Lemma 105.13.10. Let $h : \mathcal{X}' \to \mathcal{X}$ be a morphism of algebraic stacks. Assume

1. $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is finite,

2. $h$ is étale, separated, and induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 100.19.5).

Then there exists a cartesian diagram

$\xymatrix{ \mathcal{X}' \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ M' \ar[r] & M }$

where $M' \to M$ is a separated étale morphism of algebraic spaces and the vertical arrows are the moduli spaces constructed in Theorem 105.13.9.

Proof. By Morphisms of Stacks, Lemma 100.45.3 we see that $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}' \times _\mathcal {X} \mathcal{I}_\mathcal {X}$ is an isomorphism. Hence $\mathcal{I}_{\mathcal{X}'} \to \mathcal{X}'$ is finite as a base change of $\mathcal{I}_\mathcal {X} \to \mathcal{X}$. Let $f' : \mathcal{X}' \to M'$ and $f : \mathcal{X} \to M$ be as in Theorem 105.13.9. We obtain a commutative diagram as in the lemma because $f'$ is categorical moduli space. Choose $I$ and $g'_ i : \mathcal{X}'_ i \to \mathcal{X}'$ as in Lemma 105.13.8. Observe that $g_ i = h \circ g'_ i$ is étale, separated, and induces isomorphisms on automorphism groups (Morphisms of Stacks, Remark 100.19.5). Let $f'_ i : \mathcal{X}'_ i \to M'_ i$ be as in Lemma 105.13.4. In the proof of Theorem 105.13.9 we have seen that the diagrams

$\xymatrix{ \mathcal{X}'_ i \ar[d]_{f'_ i} \ar[r]_{g'_ i} & \mathcal{X}' \ar[d]^{f'} \\ M'_ i \ar[r] & M' } \quad \text{and}\quad \xymatrix{ \mathcal{X}'_ i \ar[d]_{f'_ i} \ar[r]_{g_ i} & \mathcal{X} \ar[d]^ f \\ M'_ i \ar[r] & M }$

are cartesian and that $M'_ i \to M'$ and $M'_ i \to M$ are étale (this also follows directly from the properties of the morphisms $g'_ i, g_ i, f', f'_ i, f$ listed sofar by arguing in exactly the same way). This first implies that $M' \to M$ is étale and second that the diagram in the lemma is cartesian. We still need to show that $M' \to M$ is separated. To do this we contemplate the diagram

$\xymatrix{ \mathcal{X}' \ar[r] \ar[d] & \mathcal{X}' \times _\mathcal {X} \mathcal{X}' \ar[d] \\ M' \ar[r] & M' \times _ M M' }$

The top horizontal arrow is universally closed as $\mathcal{X}' \to \mathcal{X}$ is separated. The vertical arrows are as in Theorem 105.13.9 (as flat base changes of $\mathcal{X} \to M$) hence universal homeomorphisms. Thus the lower horizontal arrow is universally closed. This (combined with it being an étale monomorphism of algebraic spaces) proves it is a closed immersion as desired. $\square$

Comment #2549 by Matthew Emerton on

The statement of the lemma doesn't say that $\mathcal X' \rightarrow M'$ is as in the Keel--Mori theorem, although the proof starts with that set-up. Maybe this could be put into the lemma statement too? (I guess it's automatic, given the rest of the lemma, b/c the base-change of a uniform categorical moduli space is again such a thing, but might be worth making explicit?)

Comment #2550 by on

Exactly what you say happens a few times in this section: given a cartesian square with a uniform categorical moduli space for the downward right vertical arrow and the lower rightwards horizontal arrow is flat and representable by algebraic spaces, then the left downward vertical is a uniform categorical moduli space too. Although I think we need to leave this kind of stuff to the reader in general, I've made the change you requested in this case. See this commit.

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