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). 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. $\square$

## 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 (0)