Timeline for Not especially famous, long-open problems which anyone can understand
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2020 at 23:29 | comment | added | user142929 | (2/2) where $f(x)=(x-a)^d$ is the polynomial in Casas-Alvero conjecture and $a\in k$, and I try to evoke/transfer conditions to state the equivalent form as the statement A suitable set of conditions for the functions $R_k(x)$ implies (say) $Q_1(x)=(x-a)^d$. This can be a naive strategy, my informal reasonings were of the kind taking products or realize those keywords in the conjecture: monic polynomials and degrees; and I don't know if an equivalent formulation in terms of rational functions can be exploited as advantage. My apologizes and thanks to the OP. | |
| Jan 25, 2020 at 23:29 | comment | added | user142929 | (1/2) Do you know if the conjecture can be paraphrased in terms of certain rational functions satisfying certain conditions? I'm not a professional mathematician, but I wondered if you @FrançoisBrunault can/want tell me if it is potentially interesting, after I known this nice post from the professor. I was trying to propose informal conditions for $d-1$ rational functions $R_k(x)=\frac{P_k(x)}{Q_k(x)}$, thus $1\leq k\leq d-1$ to evoke an equivalent form of Casas-Alvero conjecture. My toy model are these $\frac{f'(x)}{f(x)},\frac{f''(x)}{f'(x)},\ldots ,\frac{f^{(d-1)}(x)}{f^{(d-2)}(x)}$, | |
| Jul 31, 2017 at 2:51 | comment | added | user49512 | I think this squarely fails the 'anyone can understand it' requirement. | |
| Oct 31, 2016 at 8:06 | history | edited | Martin Sleziak | CC BY-SA 3.0 |
added Wikipedia link
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| Apr 24, 2016 at 13:49 | comment | added | Amr | Oh ok I misunderstood the conjecture the first time I read it | |
| Apr 24, 2016 at 13:29 | comment | added | Amr | Whaaat? I thought this was a trivially true statement | |
| Nov 20, 2015 at 18:23 | comment | added | Matheus | @FrançoisBrunault: This paper has been withdrawn today (see arxiv.org/abs/1511.04932v2) | |
| Nov 19, 2015 at 7:28 | comment | added | François Brunault | A proof of the conjecture is on arxiv arxiv.org/abs/1511.04932v1 | |
| Nov 13, 2015 at 16:27 | comment | added | Andreas Thom | @DenisSerre: The title was changed to "Towards the Casas-Alvero Conjecture". | |
| Jul 31, 2015 at 19:49 | comment | added | Denis Serre | @AndreasThom. Not yet. | |
| Jul 31, 2015 at 16:33 | comment | added | Andreas Thom | @DenisSerre. Did you check the proof? | |
| Jul 31, 2015 at 14:51 | comment | added | Denis Serre | @Andreas. Strange that this paper deals only with $k={\mathbb C}$. | |
| Jul 31, 2015 at 7:34 | comment | added | Andreas Thom | arxiv.org/abs/1504.00274 | |
| Jun 25, 2012 at 6:22 | comment | added | js21 | @François Brunault. Some months ago I asked this question mathoverflow.net/questions/94838/… with the Casas-Alvero conjecture in mind. It appeared from the answers that instead of the argument using scheme theory, the simpler Lefschetz principle ( proofwiki.org/wiki/Lefschetz_Principle_(First-Order) ) can be used. (answering to my question, Qiaochu Yuan also indicated an ultraproduct construction which is even simpler than the Lefschetz Principle, since no completeness result is used). | |
| Jun 22, 2012 at 22:57 | comment | added | François Brunault | For those interested in this conjecture, here is what I believe the current state of knowledge on the conjecture : arxiv.org/abs/math/0605090 The first open case is $n=12$. Interestingly, the proofs in the known cases use scheme theory (over $\mathbf{Z}$). | |
| Jun 22, 2012 at 20:39 | history | edited | Denis Serre | CC BY-SA 3.0 |
added 48 characters in body
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| Jun 22, 2012 at 20:38 | comment | added | Denis Serre | @Joel. Right! If $k$ is of finite characteristic $p$, then $X^{2p}+X^p$ does share a root with every derivative, but is not a monomial. | |
| Jun 22, 2012 at 19:35 | comment | added | Joël | I guess $k$ must be of characteristic $0$ | |
| Jun 22, 2012 at 19:19 | comment | added | Joël | $k$ is any field? | |
| Jun 22, 2012 at 6:39 | history | answered | Denis Serre | CC BY-SA 3.0 |