The Stacks project

Lemma 15.58.1. The construction above is independent of choices and defines an exact functor of triangulated categories $- \otimes _ R^\mathbf {L} N^\bullet : D(R) \to D(A)$. There is a functorial isomorphism

\[ E^\bullet \otimes _ R^\mathbf {L} N^\bullet = (E^\bullet \otimes _ R^\mathbf {L} A) \otimes _ A^\mathbf {L} N^\bullet \]

for $E^\bullet $ in $D(R)$.

Proof. To prove the existence of the derived functor $- \otimes _ R^\mathbf {L} N^\bullet $ we use the general theory developed in Derived Categories, Section 13.15. Set $\mathcal{D} = K(\text{Mod}_ R)$ and $\mathcal{D}' = D(A)$. Let us write $F : \mathcal{D} \to \mathcal{D}'$ the exact functor of triangulated categories defined by the rule $F(M^\bullet ) = \text{Tot}(M^\bullet \otimes _ R N^\bullet )$. To prove the stated properties of $F$ use Lemmas 15.57.1 and 15.57.2. We let $S$ be the set of quasi-isomorphisms in $\mathcal{D} = K(\text{Mod}_ R)$. This gives a situation as in Derived Categories, Situation 13.15.1 so that Derived Categories, Definition 13.15.2 applies. We claim that $LF$ is everywhere defined. This follows from Derived Categories, Lemma 13.15.15 with $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the collection of K-flat complexes: (1) follows from Lemma 15.57.12 and (2) follows from Lemma 15.57.14. Thus we obtain a derived functor

\[ LF : D(R) = S^{-1}\mathcal{D} \longrightarrow \mathcal{D}' = D(A) \]

see Derived Categories, Equation (13.15.9.1). Finally, Derived Categories, Lemma 13.15.15 guarantees that $LF(K^\bullet ) = F(K^\bullet ) = \text{Tot}(K^\bullet \otimes _ R N^\bullet )$ when $K^\bullet $ is K-flat, i.e., $LF$ is indeed computed in the way described above. Moreover, by Lemma 15.57.5 the complex $K^\bullet \otimes _ R A$ is a K-flat complex of $A$-modules. Hence

\[ (K^\bullet \otimes _ R^\mathbf {L} A) \otimes _ A^\mathbf {L} N^\bullet = \text{Tot}((K^\bullet \otimes _ R A) \otimes _ A N^\bullet ) = \text{Tot}(K^\bullet \otimes _ A N^\bullet ) = K^\bullet \otimes _ A^\mathbf {L} N^\bullet \]

which proves the final statement of the lemma. $\square$


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Your email address will not be published.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 06Y6. Beware of the difference between the letter 'O' and the digit '0'.