I read "Undergraduate algebraic geometry" by Miles Reid. I have a question about exercise 7.2 in it.
Exercise states that for any $3$ disjoint lines $l_1, l_2, l_3 \in \mathbb{P}^3(\mathbb{C})$ there exists a nonsingular quadric $Q\supset l_1, l_2, l_3$.
Over $\mathbb{C}$ there are four series of quadrics:
$(x_0)^2+(x_1)^2 + (x_2)^2+(x_3)^2=0$
$(x_0)^2+(x_1)^2 + (x_2)^2=0$
$(x_0)^2+(x_1)^2 = 0$
$(x_0)^2=0$
- and 4. are pair of projective planes and a projective plane, respectively. But one plane can not contain $2$ skew lines. If I prove that 2. can not contain $3$ disjoint lines then only 1. can. But I know there is at least one quadric for dimensional reasons. So it must be nonsingular.
Could you give any ideas why 2. can not contain $3$ disjoint lines?
A quadric $Q$ of type 2 (ie of rank $3$) is a cone on a nonsingular quadric $Q'$ in a plane. Explicitly, the plane is $H = \{x_3 = 0\}$ and the cone point is $P = [0: 0: 0: 1]$. If it contained a line $\ell$ that did not pass through $P$ then by projecting to $H$ you would get a line $\ell'$ that is contained in $Q'$. But $Q'$ is non-singular so this is not possible. In other words, all lines on $Q$ pass through $P$, in particular any two of them intersect. In fact the lines exactly correspond to points of $Q'$.