Lemma 101.40.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\Delta _ f$ is quasi-separated and if for every diagram (101.39.1.1) and choice of $\gamma $ as in Definition 101.39.1 the category of dotted arrows is a setoid, then $\Delta _ f$ is separated.

## 101.40 Valuative criterion for second diagonal

The converse statement has already been proved in Lemma 101.39.2. The criterion itself is the following.

**Proof.**
We are going to write out a detailed proof, but we strongly urge the reader to find their own proof, inspired by reading the argument given in the proof of Lemma 101.39.2.

Assume $\Delta _ f$ is quasi-separated and for every diagram (101.39.1.1) and choice of $\gamma $ as in Definition 101.39.1 the category of dotted arrows is a setoid. By Lemma 101.6.1 it suffices to show that $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion. By Lemma 101.6.4 it in fact suffices to show that $e = \Delta _{f, 2}$ is universally closed. Either of these lemmas tells us that $e = \Delta _{f, 2}$ is quasi-compact by our assumption that $\Delta _ f$ is quasi-separated.

In this paragraph we will show that $e$ satisfies the existence part of the valuative criterion. Consider a $2$-commutative solid diagram

and let $\alpha : (a, \theta ) \circ j \to e \circ x$ be any $2$-morphism witnessing the $2$-commutativity of the diagram (we use $\alpha $ instead of the letter $\gamma $ used in Definition 101.39.1). Note that $f \circ \theta = \text{id}$; we will use this below. Observe that $e \circ x = (x, \text{id}_ x)$ and $(a, \theta ) \circ j = (a \circ j, \theta \star \text{id}_ j)$. Thus we see that $\alpha $ is a $2$-arrow $\alpha : a \circ j \to x$ compatible with $\theta \star \text{id}_ j$ and $\text{id}_ x$. Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$. Reading the arguments given in the proof of Lemma 101.39.2 backwards, we see that $\theta $ is an automorphism of the dotted arrow $(a, \alpha , \beta )$ with

On the other hand, $\text{id}_ a$ is an automorphism too, hence we conclude $\theta = \text{id}_ a$ from the assumption on $f$. Then we can take as dotted arrow for the displayed diagram above the morphism $a : \mathop{\mathrm{Spec}}(A) \to \mathcal{X}$ with $2$-morphisms $(a, \text{id}_ a) \circ j \to (x, \text{id}_ x)$ given by $\alpha $ and $(a, \theta ) \to e \circ a$ given by $\text{id}_ a$.

By Lemma 101.39.11 any base change of $e$ satisfies the existence part of the valuative criterion. Since $e$ is representable by algebraic spaces, it suffices to show that $e$ is universally closed after a base change by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ which is surjective and smooth and with $I$ an algebraic space (see Properties of Stacks, Section 100.3). This base change $e' : X' \to I'$ is a quasi-compact morphism of algebraic spaces which satisfies the existence part of the valuative criterion and hence is universally closed by Morphisms of Spaces, Lemma 67.42.1. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2100 by Matthew Emerton on

Comment #2144 by Johan on