Lemma 115.9.1. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras. Let $N$ be a differential graded $(A, B)$-bimodule with property (P). Let $M$ be a differential graded $A$-module with property (P). Then $Q = M \otimes _ A N$ is a differential graded $B$-module which represents $M \otimes _ A^\mathbf {L} N$ in $D(B)$ and which has a filtration
\[ 0 = F_{-1}Q \subset F_0Q \subset F_1Q \subset \ldots \subset Q \]
by differential graded submodules such that $Q = \bigcup F_ pQ$, the inclusions $F_ iQ \to F_{i + 1}Q$ are admissible monomorphisms, the quotients $F_{i + 1}Q/F_ iQ$ are isomorphic as differential graded $B$-modules to a direct sum of $(A \otimes _ R B)[k]$.
Proof.
Choose filtrations $F_\bullet $ on $M$ and $N$. Then consider the filtration on $Q = M \otimes _ A N$ given by
\[ F_ n(Q) = \sum \nolimits _{i + j = n} F_ i(M) \otimes _ A F_ j(N) \]
This is clearly a differential graded $B$-submodule. We see that
\[ F_ n(Q)/F_{n - 1}(Q) = \bigoplus \nolimits _{i + j = n} F_ i(M)/F_{i - 1}(M) \otimes _ A F_ j(N)/F_{j - 1}(N) \]
for example because the filtration of $M$ is split in the category of graded $A$-modules. Since by assumption the quotients on the right hand side are isomorphic to direct sums of shifts of $A$ and $A \otimes _ R B$ and since $A \otimes _ A (A \otimes _ R B) = A \otimes _ R B$, we conclude that the left hand side is a direct sum of shifts of $A \otimes _ R B$ as a differential graded $B$-module. (Warning: $Q$ does not have a structure of $(A, B)$-bimodule.) This proves the first statement of the lemma. The second statement is immediate from the definition of the functor in Differential Graded Algebra, Lemma 22.33.2.
$\square$
Comments (0)