TITLE: Unique rational normal curve through d+3 points
QUESTION [7 upvotes]: We define a rational normal curve to be the image of a map 
$$\mathbb P^1\rightarrow \mathbb P^d, [x:y]\mapsto [P_0(x,y):P_1(x,y): \ldots :P_d(x,y)]$$
where $P_0(x,y),P_1(x,y), \ldots P_d(x,y)$ are linearly independent homogeneous degree $d$ polynomials.
Prove that through any $d+3$ points in $\mathbb P^d$ in general position (i.e. any $d+1$ of them span $\mathbb P^d$) there exists a unique rational curve passing through them.
While I am able to prove the existence, I don't know how to prove the uniqueness part.

REPLY [9 votes]: Without a loss of generality one can assume that the first $d+1$ points are $(1,0,\dots,0)$, $(0,1,\dots,0)$, \dots, $(0,0,\dots,1)$. Consider the birational isomorphism
$$
(x_0,x_1,\dots,x_n) \mapsto  (x_0^{-1},x_1^{-1},\dots,x_n^{-1}).
$$
It is easy to see that it takes a rational normal curve through these $(d+1)$ points to a line, and its inverse (which is given by the same formula) takes a line to a rational normal cuve through these points. So, the result follows from the fact that there is a unique line through two given points (the images of the last two of the $(d+3)$ points).