Lemma 47.12.5. With notation and hypotheses as in Lemma 47.12.4 assume $A$ is $I$-adically complete. Then

$H^0(R\Gamma _ Z(M)^\wedge ) = \mathop{\mathrm{colim}}\nolimits H^0_{V(J')}(M)$

where the filtered colimit is over $J' \subset J$ such that $V(J') \cap V(I) = V(J) \cap V(I)$.

Proof. Since $M$ is a finite $A$-module, we have that $M$ is $I$-adically complete. The proof of Lemma 47.12.4 shows that

$H^0(R\Gamma _ Z(M)^\wedge ) = \mathop{\mathrm{Ker}}(M^\wedge \to \prod M_{g_ j}^\wedge ) = \mathop{\mathrm{Ker}}(M \to \prod M_{g_ j}^\wedge )$

where on the right hand side we have usual $I$-adic completion. The kernel $K_ j$ of $M_{g_ j} \to M_{g_ j}^\wedge$ is $\bigcap I^ n M_{g_ j}$. By Algebra, Lemma 10.50.5 for every $\mathfrak p \in V(IA_{g_ j})$ we find an $f \in A_{g_ j}$, $f \not\in \mathfrak p$ such that $(K_ j)_ f = 0$.

Let $s \in H^0(R\Gamma _ Z(M)^\wedge )$. By the above we may think of $s$ as an element of $M$. The support $Z'$ of $s$ intersected with $D(g_ j)$ is disjoint from $D(g_ j) \cap V(I)$ by the arguments above. Thus $Z'$ is a closed subset of $\mathop{\mathrm{Spec}}(A)$ with $Z' \cap V(I) \subset V(J)$. Then $Z' \cup V(J) = V(J')$ for some ideal $J' \subset J$ with $V(J') \cap V(I) \subset V(J)$ and we have $s \in H^0_{V(J')}(M)$. Conversely, any $s \in H^0_{V(J')}(M)$ with $J' \subset J$ and $V(J') \cap V(I) \subset V(J)$ maps to zero in $M_{g_ j}^\wedge$ for all $j$. This proves the lemma. $\square$

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