96.14 Finite Hilbert stacks of spaces

The finite Hilbert stack of an algebraic space is an algebraic stack.

Lemma 96.14.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Then $\mathcal{H}_ d(X)$ is an algebraic stack.

Proof. The $1$-morphism

$\mathcal{H}_ d(X) \longrightarrow \mathcal{H}_ d$

is representable by algebraic spaces according to Lemma 96.12.6. The stack $\mathcal{H}_ d$ is an algebraic stack according to Proposition 96.13.2. Hence $\mathcal{H}_ d(X)$ is an algebraic stack by Algebraic Stacks, Lemma 93.15.4. $\square$

This lemma allows us to bootstrap.

Lemma 96.14.2. Let $S$ be a scheme. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$ such that

1. $\mathcal{X}$ is representable by an algebraic space, and

2. $F$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation.

Then $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is an algebraic stack.

Proof. Choose a representable stack in groupoids $\mathcal{U}$ over $S$ and a $1$-morphism $f : \mathcal{U} \to \mathcal{H}_ d(\mathcal{X})$ which is representable by algebraic spaces, smooth, and surjective. This is possible because $\mathcal{H}_ d(\mathcal{X})$ is an algebraic stack by Lemma 96.14.1. Consider the $2$-fibre product

$\mathcal{W} = \mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \times _{\mathcal{H}_ d(\mathcal{X}), f} \mathcal{U}.$

Since $\mathcal{U}$ is representable (in particular a stack in setoids) it follows from Examples of Stacks, Lemma 94.18.3 and Stacks, Lemma 8.6.7 that $\mathcal{W}$ is a stack in setoids. The $1$-morphism $\mathcal{W} \to \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ is representable by algebraic spaces, smooth, and surjective as a base change of the morphism $f$ (see Algebraic Stacks, Lemmas 93.9.7 and 93.10.6). Thus, if we can show that $\mathcal{W}$ is representable by an algebraic space, then the lemma follows from Algebraic Stacks, Lemma 93.15.3.

The diagonal of $\mathcal{Y}$ is representable by algebraic spaces according to Lemma 96.4.3. We may apply Lemma 96.12.5 to see that the $1$-morphism

$\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \longrightarrow \mathcal{H}_ d(\mathcal{X}) \times \mathcal{Y}$

is representable by algebraic spaces. Consider the $2$-fibre product

$\mathcal{V} = \mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \times _{(\mathcal{H}_ d(\mathcal{X}) \times \mathcal{Y}), f \times F} (\mathcal{U} \times \mathcal{X}).$

The projection morphism $\mathcal{V} \to \mathcal{U} \times \mathcal{X}$ is representable by algebraic spaces as a base change of the last displayed morphism. Hence $\mathcal{V}$ is an algebraic space (see Bootstrap, Lemma 79.3.6 or Algebraic Stacks, Lemma 93.9.8). The $1$-morphism $\mathcal{V} \to \mathcal{U}$ fits into the following $2$-cartesian diagram

$\xymatrix{ \mathcal{V} \ar[d] \ar[r] & \mathcal{X} \ar[d]^ F \\ \mathcal{W} \ar[r] & \mathcal{Y} }$

because

$\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \times _{(\mathcal{H}_ d(\mathcal{X}) \times \mathcal{Y}), f \times F} (\mathcal{U} \times \mathcal{X}) = (\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \times _{\mathcal{H}_ d(\mathcal{X}), f} \mathcal{U}) \times _{\mathcal{Y}, F} \mathcal{X}.$

Hence $\mathcal{V} \to \mathcal{W}$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation as a base change of $F$. It follows that the same thing is true for the corresponding sheaves of sets associated to $\mathcal{V}$ and $\mathcal{W}$, see Algebraic Stacks, Lemma 93.10.4. Thus we conclude that the sheaf associated to $\mathcal{W}$ is an algebraic space by Bootstrap, Theorem 79.10.1. $\square$

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).