Hilbert polynomial for a dimension zero projective variety by taking an affine chart I am looking at exercise 12.21 from Gathmann's notes on algebraic geometry. I am given a homogeneous ideal
$$I \unlhd k[x, y, z]  $$ with a dimension $0$ projective locus. WLOG, we assume that this has non-vanishing $z$ coordinate, and hence we can define the ideal
$$ J = \left\lbrace f(x, y, 1) : f \in I \right\rbrace  \unlhd k[x,y].$$ The task is to show that $$ \deg I = \chi_{I} = \dim_{k}k[x, y]/J $$ where $\chi_{I}$ is the Hilbert polynomial of $I$. In other words, I need to show that I can determine the Hilbert polynomial of a dimension $0$ projective set by taking an affine chart. I feel like this should not be a difficult question. Any help would be appreciated.
Edit. $\deg I$ in this case is the degree $0$ Hilbert polynomial. In other words, it is the unique natural number $n$ such that $n=\dim_kk[x,y,z]_d/I_d$ for "almost all" values of $n$.
Thanks.
 A: Let $S=k[x,y,z]$ and let $I$ be a homogeneous ideal such that the projective variety defined by $I$ has dimension $0$. Then $S/I$ has Krull dimension $1$ and so its Hilbert polynomial is a constant, i.e., the Hilbert function is eventually a constant. Suppose that $H_{S/I}(n) = H_{S/I}(n_0), \, \forall n \ge n_0$. Now let $R = k[x,y]$ and let $J$ be the dehomogenization of $I$ with respect to $z$, i.e., take all polynomials in $I$ and set $z=1$. Let $\mathcal{B}$ be a $k$-basis for the vector space $J_{\le n_0}$, i.e., the vector space of all polynomials in $J$ of degree $\le n_0$. Then $\mathcal{B}^h = \left\{z^{n_0} p(x/z,y/z): \, p \in \mathcal{B} \right\}$ is a $k$-basis for the vector space $I_{n_0}$, i.e., $\dim_k J_{\le n_0} = \dim_k I_{n_0}$. Since $\dim_k S_{n_0} = \dim_k R_{\le n_0}$, we have $\dim_k (S/I)_{n_0} = \dim_k R_{\le n_0} / J_{\le n_0}$. In fact, we have that $\dim_k R_{\le n_0} / J_{\le n_0} = \dim_k R_{\le n} / J_{\le n}, \, \forall n \ge n_0$.
We next show that $\dim_k R/J =  \dim_k R_{\le n} / J_{\le n}$ for all sufficiently large $n$. 
First, notice that for any $n$ we have 
a morphism of $k$-vector spaces 
\begin{align}
R_{\le n} \rightarrow \frac{R}{J}, \, \, \, (\dagger)
\end{align} which takes an element $p \in R_{\le n}$ to its class in $R/J$. The kernel of this morphism is clearly $J_{\le n}$. Consequently, we have a monomorphism 
\begin{align}
\frac{R_{\le n}}{J_{\le n}} \hookrightarrow \frac{R}{J}, (\ddagger)
\end{align} and so $\dim_k R_{\le n} / J_{\le n} \le \dim_k R/J$. On the other hand, recall
that the affine variety defined by $J$ has the same dimension as the projective variety defined by $I$ (the latter is the projective closure of the former), the ring $R/J$ must have Krull dimension zero (why?), and so it must be a finite dimensional $k$-vector space. Let $p_1,\dots,p_s$ be elements of $R$ such that their classes in $R/J$ form a $k$-basis for $R/J$. Let $d$ be the maximal degree among $\deg(p_1),\dots,\deg(p_s)$. Then, for $n \ge d$ the morphism $(\dagger)$ becomes surjective and so the embedding $(\ddagger)$ actually becomes an isomoprhism. Thus
\begin{align}
H_{S/I}(n_0) =  \dim_k R_{\le \max(n_0,d)} / J_{\le \max(n_0,d)}=\dim_k R/J.
\end{align}
