# 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 [1] depends on some non-trivial structure theorems for division rings.

References

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

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

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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
The standard notation for the centre of the ring $R$ is $Z(R)$ (the $Z$ not usually bold). My idle speculation is that there should be a "calculational" proof like the given one for any particular $n$, but there being such a proof for all $n$ seems unlikely. Treating general $n$ I suspect needs "second-order" concepts: subrings, ideals, quotients, stuff like that. –  Robin Chapman Jun 26 '10 at 18:55
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." –  J. H. S. 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

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]