The Stacks project

Lemma 87.29.6. Let $B \to A$ be an arrow of $\textit{WAdm}^{count}$, see Section 87.21. The following are equivalent

  1. $B \to A$ is taut and $B/J \to A/I$ is of finite type for every weak ideal of definition $J \subset B$ where $I \subset A$ is the closure of $JA$,

  2. $B \to A$ is taut and $B/J_\lambda \to A/I_\lambda $ is of finite type for a cofinal system $(J_\lambda )$ of weak ideals of definition of $B$ where $I_\lambda \subset A$ is the closure of $J_\lambda A$,

  3. $B \to A$ is taut and $A$ is topologically of finite type over $B$,

  4. $A$ is isomorphic as a topological $B$-algebra to a quotient of $B\{ x_1, \ldots , x_ n\} $ by a closed ideal.

Moreover, these equivalent conditions define a local property, i.e., they satisfy Axioms (1), (2), (3).

Proof. The implications (a) $\Rightarrow $ (b), (c) $\Rightarrow $ (a), (d) $\Rightarrow $ (c) are straightforward from the definitions. Assume (b) holds and let $J \subset B$ and $I \subset A$ be as in (a). Choose a commutative diagram

\[ \xymatrix{ A \ar[r] & \ldots \ar[r] & A_3 \ar[r] & A_2 \ar[r] & A_1 \\ B \ar[r] \ar[u] & \ldots \ar[r] & B/J_3 \ar[r] \ar[u] & B/J_2 \ar[r] \ar[u] & B/J_1 \ar[u] } \]

such that $A_{n + 1}/J_ nA_{n + 1} = A_ n$ and such that $A = \mathop{\mathrm{lim}}\nolimits A_ n$ as in Lemma 87.22.1. For every $m$ there exists a $\lambda $ such that $J_\lambda \subset J_ m$. Since $B/J_\lambda \to A/I_\lambda $ is of finite type, this implies that $B/J_ m \to A/I_ m$ is of finite type. Let $\alpha _1, \ldots , \alpha _ n \in A_1$ be generators of $A_1$ over $B/J_1$. Since $A$ is a countable limit of a system with surjective transition maps, we can find $a_1, \ldots , a_ n \in A$ mapping to $\alpha _1, \ldots , \alpha _ n$ in $A_1$. By Remark 87.28.1 we find a continuous map $B\{ x_1, \ldots , x_ n\} \to A$ mapping $x_ i$ to $a_ i$. This map induces surjections $(B/J_ m)[x_1, \ldots , x_ n] \to A_ m$ by Algebra, Lemma 10.126.9. For $m \geq 1$ we obtain a short exact sequence

\[ 0 \to K_ m \to (B/J_ m)[x_1, \ldots , x_ n] \to A_ m \to 0 \]

The induced transition maps $K_{m + 1} \to K_ m$ are surjective because $A_{m + 1}/J_ mA_{m + 1} = A_ m$. Hence the inverse limit of these short exact sequences is exact, see Algebra, Lemma 10.86.4. Since $B\{ x_1, \ldots , x_ n\} = \mathop{\mathrm{lim}}\nolimits (B/J_ m)[x_1, \ldots , x_ n]$ and $A = \mathop{\mathrm{lim}}\nolimits A_ m$ we conclude that $B\{ x_1, \ldots , x_ n\} \to A$ is surjective and open. As $A$ is complete the kernel is a closed ideal. In this way we see that (a), (b), (c), and (d) are equivalent.

Let a diagram ( as in Situation 87.21.2 be given. By Example 87.24.7 the maps $A \to (A')^\wedge $ and $B \to (B')^\wedge $ satisfy (a), (b), (c), and (d). Moreover, by Lemma 87.22.1 in order to prove Axioms (1) and (2) we may assume both $B \to A$ and $(B')^\wedge \to (A')^\wedge $ are taut. Now pick a weak ideal of definition $J \subset B$. Let $J' \subset (B')^\wedge $, $I \subset A$, $I' \subset (A')^\wedge $ be the closure of $J(B')^\wedge $, $JA$, $J(A')^\wedge $. By what was said above, it suffices to consider the commutative diagram

\[ \xymatrix{ A/I \ar[r] & (A')^\wedge /I' \\ B/J \ar[r] \ar[u]^{\overline{\varphi }} & (B')^\wedge /J' \ar[u]_{\overline{\varphi }'} } \]

and to show (1) $\overline{\varphi }$ finite type $\Rightarrow \overline{\varphi }'$ finite type, and (2) if $A \to A'$ is faithfully flat, then $\overline{\varphi }'$ finite type $\Rightarrow \overline{\varphi }$ finite type. Note that $(B')^\wedge /J' = B'/JB'$ and $(A')^\wedge /I' = A'/IA'$ by the construction of the topologies on $(B')^\wedge $ and $(A')^\wedge $. In particular the horizontal maps in the diagram are ├ętale. Part (1) now follows from Algebra, Lemma 10.6.2 and part (2) from Descent, Lemma 35.14.2 as the ring map $A/I \to (A')^\wedge /I' = A'/IA'$ is faithfully flat and ├ętale.

We omit the proof of Axiom (3). $\square$

Comments (4)

Comment #1973 by Brian Conrad on

Replace with for clarity (in three places), and near the end of the first paragraph replace "is surjective" with "is surjective and open" (to convey that in particular the topology of is the quotient topology modulo the closed kernel). This is important because in (d) there is the implicit assertion that has the quotient topology (and probably the statement of (d) should explicitly record this aspect for clarity).

In the first paragraph, the proof of surjectivity of uses a Lemma which contains a finite-type hypothesis, which in the present circumstances requires confirming that is finite type over (perhaps without knowing the more precise generation by the images of the 's that we're trying to prove). For this is rigged by design of , but without some finiteness conditions on the 's how do we infer the case ? Maybe I am overlooking something, in which case I suppose I am asking for a clearer explanation.

For the start of the 2nd paragraph, if someone is reading this on the computer then the tag-link for the Diagram (0ANC) will be confusing because that link doesn't include the hypotheses (1)--(4) implicit on the data in that diagram. I strongly recommend that Tag 0ANC include the text with those hypotheses. Indeed, I was reading the above proof on the computer, followed the link to that diagram, and was totally mystified (e.g., seemed one has to assume at least and are finite type). Only by looking at my hard copy of the section on formal algebraic spaces did I see all of the hypotheses imposed on the data in Tag 0ANC (such as being -etale and being -etale).

On line 4 of paragraph 2, replace with .

Comment #1977 by on

OK, yes, this is an idiotic mistake. So I've fixed it by replacing (b) by the condition that there should exist a cofinal system of weak ideals of definition with the desired property as you suggested in an email.

It turns out the mistake can be fixed when the rings in question are admissible and not just weakly admissible. To this end I have inserted a new lemma stating that if we have, in the situation of the lemma that is admissible and is of finite type for one ideal of definition , then this implies condition (a) of the lemma. The proof is quite simple and avoids the mistake you pointed out in your comment: namely, we can use Nakayama's lemma 10.20.1 part (11) in stead of Lemma 10.126.9.

The latex changes are here. I will have this on the website soon.

I have not yet fixed the problem with Diagram 0ANC. I will do so soon.

Comment #2024 by on

Also, for readers who get to this tag directly, the new lemma mentioned in #1977 which partially fixes the mistake found by Brian, is Lemma 87.29.7.

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