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

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

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 15.73.3) and the second by More on Algebra, Lemma 15.69.2. 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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)