28.24 Sections with support in a closed subset
Given any topological space $X$, a closed subset $Z \subset X$, and an abelian sheaf $\mathcal{F}$ you can take the subsheaf of sections whose support is contained in $Z$. If $X$ is a scheme, $Z$ a closed subscheme, and $\mathcal{F}$ a quasi-coherent module there is a variant where you take sections which are scheme theoretically supported on $Z$. However, in the scheme setting you have to be careful because the resulting $\mathcal{O}_ X$-module may not be quasi-coherent.
Lemma 28.24.1. Let $X$ be a quasi-compact and quasi-separated scheme. Let $U \subset X$ be an open subscheme. The following are equivalent:
$U$ is retrocompact in $X$,
$U$ is quasi-compact,
$U$ is a finite union of affine opens, and
there exists a finite type quasi-coherent sheaf of ideals $\mathcal{I} \subset \mathcal{O}_ X$ such that $X \setminus U = V(\mathcal{I})$ (set theoretically).
Proof.
The equivalence of (1), (2), and (3) follows from Lemma 28.2.3. Assume (1), (2), (3). Let $T = X \setminus U$. By Schemes, Lemma 26.12.4 there exists a unique quasi-coherent sheaf of ideals $\mathcal{J}$ cutting out the reduced induced closed subscheme structure on $T$. Note that $\mathcal{J}|_ U = \mathcal{O}_ U$ which is an $\mathcal{O}_ U$-modules of finite type. By Lemma 28.22.2 there exists a quasi-coherent subsheaf $\mathcal{I} \subset \mathcal{J}$ which is of finite type and has the property that $\mathcal{I}|_ U = \mathcal{J}|_ U$. Then $X \setminus U = V(\mathcal{I})$ and we obtain (4). Conversely, if $\mathcal{I}$ is as in (4) and $W = \mathop{\mathrm{Spec}}(R) \subset X$ is an affine open, then $\mathcal{I}|_ W = \widetilde{I}$ for some finitely generated ideal $I \subset R$, see Lemma 28.16.1. It follows that $U \cap W = \mathop{\mathrm{Spec}}(R) \setminus V(I)$ is quasi-compact, see Algebra, Lemma 10.29.1. Hence $U \subset X$ is retrocompact by Lemma 28.2.6.
$\square$
Lemma 28.24.2. Let $X$ be a scheme. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Consider the sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}'$ which associates to every open $U \subset X$
\[ \mathcal{F}'(U) = \{ s \in \mathcal{F}(U) \mid \mathcal{I}s = 0\} \]
Assume $\mathcal{I}$ is of finite type. Then
$\mathcal{F}'$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-modules,
on any affine open $U \subset X$ we have $\mathcal{F}'(U) = \{ s \in \mathcal{F}(U) \mid \mathcal{I}(U)s = 0\} $, and
$\mathcal{F}'_ x = \{ s \in \mathcal{F}_ x \mid \mathcal{I}_ x s = 0\} $.
Proof.
It is clear that the rule defining $\mathcal{F}'$ gives a subsheaf of $\mathcal{F}$ (the sheaf condition is easy to verify). Hence we may work locally on $X$ to verify the other statements. In other words we may assume that $X = \mathop{\mathrm{Spec}}(A)$, $\mathcal{F} = \widetilde{M}$ and $\mathcal{I} = \widetilde{I}$. It is clear that in this case $\mathcal{F}'(U) = \{ x \in M \mid Ix = 0\} =: M'$ because $\widetilde{I}$ is generated by its global sections $I$ which proves (2). To show $\mathcal{F}'$ is quasi-coherent it suffices to show that for every $f \in A$ we have $\{ x \in M_ f \mid I_ f x = 0\} = (M')_ f$. Write $I = (g_1, \ldots , g_ t)$, which is possible because $\mathcal{I}$ is of finite type, see Lemma 28.16.1. If $x = y/f^ n$ and $I_ fx = 0$, then that means that for every $i$ there exists an $m \geq 0$ such that $f^ mg_ ix = 0$. We may choose one $m$ which works for all $i$ (and this is where we use that $I$ is finitely generated). Then we see that $f^ mx \in M'$ and $x/f^ n = f^ mx/f^{n + m}$ in $(M')_ f$ as desired. The proof of (3) is similar and omitted.
$\square$
Definition 28.24.3. Let $X$ be a scheme. Let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals of finite type. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. The subsheaf $\mathcal{F}' \subset \mathcal{F}$ defined in Lemma 28.24.2 above is called the subsheaf of sections annihilated by $\mathcal{I}$.
Lemma 28.24.4. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of schemes. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be a quasi-coherent sheaf of ideals of finite type. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{F}' \subset \mathcal{F}$ be the subsheaf of sections annihilated by $f^{-1}\mathcal{I}\mathcal{O}_ X$. Then $f_*\mathcal{F}' \subset f_*\mathcal{F}$ is the subsheaf of sections annihilated by $\mathcal{I}$.
Proof.
Omitted. (Hint: The assumption that $f$ is quasi-compact and quasi-separated implies that $f_*\mathcal{F}$ is quasi-coherent so that Lemma 28.24.2 applies to $\mathcal{I}$ and $f_*\mathcal{F}$.)
$\square$
For an abelian sheaf on a topological space we have discussed the subsheaf of sections with support in a closed subset in Modules, Remark 17.6.2. For quasi-coherent modules this submodule isn't always a quasi-coherent module, but if the closed subset has a retrocompact complement, then it is.
Lemma 28.24.5. Let $X$ be a scheme. Let $Z \subset X$ be a closed subset. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Consider the sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}'$ which associates to every open $U \subset X$
\[ \mathcal{F}'(U) = \{ s \in \mathcal{F}(U) \mid \text{the support of }s\text{ is contained in }Z \cap U\} \]
If $X \setminus Z$ is a retrocompact open of $X$, then
for an affine open $U \subset X$ there exist a finitely generated ideal $I \subset \mathcal{O}_ X(U)$ such that $Z \cap U = V(I)$,
for $U$ and $I$ as in (1) we have $\mathcal{F}'(U) = \{ x \in \mathcal{F}(U) \mid I^ nx = 0 \text{ for some } n\} $,
$\mathcal{F}'$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-modules.
Proof.
Part (1) is Algebra, Lemma 10.29.1. Let $U = \mathop{\mathrm{Spec}}(A)$ and $I$ be as in (1). Then $\mathcal{F}|_ U$ is the quasi-coherent sheaf associated to some $A$-module $M$. We have
\[ \mathcal{F}'(U) = \{ x \in M \mid x = 0\text{ in }M_\mathfrak p \text{ for all }\mathfrak p \not\in Z\} . \]
by Modules, Definition 17.5.1. Thus $x \in \mathcal{F}'(U)$ if and only if $V(\text{Ann}(x)) \subset V(I)$, see Algebra, Lemma 10.40.7. Since $I$ is finitely generated this is equivalent to $I^ n x = 0$ for some $n$. This proves (2).
Proof of (3). Observe that given $U \subset X$ open there is an exact sequence
\[ 0 \to \mathcal{F}'(U) \to \mathcal{F}(U) \to \mathcal{F}(U \setminus Z) \]
If we denote $j : X \setminus Z \to X$ the inclusion morphism, then we observe that $\mathcal{F}(U \setminus Z)$ is the sections of the module $j_*(\mathcal{F}|_{X \setminus Z})$ over $U$. Thus we have an exact sequence
\[ 0 \to \mathcal{F}' \to \mathcal{F} \to j_*(\mathcal{F}|_{X \setminus Z}) \]
The restriction $\mathcal{F}|_{X \setminus Z}$ is quasi-coherent. Hence $j_*(\mathcal{F}|_{X \setminus Z})$ is quasi-coherent by Schemes, Lemma 26.24.1 and our assumption that $j$ is quasi-compact (any open immersion is separated). Hence $\mathcal{F}'$ is quasi-coherent as a kernel of a map of quasi-coherent modules, see Schemes, Section 26.24.
$\square$
Definition 28.24.6. Let $X$ be a scheme. Let $T \subset X$ be a closed subset whose complement is retrocompact in $X$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. The quasi-coherent subsheaf $\mathcal{F}' \subset \mathcal{F}$ defined in Lemma 28.24.5 is called the subsheaf of sections supported on $T$.
Lemma 28.24.7. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of schemes. Let $Z \subset Y$ be a closed subset such that $Y \setminus Z$ is retrocompact in $Y$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\mathcal{F}' \subset \mathcal{F}$ be the subsheaf of sections supported in $f^{-1}Z$. Then $f_*\mathcal{F}' \subset f_*\mathcal{F}$ is the subsheaf of sections supported in $Z$.
Proof.
Omitted. (Hint: First show that $X \setminus f^{-1}Z$ is retrocompact in $X$ as $Y \setminus Z$ is retrocompact in $Y$. Hence Lemma 28.24.5 applies to $f^{-1}Z$ and $\mathcal{F}$. As $f$ is quasi-compact and quasi-separated we see that $f_*\mathcal{F}$ is quasi-coherent. Hence Lemma 28.24.5 applies to $Z$ and $f_*\mathcal{F}$. Finally, match the sheaves directly.)
$\square$
Comments (2)
Comment #120 by Pieter Belmans on
Comment #121 by Johan on