The Stacks project

Lemma 69.17.2. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $\mathcal{F}$ be a quasi-coherent sheaf on $Y$. Let $\mathcal{I}$ be a quasi-coherent sheaf of ideals on $X$. If $f$ is affine then $\mathcal{I}f_*\mathcal{F} = f_*(f^{-1}\mathcal{I}\mathcal{F})$ (with notation as explained in the proof).

Proof. The notation means the following. Since $f^{-1}$ is an exact functor we see that $f^{-1}\mathcal{I}$ is a sheaf of ideals of $f^{-1}\mathcal{O}_ X$. Via the map $f^\sharp : f^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ on $Y_{\acute{e}tale}$ this acts on $\mathcal{F}$. Then $f^{-1}\mathcal{I}\mathcal{F}$ is the subsheaf generated by sums of local sections of the form $as$ where $a$ is a local section of $f^{-1}\mathcal{I}$ and $s$ is a local section of $\mathcal{F}$. It is a quasi-coherent $\mathcal{O}_ Y$-submodule of $\mathcal{F}$ because it is also the image of a natural map $f^*\mathcal{I} \otimes _{\mathcal{O}_ Y} \mathcal{F} \to \mathcal{F}$.

Having said this the proof is straightforward. Namely, the question is ├ętale local on $X$ and hence we may assume $X$ is an affine scheme. In this case the result is a consequence of the corresponding result for schemes, see Cohomology of Schemes, Lemma 30.13.2. $\square$

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Your email address will not be published.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 0GF8. Beware of the difference between the letter 'O' and the digit '0'.