The Stacks project

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.8.8. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is quasi-compact. We have to show that $f$ is quasi-compact. To see this, using Morphisms of Spaces, Lemma 67.8.8 again, it is enough to show that for every affine scheme $Y$ and morphism $Y \to Z$ the fibre product $Y \times _ Z X$ is quasi-compact. Here is a picture:

Note that all squares are cartesian and the bottom square consists of affine schemes. The assumption that $f'$ is quasi-compact combined with the fact that $Y \times _ Z Z'$ is affine implies that $Y \times _ Z Z' \times _ Z X$ is quasi-compact. Since

is surjective as a base change of $Z' \to Z$ we conclude that $Y \times _ Z X$ is quasi-compact, see Morphisms of Spaces, Lemma 67.8.6. This finishes the proof. $\square$

Proof. A base change of a quasi-separated morphism is quasi-separated, see Morphisms of Spaces, Lemma 67.4.4. Hence the direct implication in Definition 74.10.1.

Let $\{ Y_ i \to Y\} _{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_ i := Y_ i \times _ Y X \to Y_ i$ is quasi-separated. This means that each of the morphisms

is quasi-compact. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 73.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} $ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 74.11.1 that $\Delta $ is quasi-compact, i.e., $f$ is quasi-separated. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.9.5. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is universally closed. We have to show that $f$ is universally closed. To see this, using Morphisms of Spaces, Lemma 67.9.5 again, it is enough to show that for every affine scheme $Y$ and morphism $Y \to Z$ the map $|Y \times _ Z X| \to |Y|$ is closed. Consider the cube (74.11.1.1). The assumption that $f'$ is universally closed implies that $|Y \times _ Z Z' \times _ Z X| \to |Y \times _ Z Z'|$ is closed. As $Y \times _ Z Z' \to Y$ is quasi-compact, surjective, and flat as a base change of $Z' \to Z$ we see the map $|Y \times _ Z Z'| \to |Y|$ is submersive, see Morphisms, Lemma 29.25.12. Moreover the map

is surjective, see Properties of Spaces, Lemma 66.4.3. It follows by elementary topology that $|Y \times _ Z X| \to |Y|$ is closed. $\square$

Proof. The proof is the same as the proof of Lemma 74.11.3. $\square$

Proof. The proof is the same as the proof of Lemma 74.11.3. $\square$

Proof. Omitted. (Hint: Use Properties of Spaces, Lemma 66.4.3.) $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.9.5. Let $Z' \to Z$ be a flat surjective morphism of affine schemes over $S$ and let $f : X \to Z$ be a morphism from an algebraic space to $Z$. Assume that the base change $f' : X' \to Z'$ is universally injective. Let $K$ be a field, and let $a, b : \mathop{\mathrm{Spec}}(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. As $Z' \to Z$ is surjective there exists a field extension $K'/K$ and a morphism $\mathop{\mathrm{Spec}}(K') \to Z'$ such that the following solid diagram commutes

As the square is cartesian we get the two dotted arrows $a'$, $b'$ making the diagram commute. Since $X' \to Z'$ is universally injective we get $a' = b'$. This forces $a = b$ as $\{ \mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)\} $ is an fpqc covering, see Properties of Spaces, Proposition 66.17.1. Hence $f$ is universally injective as desired. $\square$

Proof. This can be proved in exactly the same manner as Lemma 74.11.3. Alternatively, one can use that a map of topological spaces is a homeomorphism if and only if it is injective, surjective, and open. Thus a universal homeomorphism is the same thing as a surjective, universally injective, and universally open morphism. See Morphisms of Spaces, Lemma 67.5.5 and Morphisms of Spaces, Definitions 67.19.3, 67.5.2, 67.6.2, 67.53.2. Thus the lemma follows from Lemmas 74.11.6, 74.11.7, and 74.11.4. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.23.4. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is locally of finite type. We have to show that $f$ is locally of finite type. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.23.4 again, it is enough to show that $U \to Z$ is locally of finite type. Since $f'$ is locally of finite type, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is locally of finite type. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is locally of finite type by Descent, Lemma 35.23.10 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.28.4. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is locally of finite presentation. We have to show that $f$ is locally of finite presentation. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.28.4 again, it is enough to show that $U \to Z$ is locally of finite presentation. Since $f'$ is locally of finite presentation, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is locally of finite presentation. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is locally of finite presentation by Descent, Lemma 35.23.11 as desired. $\square$

Proof. Combine Lemmas 74.11.1 and 74.11.9. $\square$

Proof. Combine Lemmas 74.11.1, 74.11.2 and 74.11.10. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.30.5. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is flat. We have to show that $f$ is flat. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.30.5 again, it is enough to show that $U \to Z$ is flat. Since $f'$ is flat, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is flat. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is flat by Descent, Lemma 35.23.15 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.12.1. Consider a cartesian diagram

of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is an open immersion. We have to show that $X \to Z$ is an open immersion. Note that $|X'| \subset |Z'|$ corresponds to an open subscheme $U' \subset Z'$ (isomorphic to $X'$) with the property that $\text{pr}_0^{-1}(U') = \text{pr}_1^{-1}(U')$ as open subschemes of $Z' \times _ Z Z'$. Hence there exists an open subscheme $U \subset Z$ such that $X' = (Z' \to Z)^{-1}(U)$, see Descent, Lemma 35.13.6. By Properties of Spaces, Proposition 66.17.1 we see that $X$ satisfies the sheaf condition for the fpqc topology. Now we have the fpqc covering $\mathcal{U} = \{ U' \to U\} $ and the element $U' \to X' \to X \in \check{H}^0(\mathcal{U}, X)$. By the sheaf condition we obtain a morphism $U \to X$ such that

is commutative. On the other hand, we know that for any scheme $T$ over $S$ and $T$-valued point $T \to X$ the composition $T \to X \to Z$ is a morphism such that $Z' \times _ Z T \to Z'$ factors through $U'$. Clearly this means that $T \to Z$ factors through $U$. In other words the map of sheaves $U \to X$ is bijective and we win. $\square$

Proof. Combine Lemmas 74.11.6 and 74.11.14. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.20.3. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is affine. Let $X'$ be a scheme representing $Z' \times _ Z X$. We obtain a canonical isomorphism

since both schemes represent the algebraic space $Z' \times _ Z Z' \times _ Z X$. This is a descent datum for $X'/Z'/Z$, see Descent, Definition 35.34.1 (verification omitted, compare with Descent, Lemma 35.39.1). Since $X' \to Z'$ is affine this descent datum is effective, see Descent, Lemma 35.37.1. Thus there exists a scheme $Y \to Z$ over $Z$ and an isomorphism $\psi : Z' \times _ Z Y \to X'$ compatible with descent data. Of course $Y \to Z$ is affine (by construction or by Descent, Lemma 35.23.18). Note that $\mathcal{Y} = \{ Z' \times _ Z Y \to Y\} $ is a fpqc covering, and interpreting $\psi $ as an element of $X(Z' \times _ Z Y)$ we see that $\psi \in \check{H}^0(\mathcal{Y}, X)$. By the sheaf condition for $X$ with respect to this covering (see Properties of Spaces, Proposition 66.17.1) we obtain a morphism $Y \to X$. By construction the base change of this to $Z'$ is an isomorphism, hence an isomorphism by Lemma 74.11.15. This proves that $X$ is representable by an affine scheme and we win. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.12.1. Consider a cartesian diagram

of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is a closed immersion. We have to show that $X \to Z$ is a closed immersion. The morphism $X' \to Z'$ is affine. Hence by Lemma 74.11.16 we see that $X$ is a scheme and $X \to Z$ is affine. It follows from Descent, Lemma 35.23.19 that $X \to Z$ is a closed immersion as desired. $\square$

Proof. A base change of a separated morphism is separated, see Morphisms of Spaces, Lemma 67.4.4. Hence the direct implication in Definition 74.10.1.

Let $\{ Y_ i \to Y\} _{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_ i := Y_ i \times _ Y X \to Y_ i$ is separated. This means that each of the morphisms

is a closed immersion. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 73.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} $ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 74.11.17 that $\Delta $ is a closed immersion, i.e., $f$ is separated. $\square$

Proof. The lemma follows by combining Lemmas 74.11.3, 74.11.18 and 74.11.11. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.21.3. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is quasi-affine. Let $X'$ be a scheme representing $Z' \times _ Z X$. We obtain a canonical isomorphism

since both schemes represent the algebraic space $Z' \times _ Z Z' \times _ Z X$. This is a descent datum for $X'/Z'/Z$, see Descent, Definition 35.34.1 (verification omitted, compare with Descent, Lemma 35.39.1). Since $X' \to Z'$ is quasi-affine this descent datum is effective, see Descent, Lemma 35.38.1. Thus there exists a scheme $Y \to Z$ over $Z$ and an isomorphism $\psi : Z' \times _ Z Y \to X'$ compatible with descent data. Of course $Y \to Z$ is quasi-affine (by construction or by Descent, Lemma 35.23.20). Note that $\mathcal{Y} = \{ Z' \times _ Z Y \to Y\} $ is a fpqc covering, and interpreting $\psi $ as an element of $X(Z' \times _ Z Y)$ we see that $\psi \in \check{H}^0(\mathcal{Y}, X)$. By the sheaf condition for $X$ (see Properties of Spaces, Proposition 66.17.1) we obtain a morphism $Y \to X$. By construction the base change of this to $Z'$ is an isomorphism, hence an isomorphism by Lemma 74.11.15. This proves that $X$ is representable by a quasi-affine scheme and we win. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemmas 67.12.1 and 67.8.8. Consider a cartesian diagram

of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is a quasi-compact immersion. We have to show that $X \to Z$ is a closed immersion. The morphism $X' \to Z'$ is quasi-affine. Hence by Lemma 74.11.20 we see that $X$ is a scheme and $X \to Z$ is quasi-affine. It follows from Descent, Lemma 35.23.21 that $X \to Z$ is a quasi-compact immersion as desired. $\square$

Proof. An integral morphism is the same thing as an affine, universally closed morphism. See Morphisms of Spaces, Lemma 67.45.7. Hence the lemma follows on combining Lemmas 74.11.3 and 74.11.16. $\square$

Proof. An finite morphism is the same thing as an integral, morphism which is locally of finite type. See Morphisms of Spaces, Lemma 67.45.6. Hence the lemma follows on combining Lemmas 74.11.9 and 74.11.22. $\square$

Proof. We have already seen that “quasi-compact” is fpqc local on the base, see Lemma 74.11.1. Hence it is enough to prove the lemma for “locally quasi-finite”. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.27.6. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is locally quasi-finite. We have to show that $f$ is locally quasi-finite. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.27.6 again, it is enough to show that $U \to Z$ is locally quasi-finite. Since $f'$ is locally quasi-finite, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is locally quasi-finite. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is locally quasi-finite by Descent, Lemma 35.23.24 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.36.4. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is syntomic. We have to show that $f$ is syntomic. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.36.4 again, it is enough to show that $U \to Z$ is syntomic. Since $f'$ is syntomic, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is syntomic. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is syntomic by Descent, Lemma 35.23.26 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.37.4. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is smooth. We have to show that $f$ is smooth. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.37.4 again, it is enough to show that $U \to Z$ is smooth. Since $f'$ is smooth, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is smooth. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is smooth by Descent, Lemma 35.23.27 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.38.5. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is unramified. We have to show that $f$ is unramified. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.38.5 again, it is enough to show that $U \to Z$ is unramified. Since $f'$ is unramified, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is unramified. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is unramified by Descent, Lemma 35.23.28 as desired. $\square$

Proof. We will use Lemma 74.10.4 to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma 67.39.2. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times _ Z X \to Z'$ is étale. We have to show that $f$ is étale. Let $U$ be a scheme and let $U \to X$ be surjective and étale. By Morphisms of Spaces, Lemma 67.39.2 again, it is enough to show that $U \to Z$ is étale. Since $f'$ is étale, and since $Z' \times _ Z U$ is a scheme étale over $Z' \times _ Z X$ we conclude (by the same lemma again) that $Z' \times _ Z U \to Z'$ is étale. As $\{ Z' \to Z\} $ is an fpqc covering we conclude that $U \to Z$ is étale by Descent, Lemma 35.23.29 as desired. $\square$

Proof. Being finite locally free is equivalent to being finite, flat and locally of finite presentation (Morphisms of Spaces, Lemma 67.46.6). Hence this follows from Lemmas 74.11.23, 74.11.13, and 74.11.10. $\square$

Proof. Let $f : X \to Y$ be a morphism of algebraic spaces. Let $\{ Y_ i \to Y\} $ be an fpqc covering, and assume each of the base changes $f_ i : X_ i \to Y_ i$ of $f$ is a monomorphism. We have to show that $f$ is a monomorphism.

First proof. Note that $f$ is a monomorphism if and only if $\Delta : X \to X \times _ Y X$ is an isomorphism. By applying this to $f_ i$ we see that each of the morphisms

is an isomorphism. The base change of an fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma 73.9.3 hence $\{ Y_ i \times _ Y (X \times _ Y X) \to X \times _ Y X\} $ is an fpqc covering of algebraic spaces. Moreover, each $\Delta _ i$ is the base change of the morphism $\Delta : X \to X \times _ Y X$. Hence it follows from Lemma 74.11.15 that $\Delta $ is an isomorphism, i.e., $f$ is a monomorphism.

Second proof. Let $V$ be a scheme, and let $V \to Y$ be a surjective étale morphism. If we can show that $V \times _ Y X \to V$ is a monomorphism, then it follows that $X \to Y$ is a monomorphism. Namely, given any cartesian diagram of sheaves

if $c$ is a surjection of sheaves, and $a$ is injective, then also $d$ is injective. This reduces the problem to the case where $Y$ is a scheme. Moreover, in this case we may assume that the algebraic spaces $Y_ i$ are schemes also, since we can always refine the covering to place ourselves in this situation, see Topologies on Spaces, Lemma 73.9.5.

Assume $\{ Y_ i \to Y\} $ is an fpqc covering of schemes. Let $a, b : T \to X$ be two morphisms such that $f \circ a = f \circ b$. We have to show that $a = b$. Since $f_ i$ is a monomorphism we see that $a_ i = b_ i$, where $a_ i, b_ i : Y_ i \times _ Y T \to X_ i$ are the base changes. In particular the compositions $Y_ i \times _ Y T \to T \to X$ are equal. Since $\{ Y_ i \times _ Y T \to T\} $ is an fpqc covering we deduce that $a = b$ from Properties of Spaces, Proposition 66.17.1. $\square$