## 80.11 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 65.16.4. Assumption (3) is a set theoretic condition which may be ignored by those not worried about set theoretic questions.

Lemma 80.11.1. 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 65.8.4).

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 80.4.2. Thus Theorem 80.10.1 applies to show that $F$ is an algebraic space. $\square$

Here is a special case of Lemma 80.11.1 where we do not need to worry about set theoretical issues.

Lemma 80.11.2. 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 80.11.1 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 65.7.3 and Morphisms of Spaces, Lemma 67.23.3. 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 66.6.3 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 65.8.4 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 80.11.3. 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 74.22.3) with $Y_ i \to X_ i$ of finite type is effective.

Proof. Proof of (1). By Descent on Spaces, Lemma 74.23.1 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 65.8.4 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 80.3.3), and surjective, flat, and locally of finite presentation (as the base change of $\coprod X_ i \to X$, see Lemma 80.4.2). Hence part (1) follows from Theorem 80.10.1.

Proof of (2). First we apply Descent on Spaces, Lemma 74.23.1 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 80.10.1 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 73.7.4. 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 80.11.2. $\square$

Here is a different type of application.

Lemma 80.11.4. 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 80.4.2). Thus the result follows from Theorem 80.10.1. $\square$

Lemma 80.11.5. Assume $B \to S$ and $(U, R, s, t, c)$ are as in Groupoids in Spaces, Definition 78.20.1 (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 78.22.3 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 65.16.4 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 80.11.1 above to conclude. To do this we use Spaces, Lemma 65.8.4 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 78.22.3 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 65.8.4 and we win. $\square$

Lemma 80.11.6. 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 78.19.5 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 67.30.4 and 67.28.3. Hence by Descent on Spaces, Lemmas 74.11.13 and 74.11.10 we see that $U \to F$ is flat and locally of finite presentation. It is surjective by Spaces, Remark 65.5.2. $\square$

Lemma 80.11.7. 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 78.19.1) is an algebraic space, the morphism $X \to X/G$ is surjective, flat, and locally of finite presentation, and $X$ is an fppf $G$-torsor over $X/G$.

Proof. The fact that $X/G$ is an algebraic space is immediate from Theorem 80.10.1 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 78.8.3 as the action is free. The morphism $X \to X/G$ is surjective, flat, and locally of finite presentation by Lemma 80.11.6. To see that $X \to X/G$ is an fppf $G$-torsor (Groupoids in Spaces, Definition 78.9.3) we have to show that $G \times _ S X \to X \times _{X/G} X$ is an isomorphism and that $X \to X/G$ fppf locally has sections. The second part is clear from the properties of $X \to X/G$ already shown. The map $G \times _ S X \to X \times _{X/G} X$ is injective (as a map of fppf sheaves) as the action is free. Finally, the map is also surjective as a map of sheaves by Groupoids in Spaces, Lemma 78.19.5. This finishes the proof. $\square$

Lemma 80.11.8. 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 65.16. 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 80.11.1 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 78.9.3). 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 34.7.7 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 65.8.4 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 80.11.2.

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