SubdirectProductOfAlgebraicSystems 2007-02-24 16:56:58 2007-08-05 00:08:06 Definition direct product of algebras subdirect product subdirect power subdirectly irreducible trivial subdirect product \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} % 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{ex}{Example} \newcommand{\real}{\mathbb{R}} In this entry, all algebraic systems are of the same type. For each algebraic system, we drop the associated operator set for simplicity. Let $A_i$ be algebraic systems indexed by $i\in I$. $B$ is called a \emph{subdirect product} of $A_i$ if \begin{enumerate} \item $B$ is a subalgebra of the direct product of $A_i$. \item for each $i\in I$, $\pi_i(B)=A_i$. \end{enumerate} In the second condition, $\pi_i$ denotes the projection homomorphism $\prod A_i \to A_i$. By restriction, we may consider $\pi_i$ as homomorphisms $B\to A_i$. When $B$ is isomorphic to $\prod A_i$, then $B$ is a \emph{trivial subdirect product} of $A_i$. This generalizes the notion of a direct product, since in many instances, an algebraic system can not be decomposed into a direct product of algebras. When all $A_i=C$ for some algebraic system $C$ of the same type, then $B$ is called a \emph{subdirect power} of $C$. \textbf{Remarks}. \begin{enumerate} \item A very simple example of a subdirect product is the following: let $A_1=A_2=\lbrace 1,2,3\rbrace$. Then the subset $B=\lbrace (x,y)\in A_1\times A_2 \mid x\le y \rbrace$ is a subdirect product of the sets $A_1$ and $A_2$ (considered as algebraic systems with no operators). \item Let $B$ is a subdirect product of $A_i$, and $p_i:=(\pi_i)_B$, the restriction of $\pi_i$ to $B$. Then $B/\ker(p_i)\cong A_i$. In addition, $$\bigcap \lbrace \ker(p_i)\mid i\in I\rbrace=\Delta,$$ where $\Delta$ is the diagonal relation. To see the last equality, suppose $a,b\in B$ with $a\equiv b \pmod {p_i}$. Then $a(i)=\pi_i(a)=p_i(a)=p_i(b)=\pi_i(b)=b(i)$. Since this is true for every $i\in I$, $a=b$. \item Conversely, if $A$ is an algebraic system and $\lbrace \mathfrak{C}_i\mid i\in I\rbrace$ is a set of congruences on $A$ such that $$\bigcap \lbrace \mathfrak{C}_i\mid i\in I\rbrace=\Delta.$$ Then $A$ is isomorphic to a subdirect product of $A/\mathfrak{C}_i$. \item An algebraic system is said to be \emph{subdirectly irreducible} if, whenever $\mathfrak{C}_i$ are congruences on $A$ and $\bigcap \lbrace \mathfrak{C}_i\mid i\in I\rbrace=\Delta$, then one of $\mathfrak{C}_i=\Delta$. \item \textbf{Birkhoff's Theorem on the Decomposition of an Algebraic System}. Every algebraic system is isomorphic to a subdirect product of subdirectly irreducible algebraic systems. This works only when the algebraic system is finitary. \end{enumerate}