The Stacks project

15.56 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 is denoted $\text{Mod}_ A$. We will use the symbol $K(A)$ to denote the homotopy category of complexes of $A$-modules, i.e., we set $K(A) = K(\text{Mod}_ A)$ as a category, see Derived Categories, Section 13.8. The bounded versions are $K^+(A)$, $K^-(A)$, and $K^ b(A)$. We view $K(A)$ as a triangulated category as in Derived Categories, Section 13.10. The derived category of $A$, denoted $D(A)$, is the category obtained from $K(A)$ by inverting quasi-isomorphisms, i.e., we set $D(A) = D(\text{Mod}_ A)$, see Derived Categories, Section 13.11¹. The bounded versions are $D^+(A)$, $D^-(A)$, and $D^ b(A)$.

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 15.55.9. Hence every complex of $A$-modules is quasi-isomorphic to a K-injective complex (Derived Categories, Lemma 13.34.6). It follows that $D(A)$ has countable products (Derived Categories, Lemma 13.34.2) 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 13.34.

Lemma 15.56.1. 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 10.14.3 and the fact that tensoring with $S$ is exact. $\square$

Lemma 15.56.2. 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 10.107.14. $\square$

Lemma 15.56.3. 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 10.14.4. $\square$

[1] See also Injectives, Remark 19.13.3.