Timeline for A seemingly Groebnerizable problem
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 18, 2018 at 21:35 | vote | accept | Igor Rivin | ||
| Apr 18, 2018 at 19:29 | answer | added | Zach Teitler | timeline score: 5 | |
| Apr 16, 2018 at 22:25 | comment | added | Igor Rivin | @ZachTeitler Actually, never mind, I see now. | |
| Apr 16, 2018 at 22:04 | comment | added | Igor Rivin | @ZachTeitler What I had meant was: is there any hypersurface for which $\mathcal{Q}$ was non-empty. Will's (and yours) answer pretty clearly says "no" (but I wasn't sure if I was missing anything...) | |
| Apr 16, 2018 at 21:29 | comment | added | R. van Dobben de Bruyn | @IgorRivin: an example of a polynomial satisfying the condition is any polynomial $p$ divisible by $x_ix_j$ for $i \neq j$, and linear combinations of such. | |
| Apr 16, 2018 at 21:29 | comment | added | Igor Rivin | @ZachTeitler Fair enough. However, I don't believe any such hypersurface exists :) | |
| Apr 16, 2018 at 20:57 | comment | added | Zach Teitler | @IgorRivin The line through $a=(a_1,\dotsc,a_n)$ parallel to the $k$th coordinate axis is parametrized by, say, $(a_1,\dotsc,a_k+t,\dotsc,a_n)$. The restriction of $p$ to this line is given by $q(t)=p(a_1,\dotsc,a_k+t,\dotsc,a_n)$. Note that $q$ is a polynomial, since $p$ is. The Taylor coefficients of $q$ are exactly the "pure" partial derivatives $\partial^i p/\partial x_k^i$ (perhaps with some factorial factors). So $p$ vanishes on the line $\iff$ $q$ vanishes identically $\iff$ the pure partials vanish at point $a$. | |
| Apr 16, 2018 at 20:53 | comment | added | Igor Rivin | @aginensky I am not wise enough in the ways of singularities to confirm or deny what you are saying. Perhaps you could elaborate? Thanks! | |
| Apr 16, 2018 at 20:52 | comment | added | Igor Rivin | @ZachTeitler You have just paraphrased Will's statement. Why is what you say true? | |
| Apr 16, 2018 at 20:48 | comment | added | Zach Teitler | @IgorRivin Will Sawin's statement follows from the observation that the derivatives $\partial^i p/\partial x_k^i$ vanish at a point $a$ for all $i \geq 0$ if and only if $p$ vanishes identically on the line through $a$ parallel to the $k$th coordinate axis. So $a \in \mathcal{Q}$ if and only if $p$ vanishes identically on the union of the axis-parallel lines through $a$. | |
| Apr 16, 2018 at 20:29 | comment | added | meh | Apologies if I am misreading what you are saying, but to say that a point is in that locus is to say that the hypersurface has a singularity of at least multiplicity k at that point. Aren't there bounds on the singularity of hypersurfaces ? At the least projection from the point bounds k at deg(p) - 1. | |
| Apr 16, 2018 at 20:01 | comment | added | Igor Rivin | @WillSawin Also, what is an example of a $p$ which satisfies your condition? | |
| Apr 16, 2018 at 19:59 | comment | added | Igor Rivin | @WillSawin Why? | |
| Apr 16, 2018 at 19:55 | comment | added | Will Sawin | In characteristic zero, $(x_1,\dots,x_n)$ is in $\mathcal Q$ if and only if $p$ vanishes on all the coordinate axes through $(x_1,\dots,x_n)$, i.e. all the lines given by fixing all but one of the coordinates and varying the last. Unlike the classical notion of singularity this depends only on the vanishing locus of $p$. | |
| Apr 16, 2018 at 19:31 | comment | added | Igor Rivin | @DimaPasechnik No, I am interested more in $\mathbb{R}, \mathbb{C}.$ | |
| Apr 16, 2018 at 18:58 | comment | added | Dima Pasechnik | Are you only interested in characteristic p case? | |
| Apr 16, 2018 at 18:40 | history | asked | Igor Rivin | CC BY-SA 3.0 |