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Let $k = \bar{k}$ a fixed field. I would like to know if there exist hypersurfaces $X \subset \mathbb{A}_k^n$ that contain no lines. By line I really mean line, and not just rational curve.

I haven't put any restrictions on $X$, but it's still not clear to me that such things exist. Most likely they form a nonempty open subset of some Hilbert scheme if the degree is large enough.

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If $k$ is the field of complex numbers, then generic hypersurface of sufficiently high degree does not contain even rational curves. –  Alexandre Eremenko Nov 3 '12 at 4:22
Alexandre, Is this clear, or more like general knowledge? –  LMN Nov 3 '12 at 4:30
A line can be parameterized by $x_i=a_it+b_i$ for $i=1,\ldots,n$ where $t\in k$. If you write this out in terms of the defining polynomial you'll see that if the degree is large, there will not be any lines there. –  J.C. Ottem Nov 3 '12 at 7:09
LMN: No, this is a deep result:-) But you already have an answer on your much simpler question. –  Alexandre Eremenko Nov 3 '12 at 13:54

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up vote 7 down vote accepted

First, consider the projective space $P^n$ instead of affine --- if a hypersurface in affine space contains a line then its closure contains the closure of the line which is a line in the projective space. Consider the Grassmannian $G = Gr(2,n+1)$ parameterizing those lines and let $U$ be the tautological rank 2 subbundle on it. The equation of a hypersurface of degree $d$ gives a section of the vector bundle $S^dU^*$ on $G$ (and vice versa). Lines on the hypersurface are parameterized by the zero locus of the corresponding section. When $$ r(S^dU^*) = d + 1 > 2(n-1) = \dim G $$ the zero locus of a general section is empty, since the vector bundle is generated by global sections.

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Great! How do hypersurfaces of degree $d$ give sections of $S^dU^*$ on $G$? –  LMN Nov 3 '12 at 4:50
A polynomial of degree $d$ on $V$ is an element of $S^dV^*$. The tautological bundle is a subbundle in $V\otimes O_G$, so there is a map $V^*\otimes O_G \to U^*$ (restricting linear functions to subspaces), its symmetric power is a map $S^dV^*\otimes O_G \to S^dU^*$. –  Sasha Nov 3 '12 at 6:36

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