Lemma 47.12.1. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $R\Gamma _ Z$ be as in Lemma 47.9.1. Let ${\ }^\wedge$ denote derived completion as in More on Algebra, Lemma 15.85.10. For an object $K$ in $D(A)$ we have

$R\Gamma _ Z(K^\wedge ) = R\Gamma _ Z(K) \quad \text{and}\quad (R\Gamma _ Z(K))^\wedge = K^\wedge$

in $D(A)$.

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

$K^\wedge = R\mathop{\mathrm{Hom}}\nolimits _ A\left((A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right)$

by More on Algebra, Lemma 15.85.10. Hence the cone $C = \text{Cone}(K \to K^\wedge )$ is given by

$R\mathop{\mathrm{Hom}}\nolimits _ A\left((\prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right)$

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

$R\mathop{\mathrm{Hom}}\nolimits _ A(A_{f_{i_0} \ldots f_{i_ p}}, K), \quad p > 0$

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

Recall that

$R\Gamma _ Z(K) = K \otimes ^\mathbf {L} (A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r})$

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

$K \otimes _ A A_{f_{i_0} \ldots f_{i_ p}}, \quad p > 0$

These complexes vanish on applying ${\ }^\wedge$, see More on Algebra, Lemma 15.85.11. Applying derived completion to the distinguished triangle $R\Gamma _ Z(K) \to K \to C \to R\Gamma _ Z(K)$ we see that the second formula of the lemma is correct. $\square$

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