Lemma 22.33.8. Let $R \to R'$ be a ring map. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $(A', \text{d})$ be the base change, i.e., $A' = A \otimes _ R R'$. If $A$ is K-flat as a complex of $R$-modules, then
$- \otimes _ A^\mathbf {L} A' : D(A, \text{d}) \to D(A', \text{d})$ is equal to the right derived functor of
\[ K(A, \text{d}) \longrightarrow K(A', \text{d}),\quad M \longmapsto M \otimes _ R R' \]
the diagram
\[ \xymatrix{ D(A, \text{d}) \ar[r]_{- \otimes _ A^\mathbf {L} A'} \ar[d]_{restriction} & D(A', \text{d}) \ar[d]^{restriction} \\ D(R) \ar[r]^{- \otimes _ R^\mathbf {L} R'} & D(R') } \]
commutes, and
if $M$ is K-flat as a complex of $R$-modules, then the differential graded $A'$-module $M \otimes _ R R'$ represents $M \otimes _ A^\mathbf {L} A'$.
Proof.
For any differential graded $A$-module $M$ there is a canonical map
\[ c_ M : M \otimes _ R R' \longrightarrow M \otimes _ A A' \]
Let $P$ be a differential graded $A$-module with property (P). We claim that $c_ P$ is an isomorphism and that $P$ is K-flat as a complex of $R$-modules. This will prove all the results stated in the lemma by formal arguments using the definition of derived tensor product in Lemma 22.33.2 and More on Algebra, Section 15.59.
Let $F_\bullet $ be the filtration on $P$ showing that $P$ has property (P). Note that $c_ A$ is an isomorphism and $A$ is K-flat as a complex of $R$-modules by assumption. Hence the same is true for direct sums of shifts of $A$ (you can use More on Algebra, Lemma 15.59.8 to deal with direct sums if you like). Hence this holds for the complexes $F_{p + 1}P/F_ pP$. Since the short exact sequences
\[ 0 \to F_ pP \to F_{p + 1}P \to F_{p + 1}P/F_ pP \to 0 \]
are split exact as sequences of graded modules, we can argue by induction that $c_{F_ pP}$ is an isomorphism for all $p$ and that $F_ pP$ is K-flat as a complex of $R$-modules (use More on Algebra, Lemma 15.59.5). Finally, using that $P = \mathop{\mathrm{colim}}\nolimits F_ pP$ we conclude that $c_ P$ is an isomorphism and that $P$ is K-flat as a complex of $R$-modules (use More on Algebra, Lemma 15.59.8).
$\square$
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