The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

15.87 Base change for derived hom

We have already seen some material discussing this in Remark 15.62.21 and in Algebra, Section 10.72.

Lemma 15.87.1. 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 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 $D(R') \to D(R)$. $\square$

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

15.87.1.1
\begin{equation} \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') \end{equation}

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

Lemma 15.87.2. Let $R \to R'$ be a ring map. Let $K, M \in D(R)$. The map (15.87.1.1)

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

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 15.87.1. Thus we conclude by Lemma 15.86.3. $\square$


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