## 86.27 Closed immersions

Here is the definition.

Definition 86.27.1. 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 79.4.1.

Lemma 86.27.2. The composition of two closed immersions is a closed immersion.

Proof. Omitted. $\square$

Lemma 86.27.3. A base change of a closed immersion is a closed immersion.

Proof. Omitted. $\square$

Lemma 86.27.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The following are equivalent

1. $f$ is a closed immersion,

2. for every scheme $T$ and morphism $T \to Y$ the projection $X \times _ Y T \to T$ is a closed immersion,

3. for every affine scheme $T$ and morphism $T \to Y$ the projection $X \times _ Y T \to T$ is a closed immersion,

4. there exists a covering $\{ Y_ j \to Y\}$ as in Definition 86.11.1 such that each $X \times _ Y Y_ j \to Y_ j$ is a closed immersion, and

5. 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

Proof. Omitted. $\square$

Lemma 86.27.5. 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 86.27.1). 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 86.9.6. The ring map $A \to B$ is taut by Lemma 86.19.8. 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 86.29.4.

Example 86.27.6. Let $S$ be a scheme. Let $A$ be a weakly admissible topological ring over $S$. Let $K \subset A$ be a closed ideal. Setting

$B = (A/K)^\wedge = \mathop{\mathrm{lim}}\nolimits _{I \subset A\ w.i.d.} A/(I + K)$

the morphism $\text{Spf}(B) \to \text{Spf}(A)$ is representable, see Example 86.19.11. 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 86.9.4). 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.

Lemma 86.27.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces. Assume

1. $f$ is representable by algebraic spaces,

2. $f$ is a monomorphism,

3. the inclusion $Y_{red} \to Y$ factors through $f$, and

4. $f$ 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 86.27.4 we may assume $Y$ is an affine formal algebraic space.

Say $Y = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } Y_\lambda$ as in Definition 86.9.1. 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 69.15.2 (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 109.35. 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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