Does every commutative ring have $2^n$ idempotents?

Question

I've spent a lot of time looking for examples, and I can't find any commutative rings which have a finite number of idempotents other than a power of $2$. Intuitively, adjoining an extra idempotent $a$ always seems to double the number, as for every other idempotent $b$, $ab$ is a new idempotent. But there doesn't seem to be any way to turn this into a proof. I'm not sure whether it's even true.

I have managed to prove the characteristic $2$ case, as there the idempotents form a subring, and hence $\mathbb{Z}_2$-Algebra. But this doesn't seem to be any help for the other cases. I'm sorry I can't show more of an attempt, but this really has me stumped.

So are there commutative rings with a finite number of idempotents which don't have $2^n$ idempotents for some $n$?

I also don't know of any rings that have $k \ne 2^n$ idempotents in the noncommutative case, but I haven't investigated that much.

Answer 1

Let $(A,+,\cdot)$ be a commutative ring and let $E$ be its set of idempotents. Define a binary operation $\&$ on $E$ by

$$e_1 \& e_2 = e_1 + e_2 - 2e_1e_2.$$

Then (exercise!) $(E,\&,\cdot)$ is a well-defined commutative ring of characteristic $2$. If $E$ is finite, it follows that $\lvert E \rvert$ is a power of $2$.

Answer 2

I believe this is another way to prove this. Let $r \neq 0, 1$ be an idempotent in the (assumed unital) ring $R$. Then, $\langle r \rangle$ forms a ring with multiplicative identity $r$. It is easily verified that $(1-r)$ is also idempotent, so the same is true for $\langle 1-r \rangle $. We now use that there is an isomorphism:

$$R \cong \langle r \rangle \times \langle 1-r \rangle$$

Note that the number of idempotents in a product ring is the product of the number of idempotents in each factor. We can now apply induction on the number of idempotents since, as unital rings, both $\langle r \rangle$ and $\langle 1-r \rangle$ have at least two idempotents.

Answer 3

From the perspective of algebraic geometry, this is obvious. Idempotent elements of a commutative ring $A$ are in bijection with clopen subsets of $\operatorname{Spec} A$. So if $A$ has finitely many idempotents, $\operatorname{Spec} A$ has finitely many clopen subsets, and the minimal nonempty clopen subsets are easily seen to be the connected components of $\operatorname{Spec} A$. The clopen subsets of $\operatorname{Spec} A$ are then just the unions of connected components, so they are in bijection with the power set of the set of connected components.

(To connect this to diracdeltafunk's answer, his operation $\&$ corresponds to the symmetric difference operation on clopen subsets, and multiplication corresponds to intersection. So his ring structure on idempotents is just the usual Boolean ring structure on the clopen subsets. Of course, the real work here is being hidden in the result that idempotent elements are in bijection with clopen subsets of the spectrum.)