# 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 $eRe$ and $(1-e)R(1-e)$ are subrings, $eR(1-e)$ is an $eRe$-$(1-e)R(1-e)$-bimodule, and $(1-e)Re$ is a $(1-e)R(1-e)$-$eRe$-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 PeirceDecomposition 2013-03-22 14:39:17 2013-03-22 14:39:17 mclase (549) mclase (549) 7 mclase (549) Definition msc 16S99