Proposition 52.12.2. 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 $z \in \overline{\{ x\} } \cap V(\mathfrak a)$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$.

Then we obtain an isomorphism

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

where the colimit is over opens $V \subset U$ containing $U \cap V(I)$.

Proof. Let $T \subset U$ be the set of points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$. Let $\mathcal{F} \to \mathcal{F}'$ be the surjection of coherent modules on $U$ constructed in Local Cohomology, Lemma 51.15.1. Since $\mathcal{F} \to \mathcal{F}'$ is an isomorphism over an open $V \subset U$ containing $U \cap V(I)$ it suffices to prove the lemma with $\mathcal{F}$ replaced by $\mathcal{F}'$. Hence we may and do assume for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 1$.

Let $\mathcal{V}$ be the set of open subschemes $V \subset U$ containing $U \cap V(I)$ ordered by reverse inclusion. This is a directed set. We first claim that

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

is injective for any $V \in \mathcal{F}$ (and in particular the map of the lemma is injective). Namely, an associated point $x$ of $\mathcal{F}$ must have $\overline{\{ x\} } \cap U \cap Y \not= \emptyset$ by the previous paragraph. If $y \in \overline{\{ x\} } \cap U \cap Y$ then $\mathcal{F}_ x$ is a localization of $\mathcal{F}_ y$ and $\mathcal{F}_ y \subset \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ y/I^ n \mathcal{F}_ y$ by Krull's intersection theorem (Algebra, Lemma 10.51.4). This proves the claim as a section $s \in \mathcal{F}(V)$ in the kernel would have to have empty support, hence would have to be zero.

Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$, see Local Cohomology, Lemma 51.8.2. We may and do assume that $H^0_\mathfrak a(M) = 0$. Let $\text{Ass}(M) \setminus V(I) = \{ \mathfrak p_1, \ldots , \mathfrak p_ n\}$. We will prove the lemma by induction on $n$. After reordering we may assume that $\mathfrak p_ n$ is a minimal element of the set $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\}$ with respect to inclusion, i.e, $\mathfrak p_ n$ is a generic point of the support of $M$. Set

$M' = H^0_{\mathfrak p_1 \ldots \mathfrak p_{n - 1} I}(M)$

and $M'' = M/M'$. Let $\mathcal{F}'$ and $\mathcal{F}''$ be the coherent $\mathcal{O}_ U$-modules corresponding to $M'$ and $M''$. Dualizing Complexes, Lemma 47.11.6 implies that $M''$ has only one associated prime, namely $\mathfrak p_ n$. On the other hand, since $\mathfrak p_ n \not\in V(\mathfrak p_1 \ldots \mathfrak p_{n - 1} I)$ we see that $\mathfrak p_ n$ is not an associated prime of $M'$. Hence the induction hypothesis applies to $M'$; note that since $\mathcal{F}' \subset \mathcal{F}$ the condition $\text{depth}(\mathcal{F}'_ x) \geq 1$ at points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ holds, see Algebra, Lemma 10.72.6.

Let $\hat s$ be an element of $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. Let $\hat s''$ be the image in $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}''/I^ n\mathcal{F}'')$. Since $\mathcal{F}''$ has only one associated point, namely the point corresponding to $\mathfrak p_ n$, we see that Lemma 52.12.1 applies and we find an open $U \cap V(I) \subset V \subset U$ and a section $s'' \in \mathcal{F}''(V)$ mapping to $\hat s''$. Let $J \subset A$ be an ideal such that $V(J) = \mathop{\mathrm{Spec}}(A) \setminus V$. By Cohomology of Schemes, Lemma 30.10.5 after replacing $J$ by a power, we may assume there is an $A$-linear map $\varphi : J \to M''$ corresponding to $s''$. Since $M \to M''$ is surjective, for each $g \in J$ we can choose $m_ g \in M$ mapping to $\varphi (g) \in M''$. Then $\hat s'_ g = g \hat s - m_ g$ is in $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}')$. By induction hypothesis there is a $V' \geq V$ section $s'_ g \in \mathcal{F}'(V')$ mapping to $\hat s'_ g$. All in all we conclude that $g \hat s$ is in the image of $\mathcal{F}(V') \to \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ for some $V' \subset V$ possibly depending on $g$. However, since $J$ is finitely generated we can find a single $V' \in \mathcal{V}$ which works for each of the generators and it follows that $V'$ works for all $g$.

Combining the previous paragraph with the injectivity shown in the second paragraph we find there exists a $V' \geq V$ and an $A$-module map $\psi : J \to \mathcal{F}(V')$ such that $\psi (g)$ maps to $g\hat s$. This determines a map $\widetilde{J} \to (V' \to \mathop{\mathrm{Spec}}(A))_*\mathcal{F}|_{V'}$ whose restriction to $V'$ provides an element $s \in \mathcal{F}(V')$ mapping to $\hat s$. This finishes the proof. $\square$

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