
## 17.13 Closed immersions of ringed spaces

When do we declare a morphism of ringed spaces $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ to be a closed immersion?

Motivated by the example of a closed immersion of normal topological spaces (ringed with the sheaf of continuous functors), or differential manifolds (ringed with the sheaf of differentiable functions), it seems natural to assume at least:

1. The map $i$ is a closed immersion of topological spaces.

2. The associated map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Denote the kernel by $\mathcal{I}$.

Already these conditions imply a number of pleasing results: For example we prove that the category of $\mathcal{O}_ Z$-modules is equivalent to the category of $\mathcal{O}_ X$-modules annihilated by $\mathcal{I}$ generalizing the result on abelian sheaves of Section 17.6

However, in the Stacks project we choose the definition that guarantees that if $i$ is a closed immersion and $(X, \mathcal{O}_ X)$ is a scheme, then also $(Z, \mathcal{O}_ Z)$ is a scheme. Moreover, in this situation we want $i_*$ and $i^*$ to provide an equivalence between the category of quasi-coherent $\mathcal{O}_ Z$-modules and the category of quasi-coherent $\mathcal{O}_ X$-modules annihilated by $\mathcal{I}$. A minimal condition is that $i_*\mathcal{O}_ Z$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-modules. A good way to guarantee that $i_*\mathcal{O}_ Z$ is a quasi-coherent $\mathcal{O}_ X$-module is to assume that $\mathcal{I}$ is locally generated by sections. We can interpret this condition as saying “$(Z, \mathcal{O}_ Z)$ is locally on $(X, \mathcal{O}_ X)$ defined by setting some regular functions $f_ i$, i.e., local sections of $\mathcal{O}_ X$, equal to zero”. This leads to the following definition.

Definition 17.13.1. A closed immersion of ringed spaces1 is a morphism $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ with the following properties:

1. The map $i$ is a closed immersion of topological spaces.

2. The associated map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Denote the kernel by $\mathcal{I}$.

3. The $\mathcal{O}_ X$-module $\mathcal{I}$ is locally generated by sections.

Actually, this definition still does not guarantee that $i_*$ of a quasi-coherent $\mathcal{O}_ Z$-module is a quasi-coherent $\mathcal{O}_ X$-module. The problem is that it is not clear how to convert a local presentation of a quasi-coherent $\mathcal{O}_ Z$-module into a local presentation for the pushforward. However, the following is trivial.

Lemma 17.13.2. Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a closed immersion of ringed spaces. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ Z$-module. Then $i_*\mathcal{F}$ is locally on $X$ the cokernel of a map of quasi-coherent $\mathcal{O}_ X$-modules.

Proof. This is true because $i_*\mathcal{O}_ Z$ is quasi-coherent by definition. And locally on $Z$ the sheaf $\mathcal{F}$ is a cokernel of a map between direct sums of copies of $\mathcal{O}_ Z$. Moreover, any direct sum of copies of the the same quasi-coherent sheaf is quasi-coherent. And finally, $i_*$ commutes with arbitrary colimits, see Lemma 17.6.3. Some details omitted. $\square$

Lemma 17.13.3. Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces. Assume $i$ is a homeomorphism onto a closed subset of $X$ and that $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Let $\mathcal{F}$ be an $\mathcal{O}_ Z$-module. Then $i_*\mathcal{F}$ is of finite type if and only if $\mathcal{F}$ is of finite type.

Proof. Suppose that $\mathcal{F}$ is of finite type. Pick $x \in X$. If $x \not\in Z$, then $i_*\mathcal{F}$ is zero in a neighbourhood of $x$ and hence finitely generated in a neighbourhood of $x$. If $x = i(z)$, then choose an open neighbourhood $z \in V \subset Z$ and sections $s_1, \ldots , s_ n \in \mathcal{F}(V)$ which generate $\mathcal{F}$ over $V$. Write $V = Z \cap U$ for some open $U \subset X$. Note that $U$ is a neighbourhood of $x$. Clearly the sections $s_ i$ give sections $s_ i$ of $i_*\mathcal{F}$ over $U$. The resulting map

$\bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ U \longrightarrow i_*\mathcal{F}|_ U$

is surjective by inspection of what it does on stalks (here we use that $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective). Hence $i_*\mathcal{F}$ is of finite type.

Conversely, suppose that $i_*\mathcal{F}$ is of finite type. Choose $z \in Z$. Set $x = i(z)$. By assumption there exists an open neighbourhood $U \subset X$ of $x$, and sections $s_1, \ldots , s_ n \in (i_*\mathcal{F})(U)$ which generate $i_*\mathcal{F}$ over $U$. Set $V = Z \cap U$. By definition of $i_*$ the sections $s_ i$ correspond to sections $s_ i$ of $\mathcal{F}$ over $V$. The resulting map

$\bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ V \longrightarrow \mathcal{F}|_ V$

is surjective by inspection of what it does on stalks. Hence $\mathcal{F}$ is of finite type. $\square$

Lemma 17.13.4. Let $i : (Z, \mathcal{O}_ Z) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces. Assume $i$ is a homeomorphism onto a closed subset of $X$ and $i^\sharp : \mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Denote $\mathcal{I} \subset \mathcal{O}_ X$ the kernel of $i^\sharp$. The functor

$i_* : \textit{Mod}(\mathcal{O}_ Z) \longrightarrow \textit{Mod}(\mathcal{O}_ X)$

is exact, fully faithful, with essential image those $\mathcal{O}_ X$-modules $\mathcal{G}$ such that $\mathcal{I}\mathcal{G} = 0$.

Proof. We claim that for a $\mathcal{O}_ Z$-module $\mathcal{F}$ the canonical map

$i^*i_*\mathcal{F} \longrightarrow \mathcal{F}$

is an isomorphism. We check this on stalks. Say $z \in Z$ and $x = i(z)$. We have

$(i^*i_*\mathcal{F})_ z = (i_*\mathcal{F})_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, z} = \mathcal{F}_ z \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, z} = \mathcal{F}_ z$

by Sheaves, Lemma 6.26.4, the fact that $\mathcal{O}_{Z, z}$ is a quotient of $\mathcal{O}_{X, x}$, and Sheaves, Lemma 6.32.1. It follows that $i_*$ is fully faithful.

Let $\mathcal{G}$ be a $\mathcal{O}_ X$-module with $\mathcal{I}\mathcal{G} = 0$. We will prove the canonical map

$\mathcal{G} \longrightarrow i_*i^*\mathcal{G}$

is an isomorphism. This proves that $\mathcal{G} = i_*\mathcal{F}$ with $\mathcal{F} = i^*\mathcal{G}$ which finishes the proof. We check the displayed map induces an isomorphism on stalks. If $x \in X$, $x \not\in i(Z)$, then $\mathcal{G}_ x = 0$ because $\mathcal{I}_ x = \mathcal{O}_{X, x}$ in this case. As above $(i_*i^*\mathcal{G})_ x = 0$ by Sheaves, Lemma 6.32.1. On the other hand, if $x \in Z$, then we obtain the map

$\mathcal{G}_ x \longrightarrow \mathcal{G}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{Z, x}$

by Sheaves, Lemmas 6.26.4 and 6.32.1. This map is an isomorphism because $\mathcal{O}_{Z, x} = \mathcal{O}_{X, x}/\mathcal{I}_ x$ and because $\mathcal{G}_ x$ is annihilated by $\mathcal{I}_ x$ by assumption. $\square$

[1] This is nonstandard notation; see discussion above.

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