Let $d \geq 4$. I'm interested by know if there is a surface $S$ of degree $d$ in $\mathbb P^3_{\mathbb C}$ such that $S$ does not contains a line. I know I have no idea how to do it.
-
1$\begingroup$ The following link might be useful: math.stackexchange.com/questions/549882/… $\endgroup$ – Krish Mar 19 '15 at 21:08
This is the famous Noether-Lefschetz theorem. The answer is that for a "very general" such surface -- meaning away from a countable union of proper closed subsets in the parameter space of all degree-d surfaces -- the only algebraic curves are complete intersections with other surfaces. In particular, there are no lines, no conics, etc. on such a surface. While this is true in general, Mumford in the 60s or maybe 70s gave a challenge to come up with a specific example of even one quartic surface not containing a line, a challenge that was not met until just a few years ago.
-
$\begingroup$ Do you understand how this is a "challenge"? (I don't know, that's why I ask.) I mean, if most quartics do not contain lines, might as well pick one where one can compute stuff and just try one's luck.. is it really hard to compute the number of lines in a quartic? I don't know what are the involved computations. $\endgroup$ – Patrick Da Silva Mar 18 '15 at 5:30
-
$\begingroup$ That's it exactly! Choose one randomly enough and you're almost certain that it won't have a line, but how do you go about proving that there is no line on a given surface? (I actually never looked into how the problem was finally solved, so I still don't know!) $\endgroup$ – John Brevik Mar 18 '15 at 11:21
I think that
$$f(x,y,z) = x^4-3 x y+x^2 z+z^3+y^2+5$$
provides an example of a surface in ${\mathbb A}^3$ that does not contain a line in in ${\mathbb A}^3$.
Substitute $x=u_0 + t p_0$, $y = v_0 + t q_0$ and $z = w_0 + t r_0$ in $f(x,y,z)$. Collect coefficients of powers of $t$:
$$C_4 t^4 + C_3 t^3 + C_2 t^2 + C_1 t + C_0$$
where the $C_i \in {\mathbb Q}[u_0,v_0,w_0,p_0,q_0,r_0]$
Now calculate three grobner bases of $\{C_i\}_{i=0..4} \cup \{h\}$ where $h$ is either $p_0 - 1$, $q_0 - 1$ or $r_0 -1$. All three reduce to the unit ideal, proving that $V(f(x,y,z))$ does not contain a line in ${\mathbb A}^3$.
-
$\begingroup$ $34 y-68 z-60 x^2+16 x^3+52 x^3 y-20 x y^3-4 x y^2 z-89 x y z^2-77 x z^3+69 y z^3$ is a polynomial that when homogenized produces a degree $4$ surface in ${\mathbb P}^3$ which has no line in any affine patch $(D(x_i))_{i=0..3}$ and therefore no line in ${\mathbb P}^3$. $\endgroup$ – Jürgen Böhm Mar 18 '15 at 17:39
-
$\begingroup$ $21 z-14 x^3-15 y^3-80 y z^2+13 x^2 y^2-15 z^4$ is another (a bit shorter) example for a quartic in ${\mathbb P}^3$ which is smooth and contains no line. $\endgroup$ – Jürgen Böhm Mar 19 '15 at 21:07