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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.

user223794
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    I looked this up in Eisenbud's Geometry of Syzygies. This is Exercise 5(b) in Chapter 6. He gives a bunch of hints but it doesn't seem very easy (involving intersection theory on a blown-up surface etc) so I don't know if there's a really simple answer – Jay Aug 17 '16 at 19:48

1 Answers1

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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).

Sasha
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    Wow what a great answer! It took me a minute to realize that the general position assumption means the last two points have all nonzero coordinates. So although the birational map isn't defined at the first $d+1$ points it is defined on the last two. +1! – Jay Aug 19 '16 at 09:48
  • How did you see that this is a line? – user7090 Jul 28 '17 at 00:06
  • @Jadwiga Here is my understanding: the birational morphism corresponds to blow up and down, where any blown up divisor become hyperplane sections. Then note that the curve intersects exceptional divisor only once, hence it becomes degree $1$. –  Mar 22 '18 at 21:15
  • @Sasha Is there a reference for such a birational map? Suppse the $n+1$ points are not rational, but are $n+1$ Galois conjugate points, do we still have the transformation? In view of blowing up and down, I think the indeterminacy locus is the $n-2$ skeleton of the simplex of the $n+1$ points, then we blow down the $n-1$ faces. It seems weird that the blowing up and down are not symmetrical (blowing up more) as we do this in projective plane. –  Mar 25 '18 at 17:21