The Stacks project

Lemma 30.5.1. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume $f$ is affine. In this case $f_*\mathcal{F} \cong Rf_*\mathcal{F}$ is a quasi-coherent sheaf, and for every base change diagram (30.5.0.1) we have

\[ g^*f_*\mathcal{F} = f'_*(g')^*\mathcal{F}. \]

Proof. The vanishing of higher direct images is Lemma 30.2.3. The statement is local on $S$ and $S'$. Hence we may assume $X = \mathop{\mathrm{Spec}}(A)$, $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$ and $\mathcal{F} = \widetilde{M}$ for some $A$-module $M$. We use Schemes, Lemma 26.7.3 to describe pullbacks and pushforwards of $\mathcal{F}$. Namely, $X' = \mathop{\mathrm{Spec}}(R' \otimes _ R A)$ and $\mathcal{F}'$ is the quasi-coherent sheaf associated to $(R' \otimes _ R A) \otimes _ A M$. Thus we see that the lemma boils down to the equality

\[ (R' \otimes _ R A) \otimes _ A M = R' \otimes _ R M \]

as $R'$-modules. $\square$


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Your email address will not be published.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 02KG. Beware of the difference between the letter 'O' and the digit '0'.