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
\[ \mathcal{O}_ X \longrightarrow \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) \]
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 $\mathcal{O}_{X, x}$:
17.23.1.1
\begin{equation} \label{modules-equation-inclusion-of-annihilator-ideals} (\text{Ann}_{\mathcal{O}_ X}(\mathcal{F}))_ x \subset \text{Ann}_{\mathcal{O}_{X, x}}(\mathcal{F}_ x) \end{equation}
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
\[ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F})_ x \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X,x}}(\mathcal{F}_ x, \mathcal{F}_ x) \]
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
\[ \xymatrix{ \mathcal{O}_ X \ar[r] \ar[d] & \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{F}) \\ \mathcal{O}_ X/\mathcal{I} \ar@{-->}[ur] } \]
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
\[ \mathcal{O}_ X/\mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \longrightarrow \mathcal{F} \]
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
Comment #8941 by Stacks project on