$x^n = x$ implies commutativity, a universal algebraic proof?

I read in an answer on MO that Nathan Jacobson had given a universal algebraic proof that a ring satisfying the equation $x^n=x$ is commutative.

The sketch given in the answer is very clear : wlog one may assume that $R$ is subdirectly irreducible, as a result of a general result in universal algebra (that is, it has a minimum nonzero ideal).

Then one proves that a subdirectly irreducible ring satisfying the equation is a finite (skew, a priori) field, and one concludes from Wedderburn's theorem that it's commutative.

But I'm having trouble with the interesting step, that is

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

I had the folliwing idea : since $R$ has no nilpotent elements it should be a subdirect product of integral domains satisfying the same equations - however I know this property for commutative rings, and it relies on the well-known fact that $\displaystyle\bigcap\{p, p\in \mathrm{Spec}R\} = \{x, x$ is nilpotent $\}$ - and I don't know whether this is true for noncommutative rings.

As a matter of fact I'm pretty much convinced that it's not true (in $M_n(K)$, $K$ a field, $n\geq 2$, the set of nilpotent elements isn't a bilateral ideal - indeed there are no nontrivial ones). So unless this idea can be saved by the specifics of the situation, I can't go any further with it.

What I also noticed (I don't know if that can help though) is that $I^2 = I$, if $I$ denotes the minimum nonzero ideal.

I can also sort of make a connection with Wedderburn's theorem by studying the case where $Z(R)$ (the center) is a field; and so $R$ is a $Z(R)$-vector space. Then, it's a finite field. I can't yet see why $Z(R)$ would be finite dimensional (this would probably help a lot).

Am I anywhere near the right direction ? Can anyone give some hints to solve this ? (If possible - I know sometimes it's not- I'd rather see some hints than a full solution; and also if someone has read the article in question and saw that the proof in question was longer than what an MS answer can suggest, I'd also like to know haha)

EDIT: Here's the MO question : https://mathoverflow.net/questions/30220/abstract-thought-vs-calculation The answer I'm mentioning should be recognizable

Solution 1:

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

First let's consider the elementary part of this statement. Observe

Lemma. If $R$ is a subdirectly irreducible ring with no nonzero elements that square to zero, then $R$ has no nontrivial zero divisors.

Reasoning: Assume not. Then $R$ has elements $a\neq 0\neq b$ such that $ab=0$. The set $bRa$ consists of elements that square to zero, so $bRa=\{0\}$. Then $(b)(a)=RbRaR=R\{0\}R=\{0\}=(0)$. If $I$ is the least nonzero ideal of $R$, then $I\subseteq (b)$ and $I\subseteq (a)$, so $I^2\subseteq (b)(a)=(0)$. This shows that all elements of $I$ square to zero, a contradiction. \\\

Here is how the lemma applies. No nonzero element $a\in R$ which squares to zero can satisfy the equation $x^n=x$, so the lemma implies that if $R$ is subdirectly irreducible and satisfies $x^n=x$, then $R$ has no nontrivial zero divisors. Given any nonzero $a\in R$ we have $0=a^n-a = a(a^{n-1}-1)$. Since $a\neq 0$, it must be that $a^{n-1}-1=0$. Hence, any nonzero $a\in R$ satisfies $a^{n-1}=1$, showing that nonzero elements are units. Thus, $R$ is a division ring.

The non-elementary part of the statement

A subdirectly irreducible ring satisfying the equation $x^n=x$ for some $n\geq 2$ is a finite division ring

is that a division $R$ whose nonzero elements satisfy $x^{n-1}=1$ must be finite. The idea for this is to first prove that any maximal subfield of $R$ is finite (which is easy), then to prove that any finite maximal subfield of $R$ has finite index in $R$ (see Lam's A First Course in Noncommutative Rings, Theorem 15.4).

Solution 2:

I had the following idea: since $R$ has no nilpotent elements it should be a subdirect product of integral domains satisfying the same equations - however I know this property for commutative rings, and it relies on the well-known fact that $\displaystyle\bigcap\{p, p\in \mathrm{Spec}R\} = \{x, x \text{ is nilpotent}\}$ - and I don't know whether this is true for noncommutative rings.

That much is true for noncommutative rings.

The condition that $x^n = x$ for some $n$ implies $R$ is von Neumann regular, and it is well-known that a reduced VNR ring (which is called a strongly regular ring) is a subdirect product of division rings. Therefore a subdirectly irreducible ring with these properties is a division ring.

The thing working for us here is that in a strongly regular ring, quotients by prime ideals are division rings, so the prime ideals actually satisfy the commutative definition of "prime." As such, their intersection contains all nilpotent elements, and their intersection is zero. So, you can rely on a similar argument.