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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
$$ J = leftlbrace f(x, y, 1) : f in I rightrbrace unlhd k[x,y].$$ 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}