ReducedDirectProduct 2007-05-28 18:26:50 2007-05-31 14:53:36 Definition prime product \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{lem}{Lemma} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} Let $\lbrace A_i\mid i\in I\rbrace$ be a set of algebraic systems of the same type, indexed by $I$. Let $A$ be the direct product of the $A_i$'s. For any $a,b\in A$, set $$\operatorname{supp}(a,b):=\lbrace k\in I\mid a(k)\ne b(k)\rbrace.$$ Consider a Boolean ideal $L$ of the Boolean algebra $P(I)$ of $I$. Define a binary relation $\Theta_L$ on $A$ as follows: $$(a,b)\in \Theta_L\quad \mbox{ iff }\quad \operatorname{supp}(a,b)\in L.$$ \begin{lem} $\Theta_L$ defined above is a congruence relation on $A$. \end{lem} \begin{proof} Since $L$ is an ideal $\varnothing\in L$. Therefore, $(a,a)\in \Theta_L$, since $\lbrace k\in I\mid a(k)\ne a(k)\rbrace=\varnothing$. Clearly, $\Theta_L$ is symmetric. For transitivity, suppose $(a,b,(b,c)\in \Theta_L$. If $a(k)\ne c(k)$ for some $k\in I$, then either $a(k)\ne b(k)$ or $b(k)\ne c(k)$ (a contrapositive argument). So $$\operatorname{supp}(a,c) \subseteq \operatorname{supp}(a,b) \cup \operatorname{supp}(b,c).$$ Since $L$ is an ideal, $\operatorname{supp}(a,c)\in L$, so $(a,c)\in \Theta_L$, and $\Theta_L$ is an equivalence relation on $A$. Next, let $\omega$ be an $n$-ary operator on $A$ and $a_j\equiv b_j\pmod {\Theta_L}$, where $j=1,\ldots,n$. We want to show that $\omega(a_1,\ldots,a_n)\equiv \omega(b_1,\ldots,b_n)\pmod {\Theta_L}$. Let $\omega_i$ be the associated $n$-ary operators on $A_i$. If $\omega(a_1,\ldots,a_n)(k)\ne \omega(b_1,\ldots,b_n)(k)$, then $\omega_k(a_1(k),\ldots,a_n(k))\ne \omega_k(b_1(k),\ldots,b_n(k))$, which implies that $a_j(k)\ne b_j(k)$ for some $j=1,\ldots,n$. This implies that $$\operatorname{supp}(\omega(a_1,\ldots,a_n),\omega(b_1,\ldots,b_n))\subseteq \bigcup_{j=1}^n \operatorname{supp}(a_j,b_j).$$ Since $L$ is an ideal, and each $\operatorname{supp}(a_j,b_j)\in L$, we have that $\operatorname{supp}(\omega(a_1,\ldots,a_n),\omega(b_1,\ldots,b_n))\in L$ as well, this means that $\omega(a_1,\ldots,a_n)\equiv \omega(b_1,\ldots,b_n)\pmod {\Theta_L}$. \end{proof} \textbf{Definition}. Let $A=\prod \lbrace A_i\mid i\in I\rbrace$, $L$ be a Boolean ideal of $P(I)$ and $\Theta_L$ be defined as above. The quotient algebra $A/\Theta_L$ is called the $L$-\emph{reduced direct product} of $A_i$. The $L$-reduced direct product of $A_i$ is denoted by $\prod_L \lbrace A_i\mid i\in I\rbrace$. Given any element $a\in A$, its image in the reduced direct product $\prod_L \lbrace A_i\mid i\in I\rbrace$ is given by $[a]\Theta_L$, or $[a]$ for short. \textbf{Example}. Let $A=A_1\times \cdots \times A_n$, and let $L$ be the principal ideal generated by $1$. Then $L=\lbrace \varnothing, \lbrace 1\rbrace\rbrace$. The congruence $\Theta_L$ is given by $(a_1,\ldots,a_n)\equiv (b_1,\ldots, b_n)\pmod {\Theta_L}$ iff $\lbrace i\mid a_i\ne b_i\rbrace =\varnothing$ or $\lbrace 1\rbrace$. This implies that $a_i=b_i$ for all $i=2,\ldots,n$. In other words, $\Theta_L$ is isomorphic to the direct product of $A_2\times\cdots\times A_n$. Therefore, the $L$-reduced direct product of $A_i$ is isomorphic to $A_1$. The example above can be generalized: if $J\subseteq I$, then $$\prod {}_{P(J)} \lbrace A_i\mid i\in I\rbrace \cong \prod \lbrace A_i\mid i\in I-J\rbrace.$$ For $a\in A=\prod \lbrace A_i\mid i\in I\rbrace$, write $a=(a_i)_{i\in I}$. It is not hard to see that the map $f:\prod_{P(J)} \lbrace A_i\mid i\in I\rbrace \to \prod \lbrace A_i\mid i\in I-J\rbrace$ given by $f([a])=(a_i)_{i\in I-J}$ is the required isomorphism. \textbf{Remark}. The definition of a reduced direct product in terms of a Boolean ideal can be equivalently stated in terms of a Boolean filter $F$. All there is to do is to replace $\operatorname{supp}(a,b)$ by its complement: $\operatorname{supp}(a,b)^c:=\lbrace k\in I\mid a(k)=b(k)\rbrace$. The congruence relation is now $\Theta_{F'}$, where $F'=\lbrace I-J \mid J\in F\rbrace$ is the ideal complement of $F$. When $F$ is prime, the $F'$-reduced direct product is called a \emph{prime product}, or an \emph{ultraproduct}, since any prime filter is also called an ultrafilter. Ultraproducts can be more generally defined over arbitrary structures. \begin{thebibliography}{7} \bibitem{gg} G. Gr\"{a}tzer: {\em Universal Algebra}, 2nd Edition, Springer, New York (1978). \end{thebibliography}