summaryrefslogtreecommitdiff
path: root/Master/texmf-dist/doc/latex/mil3/sampartb.tex
blob: 6515da73e0545357b3a7304932573e2c9826fb51 (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
% 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}