IsomorphismTheoremsOnAlgebraicSystems 2007-02-25 00:30:20 2007-03-03 09:51:10 Theorem homomorphism between algebraic systems \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; they are all $O$-algebras. We list the generalizations of three famous isomorphism theorems, familiar to those who have studied abstract algebra in college. \begin{thm} If $f:A\to B$ is a homomorphism from algebras $A$ and $B$. Then $$A/\ker(f)\cong f(A).$$ \end{thm} \begin{thm} If $B\subseteq A$ are algebras and $\mathfrak{C}$ is a \PMlinkname{congruence}{CongruenceRelationOnAnAlgebraicSystem} on $A$, then $$B/\mathfrak{C}_B\cong B^{\mathfrak{C}}/\mathfrak{C},$$ where $\mathfrak{C}_B$ is the congruence restricted to $B$, and $B^{\mathfrak{C}}$ is the extension of $B$ by $\mathfrak{C}$. \end{thm} \begin{thm} If $A$ is an algebra and $\mathfrak{C}\subseteq \mathfrak{D}$ are congruences on $A$. Then \begin{enumerate} \item there is a unique homomorphism $f:A/\mathfrak{C}\to A/\mathfrak{D}$ such that $$\xymatrix{ & A \ar[dl]_{[\cdot]_{\mathfrak{C}}} \ar[dr]^{[\cdot]_{\mathfrak{D}}} & \\ A/\mathfrak{C} \ar[rr]^f && A/\mathfrak{D} } $$ where the downward pointing arrows are the natural projections of $A$ onto the quotient algebras (induced by the respective congruences). \item Furthermore, if $ker(f)=\mathfrak{D}/\mathfrak{C}$, then \begin{itemize} \item $\mathfrak{D}/\mathfrak{C}$ is a congruence on $A/\mathfrak{C}$, and \item there is a unique isomorphism $f':A/\mathfrak{C} \to (A/\mathfrak{C})/(\mathfrak{D}/\mathfrak{C})$ satisfying the equation $f=[\cdot]_{\mathfrak{D}/\mathfrak{C}}\circ f'$. In other words, $$(A/\mathfrak{C})/(\mathfrak{D}/\mathfrak{C})\cong A/\mathfrak{D}.$$ \end{itemize} \end{enumerate} \end{thm}