Lemma 52.12.4. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

1. $A$ is $I$-adically complete and has a dualizing complex,

2. if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $\overline{\{ x\} } \cap V(I) \not\subset V(\mathfrak a)$, and $z \in V(\mathfrak a) \cap \overline{\{ x\} }$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$,

3. for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 2$,

Then we obtain an isomorphism

$H^0(U, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$

Proof. Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. By Proposition 52.12.3 we find that $\hat s$ is the image of an element $s \in \mathcal{F}(V)$ for some $V \subset U$ open containing $U \cap V(I)$. However, condition (3) shows that $\text{depth}(\mathcal{F}_ x) \geq 2$ for all $x \in U \setminus V$ and hence we find that $\mathcal{F}(V) = \mathcal{F}(U)$ by Divisors, Lemma 31.5.11 and the proof is complete. $\square$

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