## 76.11 Applications

As a first application we obtain the following fundamental fact:

$\fbox{A sheaf which is fppf locally an algebraic space is an algebraic space.}$

This is the content of the following lemma. Note that assumption (2) is equivalent to the condition that $F|_{(\mathit{Sch}/S_ i)_{fppf}}$ is an algebraic space, see Spaces, Lemma 61.16.4. Assumption (3) is a set theoretic condition which may be ignored by those not worried about set theoretic questions.

Lemma 76.11.1. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Let $\{ S_ i \to S\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Assume that

1. $F$ is a sheaf,

2. each $F_ i = h_{S_ i} \times F$ is an algebraic space, and

3. $\coprod _{i \in I} F_ i$ is an algebraic space (see Spaces, Lemma 61.8.4).

Then $F$ is an algebraic space.

Proof. Consider the morphism $\coprod F_ i \to F$. This is the base change of $\coprod S_ i \to S$ via $F \to S$. Hence it is representable, locally of finite presentation, flat and surjective by our definition of an fppf covering and Lemma 76.4.2. Thus Theorem 76.10.1 applies to show that $F$ is an algebraic space. $\square$

Here is a special case of Lemma 76.11.1 where we do not need to worry about set theoretical issues.

Lemma 76.11.2. Let $S$ be a scheme. Let $F : (\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$ be a functor. Let $\{ S_ i \to S\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Assume that

1. $F$ is a sheaf,

2. each $F_ i = h_{S_ i} \times F$ is an algebraic space, and

3. the morphisms $F_ i \to S_ i$ are of finite type.

Then $F$ is an algebraic space.

Proof. We will use Lemma 76.11.1 above. To do this we will show that the assumption that $F_ i$ is of finite type over $S_ i$ to prove that the set theoretic condition in the lemma is satisfied (after perhaps refining the given covering of $S$ a bit). We suggest the reader skip the rest of the proof.

If $S'_ i \to S_ i$ is a morphism of schemes then

$h_{S'_ i} \times F = h_{S'_ i} \times _{h_{S_ i}} h_{S_ i} \times F = h_{S'_ i} \times _{h_{S_ i}} F_ i$

is an algebraic space of finite type over $S'_ i$, see Spaces, Lemma 61.7.3 and Morphisms of Spaces, Lemma 63.23.3. Thus we may refine the given covering. After doing this we may assume: (a) each $S_ i$ is affine, and (b) the cardinality of $I$ is at most the cardinality of the set of points of $S$. (Since to cover all of $S$ it is enough that each point is in the image of $S_ i \to S$ for some $i$.)

Since each $S_ i$ is affine and each $F_ i$ of finite type over $S_ i$ we conclude that $F_ i$ is quasi-compact. Hence by Properties of Spaces, Lemma 62.6.3 we can find an affine $U_ i \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $U_ i \to F_ i$. The fact that $F_ i \to S_ i$ is locally of finite type then implies that $U_ i \to S_ i$ is locally of finite type, and in particular $U_ i \to S$ is locally of finite type. By Sets, Lemma 3.9.7 we conclude that $\text{size}(U_ i) \leq \text{size}(S)$. Since also $|I| \leq \text{size}(S)$ we conclude that $\coprod _{i \in I} U_ i$ is isomorphic to an object of $(\mathit{Sch}/S)_{fppf}$ by Sets, Lemma 3.9.5 and the construction of $\mathit{Sch}$. This implies that $\coprod F_ i$ is an algebraic space by Spaces, Lemma 61.8.4 and we win. $\square$

As a second application we obtain

$\fbox{Any fppf descent datum for algebraic spaces is effective.}$

This holds modulo set theoretical difficulties; as an example result we offer the following lemma.

Lemma 76.11.3. Let $S$ be a scheme. Let $\{ X_ i \to X\} _{i \in I}$ be an fppf covering of algebraic spaces over $S$.

1. If $I$ is countable1, then any descent datum for algebraic spaces relative to $\{ X_ i \to X\}$ is effective.

2. Any descent datum $(Y_ i, \varphi _{ij})$ relative to $\{ X_ i \to X\} _{i \in I}$ (Descent on Spaces, Definition 70.21.3) with $Y_ i \to X_ i$ of finite type is effective.

Proof. Proof of (1). By Descent on Spaces, Lemma 70.22.1 this translates into the statement that an fppf sheaf $F$ endowed with a map $F \to X$ is an algebraic space provided that each $F \times _ X X_ i$ is an algebraic space. The restriction on the cardinality of $I$ implies that coproducts of algebraic spaces indexed by $I$ are algebraic spaces, see Spaces, Lemma 61.8.4 and Sets, Lemma 3.9.9. The morphism

$\coprod F \times _ X X_ i \longrightarrow F$

is representable by algebraic spaces (as the base change of $\coprod X_ i \to X$, see Lemma 76.3.3), and surjective, flat, and locally of finite presentation (as the base change of $\coprod X_ i \to X$, see Lemma 76.4.2). Hence part (1) follows from Theorem 76.10.1.

Proof of (2). First we apply Descent on Spaces, Lemma 70.22.1 to obtain an fppf sheaf $F$ endowed with a map $F \to X$ such that $F \times _ X X_ i = Y_ i$ for all $i \in I$. Our goal is to show that $F$ is an algebraic space. Choose a scheme $U$ and a surjective étale morphism $U \to X$. Then $F' = U \times _ X F \to F$ is representable, surjective, and étale as the base change of $U \to X$. By Theorem 76.10.1 it suffices to show that $F' = U \times _ X F$ is an algebraic space. We may choose an fppf covering $\{ U_ j \to U\} _{j \in J}$ where $U_ j$ is a scheme refining the fppf covering $\{ X_ i \times _ X U \to U\} _{i \in I}$, see Topologies on Spaces, Lemma 69.7.4. Thus we get a map $a : J \to I$ and for each $j$ a morphism $U_ j \to X_{a(j)}$ over $X$. Then we see that $U_ j \times _ U F' = U_ j \times _{X_{a(j)}} Y_{a(j)}$ is of finite type over $U_ j$. Hence $F'$ is an algebraic space by Lemma 76.11.2. $\square$

Here is a different type of application.

Lemma 76.11.4. Let $S$ be a scheme. Let $a : F \to G$ and $b : G \to H$ be transformations of functors $(\mathit{Sch}/S)_{fppf}^{opp} \to \textit{Sets}$. Assume

1. $F, G, H$ are sheaves,

2. $a : F \to G$ is representable by algebraic spaces, flat, locally of finite presentation, and surjective, and

3. $b \circ a : F \to H$ is representable by algebraic spaces.

Then $b$ is representable by algebraic spaces.

Proof. Let $U$ be a scheme over $S$ and let $\xi \in H(U)$. We have to show that $U \times _{\xi , H} G$ is an algebraic space. On the other hand, we know that $U \times _{\xi , H} F$ is an algebraic space and that $U \times _{\xi , H} F \to U \times _{\xi , H} G$ is representable by algebraic spaces, flat, locally of finite presentation, and surjective as a base change of the morphism $a$ (see Lemma 76.4.2). Thus the result follows from Theorem 76.10.1. $\square$

Lemma 76.11.5. Assume $B \to S$ and $(U, R, s, t, c)$ are as in Groupoids in Spaces, Definition 74.19.1 (1). For any scheme $T$ over $S$ and objects $x, y$ of $[U/R]$ over $T$ the sheaf $\mathit{Isom}(x, y)$ on $(\mathit{Sch}/T)_{fppf}$ is an algebraic space.

Proof. By Groupoids in Spaces, Lemma 74.21.3 there exists an fppf covering $\{ T_ i \to T\} _{i \in I}$ such that $\mathit{Isom}(x, y)|_{(\mathit{Sch}/T_ i)_{fppf}}$ is an algebraic space for each $i$. By Spaces, Lemma 61.16.4 this means that each $F_ i = h_{S_ i} \times \mathit{Isom}(x, y)$ is an algebraic space. Thus to prove the lemma we only have to verify the set theoretic condition that $\coprod F_ i$ is an algebraic space of Lemma 76.11.1 above to conclude. To do this we use Spaces, Lemma 61.8.4 which requires showing that $I$ and the $F_ i$ are not “too large”. We suggest the reader skip the rest of the proof.

Choose $U' \in \mathop{\mathrm{Ob}}\nolimits (\mathit{Sch}/S)_{fppf}$ and a surjective étale morphism $U' \to U$. Let $R'$ be the restriction of $R$ to $U'$. Since $[U/R] = [U'/R']$ we may, after replacing $U$ by $U'$, assume that $U$ is a scheme. (This step is here so that the fibre products below are over a scheme.)

Note that if we refine the covering $\{ T_ i \to T\}$ then it remains true that each $F_ i$ is an algebraic space. Hence we may assume that each $T_ i$ is affine. Since $T_ i \to T$ is locally of finite presentation, this then implies that $\text{size}(T_ i) \leq \text{size}(T)$, see Sets, Lemma 3.9.7. We may also assume that the cardinality of the index set $I$ is at most the cardinality of the set of points of $T$ since to get a covering it suffices to check that each point of $T$ is in the image. Hence $|I| \leq \text{size}(T)$. Choose $W \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $W \to R$. Note that in the proof of Groupoids in Spaces, Lemma 74.21.3 we showed that $F_ i$ is representable by $T_ i \times _{(y_ i, x_ i), U \times _ B U} R$ for some $x_ i, y_ i : T_ i \to U$. Hence now we see that $V_ i = T_ i \times _{(y_ i, x_ i), U \times _ B U} W$ is a scheme which comes with an étale surjection $V_ i \to F_ i$. By Sets, Lemma 3.9.6 we see that

$\text{size}(V_ i) \leq \max \{ \text{size}(T_ i), \text{size}(W)\} \leq \max \{ \text{size}(T), \text{size}(W)\}$

Hence, by Sets, Lemma 3.9.5 we conclude that

$\text{size}(\coprod \nolimits _{i \in I} V_ i) \leq \max \{ |I|, \text{size}(T), \text{size}(W)\} .$

Hence we conclude by our construction of $\mathit{Sch}$ that $\coprod _{i \in I} V_ i$ is isomorphic to an object $V$ of $(\mathit{Sch}/S)_{fppf}$. This verifies the hypothesis of Spaces, Lemma 61.8.4 and we win. $\square$

Lemma 76.11.6. Let $S$ be a scheme. Consider an algebraic space $F$ of the form $F = U/R$ where $(U, R, s, t, c)$ is a groupoid in algebraic spaces over $S$ such that $s, t$ are flat and locally of finite presentation, and $j = (t, s) : R \to U \times _ S U$ is an equivalence relation. Then $U \to F$ is surjective, flat, and locally of finite presentation.

Proof. This is almost but not quite a triviality. Namely, by Groupoids in Spaces, Lemma 74.18.5 and the fact that $j$ is a monomorphism we see that $R = U \times _ F U$. Choose a scheme $W$ and a surjective étale morphism $W \to F$. As $U \to F$ is a surjection of sheaves we can find an fppf covering $\{ W_ i \to W\}$ and maps $W_ i \to U$ lifting the morphisms $W_ i \to F$. Then we see that

$W_ i \times _ F U = W_ i \times _ U U \times _ F U = W_ i \times _{U, t} R$

and the projection $W_ i \times _ F U \to W_ i$ is the base change of $t : R \to U$ hence flat and locally of finite presentation, see Morphisms of Spaces, Lemmas 63.30.4 and 63.28.3. Hence by Descent on Spaces, Lemmas 70.10.13 and 70.10.10 we see that $U \to F$ is flat and locally of finite presentation. It is surjective by Spaces, Remark 61.5.2. $\square$

Lemma 76.11.7. Let $S$ be a scheme. Let $X \to B$ be a morphism of algebraic spaces over $S$. Let $G$ be a group algebraic space over $B$ and let $a : G \times _ B X \to X$ be an action of $G$ on $X$ over $B$. If

1. $a$ is a free action, and

2. $G \to B$ is flat and locally of finite presentation,

then $X/G$ (see Groupoids in Spaces, Definition 74.18.1) is an algebraic space and $X \to X/G$ is surjective, flat, and locally of finite presentation.

Proof. The fact that $X/G$ is an algebraic space is immediate from Theorem 76.10.1 and the definitions. Namely, $X/G = X/R$ where $R = G \times _ B X$. The morphisms $s, t : G \times _ B X \to X$ are flat and locally of finite presentation (clear for $s$ as a base change of $G \to B$ and by symmetry using the inverse it follows for $t$) and the morphism $j : G \times _ B X \to X \times _ B X$ is a monomorphism by Groupoids in Spaces, Lemma 74.8.3 as the action is free. The assertions about the morphism $X \to X/G$ follow from Lemma 76.11.6. $\square$

Lemma 76.11.8. Let $\{ S_ i \to S\} _{i \in I}$ be a covering of $(\mathit{Sch}/S)_{fppf}$. Let $G$ be a group algebraic space over $S$, and denote $G_ i = G_{S_ i}$ the base changes. Suppose given

1. for each $i \in I$ an fppf $G_ i$-torsor $X_ i$ over $S_ i$, and

2. for each $i, j \in I$ a $G_{S_ i \times _ S S_ j}$-equivariant isomorphism $\varphi _{ij} : X_ i \times _ S S_ j \to S_ i \times _ S X_ j$ satisfying the cocycle condition over every $S_ i \times _ S S_ j \times _ S S_ j$.

Then there exists an fppf $G$-torsor $X$ over $S$ whose base change to $S_ i$ is isomorphic to $X_ i$ such that we recover the descent datum $\varphi _{ij}$.

Proof. We may think of $X_ i$ as a sheaf on $(\mathit{Sch}/S_ i)_{fppf}$, see Spaces, Section 61.16. By Sites, Section 7.26 the descent datum $(X_ i, \varphi _{ij})$ is effective in the sense that there exists a unique sheaf $X$ on $(\mathit{Sch}/S)_{fppf}$ which recovers the algebraic spaces $X_ i$ after restricting back to $(\mathit{Sch}/S_ i)_{fppf}$. Hence we see that $X_ i = h_{S_ i} \times X$. By Lemma 76.11.1 we see that $X$ is an algebraic space, modulo verifying that $\coprod X_ i$ is an algebraic space which we do at the end of the proof. By the equivalence of categories in Sites, Lemma 7.26.5 the action maps $G_ i \times _{S_ i} X_ i \to X_ i$ glue to give a map $a : G \times _ S X \to X$. Now we have to show that $a$ is an action and that $X$ is a pseudo-torsor, and fppf locally trivial (see Groupoids in Spaces, Definition 74.9.3). These may be checked fppf locally, and hence follow from the corresponding properties of the actions $G_ i \times _{S_ i} X_ i \to X_ i$. Hence the lemma is true.

We suggest the reader skip the rest of the proof, which is purely set theoretical. Pick coverings $\{ S_{ij} \to S_ j\} _{j \in J_ i}$ of $(\mathit{Sch}/S)_{fppf}$ which trivialize the $G_ i$ torsors $X_ i$ (possible by assumption, and Topologies, Lemma 33.7.7 part (1)). Then $\{ S_{ij} \to S\} _{i \in I, j \in J_ i}$ is a covering of $(\mathit{Sch}/S)_{fppf}$ and hence we may assume that each $X_ i$ is the trivial torsor! Of course we may also refine the covering further, hence we may assume that each $S_ i$ is affine and that the index set $I$ has cardinality bounded by the cardinality of the set of points of $S$. Choose $U \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$ and a surjective étale morphism $U \to G$. Then we see that $U_ i = U \times _ S S_ i$ comes with an étale surjective morphism to $X_ i \cong G_ i$. By Sets, Lemma 3.9.6 we see $\text{size}(U_ i) \leq \max \{ \text{size}(U), \text{size}(S_ i)\}$. By Sets, Lemma 3.9.7 we have $\text{size}(S_ i) \leq \text{size}(S)$. Hence we see that $\text{size}(U_ i) \leq \max \{ \text{size}(U), \text{size}(S)\}$ for all $i \in I$. Together with the bound on $|I|$ we found above we conclude from Sets, Lemma 3.9.5 that $\text{size}(\coprod U_ i) \leq \max \{ \text{size}(U), \text{size}(S)\}$. Hence Spaces, Lemma 61.8.4 applies to show that $\coprod X_ i$ is an algebraic space which is what we had to prove. $\square$

[1] The restriction on countablility can be ignored by those who do not care about set theoretical issues. We can allow larger index sets here if we can bound the size of the algebraic spaces which we are descending. See for example Lemma 76.11.2.

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