86.16 Types of formal algebraic spaces

In this section we define Noetherian, adic*, and countably indexed formal algebraic spaces. The types adic, classical, and McQuillan are missing as we do not know how to prove the analogue of the following lemmas for those cases.

Lemma 86.16.1. Let $S$ be a scheme. Let $X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, and flat. Then $X$ is countably indexed if and only if $Y$ is countably indexed.

Proof. Assume $X$ is countably indexed. We write $X = \mathop{\mathrm{colim}}\nolimits X_ n$ as in Lemma 86.6.1. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda$ as in Definition 86.5.1. For every $n$ we can pick a $\lambda _ n$ such that $X_ n \to Y$ factors through $Y_{\lambda _ n}$, see Lemma 86.5.4. On the other hand, for every $\lambda$ the scheme $Y_\lambda \times _ Y X$ is affine (Lemma 86.15.7) and hence $Y_\lambda \times _ Y X \to X$ factors through $X_ n$ for some $n$ (Lemma 86.5.4). Picture

$\xymatrix{ Y_\lambda \times _ Y X \ar[r] \ar[d] & X_ n \ar[r] \ar[d] & X \ar[d] \\ Y_\lambda \ar@{..>}[r] \ar@/_1pc/[rr] & Y_{\lambda _ n} \ar[r] & Y }$

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 86.15.9. $\square$

Lemma 86.16.2. Let $S$ be a scheme. Let $X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, and flat. Then $X$ is adic* if and only if $Y$ is adic*.

Proof. Assume $Y$ is adic*. Then $X$ is adic* by Lemma 86.15.9.

Assume $X$ is adic*. Write $X = \text{Spf}(A)$ for some adic ring $A$ which has a finitely generated ideal $I$ such that $\{ I^ n\}$ is a fundamental system of open ideals. By Lemmas 86.16.1 we see that $Y$ is countably indexed. Thus, by Lemma 86.6.4, we can write $Y = \text{Spf}(B)$ where $B$ is a weakly admissible topological ring with a countable fundamental system $\{ J_ m\}$ of weak ideals of definition. By Lemma 86.5.10 the morphism $X \to Y$ corresponds to a continuous ring map $B \to A$ which is taut by Lemma 86.15.10. Our first goal is to reduce to the case where $J_ m A$ is equal to $I^ m$.

Set $Y_ m = \mathop{\mathrm{Spec}}(B/J_ m)$ so that $Y = \mathop{\mathrm{colim}}\nolimits Y_ m$. The scheme $Y_ m \times _ Y X$ is affine (Lemma 86.15.7) and we have $X = \mathop{\mathrm{colim}}\nolimits Y_ m \times _ Y X$. Say $Y_ m \times _ Y X = \mathop{\mathrm{Spec}}(A_ m)$ so that $B/J_ m \to A_ m$ is a faithfully flat ring map. It follows from Lemma 86.12.4 that $\mathop{\mathrm{Ker}}(A \to A_ m)$ is the closure of $J_ mA$.

Choose $n \geq 1$. There exists an $m$ such that $\mathop{\mathrm{Spec}}(A/I^ n) \to Y$ factors through $Y_ m$. In terms of ideals

86.16.2.1
\begin{equation} \label{formal-spaces-equation-first} \forall n\ \exists m,\ J_ m A \subset I^ n. \end{equation}

Choose $m \geq 1$. We can find an $n$ such that the morphism $\mathop{\mathrm{Spec}}(A_ m) \to X$ factors through $\mathop{\mathrm{Spec}}(A/I^ n)$. In terms of ideals

86.16.2.2
\begin{equation} \label{formal-spaces-equation-second} \forall m\ \exists n,\ I^ n \subset \mathop{\mathrm{Ker}}(A \to A_ m). \end{equation}

Given an $m$ we can pick an $n = n(m)$ such that $I^ n \subset \mathop{\mathrm{Ker}}(A \to A_ m)$ by (86.16.2.2). Choose generators $f_1, \ldots , f_ r$ of $I$. For any $E = (e_1, \ldots , e_ r)$ with $|E| = \sum e_ i = n$ write

$f_1^{e_1} \ldots f_ r^{e_ r} = \sum g_{E, j} a_{E, j} + \delta _ E$

with $g_{E, j} \in J_ m$, $a_{E, j} \in A$, and $\delta _ E \in I^{n + 1}$ (possible by the above). Let $J = (g_{E, j}) \subset J_ m \subset B$. Then we see that

$I^ n \subset J A + I^{n + 1}$

As $I$ is contained in the Jacobson radical of $A$ and $I^ n$ is finitely generated we see that $I^ n \subset JA$ by Nakayama's lemma. More precisely apply part (2) of Algebra, Lemma 10.20.1 to see that $M = (I^ n + JA)/JA$ is zero.

We first apply what we just proved as follows: since for every $m$ there is an $n(m)$ with $I^{n(m)} \subset J_ mA$ we see that $J_ mA$ is open in $A$, hence closed, hence $\mathop{\mathrm{Ker}}(A \to A_ m) = J_ mA$, in other words, $A_ m = A/J_ mA$. This holds for every $m$.

Next, we pick $m$ with $J_ mA \subset I$ (86.16.2.1). Then choose $n = n(m)$ and finitely generated ideal $J \subset J_ m$ with $I^ n \subset JA \subset I$ as above. For every $k \geq 1$ we define $\mathfrak b_ k = \mathop{\mathrm{Ker}}(B \to A/J^ kA)$. Observe that $\mathfrak b_ k \supset \mathfrak b_{k + 1}$. For every $k$ there exists an $m'$ with $J_{m'} \subset \mathfrak b_ k$ as we have $I^{nk} \subset J^ kA$ and we can apply (86.16.2.1). On the other hand, for every $m'$ there exists a $k$ such that $I^ k \subset J_{m'}A$ because $J_{m'}A$ is open. Then $\mathfrak b_ k$ maps to zero in $A/J_{m'}A$ which is faithfully flat over $B/J_{m'}$. Hence $\mathfrak b_ k \subset J_{m'}$. In other words, we see that the topology on $B$ is defined by the sequence of ideals $\mathfrak b_ k$. Note that $J^ k \subset \mathfrak b_ k$ which implies that $\mathfrak b_ k A = J^ kA$. In other words, we have reduced the problem to the situation discussed in the following paragraph.

We are given a ring map $B \to A$ where

1. $B$ is a weakly admissible topological ring with a fundamental system $J_1 \supset J_2 \supset J_3 \supset \ldots$ of ideals of definition,

2. $A$ is a ring complete with respect to a finitely generated ideal $I$,

3. we have $J_ k A = I^ k$ for all $k$, and

4. $B/J_ k \to A/I^ k$ is faithfully flat.

We want to deduce that $B$ is adic*. Pick $g_1, \ldots , g_ r \in J_1$ whose images in $A/I^2$ generate $I/I^2$; this is possible because $J_1A/J_2A = I/I^2$. Then for all $k \geq 1$ we see that the elements $g^ E = g_1^{e_1} \ldots g_ r^{e_ r}$ with $|E| = k$ are in $J_ k$ because $B/J_ k \to A/I^ k$ is faithfully flat and $J_1A = I$. Also we have $J_1 J_ k \subset J_{k + 1}$ by similar reasoning. The classes of $g^ E$ with $|E| = k$ in $J_ k/J_{k + 1}$ map to generators of $I^ k/I^{k + 1}$ because the images of $g_1, \ldots , g_ r$ generate $I/I^2$. Since $B/J_{k + 1} \to A/I^{k + 1}$ is flat we see that

$J_ k/J_{k + 1} \otimes _{B/J_1} A/I = J_ k/J_{k + 1} \otimes _{B/J_{k + 1}} A/I^{k + 1} \to I^ k/I^{k + 1}$

is an isomorphism (see More on Morphisms, Lemma 37.10.1). Since $B/J_1 \to A/I$ is faithfully flat, we conclude that the classes of the elements $g^ E$, $|E| = k$ generate $J_ k/J_{k + 1}$. We claim that $J_ k = (g^ E, |E| = k)$. Namely, suppose that $x_ k \in J_ k$. By the above we can write

$x_ k = \sum \nolimits _{|E| = k} b_{E, 0} g^ E + x_{k + 1}$

with $x_{k + 1} \in J_{k + 1}$ and some $b_{E, 0} \in B$. Now we can write $x_{k + 1}$ as follows

$x_{k + 1} = \sum \nolimits _{|E| = k} \left(\sum \nolimits _{|E'| = 1} b_{E, E'}g^{E'}\right) g^ E + x_{k + 2}$

because every multi-index of degree $k + 1$ is a sum of a multi-index of degree $k$ and a multi-index of degree $1$. Continuing in this manner we can find $b_{E, E'} \in B$ such that for every $l > 1$ we have

$x_ k = \sum \nolimits _{|E| = k} \left(\sum \nolimits _{0 \leq |E'| < l} b_{E, E'} g^{E'}\right) g^ E + x_{k + l}$

with some $x_{k + l} \in J_{k + l}$. Then we can finally define

$b_ E = \sum \nolimits _{E'} b_{E, E'} g^{E'}$

as an element in $B$ and we see that $x_ k = \sum b_ E g^ E$ as desired. This finishes the proof as now $J_1$ is finitely generated and $J_ k = J_1^ k$ for all $k \geq 1$. $\square$

Lemma 86.16.3. Let $S$ be a scheme. Let $X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces, surjective, flat, and (locally) of finite type. Then $X$ is Noetherian if and only if $Y$ is Noetherian.

Proof. Observe that a Noetherian affine formal algebraic space is adic*, see Lemma 86.6.3. Thus by Lemma 86.16.2 we may assume that both $X$ and $Y$ are adic*. We will use the criterion of Lemma 86.6.5 to see that the lemma holds. Namely, write $Y = \mathop{\mathrm{colim}}\nolimits Y_ n$ as in Lemma 86.6.1. For each $n$ set $X_ n = Y_ n \times _ Y X$. Then $X_ n$ is an affine scheme (Lemma 86.15.7) 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). $\square$

Lemma 86.16.4. Let $S$ be a scheme. Let $P \in \{ countably\ indexed, adic*, Noetherian\}$. Let $X$ be a formal algebraic space over $S$. The following are equivalent

1. if $Y$ is an affine formal algebraic space and $f : Y \to X$ is representable by algebraic spaces and étale, then $Y$ has property $P$,

2. for some $\{ X_ i \to X\} _{i \in I}$ as in Definition 86.7.1 each $X_ i$ has property $P$.

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 86.11.2) we can pick coverings $\{ X_{ij} \to X_ i \times _ X Y\}$ as in Definition 86.7.1. Since $Y$ is quasi-compact, there exist $(i_1, j_1), \ldots , (i_ n, j_ n)$ such that

$X_{i_1 j_1} \amalg \ldots \amalg X_{i_ n j_ n} \longrightarrow Y$

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 86.15.9. 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 86.16.1, 86.16.2, and 86.16.3. $\square$

The previous lemma clears the way for the following definition.

Definition 86.16.5. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. We say $X$ is locally countably indexed, locally adic*, or locally Noetherian if the equivalent conditions of Lemma 86.16.4 hold for the corresponding property.

The formal completion of a locally Noetherian algebraic space along a closed subset is a locally Noetherian formal algebraic space.

Lemma 86.16.6. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. Let $X_{/T}$ be the formal completion of $X$ along $T$.

1. If $X \setminus T \to X$ is quasi-compact, then $X_{/T}$ is locally adic*.

2. If $X$ is locally Noetherian, then $X_{/T}$ is locally Noetherian.

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 65.6.1. 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 86.10.4). Hence $\{ (U_ i)_{/T_ i} \to X_{/T}\}$ is a covering as in Definition 86.7.1. 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 86.10.6. $\square$

Remark 86.16.7 (Warning). Suppose $X = \mathop{\mathrm{Spec}}(A)$ and $T \subset X$ is the zero locus of a finitely generated ideal $I \subset A$. Let $J = \sqrt{I}$ be the radical of $I$. Then from the definitions we see that $X_{/T} = \text{Spf}(A^\wedge )$ where $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/I^ n$ is the $I$-adic completion of $A$. On the other hand, the map $A^\wedge \to \mathop{\mathrm{lim}}\nolimits A/J^ n$ from the $I$-adic completion to the $J$-adic completion can fail to be a ring isomorphisms. As an example let

$A = \bigcup \nolimits _{n \geq 1} \mathbf{C}[t^{1/n}]$

and $I = (t)$. Then $J = \mathfrak m$ is the maximal ideal of the valuation ring $A$ and $J^2 = J$. Thus the $J$-adic completion of $A$ is $\mathbf{C}$ whereas the $I$-adic completion is the valuation ring described in Example 86.9.2 (but in particular it is easy to see that $A \subset A^\wedge$).

Lemma 86.16.8. Let $S$ be a scheme. Let $X \to Y$ and $Z \to Y$ be morphisms of formal algebraic space over $S$. Then

1. If $X$ and $Z$ are locally countably indexed, then $X \times _ Y Z$ is locally countably indexed.

2. If $X$ and $Z$ are locally adic*, then $X \times _ Y Z$ is locally adic*.

3. If $X$ and $Z$ are locally Noetherian and $X_{red} \to Y_{red}$ is locally of finite type, then $X \times _ Y Z$ is locally Noetherian.

Proof. Choose a covering $\{ Y_ j \to Y\}$ as in Definition 86.7.1. For each $j$ choose a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 86.7.1. For each $j$ choose a covering $\{ Z_{jk} \to Y_ j \times _ Y Z\}$ as in Definition 86.7.1. Observe that $X_{ji} \times _{Y_ j} Z_{jk}$ is an affine formal algebraic space by Lemma 86.12.4. Hence

$\{ X_{ji} \times _{Y_ j} Z_{jk} \to X \times _ Y Z\}$

is a covering as in Definition 86.7.1. Thus it suffices to prove (1) and (2) in case $X$, $Y$, and $Z$ are affine formal algebraic spaces.

Assume $X$, $Y$, and $Z$ are affine formal algebraic spaces and McQuillan. Then we can write $X = \text{Spf}(A)$, $Y = \text{Spf}(B)$, $Z = \text{Spf}(C)$ for some weakly admissible topological rings $A$, $B$, and $C$ and the morphsms $X \to Y$ and $Z \to Y$ are given by continuous ring maps $B \to A$ and $B \to C$, see Definition 86.5.7 and Lemma 86.5.10 By Lemma 86.12.4 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. This reduces cases (1) and (2) of our lemma to parts (3) and (4) of Lemma 86.4.16.

To deduce case (3) from Lemma 86.4.16 part (5) we need to match the hypotheses. First, we observe that taking the reduction corresponds to dividing by the ideal of topologically nilpotent elements (Example 86.8.2). Second, if $X_{red} \to Y_{red}$ is locally of finite type, then $(X_{ji})_{red} \to (Y_ j)_{red}$ is locally of finite type (and hence correspond to finite type homomorphisms of rings). This follows Morphisms of Spaces, Lemma 66.23.4 and the fact that in the commutative diagram

$\xymatrix{ (X_{ji})_{red} \ar[d] \ar[r] & (Y_ j)_{red} \ar[d] \\ X_{red} \ar[r] & Y_{red} }$

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 86.8.3. $\square$

Lemma 86.16.9. Let $S$ be a scheme. Let $X$ be a locally Noetherian formal algebraic space over $S$. Then $X = \mathop{\mathrm{colim}}\nolimits X_ n$ for a system $X_1 \to X_2 \to X_3 \to \ldots$ of finite order thickenings of locally Noetherian algebraic spaces over $S$ where $X_1 = X_{red}$ and $X_ n$ is the $n$th infinitesimal neighbourhood of $X_1$ in $X_ m$ for all $m \geq n$.

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. $\square$

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