The Stacks project

Lemma 15.90.6. Let $I$ be an ideal of a ring $A$.

  1. The derived complete $A$-modules form a weak Serre subcategory $\mathcal{C}$ of $\text{Mod}_ A$.

  2. $D_\mathcal {C}(A) \subset D(A)$ is the full subcategory of derived complete objects.

Proof. Part (2) is immediate from Lemma 15.90.1 and the definitions. For part (1), suppose that $M \to N$ is a map of derived complete modules. Denote $K = (M \to N)$ the corresponding object of $D(A)$. Pick $f \in I$. Then $\mathop{\mathrm{Ext}}\nolimits _ A^ n(A_ f, K)$ is zero for all $n$ because $\mathop{\mathrm{Ext}}\nolimits _ A^ n(A_ f, M)$ and $\mathop{\mathrm{Ext}}\nolimits _ A^ n(A_ f, N)$ are zero for all $n$. Hence $K$ is derived complete. By (2) we see that $\mathop{\mathrm{Ker}}(M \to N)$ and $\mathop{\mathrm{Coker}}(M \to N)$ are objects of $\mathcal{C}$. Finally, suppose that $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence of $A$-modules and $M_1$, $M_3$ are derived complete. Then it follows from the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits $'s that $M_2$ is derived complete. Thus $\mathcal{C}$ is a weak Serre subcategory by Homology, Lemma 12.10.3. $\square$


Comments (0)

There are also:

  • 11 comment(s) on Section 15.90: 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Your email address will not be published.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 091U. Beware of the difference between the letter 'O' and the digit '0'.