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All cubic hypersurfaces having at least one double point are birational to some $P^n$ over an algebraically closed field. How does the statement change as I pass to non alg closed fields? Does it hold at all? Or do I need to change hypothesis?

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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. –  Francesco Polizzi Sep 28 '12 at 11:11
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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 –  Jérémy Blanc Sep 28 '12 at 11:30
    
Cool, I am fine with that. What if I have MORE than one double point? Should I go to an extension of $k$? –  IMeasy Sep 28 '12 at 11:49
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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). –  Francesco Polizzi Sep 28 '12 at 11:59
    
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! –  IMeasy Sep 28 '12 at 13:00
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