Background:Let $X =\{P_1,...P_s\}$ be a set of $s$ distinct points of $\mathbb{P}^{n}= \operatorname{Proj}(K[x_0,....,x_n])$, then we know that $P_i$ are homogeneous prime ideals $\mathcal P_i$ which does not contain irrelevant maximal ideal.
Question: why $\operatorname{ht}(\mathcal P_i) = n $?
Guess: we know that $$\operatorname{ht}(\mathcal P_i) + \dim K[x_0,....,x_n]/\mathcal P_i = \dim K[x_0,....,x_n] =n+1,$$ but why $\dim K[x_0,....,x_n]/\mathcal P_i = 1$?
Any help from anyone is welcome.
Consider the affine cone $\Bbb A^{n+1}$ on $\Bbb P^n$. Here, the $\mathcal{P}_i$ define lines through the origin, which are schemes of dimension one (they're isomorphic to $\operatorname{Spec} F[x]$ for $K\subset F$ a possibly-trivial algebraic extension). So $\dim K[x_0,\cdots,x_n]/\mathcal{P_i} = \dim F[x] = 1$.
You can also show this directly: if $\mathcal{P}_i$ is the ideal associated to the point $[a_0:a_1:\cdots:a_n]$, you can pick generators for the ideal as $a_ix_j-a_jx_i$. Pick some $i$ so that $a_i$ is nonzero (there must be at least one by the definition of projective space), then enforce that relation to see that you can eliminate the $x_j$ for $j\neq i$ and end up with the ring $K[x_i]$. This requires modest modifications for a non-$K$-rational point, but the idea is the same.