The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

15.55 Derived categories of modules

In this section we put some generalities concerning the derived category of modules over a ring.

Let $A$ be a ring. The category of $A$-modules has products and products are exact. The category of $A$-modules has enough injectives by Lemma 15.54.9. Hence every complex of $A$-modules is quasi-isomorphic to a K-injective complex (Derived Categories, Lemma 13.32.5). It follows that $D(A)$ has countable products (Derived Categories, Lemma 13.32.2) and in fact arbitrary products (Injectives, Lemma 19.13.4). This implies that every inverse system of objects of $D(A)$ has a derived limit (well defined up to isomorphism), see Derived Categories, Section 13.32.

Lemma 15.55.1. Let $R \to S$ be a flat ring map. If $I^\bullet $ is a K-injective complex of $S$-modules, then $I^\bullet $ is K-injective as a complex of $R$-modules.

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _{K(R)}(M^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(S)}(M^\bullet \otimes _ R S, I^\bullet )$ by Algebra, Lemma 10.13.3 and the fact that tensoring with $S$ is exact. $\square$

Lemma 15.55.2. Let $R \to S$ be an epimorphism of rings. Let $I^\bullet $ be a complex of $S$-modules. If $I^\bullet $ is K-injective as a complex of $R$-modules, then $I^\bullet $ is a K-injective complex of $S$-modules.

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _{K(R)}(N^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(S)}(N^\bullet , I^\bullet )$ for any complex of $S$-modules $N^\bullet $, see Algebra, Lemma 10.106.14. $\square$

Lemma 15.55.3. Let $A \to B$ be a ring map. If $I^\bullet $ is a K-injective complex of $A$-modules, then $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a K-injective complex of $B$-modules.

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _{K(B)}(N^\bullet , \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{K(A)}(N^\bullet , I^\bullet )$ by Algebra, Lemma 10.13.4. $\square$


Comments (0)


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 0914. Beware of the difference between the letter 'O' and the digit '0'.