Definition 17.23.1. Let (X, \mathcal{O}_ X) be a ringed space and let \mathcal{F} be an \mathcal{O}_ X-module. The annihilator of \mathcal{F}, denoted \text{Ann}_{\mathcal{O}_ X}(\mathcal{F}) is the kernel of the map \mathcal{O}_ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) discussed above.
17.23 The annihilator of a sheaf of modules
Let (X, \mathcal{O}_ X) be a ringed space. Let \mathcal{F} be an \mathcal{O}_ X-module. There is a canonical map of sheaves of \mathcal{O}_ X-modules
which sends a local section f \in \mathcal{O}_ X(U) to the map f : \mathcal{F}|_ U \to \mathcal{F}|_ U given by multiplication by f.
For each x\in X, there is an inclusion of ideals of \mathcal{O}_{X, x}:
since after all any section of \text{Ann}_{\mathcal{O}_ X}(\mathcal{F}) will annihilate the stalks of \mathcal{F} at all points at which it is defined. Here is a simple situation in which (17.23.1.1) becomes an equality.
Lemma 17.23.2. Let (X, \mathcal{O}_ X) be a ringed space and let \mathcal{F} be a sheaf of \mathcal{O}_ X-modules. If \mathcal{F} is of finite type, then (\text{Ann}_{\mathcal{O}_ X}(\mathcal{F}))_ x = \text{Ann}_{\mathcal{O}_{X, x}}(\mathcal{F}_ x).
Proof. By Lemma 17.22.4 the map
is injective. Thus any section f of \mathcal{O}_ X over an open neighbourhood U of x which acts as zero on \mathcal{F}_ x will act as zero on \mathcal{F}|_ V for some U \supset V \ni x open. Hence the inclusion (17.23.1.1) is an equality. \square
Lemma 17.23.3. Let (X, \mathcal{O}_ X) be a ringed space, let \mathcal{F} be an \mathcal{O}_ X-module and let \mathcal{I} \subset \mathcal{O}_ X be an ideal sheaf. If \mathcal{I} \subset \text{Ann}_{\mathcal{O}_ X}(\mathcal{F}), then \mathcal{F} has a natural \mathcal{O}_ X/\mathcal{I}-module structure which agrees with the usual commutative algebra construction on stalks.
Proof. Applying the universal property of the cokernel of the inclusion \mathcal{I} \to \mathcal{O}_ X, we obtain a commutative diagram
of \mathcal{O}_ X-modules. By Lemma 17.22.1 the resulting map \mathcal{O}_ X/\mathcal{I} \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) corresponds to a map of \mathcal{O}_ X-modules
which means we have an \mathcal{O}_ X/\mathcal{I}-module structure on \mathcal{F} compatible with the given \mathcal{O}_ X-module structure. We omit the verification of the statement on stalks. \square
Lemma 17.23.4. Let (X,\mathcal{O}_ X) be a ringed space. If \mathcal{O}_ X and \mathcal{F} are coherent, then so is \text{Ann}_{\mathcal{O}_ X}(\mathcal{F}).
Proof. Since \text{Ann}_{\mathcal{O}_ X}(\mathcal{F}) is the kernel of \mathcal{O}_ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) by Lemma 17.12.4 it suffices to show that \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) is coherent. This follows from Lemma 17.22.6 and the fact that \mathcal{F} is coherent and a fortiori finitely presented (Lemma 17.12.2). \square
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Comment #8325 by Elías Guisado on
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