The Stacks project

86.15 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 86.15.1. The composition of morphisms representable by algebraic spaces is representable by algebraic spaces. The same holds for representable (by schemes).

Proof. See Bootstrap, Lemma 79.3.8. $\square$

Lemma 86.15.2. 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 79.3.3. $\square$

Lemma 86.15.3. 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 86.11.5. Thus $X \to Y$ is representable by algebraic spaces by Bootstrap, Lemma 79.3.10. $\square$

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

Lemma 86.15.4. 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 86.7.1 and for each $j$ a covering $\{ X_{ji} \to Y_ j \times _ Y X\} $ as in Definition 86.7.1 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 86.7.1 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

  6. add more here.

Proof. It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, (2) implies (1) by Bootstrap, Lemma 79.11.4 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 79.3.8) and factors through $V$. Thus $U \to V$ is representable by algebraic spaces by Lemma 86.15.3.

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 86.11.3. 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 86.7.1 and for each $j$ any covering $\{ X_{ji} \to Y_ j \times _ Y X\} $ as in Definition 86.7.1. 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 86.15.5. 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 86.5.1. 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 86.9.1. $\square$

Lemma 86.15.6. 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 86.7.1. 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 86.11.1. This follows from Lemma 86.15.5. $\square$

Lemma 86.15.7. 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 66.20.3. Write $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 86.5.1. 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 79.4.1. 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 86.5.4. Since $f$ is quasi-compact we see that $X \times _ Y T$ is quasi-compact (Lemma 86.13.3). 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 86.15.8. 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 adic*, then $X$ is adic*,

  4. 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 86.5.1. 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). Assume $Y$ is adic*. Then $Y = \text{Spf}(B)$ for some adic topological ring $B$ which has a finitely generated ideal $J$ such that $\{ J^ n\} $ is a fundamental system of open ideals. Of course, then $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J^ n)$. The schemes $X \times _ Y \mathop{\mathrm{Spec}}(B/J^ n)$ are affine and we can write $X \times _ Y \mathop{\mathrm{Spec}}(B/J^ n) = \mathop{\mathrm{Spec}}(A_ n)$. Then $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A_ n)$. The $B$-algebra maps $A_{n + 1} \to A_ n$ are surjective and induce isomorphisms $A_{n + 1}/J^ nA_{n + 1} \to A_ n$. By Algebra, Lemma 10.98.2 the ring $A = \mathop{\mathrm{lim}}\nolimits A_ n$ is $J$-adically complete and $A/J^ n A = A_ n$. Hence $X = \text{Spf}(A^\wedge )$ is adic*.

Proof of (4). Combining (3) with Lemma 86.6.3 we see that $X$ is adic*. Thus we can use the criterion of Lemma 86.6.5. 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 86.15.9. 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 adic*, then $X$ is adic*,

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

Lemma 86.15.10. 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 86.4.11.

Proof. Parts (1) and (2) are equivalent by Lemma 86.15.7.

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 86.5.6). 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 86.5.6 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 86.12.4 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 86.4.13. 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 86.5.6). 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 86.12.4 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 86.4.2. 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 86.5.4) we conclude that $\text{Spf}(\varphi )$ is representable, i.e., (2) holds. This finishes the proof. $\square$

Example 86.15.11. 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 86.4.2. By Lemma 86.4.10 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 86.15.10.

Remark 86.15.12 (Warning). Lemma 86.15.10 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 109.73.

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

Lemma 86.15.13. 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 86.5.10. 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 86.15.7). 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 86.15.14. With notation and assumptions as in Lemma 86.15.13 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 86.5.4). 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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