# 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 $eRe$ 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 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

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

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

2. (ii)

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

3. (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 IdempotentClassifications 2013-03-22 16:48:43 2013-03-22 16:48:43 Algeboy (12884) Algeboy (12884) 9 Algeboy (12884) Definition msc 16U99 msc 20M99 division idempotent local idempotent