The Stacks project

Lemma 85.16.2. 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 85.15.9.

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 85.16.1 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 85.15.10. 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 85.15.7) 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 85.12.4 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
\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 10.20.1 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 $J_ mA \subset I$ ( 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 $k \geq 1$. $\square$

Comments (2)

Comment #1956 by Brian Conrad on

This proof would benefit from some more explanation to the reader as to where it is going before one dives into the various constructions from (15.2.1) onwards. For example, at least tell the reader the nature of the target reduction to the special case where (1)--(4) hold before one proves that.

When is defined, would help to note that , say is contained in the "Jacobson radical" of rather than just the "radical" of , and when proving it should be made clearer that this depends on (even itself depends on which was chosen depending on ). Since Nakayama Lemma 10.19.1 has 10 zillion parts, when appealing to that it could help to say that one uses part (1) with taking to be the -span of the finite set of elements of arising as...

For the paragraph beginning "Next, we pick ...", perhaps note at the end of the 2nd sentence that there depends on . When is defined, perhaps note that (implicitly used later).

After conditions (1)--(4) are stated, tell the reader "We want to deduce that is adic (so the reader can orient themselves around what is going on). For the statement that the elements () are in , perhaps note that we're using that is faithfully flat and that , and likewise note that by similar reasoning (as this is used later in the argument). A few lines later it is mentioned that and are flat, but it seems only the first of these two flatness properties is being used right there.

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