# 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.

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}