The Stacks project

37.66 Contracting rational curves

In this section we study proper morphisms $f : X \to Y$ whose fibres have dimension $\leq 1$ having $R^1f_*\mathcal{O}_ X = 0$. To understand the title of this section, please take a look at Algebraic Curves, Sections 53.22, 53.23, and 53.24.

Lemma 37.66.1. Let $f : X \to Y$ be a proper morphism of schemes. Let $y \in Y$ be a point with $\dim (X_ y) \leq 1$. If

  1. $R^1f_*\mathcal{O}_ X = 0$, or more generally

  2. there is a morphism $g : Y' \to Y$ such that $y$ is in the image of $g$ and such that $R'f'_*\mathcal{O}_{X'} = 0$ where $f' : X' \to Y'$ is the base change of $f$ by $g$.

Then $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$.

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 32.17.3 and 32.17.2. The lemma follows immediately. $\square$

Lemma 37.66.2. 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 32.17.3 and 32.17.2. 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. $\square$

Lemma 37.66.3. Let $f : X \to Y$ be a proper morphism of schemes such that $\dim (X_ y) \leq 1$ and $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$ for all $y \in Y$. Let $\mathcal{F}$ be quasi-coherent on $X$. Then

  1. $R^ pf_*\mathcal{F} = 0$ for $p > 1$, and

  2. $R^1f_*\mathcal{F} = 0$ if there is a surjection $f^*\mathcal{G} \to \mathcal{F}$ with $\mathcal{G}$ quasi-coherent on $Y$.

If $Y$ is affine, then we also have

  1. $H^ p(X, \mathcal{F}) = 0$ for $p \not\in \{ 0, 1\} $, and

  2. $H^1(X, \mathcal{F}) = 0$ if $\mathcal{F}$ is globally generated.

Proof. The vanishing in (1) is Limits, Lemma 32.17.2. To prove (2) we may work locally on $Y$ and assume $Y$ is affine. Then $R^1f_*\mathcal{F}$ is the quasi-coherent module on $Y$ associated to the module $H^1(X, \mathcal{F})$. Here we use that $Y$ is affine, quasi-coherence of higher direct images (Cohomology of Schemes, Lemma 30.4.5), and Cohomology of Schemes, Lemma 30.4.6. Since $Y$ is affine, the quasi-coherent module $\mathcal{G}$ is globally generated, and hence so is $f^*\mathcal{G}$ and $\mathcal{F}$. In this way we see that (4) implies (2). Part (3) follows from (1) as well as the remarks on quasi-coherence of direct images just made. Thus all that remains is the prove (4). If $\mathcal{F}$ is globally generated, then there is a surjection $\bigoplus _{i \in I} \mathcal{O}_ X \to \mathcal{F}$. By part (1) and the long exact sequence of cohomology this induces a surjection on $H^1$. Since $H^1(X, \mathcal{O}_ X) = 0$ because $R^1f_*\mathcal{O}_ X = 0$ by Lemma 37.66.2, and since $H^1(X, -)$ commutes with direct sums (Cohomology, Lemma 20.19.1) we conclude. $\square$

Lemma 37.66.4. Let $f : X \to Y$ be a proper morphism of schemes. Assume

  1. for all $y \in Y$ we have $\dim (X_ y) \leq 1$ and $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$, and

  2. $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective.

Then $\mathcal{O}_{Y'} \to f'_*\mathcal{O}_{X'}$ is surjective for any base change $f' : X' \to Y'$ of $f$.

Proof. We may assume $Y$ and $Y'$ affine. Then we can choose a closed immersion $Y' \to Y''$ with $Y'' \to Y$ a flat morphism of affines. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we see that the result holds for $X'' \to Y''$. Thus we may assume $Y'$ is a closed subscheme of $Y$. Let $\mathcal{I} \subset \mathcal{O}_ Y$ be the ideal cutting out $Y'$. Then there is a short exact sequence

\[ 0 \to \mathcal{I}\mathcal{O}_ X \to \mathcal{O}_ X \to \mathcal{O}_{X'} \to 0 \]

where we view $\mathcal{O}_{X'}$ as a quasi-coherent module on $X$. By Lemma 37.66.3 we have $H^1(X, \mathcal{I}\mathcal{O}_ X) = 0$. It follows that

\[ H^0(Y, \mathcal{O}_ Y) \to H^0(Y, f_*\mathcal{O}_ X) = H^0(X, \mathcal{O}_ X) \to H^0(X, \mathcal{O}_{X'}) \]

is surjective as desired. The first arrow is surjective as $Y$ is affine and since we assumed $\mathcal{O}_ Y \to f_*\mathcal{O}_ X$ is surjective and the second by the long exact sequence of cohomology associated to the short exact sequence above and the vanishing just proved. $\square$

Lemma 37.66.5. 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. $f$ is proper,

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

  4. $\mathcal{O}_{Y_ s} \to f_{s, *}\mathcal{O}_{X_ s}$ is surjective.

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

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 37.66.2 so that $R^1f_*\mathcal{O}_ X$ is zero universally (the hypothesis of the lemma holds by Lemma 37.66.1). 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 37.66.4 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 37.63.1 applied to the square with corners $X, Y, S, S$ and its base change by $s \to S$. Hence

\[ Rf_{s, *}\mathcal{O}_{X_ s} = L(Y_ s \to Y)^*Rf_*\mathcal{O}_ X \]

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

\[ \xymatrix{ X \ar[d] \ar[r]_{f'} & Y' \ar[d] \ar[r] & S \ar[d] \\ X_ i \ar[r]^{f'_ i} & Y'_ i \ar[r] & S_ i } \]

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

\[ R\Gamma (X, \mathcal{O}_ X) = R\Gamma (X_ i, \mathcal{O}_{X_ i}) \otimes ^\mathbf {L}_{R_ i[x_ e; e \in E']} R[x_ e; e \in E'] \]

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 15.63.17). 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 15.63.12. 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 15.63.4. This concludes the proof. $\square$

Lemma 37.66.6. Consider a commutative diagram

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

of morphisms of schemes. Assume $X \to S$ is flat, $f$ is proper, $\dim (X_ y) \leq 1$ for $y \in Y$, and $R^1f_*\mathcal{O}_ X = 0$. Then $f_*\mathcal{O}_ X$ is $S$-flat and formation of $f_*\mathcal{O}_ X$ commutes with arbitrary base change $S' \to S$.

Proof. We may assume $Y$ and $S$ are affine, say $S = \mathop{\mathrm{Spec}}(A)$. To show the quasi-coherent $\mathcal{O}_ Y$-module $f_*\mathcal{O}_ X$ is flat relative to $S$ it suffices to show that $H^0(X, \mathcal{O}_ X)$ is flat over $A$ (some details omitted). By Lemma 37.66.3 we have $H^1(X, \mathcal{O}_ X \otimes _ A M) = 0$ for every $A$-module $M$. Since also $\mathcal{O}_ X$ is flat over $A$ we deduce the functor $M \mapsto H^0(X, \mathcal{O}_ X \otimes _ A M)$ is exact. Moreover, this functor commutes with direct sums by Cohomology, Lemma 20.19.1. Then it is an exercise to see that $H^0(X, \mathcal{O}_ X \otimes _ A M) = M \otimes _ A H^0(X, \mathcal{O}_ X)$ functorially in $M$ and this gives the desired flatness. Finally, if $S' \to S$ is a morphism of affines given by the ring map $A \to A'$, then in the affine case just discussed we see that

\[ H^0(X \times _ S S', \mathcal{O}_{X \times _ S S'}) = H^0(X, \mathcal{O}_ X \otimes _ A A') = H^0(X, \mathcal{O}_ X) \otimes _ A A' \]

This shows that formation of $f_*\mathcal{O}_ X$ commutes with any base change $S' \to S$. Some details omitted. $\square$

Lemma 37.66.7. 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 37.66.5. (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 37.66.6. 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 37.66.6.

Proof of the claim. We already know that $\mathcal{O}_{Y, y} \longrightarrow (f_*\mathcal{O}_ X)_ y$ is surjective (Lemma 37.66.5) and that $(f_*\mathcal{O}_ X)_ y$ is $\mathcal{O}_{S, s}$-flat (Lemma 37.66.6) 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. $\square$

Lemma 37.66.8. Let $f : X \to Y$ be a proper morphism of Noetherian schemes such that $f_*\mathcal{O}_ X = \mathcal{O}_ Y$, such that the fibres of $f$ have dimension $\leq 1$, and such that $H^1(X_ y, \mathcal{O}_{X_ y}) = 0$ for $y \in Y$. Then $f^* : \mathop{\mathrm{Pic}}\nolimits (Y) \to \mathop{\mathrm{Pic}}\nolimits (X)$ is a bijection onto the subgroup of $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ with $\mathcal{L}|_{X_ y} \cong \mathcal{O}_{X_ y}$ for all $y \in Y$.

Proof. By the projection formula (Cohomology, Lemma 20.51.2) we see that $f_*f^*\mathcal{N} \cong \mathcal{N}$ for $\mathcal{N} \in \mathop{\mathrm{Pic}}\nolimits (Y)$. We claim that for $\mathcal{L} \in \mathop{\mathrm{Pic}}\nolimits (X)$ with $\mathcal{L}|_{X_ y} \cong \mathcal{O}_{X_ y}$ for all $y \in Y$ we have $\mathcal{N} = f_*\mathcal{L}$ is invertible and $\mathcal{L} \cong f^*\mathcal{N}$. This will finish the proof.

The $\mathcal{O}_ Y$-module $\mathcal{N} = f_*\mathcal{L}$ is coherent by Cohomology of Schemes, Proposition 30.19.1. Thus to see that it is an invertible $\mathcal{O}_ Y$-module, it suffices to check on stalks (Algebra, Lemma 10.78.2). Since the map from a Noetherian local ring to its completion is faithfully flat, it suffices to check the completion $(f_*\mathcal{L})_ y^\wedge $ is free (see Algebra, Section 10.97 and Lemma 10.78.6). For this we will use the theorem of formal functions as formulated in Cohomology of Schemes, Lemma 30.20.7. Since $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ and hence $(f_*\mathcal{O}_ X)_ y^\wedge \cong \mathcal{O}_{Y, y}^\wedge $, it suffices to show that $\mathcal{L}|_{X_ n} \cong \mathcal{O}_{X_ n}$ for each $n$ (compatibly for varying $n$. By Lemma 37.4.1 we have an exact sequence

\[ H^1(X_ y, \mathfrak m_ y^ n\mathcal{O}_ X/\mathfrak m_ y^{n + 1}\mathcal{O}_ X) \to \mathop{\mathrm{Pic}}\nolimits (X_{n + 1}) \to \mathop{\mathrm{Pic}}\nolimits (X_ n) \]

with notation as in the theorem on formal functions. Observe that we have a surjection

\[ \mathcal{O}_{X_ y}^{\oplus r_ n} \cong \mathfrak m_ y^ n/\mathfrak m_ y^{n + 1} \otimes _{\kappa (y)} \mathcal{O}_{X_ y} \longrightarrow \mathfrak m_ y^ n\mathcal{O}_ X/\mathfrak m_ y^{n + 1}\mathcal{O}_ X \]

for some integers $r_ n \geq 0$. Since $\dim (X_ y) \leq 1$ this surjection induces a surjection on first cohomology groups (by the vanishing of cohomology in degrees $\geq 2$ coming from Cohomology, Proposition 20.20.7). Hence the $H^1$ in the sequence is zero and the transition maps $\mathop{\mathrm{Pic}}\nolimits (X_{n + 1}) \to \mathop{\mathrm{Pic}}\nolimits (X_ n)$ are injective as desired.

We still have to show that $f^*\mathcal{N} \cong \mathcal{L}$. This is proved by the same method and we omit the details. $\square$


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