Lemma 52.28.2. Let k be a field. Let X be a proper scheme over k. Let \mathcal{L} be an ample invertible \mathcal{O}_ X-module. Let s \in \Gamma (X, \mathcal{L}). Let Y = Z(s) be the zero scheme of s with nth infinitesimal neighbourhood Y_ n = Z(s^ n). Let \mathcal{F} be a coherent \mathcal{O}_ X-module. Assume that for all x \in X \setminus Y we have
\text{depth}(\mathcal{F}_ x) + \dim (\overline{\{ x\} }) > 1
Then \Gamma (V, \mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits \Gamma (Y_ n, \mathcal{F}|_{Y_ n}) is an isomorphism for any open subscheme V \subset X containing Y.
Proof.
By Proposition 52.28.1 this is true for V = X. Thus it suffices to show that the map \Gamma (V, \mathcal{F}) \to \mathop{\mathrm{lim}}\nolimits \Gamma (Y_ n, \mathcal{F}|_{Y_ n}) is injective. If \sigma \in \Gamma (V, \mathcal{F}) maps to zero, then its support is disjoint from Y (details omitted; hint: use Krull's intersection theorem). Then the closure T \subset X of \text{Supp}(\sigma ) is disjoint from Y. Whence T is proper over k (being closed in X) and affine (being closed in the affine scheme X \setminus Y, see Morphisms, Lemma 29.43.18) and hence finite over k (Morphisms, Lemma 29.44.11). Thus T is a finite set of closed points of X. Thus \text{depth}(\mathcal{F}_ x) \geq 2 is at least 1 for x \in T by our assumption. We conclude that \Gamma (V, \mathcal{F}) \to \Gamma (V \setminus T, \mathcal{F}) is injective and \sigma = 0 as desired.
\square
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