Results of this nature are sometimes referred to as Greenlees-May duality.
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 15.91.10. For an object $K$ in $D(A)$ we have in $D(A)$.
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.
$\square$
\[ R\Gamma _ Z(K^\wedge ) = R\Gamma _ Z(K) \quad \text{and}\quad (R\Gamma _ Z(K))^\wedge = K^\wedge \]
\[ K^\wedge = R\mathop{\mathrm{Hom}}\nolimits _ A\left((A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right) \]
\[ R\mathop{\mathrm{Hom}}\nolimits _ A\left((\prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}), K\right) \]
\[ R\mathop{\mathrm{Hom}}\nolimits _ A(A_{f_{i_0} \ldots f_{i_ p}}, K), \quad p > 0 \]
\[ R\Gamma _ Z(K) = K \otimes ^\mathbf {L} (A \to \prod A_{f_{i_0}} \to \prod A_{f_{i_0i_1}} \to \ldots \to A_{f_1 \ldots f_ r}) \]
\[ K \otimes _ A A_{f_{i_0} \ldots f_{i_ p}}, \quad p > 0 \]
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