The Stacks project

Lemma 75.26.8. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume

  1. $f$ is proper, flat, and of finite presentation, and

  2. the geometric fibres of $f$ are reduced and connected.

Then $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and this holds after any base change.

Proof. By Lemma 75.26.7 it suffices to show that $k = H^0(X_ k, \mathcal{O}_{X_ k})$ for all morphisms $\mathop{\mathrm{Spec}}(k) \to Y$ where $k$ is a field. This follows from Spaces over Fields, Lemma 72.14.3 and the fact that $X_ k$ is geometrically connected and geometrically reduced. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 75.26: Perfect complexes

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Your email address will not be published.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 0E0S. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 75.26.8 as pdf