Lemma 105.10.1. 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 $f$.

## 105.10 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 69.16.1. $\square$

Lemma 105.10.2. 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 69.16.1 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 105.9.1 (and Morphisms of Stacks, Lemma 100.28.8 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 105.10.1 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 100.28.10). We apply Lemma 105.10.1 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 100.6.6). 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 75.39.6 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 66.17.8 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 100.4.9). Thus the lemma now holds with $\overline{Z} = Y$, $Z = W$, $\overline{g} = \text{id}_ Y$, $g = \text{id}_ W$, $h = x$. $\square$

Theorem 105.10.3 (Chow's lemma). 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 100.29.1) and appeal to Lemma 105.10.2. 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 101.6.2. 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 69.8.1 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 101.5.1 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 101.4.2. 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 99.10.4). 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 100.29.1 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 100.28.2 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 100.28.8) 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 100.27.12). Note that $\mathcal{V} \to V$ is a universal homeomorphism (Morphisms of Stacks, Lemma 100.28.13). Since $\mathcal{V}$ is quasi-compact (see Morphisms of Stacks, Lemma 100.8.2) 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 100.27.17 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 69.13.3 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 100.38.4). We claim that $\mathcal{V} = V \times _{Y'} \mathcal{X}'$ and that Lemma 105.10.2 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 100.38.7. 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 100.38.5. 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 100.37.5 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 100.38.6) 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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