Lemma 15.89.2. Let $R \to R'$ be a ring map. Let $K, M \in D(R)$. The map (15.89.1.1)

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

is an isomorphism in $D(R')$ in the following cases

1. $K$ is perfect,

2. $R'$ is perfect as an $R$-module,

3. $R \to R'$ is flat, $K$ is pseudo-coherent, and $M \in D^{+}(R)$, or

4. $R'$ has finite tor dimension as an $R$-module, $K$ is pseudo-coherent, and $M \in D^{+}(R)$

Proof. We may check the map is an isomorphism after applying the restriction functor $D(R') \to D(R)$. After applying this functor our map becomes the map

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

of Lemma 15.68.5. See discussion above the lemma to match the left and right hand sides; in particular, this uses Lemma 15.89.1. Thus we conclude by Lemma 15.88.3. $\square$

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