Timeline for Bounding the degree of an algebraic extension containing solutions to polynomials
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2018 at 23:49 | answer | added | Dillon M | timeline score: 1 | |
| Jun 10, 2016 at 0:18 | answer | added | Mohan | timeline score: 2 | |
| Jun 8, 2016 at 3:14 | comment | added | Dillon M | Thanks for the comments, @Mohan. The natural notion of degree for me is the one given in the OP, or on MSE: that is, any exponent in any term of the polynomial is bounded by $d$. I would like to apply the lemma to the rational numbers, and to prime fields, so not necessarily finite fields. Can you say something about how you arrived at your bound? | |
| Jun 7, 2016 at 0:42 | comment | added | Mohan | As I commented in MSE, at least for an infinite field, an effective bound is $d^t$, except, I am thinking of $d$ as the maximum degree of the polynomials, but here you have made a different notion of degree and then replace $d$ with $dt$. Are you interested primarily in finite fields? | |
| May 29, 2016 at 21:25 | comment | added | Dillon M | Thank you all for the comments and advice. I'm not an algebraist, so I will need some time to digest this material. I may well be back with follow-up comments and queries! | |
| May 28, 2016 at 23:20 | comment | added | R. van Dobben de Bruyn | A possible approach for finite fields is to try to use Chevalley–Warning, but I can't quite get it to work. The idea is to replace, for each $n$, the variables $x_i$ by variables $x_{1,1}, \ldots, x_{t,n}$, corresponding to $x_i = \sum x_{i,j} \alpha^j$, where $\alpha \in \mathbb F_{q^n}$ is a primitive element. Then the total degree of $f_i$ stays the same, but the number of variables grows. Unfortunately, the conclusion of Chevalley–Warning is that the number of points is $0 \pmod{p}$, which is not useful if we don't already know a solution. There might be a trick to make this into a proof. | |
| May 28, 2016 at 23:01 | comment | added | R. van Dobben de Bruyn | A very crude method is to follow the proof of Noether normalisation to obtain a bound on the degree $d$ of a finite morphism $X \to \mathbb A^m$. By generic flatness, $\mathcal O_X$ is locally free of rank $d$ over some open $U \subseteq \mathbb A^m$ (you can use this as the definition of degree when $X$ is not integral). Over an infinite field, $U$ cannot contain all rational points of $\mathbb A^m$, and any point in a fibre over a $k$-point of $U$ will have degree $\leq d$. (Again, this method fails if $k$ is finite...) | |
| May 28, 2016 at 18:05 | comment | added | Oleg Eroshkin | @JasonStarr Right, that is for infinite field. I don't know good bounds for finite fields. | |
| May 28, 2016 at 15:29 | comment | added | Jason Starr | @OlegEroshkin. The field $F$ might be finite, and then you cannot necessarily reduce to the zero-dimensional case by fixing some values of some $x_i$. However, I do agree with your upper bound over any infinite field. | |
| May 28, 2016 at 14:16 | comment | added | Oleg Eroshkin | Let $m=\min(s,t)$. Then Bezout theorem gives bound $D\leq m! d^m$. | |
| May 28, 2016 at 13:17 | comment | added | Jason Starr | Even if the zero set does not have dimension zero, by existence of the flattening stratification, there are only finitely many possible Hilbert polynomials of zero sets (after homogenizing). The maximum of the degrees (leading coefficients, basically) of those Hilbert polynomials is an upper bound. | |
| May 28, 2016 at 13:09 | comment | added | Oleg Eroshkin | If the zero set $f_1=f_2=\dots=0$ has dimension zero, then Bezout provides a bound (at least if $F$ is infinite). For the degree restrictions given by OP (degree in each variable) it is natural to consider Bezout in the product of projective spaces $(P^1)^t$. I think this bound is sharp in general, but I don't have an example. It's trivial to reduce to zero-dimensional case by fixing the values of some $x_i$. | |
| May 28, 2016 at 3:37 | answer | added | Joe Silverman | timeline score: 0 | |
| May 28, 2016 at 1:29 | comment | added | Jason Starr | The existence of a bound is a formal consequence of "quasi-compactness" and "flattening stratifications". However, I do not know an effective bound. Probably an effective bound can be derived from the theorem about flattening stratifications, but it is likely to be a wild overestimate of the true bound (something like an interated exponential in $d$). | |
| May 28, 2016 at 0:14 | comment | added | Gerry Myerson | math.stackexchange.com/questions/1801655/… | |
| May 27, 2016 at 21:58 | review | First posts | |||
| May 27, 2016 at 22:13 | |||||
| May 27, 2016 at 21:52 | history | asked | Dillon M | CC BY-SA 3.0 |