Remark 15.59.11. In fact, we can do better than Lemma 15.59.10. Namely, we can find a quasi-isomorphism $P^\bullet \to M^\bullet$ where $P^\bullet$ is a complex of $R$-modules endowed with a filtration

$0 = F_{-1}P^\bullet \subset F_0P^\bullet \subset F_1P^\bullet \subset \ldots \subset P^\bullet$

by subcomplexes such that

1. $P^\bullet = \bigcup F_ pP^\bullet$,

2. the inclusions $F_ iP^\bullet \to F_{i + 1}P^\bullet$ are termwise split injections,

3. the quotients $F_{i + 1}P^\bullet /F_ iP^\bullet$ are isomorphic to direct sums of shifts $R[k]$ (as complexes, so differentials are zero).

This will be shown in Differential Graded Algebra, Lemma 22.20.4. Moreover, given such a complex we obtain a distinguished triangle

$\bigoplus F_ iP^\bullet \to \bigoplus F_ iP^\bullet \to M^\bullet \to \bigoplus F_ iP^\bullet [1]$

in $D(R)$. Using this we can sometimes reduce statements about general complexes to statements about $R[k]$ (this of course only works if the statement is preserved under taking direct sums). More precisely, let $T$ be a property of objects of $D(R)$. Suppose that

1. if $K_ i \in D(R)$, $i \in I$ is a family of objects with $T(K_ i)$ for all $i \in I$, then $T(\bigoplus K_ i)$,

2. if $K \to L \to M \to K[1]$ is a distinguished triangle and $T$ holds for two, then $T$ holds for the third object,

3. $T(R[k])$ holds for all $k$.

Then $T$ holds for all objects of $D(R)$.

There are also:

• 4 comment(s) on Section 15.59: Derived tensor product

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).