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 Kolmogorov^{1}.

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

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

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

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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