example. }

\emph{By \cite[(11.11)]{NAG}, each of the $D_{n}$ is a Bezout domain. 
Therefore
$D$ is also a Bezout domain. }
\end{example}

If we strengthen the assumption on the factors, we get stronger conditions on
the domain. In addition, we also get that the condition on all nonzero ideals
is equivalent to the condition on all principal ideals.

\begin{theorem}
\label{primary}Let $D$ be an integral domain. The following are equivalent:

\begin{enumerate}
\item  Each nonzero principal ideal $aD$ of $D$ can be written in the form
$aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ is primary and
the $Q_{i}$'s are pairwise comaximal.

\item $D$ has the following properties:

\begin{enumerate}
\item $D$ is one-dimensional.

\item $D$ has Noetherian spectrum.
\end{enumerate}

\item  Each nonzero ideal $A$ of $D$ can be written in the form $A=Q_{1}\cdot
Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ is primary and the $Q_{i}$'s
are pairwise comaximal.
\end{enumerate}
\end{theorem}

\begin{proof}
$(1)$ $\Rightarrow(2)$: Let $P\ $be a nonzero prime ideal of $D$. If $P$ is
not maximal, let $M$ be a maximal ideal of $D$ that properly contains $P$ and
let $a$ be a nonzero element of $P$. Let $b$ be an element of $M$ not in $P$. 
Set
$x=ab$. By hypothesis, $xD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each
$Q_{i}$ is primary and the $Q_{i}$'s are pairwise comaximal. Now, $P$ contains
$x$ and so $P$ contains some $Q$, say $Q_{1}$.
Let $P_1 = \sqrt{Q_1}$. Thus $Q_1$ is $P_1$-primary and $P_1 \subseteq P$.
Also, because the $Q_{i}$'s are pairwise
comaximal, $M$ contains at most one minimal prime ideal of $xD$ and that
minimal prime must be $P_1.$ Localize at the maximal ideal $M.$ This forces
$xD_{M}=Q_{1}D_{M}$ to be $P_1D_{M}$-primary. But this is impossible, for $b$ 
is
not in $P_1D_{M}$, but $ab=x$ is in $xD_{M}$. If $xD_{M}$ were 
$P_1D_{M}$-primary,
then $a$ must be in $xD_{M}$. But $a=x(\frac{d}{s})$, for some $d\in D,\;s\in
D\setminus M$ implies $x=x(\frac{d}{s})b$, so $s=db$. This contradicts the
fact that $b$ is in $M$. Hence, $D$ is one dimensional.

To see that $D$ has Noetherian spectrum, we first show that each nonzero ideal
of $D$ is contained in only finitely maximal ideals. Let $a$ be a nonzero
element of $D$. Then we can write $aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$,
where $\sqrt{Q_{i}}=M_{i}$ is prime; in fact, $M_{i}$ is a maximal ideal since
$D$ is one-dimensional. If $M$ is any maximal ideal of $D $ of which $a$ is a
member, then $M\supseteq M_{j}$ for some $j$. Thus, $\{M_{1},M_{2}%
,\ldots,M_{n}\}$ is the set  of maximal ideals of $D$ containing $a$. If $A$
is any nonzero ideal of $D$, then the set of maximal ideals containing $A$ is
a subset of the set of maximal ideals containing any nonzero element inside
$A$. In particular, the set of maximal ideals containing $\ A$ is finite.

To show that $D$ has Noetherian spectrum,we must prove that $D$ satisfies the
ascending chain condition on radical ideals. Thus, let $A_{1}\subsetneqq
A_{2}\subsetneqq\ldots\subsetneqq$ be an ascending sequence of radical ideals.
Now, $\sqrt{A_{1}}=M_{1}\cap M_{2}\cap\cdots\cap M_{k}$ for some maximal
ideals $M_{1},M_{2},\ldots,M_{k}$ of $D$. Since $A_{1}\subsetneqq A_{2}$, the
set of maximal ideals of $D$ that contain $A_{2}$ is a proper subset of
$\{M_{1},M_{2},\ldots,M_{k}\}$. It follows that the sequence must be finite.

$(2)\Rightarrow(3):$  Let $A$ be a nonzero ideal of $D$, with $M_{1}%
,M_{2},\ldots M_{n}$ the maximal ideals containing $A$. That there are only
finitely many follows from the fact that $D$ is one-dimensional and the fact
that in a ring with Noetherian spectrum, each nonzero ideal has only finitely
many minimal primes. Set $Q_{i}=AD_{M_{i}}\cap D$. Then $Q_{i}$ has radical
$M_{i}$ and $A$ is the intersection of the $Q_{i}$'s. Since the $M_{i}$'s are
maximal, the $Q_{i}$'s are primary and comaximal. So, $A$ is also the product
of the $Q_{i}$'s.

$(3)\Rightarrow(1):$ This is obvious.\medskip
\end{proof}

Finally, we treat the case where the $Q$'s are prime powers.

\begin{theorem}
\label{prime powers}Let $D$ be an integral domain. The following are 
equivalent:

\begin{enumerate}
\item  Each nonzero principal ideal $aD$ of $D$ can be written in the form
$aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ is a prime
power and the $Q_{i}$'s are pairwise comaximal.

\item $D$ has the following properties:

\begin{enumerate}
\item $D$ is one-dimensional.

\item $D$ is Noetherian.

\item $D$ is integrally closed.
\end{enumerate}

\item  Each nonzero ideal $A$ of $D$ can be written as a product of prime 
ideals.

\item $D$ is a Dedekind domain.

\item  Each nonzero ideal $A$ of $D$ can be written in the form $A=Q_{1}\cdot
Q_{2}\cdot\cdots\cdot Q_{n}$, where each $Q_{i}$ is a prime power and the
$Q_{i}$'s are pairwise comaximal.
\end{enumerate}
\end{theorem}

\begin{proof}
The equivalence of properties (2) - (5) is
standard, see for example \cite[Theorem 37.8]{GIL}. In
particular,
$(3)\Rightarrow(4)$ is a theorem of Matsusita
\cite[Page 68]{KAP}. It is obvious that
$(5)\Rightarrow(1)$ Thus it suffices to show:

\noindent
$(1)\Rightarrow(2):$ We first prove that $D$ is one-dimensional.
Let $M$ be a maximal ideal of $D$ and suppose there exists a
nonzero prime ideal $P$ properly contained in $M$. Let $x \in M$,
$x \not\in P$ and write $xD = P_1^{e_1}\cdots P_n^{e_n}$, where
the $P_i$ are pairwise comaximal prime ideals. It follows that the
$P_i$ are invertible and $M$ contains one, and only one, of the
$P_i$, say $P_1$. Because $x \not\in P$, $P_1$ is not contained
in $P$. Let $y$ be a nonzero element of $P$ and write
$yD = N_1^{f_1} \cdots N_m^{f_m}$, where the $N_j$ are pairwise
comaximal prime ideals. Thus the $N_j$ are invertible and $P$
contains one, and only one, of the $N_j$, say $N_1$.  By
Theorem 1, the prime ideals contained in $M$ are linearly
ordered with respect to inclusion. Since $N_1 \subseteq P$,
we have $N_1 \subsetneq P_1$. It follows that
$N_1D_{P_1}  \subsetneq P_1D_{P_1}$ are nonzero principal
prime ideals of $D_{P_1}$. This is a contradiction. (Suppose
$R$ is a quasilocal domain with principal maximal ideal
$xR$ and that $yR$ is a principal prime ideal with $yR$
strictly smaller than $xR$. Write $y = xz$ for some $z \in R$.
Since $yR$ is a prime ideal and $x \not\in yR$, it follows
that $z \in yR$, say $z = yr$. Then $y = xyr$, so
$y(1 - xr) = 0$. Since $1-xr$ is a unit of $R$, $y = 0$).


Let $M$ be a maximal ideal of $D$ and let $a$ be a nonzero
element of $M$. Then $aD=Q_{1}\cdot Q_{2}\cdot\cdots\cdot Q_{n}$, where each
$Q_{i}=M_{i}^{e_{i}}$ for some maximal ideal $M_{i}$ of $D$. Thus, $M=M_{j}$
for some $j$. Since a product of ideals is invertible if and only if each
factor is invertible, it follows that $M$ is invertible. In particular,$\ M$
is finitely generated. So, all prime ideals of $D$ are finitely generated and
$D$ is Noetherian by Cohen's Theorem \cite[Theorem 8]{KAP}. Finally, $MD_{M}$
is principal since $M$ is invertible. Therefore, each prime ideal of the
quasi-local ring $D_{M}$ is principal and by an analog to Cohen's Theorem,
\cite[Page 8]{KAP} $D_{M}$ is a principal ideal domain. It follows that $D$ is
integrally closed.
\end{proof}

\smallskip

This paper was written, composed, and typeset electronically. The difficulty
posed by two authors working in different locations and using different
interfaces to Tex would have been insurmontable without the help of Fred
Richman and to Fred we offer our sincere thanks.

\baselineskip 15pt

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\end{document}

