This section is about “absolute” separation conditions on formal algebraic spaces. We will discuss separation conditions for morphisms of formal algebraic spaces later.
Lemma 87.16.1. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. The following are equivalent
the reduction of $X$ (Lemma 87.12.1) is a quasi-separated algebraic space,
for $U \to X$, $V \to X$ with $U$, $V$ quasi-compact schemes the fibre product $U \times _ X V$ is quasi-compact,
for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is quasi-compact.
Proof. Observe that $U \times _ X V$ is a scheme by Lemma 87.11.2. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then
is a thickening of schemes. From this the equivalence of (1) and (2) is clear, keeping in mind the analogous lemma for algebraic spaces, see Properties of Spaces, Lemma 66.3.3. We omit the proof of the equivalence of (2) and (3). $\square$
Lemma 87.16.2. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. The following are equivalent
the reduction of $X$ (Lemma 87.12.1) is a separated algebraic space,
for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is affine and
is surjective.
\[ \mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times _ X V) \]
Proof. If (2) holds, then $X_{red}$ is a separated algebraic space by applying Properties of Spaces, Lemma 66.3.3 to morphisms $U \to X_{red}$ and $V \to X_{red}$ with $U, V$ affine and using that $U \times _{X_{red}} V = U \times _ X V$.
Assume (1). Let $U \to X$ and $V \to X$ be as in (2). Observe that $U \times _ X V$ is a scheme by Lemma 87.11.2. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then
is a thickening of schemes. It follows that $(U \times _ X V)_{red} = (U_{red} \times _{X_{red}} V_{red})_{red}$. In particular, we see that $(U \times _ X V)_{red}$ is an affine scheme and that
is surjective, see Properties of Spaces, Lemma 66.3.3. Then $U \times _ X V$ is affine by Limits of Spaces, Proposition 70.15.2. On the other hand, the morphism $U \times _ X V \to U \times V$ of affine schemes is the composition
The first morphism is a monomorphism and locally of finite type (Lemma 87.11.2). The second morphism is an immersion (Schemes, Lemma 26.21.9). Hence the composition is a monomorphism which is locally of finite type. On the other hand, the composition is integral as the map on underlying reduced affine schemes is a closed immersion by the above and hence universally closed (use Morphisms, Lemma 29.44.7). Thus the ring map
is an epimorphism which is integral of finite type hence finite hence surjective (use Morphisms, Lemma 29.44.4 and Algebra, Lemma 10.107.6). $\square$
Definition 87.16.3. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. We say
The following lemma implies in particular that the completed tensor product of weakly admissible topological rings is a weakly admissible topological ring.
Lemma 87.16.4. Let $S$ be a scheme. Let $X \to Z$ and $Y \to Z$ be morphisms of formal algebraic spaces over $S$. Assume $Z$ separated.
If $X$ and $Y$ are affine formal algebraic spaces, then so is $X \times _ Z Y$.
If $X$ and $Y$ are McQuillan affine formal algebraic spaces, then so is $X \times _ Z Y$.
If $X$, $Y$, and $Z$ are McQuillan affine formal algebraic spaces corresponding to the weakly admissible topological $S$-algebras $A$, $B$, and $C$, then $X \times _ Z Y$ corresponds to $A \widehat{\otimes }_ C B$.
Proof. Write $X = \mathop{\mathrm{colim}}\nolimits X_\lambda $ and $Y = \mathop{\mathrm{colim}}\nolimits Y_\mu $ as in Definition 87.9.1. Then $X \times _ Z Y = \mathop{\mathrm{colim}}\nolimits X_\lambda \times _ Z Y_\mu $. Since $Z$ is separated the fibre products are affine, hence we see that (1) holds. Assume $X$ and $Y$ corresponds to the weakly admissible topological $S$-algebras $A$ and $B$ and $X_\lambda = \mathop{\mathrm{Spec}}(A/I_\lambda )$ and $Y_\mu = \mathop{\mathrm{Spec}}(B/J_\mu )$. Then
is a closed immersion. Thus one of the conditions of Lemma 87.9.6 holds and we conclude that $X \times _ Z Y$ is McQuillan. If also $Z$ is McQuillan corresponding to $C$, then
hence we see that the weakly admissible topological ring corresponding to $X \times _ Z Y$ is the completed tensor product (see Definition 87.4.7). $\square$
Lemma 87.16.5. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Let $U \to X$ be a morphism where $U$ is a separated algebraic space over $S$. Then $U \to X$ is separated.
Proof. The statement makes sense because $U \to X$ is representable by algebraic spaces (Lemma 87.11.3). Let $T$ be a scheme and $T \to X$ a morphism. We have to show that $U \times _ X T \to T$ is separated. Since $U \times _ X T \to U \times _ S T$ is a monomorphism, it suffices to show that $U \times _ S T \to T$ is separated. As this is the base change of $U \to S$ this follows. We used in the argument above: Morphisms of Spaces, Lemmas 67.4.4, 67.4.8, 67.10.3, and 67.4.11. $\square$
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