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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?

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  • $\begingroup$ 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
    Commented Feb 4, 2014 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
    Commented Feb 4, 2014 at 20:17
  • $\begingroup$ Since you loose the requirement of the set to be a ring, infinity $\infty$ is an idempotent in $\overline{\mathbb{R}}$ and $\tilde{\infty}$ is an idempotent in ${\widehat {\mathbb {C} }}$. $\endgroup$
    – Anixx
    Commented Jun 22, 2022 at 18:39

1 Answer 1

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Proposition: In 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.

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  • $\begingroup$ 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
    Commented Feb 4, 2014 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
    Commented Feb 4, 2014 at 20:11
  • $\begingroup$ @DietrichBurde Yup $\endgroup$
    – rschwieb
    Commented Dec 17, 2023 at 2:27

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