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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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    $\begingroup$ I was about to post a (completely trivial) answer but then realized this is probably homework. $\endgroup$
    – Steven Landsburg
    Commented Feb 6, 2011 at 16:16
  • $\begingroup$ 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 :) $\endgroup$
    – ingrem
    Commented Feb 6, 2011 at 16:30
  • $\begingroup$ @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. $\endgroup$
    – Did
    Commented Feb 6, 2011 at 16:35
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    $\begingroup$ Darij: Why not just take R = Z/6Z ? $\endgroup$
    – Steven Landsburg
    Commented Feb 6, 2011 at 17:26
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    $\begingroup$ There is no kill like overkill. :) $\endgroup$
    – darij grinberg
    Commented Feb 6, 2011 at 17:27

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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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