94.8 The stack of finite type algebraic spaces

It turns out that we can get a stack of spaces if we only consider spaces of finite type. Let us denote

$p_{ft} : \mathcal{S}\! \mathit{paces}_{ft} \to (\mathit{Sch}/S)_{fppf}$

the full subcategory of $\mathcal{S}\! \mathit{paces}$ over $(\mathit{Sch}/S)_{fppf}$ consisting of pairs $X/U$ such that $X \to U$ is a morphism of finite type.

Lemma 94.8.1. The functor $p_{ft} : \mathcal{S}\! \mathit{paces}_{ft} \to (\mathit{Sch}/S)_{fppf}$ satisfies the conditions (1), (2) and (3) of Stacks, Definition 8.4.1.

Proof. We are going to write this out in ridiculous detail (which may make it hard to see what is going on).

We have seen in Lemma 94.7.1 that a morphism $(f, g) : X/U \to Y/V$ of $\mathcal{S}\! \mathit{paces}$ is strongly cartesian if the induced morphism $f : X \to U \times _ V Y$ is an isomorphism. Note that if $Y \to V$ is of finite type then also $U \times _ V Y \to U$ is of finite type, see Morphisms of Spaces, Lemma 66.23.3. So if $(f, g) : X/U \to Y/V$ of $\mathcal{S}\! \mathit{paces}$ is strongly cartesian in $\mathcal{S}\! \mathit{paces}$ and $Y/V$ is an object of $\mathcal{S}\! \mathit{paces}_{ft}$ then automatically also $X/U$ is an object of $\mathcal{S}\! \mathit{paces}_{ft}$, and of course $(f, g)$ is also strongly cartesian in $\mathcal{S}\! \mathit{paces}_{ft}$. In this way we conclude that $\mathcal{S}\! \mathit{paces}_{ft}$ is a fibred category over $(\mathit{Sch}/S)_{fppf}$. This proves (1).

The argument above also shows that the inclusion functor $\mathcal{S}\! \mathit{paces}_{ft} \to \mathcal{S}\! \mathit{paces}$ transforms strongly cartesian morphisms into strongly cartesian morphisms. In other words $\mathcal{S}\! \mathit{paces}_{ft} \to \mathcal{S}\! \mathit{paces}$ is a $1$-morphism of fibred categories over $(\mathit{Sch}/S)_{fppf}$.

Let $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$. Let $X, Y$ be algebraic spaces of finite type over $U$. By Stacks, Lemma 8.2.3 we obtain a map of presheaves

$\mathit{Mor}_{\mathcal{S}\! \mathit{paces}_{ft}}(X, Y) \longrightarrow \mathit{Mor}_{\mathcal{S}\! \mathit{paces}}(X, Y)$

which is an isomorphism as $\mathcal{S}\! \mathit{paces}_{ft}$ is a full subcategory of $\mathcal{S}\! \mathit{paces}$. Hence the left hand side is a sheaf, because in Lemma 94.7.2 we showed the right hand side is a sheaf. This proves (2).

To prove condition (3) of Stacks, Definition 8.4.1 we have to show the following: Given

1. a covering $\{ U_ i \to U\} _{i \in I}$ of $(\mathit{Sch}/S)_{fppf}$,

2. for each $i \in I$ an algebraic space $X_ i$ of finite type over $U_ i$, and

3. for each $i, j \in I$ an isomorphism $\varphi _{ij} : X_ i \times _ U U_ j \to U_ i \times _ U X_ j$ of algebraic spaces over $U_ i \times _ U U_ j$ satisfying the cocycle condition over $U_ i \times _ U U_ j \times _ U U_ k$,

there exists an algebraic space $X$ of finite type over $U$ and isomorphisms $X_{U_ i} \cong X_ i$ over $U_ i$ recovering the isomorphisms $\varphi _{ij}$. This follows from Bootstrap, Lemma 79.11.3 part (2). By Descent on Spaces, Lemma 73.11.10 we see that $X \to U$ is of finite type which concludes the proof. $\square$

Lemma 94.8.2. There exists a subcategory $\mathcal{S}\! \mathit{paces}_{ft, small} \subset \mathcal{S}\! \mathit{paces}_{ft}$ with the following properties:

1. the inclusion functor $\mathcal{S}\! \mathit{paces}_{ft, small} \to \mathcal{S}\! \mathit{paces}_{ft}$ is fully faithful and essentially surjective, and

2. the functor $p_{ft, small} : \mathcal{S}\! \mathit{paces}_{ft, small} \to (\mathit{Sch}/S)_{fppf}$ turns $\mathcal{S}\! \mathit{paces}_{ft, small}$ into a stack over $(\mathit{Sch}/S)_{fppf}$.

Proof. We have seen in Lemmas 94.8.1 that $p_{ft} : \mathcal{S}\! \mathit{paces}_{ft} \to (\mathit{Sch}/S)_{fppf}$ satisfies (1), (2) and (3) of Stacks, Definition 8.4.1. The additional condition (4) of Stacks, Remark 8.4.9 holds because every algebraic space $X$ over $S$ is of the form $U/R$ for $U, R \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, see Spaces, Lemma 64.9.1. Thus there is only a set worth of isomorphism classes of objects. Hence we obtain $\mathcal{S}\! \mathit{paces}_{ft, small}$ from the discussion in that remark. $\square$

We will often perform the replacement

$\mathcal{S}\! \mathit{paces}_{ft} \leadsto \mathcal{S}\! \mathit{paces}_{ft, small}$

without further remarking on it, and by abuse of notation we will simply denote $\mathcal{S}\! \mathit{paces}_{ft}$ this replacement.

Remark 94.8.3. Note that the whole discussion in this section works if we want to consider those algebraic spaces $X/U$ which are locally of finite type such that the inverse image in $X$ of an affine open of $U$ can be covered by countably many affines. If needed we can also introduce the notion of a morphism of $\kappa$-type (meaning some bound on the number of generators of ring extensions and some bound on the cardinality of the affines over a given affine in the base) where $\kappa$ is a cardinal, and then we can produce a stack

$\mathcal{S}\! \mathit{paces}_\kappa \longrightarrow (\mathit{Sch}/S)_{fppf}$

in exactly the same manner as above (provided we make sure that $\mathit{Sch}$ is large enough depending on $\kappa$).

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