I'm beginning to learn about Hilbert polynomials and I'm trying to find it for the variety $X=Z(x^2+y^2+z^2+w^2)\subset \mathbb{P}^3$.
I know that the leader term must be of the form $\frac{2}{2!}t^2$, since $\dim X=2$ and the degree of $x^2+y^2+z^2+w^2$ is $2$. I tried to look at the vector spaces $S(X)_m$ explicitly, for example, for $m=2$:
$$S(X)_2=(x^2,y^2,z^2,w^2, xy,xz,xw,yz,yw,zw)/(x^2+y^2+z^2+w^2)$$
Since $\overline{w}^2=-\overline{x}^2-\overline{y}^2-\overline{z}^2$, then $\{\overline{x}^2, \overline{y}^2,\overline{z}^2, \overline{x}\overline{y},\overline{x}\overline{z},\overline{x}\overline{w}, \overline{y}\overline{z}, \overline{y}\overline{w}, \overline{z}\overline{w}\}$ is a $k$-basis for $S(X)_2$ therefore $\dim_k S(X)_2=9$.
But I don't know where to go from here. What is the idea?