Timeline for Quick proofs of hard theorems
Current License: CC BY-SA 2.5
4 events
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| Jul 17, 2010 at 21:19 | comment | added | Abdelmalek Abdesselam | Isn't the resultant proof in CP^n trivial? If you have n hypersurfaces of deg d_1,...,d_n and look at their resultant with an extra linear form u, saying Bezout's thm is the same as saying that the degree of this resultant wrt u is the product d_1...d_n. Moreover this polynomial in u is in the Brill locus of completely factorizable homogeneous polynomials and the linear factors are the intersection points of the n hypersurfaces. | |
| May 20, 2010 at 13:42 | comment | added | Paul Siegel | I think I've only ever seen the nasty resultant version. How does the cohomological proof go? If the cohomological machinery is hard to develop from scratch but Bezout's theorem falls out easily once you have it, this would be a great example. | |
| May 18, 2010 at 3:26 | comment | added | Greg Kuperberg | I think that this one is only a half victory. When I went through the cohomological machinery use to prove Bezout's theorem, it wasn't as quick as it first seemed. At the same time, I think that there is a resultant-ish proof based on the Hilbert-Poincare series of the projective varieties involved, which is comparable in difficulty to the homological proof. Note that if you want Bezout's theorem in positive characteristic, you have to either develop etale cohomology or do something more direct. | |
| May 17, 2010 at 1:13 | history | answered | Steven Gubkin | CC BY-SA 2.5 |