Lemma 37.69.2. 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 32.18.3 and 32.18.2. 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. $\square$

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