The Stacks project

Proof. See Bootstrap, Lemma 80.3.8. $\square$

Proof. See Bootstrap, Lemma 80.3.3. $\square$

Proof. Note that the diagonal of $Y \to Z$ is representable by Lemma 87.15.5. Thus $X \to Y$ is representable by algebraic spaces by Bootstrap, Lemma 80.3.10. $\square$

Proof. It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, (2) implies (1) by Bootstrap, Lemma 80.11.4 applied to $U \to X \to Y$.

Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is representable by algebraic spaces (as the composition $U \to X \to Y$, see Bootstrap, Lemma 80.3.8) and factors through $V$. Thus $U \to V$ is representable by algebraic spaces by Lemma 87.19.3.

It is clear that (3) implies (2). Thus now (1) – (3) are equivalent.

Observe that the condition in (4) makes sense as the fibre product $Y_ j \times _ Y X$ is a formal algebraic space by Lemma 87.15.3. It is clear that (4) implies (5).

Assume $X_ i \to Y_ i \to Y$ as in (5). Then we set $V = \coprod Y_ i$ and $U = \coprod X_ i$ to see that (5) implies (2).

Finally, assume (1) – (3) are true. Thus we can choose any covering $\{ Y_ j \to Y\} $ as in Definition 87.11.1 and for each $j$ any covering $\{ X_{ji} \to Y_ j \times _ Y X\} $ as in Definition 87.11.1. Then $X_{ij} \to Y_ j$ is representable by algebraic spaces by (3) and we see that (4) is true. This concludes the proof. $\square$

Proof. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda $ as in Definition 87.9.1. For each $\lambda $ the fibre product $X \times _ Y Y_\lambda $ is an algebraic space. Hence $X = \mathop{\mathrm{colim}}\nolimits X \times _ Y Y_\lambda $ is a formal algebraic space by Lemma 87.13.1. $\square$

Proof. Let $\{ Y_ i \to Y\} $ be as in Definition 87.11.1. Then $X \times _ Y Y_ i \to X$ is a family of morphisms representable by algebraic spaces, étale, and jointly surjective. Thus it suffices to show that $X \times _ Y Y_ i$ is a formal algebraic space, see Lemma 87.15.1. This follows from Lemma 87.19.5. $\square$

Proof. We will show that $f$ is affine; it will then follow that $f$ is representable and affine by Morphisms of Spaces, Lemma 67.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 87.9.1. Let $T \to Y$ be a morphism where $T$ is a scheme over $S$. We have to show that $X \times _ Y T \to T$ is affine, see Bootstrap, Definition 80.4.1. To do this we may assume that $T$ is affine and we have to prove that $X \times _ Y T$ is affine. In this case $T \to Y$ factors through $Y_\mu \to Y$ for some $\mu $, see Lemma 87.9.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 87.17.3). Hence $X \times _ Y T \to X$ factors through $X_\lambda $ for some $\lambda $. Similarly $X_\lambda \to Y$ factors through $Y_\mu $ after increasing $\mu $. Then $X \times _ Y T = X_\lambda \times _{Y_\mu } T$. We conclude as fibre products of affine schemes are affine. $\square$

Proof. Parts (1) and (2) are equivalent by Lemma 87.19.7.

Assume the equivalent conditions (1) and (2) hold. If $I \subset A$ is a weak ideal of definition, then $\mathop{\mathrm{Spec}}(A/I) \to \text{Spf}(A)$ is representable and a thickening (this is clear from the construction of the formal spectrum but it also follows from Lemma 87.9.6). Then $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) \to \text{Spf}(B)$ is representable and a thickening as a base change. Hence by Lemma 87.9.6 there is a weak ideal of definition $J(I) \subset B$ such that $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) = \mathop{\mathrm{Spec}}(B/J(I))$ as subfunctors of $\text{Spf}(B)$. We obtain a cartesian diagram

By Lemma 87.16.4 we see that $B/J(I) = B \widehat{\otimes }_ A A/I$. It follows that $J(I)$ is the closure of the ideal $\varphi (I)B$, see Lemma 87.4.11. Since $\text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ with $I$ as above, we find that $\text{Spf}(B) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J(I))$. Thus the ideals $J(I)$ form a fundamental system of weak ideals of definition (see Lemma 87.9.6). Hence (3) holds.

Assume (3) holds. We are essentially just going to reverse the arguments given in the previous paragraph. Let $I \subset A$ be a weak ideal of definition. By Lemma 87.16.4 we get a cartesian diagram

If $J(I)$ is the closure of $IB$, then $J(I)$ is open in $B$ by tautness of $\varphi $. Hence if $J$ is open in $B$ and $J \subset J(B)$, then $B/J \otimes _ A A/I = B/(IB + J) = B/J(I)$ because $J(I) = \bigcap _{J \subset B\text{ open}} (IB + J)$ by Lemma 87.4.2. Hence the limit defining the completed tensor product collapses to give $B \widehat{\otimes }_ A A/I = B/J(I)$. Thus $\text{Spf}(B \widehat{\otimes }_ A A/I) = \mathop{\mathrm{Spec}}(B/J(I))$. This proves that $\text{Spf}(B) \times _{\text{Spf}(A)} \mathop{\mathrm{Spec}}(A/I)$ is representable for every weak ideal of definition $I \subset A$. Since every morphism $T \to \text{Spf}(A)$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(A/I)$ for some weak ideal of definition $I$ (Lemma 87.9.4) we conclude that $\text{Spf}(\varphi )$ is representable, i.e., (2) holds. This finishes the proof. $\square$

Proof. Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 87.9.1. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times _ Y X$ are affine. And $X = \mathop{\mathrm{colim}}\nolimits Y_\lambda \times _ Y X$. Thus $X$ is an affine formal algebraic space.

Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda $ is countable. Then we immediately see that $X$ is countably indexed too.

Proof of (3), (4), and (5). In each of these cases the assumptions imply that $Y$ is a countably indexed affine formal algebraic space (Lemma 87.10.3) and hence $X$ is too by (2). Thus we may write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebras $A$ and $B$, see Lemma 87.10.4. By Lemma 87.9.10 the morphism $f$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 87.19.8 that $\varphi $ is taut. We conclude that (3) follows from Lemma 87.5.9, (4) follows from Lemma 87.7.5, and (5) follows from Lemma 87.6.5.

Proof of (6). Combining (3) with Lemma 87.10.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 87.10.5. First, it tells us the affine schemes $Y_\lambda $ are Noetherian. Then $X_\lambda \to Y_\lambda $ is of finite type, hence $X_\lambda $ is Noetherian too (Morphisms, Lemma 29.15.6). Then the criterion tells us $X$ is Noetherian and the proof is complete. $\square$

Proof. Combine Lemmas 87.19.7 and 87.19.9. $\square$

Proof. Let $A$ be an étale $B$-algebra. Then $B/J \to A/JA$ is étale for every open ideal $J \subset B$. Hence the morphism $\text{Spf}(A^\wedge ) \to Y$ is representable and étale. The functor $\text{Spf}$ is fully faithful by Lemma 87.9.10. To finish the proof we will show in the next paragraph that any $X \to Y$ as in (1) is in the essential image.

Choose a weak ideal of definition $J_0 \subset B$. Set $Y_0 = \mathop{\mathrm{Spec}}(B/J_0)$ and $X_0 = Y_0 \times _ Y X$. Then $X_0 \to Y_0$ is an étale morphism of affine schemes (see Lemma 87.19.7). Say $X_0 = \mathop{\mathrm{Spec}}(A_0)$. By Algebra, Lemma 10.143.10 we can find an étale algebra map $B \to A$ such that $A_0 \cong A/J_0A$. Consider an ideal of definition $J \subset J_0$. As above we may write $\mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{Spec}}(\bar A)$ for some étale ring map $B/J \to \bar A$. Then both $B/J \to \bar A$ and $B/J \to A/JA$ are étale ring maps lifting the étale ring map $B/J_0 \to A_0$. By More on Algebra, Lemma 15.11.2 there is a unique $B/J$-algebra isomorphism $\varphi _ J : A/JA \to \bar A$ lifting the identification modulo $J_0$. Since the maps $\varphi _ J$ are unique they are compatible for varying $J$. Thus

and we see that the lemma holds. $\square$

Proof. Let $J \subset B$ be a weak ideal of definition. As every element of $J$ is topologically nilpotent, we see that every element of $1 + J$ is a unit. It follows that $J$ is contained in the Jacobson radical of $B$ (Algebra, Lemma 10.19.1). Hence a flat ring map $B \to A$ is faithfully flat if and only if $B/J \to A/JA$ is faithfully flat (Algebra, Lemma 10.39.16). In this way we see that (2) – (4) are equivalent. If (1) holds, then for every weak ideal of definition $J \subset B$ the morphism $\mathop{\mathrm{Spec}}(A/JA) = \mathop{\mathrm{Spec}}(B/J) \times _ Y X \to \mathop{\mathrm{Spec}}(B/J)$ is surjective which implies (3). Conversely, assume (3). A morphism $T \to Y$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(B/J)$ for some ideal of definition $J$ of $B$ (Lemma 87.9.4). Hence $X \times _ Y T = \mathop{\mathrm{Spec}}(A/JA) \times _{\mathop{\mathrm{Spec}}(B/J)} T \to T$ is surjective as a base change of the surjective morphism $\mathop{\mathrm{Spec}}(A/JA) \to \mathop{\mathrm{Spec}}(B/J)$. Thus (1) holds. $\square$