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If $e$ and $f$ are idempotents, in a ring $R$. My question is how I can show $Re+Rf=Re+R(f-fe)$.

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up vote 2 down vote accepted

You have to prove that $Re+Rf \subseteq Re+R(f-fe)$ and $Re+Rf \supseteq Re+R(f-fe)$.

The second inclusion is clear because $xe+y(f-fe) = (x-yf)e+yf \in Re+Rf $.

For the first inclusion, it suffices to prove that $f \in Re+R(f-fe)$. But this is easy because $Re+R(f-fe) \ni fe+f(f-fe)=f$. In other words, $xe+yf = (x+yf)e+(yf)(f-fe)$.

I don't think that $e$ has to be idempotent.

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yes I think too –  Vahid Dec 21 '11 at 18:50
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