Given a curve $C\subset P^n$ in projective space such that any $n+1$ points on $C$ are linearly independent. I've heard from multiple people that this implies that $C$ is a rational normal curve, some people said that one needs that $C$ is rational for that.

I would like to know what precisely is true and how I can prove it, google is not my friend here...

Thanks for your help!

  • $\begingroup$ This is certainly not true for $n=2$: any two points on an any plane curve are linearly independent. $\endgroup$ Aug 5, 2012 at 8:50
  • $\begingroup$ Sorry- I of course mean "any n+1 points" $\endgroup$
    – Miklos
    Aug 5, 2012 at 9:05
  • $\begingroup$ Hint: project from a point on the curve and proceed by induction. $\endgroup$
    – M P
    Aug 5, 2012 at 9:41

2 Answers 2


You don't need to know in advance that $C$ is rational: here is a proof that an irreducible curve $C\subset \mathbb P^n$ is normal and rational as soon as all its $(n+1)$-tuples of points are linearly independent.

Step 1: The curve $C$ has degree exactly $n$
Choose $n$ points $ p_1,...,p_n$ on your curve. They lie on a unique hyperplane and this hyperplane cannot contain $C$ because of the hypothesis, hence the hyperplane contains $n $ points of $C$ (perhaps with multiplicity $\gt 1$) and so $deg(C)\geq n$.
Conversely, a generic hyperplane will cut $C$ tranversally and contain at most $n$ points of $C$ (by linear independence again) so that $deg(C)\leq n$.
These two inequalities show that $deg(C)=n$

Step 2: The curve $C$ is normal (=non-singular)
If it had a singular point $s$, the hyperplane through $s$ and $n-1$ smooth other points would cut it in a divisor of degree at least $n+1$, which is impossible for a curve of degree $n$.

Step 3: The curve $C$ is rational
Choose $n-1$ points $ q_1,...,q_{n-1}$ on your curve and let $P\cong \mathbb P^{n-2}$ be the projective linear space they span.
Consider the pencil $\Pi(\lambda) \;(\lambda \in \mathbb P^1)$ of hyperplanes $\Pi(\lambda) \subset \mathbb P^n$ through $P$.
Since $C$ has degree $n$, each $\Pi(\lambda)$ will cut the curve $C$ in a new point point $q_n(\lambda)$ and we obtain the required rational parametrization $\mathbb P^1\xrightarrow {\cong } C:\lambda \mapsto q_n(\lambda)$


Let me elaborate on my hint, since there was some skepticism! ;)

Let $p$ be a point on $C$ and let $C' \subset \mathbb{P}^{n-1}$ denote the projection of $C$ from $p$. Think of $C'$ as a curve contained in a hyperplane in $\mathbb{P}^n$. Suppose that there are $n$ points of $C'$ not in general position; this means that there is a hyperplane of $\mathbb{P}^{n-1}$ containing them all. Then the hyperplane of $\mathbb{P}^n$ containing those $n$ points of $C'$ and the point $p$ is a hyperplane intersecting $C$ at $n+1$ points, contrary to the independence condition. Therefore, also the curve $C'$ has the independence property.

If we are very formal, then we reduce to the case $n=1$, where the assertion is a tautology. Otherwise, we stop at the case $n=2$ and the assumption of independence implies that no line in $\mathbb{P}^2$ intersects the curve in 3 points: $C$ is a conic.

Finally, to show that $C$ is a rational normal curve, you would need to specify what is the definition of rational normal curve you use: some people say it is a smooth connected curve of degree $n$ in $\mathbb{P}^n$, others say that it is the image of $\mathbb{P}^1$ under the complete linear system $\mathcal{O}(d)$, still others might even define it by the property you mention. What is your definition?

Comment. The argument given here is not especially different from the one given by Georges Elencwajg, except that he did not use the more geometric terminology: for instance, his parameterization is the result of projecting the curve away from the $n-1$ points $q_1,\ldots,q_{n-1}$.

  • $\begingroup$ When I say "rational normal curve", i mean projectively equivalent to $[X_0^n: X_0 ^{n-1} X_1^1:...:X_1^n]$. I think this is some really old use of the word "normal" that doesn't have much to do with the scheme-theoretic word normal. It's clear why it is smooth and rational, just "rational normal" is not clear to me. $\endgroup$
    – Miklos
    Aug 5, 2012 at 15:01
  • 1
    $\begingroup$ Here is then a quick, although not entirely elementary, way to show what you want. Once you know that the curve is smooth and rational, observe that the linear system used to embed it is the complete linear system determined by $\mathcal{O}(n)$: it has degree $n$ and if it were not complete, then there would be some linear dependence among some $n+1$ points on the curve. $\endgroup$
    – M P
    Aug 5, 2012 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.