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

1. $\mathcal{F}'$ is a quasi-coherent sheaf of $\mathcal{O}_ X$-modules,

2. on any affine open $U \subset X$ we have $\mathcal{F}'(U) = \{ s \in \mathcal{F}(U) \mid \mathcal{I}(U)s = 0\}$, and

3. $\mathcal{F}'_ x = \{ s \in \mathcal{F}_ x \mid \mathcal{I}_ x s = 0\}$.

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. $\square$

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