Lemma 15.60.5. Assumptions as in Lemma 15.60.3. For $M \in D(A)$ there are canonical isomorphisms

of $A' \otimes _{R'} B'$-modules.

Lemma 15.60.5. Assumptions as in Lemma 15.60.3. For $M \in D(A)$ there are canonical isomorphisms

\[ H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B') = H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') \]

of $A' \otimes _{R'} B'$-modules.

**Proof.**
Let us elucidate the two sides of the equation. On the left hand side we have the composition of the functors $D(A) \to D(A') \to D(R') \to D(B')$ with the functor $H^ i : D(B') \to \text{Mod}_{B'}$. Since there is a map from $A'$ to the endomorphisms of the object $(M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B'$ in $D(B')$, we see that the left hand side is indeed an $A' \otimes _{R'} B'$-module. By the same arguments we see that $H^ i(M \otimes _ R^\mathbf {L} B)$ has an $A \otimes _ R B$-module structure.

We first prove the result in case $B' = R' \otimes _ R B$. In this case we choose a resolution $F^\bullet \to B$ by free $R$-modules. We also choose a K-flat complex $M^\bullet $ of $A$-modules representing $M$. Then the left hand side is represented by

\begin{align*} H^ i(\text{Tot}((M^\bullet \otimes _ A A') \otimes _{R'} (R' \otimes _ R F^\bullet ))) & = H^ i(\text{Tot}(M^\bullet \otimes _ A A' \otimes _ R F^\bullet )) \\ & = H^ i(\text{Tot}(M^\bullet \otimes _ R F^\bullet ) \otimes _ A A') \\ & = H^ i(M \otimes _ R^\mathbf {L} B) \otimes _ A A' \end{align*}

The final equality because $A \to A'$ is flat. The final module is the desired module because $A' \otimes _{R'} B' = A' \otimes _ R B$ since we've assumed $B' = R' \otimes _ R B$ in this paragraph.

General case. Suppose that $B' \to B''$ is a flat ring map. Then it is easy to see that

\[ H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B'') = H^ i((M \otimes _ A^\mathbf {L} A') \otimes _{R'}^\mathbf {L} B') \otimes _{B'} B'' \]

and

\[ H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B'') = \left( H^ i(M \otimes _ R^\mathbf {L} B) \otimes _{(A \otimes _ R B)} (A' \otimes _{R'} B') \right) \otimes _{B'} B'' \]

Thus the result for $B'$ implies the result for $B''$. Since we've proven the result for $R' \otimes _ R B$ in the previous paragraph, this implies the result in general. $\square$

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