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Let $R$ be an associative ring with identity and let $x$ be an arbitrary element from the ring $R$. Could you please help me to prove that $x=ye$, where $y$ is some element in $R$ and $e$ is some primitive central idempotent in $R$. In other words, I need to prove that any element in $R$ is representable as a product of some element and some primitive central idempotent in $R$.

Thanks for the answers! Answers like "you are not right, for example ..." and "see /book/, p. /page number/" are also OK.

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I was about to post a (completely trivial) answer but then realized this is probably homework. –  Steven Landsburg Feb 6 '11 at 16:16
    
To Steven Landsburg: I'm 30 (probably, too old for homeworks :)) I'm a installation developer in a small IT company, rings theory is just not my field. However, if you tell me the answer is simple, I probably can easily find it in a book. I'll try. Anyway, thanks for your answer :) –  ingrem Feb 6 '11 at 16:30
    
@ingrem: Why is a 30 years old installation developer in a small IT company interested in this? This is not a rhetorical question, I am genuinely curious. Related is mathoverflow.net/howtoask#motivation. –  Did Feb 6 '11 at 16:35
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Darij: Why not just take R = Z/6Z ? –  Steven Landsburg Feb 6 '11 at 17:26
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There is no kill like overkill. :) –  darij grinberg Feb 6 '11 at 17:27
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1 Answer

up vote 0 down vote accepted

Consider the direct product $k\times k$ of two copies of some field and the element $x=(1,1)$. In this simple example you can describe all primitive central idempotents.

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Thanks! :) Your example is very clear. –  ingrem Feb 6 '11 at 17:36
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