Lemma 47.13.9. In Situation 47.13.6 assume $A$ is a perfect $R$-module. Then

$R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(R) \to D(A)$

is given by $K \mapsto K \otimes _ R^\mathbf {L} M$ where $M = R\mathop{\mathrm{Hom}}\nolimits (A, R) \in D(A)$.

Proof. We apply Divided Power Algebra, Lemma 23.6.10 to choose a Tate resolution $(E, \text{d})$ of $A$ over $R$. Note that $E^ i = 0$ for $i > 0$, $E^0 = R[x_1, \ldots , x_ n]$ is a polynomial algebra, and $E^ i$ is a finite free $E^0$-module for $i < 0$. It follows that $E$ viewed as a complex of $R$-modules is a bounded above complex of free $R$-modules. We check the assumptions of Lemma 47.13.8. The first holds because $A$ is perfect (hence compact by More on Algebra, Proposition 15.74.3) and the second by More on Algebra, Lemma 15.70.2. From the lemma conclude that $K \mapsto R\mathop{\mathrm{Hom}}\nolimits (E, K)$ is isomorphic to $K \mapsto K \otimes _ R^\mathbf {L} N$ for some differential graded $E$-module $N$. Observe that

$(R \otimes _ R E) \otimes _ E^\mathbf {L} A = R \otimes _ E E \otimes _ E A$

in $D(A)$. Hence by Differential Graded Algebra, Lemma 22.34.2 we conclude that the composition of $- \otimes _ R^\mathbf {L} N$ and $- \otimes _ R^\mathbf {L} A$ is of the form $- \otimes _ R M$ for some $M \in D(A)$. To finish the proof we apply Lemma 47.13.7. $\square$

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