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The Stacks project

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

  1. - \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'
  2. 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

  3. 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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