## 87.19 Morphisms representable by algebraic spaces

Let $f : X \to Y$ be a morphism of formal algebraic spaces which is representable by algebraic spaces. For these types of morphisms we have a lot of theory at our disposal, thanks to the work done in the chapters on algebraic spaces.

Lemma 87.19.1. The composition of morphisms representable by algebraic spaces is representable by algebraic spaces. The same holds for representable (by schemes).

Proof. See Bootstrap, Lemma 80.3.8. $\square$

Lemma 87.19.2. A base change of a morphism representable by algebraic spaces is representable by algebraic spaces. The same holds for representable (by schemes).

Proof. See Bootstrap, Lemma 80.3.3. $\square$

Lemma 87.19.3. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of formal algebraic spaces over $S$. If $g \circ f : X \to Z$ is representable by algebraic spaces, then $f : X \to Y$ is representable by algebraic spaces.

Proof. Note that the diagonal of $Y \to Z$ is representable by Lemma 87.15.5. Thus $X \to Y$ is representable by algebraic spaces by Bootstrap, Lemma 80.3.10. $\square$

The property of being representable by algebraic spaces is local on the source and the target.

Lemma 87.19.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The following are equivalent:

1. the morphism $f$ is representable by algebraic spaces,

2. there exists a commutative diagram

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

where $U$, $V$ are formal algebraic spaces, the vertical arrows are representable by algebraic spaces, $U \to X$ is surjective étale, and $U \to V$ is representable by algebraic spaces,

3. for any commutative diagram

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

where $U$, $V$ are formal algebraic spaces and the vertical arrows are representable by algebraic spaces, the morphism $U \to V$ is representable by algebraic spaces,

4. there exists a covering $\{ Y_ j \to Y\}$ as in Definition 87.11.1 and for each $j$ a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 87.11.1 such that $X_{ji} \to Y_ j$ is representable by algebraic spaces for each $j$ and $i$,

5. there exist a covering $\{ X_ i \to X\}$ as in Definition 87.11.1 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, such that $X_ i \to Y_ i$ is representable by algebraic spaces, and

Proof. It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, (2) implies (1) by Bootstrap, Lemma 80.11.4 applied to $U \to X \to Y$.

Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is representable by algebraic spaces (as the composition $U \to X \to Y$, see Bootstrap, Lemma 80.3.8) and factors through $V$. Thus $U \to V$ is representable by algebraic spaces by Lemma 87.19.3.

It is clear that (3) implies (2). Thus now (1) – (3) are equivalent.

Observe that the condition in (4) makes sense as the fibre product $Y_ j \times _ Y X$ is a formal algebraic space by Lemma 87.15.3. It is clear that (4) implies (5).

Assume $X_ i \to Y_ i \to Y$ as in (5). Then we set $V = \coprod Y_ i$ and $U = \coprod X_ i$ to see that (5) implies (2).

Finally, assume (1) – (3) are true. Thus we can choose any covering $\{ Y_ j \to Y\}$ as in Definition 87.11.1 and for each $j$ any covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 87.11.1. Then $X_{ij} \to Y_ j$ is representable by algebraic spaces by (3) and we see that (4) is true. This concludes the proof. $\square$

Lemma 87.19.5. Let $S$ be a scheme. Let $Y$ be an affine formal algebraic space over $S$. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable by algebraic spaces. Then $X$ is a formal algebraic space.

Proof. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\lambda$ as in Definition 87.9.1. For each $\lambda$ the fibre product $X \times _ Y Y_\lambda$ is an algebraic space. Hence $X = \mathop{\mathrm{colim}}\nolimits X \times _ Y Y_\lambda$ is a formal algebraic space by Lemma 87.13.1. $\square$

Lemma 87.19.6. Let $S$ be a scheme. Let $Y$ be a formal algebraic space over $S$. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable by algebraic spaces. Then $X$ is a formal algebraic space.

Proof. Let $\{ Y_ i \to Y\}$ be as in Definition 87.11.1. Then $X \times _ Y Y_ i \to X$ is a family of morphisms representable by algebraic spaces, étale, and jointly surjective. Thus it suffices to show that $X \times _ Y Y_ i$ is a formal algebraic space, see Lemma 87.15.1. This follows from Lemma 87.19.5. $\square$

Lemma 87.19.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces. Then $f$ is representable (by schemes) and affine.

Proof. We will show that $f$ is affine; it will then follow that $f$ is representable and affine by Morphisms of Spaces, Lemma 67.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu$ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 87.9.1. Let $T \to Y$ be a morphism where $T$ is a scheme over $S$. We have to show that $X \times _ Y T \to T$ is affine, see Bootstrap, Definition 80.4.1. To do this we may assume that $T$ is affine and we have to prove that $X \times _ Y T$ is affine. In this case $T \to Y$ factors through $Y_\mu \to Y$ for some $\mu$, see Lemma 87.9.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 87.17.3). Hence $X \times _ Y T \to X$ factors through $X_\lambda$ for some $\lambda$. Similarly $X_\lambda \to Y$ factors through $Y_\mu$ after increasing $\mu$. Then $X \times _ Y T = X_\lambda \times _{Y_\mu } T$. We conclude as fibre products of affine schemes are affine. $\square$

Lemma 87.19.8. 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. $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is representable by algebraic spaces,

2. $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is representable (by schemes),

3. $\varphi$ is taut, see Definition 87.5.1.

Proof. Parts (1) and (2) are equivalent by Lemma 87.19.7.

Assume the equivalent conditions (1) and (2) hold. If $I \subset A$ is a weak ideal of definition, then $\mathop{\mathrm{Spec}}(A/I) \to \text{Spf}(A)$ is representable and a thickening (this is clear from the construction of the formal spectrum but it also follows from Lemma 87.9.6). Then $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) \to \text{Spf}(B)$ is representable and a thickening as a base change. Hence by Lemma 87.9.6 there is a weak ideal of definition $J(I) \subset B$ such that $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) = \mathop{\mathrm{Spec}}(B/J(I))$ as subfunctors of $\text{Spf}(B)$. We obtain a cartesian diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(B/J(I)) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) }$

By Lemma 87.16.4 we see that $B/J(I) = B \widehat{\otimes }_ A A/I$. It follows that $J(I)$ is the closure of the ideal $\varphi (I)B$, see Lemma 87.4.11. Since $\text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ with $I$ as above, we find that $\text{Spf}(B) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J(I))$. Thus the ideals $J(I)$ form a fundamental system of weak ideals of definition (see Lemma 87.9.6). Hence (3) holds.

Assume (3) holds. We are essentially just going to reverse the arguments given in the previous paragraph. Let $I \subset A$ be a weak ideal of definition. By Lemma 87.16.4 we get a cartesian diagram

$\xymatrix{ \text{Spf}(B \widehat{\otimes }_ A A/I) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) }$

If $J(I)$ is the closure of $IB$, then $J(I)$ is open in $B$ by tautness of $\varphi$. Hence if $J$ is open in $B$ and $J \subset J(B)$, then $B/J \otimes _ A A/I = B/(IB + J) = B/J(I)$ because $J(I) = \bigcap _{J \subset B\text{ open}} (IB + J)$ by Lemma 87.4.2. Hence the limit defining the completed tensor product collapses to give $B \widehat{\otimes }_ A A/I = B/J(I)$. Thus $\text{Spf}(B \widehat{\otimes }_ A A/I) = \mathop{\mathrm{Spec}}(B/J(I))$. This proves that $\text{Spf}(B) \times _{\text{Spf}(A)} \mathop{\mathrm{Spec}}(A/I)$ is representable for every weak ideal of definition $I \subset A$. Since every morphism $T \to \text{Spf}(A)$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(A/I)$ for some weak ideal of definition $I$ (Lemma 87.9.4) we conclude that $\text{Spf}(\varphi )$ is representable, i.e., (2) holds. This finishes the proof. $\square$

Lemma 87.19.9. Let $S$ be a scheme. Let $Y$ be an affine formal algebraic space. Let $f : X \to Y$ be a map of sheaves on $(\mathit{Sch}/S)_{fppf}$ which is representable and affine. Then

1. $X$ is an affine formal algebraic space,

2. if $Y$ is countably indexed, then $X$ is countably indexed,

3. if $Y$ is countably indexed and classical, then $X$ is countably indexed and classical,

4. if $Y$ is weakly adic, then $X$ is weakly adic,

5. if $Y$ is adic*, then $X$ is adic*, and

6. if $Y$ is Noetherian and $f$ is (locally) of finite type, then $X$ is Noetherian.

Proof. Proof of (1). Write $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda$ as in Definition 87.9.1. Since $f$ is representable and affine, the fibre products $X_\lambda = Y_\lambda \times _ Y X$ are affine. And $X = \mathop{\mathrm{colim}}\nolimits Y_\lambda \times _ Y X$. Thus $X$ is an affine formal algebraic space.

Proof of (2). If $Y$ is countably indexed, then in the argument above we may assume $\Lambda$ is countable. Then we immediately see that $X$ is countably indexed too.

Proof of (3), (4), and (5). In each of these cases the assumptions imply that $Y$ is a countably indexed affine formal algebraic space (Lemma 87.10.3) and hence $X$ is too by (2). Thus we may write $X = \text{Spf}(A)$ and $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebras $A$ and $B$, see Lemma 87.10.4. By Lemma 87.9.10 the morphism $f$ corresponds to a continuous $S$-algebra homomorphism $\varphi : B \to A$. We see from Lemma 87.19.8 that $\varphi$ is taut. We conclude that (3) follows from Lemma 87.5.9, (4) follows from Lemma 87.7.5, and (5) follows from Lemma 87.6.5.

Proof of (6). Combining (3) with Lemma 87.10.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 87.10.5. First, it tells us the affine schemes $Y_\lambda$ are Noetherian. Then $X_\lambda \to Y_\lambda$ is of finite type, hence $X_\lambda$ is Noetherian too (Morphisms, Lemma 29.15.6). Then the criterion tells us $X$ is Noetherian and the proof is complete. $\square$

Lemma 87.19.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of affine formal algebraic spaces which is representable by algebraic spaces. Then

1. if $Y$ is countably indexed, then $X$ is countably indexed,

2. if $Y$ is countably indexed and classical, then $X$ is countably indexed and classical,

3. if $Y$ is weakly adic, then $X$ is weakly adic,

4. if $Y$ is adic*, then $X$ is adic*, and

5. if $Y$ is Noetherian and $f$ is (locally) of finite type, then $X$ is Noetherian.

Example 87.19.11. Let $B$ be a weakly admissible topological ring. Let $B \to A$ be a ring map (no topology). Then we can consider

$A^\wedge = \mathop{\mathrm{lim}}\nolimits A/JA$

where the limit is over all weak ideals of definition $J$ of $B$. Then $A^\wedge$ (endowed with the limit topology) is a complete linearly topologized ring. The (open) kernel $I$ of the surjection $A^\wedge \to A/JA$ is the closure of $JA^\wedge$, see Lemma 87.4.2. By Lemma 87.4.10 we see that $I$ consists of topologically nilpotent elements. Thus $I$ is a weak ideal of definition of $A^\wedge$ and we conclude $A^\wedge$ is a weakly admissible topological ring. Thus $\varphi : B \to A^\wedge$ is taut map of weakly admissible topological rings and

$\text{Spf}(A^\wedge ) \longrightarrow \text{Spf}(B)$

is a special case of the phenomenon studied in Lemma 87.19.8.

Remark 87.19.12 (Warning). The discussion in Lemmas 87.19.8, 87.19.9, and 87.19.10 is sharp in the following two senses:

1. If $A$ and $B$ are weakly admissible rings and $\varphi : A \to B$ is a continuous map, then $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is in general not representable.

2. If $f : Y \to X$ is a representable morphism of affine formal algebraic spaces and $X = \text{Spf}(A)$ is McQuillan, then it does not follow that $Y$ is McQuillan.

An example for (1) is to take $A = k$ a field (with discrete topology) and $B = k[[t]]$ with the $t$-adic topology. An example for (2) is given in Examples, Section 110.75.

The warning above notwithstanding, we do have the following result.

Lemma 87.19.13. Let $S$ be a scheme. Let $Y$ be a McQuillan affine formal algebraic space over $S$, i.e., $Y = \text{Spf}(B)$ for some weakly admissible topological $S$-algebra $B$. Then there is an equivalence of categories between

1. the category of morphisms $f : X \to Y$ of affine formal algebraic spaces which are representable by algebraic spaces and étale, and

2. the category of topological $B$-algebras of the form $A^\wedge$ where $A$ is an étale $B$-algebra and $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/JA$ with $J \subset B$ running over the weak ideals of definition of $B$.

The equivalence is given by sending $A^\wedge$ to $X = \text{Spf}(A^\wedge )$. In particular, any $X$ as in (1) is McQuillan.

Proof. Let $A$ be an étale $B$-algebra. Then $B/J \to A/JA$ is étale for every open ideal $J \subset B$. Hence the morphism $\text{Spf}(A^\wedge ) \to Y$ is representable and étale. The functor $\text{Spf}$ is fully faithful by Lemma 87.9.10. To finish the proof we will show in the next paragraph that any $X \to Y$ as in (1) is in the essential image.

Choose a weak ideal of definition $J_0 \subset B$. Set $Y_0 = \mathop{\mathrm{Spec}}(B/J_0)$ and $X_0 = Y_0 \times _ Y X$. Then $X_0 \to Y_0$ is an étale morphism of affine schemes (see Lemma 87.19.7). Say $X_0 = \mathop{\mathrm{Spec}}(A_0)$. By Algebra, Lemma 10.143.10 we can find an étale algebra map $B \to A$ such that $A_0 \cong A/J_0A$. Consider an ideal of definition $J \subset J_0$. As above we may write $\mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{Spec}}(\bar A)$ for some étale ring map $B/J \to \bar A$. Then both $B/J \to \bar A$ and $B/J \to A/JA$ are étale ring maps lifting the étale ring map $B/J_0 \to A_0$. By More on Algebra, Lemma 15.11.2 there is a unique $B/J$-algebra isomorphism $\varphi _ J : A/JA \to \bar A$ lifting the identification modulo $J_0$. Since the maps $\varphi _ J$ are unique they are compatible for varying $J$. Thus

$X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J) \times _ Y X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/JA) = \text{Spf}(A)$

and we see that the lemma holds. $\square$

Lemma 87.19.14. With notation and assumptions as in Lemma 87.19.13 let $f : X \to Y$ correspond to $B \to A^\wedge$. The following are equivalent

1. $f : X \to Y$ is surjective,

2. $B \to A$ is faithfully flat,

3. for every weak ideal of definition $J \subset B$ the ring map $B/J \to A/JA$ is faithfully flat, and

4. for some weak ideal of definition $J \subset B$ the ring map $B/J \to A/JA$ is faithfully flat.

Proof. Let $J \subset B$ be a weak ideal of definition. As every element of $J$ is topologically nilpotent, we see that every element of $1 + J$ is a unit. It follows that $J$ is contained in the Jacobson radical of $B$ (Algebra, Lemma 10.19.1). Hence a flat ring map $B \to A$ is faithfully flat if and only if $B/J \to A/JA$ is faithfully flat (Algebra, Lemma 10.39.16). In this way we see that (2) – (4) are equivalent. If (1) holds, then for every weak ideal of definition $J \subset B$ the morphism $\mathop{\mathrm{Spec}}(A/JA) = \mathop{\mathrm{Spec}}(B/J) \times _ Y X \to \mathop{\mathrm{Spec}}(B/J)$ is surjective which implies (3). Conversely, assume (3). A morphism $T \to Y$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(B/J)$ for some ideal of definition $J$ of $B$ (Lemma 87.9.4). Hence $X \times _ Y T = \mathop{\mathrm{Spec}}(A/JA) \times _{\mathop{\mathrm{Spec}}(B/J)} T \to T$ is surjective as a base change of the surjective morphism $\mathop{\mathrm{Spec}}(A/JA) \to \mathop{\mathrm{Spec}}(B/J)$. Thus (1) holds. $\square$

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