Lemma 15.99.1. Let $R \to R'$ be a ring map. For $K \in D(R)$ and $M \in D(R')$ there is a canonical isomorphism

$R\mathop{\mathrm{Hom}}\nolimits _ R(K, M) = R\mathop{\mathrm{Hom}}\nolimits _{R'}(K \otimes _ R^\mathbf {L} R', M)$

Proof. Choose a K-injective complex of $R'$-modules $J^\bullet$ representing $M$. Choose a quasi-isomorphism $J^\bullet \to I^\bullet$ where $I^\bullet$ is a K-injective complex of $R$-modules. Choose a K-flat complex $K^\bullet$ of $R$-modules representing $K$. Consider the map

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet \otimes _ R R', J^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , I^\bullet )$

The map on degree $n$ terms is given by the map

$\prod \nolimits _{n = p + q} \mathop{\mathrm{Hom}}\nolimits _{R'}(K^{-q} \otimes _ R R', J^ p) \longrightarrow \prod \nolimits _{n = p + q} \mathop{\mathrm{Hom}}\nolimits _ R(K^{-q}, I^ p)$

coming from precomposing by $K^{-q} \to K^{-q} \otimes _ R R'$ and postcomposing by $J^ p \to I^ p$. To finish the proof it suffices to show that we get isomorphisms on cohomology groups:

$\mathop{\mathrm{Hom}}\nolimits _{D(R)}(K, M) = \mathop{\mathrm{Hom}}\nolimits _{D(R')}(K \otimes _ R^\mathbf {L} R', M)$

which is true because base change $- \otimes _ R^\mathbf {L} R' : D(R) \to D(R')$ is left adjoint to the restriction functor $D(R') \to D(R)$ by Lemma 15.60.3. $\square$

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