Lemma 66.18.8. Let $X$ be an algebraic space locally of finite type over a field $k$. Consider the conditions

1. $|X|$ is a finite set,

2. $|X|$ is a discrete space,

3. $\dim (|X|) = 0$,

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

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

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

Proof. Parts (4) and (5) are equivalent for example by Morphisms of Spaces, Lemma 65.34.7.

Let $U \to X$ be a surjective étale morphism where $U$ is a scheme.

If $\dim (U) > 0$, then choose a nontrivial specialization $u \leadsto u'$ in $U$ and 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 $x'/k$ is well defined, see Morphisms of Spaces, Definition 65.33.1. We conclude that (2) and (3) imply (4).

Conversely, if $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite, then $U$ is locally Noetherian (Morphisms, Lemma 29.15.6) of dimension $0$ (Morphisms, Lemma 29.29.5) and hence is a disjoint union of spectra of Artinian local rings (Properties, Lemma 28.10.5). Hence $U$ is a discrete topological space, and since $|U| \to |X|$ is continuous and open, the same is true for $|X|$. In other words, (4) implies (2) and (3).

Assume $X$ is decent and (1) holds. Then we may choose $U$ above to be affine. The fibres of $|U| \to |X|$ are finite (this is a part of the defining property of decent spaces). Hence $U$ is a finite type scheme over $k$ with finitely many points. Hence $U$ is quasi-finite over $k$ (Morphisms, Lemma 29.20.7) which by definition means that $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. $\square$

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