The Stacks project

Proof. Let $y \in Y$ be a point over $s$. It suffices to find an open neighbourhood of $y$ with the desired properties. As a first step, we replace $Y$ by the open $V$ found in Lemma 37.72.2 so that $R^1f_*\mathcal{O}_ X$ is zero universally (the hypothesis of the lemma holds by Lemma 37.72.1). We also shrink $Y$ so that all fibres of $f$ have dimension $\leq 1$ (use Morphisms, Lemma 29.28.4 and properness of $f$). Thus we may assume we have (b) and (c) with $V = Y$ and after any base change $Y' \to Y$. Thus by Lemma 37.72.4 it now suffices to show (d) over $Y$. We may still shrink $Y$ further; for example, we may and do assume $Y$ and $S$ are affine.

By Theorem 37.15.1 there is an open subset $U \subset X$ where $X \to S$ is flat which contains $X_ s$ by hypothesis. Then $f(X \setminus U)$ is a closed subset not containing $y$. Thus after shrinking $Y$ we may assume $X$ is flat over $S$.

Say $S = \mathop{\mathrm{Spec}}(R)$. Choose a closed immersion $Y \to Y'$ where $Y'$ is the spectrum of a polynomial ring $R[x_ e; e \in E]$ on a set $E$. Denote $f' : X \to Y'$ the composition of $f$ with $Y \to Y'$. Then the hypotheses (1) – (4) as well as (b) and (c) hold for $f'$ and $s$. If we we show $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_ X$ is surjective in an open neighbourhood of $y$, then the same is true for $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$. Thus we may assume $Y$ is the spectrum of $R[x_ e; e \in E]$.

At this point $X$ and $Y$ are flat over $S$. Then $Y_ s$ and $X$ are tor independent over $Y$. We urge the reader to find their own proof, but it also follows from Lemma 37.69.1 applied to the square with corners $X, Y, S, S$ and its base change by $s \to S$. Hence

by Derived Categories of Schemes, Lemma 36.22.5. Because of the vanishing already established this implies $f_{s, *}\mathcal{O}_{X_ s} = (Y_ s \to Y)^*f_*\mathcal{O}_ X$. We conclude that $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is a map of quasi-coherent $\mathcal{O}_ Y$-modules whose pullback to $Y_ s$ is surjective. We claim $f_*\mathcal{O}_ X$ is a finite type $\mathcal{O}_ Y$-module. If true, then the cokernel $\mathcal{F}$ of $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is a finite type quasi-coherent $\mathcal{O}_ Y$-module such that $\mathcal{F}_ y \otimes \kappa (y) = 0$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we have $\mathcal{F}_ y = 0$. Thus $\mathcal{F}$ is zero in an open neighbourhood of $y$ (Modules, Lemma 17.9.5) and the proof is complete.

Proof of the claim. For a finite subset $E' \subset E$ set $Y' = \mathop{\mathrm{Spec}}(R[x_ e; e \in E'])$. For large enough $E'$ the morphism $f' : X \to Y \to Y'$ is proper, see Limits, Lemma 32.13.4. We fix $E'$ and $Y'$ in the following. Write $R = \mathop{\mathrm{colim}}\nolimits R_ i$ as the colimit of its finite type $\mathbf{Z}$-subalgebras. Set $S_ i = \mathop{\mathrm{Spec}}(R_ i)$ and $Y'_ i = \mathop{\mathrm{Spec}}(R_ i[x_ e; e \in E'])$. For $i$ large enough we can find a diagram

with cartesian squares such that $X_ i$ is flat over $S_ i$ and $X_ i \to Y'_ i$ is proper. See Limits, Lemmas 32.10.1, 32.8.7, and 32.13.1. The same argument as above shows $Y'$ and $X_ i$ are tor independent over $Y'_ i$ and hence

by the same reference as above. By Cohomology of Schemes, Lemma 30.19.2 the complex $R\Gamma (X_ i, \mathcal{O}_{X_ i})$ is pseudo-coherent in the derived category of the Noetherian ring $R_ i[x_ e; e \in E']$ (see More on Algebra, Lemma 15.64.17). Hence $R\Gamma (X, \mathcal{O}_ X)$ is pseudo-coherent in the derived category of $R[x_ e; e \in E']$, see More on Algebra, Lemma 15.64.12. Since the only nonvanishing cohomology module is $H^0(X, \mathcal{O}_ X)$ we conclude it is a finite $R[x_ e; e \in E']$-module, see More on Algebra, Lemma 15.64.4. This concludes the proof. $\square$