idempotent classifications

An a unital ring $R$ , an idempotent $e\in R$ is called a division idempotent if $eRe=\{ere:r\in R\}$ , with the product of $R$ , forms a division ring. If instead $e R e$ 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 local idempotent.

Lemma 1.

Any integral domain $R$ has only the trivial idempotents $0$ and $1$ . In particular, every division ring has only trivial idempotents.

Proof.

Suppose $e\in R$ with $e\neq 0$ and $e^{2}=e=1e$ . Then by cancellation $e=1$ . ∎

The integers are an integral domain which is not a division ring and they serve as a counter-example 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.

Theorem 2.

Every local ring $R$ has only trivial idempotents $0$ and $1$ .

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 element characterization of Jacobson radicals (http://planetmath.org/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 $e=1$ . ∎

Corollary 3.

Every division idempotent is a local idempotent, and every local idempotent is a primitive idempotent.

Example 4.

Let $R$ be a unital ring. Then in $M_{n}(R)$ the standard idempotents are the matrices

$E_{ii}=\begin{bmatrix}0&\\ &\ddots&\\ &&1\\ &&&\ddots\\ &&&&0\end{bmatrix},\qquad 1\leq i\leq n.$

(i)

If $R$ has only trivial idempotents (i.e.: $0$ and $1$ ) then each $E_{ii}$ is a primitive idempotent of $M_{n}(R)$ .
(ii)

If $R$ is a local ring then each $E_{ii}$ is a local idempotent.
(iii)

If $R$ is a division ring then each $E_{ii}$ is a division idempotent.

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.

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 $\mathbb{Z}$ .

Title	idempotent classifications
Canonical name	IdempotentClassifications
Date of creation	2013-03-22 16:48:43
Last modified on	2013-03-22 16:48:43
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	9
Author	Algeboy (12884)
Entry type	Definition
Classification	msc 16U99
Classification	msc 20M99
Defines	division idempotent
Defines	local idempotent