Definition 85.20.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of formal algebraic spaces over $S$. We say $f$ is a *closed immersion* if $f$ is representable by algebraic spaces and a closed immersion in the sense of Bootstrap, Definition 78.4.1.

## 85.20 Closed immersions

Here is the definition.

Lemma 85.20.2. Let $S$ be a scheme. Let $X$ be a McQuillan affine formal algebraic space over $S$. Let $f : Y \to X$ be a closed immersion of formal algebraic spaces over $S$. Then $Y$ is a McQuillan affine formal algebraic space and $f$ corresponds to a continuous homomorphism $A \to B$ of weakly admissible topological $S$-algebras which is taut, has closed kernel, and has dense image.

**Proof.**
Write $X = \text{Spf}(A)$ where $A$ is a weakly admissible topological ring. Let $I_\lambda $ be a fundamental system of weakly admissible ideals of definition in $A$. Then $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda )$ is a closed subscheme of $\mathop{\mathrm{Spec}}(A/I_\lambda )$ and hence affine (Definition 85.20.1). Say $Y \times _ X \mathop{\mathrm{Spec}}(A/I_\lambda ) = \mathop{\mathrm{Spec}}(B_\lambda )$. The ring map $A/I_\lambda \to B_\lambda $ is surjective. Hence the projections

are surjective as the compositions $A \to B \to B_\lambda $ are surjective. It follows that $Y$ is McQuillan by Lemma 85.5.6. The ring map $A \to B$ is taut by Lemma 85.14.10. The kernel is closed because $B$ is complete and $A \to B$ is continuous. Finally, as $A \to B_\lambda $ is surjective for all $\lambda $ we see that the image of $A$ in $B$ is dense. $\square$

Even though we have the result above, in general we do not know how closed immersions behave when the target is a McQuillan affine formal algebraic space, see Remark 85.22.4.

Example 85.20.3. Let $S$ be a scheme. Let $A$ be a weakly admissible topological ring over $S$. Let $K \subset A$ be a closed ideal. Setting

the morphism $\text{Spf}(B) \to \text{Spf}(A)$ is representable, see Example 85.14.11. If $T \to \text{Spf}(A)$ is a morphism where $T$ is a quasi-compact scheme, then this factors through $\mathop{\mathrm{Spec}}(A/I)$ for some weak ideal of definition $I \subset A$ (Lemma 85.5.4). Then $T \times _{\text{Spf}(A)} \text{Spf}(B)$ is equal to $T \times _{\mathop{\mathrm{Spec}}(A/I)} \mathop{\mathrm{Spec}}(A/(K + I))$ and we see that $\text{Spf}(B) \to \text{Spf}(A)$ is a closed immersion. The kernel of $A \to B$ is $K$ as $K$ is closed, but beware that in general the ring map $A \to B = (A/K)^\wedge $ need not be surjective.

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