Hilbert polynomial Twisted cubic

Solution 1:

As I explained it here one way is to first find the Hilbert function $h_C(d)$ and see what polynomial it is when $d \gg 0.$

Let $H=\{w=0\} \subset \mathbb{P}^3$ and consider $X=C \cap H=V(w,xz-y^2,z^2,yz).$ Then the homogeneous coordinate ring of $X$ is (isomorphic to) $S(X)=k[x,y,z]/(xz-y^2,z^2,yz).$

Exercise: Show that its degree $d$ part, $S(X)^{(d)},$ is generated by $x^d,x^{d-1}y, x^{d-1}z$ as a vector space over $k,$ and conclude that $h_X(d)=\dim_k S(X)^{(d)}=3,$ so is $p_X(d)=3.$

Now the question is, how to relate $C$ and $X?$

The answer is the following SES

$$0\longrightarrow k[x,y,z,w]/I \stackrel{.w}{\longrightarrow} k[x,y,z,w]/I \longrightarrow k[x,y,z,w]/(I+(w)) \longrightarrow 0$$

where $I=(y^2-xz,z^2-yw,xw-yz).$ (It would be easy to run the cohomological approach if you like, using this SES.)

Take the $d$ graded pieces of the SES to get $h_C(d)=h_C(d-1)+h_X(d)$ and note that $h_C(0)=1,$ since $S(C)^{(0)}=k.$ Finally, prove that $h_C(d)=3d+1$ which is a polynomial, thus $p_C(d)=3d+1.$

In general, the degree of the Hilbert polynomial of a projective subscheme $X$ of $\mathbb{P}^n$ coincides with the dimension of $X$, so there should be no surprise that why in this case is a degree one polynomial.

Moreover, the degree of $X$ is defined to be $(\dim X)! \times \text{leading coefficient of} \; p_X(d),$ which in our case is $3.$

Solution 2:

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 $d n + 1 - g$.