Question: Given a ring without unity and with no zero-divisors, is it possible that there are idempotents other than zero?

Def: $a$ is idempotent if $a^2 = a$.

Originally the problem was to show that $1$ and $0$ are the only idempotents in a ring with unity and no zero-divisors, but I wonder what happens if we remove the unity condition.

I am trying to find a ring with idempotents not equal to $0$ or $1$. So far my biggest struggle has been coming up with examples of rings with the given properties.

Does anyone have any hints? How should I attack this problem?

  • $\begingroup$ Every rng (a ring without unity) can be embedded in a ring, but I don't know if this can be done while preserving the no zero divisors property. $\endgroup$ – Jim Feb 4 '14 at 16:59
  • $\begingroup$ @Jim It is actually mentioned here that one might have trouble preserving the no zero divisor property when embedding a rng into a ring. $\endgroup$ – Improve Feb 4 '14 at 20:17

Proposition: If a rng $R$ which does not have nonzero zero divisors, a nonzero idempotent of $R$ must be an identity for the ring.

Proof: Let $e$ be a nonzero idempotent. Since $e(er-r)=0=(re-r)e$ for all $r\in R$ and $e$ is nonzero, we conclude $er-r=0=re-r$, and so $e$ is an identity element.

  • $\begingroup$ This is exactly what I'm looking for! I was too focused on trying to find a counterexample. I also see that I can adapt this technique for other problems:) Thanks! $\endgroup$ – Improve Feb 4 '14 at 20:10
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    $\begingroup$ @Improve Yeah, sometimes it's hard to remember to switch back and forth between trying to prove/trying to disprove something. Glad you found more uses than one for my answer :) $\endgroup$ – rschwieb Feb 4 '14 at 20:11

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