The Stacks project

Lemma 15.91.21. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet $ be a filtered complex of $A$-modules. There exists a canonical spectral sequence $(E_ r, \text{d}_ r)_{r \geq 1}$ of bigraded derived complete $A$-modules with $d_ r$ of bidegree $(r, -r + 1)$ and with

\[ E_1^{p, q} = H^{p + q}((\text{gr}^ pK^\bullet )^\wedge ) \]

If the filtration on each $K^ n$ is finite, then the spectral sequence is bounded and converges to $H^*((K^\bullet )^\wedge )$.

Proof. By Lemma 15.91.10 we know that derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ for some $C \in D^ b(A)$. By Lemmas 15.91.20 and 15.68.2 we see that $C$ has finite projective dimension. Thus we may choose a bounded complex of projective modules $P^\bullet $ representing $C$. Then

\[ M^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet ) \]

is a complex of $A$-modules representing $(K^\bullet )^\wedge $. It comes with a filtration given by $F^ pM^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , F^ pK^\bullet )$. We see that $F^ pM^\bullet $ represents $(F^ pK^\bullet )^\wedge $ and hence $\text{gr}^ pM^\bullet $ represents $(\text{gr}K^\bullet )^\wedge $. Thus we find our spectral sequence by taking the spectral sequence of the filtered complex $M^\bullet $, see Homology, Section 12.24. If the filtration on each $K^ n$ is finite, then the filtration on each $M^ n$ is finite because $P^\bullet $ is a bounded complex. Hence the final statement follows from Homology, Lemma 12.24.11. $\square$


Comments (0)

There are also:

  • 14 comment(s) on Section 15.91: Derived Completion

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Your email address will not be published.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 0BKD. Beware of the difference between the letter 'O' and the digit '0'.