Definition 87.27.1. 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 80.4.1.
87.27 Closed immersions
Here is the definition.
Please skip the initial the obligatory lemmas when reading this section.
Lemma 87.27.2. The composition of two closed immersions is a closed immersion.
Proof. Omitted. $\square$
Lemma 87.27.3. A base change of a closed immersion is a closed immersion.
Proof. Omitted. $\square$
Lemma 87.27.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The following are equivalent
$f$ is a closed immersion,
for every scheme $T$ and morphism $T \to Y$ the projection $X \times _ Y T \to T$ is a closed immersion,
for every affine scheme $T$ and morphism $T \to Y$ the projection $X \times _ Y T \to T$ is a closed immersion,
there exists a covering $\{ Y_ j \to Y\} $ as in Definition 87.11.1 such that each $X \times _ Y Y_ j \to Y_ j$ is a closed immersion, and
there exists a morphism $Z \to Y$ of formal algebraic spaces which is representable by algebraic spaces, surjective, flat, and locally of finite presentation such that $X \times _ Y Z \to X$ is a closed immersion, and
add more here.
Proof. Omitted. $\square$
Lemma 87.27.5. 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 87.27.1). 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
\[ B = \mathop{\mathrm{lim}}\nolimits B_\lambda \longrightarrow B_\lambda \]
are surjective as the compositions $A \to B \to B_\lambda $ are surjective. It follows that $Y$ is McQuillan by Lemma 87.9.6. The ring map $A \to B$ is taut by Lemma 87.19.8. 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 87.29.4.
Example 87.27.6. 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 87.19.11. 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 87.9.4). 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.
\[ B = (A/K)^\wedge = \mathop{\mathrm{lim}}\nolimits _{I \subset A\ w.i.d.} A/(I + K) \]
Lemma 87.27.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume
$f$ is representable by algebraic spaces,
$f$ is a monomorphism,
the inclusion $Y_{red} \to Y$ factors through $f$, and
$f$ is locally of finite type or $Y$ is locally Noetherian.
Then $f$ is a closed immersion.
Proof. Assumptions (2) and (3) imply that $X_{red} = X \times _ Y Y_{red} = Y_{red}$. We will use this without further mention.
If $Y' \to Y$ is an étale morphism of formal algebraic spaces over $S$, then the base change $f' : X \times _ Y Y' \to Y'$ satisfies conditions (1) – (4). Hence by Lemma 87.27.4 we may assume $Y$ is an affine formal algebraic space.
Say $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda $ as in Definition 87.9.1. Then $X_\lambda = X \times _ Y Y_\lambda $ is an algebraic space endowed with a monomorphism $f_\lambda : X_\lambda \to Y_\lambda $ which induces an isomorphism $X_{\lambda , red} \to Y_{\lambda , red}$. Thus $X_\lambda $ is an affine scheme by Limits of Spaces, Proposition 70.15.2 (as $X_{\lambda , red} \to X_\lambda $ is surjective and integral). To finish the proof it suffices to show that $X_\lambda \to Y_\lambda $ is a closed immersion which we will do in the next paragraph.
Let $X \to Y$ be a monomorphism of affine schemes such that $X_{red} = X \times _ Y Y_{red} = Y_{red}$. In general, this does not imply that $X \to Y$ is a closed immersion, see Examples, Section 110.36. However, under our assumption (4) we know that in the previous parapgrah either $X_\lambda \to Y_\lambda $ is of finite type or $Y_\lambda $ is Noetherian. This means that $X \to Y$ corresponds to a ring map $R \to A$ such that $R/I \to A/IA$ is an isomorphism where $I \subset R$ is the nil radical (ie., the maximal locally nilpotent ideal of $R$) and either $R \to A$ is of finite type or $R$ is Noetherian. In the first case $R \to A$ is surjective by Algebra, Lemma 10.126.9 and in the second case $I$ is finitely generated, hence nilpotent, hence $R \to A$ is surjective by Nakayama's lemma, see Algebra, Lemma 10.20.1 part (11). $\square$
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