# Idempotents in a ring without unity (rng) and no zero divisors.

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?

• 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.
– Jim
Commented Feb 4, 2014 at 16:59
• @Jim It is actually mentioned here that one might have trouble preserving the no zero divisor property when embedding a rng into a ring. Commented Feb 4, 2014 at 20:17
• 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 ${\widehat {\mathbb {C} }}$. Commented Jun 22, 2022 at 18:39

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.