Lemma 47.14.1. In the situation above, the map (47.14.0.1) is an isomorphism if and only if the map

of More on Algebra, Lemma 15.73.5 is an isomorphism.

In this section we consider a cocartesian square of rings

\[ \xymatrix{ A \ar[r]_\alpha & A' \\ R \ar[u]^\varphi \ar[r]^\rho & R' \ar[u]_{\varphi '} } \]

In other words, we have $A' = A \otimes _ R R'$. If $A$ and $R'$ are **tor independent over** $R$ then there is a canonical base change map

47.14.0.1

\begin{equation} \label{dualizing-equation-base-change} R\mathop{\mathrm{Hom}}\nolimits (A, K) \otimes _ A^\mathbf {L} A' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (A', K \otimes _ R^\mathbf {L} R') \end{equation}

in $D(A')$ functorial for $K$ in $D(R)$. Namely, by the adjointness of Lemma 47.13.1 such an arrow is the same thing as a map

\[ \varphi '_*\left(R\mathop{\mathrm{Hom}}\nolimits (A, K) \otimes _ A^\mathbf {L} A'\right) \longrightarrow K \otimes _ R^\mathbf {L} R' \]

in $D(R')$ where $\varphi '_* : D(A') \to D(R')$ is the restriction functor. We may apply More on Algebra, Lemma 15.61.2 to the left hand side to get that this is the same thing as a map

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

in $D(R')$ where $\varphi _* : D(A) \to D(R)$ is the restriction functor. For this we can choose $can \otimes ^\mathbf {L} \text{id}_{R'}$ where $can : \varphi _*(R\mathop{\mathrm{Hom}}\nolimits (A, K)) \to K$ is the counit of the adjunction between $R\mathop{\mathrm{Hom}}\nolimits (A, -)$ and $\varphi _*$.

Lemma 47.14.1. In the situation above, the map (47.14.0.1) is an isomorphism if and only if the map

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

of More on Algebra, Lemma 15.73.5 is an isomorphism.

**Proof.**
To see that the map is an isomorphism, it suffices to prove it is an isomorphism after applying $\varphi '_*$. Applying the functor $\varphi '_*$ to (47.14.0.1) and using that $A' = A \otimes _ R^\mathbf {L} R'$ we obtain the base change map $R\mathop{\mathrm{Hom}}\nolimits _ R(A, K) \otimes _ R^\mathbf {L} R' \to R\mathop{\mathrm{Hom}}\nolimits _{R'}(A \otimes _ R^\mathbf {L} R', K \otimes _ R^\mathbf {L} R')$ for derived hom of More on Algebra, Equation (15.99.1.1). Unwinding the left and right hand side exactly as in the proof of More on Algebra, Lemma 15.99.2 and in particular using More on Algebra, Lemma 15.99.1 gives the desired result.
$\square$

Lemma 47.14.2. Let $R \to A$ and $R \to R'$ be ring maps and $A' = A \otimes _ R R'$. Assume

$A$ is pseudo-coherent as an $R$-module,

$R'$ has finite tor dimension as an $R$-module (for example $R \to R'$ is flat),

$A$ and $R'$ are tor independent over $R$.

Then (47.14.0.1) is an isomorphism for $K \in D^+(R)$.

**Proof.**
Follows from Lemma 47.14.1 and More on Algebra, Lemma 15.98.3 part (4).
$\square$

Lemma 47.14.3. Let $R \to A$ and $R \to R'$ be ring maps and $A' = A \otimes _ R R'$. Assume

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

$A$ and $R'$ are tor independent over $R$.

Then (47.14.0.1) is an isomorphism for all $K \in D(R)$.

**Proof.**
Follows from Lemma 47.14.1 and More on Algebra, Lemma 15.98.3 part (1).
$\square$

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