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^{} $\U0001d52a$ where $eRe/\U0001d52a$ a division ring – then $e$ is called a local idempotent.
Lemma 1.
Any integral domain^{} $R$ has only the trivial idempotents $\mathrm{0}$ and $\mathrm{1}$. In particular, every division ring has only trivial idempotents.
Proof.
Suppose $e\in R$ with $e\ne 0$ 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 counterexample 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 $\mathrm{0}$ and $\mathrm{1}$.
Proof.
Let $\U0001d52a$ be the unique maximal ideal of $R$. Then $\U0001d52a$ is the Jacobson radical^{} of $R$. Now suppose $e\in \U0001d52a$ is an idempotent. Then $1e$ must be left invertible (following the element characterization of Jacobson radicals (http://planetmath.org/JacobsonRadical)). So there exists some $u\in R$ such that $1=u(1e)$. However, this produces
$$e=u(1e)e=u(e{e}^{2})=u(ee)=0.$$ 
Thus every nontrivial idempotent $e\in R$ lies outside $\U0001d52a$. As $R/\U0001d52a$ is a division ring, the only idempotents are $0$ and $1$. Thus if $e\in R$, $e\ne 0$ is an idempotent then it projects to an idempotent of $R/\U0001d52a$ and as $e\notin \U0001d52a$ it follows $e$ projects onto $1$ so that $e=1+z$ for some $z\in \U0001d52a$. As ${e}^{2}=e$ we find $0=z+{z}^{2}$ (often called an antiidempotent). Once again as $z\in \U0001d52a$ 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 ${M}_{n}\mathit{}\mathrm{(}R\mathrm{)}$ the standard idempotents are the matrices
$${E}_{ii}=\left[\begin{array}{cc}\hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \mathrm{\ddots}\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill 1\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \mathrm{\ddots}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill \end{array}\right],1\le i\le n.$$ 

(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 $R\mathrm{=}\mathrm{R}\mathrm{\oplus}\mathrm{R}$ then (i) is not satisfied and consequently neither are (ii) and (iii). When $R\mathrm{=}\mathrm{Z}$ then (i) is satisfied but not (ii) nor (iii). When $R\mathrm{=}\mathrm{R}\mathit{}\mathrm{[}\mathrm{[}x\mathrm{]}\mathrm{]}$ – the formal power series ring over $\mathrm{R}$ – then (i) and (ii) are satisfied but not (iii). Finally when $R\mathrm{=}\mathrm{R}$ then all three are satisfied.
A consequence of the WedderburnArtin 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  20130322 16:48:43 
Last modified on  20130322 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 