The Stacks project

Lemma 69.16.3. In Situation 69.16.1.

  1. Given an affine morphism $X' \to X$ of algebraic spaces, we have $H^1(X', \mathcal{F}') = 0$ for every quasi-coherent $\mathcal{O}_{X'}$-module $\mathcal{F}'$.

  2. Given an $A$-algebra $A'$ setting $X' = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A')$ the morphism $X' \to X$ is affine and $\Gamma (X', \mathcal{O}_{X'}) = A'$.

Proof. Part (1) follows from Lemma 69.8.2 and the Leray spectral sequence (Cohomology on Sites, Lemma 21.14.5). Let $A \to A'$ be as in (2). Then $X' \to X$ is affine because affine morphisms are preserved under base change (Morphisms of Spaces, Lemma 67.20.5) and the fact that a morphism of affine schemes is affine. The equality $\Gamma (X', \mathcal{O}_{X'}) = A'$ follows as $(X' \to X)_*\mathcal{O}_{X'} = A' \otimes _ A \mathcal{O}_ X$ by Lemma 69.11.1 and thus

\[ \Gamma (X', \mathcal{O}_{X'}) = \Gamma (X, (X' \to X)_*\mathcal{O}_{X'}) = \Gamma (X, A' \otimes _ A \mathcal{O}_ X) = A' \]

by Lemma 69.16.2. $\square$


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Your email address will not be published.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 07V0. Beware of the difference between the letter 'O' and the digit '0'.