From b5b3eefc8fb0473726a865efb286b15f48809537 Mon Sep 17 00:00:00 2001
From: Karl Berry <karl@freefriends.org>
Date: Thu, 12 Jan 2017 23:14:02 +0000
Subject: math-into-latex-4

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+% 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{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, 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}
+
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