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:
The map $i$ is a closed immersion of topological spaces.
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:
The map $i$ is a closed immersion of topological spaces.
The associated map $\mathcal{O}_ X \to i_*\mathcal{O}_ Z$ is surjective. Denote the kernel by $\mathcal{I}$.
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
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
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 is exact, fully faithful, with essential image those $\mathcal{O}_ X$-modules $\mathcal{G}$ such that $\mathcal{I}\mathcal{G} = 0$.
\[ i_* : \textit{Mod}(\mathcal{O}_ Z) \longrightarrow \textit{Mod}(\mathcal{O}_ X) \]
Proof. We claim that for an $\mathcal{O}_ Z$-module $\mathcal{F}$ the canonical map
is an isomorphism. We check this on stalks. Say $z \in Z$ and $x = i(z)$. We have
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
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
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. $\square$
Remark 17.13.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $Z \subset X$ be a closed subset. For an $\mathcal{O}_ X$-module $\mathcal{F}$ we can consider the submodule of sections with support in $Z$, denoted $\mathcal{H}_ Z(\mathcal{F})$, defined by the rule Observe that $\mathcal{H}_ Z(\mathcal{F})(U)$ is a module over $\mathcal{O}_ X(U)$, i.e., $\mathcal{H}_ Z(\mathcal{F})$ is an $\mathcal{O}_ X$-module. By construction $\mathcal{H}_ Z(\mathcal{F})$ is the largest $\mathcal{O}_ X$-submodule of $\mathcal{F}$ whose support is contained in $Z$. Applying Lemma 17.13.4 to the morphism of ringed spaces $(Z, \mathcal{O}_ X|_ Z) \to (X, \mathcal{O}_ X)$ we may (and we do) view $\mathcal{H}_ Z(\mathcal{F})$ as an $\mathcal{O}_ X|_ Z$-module on $Z$. Thus we obtain a functor This functor is left exact, but in general not exact. All of the statements made above follow directly from Lemma 17.5.2. Clearly the construction is compatible with the construction in Remark 17.6.2.
\[ \mathcal{H}_ Z(\mathcal{F})(U) = \{ s \in \mathcal{F}(U) \mid \text{Supp}(s) \subset U \cap Z\} \]
\[ \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X|_ Z), \quad \mathcal{F} \longmapsto \mathcal{H}_ Z(\mathcal{F}) \text{ viewed as an }\mathcal{O}_ X|_ Z\text{-module on }Z \]
Lemma 17.13.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. The functor $\mathcal{H}_ Z : \textit{Mod}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X|_ Z)$ of Remark 17.13.5 is right adjoint to $i_* : \textit{Mod}(\mathcal{O}_ X|_ Z) \to \textit{Mod}(\mathcal{O}_ X)$.
Proof. We have to show that for any $\mathcal{O}_ X$-module $\mathcal{F}$ and any $\mathcal{O}_ X|_ Z$-module $\mathcal{G}$ we have
This is clear because after all any section of $i_*\mathcal{G}$ has support in $Z$. Details omitted. $\square$
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