IdempotentClassifications 2007-03-07 20:10:21 2007-03-08 12:09:41 Definition idempotent division idempotent local idempotent primitive idempotent \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{enumerate} \usepackage{xypic} %----------------------------------------------------- % Standard theoremlike environments. % Stolen directly from AMSLaTeX sample %----------------------------------------------------- %% \theoremstyle{plain} %% This is the default \newtheorem{thm}{Theorem} \newtheorem{coro}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \newtheorem{lemma}[thm]{Lemma} \newtheorem{prop}[thm]{Proposition} \newtheorem{conjecture}[thm]{Conjecture} \newtheorem{conj}[thm]{Conjecture} \newtheorem{defn}[thm]{Definition} \newtheorem{remark}[thm]{Remark} \newtheorem{ex}[thm]{Example} %\countstyle[equation]{thm} %-------------------------------------------------- % Item references. %-------------------------------------------------- \newcommand{\exref}[1]{Example-\ref{#1}} \newcommand{\thmref}[1]{Theorem-\ref{#1}} \newcommand{\defref}[1]{Definition-\ref{#1}} \newcommand{\eqnref}[1]{(\ref{#1})} \newcommand{\secref}[1]{Section-\ref{#1}} \newcommand{\lemref}[1]{Lemma-\ref{#1}} \newcommand{\propref}[1]{Prop\-o\-si\-tion-\ref{#1}} \newcommand{\corref}[1]{Cor\-ol\-lary-\ref{#1}} \newcommand{\figref}[1]{Fig\-ure-\ref{#1}} \newcommand{\conjref}[1]{Conjecture-\ref{#1}} % Normal subgroup or equal. \providecommand{\normaleq}{\unlhd} % Normal subgroup. \providecommand{\normal}{\lhd} \providecommand{\rnormal}{\rhd} % Divides, does not divide. \providecommand{\divides}{\mid} \providecommand{\ndivides}{\nmid} \providecommand{\union}{\cup} \providecommand{\bigunion}{\bigcup} \providecommand{\intersect}{\cap} \providecommand{\bigintersect}{\bigcap} An a unital ring $R$, an idempotent $e\in R$ is called a \emph{division idempotent} if $eRe=\{ere:r\in R\}$, with the product of $R$, forms a division ring. If instead $eRe$ is a local ring -- here this means a ring with a unique maximal ideal $\mathfrak{m}$ where $eRe/\mathfrak{m}$ a division ring -- then $e$ is called a \emph{local idempotent}. \begin{lemma} Any integral domain $R$ has only the trivial idempotents $0$ and $1$. In particular, every division ring has only trivial idempotents. \end{lemma} \begin{proof} Suppose $e\in R$ with $e\neq 0$ and $e^2=e=1e$. Then by cancellation $e=1$. \end{proof} The integers are an integral domain which is not a division ring and they serve as a counter-exmple to many conjectures about idempotents of general rings as we will explore below. However, the first important result is to show the hierarchy of idempotents. \begin{thm} Every local ring $R$ has only trivial idempotents $0$ and $1$. \end{thm} \begin{proof} Let $\mathfrak{m}$ be the unique maximal ideal of $R$. Then $\mathfrak{m}$ is the Jacobson radical of $R$. Now suppose $e\in \mathfrak{m}$ is an idempotent. Then $1-e$ must be left invertible (following the \PMlinkname{element characterization of Jacobson radicals}{JacobsonRadical}). So there exists some $u\in R$ such that $1=u(1-e)$. However, this produces \[e=u(1-e)e=u(e-e^2)=u(e-e)=0.\] Thus every non-trivial idempotent $e\in R$ lies outside $\mathfrak{m}$. As $R/\mathfrak{m}$ is a division ring, the only idempotents are $0$ and $1$. Thus if $e\in R$, $e\neq 0$ is an idempotent then it projects to an idempotent of $R/\mathfrak{m}$ and as $e\notin \mathfrak{m}$ it follows $e$ projects onto $1$ so that $e=1+z$ for some $z\in\mathfrak{m}$. As $e^2=e$ we find $0=z+z^2$ (often called an anti-idempotent). Once again as $z\in\mathfrak{m}$ we know there exists a $u\in R$ such that $1=u(1+z)$ and $z=u(1+z)z=u(z+z^2)=0$ so indeed $e=1$. \end{proof} \begin{coro} Every division idempotent is a local idempotent, and every local idempotent is a primitive idempotent. \end{coro} \begin{ex} Let $R$ be a unital ring. Then in $M_n(R)$ the standard idempotents are the matrices \[E_{ii}=\begin{bmatrix} 0 & \\ & \ddots & \\ & & 1 \\ & & & \ddots\\ & & & & 0 \end{bmatrix},\qquad 1\leq i\leq n.\] \begin{enumerate}[(i)] \item If $R$ has only trivial idempotents (i.e.: $0$ and $1$) then each $E_{ii}$ is a primitive idempotent of $M_n(R)$. \item If $R$ is a local ring then each $E_{ii}$ is a local idempotent. \item If $R$ is a division ring then each $E_{ii}$ is a division idempotent. \end{enumerate} When $R=\mathbb{R}\oplus\mathbb{R}$ then (i) is not satisfied and consequently neither are (ii) and (iii). When $R=\mathbb{Z}$ then (i) is satisfied but not (ii) nor (iii). When $R=\mathbb{R}[[x]]$ -- the formal power series ring over $\mathbb{R}$ -- then (i) and (ii) are satisfied but not (iii). Finally when $R=\mathbb{R}$ then all three are satisfied. \end{ex} A consequence of the Wedderburn-Artin theorems classifies all Artinian simple rings as matrix rings over a division ring. Thus the primitive idempotents of an Artinian ring are all local idempotents. Without the Artinian assumption this may fail as we have already seen with $\mathbb{Z}$.