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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