Timeline for Bounding the degree of an algebraic extension containing solutions to polynomials
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2016 at 15:45 | comment | added | Mohan | Yes, I realized it after I posted my comment. Thanks. | |
| Jun 10, 2016 at 15:42 | comment | added | Oleg Eroshkin | Well, if the ideal contains such polynomials for every $i$, it is zero-dimensional and you don't need to cut. Otherwise, there is an $i$ and a value $c_i$ (assuming that field is infinite) that the intersection is non-empty. | |
| Jun 10, 2016 at 15:02 | comment | added | Mohan | @Oleg Eroshkin But, how do I know apriori that the ideal does not contain $P_i(x_i)$, irreducible of large degree for all $i$, in which case intersecting with $x_i=c_i$ will give empty sets. | |
| Jun 10, 2016 at 13:51 | comment | added | Oleg Eroshkin | That's why it is better to intersect with hyperplanes of the type $x_i=c_i$. Or more geometrically, embed $\mathbb{A}^t$ into a product of projective lines $(P^1)^t$ and combine with a Segre map $(P^1)^t\to P^N$. The chosen by OP notion of degree is the degree induced by this embedding. | |
| Jun 10, 2016 at 13:05 | comment | added | Mohan | The problem is, I may have to take general linear combinations and intersection with hyperplanes to reach the situation I need. For example, if I had $x_1^dx_2^d$ and I intersect with $x_1-x_2$, I end up with say $x_1^{2d}$. | |
| Jun 10, 2016 at 4:31 | comment | added | Oleg Eroshkin | The Bezout bound can be sometimes improved. For example, for the degree used by Dillon, the Bernstein-Kushnirenko theorem en.wikipedia.org/wiki/Bernstein%E2%80%93Kushnirenko_theorem gives bounds $t! d^t$ instead of $t^t d^t$ given by Bezout. | |
| Jun 10, 2016 at 0:18 | history | answered | Mohan | CC BY-SA 3.0 |