Timeline for Fields over which cubic hypersurfaces are rational
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2012 at 13:00 | comment | added | IMeasy | mmm ok I am quite convinced. thanks I guess that the only delicate point is the one that Francesco underlined. if you have 2 coinciding roots of a cubic equation the third is in the field as well. good point, thank you! | |
| Sep 28, 2012 at 11:59 | comment | added | Francesco Polizzi | Why? If the double points are all defined over $k$ and isolated, the general line through one of them will not contain any of the others, so you obtain again a rational paramatrization. The (finite number of )lines containing two double points are entirely contained in the hypersurface, since they have at least four intersection with it (counted with multiplicity). | |
| Sep 28, 2012 at 11:49 | comment | added | IMeasy | Cool, I am fine with that. What if I have MORE than one double point? Should I go to an extension of $k$? | |
| Sep 28, 2012 at 11:30 | comment | added | Jérémy Blanc | Exactly, it suffices to have one double point, defined over the field. But there are singular cubics which are not rational (with singularities consisting of distinct points not defined on the field). If there is one point defined over $k$, the surface is unirational web.math.princeton.edu/~kollar/FromMyHomePage/cubics.ps | |
| Sep 28, 2012 at 11:11 | comment | added | Francesco Polizzi | Assume that there is exactly one double point $P$, and project from this point to a hyperplane. It seems to me that this boils down to solve a cubic equation over your field $k$, which has a double root in $k$. Then also the remaining root must be in $k$, so you obtain a rational parametrization. In other words, if I'm not missing something, you should still have a birational map from your singular hypersurface to a hyperplane. | |
| Sep 28, 2012 at 10:33 | history | asked | IMeasy | CC BY-SA 3.0 |