Lemma 95.18.1. The category $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ endowed with the functor $p$ above defines a stack in groupoids over $(\mathit{Sch}/S)_{fppf}$.
95.18 Finite Hilbert stacks
We formulate this in somewhat greater generality than is perhaps strictly needed. Fix a $1$-morphism
of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$. For each integer $d \geq 1$ consider a category $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ defined as follows:
An object $(U, Z, y, x, \alpha )$ where $U, Z$ are objects of in $(\mathit{Sch}/S)_{fppf}$ and $Z$ is a finite locally free of degree $d$ over $U$, where $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$, $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ Z)$ and $\alpha : y|_ Z \to F(x)$ is an isomorphism1.
A morphism $(U, Z, y, x, \alpha ) \to (U', Z', y', x', \alpha ')$ is given by a morphism of schemes $f : U \to U'$, a morphism of schemes $g : Z \to Z'$ which induces an isomorphism $Z \to Z' \times _ U U'$, and isomorphisms $b : y \to f^*y'$, $a : x \to g^*x'$ inducing a commutative diagram
\[ \xymatrix{ y|_ Z \ar[rr]_\alpha \ar[d]_{b|_ Z} & & F(x) \ar[d]^{F(a)} \\ f^*y'|_ Z \ar[rr]^{\alpha '} & & F(g^*x') \\ } \]
It is clear from the definitions that there is a canonical forgetful functor
which assigns to the quintuple $(U, Z, y, x, \alpha )$ the scheme $U$ and to the morphism $(f, g, b, a) : (U, Z, y, x, \alpha ) \to (U', Z', y', x', \alpha ')$ the morphism $f : U \to U'$.
Proof. As usual, the hardest part is to show descent for objects. To see this let $\{ U_ i \to U\} $ be a covering of $(\mathit{Sch}/S)_{fppf}$. Let $\xi _ i = (U_ i, Z_ i, y_ i, x_ i, \alpha _ i)$ be an object of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ lying over $U_ i$, and let $\varphi _{ij} : \text{pr}_0^*\xi _ i \to \text{pr}_1^*\xi _ j$ be a descent datum. First, observe that $\varphi _{ij}$ induces a descent datum $(Z_ i/U_ i, \varphi _{ij})$ which is effective by Descent, Lemma 35.37.1 This produces a scheme $Z/U$ which is finite locally free of degree $d$ by Descent, Lemma 35.23.30. From now on we identify $Z_ i$ with $Z \times _ U U_ i$. Next, the objects $y_ i$ in the fibre categories $\mathcal{Y}_{U_ i}$ descend to an object $y$ in $\mathcal{Y}_ U$ because $\mathcal{Y}$ is a stack in groupoids. Similarly the objects $x_ i$ in the fibre categories $\mathcal{X}_{Z_ i}$ descend to an object $x$ in $\mathcal{X}_ Z$ because $\mathcal{X}$ is a stack in groupoids. Finally, the given isomorphisms
glue to a morphism $\alpha : y|_ Z \to F(x)$ as the $\mathcal{Y}$ is a stack and hence $\mathit{Isom}_\mathcal {Y}(y|_ Z, F(x))$ is a sheaf. Details omitted. $\square$
Definition 95.18.2. We will denote $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ the degree $d$ finite Hilbert stack of $\mathcal{X}$ over $\mathcal{Y}$ constructed above. If $\mathcal{Y} = S$ we write $\mathcal{H}_ d(\mathcal{X}) = \mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$. If $\mathcal{X} = \mathcal{Y} = S$ we denote it $\mathcal{H}_ d$.
Note that given $F : \mathcal{X} \to \mathcal{Y}$ as above we have the following natural $1$-morphisms of stacks in groupoids over $(\mathit{Sch}/S)_{fppf}$:
Each of the arrows is given by a "forgetful functor".
Lemma 95.18.3. The $1$-morphism $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X})$ is faithful.
Proof. To check that $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y}) \to \mathcal{H}_ d(\mathcal{X})$ is faithful it suffices to prove that it is faithful on fibre categories. Suppose that $\xi = (U, Z, y, x, \alpha )$ and $\xi ' = (U, Z', y', x', \alpha ')$ are two objects of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over the scheme $U$. Let $(g, b, a), (g', b', a') : \xi \to \xi '$ be two morphisms in the fibre category of $\mathcal{H}_ d(\mathcal{X}/\mathcal{Y})$ over $U$. The image of these morphisms in $\mathcal{H}_ d(\mathcal{X})$ agree if and only if $g = g'$ and $a = a'$. Then the commutative diagram
implies that $b|_ Z = b'|_ Z$. Since $Z \to U$ is finite locally free of degree $d$ we see $\{ Z \to U\} $ is an fppf covering, hence $b = b'$. $\square$
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Comment #194 by Pieter Belmans on
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