If $I\subset S$ is the homogeneous ideal of a finite set of points in $\mathbb P^n$, then $\dim S/I=1$.

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Let $k$ be an algebraically closed field and $X=\{p_1,\ldots,p_r\}$ be a set of $r$ distinct points in $\mathbb P^n(k)$. Write $I$ for the homogeneous ideal of $X$ in $S=k[x_0,\ldots,x_n]$. I read somewhere that $\dim S/I=1$.

Let $\newcommand{\p}{\mathfrak{p}}\p_i$ denote the prime ideal corresponding to the point $p_i$. Then $$I=\p_1\cap\cdots\cap\p_r.$$ Let $\p$ be a prime ideal containing $I$. Note that $I\subset \p$ implies $\p$ contains some $\p_i$. If we can show that the height of $\p_i$ is $n$, then $\p$ is either equal to $\p_i$ or is a maximal ideal, since the dimension of $S$ is $n+1$.

WLOG, let $p_i=[1:a_1:\cdots:a_n]$. Then $\p_i=(x_1-a_1x_0,\ldots,x_n-a_nx_0)$, so the height of $\p_i$ is at most $n$ by Krull's PIT. Clearly, we have a chain of prime ideals of length $n$ $$(0)\subset (x_1-a_1x_0)\subset\cdots\subset (x_1-a_1x_0,\ldots,x_n-a_nx_0).$$ Thus $\operatorname{height}(\p_i)=n$ and the claim follows.


Does this work? I don't have much background in algebraic geometry and I learned some of the results used above from various math.se posts. I just want to make sure that I understand these correctly.