Why is any finite set of points on a twisted cubic curve, in general position?

Question

Please, how to show that any finite set of points on a twisted cubic curve $ \nu ( \mathbb{P}^1 ) $ with $ \nu : \mathbb{P}^1 \to \mathbb{P}^3 $ defined by : $$ \nu ( [X_0 : X_1 ] ) = [ X_0^3 : X_0^2 X_1 : X_0 X_1^2 : X_1^3 ] $$ in general position ( i.e : any four of them span $ \mathbb{P}^3 $ ) ?

Thanks in advance for your help.

Answer 1

First, show that the twisted cubic is not planar.

The intersection of a hyperplane with a non-planar cubic in $\mathbb{P}^3$ is a variety of dimension zero. How many points can it have?

The answer follows from the general version of Bézout's theorem.

Answer 2

Hint: Let $p_1,\ldots,p_4$ be distinct points on the twisted cubic curve $\nu(\mathbb P^1)$. We need to show that the corresponding vectors of $p_i$ are linearly independent in $\mathbb K^4$. Suppose $$p_i=\nu([1:Y_i])=[1:Y_i:Y_i^2:Y_i^3].$$ Then consider the Vandermonde matrix $$V=\begin{bmatrix} 1 & Y_1 & Y_1^2 & Y_1^{3}\\ 1 & Y_2 & Y_2^2 & Y_2^{3}\\ 1 & Y_3 & Y_3^2 & Y_3^{3}\\ 1 & Y_4 & Y_4^2 & Y_4^3 \end{bmatrix}.$$ We have $\det V=\prod_{1 \le i < j \le 4} (Y_j - Y_i)\neq 0$. So the row vectors are independent.

To solve the general case, use the fact that $\nu(\mathbb P^1)=\{[1: a: a^2: a^3]\mid a\in\mathbb K\}\cup \{[0: 0: 0: 1]\}$.

Answer 3

Here's an easier way to see this when the twisted cubic is defined from $\mathbb R$ to $\mathbb R^3$.

If possible, let $P_i=(p_i,p_i^2,p_i^3)$ for $i=1,2,3,4$ be four distinct points on the plane $$Ax+By+Cz+D=0$$ Then, we have the four simultaneous equations $$Ap_i+Bp_i^2+Cp_i^3+D=0$$ for $i=1,2,3,4$.

But, this simply means that the equation $$AP+BP^2+CP^3+D=0$$ has four distinct solutions $p_i$, $i=1,2,3,4$.

This is a contradiction (since the last equation is a cubic).