# Clarification needed on central idempotent elements in a ring

• Nov 27th 2009, 05:53 PM
Esran
Clarification needed on central idempotent elements in a ring
I request your assistance on the following problem. I will provide everything I have so far, and indicate clearly those parts I'm having trouble with. First,to give you an idea of where I'm coming from, I'm currently enrolled in a Modern Algebra course using Hungerford as its text. The problem under discussion is number 23 on page 135.

An element $e$ in a ring $R$ is said to be idempotent if $e^{2}=e$. An element of the center of the ring $R$ is said to be central. If $e$ is central idempotent in a ring $R$ with identity, then...

a) $1_{R}-e$ is central idempotent.

First, we must demonstrate $r\left(1_{R}-e\right)=\left(1_{R}-e\right)r$ for each $r\in R$. Thus, let $r\in R$ and note:

$r\left(1_{R}-e\right)= r1_{R}-re
= 1_{R}r-er
= \left(1_{R}-e\right)r$
.

Hence, $1_{R}-e$ is central.

Now it remains to demonstrate $\left(1_{R}-e\right)^{2}=1_{R}-e$. Thus, note the following:

$\left(1_{R}-e\right)^{2}= \left(1_{R}-e\right)\left(1_{R}-e\right)
= 1_{R}^{2}-1_{R}e-e1_{R}+e^{2}
= 1_{R}-2e+e
= 1_{R}-e$

Therefore, $1_{R}-e$ is central idempotent.

b) $eR$ and $\left(1_{R}-e\right)R$ are ideals in $R$ such that $R=eR\times\left(1_{R}-e\right)R$.

Here is where I am having trouble. Is $eR$ defined as $\left\{ er:r\in R\right\}$ or something else? I've searched through the chapter and I'm having trouble locating the definition, and I suspect it might mirror the definition of multiplication of ideals in some form or fashion, but I could be totally wrong.

Assuming I'm wrong, and $eR=\left\{ er:r\in R\right\}$, should I just take the usual approach of defining an isomorphism of rings? Or is there some special trick I'm missing?
• Nov 28th 2009, 01:07 AM
clic-clac
Hi

Yes, $eR=\{er\ ;\ r\in R\}.$

Note that $eR$ and $(1-e)R$ are rings with identity (respectively $e$ and $(1-e)$). Finding an isomorphism between $R$ and $eR\times (1-e)R$ works, any idea? (try something that looks natural, and use question 1) )