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-% Sample file: sampartb.tex
-% The sample article for the amsart document class with BibTeX
-% Typeset with LaTeX format
-
-\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{menuhin@ccw.uwinnebago.edu}
-\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{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;
- \item $v$ is join-irreducible but $v$ is \emph{not}
- completely join-irreducible;
- \item $[u, v]$ 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
- $D^{\langle 2 \rangle}$ 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 condition \textup{(1)} and~\textup{(2)}.
-\end{lemma}
-
-\begin{proof}
- By conditions (1) and~(2), $D^{\langle 2 \rangle}$ 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. Condition (1)
- and~(2) 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~\textup{(2)}. 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~\textup{(2)}. 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~(3) 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 \textup{(2)} and~\textup{(3)}. Then
- $\Pi^{*} ( D_{i} \mid i \in I )$ also satisfies conditions \textup{(2)}
- and~\textup{(3)}.
-\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}.
-
-\bibliographystyle{amsplain}
-\bibliography{sampartb}
-\end{document}
-