# A condition that implies commutativity

Let $R$ be a ring. A notable theorem of N. Jacobson states that if the identity $x^{n}=x$ holds for every $x \in R$ and a fixed $n \geq 2$ then $R$ is a commutative ring.

The proof of the result for the cases $n=2, 3,4$ is the subject matter of several well-known exercises in Herstein's Topics in Algebra. The corresponding proofs rely heavily on "elementary" manipulations. For instance, the proof of the case $n=3$ can be done as follows:

1) If $a, b \in R$ are such that $ab= 0$ then $ba=0$.

2) $a^{2}$ and $-a^{2}$ belong to $\mathbf{Z}(R)$ for every $a \in R$.

3) Since $(a^{2}+a)^{3} = (a^{2}+a)^{2}+(a^{2}+a)^{2}$ it follows that

$a=a+a^{2}-a^{2} = (a+a^{2})^{3}-a^{2} = (a^{2}+a)^{2}+(a^{2}+a)^{2}-a^{2}$

and whence the result. ▮

Certainly, the mind can't but boggle at the succinctness of the above solution. Actually, it is the conciseness of this argument that has prompted me to pose the present question: is an analogous demonstration of the general theorem possible? The one that appears in  depends on some non-trivial structure theorems for division rings.

References

 I. N. Herstein, Noncommutative rings, The Carus Mathematical Monographs, no. 15, Mathematical Association of America, 1968.

 I. N. Herstein, Álgebra Moderna, Ed. Trillas, págs. 112, 119, and 153.

• As a commutative reader, I'd like to learn what are your (2) and (3) above. What is **Z**$(R)$? Why $(a^2+a)^3=(a^2+a)^2+(a^2+a)^2$? Jun 26, 2010 at 8:40
• 1. $\mathbf{Z}(R)$ is the center of the ring. 2. I don't see what the problem with #3 is: $(a^2+a)^3=(a^2+a)^2(a^2+a)=(a^2+a+a^2+a)(a^2+a).$ 3. Indeed, there are several stronger results (cf. chapter 3 of Herstein's Noncommutative rings). Yitz expressed therein that the version mentioned by Pete, "as proved has one drawback; true enough, it implies commutativity but only very few commutative rings exist which satisfy its hypothesis." Jun 26, 2010 at 19:46
• $\textit{Certainly, the mind can't but boggle at the succinctness of the above solution}$ Indeed, I don't understand it at all, especially "whence the result" (step 3 involves only one ring element, whereas two are needed for commutativity). Have you not made it a bit $\textit{too}$ succint? Jun 27, 2010 at 7:00
• @Viktor- 3) gives any element 'a' as the sum/difference of squares of elements and from 2) (and the closure of the centre under addition) we have that 'a' belongs to the centre. Jun 27, 2010 at 11:15
• Thank you, Tom! That also explains why both $a^2$ and $-a^2$ were mentioned in 2 :) I find it amusing that the author expects us to see that the "identity" $\implies 1$ and $1 \implies 2$ right away, but worries that we may get lost with "center is closed under negation". Jun 28, 2010 at 5:30

For fixed $$n \in \mathbb{N}$$, Birkhoff's completeness theorem implies that such a proof must exist in the first-order equational theory of rings - as I mentioned here in a recent post. Many years ago Stan Burris told me that John Lawrence discovered such an equational proof that works uniformly for all $$n$$ (possibly also for Jacobson's form $$x^{n(x)} = x$$). I don't know if the proof is published yet, but some clues as to how it may proceed might be gleaned from their earlier joint [work ] 1

1 S. Burris and J. Lawrence, Term rewrite rules for finite fields.
International J. Algebra and Computation 1 (1991), 353-369. http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf

• Has there been any development on this subject in these years? Is the proof published now? (I can't find it) Nov 12, 2017 at 0:54
• Based on my work, I highly doubt that there is an equational proof that works uniformly in $n$. Nevertheles, I am also very interested in John Lawrence's proof. The linked paper is not very helpful for the problem here. Oct 10 at 6:16

Yes, it is possible to write down an equational proof for every $$n$$. This is covered in the preprint

Equational proofs of Jacobson's Theorem, arXiv:2310.05301 [math.RA]

The rough idea is to prove the reduction to prime characteristic $$p$$ and then to $$n=p^k$$ in a constructive way. Then, the (more or less equivalent) proofs by Dolan and Nagahara-Tominaga, which reduce the Theorem to Wedderburn's Theorem, can be made constructive as well. The paper discusses several examples. For example, every ring with $$x^{2023}=x$$ is commutative by an equational proof.

I have started this project in 2014, then continued in 2020, and now finished it (finally!) in 2023. Unfortunately, there are still some cases missing. Namely, if $$\mathrm{gcd}(k,p^k-1) > 1$$ (cf. MO/455808), the constructive Wedderburn Theorems $$W_{p,k,T^m}$$ from the paper (which are used to prove that $$p^k$$-rings of characteristic $$p$$ are commutative) can only be verified theoretically by a computer program.

The issue with using Birkhoff's Completeness Theorem (as mentioned in Bill's answer) is that it is a pure existence result (see MO/353564). It does not tell us how equational proofs look like.

• Cool! I am going to take at look at your preprint later on... Oct 10 at 16:12