Lemma 28.10.7. Let $x$ be a point of a locally Noetherian scheme $X$. Then $\dim _ x(X) = 0$ if and only if $x$ is an isolated point of $X$.
Proof. If $x$ is an isolated point, then $\{ x\} $ is an open subset of $X$ (Topology, Definition 5.27.2) and hence $\dim _ x(X) = \dim (\{ x\} ) = 0$ by definition. Conversely, if $\dim _ x(X) = 0$, then there exists an open neighbourhood $U \subset X$ of $x$ such that $\dim (U) = 0$ (Topology, Definition 5.10.1). By Lemma 28.10.5 we see that the topology on $U$ is discrete and hence $x$ is an isolated point. $\square$
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