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Math Help - Clarification needed on central idempotent elements in a ring

  1. #1
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    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<br />
=	1_{R}r-er<br />
=	\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)<br />
=	1_{R}^{2}-1_{R}e-e1_{R}+e^{2}<br />
=	1_{R}-2e+e<br />
=	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?
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  2. #2
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    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) )
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