Hilbert polynomial of twisted cubic 'by hand'? I am asked to calculate the Hilbert polynomial of the twisted cubic curve 
\begin{equation*}
C = \{(s^3 : s^2t : st^2 : t^3); (s:t) \in \mathbb{P}^1 \} \subset \mathbb{P}^3
\end{equation*}
and I know it should be $2d-1$. In the examples in my book there are some simple examples where the Hilbert polynomial is calculated by calculating the basis of $S(X)^{(d)}$ , so I also tried to do this, by using that $C$ is the zero-set of 
\begin{equation*}
\{x_0x_2 - {x_1}^2, x_1x_3 - {x_2}^2, x_0x_3 - x_1x_2 \}. 
\end{equation*}
I am trying to calculate a basis for $S(X)^{(d)}$ with $2d-1$ elements by scoring out the elements of the zero-set, but the calculations get quit complicated and I am not sure if this is the right approach.  
 A: First, the coordinate ring of the twisted cubic is $k[s^3,s^2t,st^2,t^3]$, since $[s:t] \mapsto [s^3:s^2t:st^2:t^3]$ is a parametrization of the twisted cubic. Note that in particular, requiring that the ring homomorphism $k[w,x,y,z] \to k[s^3,s^2t,st^2,t^3]$ inducing the isomorphism is in fact graded means that the generators $s^3,s^2t,st^2,t^3$ each live in the grade one piece of the algebra. See Georges Elencwajg's answer to another question for a proof.
It therefore suffices to find the Hilbert polynomial for $k[s^3,s^2t,st^2,t^3]$. But the monomials in this algebra are in one-to-one correspondence with the monomials in $k[s,t]$ living in each graded piece that is a multiple of three. Since the Hilbert polynomial for $k[s,t]$ is $d+1$, the Hilbert polynomial for $k[s^3,s^2t,st^2,t^3]$ is therefore $3d+1$.
I want to point out that the above argument is a nice way to compute other Hilbert polynomials: for example, if $X$ is the image of the Segre embedding $\mathbf{P}^1 \times \mathbf{P}^2 \to \mathbf{P}^5$, the Hilbert polynomial of $X$ is $h_X(d) = (d+1)(d/2 + 2)$.
A: Your approach is OK till this point, find a Grobner basis then use that to find a standard monomial basis. 
