Lemma 47.24.5. Let R \to R' be a homomorphism of Noetherian rings. Let \varphi : R \to A be a perfect ring map (More on Algebra, Definition 15.82.1) such that R' and A are tor independent over R. Let \varphi ' : R' \to A' = A \otimes _ R R' be the map induced by \varphi . Then we have a functorial isomorphism
\varphi ^!(K) \otimes _ A^\mathbf {L} A' = (\varphi ')^!(K \otimes _ R^\mathbf {L} R')
for K in D(R).
Proof.
We may choose a factorization R \to P \to A where P is a polynomial ring over R such that A is a perfect P-module, see More on Algebra, Lemma 15.82.2. This gives a corresponding factorization R' \to P' \to A' by base change. Since we have (K \otimes _ R^\mathbf {L} P) \otimes _ P^\mathbf {L} P' = (K \otimes _ R^\mathbf {L} R') \otimes _{R'}^\mathbf {L} P' by More on Algebra, Lemma 15.60.5 it suffices to construct maps
R\mathop{\mathrm{Hom}}\nolimits (A, K \otimes _ R^\mathbf {L} P[n]) \otimes _ A^\mathbf {L} A' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (A', (K \otimes _ R^\mathbf {L} P[n]) \otimes _ P^\mathbf {L} P')
functorial in K. We have
A \otimes _ P^\mathbf {L} P' = A \otimes _ R^\mathbf {L} R' = A'
The first equality by More on Algebra, Lemma 15.61.2 applied to R, R', P, P'. The second equality because A and R' are tor independent over R. Hence A and P' are tor independent over P and we can use the map (47.14.0.1) constructed in Section 47.14 for P, A, P', A' get the desired arrow. By Lemma 47.14.3 to finish the proof it suffices to prove that A is a perfect P-module which we saw above.
\square
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