The Stacks project

Lemma 37.69.7. Consider a commutative diagram

\[ \xymatrix{ X \ar[rr]_ f \ar[rd] & & Y \ar[ld] \\ & S } \]

of morphisms of schemes. Let $s \in S$ be a point. Assume

  1. $X \to S$ is locally of finite presentation and flat at points of $X_ s$,

  2. $Y \to S$ is locally of finite presentation,

  3. $f$ is proper,

  4. the fibres of $f_ s : X_ s \to Y_ s$ have dimension $\leq 1$ and $R^1f_{s, *}\mathcal{O}_{X_ s} = 0$,

  5. $\mathcal{O}_{Y_ s} \to f_{s, *}\mathcal{O}_{X_ s}$ is an isomorphism.

Then there is an open $Y_ s \subset V \subset Y$ such that (a) $V$ is flat over $S$, (b) $f^{-1}(V)$ is flat over $S$, (c) $\dim (X_ y) \leq 1$ for $y \in V$, (d) $R^1f_*\mathcal{O}_ X|_ V = 0$, (e) $\mathcal{O}_ V \to f_*\mathcal{O}_ X|_ V$ is an isomorphism, and (a) – (e) remain true after base change of $f^{-1}(V) \to V$ by any $S' \to S$.

Proof. Let $y \in Y_ s$. We may always replace $Y$ by an open neighbourhood of $y$. Thus we may assume $Y$ and $S$ affine. We may also assume that $X$ is flat over $S$, $\dim (X_ y) \leq 1$ for $y \in Y$, $R^1f_*\mathcal{O}_ X = 0$ universally, and that $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective, see Lemma 37.69.5. (We won't use all of this.)

Assume $S$ and $Y$ affine. Write $S = \mathop{\mathrm{lim}}\nolimits S_ i$ as a cofiltered of affine Noetherian schemes $S_ i$. By Limits, Lemma 32.10.1 there exists an element $0 \in I$ and a diagram

\[ \xymatrix{ X_0 \ar[rr]_{f_0} \ar[rd] & & Y_0 \ar[ld] \\ & S_0 } \]

of finite type morphisms of schemes whose base change to $S$ is the diagram of the lemma. After increasing $0$ we may assume $Y_0$ is affine and $X_0 \to S_0$ proper, see Limits, Lemmas 32.13.1 and 32.4.13. Let $s_0 \in S_0$ be the image of $s$. As $Y_ s$ is affine, we see that $R^1f_{s, *}\mathcal{O}_{X_ s} = 0$ is equivalent to $H^1(X_ s, \mathcal{O}_{X_ s}) = 0$. Since $X_ s$ is the base change of $X_{0, s_0}$ by the faithfully flat map $\kappa (s_0) \to \kappa (s)$ we see that $H^1(X_{0, s_0}, \mathcal{O}_{X_{0, s_0}}) = 0$ and hence $R^1f_{0, *}\mathcal{O}_{X_{0, s_0}} = 0$. Similarly, as $\mathcal{O}_{Y_ s} \to f_{s, *}\mathcal{O}_{X_ s}$ is an isomorphism, so is $\mathcal{O}_{Y_{0, s_0}} \to f_{0, *}\mathcal{O}_{X_{0, s_0}}$. Since the dimensions of the fibres of $X_ s \to Y_ s$ are at most $1$, the same is true for the morphism $X_{0, s_0} \to Y_{0, s_0}$. Finally, since $X \to S$ is flat, after increasing $0$ we may assume $X_0$ is flat over $S_0$, see Limits, Lemma 32.8.7. Thus it suffices to prove the lemma for $X_0 \to Y_0 \to S_0$ and the point $s_0$.

Combining the reduction arguments above we reduce to the case where $S$ and $Y$ affine, $S$ Noetherian, the fibres of $f$ have dimension $\leq 1$, and $R^1f_*\mathcal{O}_ X = 0$ universally. Let $y \in Y_ s$ be a point. Claim:

\[ \mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y \]

is an isomorphism. The claim implies the lemma. Namely, since $f_*\mathcal{O}_ X$ is coherent (Cohomology of Schemes, Proposition 30.19.1) the claim means we can replace $Y$ by an open neighbourhood of $y$ and obtain an isomorphism $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$. Then we conclude that $Y$ is flat over $S$ by Lemma 37.69.6. Finally, the isomorphism $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ remains an isomorphism after any base change $S' \to S$ by the final statement of Lemma 37.69.6.

Proof of the claim. We already know that $\mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y$ is surjective (Lemma 37.69.5) and that $(f_*\mathcal{O}_ X)_ y$ is $\mathcal{O}_{S, s}$-flat (Lemma 37.69.6) and that the induced map

\[ \mathcal{O}_{Y_ s, y} = \mathcal{O}_{Y, y}/\mathfrak m_ s\mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y/\mathfrak m_ s (f_*\mathcal{O}_ X)_ y \to (f_{s, *}\mathcal{O}_{X_ s})_ y \]

is injective by the assumption in the lemma. Then it follows from Algebra, Lemma 10.99.1 that $\mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y$ is injective as desired. $\square$

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