If we have a field $K$, and a homogeneous polynomial $f \in R=K[x_1, \ldots, x_n]$, then the ideal generated by $f$ is a graded module over $K$, and we can calculate its Hilbert polynomial. (I am trying to calculate $H(R/I,i)$=$H(R,i)-H(I,i)$)
With $I=(f)$, we have
$H((f),i) := \dim ((f)_i)$ for large $i$.
If $f$ is degree $d$, then
$(f)_i=\{fg \vert g \in R, \deg fg = i\}=\{fg \vert g \in R, \deg g = i-d\} \simeq R_{i-d}$
So we get that
$\dim ((f)_i)=\dim (R_{i-d})={i-d+n-1 \choose n-1}$
However, if everything reasoned above is correct, then it doesn't seem to matter what the actual polynomial $f$ is, and all homogeneous polynomials of the same degree have the same Hilbert polynomial, which I'm pretty sure is wrong.
What would be the correct way to calculate the Hilbert polynomial for $(f)=(xy+z^2)$ with $R=\mathbb{R}[x,y,z]$, for example? Where does my reasoning above fail?