Timeline for A special case of the polynomial Bézout's identity: bounding the co-factors
Current License: CC BY-SA 4.0
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| May 31, 2019 at 20:25 | comment | added | W-t-P | @alpoge: Thanks for the useful comment - and, could you elaborate on the second part of it? Do $f$ with $\deg a=\deg f-2$ exist at all? Why would this imply that one has $\deg a=\deg f-2$ for the generic $f$? | |
| May 31, 2019 at 16:59 | comment | added | alpoge | Sorry —- I ran out of characters then was distracted! I meant to simply comment that therefore once one produces one f (out of laziness/busyness I haven’t tried) for which the corresponding a has degree exactly \deg{f} - 2, it follows that the generic f does too, since having lower degree is expressed as the vanishing of a polynomial. No comment on the other questions (besides that they’re expressed as the vanishing of various polynomials from the above, not that this gives you a good answer), and do let me know if I’ve overlooked something in characteristic p!!! | |
| May 31, 2019 at 16:29 | comment | added | alpoge | The map (a,b)\mapsto af+bf’ is a linear map from \Sym^{n-1}\oplus \Sym^n\to \Sym^{n+\deg{f}-1} (\Sym^d is the space of polynomials of degree at most d). The RHS has dimension n+\deg{f}, the LHS has dimension 2n+1. Taking n = \deg{f}-1 both sides have the same dimensions and indeed the map is injective by virtue of f and f’ being coprime. Thus it is surjective. Thus you find (a,b) with \deg{a}\leq \deg{f}-2 and \deg{b}\leq \deg{f}-1 mapping to 1. [This is the usual definition of the resultant.] Anywho what this tells you is that you can compute (a,b) by inverting a matrix (in the coeffs of f). | |
| May 31, 2019 at 14:46 | history | asked | W-t-P | CC BY-SA 4.0 |