The Stacks project

85.15 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 85.15.1. 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 85.6.1. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda $ as in Definition 85.5.1. For every $n$ we can pick a $\lambda _ n$ such that $X_ n \to Y$ factors through $Y_{\lambda _ n}$, see Lemma 85.5.4. On the other hand, for every $\lambda $ the scheme $Y_\lambda \times _ Y X$ is affine (Lemma 85.14.7) and hence $Y_\lambda \times _ Y X \to X$ factors through $X_ n$ for some $n$ (Lemma 85.5.4). 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 85.14.9. $\square$

Lemma 85.15.2. 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 85.14.9.

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 85.15.1 we see that $Y$ is countably indexed. Thus, by Lemma 85.6.4, 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 85.5.10 the morphism $X \to Y$ corresponds to a continuous ring map $B \to A$ which is taut by Lemma 85.14.10. 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 85.14.7) 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 85.11.4 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
\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
\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 ( 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 10.19.1 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 $J_ mA \subset I$ ( 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 ( 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 85.15.3. 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 85.6.3. Thus by Lemma 85.15.2 we may assume that both $X$ and $Y$ are adic*. We will use the criterion of Lemma 85.6.5 to see that the lemma holds. Namely, write $Y = \mathop{\mathrm{colim}}\nolimits Y_ n$ as in Lemma 85.6.1. For each $n$ set $X_ n = Y_ n \times _ Y X$. Then $X_ n$ is an affine scheme (Lemma 85.14.7) 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 10.162.1 (to go down) and Algebra, Lemma 10.30.1 (to go up). $\square$

Lemma 85.15.4. 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 85.7.1 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 85.10.2) we can pick coverings $\{ X_{ij} \to X_ i \times _ X Y\} $ as in Definition 85.7.1. 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 85.14.9. 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 85.15.1, 85.15.2, and 85.15.3. $\square$

The previous lemma clears the way for the following definition.

Definition 85.15.5. 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 85.15.4 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 85.15.6. 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 64.6.1. Let $T_ i \subset U_ i$ be the inverse image of $T$. We have $X_{/T} \times _ X U_ i = (U_ i)_{/T_ i}$ (small detail omitted). Hence $\{ (U_ i)_{/T_ i} \to X_{/T}\} $ is a covering as in Definition 85.7.1. 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 it suffices to prove the lemma in case $X$ is affine.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine and $X \setminus T \to X$ is quasi-compact. Then there exists a finitely generated ideal $I = (f_1, \ldots , f_ r) \subset A$ cutting out $T$ (Algebra, Lemma 10.28.1). If $Z = \mathop{\mathrm{Spec}}(B)$ is an affine scheme and $g : Z \to X$ is a morphism with $g(Z) \subset T$ (set theoretically), then $g^\sharp (f_ i)$ is nilpotent in $B$ for each $i$. Thus $I^ n$ maps to zero in $B$ for some $n$. Hence we see that $X_{/T} = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I^ n)$ and $X$ is adic*.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine with $A$ Noetherian. By the above we see that $X_{/T} = \text{Spf}(A^\wedge )$ where $A^\wedge $ is the $I$-adic completion of $A$ with respect to some ideal $I \subset A$. Then $X_{/T}$ is Noetherian because $A^\wedge $ is so, see Algebra, Lemma 10.96.6. $\square$

Remark 85.15.7 (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 85.8.2 (but in particular it is easy to see that $A \subset A^\wedge $).

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