## 15.99 Base change for derived hom

We have already seen some material discussing this in Lemma 15.65.4 and in Algebra, Section 10.73.

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. $\square$

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

15.99.1.1
$$\label{more-algebra-equation-base-change-RHom} R\mathop{\mathrm{Hom}}\nolimits _ R(K, M) \otimes _ R^\mathbf {L} R' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _{R'}(K \otimes _ R^\mathbf {L} R', M \otimes _ R^\mathbf {L} R')$$

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

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

(equality by Lemma 15.99.1) for which we can use the canonical map $M \to M \otimes _ R^\mathbf {L} R'$ (unit of the adjunction).

Lemma 15.99.2. Let $R \to R'$ be a ring map. Let $K, M \in D(R)$. The map (15.99.1.1)

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

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 15.73.5. See discussion above the lemma to match the left and right hand sides; in particular, this uses Lemma 15.99.1. Thus we conclude by Lemma 15.98.3. $\square$

Comment #5381 by Will Chen on

In the proof of Lemma 0E1W, "The map on degree n terms by the map" is missing a word.

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