SubalgebraOfAPartialAlgebra 2009-01-10 02:35:31 2009-01-18 00:37:59 Definition partial algebraic system weak subalgebra relative subalgebra subalgebra \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} % define commands here \newcommand*{\abs}[1]{\left\lvert #1\right\rvert} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \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}} Unlike an algebraic system, where there is only one way to define a subalgebra, there are several ways to define a subalgebra of a partial algebra. Suppose $\boldsymbol{A}$ and $\boldsymbol{B}$ are partial algebras of type $\tau$: \begin{enumerate} \item $\boldsymbol{B}$ is a \emph{weak subalgebra} of $\boldsymbol{A}$ if $B\subseteq A$, and $f_{\boldsymbol{B}}$ is a subfunction of $f_{\boldsymbol{A}}$ for every operator symbol $f\in \tau$. In words, $\boldsymbol{B}$ is a weak subalgebra of $\boldsymbol{A}$ iff $B\subseteq A$, and for each $n$-ary symbol $f\in \tau$, if $b_1,\ldots, b_n \in B$ such that $f_B(b_1,\ldots, b_n)$ is defined, then $f_A(b_1,\ldots, b_n)$ is also defined, and is equal to $f_B(b_1, \ldots, b_n)$. \item $\boldsymbol{B}$ is a \emph{relative subalgebra} of $\boldsymbol{A}$ if $B\subseteq A$, and $f_{\boldsymbol{B}}$ is a \PMlinkname{restriction of $f_{\boldsymbol{A}}$ relative to $B$}{Subfunction} for every operator symbol $f\in \tau$. In words, $\boldsymbol{B}$ is a relative subalgebra of $\boldsymbol{A}$ iff $B\subseteq A$, and for each $n$-ary symbol $f\in \tau$, given $b_1,\ldots, b_n \in B$, $f_B(b_1,\ldots, b_n)$ is defined iff $f_A(b_1,\ldots, b_n)$ is and belongs to $B$, and they are equal. \item $\boldsymbol{B}$ is a \emph{subalgebra} of $\boldsymbol{A}$ if $B\subseteq A$, and $f_{\boldsymbol{B}}$ is a \PMlinkname{restriction}{Subfunction} of $f_{\boldsymbol{A}}$ for every operator symbol $f\in \tau$. In words, $\boldsymbol{B}$ is a subalgebra of $\boldsymbol{A}$ iff $B\subseteq A$, and for each $n$-ary symbol $f\in \tau$, given $b_1,\ldots, b_n \in B$, $f_B(b_1,\ldots, b_n)$ is defined iff $f_A(b_1,\ldots, b_n)$ is, and they are equal. \end{enumerate} Notice that if $\boldsymbol{B}$ is a weak subalgebra of $\boldsymbol{A}$, then every constant of $\boldsymbol{B}$ is a constant of $\boldsymbol{A}$, and vice versa. Every subalgebra is a relative subalgebra, and every relative subalgebra is a weak subalgebra. But the converse is false for both statements. Below are two examples. \begin{enumerate} \item Let $F$ be a field. Then every subalgebra of $F$ is a subfield, and every relative subalgebra of $F$ is a subring. \item Let $A$ be the set of all non-negative integers, and $-_A$ the ordinary subtraction on integers. Consider the partial algebra $(A, -_A)$. \begin{itemize} \item Let $B=A$ and $-_B$ the usual subtraction on integers, but $x-_B y$ is only defined when $x,y\in B$ have the same parity. Then $(B, -_B)$ is a weak subalgebra of $(A, -_A)$. \item Let $C$ be the set of all positive integers, and $-_C$ the ordinary subtraction. Then $(C,-_C)$ is a relative subalgebra of $(A,-_A)$. \item Let $D$ be the set $\lbrace 0,1,\ldots, n\rbrace$ and $-_D$ the ordinary subtraction. Then $(D,-_D)$ is a subalgebra of $(A,-_A)$. \end{itemize} Notice that $(B,-_B)$ is not a relative subalgebra of $(A,-_A)$, since $7-_B 6$ is not defined, even though $7 -A 6 = 1\in B$, and and $(C,-_C)$ is not a subalgebra of $(A,-_A)$, since $1 -_C 1$ is not defined in $C$, even though $1 -A 1$ is defined in $A$. \end{enumerate} \textbf{Remarks}. \begin{enumerate} \item A weak subalgebra $\boldsymbol{B}$ of $\boldsymbol{A}$ is a relative subalgebra iff given $b_1,\ldots, b_n\in B$ such that $f_A(b_1,\ldots, b_n)$ is defined and is in $B$, then $f_B(b_1,\ldots, b_n)$ is defined. A relative subalgebra $\boldsymbol{B}$ of $\boldsymbol{A}$ is a subalgebra iff whenever $f_A(b_1,\ldots, b_n)$ is defined for $b_i\in B$, it is in $B$. \item Let $\boldsymbol{A}$ be a partial algebra of type $\tau$, and $B\subseteq A$. For each $n$-ary function symbol $f\in \tau$, define $f_{\boldsymbol{B}}$ on $B$ as follows: $f_{\boldsymbol{B}}(b_1,\ldots, b_n)$ is defined in $B$ iff $f_{\boldsymbol{A}}(b_1,\ldots, b_n)$ is defined in $A$ and $f_{\boldsymbol{A}}(b_1,\ldots, b_n)\in B$. This turns $\boldsymbol{B}$ into a partial algebra. However, $\boldsymbol{B}$ may not be of type $\tau$, since $f_{\boldsymbol{B}}$ may not be defined at all on $B$. When $\boldsymbol{B}$ is a partial algebra of type $\tau$, it is a relative subalgebra of $\boldsymbol{A}$. \item When $\boldsymbol{A}$ is an algebra, all three notions of subalgebras are equivalent (assuming that the partial operations on a weak subalgebra are all total). \end{enumerate} \begin{thebibliography}{7} \bibitem{gg} G. Gr\"{a}tzer: {\em Universal Algebra}, 2nd Edition, Springer, New York (1978). \end{thebibliography}