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. $\square$

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0E7L. Beware of the difference between the letter 'O' and the digit '0'.