Timeline for Ways to show a system of polynomial equations has no solution
Current License: CC BY-SA 3.0
18 events
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| S Oct 7, 2014 at 17:35 | history | bounty ended | CommunityBot | ||
| S Oct 7, 2014 at 17:35 | history | notice removed | CommunityBot | ||
| Oct 7, 2014 at 8:34 | comment | added | Binzhou Xia | @NoamD.Elkies Besides your observation, there is also a change of variables of the form $X_r=\pm\zeta^rY_r$ to kill the powers of -1 in the first two lines simultaneously; here $\zeta$ is a $m/2$ or $2m$-th root of unity according to $4$ divides $m$ or not. | |
| Oct 7, 2014 at 8:20 | comment | added | Binzhou Xia | @NoamD.Elkies I came across this system in a study for power residue difference set. I found that if the subgroup of index $m$ in $\mathbb{F}_p^\times$ form a difference set of $\mathbb{F}_p^+$, then $X_r\equiv\Gamma_p(\frac{r}{m})\equiv\frac{1}{(r(p-1)/m)!}\pmod{p}$ should satisty the first line of equations. The last two lines of equations come from the own properties of $p$-adic Gamma functions (here I'm considering the case when $2$ is a $m$-th power in $\mathbb{F}_p$). My origional purpose is to exclude the existence of certain power residue difference sets by showing it has no solutions. | |
| Oct 6, 2014 at 3:19 | comment | added | Noam D. Elkies | How did you "come across" this system? The first line says that the generating polynomial $P(t) = 1 + \sum_{k=1}^{m-2} X_k t^k$ satisfies $P(t) P(-t) \equiv 1 \bmod t^m$, but the second line doesn't look like any kind of standard polynomial property (well, $P(t) P(-u)$ has an anti-diagonal of coefficients of $1$). I suppose you know already that there's an action of a cyclic group $C$ of order $m$: for each $m$-th root of unity $\zeta$ you can multiply each $X_k$ by $\zeta^k$. Plus $X_{m/2} = \pm 1$ or $\pm i$ according as $m/2$ is odd or even, with the two sign choices equivalent under $C$. | |
| S Sep 29, 2014 at 15:54 | history | bounty started | Binzhou Xia | ||
| S Sep 29, 2014 at 15:54 | history | notice added | Binzhou Xia | Draw attention | |
| Sep 24, 2014 at 11:23 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 24, 2014 at 9:36 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 21, 2014 at 4:22 | comment | added | Noam D. Elkies | For the "further question", even $p > m^2$ is likely way too optimistic. One can easily write linear equations in $m$ variables, with no coefficient larger than $2$, that have no solution in $\bf C$ but do have a solution modulo some prime $p \sim 2^m$. (Use the binary expansion of $p$ to build up to $x_m=p$, and then add the equation $x_m=0$.) | |
| Sep 21, 2014 at 3:38 | answer | added | Igor Rivin | timeline score: 4 | |
| Sep 21, 2014 at 3:33 | comment | added | Binzhou Xia | @B.Wellington Thanks to your comment. I've changed $2X_{2t}$ to $2X_{2s}$ now. | |
| Sep 21, 2014 at 3:32 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 21, 2014 at 3:13 | history | edited | Binzhou Xia | CC BY-SA 3.0 |
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| Sep 21, 2014 at 3:04 | comment | added | James Weigandt | Call the variety $V(m)$. Have you tried looking for morphisms $V(m) \to V(d)$ for divisors $d$ of $m$? Maybe $m$ will have to be shifted to $m - 1$ or $(m - 1)/2$ or something. What's the story behind this variety? Sometimes that gives rise to natural morphisms. | |
| Sep 21, 2014 at 2:19 | comment | added | Binzhou Xia | @B.Wellington The first equation is $2X_2-X_1^2=0$, so $X_2=\frac{X_1^2}{2}$. In fact, the first line of equations shows that $X_{2t}$ is a polynomial of $X_1,\dots,X_{2t-1}$ for $1\le t\le\frac{m}{2}-1$. | |
| Sep 20, 2014 at 17:59 | comment | added | Walter Neff | In your first equation, do you mean $2X_{2s}$? | |
| Sep 20, 2014 at 15:31 | history | asked | Binzhou Xia | CC BY-SA 3.0 |