4
$\begingroup$

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 $F_{\mu}=F_1$.

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Commented May 23, 2016 at 1:53
  • $\begingroup$ I forgot to add $\lambda \neq \mu$. Thanks. $\endgroup$ Commented May 23, 2016 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. $\endgroup$ Commented May 23, 2016 at 2:01
  • $\begingroup$ The book says $\lambda\neq \mu$, but maybe you found an error... $\endgroup$ Commented May 23, 2016 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. $\endgroup$ Commented May 23, 2016 at 2:08

2 Answers 2

Reset to default
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.