In this section we introduce affine formal algebraic spaces. These will in fact be the same as what are called affine formal schemes in [BVGD]. However, we will call them affine formal algebraic spaces, in order to prevent confusion with the notion of an affine formal scheme as defined in [EGA].

Recall that a thickening of schemes is a closed immersion which induces a surjection on underlying topological spaces, see More on Morphisms, Definition 37.2.1.

Definition 85.5.1. Let $S$ be a scheme. We say a sheaf $X$ on $(\mathit{Sch}/S)_{fppf}$ is an *affine formal algebraic space* if there exist

a directed set $\Lambda $,

a system $(X_\lambda , f_{\lambda \mu })$ over $\Lambda $ in $(\mathit{Sch}/S)_{fppf}$ where

each $X_\lambda $ is affine,

each $f_{\lambda \mu } : X_\lambda \to X_\mu $ is a thickening,

such that

\[ X \cong \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } X_\lambda \]

as fppf sheaves and $X$ satisfies a set theoretic condition (see Remark 85.7.6). A *morphism of affine formal algebraic spaces* over $S$ is a map of sheaves.

Observe that the system $(X_\lambda , f_{\lambda \mu })$ is not part of the data. Suppose that $U$ is a quasi-compact scheme over $S$. Since the transition maps are monomorphisms, we see that

\[ X(U) = \mathop{\mathrm{colim}}\nolimits X_\lambda (U) \]

by Sites, Lemma 7.17.5. Thus the fppf sheafification inherent in the colimit of the definition is a Zariski sheafification which does not do anything for quasi-compact schemes.

Lemma 85.5.2. Let $S$ be a scheme. If $X$ is an affine formal algebraic space over $S$, then the diagonal morphism $\Delta : X \to X \times _ S X$ is representable and a closed immersion.

**Proof.**
Suppose given $U \to X$ and $V \to X$ where $U, V$ are schemes over $S$. Let us show that $U \times _ X V$ is representable. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1. The discussion above shows that Zariski locally on $U$ and $V$ the morphisms factors through some $X_\lambda $. In this case $U \times _ X V = U \times _{X_\lambda } V$ which is a scheme. Thus the diagonal is representable, see Spaces, Lemma 63.5.10. Given $(a, b) : W \to X \times _ S X$ where $W$ is a scheme over $S$ consider the map $X \times _{\Delta , X \times _ S X, (a, b)} W \to W$. As before locally on $W$ the morphisms $a$ and $b$ map into the affine scheme $X_\lambda $ for some $\lambda $ and then we get the morphism $X_\lambda \times _{\Delta _\lambda , X_\lambda \times _ S X_\lambda , (a, b)} W \to W$. This is the base change of $\Delta _\lambda : X_\lambda \to X_\lambda \times _ S X_\lambda $ which is a closed immersion as $X_\lambda \to S$ is separated (because $X_\lambda $ is affine). Thus $X \to X \times _ S X$ is a closed immersion.
$\square$

A morphism of schemes $X \to X'$ is a thickening if it is a closed immersion and induces a surjection on underlying sets of points, see (More on Morphisms, Definition 37.2.1). Hence the property of being a thickening is preserved under arbitrary base change and fpqc local on the target, see Spaces, Section 63.4. Thus Spaces, Definition 63.5.1 applies to “thickening” and we know what it means for a representable transformation $F \to G$ of presheaves on $(\mathit{Sch}/S)_{fppf}$ to be a thickening. We observe that this does not clash with our definition (More on Morphisms of Spaces, Definition 74.9.1) of thickenings in case $F$ and $G$ are algebraic spaces.

Lemma 85.5.3. Let $X_\lambda , \lambda \in \Lambda $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be as in Definition 85.5.1. Then $X_\lambda \to X$ is representable and a thickening.

**Proof.**
The statement makes sense by the discussion in Spaces, Section 63.3 and 63.5. By Lemma 85.5.2 the morphisms $X_\lambda \to X$ are representable. Given $U \to X$ where $U$ is a scheme, then the discussion following Definition 85.5.1 shows that Zariski locally on $U$ the morphism factors through some $X_\mu $ with $\lambda \leq \mu $. In this case $U \times _ X X_\lambda = U \times _{X_\mu } X_\lambda $ so that $U \times _ X X_\lambda \to U$ is a base change of the thickening $X_\lambda \to X_\mu $.
$\square$

Lemma 85.5.4. Let $X_\lambda , \lambda \in \Lambda $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be as in Definition 85.5.1. If $Y$ is a quasi-compact algebraic space over $S$, then any morphism $Y \to X$ factors through an $X_\lambda $.

**Proof.**
Choose an affine scheme $V$ and a surjective étale morphism $V \to Y$. The composition $V \to Y \to X$ factors through $X_\lambda $ for some $\lambda $ by the discussion following Definition 85.5.1. Since $V \to Y$ is a surjection of sheaves, we conclude.
$\square$

Lemma 85.5.5. Let $S$ be a scheme. Let $X$ be a sheaf on $(\mathit{Sch}/S)_{fppf}$. Then $X$ is an affine formal algebraic space if and only if the following hold

any morphism $U \to X$ where $U$ is an affine scheme over $S$ factors through a morphism $T \to X$ which is representable and a thickening with $T$ an affine scheme over $S$, and

a set theoretic condition as in Remark 85.7.6.

**Proof.**
It follows from Lemmas 85.5.3 and 85.5.4 that an affine formal algebraic space satisfies (1) and (2). In order to prove the converse we may assume $X$ is not empty. Let $\Lambda $ be the category of representable morphisms $T \to X$ which are thickenings where $T$ is an affine scheme over $S$. This category is directed. Since $X$ is not empty, $\Lambda $ contains at least one object. If $T \to X$ and $T' \to X$ are in $\Lambda $, then we can factor $T \amalg T' \to X$ through $T'' \to X$ in $\Lambda $. Between any two objects of $\Lambda $ there is a unique arrow or none. Thus $\Lambda $ is a directed set and by assumption $X = \mathop{\mathrm{colim}}\nolimits _{T \to X\text{ in }\Lambda } T$. To finish the proof we need to show that any arrow $T \to T'$ in $\Lambda $ is a thickening. This is true because $T' \to X$ is a monomorphism of sheaves, so that $T = T \times _{T'} T' = T \times _ X T'$ and hence the morphism $T \to T'$ equals the projection $T \times _ X T' \to T'$ which is a thickening because $T \to X$ is a thickening.
$\square$

For a general affine formal algebraic space $X$ there is no guarantee that $X$ has enough functions to separate points (for example). See Examples, Section 108.67. To characterize those that do we offer the following lemma.

Lemma 85.5.6. Let $S$ be a scheme. Let $X$ be an fppf sheaf on $(\mathit{Sch}/S)_{fppf}$ which satisfies the set theoretic condition of Remark 85.7.6. The following are equivalent:

there exists a weakly admissible topological ring $A$ over $S$ (see Remark 85.2.3) such that $X = \mathop{\mathrm{colim}}\nolimits _{I \subset A\text{ weak ideal of definition}} \mathop{\mathrm{Spec}}(A/I)$,

$X$ is an affine formal algebraic space and there exists an $S$-algebra $A$ and a map $X \to \mathop{\mathrm{Spec}}(A)$ such that for a closed immersion $T \to X$ with $T$ an affine scheme the composition $T \to \mathop{\mathrm{Spec}}(A)$ is a closed immersion,

$X$ is an affine formal algebraic space and there exists an $S$-algebra $A$ and a map $X \to \mathop{\mathrm{Spec}}(A)$ such that for a closed immersion $T \to X$ with $T$ a scheme the composition $T \to \mathop{\mathrm{Spec}}(A)$ is a closed immersion,

$X$ is an affine formal algebraic space and for some choice of $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1 the projections $\mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda }) \to \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$ are surjective,

$X$ is an affine formal algebraic space and for any choice of $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1 the projections $\mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda }) \to \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$ are surjective.

Moreover, the weakly admissible topological ring is $A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$ endowed with its limit topology and the weak ideals of definition classify exactly the morphisms $T \to X$ which are representable and thickenings.

**Proof.**
It is clear that (5) implies (4).

Assume (4) for $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1. Set $A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$. Let $T \to X$ be a closed immersion with $T$ a scheme (note that $T \to X$ is representable by Lemma 85.5.2). Since $X_\lambda \to X$ is a thickening, so is $X_\lambda \times _ X T \to T$. On the other hand, $X_\lambda \times _ X T \to X_\lambda $ is a closed immersion, hence $X_\lambda \times _ X T$ is affine. Hence $T$ is affine by Limits, Proposition 32.11.2. Then $T \to X$ factors through $X_\lambda $ for some $\lambda $ by Lemma 85.5.4. Thus $A \to \Gamma (X_\lambda , \mathcal{O}) \to \Gamma (T, \mathcal{O})$ is surjective. In this way we see that (3) holds.

It is clear that (3) implies (2).

Assume (2) for $A$ and $X \to \mathop{\mathrm{Spec}}(A)$. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1. Then $A_\lambda = \Gamma (X_\lambda , \mathcal{O})$ is a quotient of $A$ by assumption (2). Hence $A^\wedge = \mathop{\mathrm{lim}}\nolimits A_\lambda $ is a complete topological ring, see discussion in More on Algebra, Section 15.35. The maps $A^\wedge \to A_\lambda $ are surjective as $A \to A_\lambda $ is. We claim that for any $\lambda $ the kernel $I_\lambda \subset A^\wedge $ of $A^\wedge \to A_\lambda $ is a weak ideal of definition. Namely, it is open by definition of the limit topology. If $f \in I_\lambda $, then for any $\mu \in \Lambda $ the image of $f$ in $A_\mu $ is zero in all the residue fields of the points of $X_\mu $. Hence it is a nilpotent element of $A_\mu $. Hence some power $f^ n \in I_\mu $. Thus $f^ n \to 0$ as $n \to 0$. Thus $A^\wedge $ is weakly admissible. Finally, suppose that $I \subset A^\wedge $ is a weak ideal of definition. Then $I \subset A^\wedge $ is open and hence there exists some $\lambda $ such that $I \supset I_\lambda $. Thus we obtain a morphism $\mathop{\mathrm{Spec}}(A^\wedge /I) \to \mathop{\mathrm{Spec}}(A_\lambda ) \to X$. Then it follows that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A^\wedge /I)$ where now the colimit is over all weak ideals of definition. Thus (1) holds.

Assume (1). In this case it is clear that $X$ is an affine formal algebraic space. Let $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be any presentation as in Definition 85.5.1. For each $\lambda $ we can find a weak ideal of definition $I \subset A$ such that $X_\lambda \to X$ factors through $\mathop{\mathrm{Spec}}(A/I) \to X$, see Lemma 85.5.4. Then $X_\lambda = \mathop{\mathrm{Spec}}(A/I_\lambda )$ with $I \subset I_\lambda $. Conversely, for any weak ideal of definition $I \subset A$ the morphism $\mathop{\mathrm{Spec}}(A/I) \to X$ factors through $X_\lambda $ for some $\lambda $, i.e., $I_\lambda \subset I$. It follows that each $I_\lambda $ is a weak ideal of definition and that they form a cofinal subset of the set of weak ideals of definition. Hence $A = \mathop{\mathrm{lim}}\nolimits A/I = \mathop{\mathrm{lim}}\nolimits A/I_\lambda $ and we see that (5) is true and moreover that $A = \mathop{\mathrm{lim}}\nolimits \Gamma (X_\lambda , \mathcal{O}_{X_\lambda })$.
$\square$

With this lemma in hand we can make the following definition.

Definition 85.5.7. Let $S$ be a scheme. Let $X$ be an affine formal algebraic space over $S$. We say $X$ is *McQuillan* if $X$ satisfies the equivalent conditions of Lemma 85.5.6. Let $A$ be the weakly admissible topological ring associated to $X$. We say

$X$ is *classical* if $X$ is McQuillan and $A$ is admissible,

$X$ is *adic* if $X$ is McQuillan and $A$ is adic,

$X$ is *adic** if $X$ is McQuillan, $A$ is adic, and $A$ has a finitely generated ideal of definition, and

$X$ is *Noetherian* if $X$ is McQuillan and $A$ is both Noetherian and adic.

In [Fujiwara-Kato] they use the terminology “of finite ideal type” for the property that an adic topological ring $A$ contains a finitely generated ideal of definition.

Having made the connection with affine formal schemes above, it seems natural to make the following definition.

Definition 85.5.9. Let $S$ be a scheme. Let $A$ be a weakly admissible topological ring over $S$, see Definition 85.4.8^{1}. The *formal spectrum* of $A$ is the affine formal algebraic space

\[ \text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I) \]

where the colimit is over the set of weak ideals of definition of $A$ and taken in the category $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.

Such a formal spectrum is McQuillan by construction and conversely every McQuillan affine formal algebraic space is isomorphic to a formal spectrum. To be sure, in our theory there exist affine formal algebraic spaces which are not the formal spectrum of any weakly admissible topological ring. Following [Yasuda] we could introduce $S$-pro-rings to be pro-objects in the category of $S$-algebras, see Categories, Remark 4.22.4. Then every affine formal algebraic space over $S$ would be the formal spectrum of such an $S$-pro-ring. We will not do this and instead we will work directly with the corresponding affine formal algebraic spaces.

The construction of the formal spectrum is functorial. To explain this let $\varphi : B \to A$ be a continuous map of weakly admissible topological rings over $S$. Then

\[ \text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A) \]

is the unique morphism of affine formal algebraic spaces such that the diagrams

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

commute for all weak ideals of definition $I \subset A$ and $J \subset B$ with $\varphi (I) \subset J$. Since continuity of $\varphi $ implies that for every weak ideal of definition $J \subset B$ there is a weak ideal of definition $I \subset A$ with the required property, we see that the required commutativities uniquely determine and define $\text{Spf}(\varphi )$.

Lemma 85.5.10. Let $S$ be a scheme. Let $A$, $B$ be weakly admissible topological rings over $S$. Any morphism $f : \text{Spf}(B) \to \text{Spf}(A)$ of affine formal algebraic spaces over $S$ is equal to $\text{Spf}(f^\sharp )$ for a unique continuous $S$-algebra map $f^\sharp : A \to B$.

**Proof.**
Let $f : \text{Spf}(B) \to \text{Spf}(A)$ be as in the lemma. Let $J \subset B$ be a weak ideal of definition. By Lemma 85.5.4 there exists a weak ideal of definition $I \subset A$ such that $\mathop{\mathrm{Spec}}(B/J) \to \text{Spf}(B) \to \text{Spf}(A)$ factors through $\mathop{\mathrm{Spec}}(A/I)$. By Schemes, Lemma 26.6.4 we obtain an $S$-algebra map $A/I \to B/J$. These maps are compatible for varying $J$ and define the map $f^\sharp : A \to B$. This map is continuous because for every weak ideal of definition $J \subset B$ there is a weak ideal of definition $I \subset A$ such that $f^\sharp (I) \subset J$. The equality $f = \text{Spf}(f^\sharp )$ holds by our choice of the ring maps $A/I \to B/J$ which make up $f^\sharp $.
$\square$

Lemma 85.5.11. Let $S$ be a scheme. Let $f : X \to Y$ be a map of presheaves on $(\mathit{Sch}/S)_{fppf}$. If $X$ is an affine formal algebraic space and $f$ is representable by algebraic spaces and locally quasi-finite, then $f$ is representable (by schemes).

**Proof.**
Let $T$ be a scheme over $S$ and $T \to Y$ a map. We have to show that the algebraic space $X \times _ Y T$ is a scheme. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ as in Definition 85.5.1. Let $W \subset X \times _ Y T$ be a quasi-compact open subspace. The restriction of the projection $X \times _ Y T \to X$ to $W$ factors through $X_\lambda $ for some $\lambda $. Then

\[ W \to X_\lambda \times _ S T \]

is a monomorphism (hence separated) and locally quasi-finite (because $W \to X \times _ Y T \to T$ is locally quasi-finite by our assumption on $X \to Y$, see Morphisms of Spaces, Lemma 65.27.8). Hence $W$ is a scheme by Morphisms of Spaces, Proposition 65.50.2. Thus $X \times _ Y T$ is a scheme by Properties of Spaces, Lemma 64.13.1.
$\square$

## Comments (0)