# Can a reduced ring have (# idempotents) $\in 3 \mathbb{Z}$?

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.)

• 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. – Gae. S. Dec 1 '19 at 0:19
• @Gae.S. Good to know. – Pascal's Wager Dec 1 '19 at 0:21
• @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. – Pascal's Wager Dec 1 '19 at 0:22
• Commutativity was unnecessary as well. – Gae. S. Dec 1 '19 at 0:24

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.