# An exercise involving the twisted cubic

One of the exercises in Hatcher's Harris's AG book goes as follows.

Let $F_0 = Z_0 Z_2 - Z_1^2$, $F_1 = Z_0 Z_3 - Z_1 Z_2$, $F_2 = Z_1 Z_3 - Z_2^2$ (s.t. $\mathbb{V}(F_0, F_1, F_2)$ is the twisted cubic). Define $F_\lambda = \lambda_0 F_0 + \lambda_1 F_1 + \lambda_2 F_2$. Prove that for any $[\lambda_0, \lambda_1, \lambda_2] \neq 0$ and $[\mu_0, \mu_1, \mu_2] \neq 0$ the zero loci of $F_\lambda$ and $F_\mu$ intersect at the union of the twisted cubic and a line through two of its points.

I wrote down the system of equations $\mathrm{d}F_\lambda = 0$ and got a system of linear equations of $Z$ with the following matrix: $$A = \begin{pmatrix} 0 & 0 & \lambda_0 & \lambda_1 \\ 0 & -2 \lambda_0 & - \lambda_1 & \lambda_2 \\ \lambda_0 & - \lambda_1 & -2 \lambda_2 & 0 \\ \lambda_1 & \lambda_2 & 0 & 0 \end{pmatrix}$$

We want non-trivial solutions, and they only exist when the determinant of this matrix is zero. I calculated that it is only zero when $\lambda_1 = 0$ and either $\lambda_0 = 0$ or $\lambda_2 = 0$, (UPD) or $\lambda_0 \lambda_2 = \lambda_1^2 \neq 0$: $$\det A = \lambda_0^2 \lambda_2^2 - 2 \lambda_0 \lambda_1^2 \lambda_2 + \lambda_1^4$$

But this seems to clearly contradict the exercise, since for it to hold there must be two families of singularities for any $F_\lambda$!

Did I make a mistake? Is my approach to this exercise sound? Should I continue trying in the same vein or look at the problem from another angle?

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Dear Alexei, aren't you confusing Hatcher and Hartshorne? –  Georges Elencwajg May 13 '12 at 12:13
@GeorgesElencwajg actually, with Harris :D –  Alexei Averchenko May 13 '12 at 12:17