Timeline for A condition that implies commutativity
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2023 at 6:14 | answer | added | Martin Brandenburg | timeline score: 14 | |
| Sep 24, 2014 at 1:20 | history | edited | José Hdz. Stgo. | CC BY-SA 3.0 |
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| Sep 18, 2010 at 23:21 | vote | accept | José Hdz. Stgo. | ||
| Jul 24, 2010 at 21:38 | answer | added | Bill Dubuque | timeline score: 36 | |
| Jul 24, 2010 at 19:09 | history | edited | Victor Protsak |
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| Jun 28, 2010 at 19:39 | comment | added | José Hdz. Stgo. | It's not that I worry about that. I was just mentioning what the ingredients of the proof are... | |
| Jun 28, 2010 at 5:30 | comment | added | Victor Protsak | 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 27, 2010 at 11:15 | comment | added | Tom Boardman | @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 7:00 | comment | added | Victor Protsak | $\textit{Certainly, the mind can't but boggle at the succinctness of the above solution}$ 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 2:45 | comment | added | Will Jagy | The first page of the Jacobson article mentioned by in one of Kap's short pieces, mentioned in turn by Pete, is at link jstor.org/pss/1969205 The second paragraph contains the result with varying exponent. | |
| Jun 27, 2010 at 1:46 | comment | added | Wadim Zudilin | Great, I now follow your problem. Thanks! | |
| Jun 26, 2010 at 23:59 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
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| Jun 26, 2010 at 23:38 | comment | added | José Hdz. Stgo. | @Wadim: My (1), (2), and (3) above are proof steps. | |
| Jun 26, 2010 at 20:23 | comment | added | Akhil Mathew | I don't think this paper gives quite the succint argument you want, but it may be an improvement over previous proofs: springerlink.com/content/p760r6271707j8q7 | |
| Jun 26, 2010 at 19:46 | comment | added | José Hdz. Stgo. | 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 19:37 | history | edited | José Hdz. Stgo. | CC BY-SA 2.5 |
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| Jun 26, 2010 at 18:55 | comment | added | Robin Chapman | 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. | |
| Jun 26, 2010 at 17:22 | comment | added | Pete L. Clark | 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 $x$. | |
| Jun 26, 2010 at 17:20 | comment | added | Pete L. Clark | 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. | |
| Jun 26, 2010 at 14:33 | comment | added | Wadim Zudilin | 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! | |
| Jun 26, 2010 at 12:04 | comment | added | M.G. | My guess is that Z(R) is the center of the ring and 3) follows from a^3=a, I think. | |
| Jun 26, 2010 at 8:40 | comment | added | Wadim Zudilin | 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 8:21 | history | asked | José Hdz. Stgo. | CC BY-SA 2.5 |