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 87.9.1. 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
as fppf sheaves and $X$ satisfies a set theoretic condition (see Remark 87.11.5). 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
by Sites, Lemma 7.17.7. 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 87.9.2. 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 87.9.1. 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 65.5.10. 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 65.4. Thus Spaces, Definition 65.5.1 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 76.9.1) of thickenings in case $F$ and $G$ are algebraic spaces.
Lemma 87.9.3. Let $X_\lambda , \lambda \in \Lambda $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be as in Definition 87.9.1. Then $X_\lambda \to X$ is representable and a thickening.
Proof. The statement makes sense by the discussion in Spaces, Section 65.3 and 65.5. By Lemma 87.9.2 the morphisms $X_\lambda \to X$ are representable. Given $U \to X$ where $U$ is a scheme, then the discussion following Definition 87.9.1 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 87.9.4. Let $X_\lambda , \lambda \in \Lambda $ and $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ be as in Definition 87.9.1. 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 87.9.1. Since $V \to Y$ is a surjection of sheaves, we conclude. $\square$
Lemma 87.9.5. 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 87.11.5.
Proof. It follows from Lemmas 87.9.3 and 87.9.4 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 110.75. To characterize those that do we offer the following lemma.
Lemma 87.9.6. Let $S$ be a scheme. Let $X$ be an fppf sheaf on $(\mathit{Sch}/S)_{fppf}$ which satisfies the set theoretic condition of Remark 87.11.5. The following are equivalent:
there exists a weakly admissible topological ring $A$ over $S$ (see Remark 87.2.3) 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 87.9.1 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 87.9.1 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 87.9.1. 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 87.9.2). 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 87.9.4. 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 87.9.1. 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 15.36. 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 87.9.1. 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 87.9.4. 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 87.9.7. 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 87.9.6. Let $A$ be the weakly admissible topological ring associated to $X$. We say
$X$ is classical if $X$ is McQuillan and $A$ is admissible (More on Algebra, Definition 15.36.1),
$X$ is weakly adic if $X$ is McQuillan and $A$ is weakly adic (Definition 87.7.1),
$X$ is adic if $X$ is McQuillan and $A$ is adic (More on Algebra, Definition 15.36.1),
$X$ 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. Given an affine formal algebraic space $X$ here are the implications among the notions introduced in the definition:
See discussion in Section 87.7 and for a precise statement see Lemma 87.10.3.
Remark 87.9.8. The classical affine formal algebraic spaces correspond to the affine formal schemes considered in EGA ([EGA]). To explain this we assume our base scheme is $\mathop{\mathrm{Spec}}(\mathbf{Z})$. Let $\mathfrak X = \text{Spf}(A)$ be an affine formal scheme. Let $h_\mathfrak X$ be its functor of points as in Lemma 87.2.1. Then $h_\mathfrak X = \mathop{\mathrm{colim}}\nolimits h_{\mathop{\mathrm{Spec}}(A/I)}$ where the colimit is over the collection of ideals of definition of the admissible topological ring $A$. This follows from (87.2.0.1) when evaluating on affine schemes and it suffices to check on affine schemes as both sides are fppf sheaves, see Lemma 87.2.2. Thus $h_\mathfrak X$ is an affine formal algebraic space. In fact, it is a classical affine formal algebraic space by Definition 87.9.7. Thus Lemma 87.2.1 tells us the category of affine formal schemes is equivalent to the category of classical affine formal algebraic spaces.
Having made the connection with affine formal schemes above, it seems natural to make the following definition.
Definition 87.9.9. Let $S$ be a scheme. Let $A$ be a weakly admissible topological ring over $S$, see Definition 87.4.81. The formal spectrum of $A$ is the affine formal algebraic space where the colimit is over the set of weak ideals of definition of $A$ and taken in the category $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.
\[ \text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I) \]
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 4.22.5. 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
is the unique morphism of affine formal algebraic spaces such that the diagrams
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 87.9.10. 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 87.9.4 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 87.9.11. 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 87.9.1. 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
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 67.27.8). Hence $W$ is a scheme by Morphisms of Spaces, Proposition 67.50.2. Thus $X \times _ Y T$ is a scheme by Properties of Spaces, Lemma 66.13.1. $\square$
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