Lemma 15.98.1. Let $R \to R'$ be a ring map. For $K \in D(R)$ and $M \in D(R')$ there is a canonical isomorphism

## 15.98 Base change for derived hom

We have already seen some material discussing this in Lemma 15.64.4 and in Algebra, Section 10.73.

**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

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

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:

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 15.59.3. $\square$

Let $R \to R'$ be a ring map. There is a base change map

in $D(R')$ functorial in $K, M \in D(R)$. Namely, by adjointness of $- \otimes _ R^\mathbf {L} R' : D(R) \to D(R')$ and the restriction functor $D(R') \to D(R)$, this is the same thing as a map

(equality by Lemma 15.98.1) for which we can use the canonical map $M \to M \otimes _ R^\mathbf {L} R'$ (unit of the adjunction).

Lemma 15.98.2. Let $R \to R'$ be a ring map. Let $K, M \in D(R)$. The map (15.98.1.1)

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

$K$ is perfect,

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

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

$R'$ 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

of Lemma 15.72.5. See discussion above the lemma to match the left and right hand sides; in particular, this uses Lemma 15.98.1. Thus we conclude by Lemma 15.97.3. $\square$

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## Comments (2)

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