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\documentclass{amsart}
\usepackage{amssymb,latexsym}
\usepackage{lattice}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem*{main}{Main~Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}

\theoremstyle{definition}
\newtheorem{definition}{Definition}

\theoremstyle{remark}
\newtheorem*{notation}{Notation}

\numberwithin{equation}{section}

\newcommand{\Prodm}[2]{\gP(\,#1\mid#2\,)}
  % product with a middle
\newcommand{\Prodsm}[2]{\gP^{*}(\,#1\mid#2\,)}
  % product * with a middle
\newcommand{\vct}[2]{\vv<\dots,0,\dots,\overset{#1}{#2},%
\dots,0,\dots>}% special vector
\newcommand{\fp}{\F{p}}% Fraktur p
\newcommand{\Ds}{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.ca/homepages/menuhin/}
\thanks{Research supported by the NSF under grant number~23466.}
\keywords{Complete lattice, distributive lattice, complete
  congruence, congruence lattice}
\subjclass{Primary: 06B10; Secondary: 06D05}
\date{March 15, 1995}

\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{main}
  There exists an infinite complete distributive lattice
  $K$ with only the two trivial complete congruence relations.
\end{main}

\section{The $\Ds$ 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 $\fp = [u, v]$ be
  an interval of $V$.  Then $\fp$ is called
  \emph{complete-prime} if the following three conditions are satisfied:
  \begin{enumerate}
     \item[(1)] $u$ is meet-irreducible but $u$ is \emph{not}
        completely meet-irreducible;
     \item[(2)] $v$ is join-irreducible but $v$ is \emph{not}
        completely join-irreducible;
     \item[(3)] $[u, v]$ is a complete-simple lattice.
  \end{enumerate}
\end{definition}

Now we prove the following result:

\begin{lemma}\label{L:ds}
  Let $D$ be a complete distributive lattice satisfying
  conditions~\textup{(1)} and~\textup{(2)}.
  Then $\Ds$ is a sublattice of $D^{2}$; hence $\Ds$ is
  a lattice, and $\Ds$ is a complete distributive lattice
  satisfying conditions~~\textup{(1)} and~~\textup{(2)}.
\end{lemma}

\begin{proof}
  By conditions~(1) and (2), $\Ds$ is a sublattice of
  $D^{2}$.  Hence, $\Ds$ is a lattice.

  Since $\Ds$ is a sublattice of a distributive lattice, $\Ds$ is
  a distributive lattice.  Using the characterization of
  standard ideals in Ernest~T. Moynahan~\cite{eM57},
  $\Ds$ has a zero and a unit element, namely,
  $\vv<0, 0>$ and $\vv<1, 1>$.  To show that $\Ds$ is
  complete, let $\es \ne A \ci \Ds$, and let $a = \JJ A$
  in $D^{2}$.  If $a \in \Ds$, then
  $a = \JJ A$ in $\Ds$; otherwise, $a$ is of the form
  $\vv<b, 1>$ for some $b \in D$ with $b < 1$.  Now
  $\JJ A = \vv<1, 1>$ in $D^{2}$, and
  the dual argument shows that $\MM A$ also exists in
  $D^{2}$.  Hence $D$ is complete. Conditions~(1) and (2)
  are obvious for $\Ds$.
\end{proof}

\begin{corollary}\label{C:prime}
  If $D$ is complete-prime, then so is $\Ds$.
\end{corollary}

The motivation for the following result comes from Soo-Key
Foo~\cite{sF90}.

\begin{lemma}\label{L:ccr}
  Let $\gQ$ be a complete congruence relation of $\Ds$ such
  that
  \begin{equation}\label{E:rigid}
     \vv<1, d> \equiv \vv<1, 1> \pod{\gQ},
  \end{equation}
  for some $d \in D$ with $d < 1$. Then $\gQ = \gi$.
\end{lemma}

\begin{proof}
  Let $\gQ$ be a complete congruence relation of $\Ds$
  satisfying \eqref{E:rigid}. Then $\gQ = \gi$.
\end{proof}

\section{The $\gP^{*}$ 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~\tup{(2)}.  Their $\gP^{*}$
  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
  \[
     \vct{i}{d}
  \]
  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~\tup{(2)}.  Let $\gQ$ 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}
     \vct{i}{d} \equiv \vct{i}{c} \pod{\gQ},
  \end{equation}
  then $\gQ = \gi$.
\end{theorem}

\begin{proof}
  Since
  \begin{equation}\label{E:cong2}
     \vct{i}{d} \equiv \vct{i}{c} \pod{\gQ},
  \end{equation}
  and $\gQ$ is a complete congruence relation, it follows
  from condition~(3) that
  \begin{align}\label{E:cong}
     &\vct{i}{d} \equiv \notag\\
     &\qq\q{\JJm{\vct{i}{c}}{d \leq c < 1}=1} \pod{\gQ}.
  \end{align}
  Let $j \in I$ for $j \neq i$, and let
  $a \in D_{j}^{-}\).  Meeting both sides of the congruence
  \eqref{E:cong} with $\vct{j}{a}$, we obtain
  \begin{align}\label{E:comp}
      0 &= \vct{i}{d} \mm \vct{j}{a}\\
          &\equiv \vct{j}{a}\pod{\gQ}. \notag
  \end{align}
  Using the completeness of $\gQ$ and \eqref{E:comp}, we get:
  \begin{equation}\label{E:cong3}
      0=\JJm{ \vct{j}{a} }{ a \in D_{j}^{-} } \equiv 1 \pod{\gQ},
  \end{equation}
  hence $\gQ = \gi$.
\end{proof}

\begin{theorem}\label{T:P*a}
  Let $D_{i}$ for $i \in I$ be complete distributive
  lattices satisfying
  conditions~\tup{(2)} and \tup{(3)}.  Then
  $\Prodsm{ D_{i} }{i \in I}$ also satisfies
  conditions~\tup{(2)} and \tup{(3)}.
\end{theorem}

\begin{proof}
  Let $\gQ$ 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{ \vct{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 $\gQ_{i}$ be the
  restriction of $\gQ$ to $\widehat{D}_{i}$.  Since
  $D_{i}$ is complete-simple, so is $\widehat{D}_{i}$,
  hence $\gQ_{i}$ is $\go$ or $\gi$.  If $\gQ_{i} = \go$
  for all $i \in I$, then $\gQ = \go$.
  If there is an $i \in I$, such that $\gQ_{i} = \gi$,
  then $0 \equiv 1 \pod{\gQ}$, and hence $\gQ = \gi$.
\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-Toronto-London-Mel\-bourne, 1968.

  \bibitem{eM57}
     Ernest~T. Moynahan, \emph{On a problem of M.H. Stone,} Acta Math.
      Acad.Sci. Hungar. \textbf{8} (1957), 455--460.

  \bibitem{eM57a}
     \bysame, \emph{Ideals and congruence relations in lattices.~II,}
    Magyar Tud. Akad. Mat. Fiz. Oszt. K\"{o}zl. \textbf{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}