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.
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.