# 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 $(a^2+a)^3=(a^2+a)^2+(a^2+a)^2$? – Wadim Zudilin Jun 26 '10 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." – José Hdz. Stgo. Jun 26 '10 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? – Victor Protsak Jun 27 '10 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. – Tom Boardman Jun 27 '10 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". – Victor Protsak Jun 28 '10 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