Lemma 68.18.7. Let $X$ be an algebraic space locally of finite type over a field $k$. Let $x \in |X|$. Consider the conditions

1. $\dim _ x(|X|) = 0$,

2. $x$ is closed in $|X|$ and if $x' \leadsto x$ in $|X|$ then $x' = x$,

3. $x$ is an isolated point of $|X|$,

4. $\dim _ x(X) = 0$,

5. $X \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at $x$.

Then (2), (3), (4), and (5) are equivalent. If $X$ is decent, then (1) is equivalent to the others.

Proof. Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemmas 67.34.7 and 67.34.8.

Let $U \to X$ be an étale morphism where $U$ is an affine scheme and let $u \in U$ be a point mapping to $x$. Moreover, if $x$ is a closed point, e.g., in case (2) or (3), then we may and do assume that $u$ is a closed point. Observe that $\dim _ u(U) = \dim _ x(X)$ by definition and that this is equal to $\dim (\mathcal{O}_{U, u})$ if $u$ is a closed point, see Algebra, Lemma 10.114.6.

If $\dim _ x(X) > 0$ and $u$ is closed, by the arguments above we can choose a nontrivial specialization $u' \leadsto u$ in $U$. Then the transcendence degree of $\kappa (u')$ over $k$ exceeds the transcendence degree of $\kappa (u)$ over $k$. It follows that the images $x$ and $x'$ in $X$ are distinct, because the transcendence degree of $x/k$ and $x'/k$ are well defined, see Morphisms of Spaces, Definition 67.33.1. This applies in particular in cases (2) and (3) and we conclude that (2) and (3) imply (4).

Conversely, if $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite at $x$, then $U \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite at $u$, hence $u$ is an isolated point of $U$ (Morphisms, Lemma 29.20.6). It follows that (5) implies (2) and (3) as $|U| \to |X|$ is continuous and open.

Assume $X$ is decent and (1) holds. Then $\dim _ x(X) = \dim _ x(|X|)$ by Lemma 68.12.5 and the proof is complete. $\square$

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