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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.

As usual, I thank you in advance for your insightful replies, reading suggestions, web links, etc...

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.

  • 3
    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 08:40
  • My guess is that Z(R) is the center of the ring and 3) follows from a^3=a, I think. – M.G. Jun 26 '10 at 12:04
  • I have a similar guess for (3) but this isn't as obvious as what follows in (3). That's why I ask the author or everybody who loves this question (and it's obviously liked!) to give some details. Thanks! – Wadim Zudilin Jun 26 '10 at 14:33
  • I just wanted to mention that there is a relevant article by Kaplansky, Commutativity Revisited. It appears (only) in his Selected Papers and Other Writings. The beginning of it is available on google books. – Pete L. Clark Jun 26 '10 at 17:20
  • I forgot to mention that Jacobson actually proved a stronger result (and Kaplansky's paper discusses for the most part yet stronger results): the conclusion still holds if the exponent $n \geq 2$ is allowed to depend on $x$. – Pete L. Clark Jun 26 '10 at 17:22
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    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
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  • $\mathbf{Z}(R)$ 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
  • I don't think this paper gives quite the succint argument you want, but it may be an improvement over previous proofs: http://www.springerlink.com/content/p760r6271707j8q7/ – Akhil Mathew Jun 26 '10 at 20:23
  • @Wadim: My (1), (2), and (3) above are proof steps. – José Hdz. Stgo. Jun 26 '10 at 23:38
  • Great, I now follow your problem. Thanks! – Wadim Zudilin Jun 27 '10 at 01:46
  • The first page of the Jacobson article mentioned by in one of Kap's short pieces, mentioned in turn by Pete, is at link
    http://www.jstor.org/pss/1969205
    The second paragraph contains the result with varying exponent.
    – Will Jagy Jun 27 '10 at 02:45
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    $\textit{Certainly, the mind can't but boggle at the succinctness of the above solution}$ 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 07:00
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    @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
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    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 05:30
  • It's not that I worry about that. I was just mentioning what the ingredients of the proof are... – José Hdz. Stgo. Jun 28 '10 at 19:39