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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) $\in 3 \mathbb{Z}$?

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    $\begingroup$ No, because idempotents come in orthogonal pairs: if $e$ is an idempotent, $1-e$ is too, and $e(1-e)=0$. $\endgroup$ – Bernard Nov 30 '19 at 23:19
  • $\begingroup$ @Bernard I understand now. Thank you so much! $\endgroup$ – Algebraist Nov 30 '19 at 23:25
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    $\begingroup$ In addition, you need that if $e$ is idempotent, then $e\neq 1-e.$ That is not hard to show. $\endgroup$ – Thomas Andrews Nov 30 '19 at 23:27
  • $\begingroup$ @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? $\endgroup$ – Algebraist Nov 30 '19 at 23:28
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    $\begingroup$ 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. $\endgroup$ – Bernard Nov 30 '19 at 23:54
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No.

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.

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