Lemma 105.5.12. If $X$ is a finite dimensional scheme, then there exists a closed (and hence finite type) point $x \in X$ such that $\dim _ x X = \dim X$.

Proof. Let $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say

105.5.12.1
$$\label{stacks-geometry-equation-maximal-chain} Z_0 \supset Z_1 \supset \ldots \supset Z_ d.$$

The subset $Z_ d$ is a minimal irreducible closed subset of $X$, and thus any point of $Z_ d$ is a generic point of $Z_ d$. Since the underlying topological space of the scheme $X$ is sober, we conclude that $Z_ d$ is a singleton, consisting of a single closed point $x \in X$. If $U$ is any neighbourhood of $x$, then the chain

$U\cap Z_0 \supset U\cap Z_1 \supset \ldots \supset U\cap Z_ d = Z_ d = \{ x\}$

is then a strictly descending chain of irreducible closed subsets of $U$, showing that $\dim U \geq d$. Thus we find that $\dim _ x X \geq d$. The other inequality being obvious, the lemma is proved. $\square$

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