# Set of zeroes of a radical ideal?

Suppose $R=F[X_0,\dots,X_r]$, where $F$ is an algebraically closed field. Now $R$ is graded, with the homogeneous polynomials of degree $n$ being the elements of degree $n$. Now suppose $I$ is a homogeneous radical ideal. As usual, we have the functions $\phi(n)=\dim_F R_n$, $\phi(n,I)=\dim_F I_n$, and $\chi(n,I)=\dim_F R_n/I_n=\dim_F R_n-\dim_F I_n=\phi(n)-\phi(n,I)$.

I've been wondering about the following. Suppose $\chi(n,I)=m$ for all large enough $n$. From this, why is it that the zeroes of $I$ in $\mathbb{P}^r$ consist of $m$ distinct points?

Now $\phi(n)=\binom{r+n}{r}$, so as $n$ increases, $\phi(n)$ increases, so necessarily $\phi(n,I)$ must get larger to balance out $\chi(n,I)$. However, I don't see how this relates to the zeroes of $I$ in $\mathbb{P}^r$.

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If $\dim R_n/I_n=m$ for $n\gg0$, then the Hilbert polynomial (see Hartshorne, section I.7) of $R/I$ is $P(t)=m$. Since this is a constant polynomial, the dimension of the variety $V(I)$ is zero, the degree of the polynomial $P$. Since $I$ is radical, $V(I)$ is therefore a finite set of points; let us say there are $k$ of them. Using Proposition I.7.6.b in Hartshorne, you can see that the degree of $V(I)$, which by definition is the leading coefficient of $P$ times $0!$, that is $m$, is equal to the sum of the degrees of each of its points. A simple computation shows that the degree of a point is $1$, so we see that $k=m$.