central idempotent

Let $R$ be a ring. An element $e\in R$ is called a central idempotent if it is an idempotent and is in the center $Z(R)$ of $R$ .

It is well-known that if $e\in R$ is an idempotent, then $e R e$ has the structure of a ring with unity, with $e$ being the unity. Thus, if $e$ is central, $eRe=eR=Re$ is a ring with unity $e$ .

It is easy to see that the operation of ring multiplication preserves central idempotency: if $e, f$ are central idempotents, so is $e f$ . In addition, if $R$ has a multiplicative identity $1$ , then $f:=1-e$ is also a central idempotent. Furthermore, we may characterize central idempotency in a ring with $1$ as follows:

Proposition 1.

An idempotent $e$ in a ring $R$ with $1$ is central iff $eRf=fRe=0$ , where $f=1-e$ .

Proof.

If $e$ is central, then clearly $eRf=fRe=0$ . Conversely, for any $r\in R$ , we have $er=er-erf=er(1-f)=ere=(1-f)re=re-fre=re$ . ∎

Another interesting fact about central idempotents in a ring with unity is the following:

Proposition 2.

The set $C$ of all central idempotents of a ring $R$ with $1$ has the structure of a Boolean ring.

Proof.

First, note that $0,1\in C$ . Next, for $e,f\in C$ , we define addition $\oplus$ and multiplication $\odot$ on $C$ as follows:

$e\oplus f:=e+f-ef\qquad\mbox{and}\qquad e\odot f:=ef.$

As discussed above, $\oplus$ and $\odot$ are well-defined (as $C$ is closed under these operations). In addition, for any $e,f,g\in C$ , we have

1.

$(C,1,\odot)$ is a commutative monoid, in which every element is an idempotent (with respect to $\odot$ ). This fact is clear.
2.

$\oplus$ is commutative, since $C\subseteq Z(R)$ .

$\oplus$ is associative:

$\displaystyle e\oplus(f\oplus g)$	$\displaystyle=$	$\displaystyle e+(f+g-fg)-e(f+g-fg)$
	$\displaystyle=$	$\displaystyle e+f+g-ef-fg-eg+efg$
	$\displaystyle=$	$\displaystyle(e+f-ef)+g-(e+f-ef)g$
	$\displaystyle=$	$\displaystyle(e\oplus f)\oplus g.$

$\odot$ distributes over $\oplus$ : we only need to show left distributivity (since $\odot$ is commutative by $1$ above):

$\displaystyle e\odot(f\oplus g)$	$\displaystyle=$	$\displaystyle e(f+g-fg)=ef+eg-efg$
	$\displaystyle=$	$\displaystyle ef+eg-eefg=ef+eg-efeg$
	$\displaystyle=$	$\displaystyle ef\oplus eg=(e\odot f)\oplus(e\odot g).$

This shows that $(C,0,1,\oplus,\odot)$ is a Boolean ring. ∎

Title	central idempotent
Canonical name	CentralIdempotent
Date of creation	2013-03-22 19:13:07
Last modified on	2013-03-22 19:13:07
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	5
Author	CWoo (3771)
Entry type	Definition
Classification	msc 16U99
Classification	msc 20M99