Lemma 65.15.1. Let $S$ be a scheme. Let $X$ be a Zariski locally quasi-separated algebraic space over $S$. Then the topological space $|X|$ is sober (see Topology, Definition 5.8.6).
65.15 Points on quasi-separated spaces
Points can behave very badly on algebraic spaces in the generality introduced in the Stacks project. However, for quasi-separated spaces their behaviour is mostly like the behaviour of points on schemes. We prove a few results on this in this section; the chapter on decent spaces contains many more results on this, see for example Decent Spaces, Section 67.12.
Proof. Combining Topology, Lemma 5.8.8 and Lemma 65.6.6 we see that we may assume that there exists an affine scheme $U$ and a surjective, quasi-compact, étale morphism $U \to X$. Set $R = U \times _ X U$ with projection maps $s, t : R \to U$. Applying Lemma 65.6.7 we see that the fibres of $s, t$ are finite. It follows all the assumptions of Topology, Lemma 5.19.8 are met, and we conclude that $|X|$ is Kolmogorov1.
It remains to show that every irreducible closed subset $T \subset |X|$ has a generic point. By Lemma 65.12.3 there exists a closed subspace $Z \subset X$ with $|Z| = |T|$. Note that $U \times _ X Z \to Z$ is a quasi-compact, surjective, étale morphism from an affine scheme to $Z$, hence $Z$ is Zariski locally quasi-separated by Lemma 65.6.6. By Proposition 65.13.3 we see that there exists an open dense subspace $Z' \subset Z$ which is a scheme. This means that $|Z'| \subset T$ is open dense. Hence the topological space $|Z'|$ is irreducible, which means that $Z'$ is an irreducible scheme. By Schemes, Lemma 26.11.1 we conclude that $|Z'|$ is the closure of a single point $\eta \in |Z'| \subset T$ and hence also $T = \overline{\{ \eta \} }$, and we win. $\square$
Lemma 65.15.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. The topological space $|X|$ is a spectral space.
Proof. By Topology, Definition 5.23.1 we have to check that $|X|$ is sober, quasi-compact, has a basis of quasi-compact opens, and the intersection of any two quasi-compact opens is quasi-compact. By Lemma 65.15.1 we see that $|X|$ is sober. By Lemma 65.5.2 we see that $|X|$ is quasi-compact. By Lemma 65.6.3 there exists an affine scheme $U$ and a surjective étale morphism $f : U \to X$. Since $|f| : |U| \to |X|$ is open and continuous and since $|U|$ has a basis of quasi-compact opens, we conclude that $|X|$ has a basis of quasi-compact opens. Finally, suppose that $A, B \subset |X|$ are quasi-compact open. Then $A = |X'|$ and $B = |X''|$ for some open subspaces $X', X'' \subset X$ (Lemma 65.4.8) and we can choose affine schemes $V$ and $W$ and surjective étale morphisms $V \to X'$ and $W \to X''$ (Lemma 65.6.3). Then $A \cap B$ is the image of $|V \times _ X W| \to |X|$ (Lemma 65.4.3). Since $V \times _ X W$ is quasi-compact as $X$ is quasi-separated (Lemma 65.3.3) we conclude that $A \cap B$ is quasi-compact and the proof is finished. $\square$
The following lemma can be used to prove that an algebraic space is isomorphic to the spectrum of a field.
Lemma 65.15.3. Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space over $S$ and assume that there exists a surjective étale morphism $\mathop{\mathrm{Spec}}(k) \to X$. If $X$ is quasi-separated, then $X \cong \mathop{\mathrm{Spec}}(k')$ where $k/k'$ is a finite separable extension.
Proof. Set $R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k)$, so that we have a fibre product diagram
\[ \xymatrix{ R \ar[r]_-s \ar[d]_-t & \mathop{\mathrm{Spec}}(k) \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & X } \]
By Spaces, Lemma 64.9.1 we know $X = \mathop{\mathrm{Spec}}(k)/R$ is the quotient sheaf. Because $\mathop{\mathrm{Spec}}(k) \to X$ is étale, the morphisms $s$ and $t$ are étale. Hence $R = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ is a disjoint union of spectra of fields, and both $s$ and $t$ induce finite separable field extensions $s, t : k \subset k_ i$, see Morphisms, Lemma 29.36.7. Because
\[ R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k) = (\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)) \times _{X \times _ S X, \Delta } X \]
and since $\Delta $ is quasi-compact by assumption we conclude that $R \to \mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)$ is quasi-compact. Hence $R$ is quasi-compact as $\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)$ is affine. We conclude that $I$ is finite. This implies that $s$ and $t$ are finite locally free morphisms. Hence by Groupoids, Proposition 39.23.9 we conclude that $\mathop{\mathrm{Spec}}(k)/R$ is represented by $\mathop{\mathrm{Spec}}(k')$, with $k' \subset k$ finite locally free where
\[ k' = \{ x \in k \mid s_ i(x) = t_ i(x)\text{ for all }i \in I\} \]
It is easy to see that $k'$ is a field. $\square$
Remark 65.15.4. Lemma 65.15.3 holds for decent algebraic spaces, see Decent Spaces, Lemma 67.12.8. In fact a decent algebraic space with one point is a scheme, see Decent Spaces, Lemma 67.14.2. This also holds when $X$ is locally separated, because a locally separated algebraic space is decent, see Decent Spaces, Lemma 67.15.2.
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