The Stacks project

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. $\square$

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

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. $\square$

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

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. $\square$

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. $\square$

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

Hence, by Sets, Lemma 3.9.5 we conclude that

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. $\square$

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

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. $\square$

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. $\square$

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. $\square$