Lemma 37.72.4. Let $f : X \to Y$ be a proper morphism of schemes. Assume
for all $y \in Y$ we have $\dim (X_ y) \leq 1$ and $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$, and
$\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective.
Then $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_{X'}$ is surjective for any base change $f' : X' \to Y'$ of $f$.
Proof. We may assume $Y$ and $Y'$ affine. Then we can choose a closed immersion $Y' \to Y''$ with $Y'' \to Y$ a flat morphism of affines. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we see that the result holds for $X'' \to Y''$. Thus we may assume $Y'$ is a closed subscheme of $Y$. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the ideal cutting out $Y'$. Then there is a short exact sequence
where we view $\mathcal{O}_{X'}$ as a quasi-coherent module on $X$. By Lemma 37.72.3 we have $H^1(X, \mathcal{I}\mathcal{O}_ X) = 0$. It follows that
is surjective as desired. The first arrow is surjective as $Y$ is affine and since we assumed $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective and the second by the long exact sequence of cohomology associated to the short exact sequence above and the vanishing just proved. $\square$
Your email address will not be published.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 (0)