Lemma 37.72.2. Let $f : X \to Y$ be a proper morphism of schemes. Let $y \in Y$ be a point with $\dim (X_ y) \leq 1$ and $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$. Then there is an open neighbourhood $V \subset Y$ of $y$ such that $R^1f_*\mathcal{O}_ X|_ V = 0$ and the same is true after base change by any $Y' \to V$.
Proof. To prove the lemma we may replace $Y$ by an open neighbourhood of $y$. Thus we may assume $Y$ is affine and that all fibres of $f$ have dimension $\leq 1$, see Morphisms, Lemma 29.28.4. In this case $R^1f_*\mathcal{O}_ X$ is a quasi-coherent $\mathcal{O}_ Y$-module of finite type and its formation commutes with arbitrary base change, see Limits, Lemmas 32.19.3 and 32.19.2. Say $Y = \mathop{\mathrm{Spec}}(A)$, $y$ corresponds to the prime $\mathfrak p \subset A$, and $R^1f_*\mathcal{O}_ X$ corresponds to the finite $A$-module $M$. Then $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$ means that $\mathfrak pM_\mathfrak p = M_\mathfrak p$ by the statement on base change. By Nakayama's lemma we conclude $M_\mathfrak p = 0$. Since $M$ is finite, we find an $f \in A$, $f \not\in \mathfrak p$ such that $M_ f = 0$. Thus taking $V$ the principal open $D(f)$ we obtain the desired result. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)