Lemma 15.86.4. Let $R$ be a ring. Let $P^\bullet $ be a bounded above complex of projective $R$-modules. Let $K^\bullet $ be a K-flat complex of $R$-modules. If $P^\bullet $ is a perfect object of $D(R)$, then $\mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet )$ is K-flat and represents $R\mathop{\mathrm{Hom}}\nolimits _ R(P^\bullet , K^\bullet )$.

**Proof.**
The last statement is Lemma 15.68.2. Since $P^\bullet $ represents a perfect object, there exists a finite complex of finite projective $R$-modules $F^\bullet $ such that $P^\bullet $ and $F^\bullet $ are isomorphic in $D(R)$, see Definition 15.69.1. Then $P^\bullet $ and $F^\bullet $ are homotopy equivalent, see Derived Categories, Lemma 13.19.8. Then $\mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet )$ and $\mathop{\mathrm{Hom}}\nolimits ^\bullet (F^\bullet , K^\bullet )$ are homotopy equivalent. Hence the first is K-flat if and only if the second is (follows from Definition 15.57.3 and Lemma 15.57.1). It is clear that

where $E^\bullet $ is the dual complex to $F^\bullet $ with terms $E^ n = \mathop{\mathrm{Hom}}\nolimits _ R(F^{-n}, R)$, see Lemma 15.69.14 and its proof. Since $E^\bullet $ is a bounded complex of projectives we find that it is K-flat by Lemma 15.57.9. Then we conclude by Lemma 15.57.6. $\square$

## Post a comment

Your email address will not be published.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)