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
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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8577 by Haohao Liu on
Comment #9156 by Stacks project on