% Sample file: sampart.tex
% The sample article for the amsart document class

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

\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}

\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, MN 53714}
\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{main}
  There exists an infinite complete distributive lattice~$K$ with only
  the two trivial complete congruence relations.
\end{main}

\section{The $D^{\langle 2 \rangle}$ 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 $\mathfrak{p} = [u, v]$ be
  an interval of $V$.  Then $\mathfrak{p}$ is called
  \emph{complete-prime} if the following three conditions are satisfied:
  \begin{enumerate}
     \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{enumerate}
\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
  $D^{\langle 2 \rangle}$ is a sublattice of $D^{2}$;
  hence $D^{\langle 2 \rangle}$ is a lattice, and
  $D^{\langle 2 \rangle}$ 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},
  $D^{\langle 2 \rangle}$ is a sublattice
  of $D^{2}$.  Hence, $D^{\langle 2 \rangle}$ is a lattice.

  Since $D^{\langle 2 \rangle}$ is a sublattice of a distributive
  lattice, $D^{\langle 2 \rangle}$ is a distributive lattice.  Using
  the characterization of standard ideals in Ernest~T. Moynahan~\cite{eM57},
  $D^{\langle 2 \rangle}$ has a zero and a unit element,
  namely, $\langle 0, 0 \rangle$ and $\langle 1, 1 \rangle$.
  To show that $D^{\langle 2 \rangle}$ is complete, let
  $\varnothing \ne A \subseteq D^{\langle 2 \rangle}$, and let
  $a = \bigvee A$ in $D^{2}$.  If
  $a \in D^{\langle 2 \rangle}$, then
  $a = \bigvee A$ in $D^{\langle 2 \rangle}$; otherwise, $a$
  is of the form $\langle b, 1 \rangle$ for some
  $b \in D$ with $b < 1$.  Now $\bigvee A = \langle 1, 1\rangle$
  in $D^{2}$ and the dual argument shows that $\bigwedge A$ also
  exists in $D^{2}$.  Hence $D$ is complete. Conditions \eqref{m-i}
  and~\eqref{j-i} are obvious for $D^{\langle 2 \rangle}$.
\end{proof}

\begin{corollary}\label{C:prime}
  If $D$ is complete-prime, then so is $D^{\langle 2 \rangle}$.
\end{corollary}

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

\begin{lemma}\label{L:ccr}
  Let $\Theta$ be a complete congruence relation of
  $D^{\langle 2 \rangle}$ such that
  \begin{equation}\label{E:rigid}
     \langle 1, d \rangle \equiv \langle 1, 1 \rangle \pmod{\Theta},
  \end{equation}
  for some $d \in D$ with $d < 1$. Then $\Theta = \iota$.
\end{lemma}

\begin{proof}
  Let $\Theta$ be a complete congruence relation of
  $D^{\langle 2 \rangle}$ satisfying \eqref{E:rigid}. Then $\Theta =
\iota$.
\end{proof}

\section{The $\Pi^{*}$ 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 $\Pi^{*}$ product is defined
as follows:
  \[
     \Pi^{*} ( D_{i} \mid i \in I ) = \Pi ( D_{i}^{-} \mid i \in I ) + 1;
  \]
  that is, $\Pi^{*} ( D_{i} \mid i \in I )$ is $\Pi ( D_{i}^{-} \mid
  i \in I )$ with a new unit element.
\end{definition}

\begin{notation}
  If $i \in I$ and $d \in D_{i}^{-}$, then
  \[
     \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0, \dots \rangle
  \]
  is the element of $\Pi^{*} ( D_{i} \mid i \in I )$ 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 $\Theta$ be a complete
congruence
  relation on $\Pi^{*} ( D_{i} \mid i \in I )$.  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}
     \langle \dots, 0, \dots,\overset{i}{d},
     \dots, 0, \dots \rangle \equiv \langle \dots, 0, \dots,
     \overset{i}{c}, \dots, 0, \dots \rangle \pmod{\Theta},
  \end{equation}
  then $\Theta = \iota$.
\end{theorem}

\begin{proof}
  Since
  \begin{equation}\label{E:cong2}
     \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0,
        \dots \rangle \equiv \langle \dots, 0, \dots,
        \overset{i}{c}, \dots, 0, \dots \rangle \pmod{\Theta},
  \end{equation}
  and $\Theta$ 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:cong2} with
  $\langle \dots, 0, \dots, \overset{j}{a}, \dots, 0, \dots \rangle$,
  we obtain
\begin{equation}\label{E:comp}
  \begin{split}
     0 &= \langle \dots, 0, \dots, \overset{i}{d}, \dots, 0, \dots
        \rangle \wedge \langle \dots, 0, \dots, \overset{j}{a}, \dots, 0,
        \dots \rangle\\
         &\equiv \langle \dots, 0, \dots, \overset{j}{a}, \dots, 0, \dots
        \rangle \pmod{\Theta}.
  \end{split}
 \end{equation}
  Using the completeness of $\Theta$ and \eqref{E:comp}, we get:
  \[
     0 \equiv \bigvee ( \langle \dots, 0, \dots, \overset{j}{a},
     \dots, 0, \dots \rangle \mid a \in D_{j}^{-} ) = 1 \pmod{\Theta},
  \]
  hence $\Theta = \iota$.
\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
  $\Pi^{*} ( D_{i} \mid i \in I )$ also satisfies
  conditions~\eqref{j-i} and \eqref{c-s}.
\end{theorem}

\begin{proof}
  Let $\Theta$ be a complete congruence on
  $\Pi^{*} ( D_{i} \mid i \in I )$. Let $i \in I$.  Define
  \[
     \widehat{D}_{i} = \{ \langle \dots, 0, \dots, \overset{i}{d},
     \dots, 0, \dots \rangle \mid d \in D_{i}^{-} \} \cup \{ 1 \}.
  \]
  Then $\widehat{D}_{i}$ is a complete sublattice of
  $\Pi^{*} ( D_{i} \mid i \in I )$, and $\widehat{D}_{i}$ is
  isomorphic to $D_{i}$.  Let $\Theta_{i}$ be the restriction of
  $\Theta$ to $\widehat{D}_{i}$.

  Since $D_{i}\) is complete-simple, so is $\widehat{D}_{i}$, and
  hence $\Theta_{i}$ is $\omega$ or $\iota$.  If
  $\Theta_{i} = \rho$ for all $i \in I$, then
  $\Theta = \omega$.  If there is an $i \in I$, such that
  $\Theta_{i} = \iota$, then $0 \equiv 1 \pmod{\Theta}$, hence
  $\Theta = \iota$.
\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. \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}

You are viewing proxied material from ftp.icm.edu.pl. The copyright of proxied material belongs to its original authors. Any comments or complaints in relation to proxied material should be directed to the original authors of the content concerned. Please see the disclaimer for more details.