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\documentclass{amsart}
\usepackage{newlattice}
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\theoremstyle{definition}
\newtheorem{definition}{Definition}
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\newcommand{\Prodm}[2]{\GrP(\,#1\mid#2\,)}
% product with a middle
\newcommand{\Prodsm}[2]{\GrP^{*}(\,#1\mid#2\,)}
% product * with a middle
\newcommand{\vectsup}[2]{\vect<\dots,0,\dots,\overset{#1}{#2},%
\dots,0,\dots>}% special vector
\newcommand{\Dsq}{D^{\langle2\rangle}}
\begin{document}
\title[Complete-simple distributive lattices]
{A construction of complete-simple\\
distributive lattices}
\author{George~A. Menuhin}
\address{Computer Science Department\\
University of Winnebago\\
Winnebago, Minnesota 23714}
\email{
[email protected]}
\urladdr{
http://math.uwinnebago.edu/homepages/menuhin/}
\thanks{Research supported by the NSF under grant number~23466.}
\keywords{Complete lattice, distributive lattice, complete
congruence, congruence lattice}
\subjclass[2000]{Primary: 06B10; Secondary: 06D05}
\date{March 15, 2006}
\begin{abstract}
In this note we prove that there exist \emph{complete-simple
distributive lattices,} that is, complete distributive
lattices in which there are only two complete congruences.
\end{abstract}
\maketitle
\section{Introduction}\label{S:intro}
In this note we prove the following result:
\begin{named}{Main Theorem}
There exists an infinite complete distributive lattice
$K$ with only the two trivial complete congruence relations.
\end{named}
\section{The $\Dsq$ construction}\label{S:Ds}
For the basic notation in lattice theory and universal algebra,
see Ferenc~R. Richardson~\cite{fR82} and George~A.
Menuhin~\cite{gM68}. We start with some definitions:
\begin{definition}\label{D:prime}
Let $V$ be a complete lattice, and let $\Frak{p} = [u, v]$ be
an interval of $V$. Then $\Frak{p}$ is called
\emph{complete-prime} if the following three conditions
are satisfied:
\begin{enumeratei}
\item $u$ is meet-irreducible but $u$ is \emph{not}
completely meet-irreducible;\label{m-i}
\item $v$ is join-irreducible but $v$ is \emph{not}
completely join-irreducible;\label{j-i}
\item $[u, v]$ is a complete-simple lattice.\label{c-s}
\end{enumeratei}
\end{definition}
Now we prove the following result:
\begin{lemma}\label{L:Dsq}
Let $D$ be a complete distributive lattice satisfying
conditions \itemref{m-i} and~\itemref{j-i}.
Then $\Dsq$ is a sublattice of $D^{2}$; hence $\Dsq$ is
a lattice, and $\Dsq$ is a complete distributive lattice
satisfying conditions \itemref{m-i} and~\itemref{j-i}.
\end{lemma}
\begin{proof}
By conditions~\itemref{m-i} and \itemref{j-i}, $\Dsq$ is a
sublattice of $D^{2}$. Hence, $\Dsq$ is a lattice.
Since $\Dsq$ is a sublattice of a distributive lattice,
$\Dsq$ is a distributive lattice. Using the characterization
of standard ideals in Ernest~T. Moynahan~\cite{eM57},
$\Dsq$ has a zero and a unit element, namely,
$\vect<0, 0>$ and $\vect<1, 1>$. To show that $\Dsq$ is
complete, let $\empset \ne A \contd \Dsq$, and let $a = \JJ A$
in $D^{2}$. If $a \in \Dsq$, then
$a = \JJ A$ in $\Dsq$; otherwise, $a$ is of the form
$\vect<b, 1>$ for some $b \in D$ with $b < 1$. Now
$\JJ A = \vect<1, 1>$ in $D^{2}$, and
the dual argument shows that $\MM A$ also exists in
$D^{2}$. Hence $D$ is complete. Conditions \itemref{m-i}
and~\itemref{j-i} are obvious for $\Dsq$.
\end{proof}
\begin{corollary}\label{C:prime}
If $D$ is complete-prime, then so is $\Dsq$.
\end{corollary}
The motivation for the following result comes from Soo-Key
Foo~\cite{sF90}.
\begin{lemma}\label{L:ccr}
Let $\GrQ$ be a complete congruence relation of $\Dsq$ such
that
\begin{equation}\label{E:rigid}
\congr \vect<1, d>=\vect<1, 1>(\GrQ),
\end{equation}
for some $d \in D$ with $d < 1$. Then $\GrQ = \Gri$.
\end{lemma}
\begin{proof}
Let $\GrQ$ be a complete congruence relation of $\Dsq$
satisfying \itemref{E:rigid}. Then $\GrQ = \Gri$.
\end{proof}
\section{The $\Grp^{*}$ construction}\label{S:P*}
The following construction is crucial to our proof of the
Main~Theorem:
\begin{definition}\label{D:P*}
Let $D_{i}$, for $i \in I$, be complete distributive
lattices satisfying condition~\itemref{j-i}. Their $\Grp^{*}$
product is defined as follows:
\[
\Prodsm{ D_{i} }{i \in I} = \Prodm{ D_{i}^{-} }{i \in I}+1;
\]
that is, $\Prodsm{ D_{i} }{i \in I}$ is
$\Prodm{ D_{i}^{-} }{i \in I}$ with a new unit element.
\end{definition}
\begin{notation}
If $i \in I$ and $d \in D_{i}^{-}$, then
\[
\vectsup{i}{d}
\]
is the element of $\Prodsm{ D_{i} }{i \in I}$ whose
$i$-th component is $d$ and all the other
components are $0$.
\end{notation}
See also Ernest~T. Moynahan~\cite{eM57a}. Next we verify:
\begin{theorem}\label{T:P*}
Let $D_{i}$, for $i \in I$, be complete distributive
lattices satisfying condition~\itemref{j-i}. Let $\GrQ$
be a complete congruence relation on
$\Prodsm{ D_{i} }{i \in I}$. If there exist
$i \in I$ and $d \in D_{i}$ with $d < 1_{i}$ such
that for all $d \leq c < 1_{i}$,
\begin{equation}\label{E:cong1}
\congr\vectsup{i}{d}=\vectsup{i}{c}(\GrQ),
\end{equation}
then $\GrQ = \Gri$.
\end{theorem}
\begin{proof}
Since
\begin{equation}\label{E:cong2}
\congr\vectsup{i}{d}=\vectsup{i}{c}(\GrQ),
\end{equation}
and $\GrQ$ is a complete congruence relation, it follows
from condition~\itemref{c-s} that
\begin{equation}\label{E:cong}
\begin{split}
&\langle \dots, \overset{i}{d}, \dots, 0,
\dots \rangle\\
&\equiv \bigvee ( \langle \dots, 0, \dots,
\overset{i}{c},\dots, 0,\dots \rangle \mid d \leq c < 1)
\equiv 1 \pmod{\Theta}.
\end{split}
\end{equation}
Let $j \in I$, for $j \neq i$, and let
$a \in D_{j}^{-}$. Meeting both sides of the congruence
\itemref{E:cong} with $\vectsup{j}{a}$, we obtain
\begin{equation}\label{E:comp}
\begin{split}
0 &= \vectsup{i}{d} \mm \vectsup{j}{a}\\
&\equiv \vectsup{j}{a}\pod{\GrQ}.
\end{split}
\end{equation}
Using the completeness of $\GrQ$ and \itemref{E:comp}, we get:
\begin{equation}\label{E:cong3}
\congr{0=\JJm{ \vectsup{j}{a} }{ a \in D_{j}^{-} }}={1}(\GrQ),
\end{equation}
hence $\GrQ = \Gri$.
\end{proof}
\begin{theorem}\label{T:P*a}
Let $D_{i}$, for $i \in I$, be complete distributive
lattices satisfying
conditions \itemref{j-i} and~\itemref{c-s}. Then
$\Prodsm{ D_{i} }{i \in I}$ also satisfies
conditions~\itemref{j-i} and \itemref{c-s}.
\end{theorem}
\begin{proof}
Let $\GrQ$ be a complete congruence on
$\Prodsm{ D_{i} }{i \in I}$. Let $i \in I$. Define
\begin{equation}\label{E:dihat}
\widehat{D}_{i} = \setm{ \vectsup{i}{d} }{ d \in D_{i}^{-} }
\uu \set{1}.
\end{equation}
Then $\widehat{D}_{i}$ is a complete sublattice of
$\Prodsm{ D_{i} }{i \in I}$, and $\widehat{D}_{i}$
is isomorphic to $D_{i}$. Let $\GrQ_{i}$ be the
restriction of $\GrQ$ to $\widehat{D}_{i}$. Since
$D_{i}$ is complete-simple, so is $\widehat{D}_{i}$,
hence $\GrQ_{i}$ is $\Gro$ or $\Gri$. If $\GrQ_{i} = \Gro$
for all $i \in I$, then $\GrQ = \Gro$.
If there is an $i \in I$, such that $\GrQ_{i} = \Gri$,
then $\congr0=1(\GrQ)$, and hence $\GrQ = \Gri$.
\end{proof}
The Main Theorem follows easily from Theorems~\ref{T:P*} and
\ref{T:P*a}.
\begin{thebibliography}{9}
\bibitem{sF90}
Soo-Key Foo, \emph{Lattice Constructions}, Ph.D. thesis,
University of Winnebago, Winnebago, MN, December, 1990.
\bibitem{gM68}
George~A. Menuhin, \emph{Universal algebra}. D.~van
Nostrand, Princeton, 1968.
\bibitem{eM57}
Ernest~T. Moynahan, \emph{On a problem of M. Stone},
Acta Math. Acad. Sci. Hungar. \tbf{8} (1957), 455--460.
\bibitem{eM57a}
\bysame, \emph{Ideals and congruence relations in
lattices}.~II, Magyar Tud. Akad. Mat. Fiz. Oszt. K\"{o}zl.
\tbf{9} (1957), 417--434 (Hungarian).
\bibitem{fR82}
Ferenc~R. Richardson, \emph{General lattice theory}. Mir,
Moscow, expanded and revised ed., 1982 (Russian).
\end{thebibliography}
\end{document}
