Peirce decomposition

Let $e$ be an idempotent of a ring $R$ , not necessarily with an identity. For any subset $X$ of $R$ , we introduce the notations:

$(1-e)X=\{x-ex\mid x\in X\}$

and

$X(1-e)=\{x-xe\mid x\in X\}.$

If it happens that $R$ has an identity element, then $1-e$ is a legitimate element of $R$ , and this notation agrees with the usual product of an element and a set.

It is easy to see that $Xe\cap X(1-e)=0=eX\cap(1-e)X$ for any set $X$ which contains $0$ .

Applying this first on the right with $X=R$ and then on the left with $X=Re$ and $X=R(1-e)$ , we obtain:

$R=eRe\oplus eR(1-e)\oplus(1-e)Re\oplus(1-e)R(1-e).$

This is called the Peirce Decompostion of $R$ with respect to $e$ .

Note that $e R e$ and $(1-e)R(1-e)$ are subrings, $eR(1-e)$ is an $e R e$ - $(1-e)R(1-e)$ -bimodule, and $(1-e)Re$ is a $(1-e)R(1-e)$ - $e R e$ -bimodule.

This is an example of a generalized matrix ring:

$R\cong\begin{pmatrix}eRe&eR(1-e)\\ (1-e)Re&(1-e)R(1-e)\\ \end{pmatrix}$

More generally, if $R$ has an identity element, and $e_{1},e_{2},\dots,e_{n}$ is a complete set of orthogonal idempotents, then

$R\cong\begin{pmatrix}e_{1}Re_{1}&e_{1}Re_{2}&\dots&e_{1}Re_{n}\\ e_{2}Re_{1}&e_{2}Re_{2}&\dots&e_{2}Re_{n}\\ \vdots&\vdots&\ddots&\vdots\\ e_{n}Re_{1}&e_{n}Re_{2}&\dots&e_{n}Re_{n}\end{pmatrix}$

is a generalized matrix ring.

Title	Peirce decomposition
Canonical name	PeirceDecomposition
Date of creation	2013-03-22 14:39:17
Last modified on	2013-03-22 14:39:17
Owner	mclase (549)
Last modified by	mclase (549)
Numerical id	7
Author	mclase (549)
Entry type	Definition
Classification	msc 16S99