MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.