The Stacks project

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

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

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 15.91.10. 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 $K \to K^\wedge \to C \to K[1]$ 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 15.91.12. 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. $\square$


Comments (0)

There are also:

  • 5 comment(s) on Section 47.12: Torsion versus complete modules

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A6W. Beware of the difference between the letter 'O' and the digit '0'.