# Is there a reduced ring with exactly $3$ idempotents?

Let $$R$$ be a reduced ring (let's assume with identity, although not necessarily commutative).

Is it possible for $$R$$ to have exactly 3 idempotent elements? If so, what would be an example?

I know that the idempotents of a reduced ring commute with everything in the ring. This implies that the product of two idempotents is idempotent. But, alas, I don't see how this would help. (After all, if $$0$$ and $$1$$ and $$u$$ are the only idempotents, we can take products of these elements but we don't get anything new.)

Edit: I see now that it is impossible for $$R$$ to have exactly $$3$$ idempotents. (Thank you to the commentors for your insight!) I am still, however, interesting in doing further investigation on the structure of reduced rings. I have now posted a more general follow-up question Can a reduced ring have (# idempotents) $\in 3 \mathbb{Z}$?

• No, because idempotents come in orthogonal pairs: if $e$ is an idempotent, $1-e$ is too, and $e(1-e)=0$. – Bernard Nov 30 '19 at 23:19
• @Bernard I understand now. Thank you so much! – Pascal's Wager Nov 30 '19 at 23:25
• In addition, you need that if $e$ is idempotent, then $e\neq 1-e.$ That is not hard to show. – Thomas Andrews Nov 30 '19 at 23:27
• @Bernard I think this would imply that there have to be an even number of idempotents (assuming there are not infinitely many). Is that right? – Pascal's Wager Nov 30 '19 at 23:28
• I think so, I'm not too sure, because I'm not a noncommutativist (or should I say I'm commuting? :) and I'm wondering whether noncommuting people will pull out of their sleeve. – Bernard Nov 30 '19 at 23:54

Assume $$\{0,1,e\}$$ are the only idempotents. (All distinct).
By Bernard's comment, $$1-e$$ is idempotent. We can't have $$1-e=0$$, otherwise $$e=1$$. We can't have $$1-e=1$$, otherwise $$e=0$$. So we must have $$e=e-1$$. So $$e^2=e^2-e=0$$. Since $$e^2=0$$ and our ring is a reduced ring, we must have $$e=0$$, a contradiction.