The Stacks project

Proof. Assume $X$ is countably indexed. We write $X = \mathop{\mathrm{colim}}\nolimits X_ n$ as in Lemma 87.10.1. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda$ as in Definition 87.9.1. For every $n$ we can pick a $\lambda _ n$ such that $X_ n \to Y$ factors through $Y_{\lambda _ n}$, see Lemma 87.9.4. On the other hand, for every $\lambda$ the scheme $Y_\lambda \times _ Y X$ is affine (Lemma 87.19.7) and hence $Y_\lambda \times _ Y X \to X$ factors through $X_ n$ for some $n$ (Lemma 87.9.4). Picture

If we can show the dotted arrow exists, then we conclude that $Y = \mathop{\mathrm{colim}}\nolimits Y_{\lambda _ n}$ and $Y$ is countably indexed. To do this we pick a $\mu$ with $\mu \geq \lambda$ and $\mu \geq \lambda _ n$. Thus both $Y_\lambda \to Y$ and $Y_{\lambda _ n} \to Y$ factor through $Y_\mu \to Y$. Say $Y_\mu = \mathop{\mathrm{Spec}}(B_\mu )$, the closed subscheme $Y_\lambda$ corresponds to $J \subset B_\mu$, and the closed subscheme $Y_{\lambda _ n}$ corresponds to $J' \subset B_\mu$. We are trying to show that $J' \subset J$. By the diagram above we know $J'A_\mu \subset JA_\mu$ where $Y_\mu \times _ Y X = \mathop{\mathrm{Spec}}(A_\mu )$. Since $X \to Y$ is surjective and flat the morphism $Y_\lambda \times _ Y X \to Y_\lambda$ is a faithfully flat morphism of affine schemes, hence $B_\mu \to A_\mu$ is faithfully flat. Thus $J' \subset J$ as desired.

Assume $Y$ is countably indexed. Then $X$ is countably indexed by Lemma 87.19.10. $\square$

Proof. We have already seen the implication in one direction in Lemma 87.19.10. For the other direction, note that by Lemma 87.20.1 we may assume both $X$ and $Y$ are countably indexed. Thus $X = \text{Spf}(A)$ 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 $X \to Y$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 87.19.8 that $\varphi$ is taut. Let $J \subset B$ be an open ideal and let $I \subset A$ be the closure of $JA$. By Lemmas 87.16.4 and 87.4.11 we see that $\mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{Spec}}(A/I)$. Hence $B/J \to A/I$ is faithfully flat (since $X \to Y$ is surjective and flat). This means that $\varphi : B \to A$ is as in Section 87.8 (with the roles of $A$ and $B$ swapped). We conclude that the lemma holds by Lemma 87.8.2. $\square$

Proof. The proof is exactly the same as the proof of Lemma 87.20.2 except that at the end we use Lemma 87.8.4. $\square$

Proof. The proof is exactly the same as the proof of Lemma 87.20.2 except that at the end we use Lemma 87.8.5. $\square$

Proof. Observe that a Noetherian affine formal algebraic space is adic*, see Lemma 87.10.3. Thus by Lemma 87.20.4 we may assume that both $X$ and $Y$ are adic*. We will use the criterion of Lemma 87.10.5 to see that the lemma holds. Namely, write $Y = \mathop{\mathrm{colim}}\nolimits Y_ n$ as in Lemma 87.10.1. For each $n$ set $X_ n = Y_ n \times _ Y X$. Then $X_ n$ is an affine scheme (Lemma 87.19.7) and $X = \mathop{\mathrm{colim}}\nolimits X_ n$. Each of the morphisms $X_ n \to Y_ n$ is faithfully flat and of finite type. Thus the lemma follows from the fact that in this situation $X_ n$ is Noetherian if and only if $Y_ n$ is Noetherian, see Algebra, Lemma 10.164.1 (to go down) and Algebra, Lemma 10.31.1 (to go up). $\square$

Proof. It is clear that (1) implies (2). Assume (2) and let $Y \to X$ be as in (1). Since the fibre products $X_ i \times _ X Y$ are formal algebraic spaces (Lemma 87.15.2) we can pick coverings $\{ X_{ij} \to X_ i \times _ X Y\}$ as in Definition 87.11.1. Since $Y$ is quasi-compact, there exist $(i_1, j_1), \ldots , (i_ n, j_ n)$ such that

is surjective and étale. Then $X_{i_ kj_ k} \to X_{i_ k}$ is representable by algebraic spaces and étale hence $X_{i_ kj_ k}$ has property $P$ by Lemma 87.19.10. Then $X_{i_1 j_1} \amalg \ldots \amalg X_{i_ n j_ n}$ is an affine formal algebraic space with property $P$ (small detail omitted on finite disjoint unions of affine formal algebraic spaces). Hence we conclude by applying one of Lemmas 87.20.1, 87.20.2, 87.20.3, 87.20.4, and 87.20.5. $\square$

Proof. Choose a surjective étale morphism $U \to X$ with $U = \coprod U_ i$ a disjoint union of affine schemes, see Properties of Spaces, Lemma 66.6.1. Let $T_ i \subset U_ i$ be the inverse image of $T$. We have $X_{/T} \times _ X U_ i = (U_ i)_{/T_ i}$ (Lemma 87.14.4). Hence $\{ (U_ i)_{/T_ i} \to X_{/T}\}$ is a covering as in Definition 87.11.1. Moreover, if $X \setminus T \to X$ is quasi-compact, so is $U_ i \setminus T_ i \to U_ i$ and if $X$ is locally Noetherian, so is $U_ i$. Thus the lemma follows from the affine case which is Lemma 87.14.6. $\square$

Proof. Choose a covering $\{ Y_ j \to Y\}$ as in Definition 87.11.1. For each $j$ choose a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 87.11.1. For each $j$ choose a covering $\{ Z_{jk} \to Y_ j \times _ Y Z\}$ as in Definition 87.11.1. Observe that $X_{ji} \times _{Y_ j} Z_{jk}$ is an affine formal algebraic space by Lemma 87.16.4. Hence

is a covering as in Definition 87.11.1. Thus it suffices to prove (1), (2), (3), and (4) in case $X$, $Y$, and $Z$ are affine formal algebraic spaces.

Assume $X$ and $Z$ are countably indexed. Say $X = \mathop{\mathrm{colim}}\nolimits X_ n$ and $Z = \mathop{\mathrm{colim}}\nolimits Z_ m$ as in Lemma 87.10.1. Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda$ as in Definition 87.9.1. For each $n$ and $m$ we can find $\lambda _{n, m} \in \Lambda$ such that $X_ n \to Y$ and $Z_ m \to Y$ factor through $Y_{\lambda _{n, m}}$ (for example see Lemma 87.9.4). Pick $\lambda _0 \in \Lambda$. By induction for $t \geq 1$ pick an element $\lambda _ t \in \Lambda$ such that $\lambda _ t \geq \lambda _{n, m}$ for all $1 \leq n, m \leq t$ and $\lambda _ t \geq \lambda _{t - 1}$. Set $Y' = \mathop{\mathrm{colim}}\nolimits Y_{\lambda _ t}$. Then $Y' \to Y$ is a monomorphism such that $X \to Y$ and $Z \to Y$ factor through $Y'$. Hence we may replace $Y$ by $Y'$, i.e., we may assume that $Y$ is countably indexed.

Assume $X$, $Y$, and $Z$ are countably indexed. By Lemma 87.10.4 we can write $X = \text{Spf}(A)$, $Y = \text{Spf}(B)$, $Z = \text{Spf}(C)$ for some weakly admissible topological rings $A$, $B$, and $C$. The morphsms $X \to Y$ and $Z \to Y$ are given by continuous ring maps $B \to A$ and $B \to C$, see Lemma 87.9.10. By Lemma 87.16.4 we see that $X \times _ Y Z = \text{Spf}(A \widehat{\otimes }_ B C)$ and that $A \widehat{\otimes }_ B C$ is a weakly admissible topological ring. In particular, we see that $X \times _ Y Z$ is countably indexed by Lemma 87.4.12 part (3). This proves (1).

Proof of (2). In this case $X$ and $Z$ are countably indexed and hence the arguments above show that $X \times _ Y Z$ is the formal spectrum of $A \widehat{\otimes }_ B C$ where $A$ and $C$ are admissible. Then $A \widehat{\otimes }_ B C$ is admissible by Lemma 87.4.12 part (2).

Proof of (3). As before we conclude that $X \times _ Y Z$ is the formal spectrum of $A \widehat{\otimes }_ B C$ where $A$ and $C$ are weakly adic. Then $A \widehat{\otimes }_ B C$ is weakly adic by Lemma 87.7.6.

Proof of (4). Arguing as above, this follows from Lemma 87.4.12 part (4).

Proof of (5). To deduce case (5) from Lemma 87.4.12 part (5) we need to show the hypotheses match. Namely, with notation as in the first parapgrah of the proof, if $X_{red} \to Y_{red}$ is locally of finite type, then $(X_{ji})_{red} \to (Y_ j)_{red}$ is locally of finite type. This follows from Morphisms of Spaces, Lemma 67.23.4 and the fact that in the commutative diagram

the vertical morphisms are étale. Namely, we have $(X_{ji})_{red} = X_{ij} \times _ X X_{red}$ and $(Y_ j)_{red} = Y_ j \times _ Y Y_{red}$ by Lemma 87.12.3. Thus as above we reduce to the case where $X$, $Y$, $Z$ are affine formal algebraic spaces, $X$, $Z$ are Noetherian, and $X_{red} \to Y_{red}$ is of finite type. Next, in the second paragraph of the proof we replaced $Y$ by $Y'$ but by construction $Y_{red} = Y'_{red}$, hence the finite type assumption is preserved by this replacement. Then we see that $X, Y, Z$ correspond to $A, B, C$ and $X \times _ Y Z$ to $A \widehat{\otimes }_ B C$ with $A$, $C$ Noetherian adic. Finally, taking the reduction corresponds to dividing by the ideal of topologically nilpotent elements (Example 87.12.2) hence the fact that $X_{red} \to Y_{red}$ is of finite type does indeed mean that $B/\mathfrak b \to A/\mathfrak a$ is of finite type and the proof is complete. $\square$

Proof. We only sketch the proof and omit some of the details. Set $X_1 = X_{red}$. Define $X_ n \subset X$ as the subfunctor defined by the rule: a morphism $f : T \to X$ where $T$ is a scheme factors through $X_ n$ if and only if the $n$th power of the ideal sheaf of the closed immersion $X_1 \times _ X T \to T$ is zero. Then $X_ n \subset X$ is a subsheaf as vanishing of quasi-coherent modules can be checked fppf locally. We claim that $X_ n \to X$ is representable by schemes, a closed immersion, and that $X = \mathop{\mathrm{colim}}\nolimits X_ n$ (as fppf sheaves). To check this we may work étale locally on $X$. Hence we may assume $X = \text{Spf}(A)$ is a Noetherian affine formal algebraic space. Then $X_1 = \mathop{\mathrm{Spec}}(A/\mathfrak a)$ where $\mathfrak a \subset A$ is the ideal of topologically nilpotent elements of the Noetherian adic topological ring $A$. Then $X_ n = \mathop{\mathrm{Spec}}(A/\mathfrak a^ n)$ and we obtain what we want. $\square$