Abstract Algebra – Rings with a^5=a are Commutative

Question

Let $R$ be a ring such that $a^5=a$ for all $a \in R$. Then it follows that $R$ is commutative.

This is part of a more general well-known theorem by Jacobson for arbitrary exponents ($a^n=a$), which appeared on math.stackexchange a couple of times. But what I would like to see is a proof which is (1) direct / elementary and (2) in equational language (in particular, first-order). (As far as I know, no published proof of Jacobson's Theorem is equational, but such a proof has to exist). See here and there for such proofs for the exponents $3$ and $4$. I hope that these proofs make the "rules" clear.

What I've done so far: Reduced to the case that $R$ has characteristic $5$. Any element of the form $a^4$ is central, because it is idempotent and $R$ is reduced. Also, it suffices to prove $(ab)^2= b^2 a^2$ for $a,b \in R$. In fact, this implies
$$ab=(ab)^5=a(ba)^4 b = (ba)^4 ab = b a b (ab)^2 a^2 b$$
$$ = b a b (b^2 a^2) a^2 b = b a b^3 a^4 b = b a a^4 b^3 b = b a b^4 = b b^4 a = ba.$$
Perhaps someone can feed automatic theorem provers with this problem. But my experience is that their proofs are not so easy to follow, quite long and not intuitive. So I am actually looking for hand-made proofs which eventually might also work for rings satisfying $a^p=a$ where $p$ is any prime.

Edit. Here is some progress, inspired by the proof by Nagahara and Tominaga. Consider $a \in R$ and decompose the finite commutative reduced ring $\langle a \rangle$ into a product of fields – this is just the Chinese Remainder Theorem. Hence, $a=a_1+\dotsc+a_n$ where each $\langle a_i \rangle$ is a field with unit $e_i=a_i^4$. The only fields satisfying the equation are $\mathbb{F}_2$, $\mathbb{F}_3$ and $\mathbb{F}_5$, which are prime fields. Hence, $a_i = z_i e_i$ for some $z \in \mathbb{Z}$ is central and therefore $a$ is central. (This incredibly quick proof for the $(a^n=a)$-problem works for all exponents $n$ where every prime power $q$ such that $q-1|n-1$ is actually a prime number.)

2nd Edit. This leads to an equational proof. For simplicity, let us assume that $R$ has characteristic $5$. (This reduction is not in equational language, but the cases of characteristic $2$ and $3$ can be dealt separately and in the end everything may be put into a single equational proof.) Then $\langle a \rangle$ is a quotient of $\mathbb{F}_5[T]/(T^4-1)$, which is isomorphic to $\mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_5$ via $f \mapsto (f(1),f(2),f(3),f(4))$. We can compute the corresponding idempotents:
$$e_1 = (t-2)^2 (t-3)(t-4), e_2 = (t-1) (t-3) (t-4)^2, \\e_3 = (t-1)^2 (t-2) (t-4), e_4 = (t-1) (t-2) (t-3)^2.$$
Back in $R$ we obtain the equation
$$a = e_1(a) + 2 e_2(a) + 3 e_3(a) + 4 e_4(a).$$
Each $e_i(a)$ is idempotent, hence central, and therefore $a$ is central. This proof may be carried out in equational language – but then we would have to verify the equation and that the $e_i(a)$ are idempotent, which is obviously very tedious.

Is there also a proof which is (3) not tedious?

Answer 1

I have found a really smart proof in these notes by Geunho Gim.

Lemma. In a reduced ring, if $e^2=ze$ for some $z \in \mathbb{Z}$, then $ze$ is central.

Proof: For a given element $x$, expand $(exe-zex)^2$, it evaluates to zero. Hence, $exe=zex$. Similarly, $exe=xze$. $\square$

Theorem. If $R$ is a ring in which every element $x$ satisfies $x^5=x$, then $R$ is commutative.

Proof. Let $x \in R$. Then $(x^4+x^2)^2=x^8+2x^6+x^4=2(x^4+x^2)$, hence $2(x^4+x^2)$ is central by the Lemma. Since $x^4$ is idempotent, it is central by the Lemma. Hence, $2x^2=2(x^4+x^2)-2 x^4$ is central. But then $4x^3=2(x^2+x)^2-2x^4-2x^2$ is central. It follows that $$2x^3+x^2+5x=(x^2+x)^5-(x^2+x) - 10 x^4 - 2 (4 x^3) - 2(2x^2)$$ is central. Therefore $$7x^4+4x^3 + 8x^2 + 11x = 2(x^2+x)^3 + (x^2+x)^2+5(x^2+x)$$ is central, too. Finally, we see that $$x = (7x^4+4x^3+8x^2+11x)-7x^4 - 3(2x^2)- 2(2x^3+x^2+5x)$$ is central. $\square$