% Sample file: sampartu.tex % The sample article with user-defined commands and environments \documentclass{amsart} \usepackage{newlattice} \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]{\GrP(\,#1\mid#2\,)} % product with a middle \newcommand{\Prodsm}[2]{\GrP^{*}(\,#1\mid#2\,)} % product * with a middle \newcommand{\vectsup}[2]{\vect<\dots,0,\dots,\overset{#1}{#2},% \dots,0,\dots>}% special vector \newcommand{\Dsq}{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{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{named}{Main Theorem} There exists an infinite complete distributive lattice $K$ with only the two trivial complete congruence relations. \end{named} \section{The $\Dsq$ 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 $\Frak{p} = [u, v]$ be an interval of $V$. Then $\Frak{p}$ 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:Dsq} Let $D$ be a complete distributive lattice satisfying conditions \itemref{m-i} and~\itemref{j-i}. Then $\Dsq$ is a sublattice of $D^{2}$; hence $\Dsq$ is a lattice, and $\Dsq$ is a complete distributive lattice satisfying conditions \itemref{m-i} and~\itemref{j-i}. \end{lemma} \begin{proof} By conditions~\itemref{m-i} and \itemref{j-i}, $\Dsq$ is a sublattice of $D^{2}$. Hence, $\Dsq$ is a lattice. Since $\Dsq$ is a sublattice of a distributive lattice, $\Dsq$ is a distributive lattice. Using the characterization of standard ideals in Ernest~T. Moynahan~\cite{eM57}, $\Dsq$ has a zero and a unit element, namely, $\vect<0, 0>$ and $\vect<1, 1>$. To show that $\Dsq$ is complete, let $\empset \ne A \contd \Dsq$, and let $a = \JJ A$ in $D^{2}$. If $a \in \Dsq$, then $a = \JJ A$ in $\Dsq$; otherwise, $a$ is of the form $\vect$ for some $b \in D$ with $b < 1$. Now $\JJ A = \vect<1, 1>$ in $D^{2}$, and the dual argument shows that $\MM A$ also exists in $D^{2}$. Hence $D$ is complete. Conditions \itemref{m-i} and~\itemref{j-i} are obvious for $\Dsq$. \end{proof} \begin{corollary}\label{C:prime} If $D$ is complete-prime, then so is $\Dsq$. \end{corollary} The motivation for the following result comes from Soo-Key Foo~\cite{sF90}. \begin{lemma}\label{L:ccr} Let $\GrQ$ be a complete congruence relation of $\Dsq$ such that \begin{equation}\label{E:rigid} \congr \vect<1, d>=\vect<1, 1>(\GrQ), \end{equation} for some $d \in D$ with $d < 1$. Then $\GrQ = \Gri$. \end{lemma} \begin{proof} Let $\GrQ$ be a complete congruence relation of $\Dsq$ satisfying \itemref{E:rigid}. Then $\GrQ = \Gri$. \end{proof} \section{The $\Grp^{*}$ 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~\itemref{j-i}. Their $\Grp^{*}$ 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 \[ \vectsup{i}{d} \] 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~\itemref{j-i}. Let $\GrQ$ 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} \congr\vectsup{i}{d}=\vectsup{i}{c}(\GrQ), \end{equation} then $\GrQ = \Gri$. \end{theorem} \begin{proof} Since \begin{equation}\label{E:cong2} \congr\vectsup{i}{d}=\vectsup{i}{c}(\GrQ), \end{equation} and $\GrQ$ is a complete congruence relation, it follows from condition~\itemref{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 \itemref{E:cong} with $\vectsup{j}{a}$, we obtain \begin{equation}\label{E:comp} \begin{split} 0 &= \vectsup{i}{d} \mm \vectsup{j}{a}\\ &\equiv \vectsup{j}{a}\pod{\GrQ}. \end{split} \end{equation} Using the completeness of $\GrQ$ and \itemref{E:comp}, we get: \begin{equation}\label{E:cong3} \congr{0=\JJm{ \vectsup{j}{a} }{ a \in D_{j}^{-} }}={1}(\GrQ), \end{equation} hence $\GrQ = \Gri$. \end{proof} \begin{theorem}\label{T:P*a} Let $D_{i}$, for $i \in I$, be complete distributive lattices satisfying conditions \itemref{j-i} and~\itemref{c-s}. Then $\Prodsm{ D_{i} }{i \in I}$ also satisfies conditions~\itemref{j-i} and \itemref{c-s}. \end{theorem} \begin{proof} Let $\GrQ$ 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{ \vectsup{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 $\GrQ_{i}$ be the restriction of $\GrQ$ to $\widehat{D}_{i}$. Since $D_{i}$ is complete-simple, so is $\widehat{D}_{i}$, hence $\GrQ_{i}$ is $\Gro$ or $\Gri$. If $\GrQ_{i} = \Gro$ for all $i \in I$, then $\GrQ = \Gro$. If there is an $i \in I$, such that $\GrQ_{i} = \Gri$, then $\congr0=1(\GrQ)$, and hence $\GrQ = \Gri$. \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}