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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