# Rings with $a^5=a$ are commutative

Update. I have generalized the method here and have shown the theorem for many more exponents: Equational proofs of Jacobson's Theorem

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?

• The paper "Automated Proof of Ring Commutativity Problems by Algebraic Methods" by H. Zhang (sciencedirect.com/science/article/pii/S074771710880020X) looks very promising, at least for the case that the exponent is even. A little bit is said about the case of $5$, however not enough (in particular, the mentioned equalities are trivial in characteristic $5$). Does someone have access to "Prove ring commutativity problems by algebraic methods" (1989) by the same author? This seems to have more details. Sep 23, 2014 at 2:50
• Related Dec 12, 2019 at 18:54

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. $\square$

• Another proof (actually for all $n \leq 25$ odd and all even $n \leq 50$) appears in Y. Morita, Elementary proofs of the commutativity of rings satisfying $x^n=x$, Memoirs Def. Acad. Jap. XVIII (1978), 1-23. Nov 16, 2014 at 13:56
• How does the second to the last equation follow? I mean, how do we know that each term in the right hand side is in the center? Apr 30, 2015 at 17:38
• In fact, how do we even show that $4x^3$ is central? How do we know that $2(x^2+x)^2$ is central? Apr 30, 2015 at 17:49
• @user134070 substitute $x^2 + x$ everywhere you see an $x$ on the LHS of the preceding line. although now that i look, you asked this five and a half years ago, so you've probably moved on :) Sep 29, 2020 at 7:29