Lemma 101.42.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume

$f$ is quasi-compact, and

$f$ satisfies the existence part of the valuative criterion.

Then $f$ is universally closed.

Lemma 101.42.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Assume

$f$ is quasi-compact, and

$f$ satisfies the existence part of the valuative criterion.

Then $f$ is universally closed.

**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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Y } \]

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

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K') \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & U \ar[d] \ar[r] & \mathcal{X} \ar[d]^ f \\ \mathop{\mathrm{Spec}}(A') \ar[r] \ar@{..>}[rrru] & \mathop{\mathrm{Spec}}(A) \ar[r] & Y \ar@{=}[r] & Y } \]

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

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