Lemma 106.10.1. Let Y be a quasi-compact and quasi-separated algebraic space. Let V \subset Y be a quasi-compact open. Let f : \mathcal{X} \to V be surjective, flat, and locally of finite presentation. Then there exists a finite surjective morphism g : Y' \to Y such that V' = g^{-1}(V) \to Y factors Zariski locally through f.
106.10 Chow's lemma for algebraic stacks
In this section we discuss Chow's lemma for algebraic stacks.
Proof. We first prove this when Y is a scheme. We may choose a scheme U and a surjective smooth morphism U \to \mathcal{X}. Then \{ U \to V\} is an fppf covering of schemes. By More on Morphisms, Lemma 37.48.6 there exists a finite surjective morphism V' \to V such that V' \to V factors Zariski locally through U. By More on Morphisms, Lemma 37.48.4 we can find a finite surjective morphism Y' \to Y whose restriction to V is V' \to V as desired.
If Y is an algebraic space, then we see the lemma is true by first doing a finite base change by a finite surjective morphism Y' \to Y where Y' is a scheme. See Limits of Spaces, Proposition 70.16.1. \square
Lemma 106.10.2. Let f : \mathcal{X} \to Y be a morphism from an algebraic stack to an algebraic space. Let V \subset Y be an open subspace. Assume
f is separated and of finite type,
Y is quasi-compact and quasi-separated,
V is quasi-compact, and
\mathcal{X}_ V is a gerbe over V.
Then there exists a commutative diagram
with j an open immersion, \overline{g} and h proper, and such that |V| is contained in the image of |g|.
Proof. Suppose we have a commutative diagram
and a quasi-compact open V' \subset Y', such that Y' \to Y is a proper morphism of algebraic spaces, \mathcal{X}' \to \mathcal{X} is a proper morphism of algebraic stacks, V' \subset Y' maps surjectively onto V, and \mathcal{X}'_{V'} is a gerbe over V'. Then it suffices to prove the lemma for the pair (f' : \mathcal{X}' \to Y', V'). Some details omitted.
Overall strategy of the proof. We will reduce to the case where the image of f is open and f has a section over this open by repeatedly applying the above remark. Each step is straightforward, but there are quite a few of them which makes the proof a bit involved.
Using Limits of Spaces, Proposition 70.16.1 we reduce to the case where Y is a scheme. (Let Y' \to Y be a finite surjective morphism where Y' is a scheme. Set \mathcal{X}' = \mathcal{X}_{Y'} and apply the initial remark of the proof.)
Using Lemma 106.9.1 (and Morphisms of Stacks, Lemma 101.28.8 to see that a gerbe is flat and locally of finite presentation) we reduce to the case where f is flat and of finite presentation.
Since f is flat and locally of finite presentation, we see that the image of |f| is an open W \subset Y. Since \mathcal{X} is quasi-compact (as f is of finite type and Y is quasi-compact) we see that W is quasi-compact. By Lemma 106.10.1 we can find a finite surjective morphism g : Y' \to Y such that g^{-1}(W) \to Y factors Zariski locally through \mathcal{X} \to Y. After replacing Y by Y' and \mathcal{X} by \mathcal{X} \times _ Y Y' we reduce to the situation described in the next paragraph.
Assume there exists n \geq 0, quasi-compact opens W_ i \subset Y, i = 1, \ldots , n, and morphisms x_ i : W_ i \to \mathcal{X} such that (a) f \circ x_ i = \text{id}_{W_ i}, (b) W = \bigcup _{i = 1, \ldots , n} W_ i contains V, and (c) W is the image of |f|. We will use induction on n. The base case is n = 0: this implies V = \emptyset and in this case we can take \overline{Z} = \emptyset . If n > 0, then for i = 1, \ldots , n consider the reduced closed subschemes Y_ i with underlying topological space Y \setminus W_ i. Consider the finite morphism
and the quasi-compact open
By the initial remark of the proof, if we can prove the lemma for the pairs
then the result is true. Here we use the settheoretic equality V = (W_1 \cap \ldots \cap W_ n \cap V) \cup \bigcup \nolimits _{i = 1, \ldots n} (V \cap Y_ i). The induction hypothesis applies to the second type of pairs above. Hence we reduce to the situation described in the next paragraph.
Assume there exists n \geq 0, quasi-compact opens W_ i \subset Y, i = 1, \ldots , n, and morphisms x_ i : W_ i \to \mathcal{X} such that (a) f \circ x_ i = \text{id}_{W_ i}, (b) W = \bigcup _{i = 1, \ldots , n} W_ i contains V, (c) W is the image of |f|, and (d) V \subset W_1 \cap \ldots \cap W_ n. The morphisms
are surjective, flat, and locally of finite presentation (Morphisms of Stacks, Lemma 101.28.10). We apply Lemma 106.10.1 to each quasi-compact open W_ i \cap W_ j \cap V and the morphisms T_{ij} \to W_ i \cap W_ j \cap V to get finite surjective morphisms Y'_{ij} \to Y. After replacing Y by the fibre product of all of the Y'_{ij} over Y we reduce to the situation described in the next paragraph.
Assume there exists n \geq 0, quasi-compact opens W_ i \subset Y, i = 1, \ldots , n, and morphisms x_ i : W_ i \to \mathcal{X} such that (a) f \circ x_ i = \text{id}_{W_ i}, (b) W = \bigcup _{i = 1, \ldots , n} W_ i contains V, (c) W is the image of |f|, (d) V \subset W_1 \cap \ldots \cap W_ n, and (e) x_ i and x_ j are Zariski locally isomorphic over W_ i \cap W_ j \cap V. Let y \in V be arbitrary. Suppose that we can find a quasi-compact open neighbourhood y \in V_ y \subset V such that the lemma is true for the pair (\mathcal{X} \to Y, V_ y), say with solution \overline{Z}_ y, Z_ y, \overline{g}_ y, g_ y, h_ y. Since V is quasi-compact, we can find a finite number y_1, \ldots , y_ m such that V = V_{y_1} \cup \ldots \cup V_{y_ m}. Then it follows that setting
is a solution for the lemma. Given y by condition (e) we can choose a quasi-compact open neighbourhood y \in V_ y \subset V and isomorphisms \varphi _ i : x_1|_{V_ y} \to x_ i|_{V_ y} for i = 2, \ldots , n. Set \varphi _{ij} = \varphi _ j \circ \varphi _ i^{-1}. This leads us to the situation described in the next paragraph.
Assume there exists n \geq 0, quasi-compact opens W_ i \subset Y, i = 1, \ldots , n, and morphisms x_ i : W_ i \to \mathcal{X} such that (a) f \circ x_ i = \text{id}_{W_ i}, (b) W = \bigcup _{i = 1, \ldots , n} W_ i contains V, (c) W is the image of |f|, (d) V \subset W_1 \cap \ldots \cap W_ n, and (f) there are isomorphisms \varphi _{ij} : x_ i|_ V \to x_ j|_ V satisfying \varphi _{jk} \circ \varphi _{ij} = \varphi _{ik}. The morphisms
are proper because f is separated (Morphisms of Stacks, Lemma 101.6.6). Observe that \varphi _{ij} defines a section V \to I_{ij} of I_{ij} \to W_ i \cap W_ j over V. By More on Morphisms of Spaces, Lemma 76.39.6 we can find V-admissible blowups p_{ij} : Y_{ij} \to Y such that s_{ij} extends to p_{ij}^{-1}(W_ i \cap W_ j). After replacing Y by the fibre product of all the Y_{ij} over Y we get to the situation described in the next paragraph.
Assume there exists n \geq 0, quasi-compact opens W_ i \subset Y, i = 1, \ldots , n, and morphisms x_ i : W_ i \to \mathcal{X} such that (a) f \circ x_ i = \text{id}_{W_ i}, (b) W = \bigcup _{i = 1, \ldots , n} W_ i contains V, (c) W is the image of |f|, (d) V \subset W_1 \cap \ldots \cap W_ n, and (g) there are isomorphisms \varphi _{ij} : x_ i|_{W_ i \cap W_ j} \to x_ j|_{W_ i \cap W_ j} satisfying
After replacing Y by another V-admissible blowup if necessary we may assume that V is dense and scheme theoretically dense in Y and hence in any open subspace of Y containing V. After such a replacement we conclude that
by appealing to Morphisms of Spaces, Lemma 67.17.8 and the fact that I_{ik} \to W_ i \cap W_ j is proper (hence separated). Of course this means that (x_ i, \varphi _{ij}) is a desent datum and we obtain a morphism x : W \to \mathcal{X} agreeing with x_ i over W_ i because \mathcal{X} is a stack. Since x is a section of the separated morphism \mathcal{X} \to W we see that x is proper (Morphisms of Stacks, Lemma 101.4.9). Thus the lemma now holds with \overline{Z} = Y, Z = W, \overline{g} = \text{id}_ Y, g = \text{id}_ W, h = x. \square
Theorem 106.10.3 (Chow's lemma).reference Let f : \mathcal{X} \to Y be a morphism from an algebraic stack to an algebraic space. Assume
Y is quasi-compact and quasi-separated,
f is separated of finite type.
Then there exists a commutative diagram
where X \to \mathcal{X} is proper surjective, X \to \overline{X} is an open immersion, and \overline{X} \to Y is proper morphism of algebraic spaces.
Proof. The rough idea is to use that \mathcal{X} has a dense open which is a gerbe (Morphisms of Stacks, Proposition 101.29.1) and appeal to Lemma 106.10.2. The reason this does not work is that the open may not be quasi-compact and one runs into technical problems. Thus we first do a (standard) reduction to the Noetherian case.
First we choose a closed immersion \mathcal{X} \to \mathcal{X}' where \mathcal{X}' is an algebraic stack separated and of finite type over Y. See Limits of Stacks, Lemma 102.6.2. Clearly it suffices to prove the theorem for \mathcal{X}', hence we may assume \mathcal{X} \to Y is separated and of finite presentation.
Assume \mathcal{X} \to Y is separated and of finite presentation. By Limits of Spaces, Proposition 70.8.1 we can write Y = \mathop{\mathrm{lim}}\nolimits Y_ i as the directed limit of a system of Noetherian algebraic spaces with affine transition morphisms. By Limits of Stacks, Lemma 102.5.1 there is an i and a morphism \mathcal{X}_ i \to Y_ i of finite presentation from an algebraic stack to Y_ i such that \mathcal{X} = Y \times _{Y_ i} \mathcal{X}_ i. After increasing i we may assume that \mathcal{X}_ i \to Y_ i is separated, see Limits of Stacks, Lemma 102.4.2. Then it suffices to prove the theorem for \mathcal{X}_ i \to Y_ i. This reduces us to the case discussed in the next paragraph.
Assume Y is Noetherian. We may replace \mathcal{X} by its reduction (Properties of Stacks, Definition 100.10.4). This reduces us to the case discussed in the next paragraph.
Assume Y is Noetherian and \mathcal{X} is reduced. Since \mathcal{X} \to Y is separated and Y quasi-separated, we see that \mathcal{X} is quasi-separated as an algebraic stack. Hence the inertia \mathcal{I}_\mathcal {X} \to \mathcal{X} is quasi-compact. Thus by Morphisms of Stacks, Proposition 101.29.1 there exists a dense open substack \mathcal{V} \subset \mathcal{X} which is a gerbe. Let \mathcal{V} \to V be the morphism which expresses \mathcal{V} as a gerbe over the algebraic space V. See Morphisms of Stacks, Lemma 101.28.2 for a construction of \mathcal{V} \to V. This construction in particular shows that the morphism \mathcal{V} \to Y factors as \mathcal{V} \to V \to Y. Picture
Since the morphism \mathcal{V} \to V is surjective, flat, and of finite presentation (Morphisms of Stacks, Lemma 101.28.8) and since \mathcal{V} \to Y is locally of finite presentation, it follows that V \to Y is locally of finite presentation (Morphisms of Stacks, Lemma 101.27.12). Note that \mathcal{V} \to V is a universal homeomorphism (Morphisms of Stacks, Lemma 101.28.13). Since \mathcal{V} is quasi-compact (see Morphisms of Stacks, Lemma 101.8.2) we see that V is quasi-compact. Finally, since \mathcal{V} \to Y is separated the same is true for V \to Y by Morphisms of Stacks, Lemma 101.27.17 applied to \mathcal{V} \to V \to Y (whose assumptions are satisfied as we've already seen).
All of the above means that the assumptions of Limits of Spaces, Lemma 70.13.3 apply to the morphism V \to Y. Thus we can find a dense open subspace V' \subset V and an immersion V' \to \mathbf{P}^ n_ Y over Y. Clearly we may replace V by V' and \mathcal{V} by the inverse image of V' in \mathcal{V} (recall that |\mathcal{V}| = |V| as we've seen above). Thus we may assume we have a diagram
where the arrow V \to \mathbf{P}^ n_ Y is an immersion. Let \mathcal{X}' be the scheme theoretic image of the morphism
and let Y' be the scheme theoretic image of the morphism V \to \mathbf{P}^ n_ Y. We obtain a commutative diagram
(See Morphisms of Stacks, Lemma 101.38.4). We claim that \mathcal{V} = V \times _{Y'} \mathcal{X}' and that Lemma 106.10.2 applies to the morphism \mathcal{X}' \to Y' and the open subspace V \subset Y'. If the claim is true, then we obtain
with X \to \overline{X} an open immersion, \overline{g} and h proper, and such that |V| is contained in the image of |g|. Then the composition X \to \mathcal{X}' \to \mathcal{X} is proper (as a composition of proper morphisms) and its image contains |\mathcal{V}|, hence this composition is surjective. As well, \overline{X} \to Y' \to Y is proper as a composition of proper morphisms.
The last step is to prove the claim. Observe that \mathcal{X}' \to Y' is separated and of finite type, that Y' is quasi-compact and quasi-separated, and that V is quasi-compact (we omit checking all the details completely). Next, we observe that b : \mathcal{X}' \to \mathcal{X} is an isomorphism over \mathcal{V} by Morphisms of Stacks, Lemma 101.38.7. In particular \mathcal{V} is identified with an open substack of \mathcal{X}'. The morphism j is quasi-compact (source is quasi-compact and target is quasi-separated), so formation of the scheme theoretic image of j commutes with flat base change by Morphisms of Stacks, Lemma 101.38.5. In particular we see that V \times _{Y'} \mathcal{X}' is the scheme theoretic image of \mathcal{V} \to V \times _{Y'} \mathcal{X}'. However, by Morphisms of Stacks, Lemma 101.37.5 the image of |\mathcal{V}| \to |V \times _{Y'} \mathcal{X}'| is closed (use that \mathcal{V} \to V is a universal homeomorphism as we've seen above and hence is universally closed). Also the image is dense (combine what we just said with Morphisms of Stacks, Lemma 101.38.6) we conclude |\mathcal{V}| = |V \times _{Y'} \mathcal{X}'|. Thus \mathcal{V} \to V \times _{Y'} \mathcal{X}' is an isomorphism and the proof of the claim is complete. \square
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