# 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. – Mohan Dec 27 '16 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? – JuliusL33t Dec 27 '16 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. – Mohan Dec 28 '16 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? – JuliusL33t Dec 28 '16 at 10:54
• Pick two of the $F_i$'s. What's their GCD? – Tabes Bridges Dec 28 '16 at 18:39