TITLE: Hilbert polynomial Twisted cubic
QUESTION [12 upvotes]: Let $C$ the twisted cubic in $\mathbb{P}^3$ defined as $V(XZ-Y^2, YW-Z^2,XW-YZ)$. I have to calculate Hilbert polynomial of $C$, that I denote $P_C(n)$. In order to calculate Hilber polynomials in general, I consider the exact twisted sequence
$$ 0 \to \mathcal{I}_C(n) \to \mathcal{O}_{\mathbb{P}^3}(n) \to \mathcal{O}_C(n) \to 0 .$$
If $C=V(F)$, where $deg(F)=m$ I use the isomophism $\mathcal{I}_C \simeq \mathcal{O}_{\mathbb{P}^3}(n-m)$ given by the multiplication for $C$. In this situation I don't know how to use this way in order to calculate Hilbert polynomial of the twisted cubic.

REPLY [8 votes]: When $n$ is large, the value of the Hilbert polynomial is just the dimension of $H^0(C,\mathcal O_C(n) ).$  
By definition $C$ is a copy of $\mathbb P^1$, embedded with degree $3$, and so $\mathcal O_C(1)$ is a degree $3$ line bundle on $C$.  Thus $\mathcal O_C(n)$ has degree $3n$, and the dimension of global sections of a degree $3n$ line bundle on $\mathbb P^1$ is $3n + 1$.  (Just the number of polys. in one variable of degree at most $3n$.)
So the Hilbert poly. is equal to $3n + 1$.

In general, a degree $d$ curve of genus $g$ (in any projective space) will have
Hilbert poly. equal to $d n + 1 - g$.