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

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\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.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, 1999}

\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 $\Ds$ 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{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:ds}
  Let $D$ be a complete distributive lattice satisfying
  conditions \eqref{m-i} and~\eqref{j-i}.
  Then $\Ds$ is a sublattice of $D^{2}$; hence $\Ds$ is
  a lattice, and $\Ds$ is a complete distributive lattice
  satisfying conditions \eqref{m-i} and~\eqref{j-i}.
\end{lemma}

\begin{proof}
  By conditions~\eqref{m-i} and \eqref{j-i}, $\Ds$ 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 \eqref{m-i}
  and~\eqref{j-i} 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}
     \con \vv<1, d>=\vv<1, 1>(\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~\eqref{j-i}.  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~\eqref{j-i}.  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}
     \con\vct{i}{d}=\vct{i}{c}(\gQ),
  \end{equation}
  then $\gQ = \gi$.
\end{theorem}

\begin{proof}
  Since
  \begin{equation}\label{E:cong2}
     \con\vct{i}{d}=\vct{i}{c}(\gQ),
  \end{equation}
  and $\gQ$ is a complete congruence relation, it follows
  from condition~\eqref{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
  \eqref{E:cong} with $\vct{j}{a}$, we obtain
  \begin{equation}\label{E:comp}
     \begin{split}
         0 &= \vct{i}{d} \mm \vct{j}{a}\\
           &\equiv \vct{j}{a}\pod{\gQ}.
    \end{split}
  \end{equation}
 Using the completeness of $\gQ$ and \eqref{E:comp}, we get:
  \begin{equation}\label{E:cong3}
      \con{0=\JJm{ \vct{j}{a} }{ a \in D_{j}^{-} }}={1}(\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 \eqref{j-i} and~\eqref{c-s}.  Then
  $\Prodsm{ D_{i} }{i \in I}$ also satisfies
  conditions~\eqref{j-i} and \eqref{c-s}.
\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 $\con0=1(\gQ)$, 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, 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}

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