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\begin{document}
\baselineskip 16 pt
\title[Commutative Ideal Theory without Finiteness Conditions]{Commutative
Ideal Theory without Finiteness Conditions: Irreducibility in the
Quotient Field}
% Information for first author
\author{Laszlo Fuchs}
% Address of record for the research reported here
\address{Department of Mathematics, Tulane University, New
Orleans, Louisiana 70118}
\email{fuchs@tulane.edu}
% Information for second author
\author{William Heinzer}
\address{Department of Mathematics, Purdue University, West
Lafayette, Indiana 47907}
\email{heinzer@math.purdue.edu}
\author{Bruce Olberding}
\address{Department of Mathematical Sciences, New Mexico State University,
Las Cruces, New Mexico 88003-8001}
\email{olberdin@emmy.nmsu.edu}
% General info
\subjclass{Primary 13A15, 13F05}
\date \today
%\dedicatory{This paper is dedicated to our authors.}
\keywords{irreducible ideal, completely irreducible ideal, injective
module, Pr\"ufer domain, $m$-canonical ideal.}
\begin{abstract} Let $R$ be an integral domain
and let $Q$ denote the quotient field of $R$. We investigate the
structure of $R$-submodules of $Q$ that are $Q$-irreducible, or
completely $Q$-irreducible. One of our goals is to describe the
integral domains that admit a completely $Q$-irreducible ideal, or
a nonzero $Q$-irreducible ideal. If $R$ has a nonzero finitely
generated $Q$-irreducible ideal, then $R$ is quasilocal. If $R$ is
integrally closed and admits a nonzero principal $Q$-irreducible
ideal, then $R$ is a valuation domain. If $R$ has an $m$-canonical
ideal and admits a completely $Q$-irreducible ideal, then $R$ is
quasilocal and all the completely $Q$-irreducible ideals of $R$ are
isomorphic. We consider the condition that every nonzero ideal of
$R$ is an irredundant intersection of completely $Q$-irreducible
submodules of $Q$ and present eleven conditions that are equivalent
to this. We classify the domains for which every nonzero ideal can
be represented uniquely as an irredundant intersection of completely
$Q$-irreducible submodules of $Q$. The domains with this property
are the Pr\"ufer domains that are almost semi-artinian, that is,
every proper homomorphic image has a nonzero socle. We characterize
the Pr\"ufer or Noetherian domains that possess a completely
$Q$-irreducible ideal or a nonzero $Q$-irreducible ideal.
\end{abstract}
\maketitle
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\section{Introduction}
This article continues a study of commutative ideal theory in rings
without finiteness conditions begun in \cite{FHO}, \cite{FHO3},
\cite{FHO2} and \cite{HO}. In \cite{FHO} and \cite{FHO3} we examine
irreducible and completely irreducible ideals of commutative rings.
In the present article we investigate stronger versions of these two
notions of irreducibility for ideals of integral domains. In
particular, we consider irreducibility of an ideal of an integral
domain when it is viewed as a submodule of the quotient field of the
domain.
All rings in this paper are commutative and contain a multiplicative
identity. Our notation is as in \cite{FS}. Let $R$ be a ring and let
$C$ be an $R$-module. An $R$-submodule $A$ of $C$ is {\it
$C$-irreducible} if $A = B_1 \cap B_2$, where $B_1$ and $B_2$ are
$R$-submodules of $C$, implies that either $B_1 = A$ or $B_2 =A$. An
$R$-submodule $A$ of $C$ is {\it completely $C$-irreducible} (or
{\it completely irreducible} when the module $C$ is clear from
context) if $A = \bigcap_{i \in I}B_i$, where $\{B_i\}_{i \in I}$ is
a family of $R$-submodules of $C$, implies $A = B_i$ for some $i \in
I$.
In the case where the module $C$ is the ring $R$, an ideal $A$ of
$R$ is $R$-irreducible as a submodule of $R$ precisely if $A$ is
{\it irreducible} as an ideal in the conventional sense that $A$ is
not the intersection of two strictly larger ideals. It is
established by Fuchs in \cite[Theorem 1]{F} that a proper
irreducible ideal $A$ of the ring $R$ is a {\it primal ideal} in the
sense that the set of elements of $R$ that are non-prime to $A$ form
an ideal $P$ that is necessarily a prime ideal and is called the
{\it adjoint prime ideal} of $A$. One then says that $A$ is
$P$-primal. For such an ideal $A$, it is the case that $A =
A_{(P)}$, where $A_{(P)} = \bigcup_{x \in R \setminus P}(A:_Rx)$.
In Remark \ref{0.1} we record several general facts about completely
$C$-irreducible submodules. The straightforward proofs are omitted.
\begin{remark} \label{0.1}
For a proper submodule $A$ of $C$ the following are equivalent:
\begin{enumerate}
\item
$A$ is completely $C$-irreducible.
\item
There exists an element $x \in C \setminus A$ such that $x \in B$
for every submodule $B$ of $C$ that properly contains $A$.
\item
$C/A$ has a simple essential socle, that is, $C/A$ is a cocyclic
$R$-module.
\item
$C/A$ is subdirectly irreducible in the sense that in any
representation of $C/A$ as a subdirect product of $R$-modules, one
of the projections to a component is an isomorphism.
\end{enumerate}
\end{remark}
It is also straightforward to see that every submodule of a module
$C$ is an intersection of completely $C$-irreducible submodules of
$C$. Thus a nonzero module $C$ contains proper completely
$C$-irreducible submodules.
The main focus of our present study is the case where $R$ is an
integral domain and $C = Q$ is the quotient field of $R$.
(Throughout this paper $Q$ is understood to be the quotient field
of the integral domain $R$.) We are thus interested in
$Q$-irreducible and completely $Q$-irreducible submodules of $Q$. We
are particularly interested in determining conditions on an integral
domain $R$ in order that $R$ admit a completely $Q$-irreducible
ideal, or a nonzero $Q$-irreducible ideal. The zero ideal of $R$ is
always $Q$-irreducible, but if $R \ne Q$, the zero ideal of $R$ is
not completely $Q$-irreducible. In the case where $R$ admits
completely $Q$-irreducible ideals, or nonzero $Q$-irreducible
ideals, we are interested in describing the structure of such
ideals. Ideals with either of these properties are necessarily
primal ideals.
It is frequently the case that an integral domain $R$ may fail to
have any fractional ideals that are completely $Q$-irreducible, or
any nonzero ideals that are $Q$-irreducible. If $R = \mathbb Z$ is
the ring of integers, then every nonzero proper $Q$-irreducible
$R$-submodule of $Q$ is completely $Q$-irreducible and has the form
$p^n\mathbb Z_{p\mathbb Z}$, where $p$ is a prime integer and $n$ is
an integer. Thus for $R = \mathbb Z$ every nonzero proper
$Q$-irreducible $R$-submodule of $Q$ is a fractional ideal of a
valuation overring of $R$. Moreover, every nonzero fractional
$R$-ideal has a unique representation as an irredundant intersection
of infinitely many completely $Q$-irreducible $R$-submodules of $Q$.
It follows that $R$ has no nonzero fractional ideal that is
$Q$-irreducible.
In Section 2 we establish basic properties of irreducible submodules
of an $R$-module $C$ with special emphasis on the case where $C =
Q$. We prove in Theorem \ref{1.2} that if $R$ admits a nonzero
principal $Q$-irreducible fractional ideal, then $R$ is quasilocal,
and $R$ is integrally closed if and only if $R$ is a valuation
domain. In Theorem~\ref{qirreducible} we give several necessary
conditions for an integral domain to possess a nonzero
$Q$-irreducible ideal. If $A$ is a nonzero $Q$-irreducible ideal, we
prove that $\End(A)$ is quasilocal, and that $A$ is a primal ideal
of $\End(A)$ with adjoint prime the maximal ideal of $\End(A)$. If
the integral domain $R$ admits a nonzero finitely generated
$Q$-irreducible ideal, we prove that $R$ is quasilocal. Moreover,
every nonzero $Q$-irreducible ideal of a Noetherian domain is
completely $Q$-irreducible.
In Section 3 we review some relevant results and examples regarding
completely $Q$-irreducible fractional ideals. Over a quasilocal
domain, an $m$-canonical ideal (if it exists) is an example of a
completely $Q$-irreducible ideal. If $R$ has an $m$-canonical ideal
and admits a completely $Q$-irreducible ideal, we prove that $R$ is
quasilocal and all completely $Q$-irreducible ideals of $R$ are
isomorphic. We classify the Noetherian domains that admit a nonzero
$Q$-irreducible ideal.
In Proposition~\ref{3.3} of Section 4 we show that a proper
submodule $A$ of the quotient field $Q$ of a domain is an
irredundant intersection of $Q$-irreducible submodules if and only
if the injective hull of $Q/A$ is an interdirect sum of
indecomposable injectives.
In Section 5 we continue to examine irredundant intersections of
$Q$-irreducible submodules in $Q$. We draw on the literature to
give in Theorem~\ref{semi-artinian} eleven different module- and
ideal-theoretic conditions that are equivalent to the assertion that
every nonzero ideal of a domain is an irredundant intersection of
completely irreducible submodules of $Q$. We show in particular
that such a domain is locally almost perfect, and from this
observation we answer in the negative a question of Bazzoni and
Salce of whether every locally almost perfect domain $R$ has the
property that $Q/R$ is semi-artinian (Example~\ref{semi-artinian
example}). In Theorem~\ref{uniqueness} we classify the domains for
which every nonzero ideal can be represented {\it uniquely} as an
irredundant intersection of completely $Q$-irreducible submodules of
$Q$. The domains having this property have Krull dimension at most
one and are necessarily Pr\"ufer, that is, every nonzero finitely
generated ideal is invertible. They may be described precisely as
the Pr\"ufer domains $R$ that are almost semi-artinian, that is,
every proper homomorphic image of $R$ has a nonzero socle.
In light of Theorem~\ref{uniqueness} it is useful to describe the
completely irreducible submodules of the quotient field of a
Pr\"ufer domain. This is done in Theorem~\ref{Prufer case}. Also in
Section 6 we characterize the Pr\"ufer domains that possess a
completely $Q$-irreducible ideal, or a nonzero $Q$-irreducible
ideal. We prove that a Pr\"ufer domain $R$ that admits a nonzero
$Q$-irreducible ideal also admits a completely $Q$-irreducible
ideal, and this holds if and only if every proper $R$-submodule of
$Q$ is a fractional $R$-ideal.
In Section 7 we discuss several open questions, and in an appendix
we correct some errors in the article \cite{FHO2} that were pointed
out to us by Jung-Chen Liu and her student Zhi-Wei Ying. We are
grateful to them for showing us these mistakes.
\section{The structure of $Q$-irreducible ideals}
We begin with several general results.
\begin{proposition}\label{characterization}
Let $R$ be a ring and $C$ an $R$-module. The following statements are
equivalent for a proper $R$-submodule $A$ of $C$.
\begin{itemize}
\item[(i)]
$A$ is a completely $C$-irreducible $R$-submodule of $C$.
\item[(ii)] There
exists $x \in C \setminus A$ such that for all $y \in C \setminus A$
we have $x \in A+Ry$.
\item[(iii)] $A$ is $C$-irreducible and there exists a maximal ideal
$M$ of $R$ such that \newline $A \subset (A:_CM)$, where
$(A:_CM)=\{y \in C : yM \subseteq A\}$.
\end{itemize}
Furthermore, if $R$ is a domain, $A$ is torsionfree and $C$ is the
divisible hull of $A$, then statements (i)-(iii) are equivalent to:
\begin{itemize}
\item[(iv)] There is a maximal ideal $M$ of $R$ such that $A = AR_M$ and
$A$ is completely $C$-irreducible as an $R_M$-submodule of $C$.
\end{itemize}
\end{proposition}
\begin{proof}
(i) $\Rightarrow$ (ii) Let $A^*$ be the intersection of all
$R$-submodules of $C$ properly containing $A$. Then $A \subset
A^*$, and $A^*/A$ is a simple $R$-module. Hence $A^* = Rx + A$ for
some $x \in Q \setminus A$, and (ii) follows.
(ii) $\Rightarrow$ (iii) By (ii) there exists $x \in C \setminus
A$ such that $A^*:=A + Rx$ is contained in every $R$-submodule of
$C$ properly containing $A$. Hence $A^*/A$ is a simple $R$-module
and $A^*/A \cong R/M$ for some maximal ideal $M$ of $R$. Thus $A^*
\subseteq (A:_CM)$ so that $(A:_CM) \ne A$.
(iii) $\Rightarrow$ (i) Since $A$ is irreducible, $(A:_CM)/A \cong
R/M$ and every proper submodule containing $A$ contains $(A:_CM)$,
proving (i).
(i) $\Rightarrow$ (iv) Since $R$ is a domain and $A$ is
torsion-free, $A = \bigcap_{M \in \Max(R)}A_M$, where each $A_M$ is
identified with its image in $C = QA$. Because $A$ is completely
$C$-irreducible, $A = A_M$ for some maximal ideal $M$ of $R$. The
assumption that $A$ is completely $C$-irreducible as an $R$-module
clearly implies $A$ is completely $C$-irreducible as an
$R_M$-submodule of $C$.
(iv) $\Rightarrow$ (iii) Since we have established the equivalence
of (i)-(iii), and since by assumption $A$ is a completely
irreducible $R_M$-submodule of $C$, we have by (iii) (applied to
the $R_M$-module $A$) that there exists $x \in (A :_C MR_M)
\setminus A$. Now since $A = A_M$, we have $A \ne (A:_C MR_M) =
(A:_C M)$. Thus it remains to observe that $A$ is $C$-irreducible.
Suppose $A = B \cap D$ for some
$R$-submodules $B$ and $D$ of $C$. Then $A = A_M = B_M \cap D_M$,
so since by assumption $A$ is irreducible as an $R_M$-submodule of
$C$, it must be that $A = B_M$ or $A = D_M$. Thus $B \subseteq A$
or $D \subseteq A$, proving that $A$ is irreducible.
\end{proof}
\begin{remark} \label{1.1}
Let $R$ be an integral domain that is properly contained in its
quotient field $Q$.
{\rm (i)} By Remark \ref{0.1}, every $R$-submodule of $Q$ is an
intersection of completely irreducible submodules of $Q$. In
particular, every ideal of $R$ is an intersection of completely
irreducible submodules of $Q$.
{\rm (ii)} A fractional ideal $A$ of $R$ is completely
$Q$-irreducible if and only if $A$ is not the intersection of
fractional $R$-ideals that properly contain $A$. If $A$ is a
fractional $R$-ideal and $A \ne Q$, then $A$ is completely
$Q$-irreducible if and only if there exists $x \in Q \setminus A$
such that $x$ is in every fractional ideal that properly contains
$A$.
{\rm (iii)} A maximal ideal $P$ of $R$ is completely $Q$-irreducible
if it is $Q$-irreducible. This is immediate from Proposition
\ref{characterization}, since $P \subsetneq R \subseteq (P:_QP)$.
\end{remark}
In Lemma \ref{2.2}, we establish several general facts about
$Q$-irreducible and completely $Q$-irreducible ideals.
\begin{lemma} \label{2.2} Let $A$ be a proper
ideal of the integral domain $R$. Then
{\rm (i)} $A$ is $Q$-irreducible if and only if for each nonzero $r
\in R$ the ideal $rA$ is irreducible.
{\rm (ii)} For a nonzero $q \in Q$, the fractional ideal $qA$ is
$Q$-irreducible if and only if $A$ is $Q$-irreducible. Therefore the
property of being $Q$-irreducible is an invariant of isomorphism
classes of fractional $R$-ideals.
{\rm (iii)} $A$ is $Q$-irreducible if and only if there is a
prime ideal $P$ of $R$ such that $A = AR_{P}$ and $A$ is a
$Q$-irreducible ideal of $R_P$. It then follows that $P$ is uniquely
determined by $A$ and $A$ is $P$-primal.
{\rm (iv)} For a nonzero $q \in Q$, the fractional ideal $qA$ is
completely $Q$-irreducible if and only if $A$ is completely
$Q$-irreducible. Therefore the property of being completely
$Q$-irreducible is an invariant of isomorphism classes of fractional
$R$-ideals.
{\rm (v)} If $A$ is completely $R$-irreducible and if for each
nonzero $r \in R$ the ideal $rA$ is irreducible, then $A$ is
completely $Q$-irreducible.
\label{ideal characterization}
\end{lemma}
\begin{proof} {\rm (i)} Assume $A$ is $Q$-irreducible
and $r$ is a nonzero element of $R$. If $rA = B \cap C$ for ideals
$B$ and $C$ of $R$, then $A = r^{-1}B \cap r^{-1}C$. Since $A$ is
$Q$-irreducible, either $A = r^{-1}B$ or $A = r^{-1}C$. Hence
either $rA = B$ or $rA = C$ and $rA$ is irreducible. Conversely,
assume $A$ is not $Q$-irreducible. Then there exist $R$-submodules
$B$ and $C$ of $Q$ that properly contain $A$ such that $A = B \cap
C$. We may assume that $B$ and $C$ are fractional ideals of $R$.
Then there exists a nonzero $r \in R$ such that $rB$ and $rC$ are
integral ideals of $R$. Moreover, $A = B \cap C$ implies $rA = rB
\cap rC$ and $A \subset B$ implies $rA \subset rB$ and similarly $A
\subset C$ implies $rA \subset rC$. Therefore $rA$ is reducible.
This completes the proof of {\rm (i)}.
Statements {\rm (ii)} and {\rm(iv)} are clear since $A = \bigcap_{i
\in I}B_i$ if and only if $qA = \bigcap_{i \in I}qB_i$ and
multiplication by $q$ (or by $q^{-1}$) preserves strict inclusion.
{\rm (iii)} Assume $A$ is $Q$-irreducible. Then $A$ is $P$-primal
for some prime ideal $P$ of $R$, so that $A = A_{(P)} = AR_P \cap
R$. Since $A$ is $Q$-irreducible, this forces $A = AR_P$. Clearly
then $A$ is $Q$-irreducible as an $R_P$-module since it is
$Q$-irreducible as an $R$-module. Conversely, suppose that $A =
AR_P$ and $A$ is $Q$-irreducible as an ideal of $R_P$. If $A = B
\cap C$ for some $R$-submodules $B$ and $C$ of $Q$, then $A = AR_P =
BR_P \cap CR_P$, and since $A$ is a $Q$-irreducible $R_P$-submodule
of $Q$, $A = BR_P$ or $A = CR_P$. Thus $B \subseteq A$ or $C
\subseteq A$, which completes the proof.
{\rm(v)} Since $A$ is completely $R$-irreducible, there exists an
element $x \in R \setminus A$ such that $x$ is in every ideal of $R$
that properly contains $A$. Let $A^* = A + xR$. If $A$ is not
completely $Q$-irreducible, then there exists an $R$-submodule $B$
of $Q$ that properly contains $A$ but does not contain $x$. Since
there are no ideals properly between $A$ and $A^*$, $A = A^* \cap B$
and this intersection is irredundant. We may assume that $B$ is a
fractional ideal of $A$. Then there exists a nonzero $r \in R$ such
that $rB$ is an integral ideal of $R$. Therefore $rA = rA^* \cap rB$
is an irredundant intersection. It follows that $rA$ is not
irreducible.
\end{proof}
\begin{remark} \label{2.44}
With regard to Lemma \ref{2.2} we have:
\begin{enumerate}
\item
If $A$ is a nonzero $Q$-irreducible ideal of $R$ and $P$ is as in
Lemma \ref{2.2}(iii), then $R_P \subseteq \End(A)$ and $rA = A$ for
each $r \in R \setminus P$. It follows that $A$ is contained in
every ideal of $R$ not contained in $P$. Thus if $P$ is a maximal
ideal of $R$ and $A$ is $P$-primary with $A = AR_P$, then $R$ is
quasilocal.
\item
It is also true that if $A$ and $B$ are isomorphic $R$-submodules of
$Q$, then $A$ is (completely) $Q$-irreducible if and only if $B$ is
(completely) $Q$-irreducible. For $A$ and $B$ are $R$-isomorphic if
and only if there exists $q \in Q$ such that $A = qB$.
\end{enumerate}
\end{remark}
\begin{theorem}\label{1.2}
If the integral domain $R$ has a nonzero principal fractional ideal
that is $Q$-irreducible, then $R$ is quasilocal and every principal
ideal of $R$ is $Q$-irreducible. If $R$ is integrally closed, then
\begin{itemize}
\item[(i)]
$R$ is $Q$-irreducible if and only if $R$ is a valuation domain, and
\item[(ii)]
$R$ is completely $Q$-irreducible if and only if $R$ is a valuation
domain with principal maximal ideal. \end{itemize}
\end{theorem}
\begin{proof}{\rm(i)} Lemma \ref{ideal characterization} implies
that $R$ has a nonzero principal fractional ideal that is
(completely) $Q$-irreducible if and only if every nonzero principal
fractional ideal of $R$ is (completely) $Q$-irreducible. Suppose $R$
has distinct maximal ideals $M$ and $N$. Then there exist $x \in M$
and $y \in N$ such that $x + y = 1$. It follows that $xyR = xR \cap
yR$ is an irredundant intersection. By Lemma~\ref{ideal
characterization}(i), $R$ is not $Q$-irreducible.
{\rm(ii)} Suppose that $R$ is integrally closed and $Q$-irreducible
but is not a valuation domain. Then there exists $x \in Q$ such
that neither $x$ nor $1/x$ is in $R$. Let ${\mathcal{F}}$ be the
set of valuation overrings of $R$ that contain $x$ and let
${\mathcal{G}}$ be the set of valuation overrings of $R$ that
contain $1/x$. Let $A = \bigcap_{V \in {\mathcal{F}}}V$ and $B =
\bigcap_{W \in {\mathcal{G}}}W$. Then $x \in A$ implies $R
\subsetneq A$ and $1/x \in B$ implies $R \subsetneq B$. Observe that
every valuation overring of $R$ is a member of at least one of the
sets ${\mathcal{F}}$ or ${\mathcal{G}}$. Since $R$ is integrally
closed, we have $R = A \cap B$, a contradiction to the assumption
that $R$ is $Q$-irreducible. Conversely, it is clear that if $R$ is
a valuation domain, then $R$ is integrally closed and
$Q$-irreducible.
{\rm(iii)} By (ii) we need only observe the well-known fact that a
valuation domain $R$ is completely $Q$-irreducible if and only if
the maximal ideal of $R$ is principal. (See for example \cite{Baz}.)
\end{proof}
\begin{remark} \label{2.4} There exist integral domains $R$
that are completely $Q$-irreducible and are not integrally closed.
If $R$ is a one-dimensional Gorenstein local domain, then $R$, and
every nonzero principal fractional ideal of $R$, is completely
$Q$-irreducible. Thus, for example, if $k$ is a field and $a$ and
$b$ are relatively prime positive integers, then the subring $R :=
k[[t^a, t^b]]$ of the formal power series ring $k[[t]]$ is
completely $Q$-irreducible.
\end{remark}
Theorem~\ref{1.2}(ii) characterizes among integrally closed domains
$R$ the ones that are valuation domains as precisely those $R$ that
are $Q$-irreducible. As a corollary to
Proposition~\ref{characterization}, we have the following additional
characterizations of the valuation property in terms of
$Q$-irreducibility.
\begin{corollary} \label{valuation criterion} The following are equivalent
for a domain $R$ with quotient field $Q$.
\begin{itemize}
\item[(i)] $R$ is a
valuation domain.
\item[(ii)] Every irreducible ideal is $Q$-irreducible.
\item[(iii)]
Every completely irreducible ideal is completely $Q$-irreducible.
\item[(iv)]
There exists a maximal ideal of $R$ that is $Q$-irreducible.
\item[(v)]
There exists a maximal ideal of $R$ that is completely
$Q$-irreducible.
\end{itemize}
\end{corollary}
\begin{proof} (i) $\Rightarrow$ (ii)
If $R$ is a valuation domain, then it is
easy to see that irreducible ideals are $Q$-irreducible since the
$R$-submodules of $Q$ are linearly ordered.
(ii) $\Rightarrow$ (iii) If $A$ is a completely irreducible ideal of
$R$, then there is a maximal ideal $M$ of $R$ such that $(A:_R M)
\ne A$. Thus $(A:_QM) \ne A$, and since $A$ is by (ii)
$Q$-irreducible, we have from Proposition~\ref{characterization}
(iii) that $A$ is $Q$-irreducible.
(iii) $\Rightarrow$ (iv) This is clear from the fact that maximal
ideals are completely irreducible.
(iv) $\Rightarrow$ (v) This follows from Remark \ref{1.1}(iii).
(v) $\Rightarrow$ (i) Let $M$ be a completely $Q$-irreducible
maximal ideal of $R$. For every nonzero $r \in R$, $rM$ is
completely irreducible by Proposition~\ref{ideal characterization}.
It is shown in Lemma 5.1 of \cite{FHO3} that this property
characterizes valuation domains, so the proof is complete.
\end{proof}
\begin{corollary} \label{prime corollary} Let $P$ be a prime ideal of a
domain $R$. Then $P$ is $Q$-irreducible if and only if $P = PR_P$
and $R_P$ is a valuation domain. Thus if $P$ is $Q$-irreducible,
then $R_P/P$ is the quotient field of $R/P$, and $R$ is a pullback
of $R/P$ and the valuation domain $R_P$. Moreover $P$ is completely
$Q$-irreducible as an ideal of $R_P$.
\end{corollary}
\begin{proof} Suppose that $P$ is $Q$-irreducible. By
Lemma~\ref{2.2}, $P=PR_P$ and $PR_P$ is a $Q$-irreducible ideal of
$R_P$. Hence, by Corollary~\ref{valuation criterion}, $R_P$ is a
valuation domain.
Conversely, assume $P = PR_P$ and $R_P$ is a valuation domain. By
Corollary~\ref{valuation criterion}, $P$ is a $Q$-irreducible ideal
of $R_P$. Hence, by Lemma~\ref{2.2}, $P$ is $Q$-irreducible. It
follows from Remark \ref{1.1}(iii) that $P = PR_P$ is a completely
$Q$-irreducible ideal of $R_P$.
\end{proof}
\begin{remark} With $P = PR_P$ as in Corollary \ref{prime
corollary}, if $R \ne R_P$, then $P$ as an ideal of $R$ is not
completely $Q$-irreducible. For Proposition \ref{characterization}
(iii) implies that a completely $Q$-irreducible prime ideal is a
maximal ideal, and by Remark \ref{2.44}(i), if $P$ is maximal and
$Q$-irreducible, then $R = R_P$. It can happen however that $P$ is
$Q$-irreducible and nonmaximal. This is the case, for example, if
$P$ is a nonmaximal prime of a valuation domain $R$.
\end{remark}
\begin{remark} Pullbacks arising as in Corollary~\ref{prime corollary}
have been well-studied; for a recent survey see \cite{GaH}. For
example, a consequence of our Corollary~\ref{prime corollary} and
Theorem 4.8 in \cite{GaH2} is that if a domain $R$ has a
$Q$-irreducible prime ideal $P$, then $R$ is coherent if and only if
$R/P$ is coherent.
\end{remark}
\begin{theorem} Assume that $A$ is a
nonzero $\, Q$-irreducible ideal of the integral domain $R$. Then
\begin{itemize}
\item[(i)] If $A$ is not principal, then $AA^{-1}$ is
contained in the Jacobson radical of $R$.
\item[(ii)] $\End(A)$ is a quasilocal integral domain.
\end{itemize}
Let $M$ denote the maximal ideal of $\End(A)$.
\begin{itemize}
\item[(iii)] $A$ is an $M$-primal ideal of $\End(A)$.
\item[(iv)] If $\,M$ is finitely generated as an ideal of $\End(A)$, then
$A$ is completely $Q$-irreducible as an ideal both of $R$ and of
$\End(A)$.
\item[(v)] If $A$ is a finitely generated ideal of $R$,
then $R$ is quasilocal and the maximal ideal of $R$ is the adjoint
prime of $A$.
\item[(vi)] If both $A$ and its adjoint prime are finitely
generated ideals, then $A$ is completely $Q$-irreducible.
\end{itemize}\label{qirreducible}
\end{theorem}
\begin{proof}
(i) Let $x \in A^{-1}$ and suppose that there is a maximal ideal
$N$ of $R$ not containing $xA$. Then there exists $y \in N$ such
that $xA + yR = R$. It follows that $xyA = xA \cap yR$. By
Lemma~\ref{ideal characterization}(ii), $xyA$ is irreducible.
Therefore either $xyA = xA$ or $xyA = yR$. If $xA = xyA$, then $xA
\subseteq yR \subseteq N$, a contradiction, while if $xyA = yR$,
then $xA = R$ and $A$ is principal. We conclude that every maximal
ideal of $R$ contains $xA$. Therefore $AA^{-1}$ is contained in
the Jacobson radical of $R$.
(ii) and (iii) Since $A$ is $Q$-irreducible as an ideal of $R$, it
is also $Q$-irreducible as an ideal of $\End(A)$. By
Lemma~\ref{ideal characterization}(iii), there is a prime ideal $M$
of $\End(A)$ such that $A = A\End(A)_M$. Thus $\End(A)_M \subseteq
\End(A)$, which implies that $M$ is the unique maximal ideal of
$\End(A)$. Also by Lemma~\ref{ideal characterization}(iii), $A$ is
$M$-primal.
(iv) Let $x_1,\ldots,x_n$ generate $M$. By
Lemma~\ref{characterization}(iii), to show that $A$ is completely
$Q$-irreducible it suffices to prove that $(A:_QM) \ne A$. Now
$(A:_QM) = x_1^{-1}A \cap \cdots \cap x_{n}^{-1}A$, so if $(A:_QM) =
A$, then the $Q$-irreduciblity of $A$ implies $x_i^{-1}A = A$ for
some $i$. In this case, $x_i^{-1} \in \End(A)$, which is impossible
since $x_i \in M$, the maximal ideal of $\End(A)$.
(v) By Lemma~\ref{ideal characterization}(ii), $A = AR_P$ for some
prime ideal $P$ of $R$. Thus $R_P \subseteq \End(A)$. But $A$ is
a finitely generated ideal of $R$ implies that $\End(A)$ is an
integral extension of $R$. This forces $R = R_P$, so that $P$ is
the unique maximal ideal of $R$.
(vi) By (v), $R$ is quasilocal with maximal ideal $M$, and $M$ is
the adjoint prime of $A$. As in the proof of (iv), we have $A
\subset (A:_QM)$. Therefore Lemma~\ref{characterization}(ii) implies
that $A$ is completely $Q$-irreducible.
\end{proof}
\begin{corollary} \label{Noetherian case} Every nonzero $Q$-irreducible
ideal over a Noetherian domain is completely $Q$-irreducible. If the
Noetherian domain $R$ admits a completely $Q$-irreducible ideal,
then $R$ is local and $\dim R \le 1$.
\end{corollary}
\begin{proof}
Suppose that $A$ is a nonzero $Q$-irreducible ideal of $R$. By
Theorem~\ref{qirreducible}(vi) $A$ is a completely $Q$-irreducible
ideal of $R$, and hence also of $\End(A)$. By
Theorem~\ref{qirreducible}(ii), $\End(A)$ is quasilocal. Since $R$
is Noetherian, $\End(A)$ is a finitely generated integral extension
of $R$. Therefore $R$ is local.
If $\dim R > 1$, then there exists a nonzero nonmaximal prime ideal
$P$ of $R$. Let $x \in P$ with $x \ne 0$. Then $xM$ is completely
irreducible by Lemma~\ref{ideal characterization}(iv). However, by
Corollary 1.4 in \cite{FHO3} a completely irreducible ideal of a
Noetherian local domain is primary for the maximal ideal,
contradicting $xM \subseteq P$. Therefore $\dim R \le 1$.
\end{proof}
\begin{corollary}
If the integral domain $R$ admits an invertible $Q$-irreducible
ideal, then every invertible ideal of $R$ is principal and
completely $Q$-irreducible.
\end{corollary}
\begin{proof} Suppose that $A$ is an invertible $Q$-irreducible ideal of $R$.
By Theorem~\ref{qirreducible}(i) $A$ is principal. Let $B$ be
an invertible ideal of $R$. Since $A$
is invertible, $A = (B:_Q:(B:_QA))$. Moreover, $(B:_Q A)$ is an
invertible, hence finitely generated, fractional ideal of $R$. Hence
there are elements $q_1,\ldots,q_k \in Q$ such that $A = (B:_Q
(q_1,\ldots,q_k)R) = q_1^{-1}B \cap \cdots \cap q_k^{-1}B$. Since
$A$ is $Q$-irreducible, there exists $i \in \{1,\ldots,k\}$ such
that $B = q_iA$. Hence $B$ is principal and $R$-isomorphic to $A$.
By Lemma~\ref{2.2}, $B$ is $Q$-irreducible.
\end{proof}
\begin{remark} Statement (ii) of Theorem~\ref{qirreducible} is true
also when $A$ is a completely irreducible submodule of $Q$. For by
Lemma~\ref{characterization}(iv) (with $A$ viewed as a completely
irreducible $\End(A)$-submodule of $Q$) there is a maximal ideal $M$
of $\End(A)$ such that $A = A_M$ This forces $\End(A)_M \subseteq
\End(A)$, so $\End(A)$ is quasilocal. \end{remark}
\section{Completely $Q$-irreducible and $m$-canonical ideals}
As noted in Remark~\ref{1.1} every ideal of a domain is the
intersection of completely irreducible submodules of the quotient
field. Thus for a given domain there exists an abundance of
completely irreducible submodules of $Q$. However, as we observe in
Section 1, a domain need not possess a completely $Q$-irreducible
ideal (see also Example~\ref{no canonical}).
In this section we examine the existence and structure of completely
$Q$-irreducible ideals. We also consider the class of
``$m$-canonical'' ideals. A nonzero fractional ideal $A$ of a
domain $R$ is an {\it $m$-canonical} fractional ideal if for all
nonzero ideals $B$ of $R$, $B = (A:_Q(A:_Q B))$. This terminology is
from \cite{BHLP} and \cite{HHP}. Different terminology is used in
\cite{Baz} and \cite{FS} to express the same concept. An ideal $A$
is, in our terminology, $m$-canonical if and only if, in the
terminology of \cite{Baz} and \cite{FS}, $R$ is an
``$A$-divisorial'' domain and $\End(A) = R$. Notice that the
property of being an $m$-canonical ideal is invariant with respect
to $R$-isomorphism for fractional ideals of $R$.
It follows from \cite[Lemma 4.1]{HHP} that an $m$-canonical ideal of
a quasilocal domain is completely $Q$-irreducible. A deeper result
is due to S. Bazzoni \cite{Baz}: {\it A fractional ideal $A$ of a
quasilocal domain $R$ is $m$-canonical if and only if $A$ is
completely $Q$-irreducible, $\End(A) =R$ and for all nonzero $r \in
R$, $A/rA$ satisfies the dual AB-$5^*$ of Grothendieck's AB-5.} (An
$R$-module $B$ satisfies AB-$5^*$ if for any submodule $C$ of $B$
and inverse system of submodules $\{B_i\}_{i \in I}$ of $B$, it is
the case that $\bigcap_{i \in I}(C + B_i) = C+\bigcap_{i \in
I}(B_i)$.)
As examples later in this section show, a domain need not possess an
$m$-canonical ideal. However if $R$ admits an $m$-canonical ideal,
then all completely $Q$-irreducible ideals of $R$ are isomorphic:
\begin{proposition}\label{criterion} Let $R$ be a domain that is not a field.
If $R$ has an $m$-canonical ideal $A$, then every completely
$Q$-irreducible ideal of $R$ is isomorphic to $A$. Consider the
following statements.
\begin{itemize}
\item[(i)] $R$ has an $m$-canonical ideal.
\item[(ii)] Any two completely
$Q$-irreducible ideals of $R$ are isomorphic.
\end{itemize}
Then $(i) \Rightarrow (ii)$. If every completely irreducible
proper submodule of $Q$ is a fractional ideal of $R$, then $(ii)
\Rightarrow (i)$.
\end{proposition}
\begin{proof} Suppose that $R$ has an $m$-canonical ideal $A$.
If $B$ is a nonzero ideal of $R$, then $B = \bigcap_{q}q^{-1}A$,
where $q$ ranges over all nonzero elements of $(A:_Q B)$. Thus if
$B$ is completely $Q$-irreducible, then $B = q^{-1}A$ for some $0
\ne q \in (A:_QB)$. Thus every proper completely $Q$-irreducible
ideal is isomorphic to $A$, and (i) $\Rightarrow$ (ii).
Assume that any two completely $Q$-irreducible ideals are
isomorphic and every completely $Q$-irreducible proper submodule of
$Q$ is a fractional ideal of $R$. Let $A$ be a completely
irreducible $R$-ideal. By Remark~\ref{0.1} every ideal of $R$ is an
intersection of completely $Q$-irreducible submodules of $Q$ and
therefore of completely $Q$-irreducible fractional ideals of $R$.
Thus every ideal of $R$ is an intersection of ideals isomorphic to
$A$; that is, for any ideal $B$, there exists a set $X \subseteq Q$
such that $B = \bigcap_{q \in X}qA$. It follows that $B =
(A:_Q(A:_QB))$. Hence $A$ is an $m$-canonical ideal.
\end{proof}
\begin{remark} An integral domain may have an $m$-canonical ideal,
but not admit a completely $Q$-irreducible fractional ideal. For
example, if $R$ is a Dedekind domain having more than one maximal
ideal, then $R$ admits an $m$-canonical ideal, but does not have any
completely $Q$-irreducible fractional ideals. Indeed, as we observe
in Proposition \ref{quasilocal}, if $R$ has an $m$-canonical ideal
and admits a completely $Q$-irreducible ideal, then $R$ is
quasilocal.
\end{remark}
\begin{proposition} \label{quasilocal}
If $R$ has an $m$-canonical ideal and a completely $Q$-irreducible
ideal, then $R$ is quasilocal.
\end{proposition}
\begin{proof} Let $A$ be a completely $Q$-irreducible ideal of $R$.
By Proposition \ref{criterion}, $A$ is an $m$-canonical ideal.
Therefore $R = \End(A)$. By Theorem \ref{qirreducible}, $\End(A)$ is
quasilocal. Therefore $R$ is quasilocal.
\end{proof}
\begin{remark} If $A$ is a proper $R$-submodule of $Q$, then $A$ is
contained in a completely irreducible proper submodule of $R$. Thus
if every completely irreducible proper submodule of $Q$ is a
fractional ideal of $R$, then every proper submodule of $Q$ is a
fractional ideal of $R$. The latter property holds for $R$ if and
only if there exists a valuation overring of $R$ which is a
fractional ideal of $R$ \cite[Theorem 79]{M2}.
\end{remark}
Routine arguments show that a nonzero fractional ideal of a
valuation domain is $m$-canonical if and only if it is completely
$Q$-irreducible. Also in the Noetherian case, the condition
AB-$5^*$ is redundant, as we note next. The following proposition
is essentially due in the case of Krull dimension 1 to Matlis
\cite{Mat} and in the general case with the assumption that $\End(A)
= R$ to Bazzoni \cite{Baz}. Bazzoni's proof shows that you can omit
in our context the assumption that $\End(A) = R$. We outline how to
do this in the proof. We also include a different proof of the step
(iv) $\Rightarrow$ (iii).
\begin{proposition} \label{Bazzoni} \label{same}
{\em (Bazzoni \cite[Theorem 3.2]{Baz}, Matlis \cite[Theorem
15.5]{Mat})} The following statements are equivalent for a nonzero
fractional ideal $A$ of a Noetherian local domain $(R,M)$ that is
not a field.
\begin{itemize}
\item[(i)] $Q/A$ is
an injective $R$-module.
\item[(ii)] $R$ has Krull dimension 1 and $(A:M)/A$ is a simple $R$-module.
\item[(iii)] $A$ is an $m$-canonical ideal.
\item[(iv)] $A$ is $Q$-irreducible.
\end{itemize}
\end{proposition}
\begin{proof}
(i) $\Rightarrow$ (ii) By Proposition 4.4 in \cite{Mati} a
Noetherian domain that admits an ideal of injective dimension 1
necessarily has Krull dimension 1. Thus $\dim(R) =1$, so we may
apply Theorem 15.5 in \cite{Mat} to obtain (ii).
(ii) $\Rightarrow$ (iii) This is contained in Theorem 15.5 of
\cite{Mat}.
(iii) $\Rightarrow$ (i) If $A$ is an $m$-canonical ideal, then
necessarily $\End(A) =R$, so Theorem 3.2 of \cite{Baz} applies.
(iii) $\Rightarrow$ (iv) An $m$-canonical ideal of a quasilocal
domain is completely $Q$-irreducible \cite[Lemma 4.1]{HHP}.
(iv) $\Rightarrow$ (iii) Suppose that $A$ is $Q$-irreducible. By
Corollary~\ref{Noetherian case} $\dim R = 1$ and $A$ is completely
$Q$-irreducible. By Theorem~\ref{qirreducible} $\End(A)$ is a
quasilocal domain. Since $R$ is Noetherian, $\End(A)$ is Noetherian.
Thus by Theorem 3.2 in \cite{Baz} $A$ is an $m$-canonical ideal of
$\End(A)$.
By \cite[Theorem 15.7]{Mat} a Noetherian local domain of Krull
dimension 1 has an $m$-canonical ideal if and only if the total
quotient ring of the completion of the domain is Gorenstein.
Therefore the total quotient ring of the completion of $\End(A)$ is
Gorenstein. Now $\End(A)$ is an overring of $R$ that is finitely
generated as a module over $R$. Hence there exists a nonzero $x \in
R$ such that $x\End(A) \subseteq R$. It follows that the total
quotient ring $T$ of the completion of $R$ coincides with the
completion of $\End(A)$. Thus $T$ is a Gorenstein ring, and by the
result cited above, $R$ has an $m$-canonical ideal, say $B$. By
Proposition~\ref{criterion} $B$ is isomorphic to $A$, so $A$ is an
$m$-canonical ideal of $R$.
\end{proof}
\begin{remark}
Let $R$ be a Noetherian domain of positive dimension. If $R$ admits
a nonzero $Q$-irreducible ideal, then $R$ is local and $\dim R = 1$.
Every proper $R$-submodule of $Q$ is a fractional $R$-ideal if and
only if the integral closure $\overline{R}$ of $R$ is local (so a
DVR) and is a finitely generated $R$-module. In this case every
proper $R$-submodule of $Q$ that is completely $Q$-irreducible is a
fractional $R$-ideal. There exist, however, other one-dimensional
Noetherian local domains $R$ that admit completely $Q$-irreducible
ideals. By Proposition \ref{same}, $R$ admits a completely
$Q$-irreducible ideal if and only if the total quotient ring of the
completion of $R$ is Gorenstein. In particular, this is true if $R$
is Gorenstein. There exist examples where $R$ is Gorenstein and
$\overline{R}$ is not local, or not a finitely generated $R$-module,
or both. For such an $R$, nonzero principal fractional ideals of
$R$ are completely $Q$-irreducible, and there also exist completely
$Q$-irreducible proper $R$-submodules of $Q$ that are not
fractional $R$-ideals.
\end{remark}
\begin{example} \label{no canonical} {\it A one-dimensional Noetherian
local domain need not possess a nonzero $Q$-irreducible ideal.} As
noted in the proof of Proposition~\ref{same} it suffices to exhibit
a Noetherian local domain $R$ of Krull dimension 1 such that the
total quotient ring of $R$ is not Gorenstein. Such examples can be
found in Proposition 3.1 of \cite{FR} and Theorem 1.26 and Corollary
1.27 of \cite{HRS}. A specific example, based on \cite{HRS} is
obtained as follows. Let $x, y, z$ be algebraically independent over
the field $k$ and let $R = k[x, y, z]_{(x, y, z)}$. Let $f, g \in
xk[[x]]$ be such that $x, f, g$ are algebraically independent over
$k$. Let $u = y - f$ and $v = z - g$. Then $P := (u, v)k[[x, y, z]]$
is a prime ideal of height 2 of the completion $\widehat R = k[[x,
y, z]]$ of $R$ having the property that $P \cap R = (0)$. If $\q$ is
a $P$-primary ideal of $\widehat R$, it follows from \cite[Theorem
1.26]{HRS} that $(\widehat R/\q) \cap k(x,y,z)$ is a one-dimensional
Noetherian local domain having $\widehat R/\q$ as its completion. If
we take $\q = P^2 = (u^2, uv, v^2)\widehat R$, then the total
quotient ring of $\widehat R/\q$ is not Gorenstein.
\end{example}
\begin{remark}
(i) It is an open question whether a completely $Q$-irreducible
ideal of a quasilocal integrally closed domain $R$ is an
$m$-canonical ideal if $\End(A) = R$ \cite[Question 5.5]{Baz}. The
answer is affirmative when $A = R$: this is Theorem 2.3 of
\cite{Baz}.
(ii) In \cite{Baz} Bazzoni relates the question in (i) to a 1968
question of Heinzer \cite{H}: If $R$ is a domain for which every
nonzero ideal is divisorial, is the integral closure of $R$ a
Pr\"ufer domain? To obtain that $R$ has a Pr\"ufer integral closure
the weaker requirement that $R$ be completely $Q$-irreducible is
not sufficient, as we note below in Example~\ref{gilmerhoffmann}.
(iii) Bazzoni constructs in Example 2.11 of \cite{Baz} an example of
a quasilocal domain $R$ such that $R$ is completely $Q$-irreducible
but not $m$-canonical. By Lemma~\ref{Bazzoni} and (i) such a
domain is neither Noetherian nor integrally closed.
\end{remark}
The $D+M$ construction provides a source of interesting examples of
completely $Q$-irreducible ideals. The following example is from
\cite[Remark 5.3]{HHP}, as strengthened in \cite{BHLP}. We recall
it here, since it is relevant to Example~\ref{gilmerhoffmann}.
\begin{example} \label{HHP example} {\rm Let $k \subset F$ be a proper
extension of fields and $V$ be a valuation domain (that is not a
field) of the form $V = F +M$, where $M$ is the maximal ideal of
$V$. Define $R = k +M$. Then $R$ is a quasilocal domain with
maximal ideal $M$. If $U$ is any $k$-subspace of $F$ of codimension
$1$, then the fractional ideal $A = U + M$ is a completely
$Q$-irreducible fractional ideal of $R$ since every $R$-submodule of
the quotient field $Q$ of $R$ that properly contains $A$ contains
also $V$.
It is proved in Theorem 3.2 of \cite{BHLP} that if $F$ is an
algebraic extension of $k$ with $[F:k]$ infinite, then there exist
codimension $1$ subspaces $U$ and $W$ of $F$ such that $U +M$ and
$W+M$ are non-isomorphic completely $Q$-irreducible fractional
ideals of $R$. Thus by Proposition~\ref{criterion} $R$ does not
possess an $m$-canonical ideal. Indeed, it is shown in Theorem 3.1
of \cite{BHLP} that $R$ has an $m$-canonical ideal if and only if
$[F:k]$ is finite. }
\end{example}
We shall see in Theorem~\ref{existence} that it is possible for a
domain $R$ to possess a completely $Q$-irreducible ideal $A$ and not
be quasilocal. It follows from this result that $\End(A)$ need not
equal $R$. However, in this situation, $R$ is not quasilocal. The
next example shows that even when $R$ is quasilocal, it is possible
for a completely $Q$-irreducible ideal to have an endomorphism ring
not equal to $R$.
Gilmer and Hoffmann in \cite{GH} establish the existence of an
integral domain $R$ that admits a unique minimal overring, but has
the property that the integral closure of $R$ is not Pr\"ufer. In
Example~\ref{gilmerhoffmann} we modify this example to establish the
existence of an integral domain $R$ that has infinitely many
distinct fractional overrings $R_t$, $t \in \mathbb N$, such that
each $R_t$ is completely $Q$-irreducible as a fractional ideal of
$R$. Since $R_t$ is a fractional overring of $R$, $\End(R_t) = R_t$.
We remark that Bazzoni in \cite[Section 4]{Baz} has abstracted and
greatly generalized the example of \cite{GH}.
\begin{example} \label{gilmerhoffmann} Let $K$ be a field and let
$L = K((X))$ be the quotient field of the formal power series ring
$K[[X]]$. Every nonzero element of $L$ has a unique expression as a
Laurent series $\sum_{n \ge k}a_nX^n$, where $k$ is an integer, the
$a_n \in K$ and $a_k \ne 0$. Let $Y$ be an indeterminate over $L$
and let $V = L[[Y]]$ denote the formal power series ring in $Y$ over
the field $L$. Thus $V$ is a rank-one discrete valuation domain
(DVR) of the form $L + M$, where $M = YL[[Y]]$ is the maximal ideal
of $V$. Let $R = K + M^2$. It is well-known and readily established
that $R$ is a one-dimensional quasilocal domain with maximal ideal
$M^2$. For $t$ a positive integer, let $W_t$ be the set of all
elements $f \in K((X))$ such that $f = 0$ or the coefficient of
$X^{-t}$ in the Laurent expansion of $f$ is $0$. Notice that $W_t$
is a $K$-subspace of $L$ and $L = W_t \oplus KX^{-t}$ as $K$ vector
spaces. Let $R_t = K + W_tY + M^2$. Then $R_t$ is an overring of $R$
and $Y^2R_t \subseteq M^2$, so $R_t$ is a fractional ideal of $R$.
We show that $R_t$ is completely $Q$ irreducible as a fractional
$R$-ideal by proving that $X^{-t}Y$ is in every fractional ideal of
$R$ that properly contains $R_t$. Let $f \in Q \setminus R_t$. Since
$Q = L((Y))$, there exists an integer $j$ such that $f = \sum_{n \ge
j}b_nY^n$, where each $b_n \in L$ and $b_j \ne 0$. Notice that $f
\not\in R_t$ implies $j \le 1$. Since $L = K((X))$, there exists an
integer $k$ such that $b_j = \sum_{n \ge k}a_nX^n$, where each $a_n
\in K$ and $a_k \ne 0$. Since $a_k$ is a unit of $R$, the fractional
ideal $R_t + Rf = R_t + a_k^{-1}f$, so we may assume that $a_k = 1$.
If $j < 0$, then $X^{-k-t}Y^{1-j} \in M^2 \subset R$ and
$X^{-k-t}Y^{1-j}f = X^{-t}Y + \alpha Y + \beta Y^2$, where $\alpha
\in K[[X]]$ and $\beta \in V = L[[Y]]$. Since $\alpha \in W_t$,
$\alpha Y + \beta Y^2 \in R_t$. Hence $X^{-t}Y \in R_t + Rf$ if $j <
0$. If $j = 0$ and $k \ne 0$, then $X^{-k-t}Y^{1-j} \in W_tY \subset
R_t$ and $X^{-k-t}Yf = X^{-t}Y + \alpha Y + \beta Y^2$, where
$\alpha Y + \beta Y^2 \in R_t$, so $X^{-t}Y \in R_t + Rf$ in this
case. If $j = 0$ and $k = 0$, replace $f$ by $f-1$ to obtain a
situation where $k
>0$ and $j \ge 0$. If $j = 1$, then $f \not\in R_t$ implies $b_1 \not\in
W_t$. Hence $b_1 = c + dX^{-t}$, where $c \in W_t$ and $0 \ne d \in
K$. Hence $f - cY = dX^{-t}Y + \alpha Y^2$, where $\alpha \in
L[[Y]]$. Therefore also in this case $X^{-t}Y \in R_t + Rf$. We
conclude that $R_t$ is completely $Q$-irreducible.
\end{example}
In Example~\ref{gilmerhoffmann} the completely $Q$-irreducible
fractional ideals that are constructed have endomorphism rings
integral over the base ring. In Example \ref{valuation example} we
exhibit a Noetherian local domain $R$ and a completely
$Q$-irreducible $R$-submodule $A$ of $Q$ such that $\End(A)$ is not
integral over $R$. We first give a partial characterization of when
valuation overrings are (completely) $Q$-irreducible.
\begin{theorem} \label{divisible} Let $V$ be a valuation overring of the
domain $R$. Then the following two statements hold for $V$.
\begin{itemize}
\item[(i)] If $V/R$ is a
divisible $R$-module, then $V$ is a $Q$-irreducible $R$-submodule of
$Q$. Moreover, $V$ has a principal maximal ideal if and only if $V$
is a completely $Q$-irreducible $R$-submodule of $Q$.
\item[(ii)] Suppose that $V$ is a DVR. Then $V$
is a completely $Q$-irreducible $R$-submodule of $Q$ if and only if
$V/R$ is a divisible $R$-module.
\end{itemize}
\end{theorem}
\begin{proof} (i) The assumption that $V/R$ is divisible implies that every
$R$-submodule of $Q$ containing $V$ is also a $V$-submodule of $Q$.
For if $x \not \in V$, then $1/x \in V$. Since $V/R$ is divisible,
$V = (1/x)V + R$. Thus $V + xR =xV$. Hence $V + xR$ is a
$V$-submodule of $Q$. This implies that any $R$-submodule of $Q$
containing $V$ is a $V$-module. Since $V$ is $Q$-irreducible as a
$V$-submodule of $Q$, it follows that $V$ is $Q$-irreducible as an
$R$-submodule of $Q$.
If the valuation domain $V$ has principal maximal ideal, then, by
Theorem~\ref{1.2}, $V$ is a completely $Q$-irreducible
$V$-submodule of $Q$. Therefore $V$ is a completely $Q$-irreducible
$R$-submodule of $Q$.
Conversely, if $V$ is a completely $Q$-irreducible $R$-submodule of
$Q$, then necessarily $V$ is a completely $Q$-irreducible
$V$-submodule of $Q$. By Corollary~\ref{valuation criterion} every
principal ideal of $V$ is $Q$-irreducible. Hence by
Theorem~\ref{1.2} $V$ has a principal maximal ideal.
(ii) Suppose that $V$ is a completely $Q$-irreducible $R$-submodule
of $Q$. Let $0 \ne x \in R$. We claim that $V = R + xV$. Consider
the ideal $C = (R+xV :_Q V)$ of $V$. Since $V$ is a DVR, $C$ is
isomorphic to $V$. Also, $C = \cap_{y \in V} y^{-1}(R+xV)$, so
since $C$ is completely $Q$-irreducible, $C$ is isomorphic to
$R+xV$. Thus $V$ and $R+xV$ are isomorphic as $R$-modules, and since
these two modules are rings, this forces $R+xV = V$, proving that
$V/R$ is divisible. The converse follows from (i).
\end{proof}
\begin{remark}\label{dvr} Let $V$ be a DVR overring of the integral domain $R$
and let $P$ be the center of $V$ on $R$. Necessary and sufficient
conditions in order that $V/R$ be a divisible $R$-module are that
(i) $PV$ is the maximal ideal of $V$, and (ii) the canonical
inclusion map of $R/P \hookrightarrow V/PV$ is an isomorphism. By
Theorem \ref{divisible}(ii), these conditions are also necessary and
sufficient in order that $V$ be completely $Q$-irreducible as an
$R$-submodule of $Q$.
\end{remark}
\begin{example} \label{valuation example}
Let $K$ be a field, and let $X$ and $Y$ be indeterminates for $K$.
Define $R = K[X,Y]_{(X,Y)}$.
We construct a valuation overring $V$ of $R$ such that $V$ is a
completely $Q$-irreducible $R$-submodule of $Q$. Let $g(X) \in
XK[[X]]$ be such that $X$ and $g(X)$ are algebraically independent
over $K$. Define a mapping $v$ on $K[X,Y] \setminus \{0\}$ by $v(\:
f(X,Y) \:) = $ smallest exponent of $X$ appearing in the power
series $f(X, g(X))$. Then $v$ extends to a rank-one discrete
valuation on $K(X,Y)$ centered on $(X,Y)R$ and having residue field
$K$. (More details regarding this construction can be found in
Chapter VI, Section 15, of \cite{ZS}.) Since the valuation ring $V$
of $v$ has maximal ideal $(X,Y)V$ and residue field $V/(X,Y)V = K$,
it follows that $V = R + (X,Y)^kV$ for all $k>0$. Since $V$ is a
DVR, $V = R+fV$ for every nonzero $f \in R$. Hence $V/R$ is a
divisible $R$-module. By Theorem~\ref{divisible}, $V$ is a
completely $Q$-irreducible $R$-submodule of $Q$.
\end{example}
\section{$Q$-irreducibility and injective modules}
Let $N$ be a submodule of the \tf\ $R$-module $M$. $N$ is said
to be an {\it RD-submodule} (relatively divisible) if $rN = N \cap
rM$ for all $r \in R$. An $R$-module $X$ is called {\it
RD-injective} if every homomorphism from an RD-submodule $N$ of any
$R$-module $M$ can be extended to a homomorphism $M \to X$. Every
$R$-module $M$ can be embedded as an RD-submodule in an RD-injective
module, and among such RD-injectives there is a minimal one, unique
up to isomorphisms over $M$, called the {\it RD-injective hull}
$\widehat M$ of $M$. If $M$ is torsion-free, then so are both
$\widehat M$ and $\widehat M/M$.
The $R$-topology of an $R$-module $M$ is defined by declaring the
submodules $rM$ for all $0 \ne r \in R$ as a subbase of open
neighborhoods of $0$. If $M$ is torsion-free, then it is Hausdorff
in the $R$-topology if and only if it is reduced (i.e. it has no
divisible submodules $\ne 0$). $M$ is $R$-complete if it is complete
(Hausdorff) in the $R$-topology. If $M$ is reduced torsion-free,
then it is an RD-submodule of its $R$-completion $\widetilde M$.
Observe that for a prime ideal $P$ the $R$-completion and
$R_P$-completion of $R_P$ are identical. The $R$-completion
$\widetilde M$ of a torsion-free $R$-module $M$ is an RD-submodule
of the RD-injective hull $\widehat M$ such that $\widehat
M/\widetilde M$ is reduced torsion-free.
\begin{lemma} For a proper $R$-submodule $A$ of $Q$ the
following conditions are equivalent:
{\rm (i)} $A$ is $Q$-irreducible;
{\rm (ii)} the injective hull $E(Q/A)$ of the $R$-module $Q/A$ is
indecomposable;
{\rm (iii)} the RD-injective hull $\widehat A$ of $A$ is
indecomposable.
\end{lemma}
\begin{proof} (i) $\Leftrightarrow$ (ii) An injective module is
indecomposable exactly if it is uniform.
(ii) $\Leftrightarrow$ (iii) This equivalence is a consequence of
Matlis' category equivalence between the category of $h$-divisible
torsion $R$-modules $T$ and the category of reduced $R$-complete
torsion-free $R$-modules $M$, given by the correspondences
$$ T \mapsto \Hom_R(Q/R, T) \quad {\rm and} \quad M \mapsto Q/R
\otimes_R M$$ which are inverse to each other. Under the category
equivalence, $Q/A$ and the $R$-completion $\widetilde A$ of $A$
correspond to each other, and so do the injective hull of $Q/A$ and
the RD-injective hull $\widehat A$ of $A$. As equivalence preserves
direct decompositions, the claim is evident.
\end {proof}
Let $I$ be an ideal of the ring $R$. It is well known that if
$E(R/I)$ is indecomposable, then $I$ is irreducible. Note that
$E(R/I)$ can also be written as $E(Q/A)$ for a $Q$-irreducible
$R$-submodule $A$ of $Q$. In fact, $E(R/I)$ is a summand of
$E(Q/I)$, so we can write: $E(Q/I) = E(R/I) \oplus E$ for an
injective $R$-module $E$. The kernel of the projection of $Q/I$ into
the first summand is of the form $A/I$ for a $Q$-irreducible
submodule $A$ of $Q$, and then $E(R/I) = E(Q/A)$.
Conversely, if $A$ is a $Q$-irreducible proper submodule of $Q$,
and $x \in Q \setminus A$, then the set $I = \{r \in R \ | \ rx \in
A\}$ is a primal ideal of $R$ such that $E(Q/A) = E(R/I)$. The
adjoint prime $P$ of the primal ideal $I$ may be called the {\it
prime associated to} $A$: this is uniquely determined by $A$, though
$I$ depends on the choice of $x$.
\begin{lemma} Every indecomposable injective $R$-module $E$
can be written as $E(Q/A)$ for a $Q$-irreducible $R$-submodule of
$A$ of $Q$. Moreover, there is a unique prime ideal $P$ of $R$ such
that $E(Q/A) \cong E(R/I)$ for a $P$-primal ideal $I$ of $R$, and
$P$ is a maximal ideal whenever $A$ is completely $Q$-irreducible.
\qed
\end{lemma}
We can add that $I$ can be replaced by $P$ if and only if $P$ is a
strong Bourbaki associated prime for $I$. Indeed, $E(R/I) = E(R/P)$
if and only if there are elements $r \in R \setminus I$ and $s \in R
\setminus P$ such that $(I:_Rr) = (P:_Rs)$. Since $(P:_Rs) = P$,
this is equivalent to $P = (I:_Rr)$, that is, $P$ is a strong
Bourbaki associated prime of $I$.
It is clear that every proper submodule of $Q$ is the intersection
of $Q$-irreducible submodules. This intersection is in general
redundant. A criterion for irredundancy is as follows.
\begin{proposition} \label{3.3} A proper submodule $A$ of $Q$ admits an
irredundant representation as an intersection of $Q$-irreducible
submodules if and only if $E(Q/A)$ is an interdirect sum of
indecomposable injectives.
\end{proposition}
\begin{proof} Suppose $A= \bigcap_{i \in I} A_i$ is an
irredundant intersection with $Q$-irreducible submodules $A_i$ of
$Q$. Setting $B_i = \bigcap_{j \in I, j \ne i} A_j$, it is clear
that the submodule generated by $B_i/A \ (i \in I)$ in $Q/A$ is
their direct sum. Hence $E(Q/A)$ contains the direct sum of the
injective hulls $E(Q/A_i) \cong E(B_i/A)$. As $Q/A$ embeds in the
direct product of the $Q/A_i$, $E(Q/A)$ embeds in the direct product
of the $E(Q/A_i)$. Thus $E(Q/A)$ is an interdirect sum of the
$E(Q/A_i)$ (these are evidently indecomposable).
Conversely, suppose $E(Q/A)$ is an interdirect sum of indecomposable
injectives $E_i \ (i \in I)$. Since $E_i$ is a uniform module, we
have $(Q/A) \cap E_i \ne 0$ for each $i \in I$. Clearly, $A_i$
(defined by $A_i/A = (Q/A) \cap \prod_{j \in I, j \ne i} E_j$) is a
submodule of $Q$, which is maximal disjoint from $E_i$, so
$Q$-irreducible. The intersection $A = \bigcap_{i \in I} A_i$ is
evidently irredundant.
\end{proof}
\section{Irredundant decompositions and semi-artinian modules}
In this section we examine domains for which every nonzero submodule
of $Q$ is an irredundant intersection of completely irreducible
submodules of $Q$. Such domains are closely related to the class of
almost perfect rings.
A ring $R$ is {\it perfect} if every $R$-module has a projective
cover; equivalently, (since our rings are assumed to be commutative)
$R$ satisfies the descending chain condition on principal ideals
\cite{Ba}. In their study \cite{BS1} of strongly flat covers of
modules, Bazzoni and Salce introduced the class of {\it almost
perfect domains}, consisting of those domains $R$ for which every
proper homomorphic image of $R$ is perfect. Every Noetherian domain
of Krull dimension 1 is almost perfect, but the class of almost
perfect domains includes also non-Noetherian non-integrally closed
domains-- see for example Section 3 of \cite{BS}.
There are a number of applications of perfect and almost perfect
domains in the literature, most of which are motivated by the rich
module theory for these classes of rings \cite{BS,BS1,EJ}. In this
section we emphasize different features of the module and ideal
theory of almost perfect domains, namely, the close connection with
irredundant decompositions into completely irreducible submodules.
If $R$ is a ring, then an $R$-module $A$ is (almost) {\it
semi-artinian} if every (proper) homomorphic image of $A$ has a
nonzero socle. In a semi-artinian module every irreducible
submodule is completely irreducible (see for example \cite[Lemma
2.4]{DC}), but this property does not characterize semi-artinian
modules \cite[Example 1.7]{FHO3}.
As indicated by Lemma~\ref{DC} below, the semi-artinian property is
both necessary and sufficient for irredundant decompositions into
completely irreducible submodules. Bazzoni and Salce note in
\cite{BS} that:
\begin{center}
$R$ almost perfect $\Rightarrow Q/R$ semi-artinian $\Rightarrow$ $R$
locally almost perfect.
\end{center}
They show also that $R$ is almost perfect if and only if $R$ is
$h$-local and every localization of $R$ at a maximal ideal is almost
perfect. In general, the first implication cannot be reversed
\cite[Example 2.1]{BS}. Smith asserts in \cite{Sm} that the
converse of the second implication is always true, but as noted in
\cite[p. 288]{BS} the proof is incorrect. Thus Bazzoni and Salce
raise the question in \cite[p. 288]{BS} of whether the converse is
always true; namely, if $R$ is locally almost perfect, is $Q/R$
necessarily semi-artinian?
We give an example in this section to show that the answer is
negative, and we characterize in Theorem~\ref{semi-artinian}(vi) and
(vii) precisely when a locally almost perfect domain $R$ has $Q/R$
semi-artinian. We collect also in this theorem a number of
different characterizations of domains $R$ for which $Q/R$ is
semi-artinian.
The following lemma is a special case of a lattice theoretic result
\cite[Theorem 4.1]{DC}. A number of other properties of irredundant
intersections of completely irreducible submodules of semi-artinian
modules can be deduced from this same article.
\begin{lemma} \label{DC} {\em (Dilworth-Crawley \cite{DC})} Let $R$ be a
ring and $A$ be an $R$-module. Then $A$ is (almost) semi-artinian
if and only if every (nonzero) submodule of $A$ is an irredundant
intersection of completely irreducible submodules of $A$.\qed
%
%\item[(ii)]
% $A$ is a distributive (almost) semi-artinian module if
%and only if every (nonzero) submodule of $A$ is a unique
%irredundant intersection of completely irreducible
%submodules of $A$.
%%
%
%\end{itemize}
\end{lemma}
% and
%statement (ii) follows from Theorem 6.1 and its second corollary in
%\cite{DC}.
In order to formulate (vii) of the next theorem, we recall that a
topological space $X$ is {\it scattered} if every nonempty subspace
of $X$ contains an isolated point.
\begin{theorem} \label{semi-artinian} The following
statements are equivalent for a domain $R$ with quotient field $Q$.
\begin{itemize}
\item[(i)] $Q/R$ is semi-artinian.
\item[(ii)] Every nonzero torsion module is semi-artinian.
\item[(iii)] $R$ is almost semi-artinian.
\item[(iv)] $Q$ is almost semi-artinian.
\item[(v)] For each nonzero proper ideal $A$ of $R$, there is
a maximal ideal that is a strong Bourbaki associated prime of $A$.
\item[(vi)] $R$ is locally almost perfect and for each nonzero
radical ideal $J$ of $R$, there is a maximal ideal of $R/J$ that is
principal.
\item[(vii)] $R$ is locally almost perfect and for each
nonzero radical ideal $J$ of $R$, $\Spec(R/J)$ is scattered.
\item[(viii)] For each torsion $R$-module $T$, every submodule
of $T$ is an irredundant intersection of completely irreducible
submodules of $T$.
\item[(ix)] For each torsion-free module $A$, every nonzero
submodule of $A$ is an irredundant intersection of completely
irreducible submodules of $QA$.
\item[(x)] Each nonzero submodule of $Q$ is an
irredundant intersection of completely irreducible submodules of
$Q$.
\item[(xi)] Each nonzero ideal of $R$ is an irredundant
intersection of completely irreducible submodules of $Q$.
\item[(xii)] Each nonzero ideal of $R$ is an irredundant
intersection of completely irreducible ideals.
\end{itemize}
\end{theorem}
\begin{proof} The equivalence of (i)-(iv) can be found in {\em
\cite[Theorem 4.4.1]{EJ}}. It follows then from Lemma~\ref{DC} that
(i) - (iv) are equivalent to (viii), (ix), (x) and (xii). The
equivalence of (vi) and (vii) is a consequence of Corollary 2.10 in
\cite{HO}. To complete the proof it is enough to show that (v) and
(vi) are equivalent to (i) and that (xi) is equivalent to (iii).
(i) $\Rightarrow$ (vi) Since $Q/R$ is semi-artinian, $R$ is locally
almost perfect. We have already established that (i) is equivalent
to (xii). That (xii) implies (vi) is a consequence of Corollary
2.10 of \cite{HO}.
(vi) $\Rightarrow$ (v) Suppose that $A$ is a proper nonzero ideal
of $R$. Since for every nonzero radical ideal $J$ of $R$, $R/J$ has
a maximal ideal of $R$ that is principal, every nonzero ideal of
$R$ has a Zariski-Samuel associated prime $M$ \cite[Theorem
2.8]{HO}; that is, $M = \sqrt{A:_Rx}$ for some $x \in R \setminus
A$. Since $R$ has Krull dimension $1$, $M$ is a maximal ideal of
$R$. By (vi) $R_M/(A_M:_{R_M} x)$ contains a simple $R_M$-module.
Thus there exists $y \in R \setminus (A_M:_{R_M}x)$ such that $MR_M
= (A_M:_{R_M}x):_{R_M}y = A_M:_{R_M}xy$. Since $A:_Rx \subseteq
A:_Rxy \subseteq M$ and $ \sqrt{A:_Rx} = M$, it follows that $M$ is
the only maximal ideal of $R$ containing $A:_Rxy$. Thus since
$A_M:_{R_M}xy = MR_M$, it is the case that $A:_Rxy = M$.
(v) $\Rightarrow$ (iii) If $A$ is a proper nonzero ideal of $R$ and
$M$ is a strong Bourbaki associated prime of $A$, then $A:_RM \ne
A$, so $R/A$ contains a simple $R$-module.
(iii) $\Rightarrow$ (xi) Since (iii) is equivalent to (x), it is
sufficient to note that (x) implies (xi).
(xi) $\Rightarrow$ (iii) Let $A$ be a proper nonzero ideal of $R$.
Then there exists a completely irreducible submodule $C$ of $Q$ such
that $A = C \cap D$ is an irredundant intersection for some
submodule $D$ of $Q$. Let $x \in D \setminus C$. Now $(C:_Q M)/C$
is the essential socle of $Q/C$, so if $y \in (C:_QM) \setminus C$,
then $y \in xR +C$. Thus $rx \in yR + C$ for some $r\in R$ such
that $rx \not \in C$. Consequently, $rxM \subseteq C$, and since $x
\in D$, it is the case that $rxM \subseteq A$ with $rx \not \in A$.
Thus $rx + A$ is a nonzero member of the socle of $R/A$. Statement
(iii) now follows.
\end{proof}
An integral domain $R$ is {\it almost Dedekind} if for each maximal
ideal $M$ of $R$, $R_M$ is a DVR. In \cite[Theorem 3.2]{O2} it is
shown that if $X$ is a Boolean (i.e. compact Hausdorff totally
disconnected) topological space, then there exists an almost
Dedekind domain $R$ with nonzero Jacobson radical such that
$\Max(R)$ is homeomorphic to $X$. Thus we obtain the following
corollary to Theorem~\ref{semi-artinian}(vii).
\begin{corollary} The following statements are equivalent for a
Boolean topological space $X$.
\begin{itemize}
\item[(i)] $X$ is a scattered space.
\item[(ii)] There exists a domain $R$ with nonzero Jacobson
radical such that $Q/R$ is semi-artinian and $\Max(R)$ is
homeomorphic to $X$. \qed
\end{itemize}
\end{corollary}
\begin{remark} In Example 2.1 of \cite{BS} an example is given of
a domain $R$ for which $Q/R$ is semi-artinian but $R$ is not almost
perfect. Using the corollary, we may obtain many such examples.
Indeed, let $X$ be an infinite Boolean scattered space. Then there
exists an almost Dedekind domain $R$ such that $\Max(R)$ is
homeomorphic to $X$ and $R$ is not a Dedekind domain. In
particular, $R$ is not $h$-local, since an $h$-local almost Dedekind
domain is Dedekind. Thus $Q/R$ is semi-artinian but $R$ is not
almost perfect.
\end{remark}
It is not difficult to exhibit infinite Boolean scattered spaces.
For example, let $X$ be a well-ordered set such that not every
element has an immediate successor. Then $X$ is a 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 $X$ (see \cite[Example 17.3, p. 272]{Ko}).
In \cite{BS} Bazzoni and Salce raise the question of whether every
locally almost perfect domain $R$ has the property that $Q/R$ is
semi-artinian. Using Theorem~\ref{semi-artinian} we give an example
to show that this is not the case.
\begin{example} \label{semi-artinian example} Let $X$ be a Boolean space
that is not scattered (e.g. let $X$ be the Stone-\^Cech
compactification of the set of natural numbers with the discrete
topology). As noted above, there exists an almost Dedekind domain
$R$ such that $\Max(R)$ is homeomorphic to $X$ and $R$ has nonzero
Jacobson radical. Then $R$ is locally almost perfect but by
Theorem~\ref{semi-artinian}(vii) $Q/R$ is not semi-artinian.
\end{example}
In \cite{FHO} it is shown that every irreducible ideal of an almost
perfect domain is primary. A similar argument yields:
\begin{lemma}
\label{primary} If $R$ is a locally almost perfect domain, then
every proper irreducible ideal is primary. \end{lemma}
\begin{proof} Let $A$ be a nonzero
irreducible ideal. Then $A$ is primary if and only if any strictly
ascending chain of the form $A \subset A:_Rb_1 \subset A:_Rb_1b_2
\subset \cdots \subset A:_Rb_1b_2 \cdots b_n \subseteq \cdots$ for
$b_1,b_2,\ldots,b_n,\ldots \in R$ terminates \cite{F}. Suppose
there is an infinite such strictly ascending chain, and let $M$ be
a maximal ideal containing every residual $A:_Rb_1b_2 \cdots b_n$.
Since $R_M$ is an almost perfect domain, $R_M/A_M$ has the
descending chain condition for principal ideals. Thus there exists
$n>0$ such that $A_M:_Rb_1b_2 \cdots b_n = A_M:_Rb_1b_2 \cdots
b_{n+1}$. If $r \in A:_Rb_1b_2 \cdots b_{n+1}$, then there exists $x
\in R \setminus M$ such that $xr \in A:_Rb_1b_2 \ldots b_{n}$. An
irreducible ideal of a domain of Krull dimension 1 is contained in a
unique maximal ideal (see for example \cite[Lemma 2.7]{HO}), so
necessarily $A$ is $M$-primal. Thus $x$ is prime to $A$ and it
follows that $r \in A:_Rb_1b_2 \cdots b_n$. However, this forces
$A:_Rb_1b_2 \cdots b_n = A:_Rb_1b_2 \cdots b_{n+1}$, contrary to
assumption. Thus $A$ is primary.
\end{proof}
\begin{theorem} If $R$ is
an almost semi-artinian domain, then every ideal of $R$ is an
irredundant intersection of primary completely irreducible ideals.
\end{theorem}
\begin{proof} The theorem follows from Lemma~\ref{primary} and
Theorem~\ref{semi-artinian}(xii).
\end{proof}
We characterize next the domains $R$ for which every nonzero
submodule of $Q$ can be represented uniquely as an irredundant
intersection of completely $Q$-irreducible $R$-submodules.
An $R$-module $B$ is {\it distributive} if for all submodules
$A_1,A_2$ and $A_3$ of $B$, $(A_1 \cap A_2) + A_3 = (A_1 + A_3) \cap
(A_2 \cap A_3)$. The module $B$ is {\it uniserial} if its
submodules are linearly ordered by inclusion. An $R$-module is
distributive if and only if for all maximal ideals $M$ of $R$, $B_M$
is a uniserial $R_M$-module \cite{LS}.
\begin{lemma} Let $R$ be a ring and $B$ be an $R$-module. Let $\AA$ be
the set of all $R$-submodules of $B$ that are finite intersections
of completely irreducible submodules of $B$. Then the module $B$ is
distributive if and only if for each $A \in \AA$, the representation
of $A$ as an irredundant intersection of completely irreducible
submodules of $B$ is unique.
Furthermore, if a submodule $B$ of a distributive $R$-module can be
represented as a (possibly infinite) irredundant intersection of
irreducible submodules, then this representation is unique.
\label{distributive}\end{lemma}
\begin{proof} Suppose that each representation of $A \in \AA$ as an
irredundant intersection of completely $Q$-irreducible submodules of
$B$ is unique. Then this property holds also for the
$R_M$-submodules of $B_M$ for each maximal ideal $M$ of $R$. Thus
by the remark preceding the theorem, to prove that $B$ is
distributive it suffices to show that $B_M$ is a uniserial
$R_M$-module. Thus we may reduce to the case where $R$ is a
quasilocal domain with maximal ideal $M$ and show that $B$ is a
uniserial $R$-module. If $B$ is not uniserial, there exist
incomparable completely $B$-irreducible submodules $C_1$ and $C_2$
of $B$. Define $A = C_1 \cap C_2$, $C_1^*= C_1:_B M$ and $C_2^* =
C_2:_B M$. By Lemma~\ref{characterization}, $C_1 \subset C_1^*$ and
$C_2 \subset C_2^*$. Now there exist $x \in (C_1^* \cap C_2)
\setminus A$ and $y \in (C_1 \cap C_2^*) \setminus A$. (This
follows from the irreduciblity of the $C_i$ and the modularity of
the lattice of submodules of $Q$; see for example Noether
\cite[Hilfssatz II]{No}.) We have Soc $B/A = (A + xR + yR)/A$ is a
2-dimensional vector space over $R/M$ and $x+y \not \in C_1 \cup
C_3$. Let $C_3$ be an $R$-submodule of $B$ containing $A + (x+y)R$
that is maximal with respect to $x \not \in C_3$. Then $C_3$ is
completely $B$-irreducible, distinct from $C_1$ and $C_2$ and $A =
C_1 \cap C_3$. Yet $A \in \AA$, so this contradiction means that
the submodules of $B$ are comparable. The converse and the last
assertion follow from the fact that in a complete distributive
lattice, an irredundant meet decomposition into meet-irreducible
elements is unique \cite[pp. 5-6]{D} .
\end{proof}
\begin{theorem} \label{uniqueness}
The following are equivalent for a domain $R$ with quotient field
$Q$.
\begin{itemize}
% \item[(ii)]
%Every nonzero ideal of $R$ is an irredundant intersection $A = \bigcap_i
%M_i^{e_i}$ of powers of maximal ideals $M_i$ of $R$.
\item[(i)] Every
nonzero submodule of $Q$ can be represented uniquely as an
irredundant intersection of completely irreducible submodules of
$Q$.
\item[(ii)] Every nonzero ideal of $R$ can be represented uniquely as an
irredundant intersection of completely irreducible submodules of
$Q$.
\item[(iii)] Every nonzero
proper ideal of $R$ can be represented uniquely as an irredundant
intersection of completely irreducible ideals of $R$.
\item[(iv)]
$R$ is an almost Dedekind domain such that for each radical ideal $J$ of
$R$, $R/J$ has a finitely generated maximal ideal.
\item[(v)] $R$ is an almost semi-artinian Pr\"ufer domain.
\end{itemize} \end{theorem}
\begin{proof}
(i) $\Rightarrow$ (ii) This is clear.
(ii) $\Rightarrow$ (iii) This follows from
Theorem~\ref{semi-artinian} and Lemma~\ref{distributive}.
(iii) $\Leftrightarrow$ (iv) This is proved in \cite[Corollaries
2.10 and 3.9]{HO}.
(iv) $\Rightarrow$ (v) This follows from
Theorem~\ref{semi-artinian}.
(v) $\Rightarrow$ (i) Since $R$ is a Pr\"ufer domain, $Q$ is a
distributive $R$-module. Thus (i) is a consequence of
Theorem~\ref{semi-artinian} and Lemma~\ref{distributive}.
\end{proof}
\section{Pr\"ufer domains}
In light of Theorem~\ref{uniqueness} it is of interest to describe
the completely irreducible submodules of the quotient field of a
Pr\"ufer domain. We do this in Theorem~\ref{Prufer case}. We need
for the proof of this theorem a description of the completely
irreducible ideals of a Pr\"ufer domain. This is a special case of
Theorem 5.3 in \cite{FHO3}: {\it A proper ideal $A$ of a Pr\"ufer
domain is completely irreducible if and only if $A = MB_{(M)}$ for
some maximal ideal $M$ and nonzero principal ideal $B$ of $R$.}
\begin{lemma}\label{flat} Let $R$ be an integral domain and
let $A$ be a flat $R$-submodule of $Q$. If $A$ is $Q$-irreducible,
then $\End(A)$ is quasilocal and is $Q$-irreducible as an
$R$-submodule of $Q$. \end{lemma}
\begin{proof} Since $A$ is a flat $R$-submodule of $Q$, it is the case that
$A(B \cap C) = AB \cap AC$ for all $R$-submodules $B$ and $C$ of $Q$
\cite[I.2, Proposition 6]{B}.
Suppose now that $\End(A) = B \cap C$ for $R$-submodules
$B$ and $C$ of $Q$. Then $A = A\End(A) = A(B \cap C) = AB \cap AC$,
and since $A$ is $Q$-irreducible, $A = AB$ or $A = AC$. Thus $B
\subseteq \End(A)$ or $C \subseteq \End(A)$, so that $\End(A)$ is
$Q$-irreducible. Finally, if $\End(A)$ is not quasilocal, then
there exist two nonzero non-units $x,y \in \End(A)$ such that
$x\End(A) + y\End(A) = \End(A)$. Thus $xy\End(A) = x\End(A) \cap y
\End(A)$, so $\End(A) = y^{-1}\End(A) \cap x^{-1}\End(A)$. Since
$\End(A)$ is $Q$-irreducible, this forces $x$ or $y$ to be a unit, a
contradiction.
\end{proof}
\begin{theorem} \label{Prufer case} Let $R$ be a Pr\"ufer domain. Then
\begin{itemize}
\item[(i)] the $Q$-irreducible $R$-submodules of $Q$ are precisely the
$R$-submodules of $Q$ that are also $R_P$-submodules for some prime
ideal $P$, and
\item[(ii)] the completely
$Q$-irreducible proper $R$-submodules of $Q$ are precisely the
$R$-submodules of $Q$ that are isomorphic to $MR_M$ for some maximal
ideal $M$ of $R$.
\end{itemize}
Conversely, either of statements (i) and (ii) characterizes among
the class of domains those that are Pr\"ufer. \end{theorem}
\begin{proof}
(i) If $A$ is $Q$-irreducible submodule of $Q$, then by
Lemma~\ref{flat} $\End(A)$ is quasilocal. Since $R$ is a Pr\"ufer
domain, there is a prime ideal $P$ of $R$ such that $R_P = \End(A)$
and $A$ is an $R_P$-submodule of $Q$.
Conversely, if $P$ is a prime
ideal of $R$, $A$ is an $R_P$-submodule of $Q$ and $A = B \cap C$
for some $R$-submodules $B$ and $C$ of $Q$, then $A = BR_{P} \cap
CR_P$. Since $R_P$ is a valuation domain $A = BR_P$ or $A = CR_P$.
Thus $A = B$ or $A = C$ and $A$ is $Q$-irreducible.
(ii) Suppose that $R$ is a Pr\"ufer domain and let $A$ be a
completely $Q$-irreducible proper $R$-submodule of $Q$. Then by
Proposition~\ref{characterization}, $A = AR_M$ for some maximal
ideal $M$ of $R$ and $A$ is a completely $Q$-irreducible submodule
of $R_M$. Since $R_M$ is a valuation domain, there exists $q \in Q$
such that $qA \subseteq R_M$. Moreover, $qA$ is a completely
irreducible ideal of $R_M$, so by Lemma 5.1 of \cite{FHO3}, $qA =
xMR_M$ for some $x \in R_M$. Hence $A$ is isomorphic to $MR_M$.
On the other hand, if $A$ is an $R$-submodule of the form $xMR_M$
for some $x \in Q$, then $A$ is a completely irreducible fractional
ideal of the valuation domain $R_M$ \cite[Lemma 5.1]{FHO3}. Since
$R_M$ is a valuation domain, $A$ is a completely $Q$-irreducible of
$R_M$. Thus by Proposition~\ref{characterization}, $A$ is a
completely $Q$-irreducible $R$-submodule of $Q$.
It is easy to see that statement (i) characterizes Pr\"ufer domains.
For let $M$ is a maximal ideal of $R$, and observe that since by (i)
the ideals of $R_M$ are irreducible, they are linearly ordered.
Finally, suppose that each completely $Q$-irreducible proper
$R$-submodule of $Q$ is isomorphic for some maximal ideal $M$ to the
maximal ideal of $R_M$. Let $M$ be a maximal ideal of $R$. Then by
assumption $rMR_M$ is an irreducible ideal of $R_M$ for all $r \in
R$. By Lemma 5.1 of \cite{FHO3}, $R_M$ must be a valuation domain.
Thus $R$ is a Pr\"ufer domain since every localization of $R$ at a
maximal ideal is a valuation domain.
\end{proof}
In Theorem \ref{existence}, we describe the Pr\"ufer domains that
have a completely $Q$-irreducible ideal.
\begin{theorem} \label{existence}The following statements are
equivalent for a Pr\"ufer domain $R$.
\begin{itemize}
\item[(i)] There exists a completely $Q$-irreducible ideal of $R$.
\item[(ii)] There exists a nonzero $Q$-irreducible ideal of $R$.
\item[(iii)] There is a nonzero prime ideal contained in the
Jacobson radical of $R$.
\item[(iv)] Every proper $R$-submodule of $Q$ is a fractional ideal of $R$.
\end{itemize}
\end{theorem}
\begin{proof} (i) $\Rightarrow$ (ii) This is clear.
(ii) $\Rightarrow$ (iii) Suppose that $A$ is a $Q$-irreducible ideal
of $R$.
By Lemma~\ref{ideal
characterization}, $A = AR_P$ for some prime ideal of $R$. If $A$
is an invertible ideal of $R$, then by Theorem~\ref{qirreducible}
$P$ is the unique maximal ideal of $R$, so that statement (iii) is
clearly true. It remains to consider the case where $A$ is not
invertible. By Theorem~\ref{qirreducible}, if $x$ is a nonzero
element in $A^{-1}$, then $xA$ is contained in the Jacobson radical
of $R$. Since by Lemma~\ref{ideal characterization}(ii), $xA$ is
$Q$-irreducible we may assume without loss of generality that $A$
itself is contained in the Jacobson radical of $R$.
Now let $\{N_i\}$ be the set of maximal ideals of $R$. Since $AR_P$
is an ideal of $R$ and $A$ is contained in each $N_i$, it follows
that for each $i$, $AR_PR_{N_i} = AR_{N_i} \subset R_{N_i}$. Thus
there is prime ideal $P_i$ contained in $P$ and $N_i$ that contains
$A$ (the ideal $P_i$ can be chosen to be the contraction of the
maximal ideal of the ring $R_PR_{N_i}$ that contains $A$). Because
$R$ is a Pr\"ufer domain, the prime ideals contained in $P$ are
linearly ordered by inclusion. Thus if $Q = \bigcap_{i} P_i$, then
$Q$ is a nonzero prime ideal of $R$ (for it contains $A$) and $Q$ is
contained in every maximal ideal of $R$.
(iii) $\Rightarrow$ (i)
Let $P$ be a nonzero prime ideal of $R$ contained in the Jacobson
radical of $R$. Since $R$ is a Pr\"ufer domain, $P = PR_P$, so if
$0 \ne x$ is in $P$ it follows that $xR_M$ is contained in $P$. Thus
$xMR_M$ is contained in $P = PR_M$. Moreover by
Proposition~\ref{Prufer case} $xMR_M$ is a completely
$Q$-irreducible $R$-submodule of $Q$.
(iii) $\Rightarrow$ (iv) Statement (iv) is equivalent to the
assertion that there exists a valuation overring $V \subset Q$ of
$R$ such that $(R:_QV) \ne 0$ \cite[Theorem 79]{M2}. If $R$
satisfies (iii), then a nonzero prime ideal $P$ contained in the
Jacobson radical of $R$ has the property that $PR_P = P$. Thus
$V$ can be chosen to be $R_P$.
(iv) $\Rightarrow$ (ii) By the theorem of Matlis cited in (iii)
$\Rightarrow$ (iv), there exists a valuation ring $V$ with $(R:_QV)
\ne 0$. Thus since $R$ is a Pr\"ufer domain there is a prime ideal
$P$ with $V = R_P$ and $rR_P \subseteq R$ for some nonzero $r \in
R$. By Proposition~\ref{Prufer case}, $rR_P$ is a $Q$-irreducible
ideal of $R$.
\end{proof}
\begin{remark}
If $R$ is a Pr\"ufer domain with nonzero Jacobson radical ideal $J$,
then there exists a unique largest prime ideal $P$ contained in $J$.
If $M$ is a maximal ideal of $R$, then $PR_M = PR_P$ since $R_M$ is
a valuation domain. Thus $P = \bigcap_{M \in \Max(R)}PR_M = PR_P$.
It follows that $R_P/P$ is the quotient field of $R/P$. Using this
observation it is not hard to see that a Pr\"ufer domain $R$
satisfies the equivalent conditions of Theorem~\ref{existence} if
and only if $R$ occurs in a pullback diagram of the form
\begin{center} $\begin{CD} R @>>> D \\ @VVV \: @V{\alpha}VV \\ V
@>{\beta}>> K \end{CD}$
\end{center}
where
\begin{itemize}
\item[$\bullet$]
$\alpha$ is injective and $D$ is a Pr\"ufer domain such that the Jacobson
radical of $D$ does not contain a nonzero prime ideal,
\item[$\bullet$] $K$ is isomorphic to the quotient field of $D$,
and
\item[$\bullet$] $\beta$ is surjective with $V$ a valuation
domain.
\end{itemize}
Thus if $D$ is any Pr\"ufer domain with quotient field $Q$ and $X$
is an indeterminate for $Q$, then $D+XQ[[X]]$ is a Pr\"ufer domain
satisfying the equivalent conditions of Theorem~\ref{existence}.
\end{remark}
%\begin{proposition}
%Let $R$ be a Pr\"ufer domain. The following statements are equivalent for
%a
%completely $Q$-irreducible $R$-submodule $A$ of $Q$ with adjoint maximal
%ideal
%$M$.
%\begin{itemize}
%\item[(i)] $Q/A$ is an injective $R$-module.
%\item[(ii)] $R_M$ is an almost maximal valuation domain.
%\item[(iii)] $R$ is $A$-reflexive.
%\end{itemize}
%\end{proposition}
%\begin{proof} (i) $\Rightarrow$ (ii) Since $Q/A$ is injective, $Q/\End(A)$
%is
%also an injective $R$-module [reference needs to be inserted]. By
%Proposition~\ref{Prufer case}, $\End(A) = R_M$. Thus $Q/R_M$ is injective
%and
%since $R_M$ is a valuation domain, $R_M$ is almost maximal \cite{FS}.%
%(ii) $\Rightarrow$ (i) Since $R_M$ is an almost maximal valuation domain
%and
%$A$ is an $R_M$-module, $A$ has injective dimension one as a module over
%$R_M$-,
%hence over $R$.
%(i) $\Leftrightarrow$ (iii) (This follows from something in one of may
%papers--
%I need to look up the reference.)
%\end{proof}
\section{Questions}
We conclude with several questions that we have not been able to
resolve. Other questions touching on similar issues can be found in
\cite{BHLP}, \cite{Baz} and \cite{HHP}.
\begin{question}\label{2} What conditions on a domain $R$ guarantee
that any two completely $Q$-irreducible fractional ideals are
necessarily isomorphic?
\end{question}
Proposition~\ref{criterion} gives an answer to this question in the
case where every proper submodule of $Q$ is a fractional $R$-ideal.
By Theorem~\ref{Prufer case} if $R$ is a valuation domain, then all
completely $Q$-irreducible ideals of $R$ are isomorphic. If $R$ is a
Noetherian local domain, then by Propositions~\ref{criterion}
and~\ref{same} any two $Q$-irreducible ideals are isomorphic.
\begin{question} \label{3} What integral domains $R$ admit a completely
$Q$-irreducible ideal? a nonzero $Q$-irreducible ideal?
\end{question}
The Noetherian and Pr\"ufer cases of Question~\ref{3} are settled in
Proposition~\ref{same} and Theorem~\ref{existence}, respectively.
\begin{question} \label{33} If $R$ admits a nonzero $Q$-irreducible
ideal, does $R$ also admit a completely $Q$-irreducible ideal?
\end{question}
The answer to Question \ref{33} is yes if $R$ is Pr\"ufer or
Noetherian.
\begin{question} \label{4} If $A$ is a (completely)
irreducible submodule of the quotient field of a quasilocal domain
$R$, what can be said about $\End(A)$? For a completely
$Q$-irreducible ideal $A$ of a quasilocal domain $R$ does it follow
that $\End(A)$ is integral over $R$?
\end{question}
Theorem \ref{existence} along with the fact that if $A$ is
completely irreducible, then $\End(A)$ is quasilocal, shows that if
$R$ is not quasilocal, then $\End(A)$ need not be integral over $R$
even if $R$ is a Pr\"ufer domain.
Theorem~\ref{qirreducible}, Example~\ref{gilmerhoffmann} and
Example~\ref{valuation example} are relevant to Question~\ref{4}.
\begin{question}\label{5} If $R$ is a (Noetherian) domain, what are the
completely irreducible submodules of $Q$?
\end{question}
Theorem~\ref{Prufer case} answers Question~\ref{5} in the case where
$R$ is Pr\"ufer.
\begin{question} \label{6} If $A$ is a completely $Q$-irreducible
$R$-submodule of $Q$, when is $A$ a fractional ideal of $R$? of
$\End(A)$?
\end{question}
If $R$ is a valuation domain, then every proper submodule of $Q$ is
a fractional ideal of $R$. The case where $R$ is a one-dimensional
Noetherian domain is deeper, but has been resolved independently by
Bazzoni and Goeters. A consequence of Theorem 3.4 of \cite{Baz} is
that if $A$ is a completely $Q$-irreducible submodule of $Q$ such
that $\End(A)$ is Noetherian and has Krull dimension 1, then (by
Theorem~\ref{qirreducible}) $\End(A)$ is local and (by the cited
result of Bazzoni) $A$ is a fractional ideal of $\End(A)$. Indeed, a
more general result due to H. P. Goeters is true: If $A$ is a
submodule of the quotient field of a local Noetherian domain of
Krull dimension 1, then $A$ is a fractional ideal of $\End(A)$
\cite[Lemma 1]{Goe}. Recently, Goeters has extended this to all
quasilocal Matlis domains \cite{Goe2}.
\section{Appendix: Corrections to \cite{FHO2}}
In this appendix we correct several mistakes from our earlier paper
\cite{FHO2}. We include also a stronger version of Lemma 3.2 of
this paper. The main corrections concern Lemmas 2.1(iv) and 3.2 of
\cite{FHO2}. The notation and terminology of this appendix is that
of \cite{FHO2}.
The proof of statement (iv) of Lemma 2.1 of \cite{FHO2} is
incorrect. Statement (iv) should be modified in the following way:
\begin{itemize}
\item[(iv)] {\it For each nonzero nonmaximal prime ideal $P$ of $R$,
if $\{M_i\}$ is the collection of maximal ideals of $R$ not
containing $P$, then $R_P \subseteq (\bigcap_{i}R_{M_i})R_M$ for
each maximal ideal $M$ of $R$ containing $P$.}
\end{itemize}
{\noindent}Having changed statement (iv), we modify now the original
proofs of (iii) $\Rightarrow$ (iv) and (iv) $\Rightarrow$ (v) in
the following way. For (iii) $\Rightarrow$ (iv) we note that by
Theorem 3.2.6 of \cite{FHP} $\End(P) = R_P \cap (\bigcap_i R_{M_i})$
and $\End(P_M) = R_P$. Thus by (iii) $R_P = \End(P_M) = \End(P)_M =
R_P \cap (\bigcap_{i}R_{M_i})R_M$, and (iv) follows.
For the proof of (iv) $\Rightarrow$ (v), we have as in the original
proof that
$$
R_P = \End(A)_M =(\bigcap_{Q \in \XX_A} R_{Q})R_M \cap
(\bigcap_{N}R_N)R_M.
$$
We claim that $\bigcap_{Q \in \XX_A} R_{Q} \subseteq R_P$. If this
is not the case then since $R_M$ is a valuation domain it must be
that $R_P \subset (\bigcap_{Q \in \XX_A} R_{Q})R_M$ (proper
containment). Hence from the above representation of $\End(A)_M$ we
deduce that since $R_P$ is a valuation domain, $R_P =
(\bigcap_{N}R_N)R_M$. Thus $(\bigcap_{N}R_N)R_M \subset (\bigcap_{Q
\in \XX_A} R_{Q})R_M$. By (iv), $R_{Q'} \subseteq
(\bigcap_{N}R_N)R_M$ since no $N$ contains $Q'$. However $Q' \in
\XX_A$, so this implies $R_{Q'} \subset R_{Q'}R_M$, but since $M$
contains $Q'$, $R_{Q'}R_M = R_{Q'}$. This contradiction implies
that $\bigcap_{Q \in \XX_A}R_Q \subseteq R_P$, so every element $r
\in P$ is contained in some $Q \in \XX_A$. Consequently, no element
of $P$ is prime to $A$.
\medskip
Reference is made in the first paragraph of the proof of Lemma 3.3
of \cite{FHO3} to the original version of statement (iv). In
particular it is claimed that since $R$ has the separation property,
$P_iS $ is a maximal ideal of $S$. This can be justified now using
the following more general fact, which does not appear explicitly in
\cite{FHO2}:
\begin{lemma} A Pr\"ufer domain $R$ has the separation property if and
only if for each collection $\{P_i:i \in I\}$ of incomparable prime
ideals, the ideals $P_i$ extend to maximal ideals of
$S:=\bigcap_{i\in I} R_{P_i}$.
\end{lemma}
\begin{proof} If $R$ has the separation property, then for each $j \in I$,
$\End(P_j) = R_{P_j} \cap (\bigcap_{N}R_N)$ by Theorem 3.2.6 of
\cite{FHP}, where $N$ ranges over the maximal ideals of $R$ that do
not contain $P_j$. Thus $\End(P_j) \subseteq S$ since the $P_i$'s
are comaximal. By Lemma 2.1(ii) of \cite{FHO2} $P_j$ is a maximal
ideal of $\End(P_j)$, and since $R$ is a Pr\"ufer domain, either
$P_j$ extends to a maximal ideal $SP_j$ of $S$ or $SP_j = S$. The
latter case is impossible since $S \subseteq R_{P_j}$. Thus $SP_j$
is a maximal ideal of $S$. The converse follows from Theorem 3.2.6
of \cite{FHO} and Lemma 2.1(ii) of \cite{FHO2}.
\end{proof}
A second reference to the original version of Lemma 2.1(iv) is made
in the first paragraph of the proof of (i) $\Rightarrow$ (ii) of
Theorem 3.7. In this paragraph it is claimed that since $\End(A)_M
= R_P$, the elements of $P$ are not prime to $A$. Since (by Theorem
2.3 of \cite{FHO2}) $R$ has the separation property, this claim is
immediate from Lemma 2.1(v) and the original argument that appealed
to Lemma 2.1(iv) is unnecessary. \medskip
The argument in the third paragraph of the proof of Lemma 3.2 of
\cite{FHO2} is incorrect, but rather than patch this argument we
give below a stronger version of this lemma. It requires a slight
strengthening of Lemma 3.1 of \cite{FHO2}.
\begin{lemma}
\label{lemma 6.6} (cf. Lemma 3.1 of \cite{FHO2}) Let $A$ be an
ideal of a Pr\"ufer domain $R$. Suppose $Q$ is a prime ideal of $R$
that contains $A$, and $P$ is a prime ideal such that $\End(A)_Q =
R_P$. If $P \in \Ass(A)$, then $\End(A)_Q = \End(A_Q)$. \end{lemma}
\begin{proof} Since $P \in \Ass(A)$, $A_{(P)}$ is a primal ideal with
adjoint prime $P$, and it follows that $A_P$ is a $P_P$-primal
ideal. By \cite[Lemma 1.4]{FHO2}, $\End(A_P) = R_P$. Thus
$\End(A_P) = \End(A)_Q$, so $A\End(A_P) = A\End(A)_Q$ implies $A_P =
A_Q$. Consequently, $ \End(A_Q) = \End(A_P) = R_P =
\End(A)_Q$.\end{proof}
\begin{lemma}
\label{lemma 6.7} (cf. Lemma 3.2 of \cite{FHO2})
Let $R$ be a Pr\"ufer domain with field of
fractions $F$, let $X$ be an $R$-submodule of $F$, and let $M$ be a
maximal ideal of $R$. Then $\End(X)_M = R_P$ for some $P \in \Spec
R$ with $P \subseteq M$. Assume that $P$ is the union of prime
ideals $P_i$, where each $P_i$ is the radical of a finitely
generated ideal. Then $\End(X)_Q = \End(X_Q)$ for all prime ideals
$Q$ such that $P \subseteq Q \subseteq M$. \end{lemma}
\begin{proof} Since $R_M \subseteq \End(X)_M$ and $R_M$ is a
valuation domain, $\End(X)_M = R_P$ for some prime ideal $P
\subseteq M$. If $\End(X)_M = F$, then clearly $\End(X)_M =
\End(X_M)$, so we assume $\End(X)_M \ne F$ and thus $P \ne (0)$. Let
$Q$ be a prime ideal of $R$ such that $P \subseteq Q \subseteq M$.
Since $\End(X)_M = R_P$, we have $\End(X)_Q = R_P$. Now
$R_P = \End(X)_Q \subseteq \End(X_Q) \subseteq
\End(X_P)$, so to prove Lemma \ref{lemma 6.7}, it suffices to show
that $\End(X_P) \subseteq R_P$.
Let $S = \End(X)$. Now $PS \subseteq PR_P$, so $PS \ne S$. Since $S$ is
an overring of the Pr\"ufer domain $R$, $S$ is a flat extension of
$R$, so $PS$ is a prime ideal of $S$ and $S_{PS} = R_P$. Also,
$PS$ is the union of the prime ideals $P_iS$, and each $P_iS$ is
the radical of a finitely generated ideal of $S$.
Let $L$ be a prime ideal of $S$ such that $L \subseteq PS$ and such
that $L = \sqrt{I}$, where $I$ is a finitely generated ideal of $S$.
We prove there exists a nonzero $q \in F$ such that $qX_L$ is an
ideal of $S_L$ that is primary for $L_L$. The invertible ideal $I^2$
of $S$ is an intersection of principal fractional ideals of $S$.
Since $\End(X) = S$, each principal fractional ideal of $S$ is an
intersection of $S$-submodules of $F$ of the form $qX$, $q \in F$.
Since $I^2 \subseteq L$, $I^2$ is an intersection of ideals of $S$
of the form $L \cap qX$, where $q \in F$. Since $I^2 \subset I
\subseteq L$ (where $\subset$ denotes proper containment), there
exists $q \in F$ such that $I^2 \subseteq L \cap qX \subset L$.
Hence there exists a maximal ideal $N$ of $S$ with $L \subseteq N$
such that $I^2_N \subseteq L_N \cap qX_N \subset L_N$. Since $S_N$
is a valuation domain, the $S_N$-modules $qX_N$ and $L_N$ are
comparable and $I^2_N \subseteq L_N \cap qX_N \subset L_N$ implies
$I^2_N \subseteq qX_N \subset L_N$. Now $\sqrt{I^2} =\sqrt{I} = L$
and $I^2 \subseteq N$ implies $L \subseteq N$. Thus $I^2_L \subseteq
qX_L \subseteq L_L$, and we conclude that $\sqrt{qX_L} = L_L$.
We observe next that $X_P \ne F$. Since $P \ne 0$, there exists $i$
such that $P_i \ne 0$ and $L := P_iS \subseteq PS$, where $L =
\sqrt{I}$ for some finitely generated ideal $I$ of $S$. As we have
established in the paragraph above, there exists a nonzero $q \in F$
such that $qX_L$ is an ideal of $S_L$. Thus $qX_P \subseteq qX_L
\subseteq S_L$, so it is not possible that $X_P =F$.
Fix some member $L$ of the chain $\{P_iS\}$. Since $X_P \ne F$, $L
\subseteq PS$ and $R_P$ is a valuation domain, there exists a
nonzero element $s$ of $S$ such that $sX \subseteq L_L$. Since
$\End(X_P) = \End(sX_P)$ and we wish to show that $\End(X_P)
\subseteq R_P$ we may assume without loss of generality that $s =
1$; that is, we assume for the rest of the proof that $X \subseteq
L_{L}$. Define $A = X \cap S$. Then $A$ is an ideal of $S$.
Moreover $A$ is contained in $L$ since $A_{L} \subseteq X_L
\subseteq L_L$.
With the aim of applying Lemma~\ref{lemma 6.6}, we show that $PS \in
\Ass(A)$. For each $i$ define $L_i = P_iS$. It suffices to show
each $L_i$ with $L \subseteq L_i \subseteq PS$ is in $\Ass(A)$,
since this implies that $PS = \bigcup_{L_i \supseteq L}L_i$ is a
union of members of $\Ass(A)$. Let $i$ be such that $L \subseteq
L_i$. Since $L_i$ is the radical of a finitely generated ideal of
$S$, there exists (as we have established above) a nonzero $q \in
F$ such that $qX_{L_i}$ is an ideal of $S_{L_i}$ that is primary for
$(L_i)_{L_i}$. Now $A_{L_i} = X_{L_i} \cap S_{L_i}$. Since $S_{L_i}$
is a valuation domain, $A_{L_i} = X_{L_i}$ or $S_{L_i} \subseteq
X_{L_i}$. By assumption, $X \subseteq L_L$. Since $L \subseteq
L_i$, it follows that $X_{L_i} \subseteq L_{L}$, so it is impossible
that $S_{L_i} \subseteq X_{L_i}$. Thus $A_{L_i} = X_{L_i}$.
Consequently, $qX_{L_i} = qA_{L_i}$ and $qA_{L_i}$ is an ideal of
$S_{L_i}$ that is primary for $(L_i)_{L_i}$. Since $S_{L_i}$ is a
valuation domain, it follows that $qA_{L_i}= A_{L_i}:s$ for some $s
\in S$. Thus $(L_i)_{L_i} \in \Ass(A_{L_i})$, so $L_i \in
\Ass(A)$. This proves $PS \in \Ass(A)$.
Since $A =X \cap S$ is an ideal of $S$, $S \subseteq \End(A)$. For
each maximal ideal $N$ of $S$, either $A_N = X_N$ or $A_N = S_N$ It
follows that $\End(A) \subseteq \End(X) = S$, so $\End(A) = S$.
Thus $\End(A)_P = S_P = R_P$, and by Lemma~\ref{lemma 6.6},
$\End(A_P) = R_P$. (We have used here that $S_{SP} = R_P$.) Now
$A_P = X_P \cap S_P = X_P \cap R_P$. Since $R_P$ is a valuation
domain, $A_P = X_P$ or $R_P \subseteq X_P$. The latter case is
impossible since $X_P \subseteq X_L \subseteq L_L$. Thus
$A_P = X_P$. We conclude that End($X_{P}) =$ End($A_{P}) = R_P$.
\end{proof}
Finally we make two corrections to the proof of Lemma 3.3. The
third paragraph should read: {\it Define $A = JR_Q \cap R$. Then $AS
= JR_Q \cap S$ is $QS$-primary. In particular, $QS$ is the unique
minimal prime of $AS$ and $A \not\subseteq P_iS \cap R = P_i$ for
each $i \ge 1$.}
Also, in the fifth paragraph an exponent is incorrect: $x_i$ needs
to be chosen in $A_i \setminus (P_1 \cup \cdots \cup P_i \cup
A^{i+1})$. Then in the eighth paragraph, we have $x_{i+1}S_N
\subset x_{i}S_N$ since $x_i \in A^i \setminus A^{i+1}$ and
$A^{i+1}S_N \cap R = A^{i+1}R_Q \cap R = A^{i+1}$.
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\bigskip
\end{document}
\section*{Questions}
{\noindent}This section will eventually be deleted.
\begin{itemize}
\item[(1)]
Is a completely $Q$-irreducible ideal of a quasilocal integrally
closed domain an m-canonical ideal?
\item[(2)] Find conditions on $R$
in order that any two completely $Q$-irreducible fractional
$R$-ideals are necessarily isomorphic.
\item[(3)] What integral domains $R$ admit a completely $Q$-irreducible
fractional ideal?
\item[(4)] If $R$ is a quasilocal domain with completely $Q$-irreducible
ideal $A$, is $\End(A) = R$?
\item[(5)] If $R$ is a Noetherian domain, then is it possible to describe
all completely $Q$-irreducible submodules of the quotient field $Q$?
(Proposition~\ref{Prufer case} takes care of the Pr\"ufer case.)
\item[(6)] If $A$ is a completely $Q$-irreducible submodule of the quotient
field of a domain, is $A$ a fractional ideal over $\End(A)$? It is
easy to see that this is true for Pr\"ufer domains. I believe
Bazzoni shows that it is also true when $\End(A)$ is a Noetherian
domain.
\item[(7)] If $A$ is a completely $Q$-irreducible $R$-submodule of the
quotient field of a quasilocal domain $R$, what can be said about
$\End(A)$? I think at the very least it is true that $\End(A)M$ is
not equal to $\End(A)$, where $M$ is the maximal ideal of R. (This
is because $\Hom(M,A)$ can be viewed as a cover of $A$.) Must
$\End(A)$ be integral over $R$?
\item[(8)] Is it possible that an ideal of a non-Dedekind (local?)
Noetherian domain $R$ is an infinite irredundant intersection of
completely $Q$-irreducible submodules of the quotient field of $R$?
\item[(9)] Can it happen that every ideal of a Noetherian domain is an
irredundant intersection of completely $Q$-irreducible submodules of
the quotient field of R? (Is it intended here to be every nonzero
ideal? For a domain that is not a field, I believe $(0)$ is never an
irredundant intersection of completely $Q$-irreducible ideals. On
the other hand we are asserting that for $R = \mathbb Z$ this is
true. I bet it is true for all Dedekind domains.)
\item[(10)] If $R$ is a quasilocal domain and $A$ is an $R$-submodule of
$Q$, is it true that: $\End(A)$ is $A$-reflexive iff $Q/A$ is
injective and $A$ is completely $Q$-irreducible. (The terminology
is from Bazzoni and Salce and is also in \cite{FS}.)
\item[(11)] Are valuation overrings useful in constructing examples of
completely $Q$-irreducible submodules of $Q$ when the base ring $R$
is Noetherian? If $V$ is a DVR overring of a quasilocal domain $R$
(with maximal ideal $M$) that is completely $Q$-irreducible as an
$R$-submodule of $Q$, then I believe $V$ is centered on a maximal
ideal of $R$. Furthermore I believe $V/VM = R/M$ so that $V$ is
zero-dimensional. For $VM = Vm$ for some $m$ in $M$. Thus $[V:M] =
[V:VM] = V(1/m)$, so $[V:M] / V = V(1/m) / V = V / Vm.$ (By
$[V:M]$ I mean the submodule of $Q$ that is identified with
$Hom(M,V)$.) If $V$ is completely $Q$-irreducible, then $[V:M] / V$,
hence $V / Vm$, must be $R/M$. \end{itemize}
\end{document}
%End Document
\end{document}
Let $A$ be a fractional ideal of a domain $R$. Then $R$ is an {\it
$A$-reflexive domain} if and only if the canonical homomorphism $$G
\rightarrow \Hom_R(\Hom_R(G,A),A)$$ is an isomorphism for all
$\End(A)$-submodules $G$ of finite rank free $\End(A)$-modules. It
is a consequence of Theorem 5.3 in \cite{FS} that if $\End(A) = R$,
then $R$ is an $A$-reflexive domain if and only if $A$ is an
$m$-canonical ideal of $R$ and $Q/A$ is an injective $R$-module.
This raises the question of whether ``$m$-canonical'' can be
weakened to ``completely $Q$-irreducible'':
\begin{question} \label{7} If $R$ is a quasilocal domain and $A$ is an
$R$-submodule of $Q$, is it true that: $\End(A)$ is $A$-reflexive if
and only if $Q/A$ is injective and $A$ is completely
$Q$-irreducible.
\end{question}
By Lemma~\ref{Bazzoni} the answer to Question~\ref{7} is affirmative
if $R$ is Noetherian. Similarly, the answer is affirmative for
valuation domains since completely $Q$-irreducible ideals of these
domains are $m$-canonical.
\begin{proposition} If $R$ is a domain and $A$ is a fractional ideal of $R$,
then $A$ is a completely $Q$-irreducible fractional ideal of $R$ iff
$A$ is an completely $Q$-irreducible fractional ideal of $R_M$ for
some maximal ideal of $M$ of $R$. \end{proposition}
\begin{proof} It suffices to prove the claim when $A$ is an ideal of $R$.
This is because a fractional ideal $B$ of a ring $S$ is completely
$Q$-irreducible if and only if every fractional ideal isomorphic to
$B$ is completely $Q$-irreducible.
Suppose that $A$ is a completely $Q$-irreducible ideal of $R$. Then $A =
\cap AR_M$, where $M$ ranges over the maximal ideals of $R$, so
since $A$ is completely $Q$-irreducible, $A = AR_M$ for some maximal
ideal $M$ of $R$. Clearly $A$ is a completely $Q$-irreducible ideal
of $R_M$.
Conversely, suppose that $A$ is a completely $Q$-irreducible ideal of$R_M$
for some maximal ideal $M$. Since $A$ is a completely
$Q$-irreducible ideal of $R_M$, there exists $x \in (A : MR_M)
\setminus A$. Write $x = \frac{c}{b}$, where $c \in R$ and $b \in R
\setminus M$. Then if $c \in A$, since $A = AR_M$ it follows that
$x \in A$, contrary to the choice of the $x$. Thus $x \in (A:M)
\setminus A$, and to prove that $A$ is a completely $Q$-irreducible
ideal of $R$, it suffices to show that $x$ is in every fractional
ideal $J$ properly containing $A$. We show in fact that $x$ is in
$JR_N$ for all maximal ideals $N$ of $R$. By the choice of $x$, we
have that $x \in (A: MR_M)$, so since $A$ is a completely
$Q$-irreducible ideal of $R_M$, $(A : MR_M)$ is contained in $JR_M$.
(Note that $A$ is properly contained in $JR_M$, since otherwise $A =
AR_M = JR_M$, and $J \subseteq A$, contrary to assumption.) Finally
if $N$ is a maximal ideal of $R$ distinct from $M$, then since $x$
is in $(A:M)$, we have $xM \subseteq AR_N$, so $x \in AR_N \subseteq
JR_N$. This shows $x \in JR_N$ for every maximal ideal $N$ of $R$,
proving the claim.
\end{proof}
\begin{corollary} There exists a completely $Q$-irreducible ideal
of $R$ if and only if there is a maximal ideal $M$ of $R$ such that
$R_M$ has an absolutely irreducible ideal and $R_M$ is a fractional
ideal of $R$. \end{corollary}
(i) $\Rightarrow$ (iii) Suppose that $A$ is a completely
$Q$-irreducible ideal of $R$. Then $A = \bigcap_{M \in \Max(R)}
AR_M$, where $M$ ranges over the maximal ideals of $R$, so since $A$
is completely $Q$-irreducible, $A = AR_M$ for some maximal ideal $M$
of $R$. Clearly $A$ is a completely $Q$-irreducible ideal of $R_M$.
(iii) $\Rightarrow$ (i) Suppose that $A$ is a completely
$Q$-irreducible ideal of $R_M$ for some maximal ideal $M$. Since
$A$ is a completely $Q$-irreducible ideal of $R_M$, there exists $x
\in (A : MR_M) \setminus A$. Write $x = \frac{c}{b}$, where $c \in
R$ and $b \in R \setminus M$. Then if $c \in A$, since $A = AR_M$
it follows that $x \in A$, contrary to the choice of the $x$.
Thus $x \in (A:M) \setminus A$, and to prove that $A$ is an absolutely
irreducible ideal of $R$, it suffices to show that $x$ is in every
fractional ideal $J$ properly containing $A$. We show in fact that
$x$ is in $JR_N$ for all maximal ideals $N$ of $R$. By the choice
of $x$, we have that $x \in (A: MR_M)$, so since $A$ is a completely
$Q$-irreducible ideal of $R_M$, $(A : MR_M)$ is contained in $JR_M$.
(Note that $A$ is properly contained in $JR_M$, since otherwise $A =
AR_M = JR_M$, and $J \subseteq A$, contrary to assumption.) Finally
if $N$ is a maximal ideal of $R$ distinct from $M$, then since $x$
is in $(A:M)$, we have $xM \subseteq AR_N$, so $x \in AR_N \subseteq
JR_N$. This shows $x \in JR_N$ for every maximal ideal $N$ of $R$,
proving the claim. No virus found in this outgoing message. Checked
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