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This is a follow-up question to Is there a reduced ring with exactly $3$ idempotents?, to which the answer was "no."

Note: In this question, 'ring' means ring with unity, but not necessarily commutative

In fact, in a (non-trivial) reduced ring, the number of idempotents is either even or $\infty$. The reason is that the idempotents come in pairs $e,1-e$. And $e \neq1-e$, otherwise $ee=e-ee$ and $e^2=0$, implying (since the ring is reduced) that $e=0$, which can't happen if $e=1-e$.

My next question is, does there exist a reduced ring whose number of idempotents is a multiple of $3$? (For example, can we find a reduced ring with $6$ idempotent elements? $12$? $18$? $3000$?)

What about rings in general? (i.e. not necessarily reduced)

Attempting the easiest case first, assume $R$ is a reduced ring and the idempotent elements are $\{0,1,a,(1-a),b,(1-b)\}$ (all distinct). I see that the product of two idempotents must be idempotent (since the idempotents commute with everything). Also, I see that the square of the difference of two idempotents must also be idempotent. So $ab \in \{0,1,a,(1-a),b,(1-b)\}$ . (I suspect that there might be a way to derive a contradiction from this, although I don't see how to do so yet.)

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    $\begingroup$ For the record, the number of idempotents of a commutative ring with $1$ is even in that generalised sense regardless of what the nilradical does. The only caveat is for the zero ring, which has exactly one idempotent. $\endgroup$ – Gae. S. Dec 1 '19 at 0:19
  • $\begingroup$ @Gae.S. Good to know. $\endgroup$ – Pascal's Wager Dec 1 '19 at 0:21
  • $\begingroup$ @Gae.S. By the way, to clear up any ambiguity, I am not assuming $R$ to be commutative but I am assuming it to have unity. $\endgroup$ – Pascal's Wager Dec 1 '19 at 0:22
  • $\begingroup$ Commutativity was unnecessary as well. $\endgroup$ – Gae. S. Dec 1 '19 at 0:24
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If there are finitely many central idempotents, then $R=\prod_{i=1}^n e_iR$ for some set of $n$ idempotents, where $e_iRe_i$ are all rings with only trivial central idempotents. Since the central idempotents of this product ring are simply described as being all possible elements of $\prod_{i=1}^n\{0, e_i\}$, we can immediately see there are $2^n$ of them.

And all such cardinalities are possible: just consider $F_2^n$ for various $n$, where $F_2$ is the field of two elements.

In your case (a reduced ring) all idempotents are central, so this applies. So the answer to your question is "no."

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