I saw a result in notes on by Olivier Debarre (Rational Curves on Hypersurfaces, Lecture notes for the II Latin American School of Algebraic Geometry and Applications 1-12 of June 2015 in Cabo Frio, Brazil) that if $ Z $ is a hypersurface in $ \mathbb{P}^{n}_{\mathbb{C}} $, of degree less than or equal to $ n $, then $ Z $ is uniruled, even if $ Z $ is not smooth. Does anyone know a reference for this fact. All of the books I have looked in use smoothness. If anyone knows a reference, I would greatly appreciate it. Thank you.
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1$\begingroup$ I'd recommend you systematically use the parent tag ag.algebraic-geometry in addition to more specific tags. $\endgroup$– YCorCommented Dec 8, 2019 at 15:59
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1$\begingroup$ This follows from uniruledness being a closed condition - this is proven e.g. somewhere in Kollar's book on rational curves. $\endgroup$– dhyCommented Dec 8, 2019 at 16:53
1 Answer
Not the most important thing, but your title does not match your question. The canonical divisor (if it exists!) of these objects is not vanishing. It is actually better to talk about the canonical sheaf than the canonical divisor and probably what you had in mind was that the canonical sheaf does not have non-zero global sections.
The statement should be made by degree, as you did in the body of the question.
As @dhy already pointed out, this is a "limit" argument. Families of rational curves can only degenerate to rational curves. So, let $X\to S$ be the family of hypersurfaces of a fixed degree which is less than the dimension of the projective space. Then, because of the previous observation, $\operatorname{RatLocus}(X/S)$ is closed. By the corresponding result for smooth hypersurfaces you know that the general fiber of this over $S$ is the entire general fiber of $X$ and hence $\operatorname{RatLocus}(X/S)$ is also dense. (For details, see Kollár's book as suggested by @dhy).
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$\begingroup$ Thank you Sandor. I did mean the canonical sheaf. Also, I appreciate the explanation of how uniruledness being a closed condition implies the result I was seeking. $\endgroup$– schemerCommented Dec 9, 2019 at 23:45