The Stacks project

Lemma 36.5.3. Let $f : X \to S$ be an affine morphism of schemes. For $E$ in $D_\mathit{QCoh}(\mathcal{O}_ S)$ we have $Rf_* Lf^* E = E \otimes ^\mathbf {L}_{\mathcal{O}_ S} f_*\mathcal{O}_ X$.

Proof. Since $f$ is affine the map $f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X$ is an isomorphism (Cohomology of Schemes, Lemma 30.2.3). There is a canonical map $E \otimes ^\mathbf {L} f_*\mathcal{O}_ X = E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X \to Rf_* Lf^* E$ adjoint to the map

\[ Lf^*(E \otimes ^\mathbf {L} Rf_*\mathcal{O}_ X) = Lf^*E \otimes ^\mathbf {L} Lf^*Rf_*\mathcal{O}_ X \longrightarrow Lf^* E \otimes ^\mathbf {L} \mathcal{O}_ X = Lf^* E \]

coming from $1 : Lf^*E \to Lf^*E$ and the canonical map $Lf^*Rf_*\mathcal{O}_ X \to \mathcal{O}_ X$. To check the map so constructed is an isomorphism we may work locally on $S$. Hence we may assume $S$ and therefore $X$ is affine. In this case the statement is clear from the description of the derived categories $D_\mathit{QCoh}(\mathcal{O}_ X)$ and $D_\mathit{QCoh}(\mathcal{O}_ S)$ and the functor $Lf^*$ given in Lemmas 36.3.5 and 36.3.8. Some details omitted. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08I9. Beware of the difference between the letter 'O' and the digit '0'.