# A problem in Harris: Equations for the twisted cubic

I came across this exercise in a book by J. Harris on Algebraic geometry, one of the first ones:

The twisted cubic is the image $C$ of the map $\nu: \mathbb{P}^1 \to \mathbb{P}^3$, given in affine coordinates as $$\nu(x) = (x,x^2,x^3)$$

or, in homogeneous coordinates as $$\nu(X_0,X_1) = (X^3_0, \ X^2_0X_1, \ X_0X^2_1, \ X_1^3)$$

(The part above is clear.)

Then $C$ = the intersection of the of the set of zeros of the quadrics:

$$F_0(Z) = Z_0Z_2 -Z_1^2 \\ F_1(Z) = Z_0Z_3 - Z_1Z_2 \\ F_2(Z) = Z_1Z_3 - Z_2^2$$ Let's denote each of these quadric surfaces as $Q_0, Q_1,Q_2$ respectively

(I think this is clear, in the sense that plugging in the points in the image of $\nu$ into these quadrics will yield $0$)

Harris' exercise: Show that for any $0 \leq i < j \leq 2$, the surfaces $Q_i, Q_j$ intersect in $C \cup L_{ij}$ where $L_{ij}$ is a line.

Here I lose Harris completely, I don't even know where to start! It seems intuitive that if three quadrics define $C$ then two of them will define $C$ and something more, and apparently, this "something more" should be a line.

Elsewhere he states:

An inclusion of vector spaces $W \simeq K^{k+1} \hookrightarrow V \simeq K^{n+1}$ induces a map $\mathbb{P}V \hookrightarrow \mathbb{P}W$. The image $\Lambda$ of such map is called a $\textit{linear subspace}$ of dimension $k$ in $\mathbb{P}V$. In case $k = 1$, we call $\Lambda$ a $\textit{line}$. I think I understand this, as the $\mathbb{P}W$ for $\dim W = 2$ would be the projective line (all subspaces $U \subset W,$ with $\dim \ U = 1$), but this doesn't help me answer the question.

Is it expected of me show that two of these quadrics define the image of such an map? How?

I am looking for guidance more than an explicit answer, how would a more experienced math person do this? And most importantly, if you are more experienced: could you say a few words about what happened in your mind when you saw this? What made you reach for a specific plan of attack? How did you translate the problem into a clear set of steps on what to do? Is there some "first thing" that came to mind? Why?

I want to improve, so I want the thought process, not the answer.

• Some thoughts. If you take two (there are more general versions) homogeneous polynomials of degrees $d,e$, you should expect to get something of degree $de$ (Bezout's theorem). So, $Q_0=0=Q_1$ (say) shoud give you a degree 4 curve, but it contains the cubic and so the rest should be degree 1, aka a line. By examination, you can see that it contains the line $Z_0=Z_1=0$ and so you should expect this is all. The rest is just verifying this is true. Dec 27, 2016 at 22:07
• Not following, in Harris' book, Bezouts theorem is covered in Lecture 18 (page 226), and this problem is from page 9. Plus there seems to be some ambiguity in "degree of a curve" as witnessed by this question: math.stackexchange.com/questions/58325/… Aslo, setting $Q_0 = 0 = Q_1$ doesn't produce a forth degree homogeneous polynomial, from what I can see. Could you please elaborate? Dec 27, 2016 at 22:58
• I did not say it produces a fourth degree polynomial, just a curve of degree 4. For example, if you take two different hyperplanes (degree 1), their intersection is a curve (line) of degree $1\cdot 1=1$ etc. Dec 28, 2016 at 4:52
• Could you (or someone else) provide a working definition of "degree of a curve"? Are hyperplanes and lines always of the same degree? How does that wok? Harris does not define this, so how can I answer his question? Dec 28, 2016 at 10:54
• Pick two of the $F_i$'s. What's their GCD? Dec 28, 2016 at 18:39

Here is the process for $$V(F_0)\cap V(F_1)$$. The others are similar.

Suppose $$F_0 = F_1 = 0$$. Then if $$z_0 \neq 0$$, we can write $$z_2 = \frac{z_1^2}{z_0}$$ and $$z_3 = \frac{z_1 z_2}{z_0} = \frac{z_1^3}{z_0^2}$$ by substitution in $$F_0=0, F_1=0$$ respectively. But now

$$[z_0 : z_1 : \frac{z_1^2}{z_0} : \frac{z_1^3}{z_0^2} ] = [z_0^3 : z_0^2 z_1 : z_0 z_1^2 : z_1^3] \in C.$$

Therefore, if $$p\in V(F_0) \cap V(F_1) \setminus C$$, we must have $$z_0=0$$. But $$z_0=0 \implies z_1=0$$ by $$F_0=0$$. Now, $$C$$ is exactly the vanishing set of all three of $$F_0,F_1, F_2$$, and to show that $$V(F_0,F_1,F_2)\subset C$$, we note that any point in $$\mathbb{P}^3$$ has some coordinate non-zero. But if $$z_0=0$$, then the defining equations $$F_0,F_1,F_2=0$$ imply that $$z_1=0$$ and $$z_2=0$$ so either $$z_0\neq 0$$ or $$z_3\neq 0$$. But if $$z_0\neq 0$$ we can write $$q \in V(F_0,F_1,F_2)$$ as $$q= \nu [z_0 : z_1]$$ and if $$z_3\neq 0$$ then $$q = \nu[z_2:z_3]$$. This shows that $$V(F_0,F_1,F_2)\subset C$$. Therefore, $$p\in V(F_0)\cap V(F_1) \setminus C$$ is exactly equivalent to $$F_0=F_1=0, F_2\neq 0$$. Now if $$F_2\neq 0$$, $$z_2\neq 0$$. This gives

$$p = [0 : 0 : \lambda_2 : z_3] = [0 : 0 : 1 : \frac{z_3}{\lambda_2} ]$$ for $$\lambda_2 \in k^*$$. If we write $$\frac{z_3}{\lambda_2}=w_3 \in k$$ (there are no restrictions on $$z_3$$), we have that

$$V(F_0) \cap V(F_1) \setminus C = \{ [0 : 0 : 1 : w_3 ] \text{ }|\text{ } w_3 \in k \} \simeq k$$ is a line.

• This solution seems a bit messy. You start out by considering the case when $z_0 \neq 0$, and then you consider this case again? You should have two cases. The first case is $z_0 \neq 0$ which is the easier case to handle and does not have subcases. The second case is when $z_0=0$. This will imply $z_1 =0$, but $z_2$ and $z_3$ can be anything. At this point we get two subcases: either $z_2 = 0$ or $z_2 \neq 0$. If the former holds, then $z_3 = 0$ and it the point reduces to $[0,0,0,1]$ (not sure what to conclude at this here). Feb 21 at 13:58
• If the latter holds, then the point is $[0,0,1,z_3/z_2]$ which is in the line segment. That seems a bit cleaner, although I don't know what to conclude from the first subcase; i.e., when the point reduces to $[0,0,0,1]$. Feb 21 at 13:59