Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $e^2=e$.

share|cite|improve this answer
Thanks! Now I see the point, further use the minimal condition. – ougao Nov 18 '12 at 2:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.