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 $n$th 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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