The Stacks project

Proof. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. Then $U$ is Jacobson (by definition) and for a morphism of schemes $V \to U$ which is locally of finite type we see that $V$ is Jacobson by the corresponding result for schemes (Morphisms, Lemma 29.16.9). Thus if $Y \to X$ is a morphism of algebraic spaces which is locally of finite type, then setting $V = U \times _ X Y$ we see that $Y$ is Jacobson by definition. $\square$

Proof. Let $U \to X$ be an étale morphism with $U$ a scheme and with $u \in U$ closed mapping to $x$, see Morphisms of Spaces, Lemma 67.25.3. Observe that $W$, $W \times _ X U$, and $U$ are Jacobson schemes by Lemma 68.23.1. Hence finite type points on these schemes are the same thing as closed points by Morphisms, Lemma 29.16.8. The inverse image $T \subset W \times _ X U$ of $u$ is a nonempty (as $x$ in the image of $W \to X$) closed subset. By Morphisms, Lemma 29.16.7 there is a closed point $t$ of $W \times _ X U$ which maps to $u$. As $W \times _ X U \to W$ is locally of finite type the image of $t$ in $W$ is closed by Morphisms, Lemma 29.16.8. $\square$

Proof. If $x \in |X|$ is closed, then we can represent $x$ by a closed immersion $\mathop{\mathrm{Spec}}(k) \to X$, see Lemma 68.14.6. Hence $x$ is certainly a finite type point.

Conversely, let $x \in |X|$ be a finite type point. We know that $x$ can be represented by a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k) \to X$ where $k$ is a field (Definition 68.6.1). On the other hand, by definition, there exists a morphism $\mathop{\mathrm{Spec}}(k') \to X$ which is locally of finite type and represents $x$ (Morphisms, Definition 29.16.3). We obtain a factorization $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k) \to X$. Let $U \to X$ be any étale morphism with $U$ affine and consider the morphisms

The quasi-compact scheme $\mathop{\mathrm{Spec}}(k) \times _ X U$ is étale over $\mathop{\mathrm{Spec}}(k)$ hence is a finite disjoint union of spectra of fields (Remark 68.4.1). Moreover, the first morphism is surjective and locally of finite type (Morphisms, Lemma 29.15.8) hence surjective on finite type points (Morphisms, Lemma 29.16.6) and the composition (which is locally of finite type) sends finite type points to closed points as $U$ is Jacobson (Morphisms, Lemma 29.16.8). Thus the image of $\mathop{\mathrm{Spec}}(k) \times _ X U \to U$ is a finite set of closed points hence closed. Since this is true for every affine $U$ and étale morphism $U \to X$, we conclude that $x \in |X|$ is closed. $\square$

Proof. Assume $X$ is Jacobson and that $T \subset |X|$ is a closed subset. By Morphisms of Spaces, Lemma 67.25.6 we see that $T \cap X_{\text{ft-pts}}$ is dense in $T$. By Lemma 68.23.3 we see that $X_{\text{ft-pts}}$ are the closed points of $|X|$. Thus $|X|$ is indeed Jacobson.

Assume $|X|$ is Jacobson. Let $f : U \to X$ be an étale morphism with $U$ an affine scheme. We have to show that $U$ is Jacobson. If $x \in |X|$ is closed, then the fibre $F = f^{-1}(\{ x\} )$ is a finite (by definition of decent) closed (by construction of the topology on $|X|$) subset of $U$. Since there are no specializations between points of $F$ (Lemma 68.12.1) we conclude that every point of $F$ is closed in $U$. If $U$ is not Jacobson, then there exists a non-closed point $u \in U$ such that $\{ u\} $ is locally closed (Topology, Lemma 5.18.3). We will show that $f(u) \in |X|$ is closed; by the above $u$ is closed in $U$ which is a contradiction and finishes the proof. To prove this we may replace $U$ by an affine open neighbourhood of $u$. Thus we may assume that $\{ u\} $ is closed in $U$. Let $R = U \times _ X U$ with projections $s, t : R \to U$. Then $s^{-1}(\{ u\} ) = \{ r_1, \ldots , r_ m\} $ is finite (by definition of decent spaces). After replacing $U$ by a smaller affine open neighbourhood of $u$ we may assume that $t(r_ j) = u$ for $j = 1, \ldots , m$. It follows that $\{ u\} $ is an $R$-invariant closed subset of $U$. Hence $\{ f(u)\} $ is a locally closed subset of $X$ as it is closed in the open $|f|(|U|)$ of $|X|$. Since $|X|$ is Jacobson we conclude that $f(u)$ is closed in $|X|$ as desired. $\square$

Proof. We may replace by any open subspace containing $x$. Thus we may assume that $X$ is quasi-compact. Then $|X|$ is a Noetherian topological space (Properties of Spaces, Lemma 66.24.2). Thus $W$ is a Noetherian topological space (Topology, Lemma 5.9.2).

Combining Lemma 68.14.1 with Properties of Spaces, Lemma 66.15.2 we see that $|X|$ is a spectral toplogical space. By Topology, Lemma 5.24.7 we see that $W \cup \{ x\} $ is a spectral topological space. Now $W$ is a quasi-compact open of $W \cup \{ x\} $ and hence $W$ is spectral by Topology, Lemma 5.23.5.

Let $E \subset W$ be an irreducible closed subset. Then if $Z \subset |X|$ is the closure of $E$ we see that $x \in Z$. There is a unique generic point $\eta \in Z$ by Proposition 68.12.4. Of course $\eta \in W$ and hence $\eta \in E$. We conclude that $E$ has a unique generic point, i.e., $W$ is sober.

Let $x' \in W$ be a point such that $\{ x'\} $ is locally closed in $W$. To finish the proof we have to show that $x'$ is a closed point of $W$. If not, then there exists a nontrivial specialization $x' \leadsto x'_1$ in $W$. Let $U$ be an affine scheme, $u \in U$ a point, and let $U \to X$ be an étale morphism mapping $u$ to $x$. By Lemma 68.12.2 we can choose specializations $u' \leadsto u'_1 \leadsto u$ mapping to $x' \leadsto x'_1 \leadsto x$. Let $\mathfrak p' \subset \mathcal{O}_{U, u}$ be the prime ideal corresponding to $u'$. The existence of the specializations implies that $\dim (\mathcal{O}_{U, u}/\mathfrak p') \geq 2$. Hence every nonempty open of $\mathop{\mathrm{Spec}}(\mathcal{O}_{U, u}/\mathfrak p')$ is infinite by Algebra, Lemma 10.61.1. By Lemma 68.12.1 we obtain a continuous map

Since the generic point of the LHS maps to $x'$ the image is contained in $\overline{\{ x'\} }$. We conclude the inverse image of $\{ x'\} $ under the displayed arrow is nonempty open hence infinite. However, the fibres of $U \to X$ are finite as $X$ is decent and we conclude that $\{ x'\} $ is infinite. This contradiction finishes the proof. $\square$