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Suppose $R$ is a ring (possibly noncommutative), $I$ is a minimal left ideal in it, and $I^2\neq 0$, show that $I=Re$ for some idemopotent $e$.

It is easy to show that we can find some $x\in I$, such that $I=Ix$, so $I=Rx=Rx^n=Ix^n$, for all $n\geq 1$, but how to construct the idemopotent $e$ using $x$, I guess I must have failed to realize some key point.

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up vote 4 down vote accepted

As $I^2\neq 0$, then there's an element $x\in I$ such $Ix=I$,also there is an element $e\in I$ with $ex=x$ with $e\neq 0$ then $e^2x=x$. Since right multiplication by $x$ is an isomorphism from $I$ to $Ix$ by minimality of $I$, this implies that $e^2=e$.

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Thanks! Now I see the point, further use the minimal condition. –  ougao Nov 18 '12 at 2:23
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