## 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.

slogan
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

$F$ is a sheaf,

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

$\coprod _{i \in I} F_ i$ 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

$F$ is a sheaf,

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

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.

slogan
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 $S$.

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

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

$F, G, H$ are sheaves,

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

$b \circ a : F \to H$ 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

$a$ is a free action, and

$G \to B$ 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

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

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.
$\square$

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