Lemma 47.12.5. With notation and hypotheses as in Lemma 47.12.4 assume A is I-adically complete. Then
where the filtered colimit is over J' \subset J such that V(J') \cap V(I) = V(J) \cap V(I).
Lemma 47.12.5. With notation and hypotheses as in Lemma 47.12.4 assume A is I-adically complete. Then
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
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 10.51.5 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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