# Thread: Clarification needed on central idempotent elements in a ring

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

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

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

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

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

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

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

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

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

Here is where I am having trouble. Is $\displaystyle eR$ defined as $\displaystyle \left\{ er:r\in R\right\}$ 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 $\displaystyle eR=\left\{ er:r\in R\right\}$, should I just take the usual approach of defining an isomorphism of rings? Or is there some special trick I'm missing?

2. Hi

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

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