## 86.29 Algebras topologically of finite type

Here is our definition. This definition is not generally agreed upon. Many authors impose further conditions, often because they are only interested in specific types of rings and not the most general case.

Definition 86.29.1. Let $A \to B$ be a continuous map of topological rings (More on Algebra, Definition 15.36.1). We say $B$ is topologically of finite type over $A$ if there exists an $A$-algebra map $A[x_1, \ldots , x_ n] \to B$ whose image is dense in $B$.

If $A$ is a complete, linearly topologized ring, then the restricted power series ring $A\{ x_1, \ldots , x_ r\}$ is topologically of finite type over $A$. If $k$ is a field, then the power series ring $k[[x_1, \ldots , x_ r]]$ is topologically of finite type over $k$.

For continuous taut maps of weakly admissible topological rings, being topologically of finite type corresponds exactly to morphisms of finite type between the associated affine formal algebraic spaces.

Lemma 86.29.2. Let $S$ be a scheme. Let $\varphi : A \to B$ be a continuous map of weakly admissible topological rings over $S$. The following are equivalent

1. $\text{Spf}(\varphi ) : Y = \text{Spf}(B) \to \text{Spf}(A) = X$ is of finite type,

2. $\varphi$ is taut and $B$ is topologically of finite type over $A$.

Proof. We can use Lemma 86.19.8 to relate tautness of $\varphi$ to representability of $\text{Spf}(\varphi )$. We will use this without further mention below. It follows that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ and $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J(I))$ where $I \subset A$ runs over the weak ideals of definition of $A$ and $J(I)$ is the closure of $IB$ in $B$.

Assume (2). Choose a ring map $A[x_1, \ldots , x_ r] \to B$ whose image is dense. Then $A[x_1, \ldots , x_ r] \to B \to B/J(I)$ has dense image too which means that it is surjective. Therefore $B/J(I)$ is of finite type over $A/I$. Let $T \to X$ be a morphism with $T$ a quasi-compact scheme. Then $T \to X$ factors through $\mathop{\mathrm{Spec}}(A/I)$ for some $I$ (Lemma 86.9.4). Then $T \times _ X Y = T \times _{\mathop{\mathrm{Spec}}(A/I)} \mathop{\mathrm{Spec}}(B/J(I))$, see proof of Lemma 86.19.8. Hence $T \times _ Y X \to T$ is of finite type as the base change of the morphism $\mathop{\mathrm{Spec}}(B/J(I)) \to \mathop{\mathrm{Spec}}(A/I)$ which is of finite type. Thus (1) is true.

Assume (1). Pick any $I \subset A$ as above. Since $\mathop{\mathrm{Spec}}(A/I) \times _ X Y = \mathop{\mathrm{Spec}}(B/J(I))$ we see that $A/I \to B/J(I)$ is of finite type. Choose $b_1, \ldots , b_ r \in B$ mapping to generators of $B/J(I)$ over $A/I$. We claim that the image of the ring map $A[x_1, \ldots , x_ r] \to B$ sending $x_ i$ to $b_ i$ is dense. To prove this, let $I' \subset I$ be a second weak ideal of definition. Then we have

$B/(J(I') + IB) = B/J(I)$

because $J(I)$ is the closure of $IB$ and because $J(I')$ is open. Hence we may apply Algebra, Lemma 10.126.9 to see that $(A/I')[x_1, \ldots , x_ r] \to B/J(I')$ is surjective. Thus (2) is true, concluding the proof. $\square$

Let $A$ be a topological ring complete with respect to a linear topology. Let $(I_\lambda )$ be a fundamental system of open ideals. Let $\mathcal{C}$ be the category of inverse systems $(B_\lambda )$ where

1. $B_\lambda$ is a finite type $A/I_\lambda$-algebra, and

2. $B_\mu \to B_\lambda$ is an $A/I_\mu$-algebra homomorphism which induces an isomorphism $B_\mu /I_\lambda B_\mu \to B_\lambda$.

Morphisms in $\mathcal{C}$ are given by compatible systems of homomorphisms.

Lemma 86.29.3. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. Assume $X$ is McQuillan and let $A$ be the weakly admissible topological ring associated to $X$. Then there is an anti-equivalence of categories between

1. the category $\mathcal{C}$ introduced above, and

2. the category of maps $Y \to X$ of finite type of affine formal algebraic spaces.

Proof. Let $(I_\lambda )$ be a fundamental system of weakly admissible ideals of definition in $A$. Consider $Y$ as in (2). Then $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ is affine (Definition 86.24.1 and Lemma 86.19.7). Say $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$. The ring map $A/I_\lambda \to B_\lambda$ is of finite type because $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(A/I_\lambda )$ is of finite type (by Definition 86.24.1). Then $(B_\lambda )$ is an object of $\mathcal{C}$.

Conversely, given an object $(B_\lambda )$ of $\mathcal{C}$ we can set $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$. This is an affine formal algebraic space. We claim that

$Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$

To show this it suffices we get the same values if we evaluate on a quasi-compact scheme $U$. A morphism $U \to \left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ comes from a morphism $U \to \mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda )$ for some $\mu \geq \lambda$ (use Lemma 86.9.4 two times). Since $\mathop{\mathrm{Spec}}(B_\mu ) \times _{\mathop{\mathrm{Spec}}(A/I_\mu )} \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$ by our second assumption on objects of $\mathcal{C}$ this proves what we want. Using this we can show the morphism $Y \to X$ is of finite type. Namely, we note that for any morphism $U \to X$ with $U$ a quasi-compact scheme, we get a factorization $U \to \mathop{\mathrm{Spec}}(A/I_\lambda ) \to X$ for some $\lambda$ (see lemma cited above). Hence

$Y \times _ X U = Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U = \mathop{\mathrm{Spec}}(B_\lambda ) \times _{\mathop{\mathrm{Spec}}(A/I_\lambda )} U$

is a scheme of finite type over $U$ as desired. Thus the construction $(B_\lambda ) \mapsto \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$ does give a functor from category (1) to category (2).

To finish the proof we show that the above constructions define quasi-inverse functors between the categories (1) and (2). In one direction you have to show that

$\left(\mathop{\mathrm{colim}}\nolimits _\mu \mathop{\mathrm{Spec}}(B_\mu )\right) \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$

for any object $(B_\lambda )$ in the category $\mathcal{C}$. This we proved above. For the other direction you have to show that

$Y = \mathop{\mathrm{colim}}\nolimits (Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ))$

given $Y$ in the category (2). Again this is true by evaluating on quasi-compact test objects and because $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I_\lambda )$. $\square$

Remark 86.29.4. Let $A$ be a weakly admissible topological ring and let $(I_\lambda )$ be a fundamental system of weak ideals of definition. Let $X = \text{Spf}(A)$, in other words, $X$ is a McQuillan affine formal algebraic space. Let $f : Y \to X$ be a morphism of affine formal algebraic spaces. In general it will not be true that $Y$ is McQuillan. More specifically, we can ask the following questions:

1. Assume that $f : Y \to X$ is a closed immersion. Then $Y$ is McQuillan and $f$ corresponds to a continuous map $\varphi : A \to B$ of weakly admissible topological rings which is taut, whose kernel $K \subset A$ is a closed ideal, and whose image $\varphi (A)$ is dense in $B$, see Lemma 86.27.5. What conditions on $A$ guarantee that $B = (A/K)^\wedge$ as in Example 86.27.6?

2. What conditions on $A$ guarantee that closed immersions $f : Y \to X$ correspond to quotients $A/K$ of $A$ by closed ideals, in other words, the corresponding continuous map $\varphi$ is surjective and open?

3. Suppose that $f : Y \to X$ is of finite type. Then we get $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B_\lambda )$ where $(B_\lambda )$ is an object of $\mathcal{C}$ by Lemma 86.29.3. In this case it is true that there exists a fixed integer $r$ such that $B_\lambda$ is generated by $r$ elements over $A/I_\lambda$ for all $\lambda$ (the argument is essentially already given in the proof of (1) $\Rightarrow$ (2) in Lemma 86.29.2). However, it is not clear that the projections $\mathop{\mathrm{lim}}\nolimits B_\lambda \to B_\lambda$ are surjective, i.e., it is not clear that $Y$ is McQuillan. Is there an example where $Y$ is not McQuillan?

4. Suppose that $f : Y \to X$ is of finite type and $Y$ is McQuillan. Then $f$ corresponds to a continuous map $\varphi : A \to B$ of weakly admissible topological rings. In fact $\varphi$ is taut and $B$ is topologically of finite type over $A$, see Lemma 86.29.2. In other words, $f$ factors as

$Y \longrightarrow \mathbf{A}^ r_ X \longrightarrow X$

where the first arrow is a closed immersion of McQuillan affine formal algebraic spaces. However, then questions (1) and (2) are in force for $Y \to \mathbf{A}^ r_ X$.

Below we will answer these questions when $X$ is countably indexed, i.e., when $A$ has a countable fundamental system of open ideals. If you have answers to these questions in greater generality, or if you have counter examples, please email stacks.project@gmail.com.

Lemma 86.29.5. Let $S$ be a scheme. Let $X$ be a countably indexed affine formal algebraic space over $S$. Let $f : Y \to X$ be a closed immersion of formal algebraic spaces over $S$. Then $Y$ is a countably indexed affine formal algebraic space and $f$ corresponds to $A \to A/K$ where $A$ is an object of $\textit{WAdm}^{count}$ (Section 86.21) and $K \subset A$ is a closed ideal.

Proof. By Lemma 86.10.4 we see that $X = \text{Spf}(A)$ where $A$ is an object of $\textit{WAdm}^{count}$. Since a closed immersion is representable and affine, we conclude by Lemma 86.19.9 that $Y$ is an affine formal algebraic space and countably index. Thus applying Lemma 86.10.4 again we see that $Y = \text{Spf}(B)$ with $B$ an object of $\textit{WAdm}^{count}$. By Lemma 86.27.5 we conclude that $f$ is given by a morphism $A \to B$ of $\textit{WAdm}^{count}$ which is taut and has dense image. To finish the proof we apply Lemma 86.5.10. $\square$

Lemma 86.29.6. Let $B \to A$ be an arrow of $\textit{WAdm}^{count}$, see Section 86.21. 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 86.22.1. 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 86.28.1 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 (86.21.2.1) as in Situation 86.21.2 be given. By Example 86.24.7 the maps $A \to (A')^\wedge$ and $B \to (B')^\wedge$ satisfy (a), (b), (c), and (d). Moreover, by Lemma 86.22.1 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 étale. 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 étale.

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

Lemma 86.29.7. In Lemma 86.29.6 if $B$ is admissible (for example adic), then the equivalent conditions (a) – (d) are also equivalent to

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

Proof. It is enough to show that (e) implies (a). Let $J' \subset B$ be a weak ideal of definition and let $I' \subset A$ be the closure of $J'A$. We have to show that $B/J' \to A/I'$ is of finite type. If the corresponding statement holds for the smaller weak ideal of definition $J'' = J' \cap J$, then it holds for $J'$. Thus we may assume $J' \subset J$. As $J$ is an ideal of definition (and not just a weak ideal of definition), we get $J^ n \subset J'$ for some $n \geq 1$. Thus we can consider the diagram

$\xymatrix{ 0 \ar[r] & I/I' \ar[r] & A/I' \ar[r] & A/I \ar[r] & 0 \\ 0 \ar[r] & J/J' \ar[r] \ar[u] & B/J' \ar[r] \ar[u] & B/J \ar[r] \ar[u] & 0 }$

with exact rows. Since $I' \subset A$ is open and since $I$ is the closure of $J A$ we see that $I/I' = (J/J') \cdot A/I'$. Because $J/J'$ is a nilpotent ideal and as $B/J \to A/I$ is of finite type, we conclude from Algebra, Lemma 10.126.8 that $A/I'$ is of finite type over $B/J'$ as desired. $\square$

Lemma 86.29.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces. Assume $Y$ countably indexed. The following are equivalent

1. $f$ is locally of finite type,

2. $f$ is of finite type,

3. $f$ corresponds to a morphism $B \to A$ of $\textit{WAdm}^{count}$ (Section 86.21) satisfying the equivalent conditions of Lemma 86.29.6.

Proof. Since $X$ and $Y$ are affine it is clear that conditions (1) and (2) are equivalent. In cases (1) and (2) the morphism $f$ is representable by algebraic spaces by definition, hence affine by Lemma 86.19.7. Thus if (1) or (2) holds we see that $X$ is countably indexed by Lemma 86.19.9. Write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for topological $S$-algebras $A$ and $B$ in $\textit{WAdm}^{count}$, see Lemma 86.10.4. By Lemma 86.9.10 we see that $f$ corresponds to a continuous map $B \to A$. Hence now the result follows from Lemma 86.29.2. $\square$

Lemma 86.29.9. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of locally countably indexed formal algebraic spaces over $S$. The following are equivalent

1. for every commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

with $U$ and $V$ affine formal algebraic spaces, $U \to X$ and $V \to Y$ representable by algebraic spaces and étale, the morphism $U \to V$ corresponds to a morphism of $\textit{WAdm}^{count}$ which is taut and topologically of finite type,

2. there exists a covering $\{ Y_ j \to Y\}$ as in Definition 86.11.1 and for each $j$ a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 86.11.1 such that each $X_{ji} \to Y_ j$ corresponds to a morphism of $\textit{WAdm}^{count}$ which is taut and topologically of finite type,

3. there exist a covering $\{ X_ i \to X\}$ as in Definition 86.11.1 and for each $i$ a factorization $X_ i \to Y_ i \to Y$ where $Y_ i$ is an affine formal algebraic space, $Y_ i \to Y$ is representable by algebraic spaces and étale, and $X_ i \to Y_ i$ corresponds to a morphism of $\textit{WAdm}^{count}$ which is, taut and topologically of finite type, and

4. $f$ is locally of finite type.

Proof. By Lemma 86.29.6 the property $P(\varphi )=$“$\varphi$ is taut and topologically of finite type” is local on $\text{WAdm}^{count}$. Hence by Lemma 86.21.3 we see that conditions (1), (2), and (3) are equivalent. On the other hand, by Lemma 86.29.8 the condition $P$ on morphisms of $\textit{WAdm}^{count}$ corresponds exactly to morphisms of countably indexed, affine formal algebraic spaces being locally of finite type. Thus the implication (1) $\Rightarrow$ (3) of Lemma 86.24.6 shows that (4) implies (1) of the current lemma. Similarly, the implication (4) $\Rightarrow$ (1) of Lemma 86.24.6 shows that (2) implies (4) of the current lemma. $\square$

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