I am trying to do part (b) of Exercise 1.11 in Harris' book Algebraic Geometry: A First Course.

Let $F_0=Z_0Z_2−Z_1^2$, $F_1=Z_0Z_3−Z_1Z_2$, $F_2=Z_1Z_3−Z_2^2$ (s.t. $V(F_0,F_1,F_2)$ is the twisted cubic). Define $F_λ=λ_0F_0+λ_1F_1+λ_2F_2$. Prove that for any $[λ_0,λ_1,λ_2]≠0$ and $[μ_0,μ_1,μ_2]≠0$ with $\lambda\neq \mu$ the zero loci of $F_λ$ and $F_μ$ intersect at the union of the twisted cubic and a line through two of its points.

This question was posted here a while ago, but no one answered.

I have no idea where to start. The fact that I have to show that something is the union of varieties is confusing to me, as it corresponds to intersecting ideals.

I would be happy with just a solution to part (a) of this exercise, which is the special case where $F_{\lambda}=F_0$ and $F_{\mu}=F_1$.

  • $\begingroup$ Let $\lambda_0 = 1$, all others zero, and let $\mu = \lambda$. Then the intersection of the zero loci of $F_\lambda$ and $F_\mu$ is $Z_0Z_2-Z_1^2=0$. That does not seem to me like the union of two curves. I don't have Harris's book but I presume the statement provided here leaves out some hypotheses present in the original. $\endgroup$ – guest May 23 '16 at 1:53
  • $\begingroup$ I forgot to add $\lambda \neq \mu$. Thanks. $\endgroup$ – Justine May 23 '16 at 1:56
  • $\begingroup$ Do you mean something like $\lambda$ and $\mu$ linearly independent? Otherwise let $\lambda_0=1$ and $\mu_0=2$ and continue as before. $\endgroup$ – guest May 23 '16 at 2:01
  • $\begingroup$ The book says $\lambda\neq \mu$, but maybe you found an error... $\endgroup$ – Justine May 23 '16 at 2:02
  • $\begingroup$ If the expression $[\lambda_0,\lambda_1,\lambda_2]$ refers to homogeneous coordinates in projective space, there is no error, and I suppose that is what the book means. I'd write it $[\lambda_0 : \lambda_1 : \lambda_2]$ to avoid confusion. $\endgroup$ – guest May 23 '16 at 2:08

I'll look at problem (a). On $U_0 = \{Z_0 \ne 0\}$ you are intersecting the affine curves $$z_2 = z_1^2, \; \; z_3 = z_1 z_2.$$

From these equations already follows $$z_2^2 = z_1^2 z_2 = z_1 z_3,$$ so it looks like the twisted cubic.

When $Z_0 = 0$, $V(F_1,F_2)$ reduces to $Z_1 = 0;$ i.e. the projective line $$\{[0 : 0 : z_2 : z_3]\}.$$

So $V(F_1,F_2)$ is the union of the twisted cubic and the line $\{[0:0:z_2:z_3]\}$. This line does not connect two distinct points of the cubic but that is not actually claimed in the exercise as written in Harris' book.


Here is a partial answer to (a):

We know $$z_0z_2=z_1^2 $$ and $$z_0z_3 = z_1z_2.$$

Suppose first that $z_0$ and $z_2$ are non-zero.

Multiply $F_2$ by $z_0z_2$ to get $$ z_0z_2F_2 = z_0z_1z_2z_3 - z_2^2z_0z_2$$ which becomes $$(z_0z_3)^2 - z_2^2z_1^2 $$ or after factoring $$z_0z_2F_2 = (z_0z_3-z_1z_2)(z_0z_3 + z_1z_2) = F_1(z_0z_3 + z_1z_2) = 0$$ since $F_1=0$ in our zero locus. Thus we have $F_2 = 0$.

Thus the part of our zero locus with $z_0z_2 \neq 0$ (or equivalently $z_1\neq 0$) is contained in the twisted cubic. Thus we now examine what happens if $z_1=0$. If $z_2=0$ we are still in the cubic, so suppose $z_2\neq 0$. Then the equation $F_0=0$ gives $z_0=0$. Therefore the rest of our zero locus is contained in the line $[0:0:S:T]$ and as we can verify immediately, that line is indeed in the zero locus.


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