Lemma 99.13.1. The category $\mathcal{S}\! \mathit{paces}'_{ft}$ is fibred in groupoids over $\mathit{Sch}_{fppf}$. The same is true for $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$.

## 99.13 The stack of algebraic spaces

This section continuous the discussion started in Examples of Stacks, Sections 95.7, 95.8, and 95.12. Working over $\mathbf{Z}$, the discussion therein shows that we have a stack in groupoids

parametrizing (nonflat) families of finite type algebraic spaces. More precisely, an object^{1} of $\mathcal{S}\! \mathit{paces}'_{ft}$ is a finite type morphism $X \to S$ from an algebraic space $X$ to a scheme $S$ and a morphism $(X' \to S') \to (X \to S)$ is given by a pair $(f, g)$ where $f : X' \to X$ is a morphism of algebraic spaces and $g : S' \to S$ is a morphism of schemes which fit into a commutative diagram

inducing an isomorphism $X' \to S' \times _ S X$, in other words, the diagram is cartesian in the category of algebraic spaces. The functor $p'_{ft}$ sends $(X \to S)$ to $S$ and sends $(f, g)$ to $g$. We define a full subcategory

consisting of objects $X \to S$ of $\mathcal{S}\! \mathit{paces}'_{ft}$ such that $X \to S$ is of finite presentation, flat, and proper. We denote

the restriction of the functor $p'_{ft}$ to the indicated subcategory. We first review the results already obtained in the references listed above, and then we start adding further results.

**Proof.**
We have seen this in Examples of Stacks, Section 95.12 for the case of $\mathcal{S}\! \mathit{paces}'_{ft}$ and this easily implies the result for the other case. However, let us also prove this directly by checking conditions (1) and (2) of Categories, Definition 4.35.1.

Condition (1). Let $X \to S$ be an object of $\mathcal{S}\! \mathit{paces}'_{ft}$ and let $S' \to S$ be a morphism of schemes. Then we set $X' = S' \times _ S X$. Note that $X' \to S'$ is of finite type by Morphisms of Spaces, Lemma 67.23.3. to obtain a morphism $(X' \to S') \to (X \to S)$ lying over $S' \to S$. Argue similarly for the other case using Morphisms of Spaces, Lemmas 67.28.3, 67.30.4, and 67.40.3.

Condition (2). Consider morphisms $(f, g) : (X' \to S') \to (X \to S)$ and $(a, b) : (Y \to T) \to (X \to S)$ of $\mathcal{S}\! \mathit{paces}'_{ft}$. Given a morphism $h : T \to S'$ with $g \circ h = b$ we have to show there is a unique morphism $(k, h) : (Y \to T) \to (X' \to S')$ of $\mathcal{S}\! \mathit{paces}'_{ft}$ such that $(f, g) \circ (k, h) = (a, b)$. This is clear from the fact that $X' = S' \times _ S X$. The same therefore works for any full subcategory of $\mathcal{S}\! \mathit{paces}'_{ft}$ satisfying (1). $\square$

Lemma 99.13.2. The diagonal

is representable by algebraic spaces.

**Proof.**
We will use criterion (2) of Algebraic Stacks, Lemma 94.10.11. Let $S$ be a scheme and let $X$ and $Y$ be algebraic spaces of finite presentation over $S$, flat over $S$, and proper over $S$. We have to show that the functor

is an algebraic space. An elementary argument shows that $\mathit{Isom}_ S(X, Y)$ sits in a fibre product

The bottom arrow sends $(\varphi , \psi )$ to $(\psi \circ \varphi , \varphi \circ \psi )$. By Proposition 99.12.3 the functors on the bottom row are algebraic spaces over $S$. Hence the result follows from the fact that the category of algebraic spaces over $S$ has fibre products. $\square$

Lemma 99.13.3. The category $\mathcal{S}\! \mathit{paces}'_{ft}$ is a stack in groupoids over $\mathit{Sch}_{fppf}$. The same is true for $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$.

**Proof.**
The reason this lemma holds is the slogan: any fppf descent datum for algebraic spaces is effective, see Bootstrap, Section 80.11. More precisely, the lemma for $\mathcal{S}\! \mathit{paces}'_{ft}$ follows from Examples of Stacks, Lemma 95.8.1 as we saw in Examples of Stacks, Section 95.12. However, let us review the proof. We need to check conditions (1), (2), and (3) of Stacks, Definition 8.5.1.

Property (1) we have seen in Lemma 99.13.1.

Property (2) follows from Lemma 99.13.2 in the case of $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$. In the case of $\mathcal{S}\! \mathit{paces}'_{ft}$ it follows from Examples of Stacks, Lemma 95.7.2 (and this is really the “correct” reference).

Condition (3) for $\mathcal{S}\! \mathit{paces}'_{ft}$ is checked as follows. Suppose given

an fppf covering $\{ U_ i \to U\} _{i \in I}$ in $\mathit{Sch}_{fppf}$,

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

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 $U_ i \times _ U U_ j \times _ U U_ k$.

We have to show 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 80.11.3 part (2). By Descent on Spaces, Lemma 74.11.11 we see that $X \to U$ is of finite type. In the case of $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ one additionally uses Descent on Spaces, Lemma 74.11.12, 74.11.13, and 74.11.19 in the last step. $\square$

Sanity check: the stacks $\mathcal{S}\! \mathit{paces}'_{ft}$ and $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ play the same role among algebraic spaces.

Lemma 99.13.4. Let $T$ be an algebraic space over $\mathbf{Z}$. Let $\mathcal{S}_ T$ denote the corresponding algebraic stack (Algebraic Stacks, Sections 94.7, 94.8, and 94.13). We have an equivalence of categories

and an equivalence of categories

**Proof.**
We are going to deduce this lemma from the fact that it holds for schemes (essentially by construction of the stacks) and the fact that fppf descent data for algebraic spaces over algerbaic spaces are effective. We strongly encourage the reader to skip the proof.

The construction from left to right in either arrow is straightforward: given $X \to T$ of finite type the functor $\mathcal{S}_ T \to \mathcal{S}\! \mathit{paces}'_{ft}$ assigns to $U/T$ the base change $X_ U \to U$. We will explain how to construct a quasi-inverse.

If $T$ is a scheme, then there is a quasi-inverse by the $2$-Yoneda lemma, see Categories, Lemma 4.41.2. Let $p : U \to T$ be a surjective étale morphism where $U$ is a scheme. Let $R = U \times _ T U$ with projections $s, t : R \to U$. Observe that we obtain morphisms

satisfying various compatibilities (on the nose).

Let $G : \mathcal{S}_ T \to \mathcal{S}\! \mathit{paces}'_{ft}$ be a functor over $\mathit{Sch}_{fppf}$. The restriction of $G$ to $\mathcal{S}_ U$ via the map displayed above corresponds to a finite type morphism $X_ U \to U$ of algebraic spaces via the $2$-Yoneda lemma. Since $p \circ s = p \circ t$ we see that $R \times _{s, U} X_ U$ and $R \times _{t, U} X_ U$ both correspond to the restriction of $G$ to $\mathcal{S}_ R$. Thus we obtain a canonical isomorphism $\varphi : X_ U \times _{U, t} R \to R \times _{s, U} X_ U$ over $R$. This isomorphism satisfies the cocycle condition by the various compatibilities of the diagram given above. Thus a descent datum which is effective by Bootstrap, Lemma 80.11.3 part (2). In other words, we obtain an object $X \to T$ of the right hand side category. We omit checking the construction $G \leadsto X$ is functorial and that it is quasi-inverse to the other construction. In the case of $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ one additionally uses Descent on Spaces, Lemma 74.11.12, 74.11.13, and 74.11.19 in the last step to see that $X \to T$ is of finite presentation, flat, and proper. $\square$

Remark 99.13.5. Let $B$ be an algebraic space over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Let $B\textit{-Spaces}'_{ft}$ be the category consisting of pairs $(X \to S, h : S \to B)$ where $X \to S$ is an object of $\mathcal{S}\! \mathit{paces}'_{ft}$ and $h : S \to B$ is a morphism. A morphism $(X' \to S', h') \to (X \to S, h)$ in $B\textit{-Spaces}'_{ft}$ is a morphism $(f, g)$ in $\mathcal{S}\! \mathit{paces}'_{ft}$ such that $h \circ g = h'$. In this situation the diagram

is $2$-fibre product square. This trivial remark will occasionally be useful to deduce results from the absolute case $\mathcal{S}\! \mathit{paces}'_{ft}$ to the case of families over a given base algebraic space. Of course, a similar construction works for $B\textit{-Spaces}'_{fp, flat, proper}$

Lemma 99.13.6. The stack $p'_{fp, flat, proper} : \mathcal{S}\! \mathit{paces}'_{fp, flat, proper} \to \mathit{Sch}_{fppf}$ is limit preserving (Artin's Axioms, Definition 98.11.1).

**Proof.**
Let $T = \mathop{\mathrm{lim}}\nolimits T_ i$ be the limits of a directed inverse system of affine schemes. By Limits of Spaces, Lemma 70.7.1 the category of algebraic spaces of finite presentation over $T$ is the colimit of the categories of algebraic spaces of finite presentation over $T_ i$. To finish the proof use that flatness and properness descends through the limit, see Limits of Spaces, Lemmas 70.6.12 and 70.6.13.
$\square$

Lemma 99.13.7. Let

be a pushout in the category of schemes where $T \to T'$ is a thickening and $T \to S$ is affine, see More on Morphisms, Lemma 37.14.3. Then the functor on fibre categories

is an equivalence.

**Proof.**
The functor is an equivalence if we drop “proper” from the list of conditions and replace “of finite presentation” by “locally of finite presentation”, see Pushouts of Spaces, Lemma 81.6.7. Thus it suffices to show that given a morphism $X' \to S'$ of an algebraic space to $S'$ which is flat and locally of finite presentation, then $X' \to S'$ is proper if and only if $S \times _{S'} X' \to S$ and $T' \times _{S'} X' \to T'$ are proper. One implication follows from the fact that properness is preserved under base change (Morphisms of Spaces, Lemma 67.40.3) and the other from the fact that properness of $S \times _{S'} X' \to S$ implies properness of $X' \to S'$ by More on Morphisms of Spaces, Lemma 76.10.2.
$\square$

Lemma 99.13.8. Let $k$ be a field and let $x = (X \to \mathop{\mathrm{Spec}}(k))$ be an object of $\mathcal{X} = \mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ over $\mathop{\mathrm{Spec}}(k)$.

If $k$ is of finite type over $\mathbf{Z}$, then the vector spaces $T\mathcal{F}_{\mathcal{X}, k, x}$ and $\text{Inf}(\mathcal{F}_{\mathcal{X}, k, x})$ (see Artin's Axioms, Section 98.8) are finite dimensional, and

in general the vector spaces $T_ x(k)$ and $\text{Inf}_ x(k)$ (see Artin's Axioms, Section 98.21) are finite dimensional.

**Proof.**
The discussion in Artin's Axioms, Section 98.8 only applies to fields of finite type over the base scheme $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Our stack satisfies (RS*) by Lemma 99.13.7 and we may apply Artin's Axioms, Lemma 98.21.2 to get the vector spaces $T_ x(k)$ and $\text{Inf}_ x(k)$ mentioned in (2). Moreover, in the finite type case these spaces agree with the ones mentioned in (1) by Artin's Axioms, Remark 98.21.7. With this out of the way we can start the proof. Observe that the first order thickening $\mathop{\mathrm{Spec}}(k) \to \mathop{\mathrm{Spec}}(k[\epsilon ]) = \mathop{\mathrm{Spec}}(k[k])$ has conormal module $k$. Hence the formula in Deformation Theory, Lemma 91.14.2 describing infinitesimal deformations of $X$ and infinitesimal automorphisms of $X$ become

By More on Morphisms of Spaces, Lemma 76.21.5 and the fact that $X$ is Noetherian, we see that $\mathop{N\! L}\nolimits _{X/k}$ has coherent cohomology sheaves zero except in degrees $0$ and $-1$. By Derived Categories of Spaces, Lemma 75.8.4 the displayed $\mathop{\mathrm{Ext}}\nolimits $-groups are finite $k$-vector spaces and the proof is complete. $\square$

Beware that openness of versality (as proved in the next lemma) is a bit strange because our stack does not satisfy formal effectiveness, see Examples, Section 110.71. Later we will apply the openness of versality to suitable substacks of $\mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ which do satisfy formal effectiveness to conclude that these stacks are algebraic.

Lemma 99.13.9. The stack in groupoids $\mathcal{X} = \mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$ satisfies openness of versality over $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Similarly, after base change (Remark 99.13.5) openness of versality holds over any Noetherian base scheme $S$.

**Proof.**
For the “usual” proof of this fact, please see the discussion in the remark following this proof. We will prove this using Artin's Axioms, Lemma 98.20.3. We have already seen that $\mathcal{X}$ has diagonal representable by algebraic spaces, has (RS*), and is limit preserving, see Lemmas 99.13.2, 99.13.7, and 99.13.6. Hence we only need to see that $\mathcal{X}$ satisfies the strong formal effectiveness formulated in Artin's Axioms, Lemma 98.20.3.

Let $(R_ n)$ be an inverse system of rings such that $R_ n \to R_ m$ is surjective with square zero kernel for all $n \geq m$. Let $X_ n \to \mathop{\mathrm{Spec}}(R_ n)$ be a finitely presented, flat, proper morphism where $X_ n$ is an algebraic space and let $X_{n + 1} \to X_ n$ be a morphism over $\mathop{\mathrm{Spec}}(R_{n + 1})$ inducing an isomorphism $X_ n = X_{n + 1} \times _{\mathop{\mathrm{Spec}}(R_{n + 1})} \mathop{\mathrm{Spec}}(R_ n)$. We have to find a flat, proper, finitely presented morphism $X \to \mathop{\mathrm{Spec}}(\mathop{\mathrm{lim}}\nolimits R_ n)$ whose source is an algebraic space such that $X_ n$ is the base change of $X$ for all $n$.

Let $I_ n = \mathop{\mathrm{Ker}}(R_ n \to R_1)$. We may think of $(X_1 \subset X_ n) \to (\mathop{\mathrm{Spec}}(R_1) \subset \mathop{\mathrm{Spec}}(R_ n))$ as a morphism of first order thickenings. (Please read some of the material on thickenings of algebraic spaces in More on Morphisms of Spaces, Section 76.9 before continuing.) The structure sheaf of $X_ n$ is an extension

over $0 \to I_ n \to R_ n \to R_1$, see More on Morphisms of Spaces, Lemma 76.18.1. Let's consider the extension

over $0 \to \mathop{\mathrm{lim}}\nolimits I_ n \to \mathop{\mathrm{lim}}\nolimits R_ n \to R_1 \to 0$. The displayed sequence is exact as the $R^1\mathop{\mathrm{lim}}\nolimits $ of the system of kernels is zero by Derived Categories of Spaces, Lemma 75.5.4. Observe that the map

induces an isomorphism upon applying the functor $DQ_ X$, see Derived Categories of Spaces, Lemma 75.25.6. Hence we obtain a unique extension

over $0 \to \mathop{\mathrm{lim}}\nolimits I_ n \to \mathop{\mathrm{lim}}\nolimits R_ n \to R_1 \to 0$ by the equivalence of categories of Deformation Theory, Lemma 91.14.4. The sheaf $\mathcal{O}'$ determines a first order thickening of algebraic spaces $X_1 \subset X$ over $\mathop{\mathrm{Spec}}(R_1) \subset \mathop{\mathrm{Spec}}(\mathop{\mathrm{lim}}\nolimits R_ n)$ by More on Morphisms of Spaces, Lemma 76.9.7. Observe that $X \to \mathop{\mathrm{Spec}}(\mathop{\mathrm{lim}}\nolimits R_ n)$ is flat by the already used More on Morphisms of Spaces, Lemma 76.18.1. By More on Morphisms of Spaces, Lemma 76.18.3 we see that $X \to \mathop{\mathrm{Spec}}(\mathop{\mathrm{lim}}\nolimits R_ n)$ is proper and of finite presentation. This finishes the proof. $\square$

Remark 99.13.10. Lemma 99.13.9 can also be shown using either Artin's Axioms, Lemma 98.24.4 (as in the first proof of Lemma 99.5.11), or using an obstruction theory as in Artin's Axioms, Lemma 98.22.2 (as in the second proof of Lemma 99.5.11). In both cases one uses the deformation and obstruction theory developed in Cotangent, Section 92.23 to translate the needed properties of deformations and obstructions into $\mathop{\mathrm{Ext}}\nolimits $-groups to which Derived Categories of Spaces, Lemma 75.23.3 can be applied. The second method (using an obstruction theory and therefore using the full cotangent complex) is perhaps the “standard” method used in most references.

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