## 85.11 Separation axioms for formal algebraic spaces

This section is about “absolute” separation conditions on formal algebraic spaces. We will discuss separation conditions for morphisms of formal algebraic spaces later.

Lemma 85.11.1. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. The following are equivalent

1. the reduction of $X$ (Lemma 85.7.2) is a quasi-separated algebraic space,

2. for $U \to X$, $V \to X$ with $U$, $V$ quasi-compact schemes the fibre product $U \times _ X V$ is quasi-compact,

3. 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 85.7.3. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then

$U_{red} \times _{X_{red}} V_{red} = U_{red} \times _ X V_{red} \to U \times _ X V$

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 64.3.3. We omit the proof of the equivalence of (2) and (3). $\square$

Lemma 85.11.2. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. The following are equivalent

1. the reduction of $X$ (Lemma 85.7.2) is a separated algebraic space,

2. for $U \to X$, $V \to X$ with $U$, $V$ affine the fibre product $U \times _ X V$ is affine and

$\mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times _ X V)$

is surjective.

Proof. If (2) holds, then $X_{red}$ is a separated algebraic space by applying Properties of Spaces, Lemma 64.3.3 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 85.7.3. Let $U_{red}, V_{red}, X_{red}$ be the reduction of $U, V, X$. Then

$U_{red} \times _{X_{red}} V_{red} = U_{red} \times _ X V_{red} \to U \times _ X V$

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

$\mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}((U \times _ X V)_{red})$

is surjective, see Properties of Spaces, Lemma 64.3.3. Then $U \times _ X V$ is affine by Limits of Spaces, Proposition 68.15.2. On the other hand, the morphism $U \times _ X V \to U \times V$ of affine schemes is the composition

$U \times _ X V = X \times _{(X \times _ S X)} (U \times _ S V) \to U \times _ S V \to U \times V$

The first morphism is a monomorphism and locally of finite type (Lemma 85.7.3). 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 29.43.7). Thus the ring map

$\mathcal{O}(U) \otimes _\mathbf {Z} \mathcal{O}(V) \longrightarrow \mathcal{O}(U \times _ X V)$

is an epimorphism which is integral of finite type hence finite hence surjective (use Morphisms, Lemma 29.43.4 and Algebra, Lemma 10.106.6). $\square$

Definition 85.11.3. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. We say

1. $X$ is quasi-separated if the equivalent conditions of Lemma 85.11.1 are satisfied.

2. $X$ is separated if the equivalent conditions of Lemma 85.11.2 are satisfied.

The following lemma implies in particular that the completed tensor product of weakly admissible topological rings is a weakly admissible topological ring.

Lemma 85.11.4. Let $S$ be a scheme. Let $X \to Z$ and $Y \to Z$ be morphisms of formal algebraic spaces over $S$. Assume $Z$ separated.

1. If $X$ and $Y$ are affine formal algebraic spaces, then so is $X \times _ Z Y$.

2. If $X$ and $Y$ are McQuillan affine formal algebraic spaces, then so is $X \times _ Z Y$.

3. 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 85.5.1. 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

$X_\lambda \times _ Z Y_\mu \to X_\lambda \times Y_\mu \to \mathop{\mathrm{Spec}}(A \otimes B)$

is a closed immersion. Thus one of the conditions of Lemma 85.5.6 holds and we conclude that $X \times _ Z Y$ is McQuillan. If also $Z$ is McQuillan corresponding to $C$, then

$X_\lambda \times _ Z Y_\mu = \mathop{\mathrm{Spec}}(A/I_\lambda \otimes _ C B/J_\mu )$

hence we see that the weakly admissible topological ring corresponding to $X \times _ Z Y$ is the completed tensor product (see Definition 85.4.7). $\square$

Lemma 85.11.5. 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 85.7.4). 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 65.4.4, 65.4.8, 65.10.3, and 65.4.11. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).