The Stacks project

Proof. Choose $f_1, \ldots , f_ r \in A$ generating $I$. Recall that

by More on Algebra, Lemma 15.91.10. Hence the cone $C = \text{Cone}(K \to K^\wedge )$ is given by

which can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to

These complexes vanish on applying $R\Gamma _ Z$, see Lemma 47.9.4. Applying $R\Gamma _ Z$ to the distinguished triangle $K \to K^\wedge \to C \to K[1]$ we see that the first formula of the lemma is correct.

Recall that

by Lemma 47.9.1. Hence the cone $C = \text{Cone}(R\Gamma _ Z(K) \to K)$ can be represented by a complex endowed with a finite filtration whose successive quotients are isomorphic to

These complexes vanish on applying ${\ }^\wedge $, see More on Algebra, Lemma 15.91.12. Applying derived completion to the distinguished triangle $R\Gamma _ Z(K) \to K \to C \to R\Gamma _ Z(K)[1]$ we see that the second formula of the lemma is correct. $\square$

Proof. Follows immediately from Lemma 47.12.1. $\square$

Proof. Say $I = (f_1, \ldots , f_ r)$. Denote $C = (A \to \prod A_{f_ i} \to \ldots \to A_{f_1 \ldots f_ r})$ the alternating Čech complex. Then derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ (More on Algebra, Lemma 15.91.10) and local cohomology by $C \otimes ^\mathbf {L} -$ (Lemma 47.9.1). Combining the isomorphism

(More on Algebra, Lemma 15.73.1) and the map

(More on Algebra, Lemma 15.73.6) we obtain a map

On the other hand, the right hand side is derived complete as it is equal to

Thus $\gamma $ factors through the derived completion of $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ by the universal property of derived completion. However, the derived completion goes inside the $R\mathop{\mathrm{Hom}}\nolimits _ A$ by More on Algebra, Lemma 15.91.13 and we obtain the desired map.

To show that the map of the lemma is an isomorphism we may assume that $K$ and $L$ are derived complete, i.e., $K = K^\wedge $ and $L = L^\wedge $. In this case we are looking at the map

By Proposition 47.12.2 we know that the cohomology groups of the left and the right hand side coincide. In other words, we have to check that the map $\gamma $ sends a morphism $\alpha : K \to L$ in $D(A)$ to the morphism $R\Gamma _ Z(\alpha ) : R\Gamma _ Z(K) \to R\Gamma _ Z(L)$. We omit the verification (hint: note that $R\Gamma _ Z(\alpha )$ is just the map $\alpha \otimes \text{id}_ C : K \otimes ^\mathbf {L} C \to L \otimes ^\mathbf {L} C$ which is almost the same as the construction of the map in More on Algebra, Lemma 15.73.6). $\square$

Proof. Suppose that $J = (g_1, \ldots , g_ m)$. Then $R\Gamma _ Z(M)$ is computed by the complex

by Lemma 47.9.1. By More on Algebra, Lemma 15.94.6 the derived $I$-adic completion of this complex is given by the complex

of usual completions. Since $R\Gamma _ Z(M/I^ nM)$ is computed by the complex $ M/I^ nM \to \prod (M/I^ nM)_{g_{j_0}} \to \ldots \to (M/I^ nM)_{g_1g_2\ldots g_ m}$ and since the transition maps between these complexes are surjective, we conclude that (1) holds by More on Algebra, Lemma 15.87.1. Part (2) then follows from More on Algebra, Lemma 15.87.4. $\square$

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. $\square$