The Stacks project

Proof. By Lemmas 101.7.3 and 101.39.11 properties (1) and (2) are preserved under any base change. By Lemma 101.13.5 we only have to show that $|T \times _\mathcal {Y} \mathcal{X}| \to |T|$ is closed, whenever $T$ is an affine scheme mapping into $\mathcal{Y}$. Hence it suffices to show: if $f : \mathcal{X} \to Y$ is a quasi-compact morphism from an algebraic stack to an affine scheme satisfying the existence part of the valuative criterion, then $|f|$ is closed. Let $T \subset |\mathcal{X}|$ be a closed subset. We have to show that $f(T)$ is closed to finish the proof.

Let $\mathcal{Z} \subset \mathcal{X}$ be the reduced induced algebraic stack structure on $T$ (Properties of Stacks, Definition 100.10.4). Then $i : \mathcal{Z} \to \mathcal{X}$ is a closed immersion and we have to show that the image of $|\mathcal{Z}| \to |Y|$ is closed. Since closed immersions are quasi-compact (Lemma 101.7.5) and satisfies the existence part of the valuative criterion (Lemma 101.39.14) and since compositions of quasi-compact morphisms are quasi-compact (Lemma 101.7.4) and since compositions preserve the property of satisfying the existence part of the valuative criterion (Lemma 101.39.12) we conclude that it suffices to show: if $f : \mathcal{X} \to Y$ is a quasi-compact morphism from an algebraic stack to an affine scheme satisfying the existence part of the valuative criterion, then $|f|(|\mathcal{X}|)$ is closed.

Since $\mathcal{X}$ is quasi-compact (being quasi-compact over the affine $Y$), we can choose an affine scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$ (Properties of Stacks, Lemma 100.6.2). Suppose that $y \in Y$ is in the closure of the image of $U \to Y$ (in other words, in the closure of the image of $|f|$). Then by Morphisms, Lemma 29.6.5 we can find a valuation ring $A$ with fraction field $K$ and a commutative diagram

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $y$. By assumption we get an extension $K'/K$ and a valuation ring $A' \subset K'$ dominating $A$ and the dotted arrow in the following diagram

Thus $y$ is in the image of $|f|$ and we win. $\square$

Proof. Consider a solid diagram

where $A$ is a valuation ring with field of fractions $K$ and $\gamma : y \circ j \longrightarrow f \circ x$ as in Definition 101.39.1. By Lemma 101.39.4 in order to find a dotted arrow (after possibly replacing $K$ by an extension and $A$ by a valuation ring dominating it) we may replace $\mathcal{Y}$ by $\mathop{\mathrm{Spec}}(A)$ and $\mathcal{X}$ by $\mathop{\mathrm{Spec}}(A) \times _\mathcal {Y} \mathcal{X}$. Of course we use here that being quasi-separated and universally closed are preserved under base change. Thus we reduce to the case discussed in the next paragraph.

Consider a solid diagram

where $A$ is a valuation ring with field of fractions $K$ as in Definition 101.39.1. By Lemma 101.7.7 and the fact that $f$ is quasi-separated we have that the morphism $x$ is quasi-compact. Since $f$ is universally closed, we have in particular that $|f|(\overline{\{ x\} })$ is closed in $\mathop{\mathrm{Spec}}(A)$. Since this image contains the generic point of $\mathop{\mathrm{Spec}}(A)$ there exists a point $x' \in |\mathcal{X}|$ in the closure of $x$ mapping to the closed point of $\mathop{\mathrm{Spec}}(A)$. By Lemma 101.7.9 we can find a commutative diagram

such that the closed point of $\mathop{\mathrm{Spec}}(A')$ maps to $x' \in |\mathcal{X}|$. It follows that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(A)$ maps the closed point to the closed point, i.e., $A'$ dominates $A$ and this finishes the proof. $\square$