# Ring where all elements are potent

An element $$a$$ of a ring $$R$$ is said to be a potent element if there exists a positive integer $$n\ge 2$$ such that $$a^n=a$$. It is obvious that all idempotent elements are potent elements but converse is not true. My question is : like we have a Boolean ring where each element is idempotent. Is there any ring where all elements are potent? Obviously all Boolean rings satisfies this condition but we have to find non-trivial example. Similar to these rings we have quasi-clean rings in which every element is a sum of a potent element and a unit. We also have tripotent rings if all its elements are tripotent. But I am finding something more general.

These are the subject of a famous theorem of Jacobson. Following Martin Brandenburg let's call a ring $$R$$ such that $$x^n = x$$ for every $$x$$ an $$n$$-ring. Jacobson showed that such a ring is necessarily commutative, so let's only discuss the commutative case from here. From the definition it's clear that

1. $$n$$-rings are reduced, and
2. quotients of $$n$$-rings are $$n$$-rings.

So an $$n$$-ring $$R$$ embeds into the product $$\prod_P R/P$$ of quotients by its prime ideals $$P$$, and the quotients $$R/P$$ are integral domains which are also $$n$$-rings. So every element of $$R/P$$ satisfies $$x^n = x$$. Since $$R/P$$ is an integral domain this means it has at most $$n$$ elements, hence is finite. If $$x \neq 0$$ then $$x^{n-1} = 1$$, so $$R/P$$ is a finite field. (This means every prime ideal of $$R$$ is maximal, so $$R$$ is von Neumann regular, which also follows from directly from the definition.)

In a finite field $$\mathbb{F}_q$$ the nonzero elements are a cyclic group of order $$q - 1$$; since the nonzero elements of $$R/P$$ must also have order dividing $$n - 1$$, it follows that $$R/P \cong \mathbb{F}_q$$ is an $$n$$-ring iff $$q - 1 \mid n - 1$$. The conclusion is:

Theorem: A commutative ring $$R$$ is an $$n$$-ring $$R$$ iff it's a subdirect product of finite fields $$\mathbb{F}_q$$ with $$q - 1 \mid n - 1$$.

This is a direct generalization of the Boolean case $$n = 2$$ where the conclusion is that $$R$$ is a subdirect product of copies of $$\mathbb{F}_2$$. For example we can take $$q$$ any prime power such that $$q - 1 \mid n -1$$ and $$X$$ any Stone space and construct the ring of continuous functions $$X \to \mathbb{F}_q$$ which will be an example; this gives a contravariant equivalence of categories between Stone spaces and $$\mathbb{F}_q$$-algebras which are also $$q$$-rings, generalizing Stone's representation theorem which is the case $$q = 2$$.

A corollary of the theorem is that an $$n$$-ring has characteristic $$\prod p$$ where $$p$$ runs over all primes s.t. $$p - 1 \mid n-1$$. The Chinese remainder theorem then implies that an $$n$$-ring canonically decomposes as a product of rings of characteristic $$p$$ for all primes $$p$$ satisfying this condition, so the classification reduces to the characteristic $$p$$ case.

Here's a little table of what prime powers $$q$$ satisfy $$q - 1 \mid n - 1$$ for the first few values of $$n$$.

$$n$$ $$q$$
$$2$$ $$2$$
$$3$$ $$2, 3$$
$$4$$ $$2, 4$$
$$5$$ $$2, 3, 5$$
$$6$$ $$2$$
$$7$$ $$2, 3, 4, 7$$
$$8$$ $$2, 8$$
$$9$$ $$2, 3, 5, 9$$
$$10$$ $$2, 4$$

This table implies some fun facts; for example, that every $$6$$-ring is a $$2$$-ring (so a Boolean ring) and that every $$10$$-ring is a $$4$$-ring. More details and examples can be found in Martin's very nice linked paper.

Edit: Okay, skimming more through Martin's paper confirmed my guess: there's a precise generalization of Stone's representation theorem available here. Namely, we can reduce to the case of prime characteristic $$p$$ and classify $$n$$-rings of characteristic $$p$$ via Galois descent; after extending scalars to $$\mathbb{F}_{p^k}$$ for a suitable $$k$$ we get a Stone space $$X$$ together with a continuous action of $$\mathbb{Z}/k$$, and our original $$n$$-ring is the ring of $$\mathbb{Z}/k$$-equivariant continuous maps $$X \to \mathbb{F}_{p^k}$$.

• What about rings where the value of $n$ varies according to the element? For example, I'd think $\overline{\mathbb{F}_p}$ would be an example of such a ring which is not an $n$-ring for any $n$. Commented Jul 25 at 18:13
• @Daniel: we could consider this case too, I just had less to say about it. I think such a ring is a filtered colimit of $n$-rings. (This isn't entirely clear, we'd need to show the value of $n$ is bounded on any finitely generated subring.) Commented Jul 25 at 18:16