53.23 Contracting rational bridges

In this section we discuss the next simplest possible case (after the case discussed in Section 53.22) of contracting a scheme to improve positivity properties of its canonical sheaf.

Example 53.23.1 (Contracting a rational bridge). Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume the singularities of $X$ are at-worst-nodal. A rational bridge will be an irreducible component $C \subset X$ (viewed as an integral closed subscheme) with the following properties

1. $X' \not= \emptyset$ where $X' \subset X$ is the scheme theoretic closure of $X \setminus C$,

2. the scheme theoretic interesection $C \cap X'$ has degree $2$ over $H^0(C, \mathcal{O}_ C)$, and

3. $C$ has genus zero.

Set $k' = H^0(C, \mathcal{O}_ C)$ and $k'' = H^0(C \cap X', \mathcal{O}_{C \cap X'})$. Then $k'$ is a field (Varieties, Lemma 33.9.3) and $\dim _{k'}(k'') = 2$. Since there are at least two irreducible components of $X$ passing through each point of $C \cap X'$, we conclude these points are nodes and smooth points on both $C$ and $X'$ (Lemma 53.19.17). Hence $k'/k$ is a finite separable extension of fields and $k''/k'$ is either a degree $2$ separable extension of fields or $k'' = k' \times k'$ (Lemma 53.19.7). By Section 53.14 there exists a pushout

$\xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^ a \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Y }$

with many good properties (all of which we will use below without futher mention). Let $y \in Y$ be the image of $\mathop{\mathrm{Spec}}(k') \to Y$. Then

$\mathcal{O}_{Y, y}^\wedge \cong k'[[s, t]]/(st) \quad \text{or}\quad \mathcal{O}_{Y, y}^\wedge \cong \{ f \in k''[[s]] : f(0) \in k'\}$

depending on whether $C \cap X'$ has $2$ or $1$ points. This follows from Lemma 53.14.1 and the fact that $\mathcal{O}_{X', p} \cong \kappa (p)[[t]]$ for $p \in C \cap X'$ by More on Algebra, Lemma 15.38.4. Thus we see that $y \in Y$ is a node, see Lemmas 53.19.7 and 53.19.4 and in particular the discussion of Case II in the proof of (2) $\Rightarrow$ (1) in Lemma 53.19.4. Thus the singularities of $Y$ are at-worst-nodal.

We can extend the commutative diagram above to a diagram

$\xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^ a \ar[r] & X \ar[ld]^ c & C \ar[ld] \ar[l] \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Y & \mathop{\mathrm{Spec}}(k') \ar[l] }$

where the two lower horizontal arrows are the same. Namely, $X$ is the scheme theoretic union of $X'$ and $C$ (thus a pushout by Morphisms, Lemma 29.4.6) and the morphisms $C \to Y$ and $X' \to Y$ agree on $C \cap X'$. Finally, we claim that

$c_*\mathcal{O}_ X = \mathcal{O}_ Y \quad \text{and}\quad R^1c_*\mathcal{O}_ X = 0$

To see this use the exact sequence

$0 \to \mathcal{O}_ X \to \mathcal{O}_ C \oplus \mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'} \to 0$

of Morphisms, Lemma 29.4.6. The long exact sequence of higher direct images is

$0 \to c_*\mathcal{O}_ X \to c_*\mathcal{O}_ C \oplus c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'} \to R^1c_*\mathcal{O}_ X \to R^1c_*\mathcal{O}_ C \oplus R^1c_*\mathcal{O}_{X'}$

Since $c|_{X'} = a$ is affine we see that $R^1c_*\mathcal{O}_{X'} = 0$. Since $c|_ C$ factors as $C \to \mathop{\mathrm{Spec}}(k') \to X$ and since $C$ has genus zero, we find that $R^1c_*\mathcal{O}_ C = 0$. Since $\mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'}$ is surjective and since $c|_{X'}$ is affine, we see that $c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'}$ is surjective. This proves that $R^1c_*\mathcal{O}_ X = 0$. Finally, we have $\mathcal{O}_ Y = c_*\mathcal{O}_ X$ by the exact sequence and the description of the structure sheaf of the pushout in More on Morphisms, Proposition 37.64.3.

All of this means that $Y$ is also a proper scheme over $k$ having dimension $1$ and $H^0(Y, \mathcal{O}_ Y) = k$ whose singularities are at-worst-nodal (Lemma 53.19.17) and that $Y$ has the same genus as $X$. We will say $c : X \to Y$ is the contraction of a rational bridge.

Lemma 53.23.2. Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume the singularities of $X$ are at-worst-nodal. Let $C \subset X$ be a rational bridge (Example 53.23.1). Then $\deg (\omega _ X|_ C) = 0$.

Proof. Let $X' \subset X$ be as in the example. Then we have a short exact sequence

$0 \to \omega _ C \to \omega _ X|_ C \to \mathcal{O}_{C \cap X'} \to 0$

See Lemmas 53.4.6, 53.19.16, and 53.19.17. With $k''/k'/k$ as in the example we see that $\deg (\omega _ C) = -2[k' : k]$ as $C$ has genus $0$ (Lemma 53.5.2) and $\deg (C \cap X') = [k'' : k] = 2[k' : k]$. Hence $\deg (\omega _ X|_ C) = 0$. $\square$

Lemma 53.23.3. Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume the singularities of $X$ are at-worst-nodal. Let $C \subset X$ be a rational bridge (Example 53.23.1). For any field extension $K/k$ the base change $C_ K \subset X_ K$ is a finite disjoint union of rational bridges.

Proof. Let $k''/k'/k$ be as in the example. Since $k'/k$ is finite separable, we see that $k' \otimes _ k K = K'_1 \times \ldots \times K'_ n$ is a finite product of finite separable extensions $K'_ i/K$. The corresponding product decomposition $k'' \otimes _ k K = \prod K''_ i$ gives degree $2$ separable algebra extensions $K''_ i/K'_ i$. Set $C_ i = C_{K'_ i}$. Then $C_ K = \coprod C_ i$ and therefore each $C_ i$ has genus $0$ (viewed as a curve over $K'_ i$), because $H^1(C_ K, \mathcal{O}_{C_ K}) = 0$ by flat base change. Finally, we have $X'_ K \cap C_ i = \mathop{\mathrm{Spec}}(K''_ i)$ has degree $2$ over $K'_ i$ as desired. $\square$

Lemma 53.23.4. Let $c : X \to Y$ be the contraction of a rational bridge (Example 53.23.1). Then $c^*\omega _ Y \cong \omega _ X$.

Proof. You can prove this by direct computation, but we prefer to use the characterization of $\omega _ X$ as the coherent $\mathcal{O}_ X$-module which represents the functor $\textit{Coh}(\mathcal{O}_ X) \to \textit{Sets}$, $\mathcal{F} \mapsto \mathop{\mathrm{Hom}}\nolimits _ k(H^1(X, \mathcal{F}), k) = H^1(X, \mathcal{F})^\vee$, see Lemma 53.4.2 or Duality for Schemes, Lemma 48.22.5.

To be precise, denote $\mathcal{C}_ Y$ the category whose objects are invertible $\mathcal{O}_ Y$-modules and whose maps are $\mathcal{O}_ Y$-module homomorphisms. Denote $\mathcal{C}_ X$ the category whose objects are invertible $\mathcal{O}_ X$-modules $\mathcal{L}$ with $\mathcal{L}|_ C \cong \mathcal{O}_ C$ and whose maps are $\mathcal{O}_ Y$-module homomorphisms. We claim that the functor

$c^* : \mathcal{C}_ Y \to \mathcal{C}_ X$

is an equivalence of categories. Namely, by More on Morphisms, Lemma 37.69.8 it is essentially surjective. Then the projection formula (Cohomology, Lemma 20.51.2) shows $c_*c^*\mathcal{N} = \mathcal{N}$ and hence $c^*$ is an equivalence with quasi-inverse given by $c_*$.

We claim $\omega _ X$ is an object of $\mathcal{C}_ X$. Namely, we have a short exact sequence

$0 \to \omega _ C \to \omega _ X|_ C \to \mathcal{O}_{C \cap X'} \to 0$

See Lemma 53.4.6. Taking degrees we find $\deg (\omega _ X|_ C) = 0$ (small detail omitted). Thus $\omega _ X|_ C$ is trivial by Lemma 53.10.1 and $\omega _ X$ is an object of $\mathcal{C}_ X$.

Since $R^1c_*\mathcal{O}_ X = 0$ the projection formula shows that $R^1c_*c^*\mathcal{N} = 0$ for $\mathcal{N} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ Y)$. Therefore the Leray spectral sequence (Cohomology, Lemma 20.13.6) the diagram

$\xymatrix{ \mathcal{C}_ Y \ar[rr]_{c^*} \ar[dr]_{H^1(Y, -)^\vee } & & \mathcal{C}_ X \ar[ld]^{H^1(X, -)^\vee } \\ & \textit{Sets} }$

of categories and functors is commutative. Since $\omega _ Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ Y)$ represents the south-east arrow and $\omega _ X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ X)$ represents the south-east arrow we conclude by the Yoneda lemma (Categories, Lemma 4.3.5). $\square$

Lemma 53.23.5. Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume

1. the singularities of $X$ are at-worst-nodal,

2. $X$ does not have a rational tail (Example 53.22.1),

3. $X$ does not have a rational bridge (Example 53.23.1),

4. the genus $g$ of $X$ is $\geq 2$.

Then $\omega _ X$ is ample.

Proof. It suffices to show that $\deg (\omega _ X|_ C) > 0$ for every irreducible component $C$ of $X$, see Varieties, Lemma 33.44.15. If $X = C$ is irreducible, this follows from $g \geq 2$ and Lemma 53.8.3. Otherwise, set $k' = H^0(C, \mathcal{O}_ C)$. This is a field and a finite extension of $k$ and $[k' : k]$ divides all numerical invariants below associated to $C$ and coherent sheaves on $C$, see Varieties, Lemma 33.44.10. Let $X' \subset X$ be the closure of $X \setminus C$ as in Lemma 53.4.6. We will use the results of this lemma and of Lemmas 53.19.16 and 53.19.17 without further mention. Then we get a short exact sequence

$0 \to \omega _ C \to \omega _ X|_ C \to \mathcal{O}_{C \cap X'} \to 0$

See Lemma 53.4.6. We conclude that

$\deg (\omega _ X|_ C) = \deg (C \cap X') + \deg (\omega _ C) = \deg (C \cap X') - 2\chi (C, \mathcal{O}_ C)$

Hence, if the lemma is false, then

$2[k' : k] \geq \deg (C \cap X') + 2\dim _ k H^1(C, \mathcal{O}_ C)$

Since $C \cap X'$ is nonempty and by the divisiblity mentioned above, this can happen only if either

1. $C \cap X'$ is a single $k'$-rational point of $C$ and $H^1(C, \mathcal{O}_ C) = 0$, and

2. $C \cap X'$ has degree $2$ over $k'$ and $H^1(C, \mathcal{O}_ C) = 0$.

The first possibility means $C$ is a rational tail and the second that $C$ is a rational bridge. Since both are excluded the proof is complete. $\square$

Lemma 53.23.6. Let $k$ be a field. Let $X$ be a proper scheme over $k$ of dimension $1$ with $H^0(X, \mathcal{O}_ X) = k$ having genus $g \geq 2$. Assume the singularities of $X$ are at-worst-nodal and that $X$ has no rational tails. Consider a sequence

$X = X_0 \to X_1 \to \ldots \to X_ n = X'$

of contractions of rational bridges (Example 53.23.1) until none are left. Then $\omega _{X'}$ ample. The morphism $X \to X'$ is independent of choices and formation of this morphism commutes with base field extensions.

Proof. We proceed by contracting rational bridges until there are none left. Then $\omega _{X'}$ is ample by Lemma 53.23.5.

Denote $f : X \to X'$ the composition. By Lemma 53.23.4 and induction we see that $f^*\omega _{X'} = \omega _ X$. We have $f_*\mathcal{O}_ X = \mathcal{O}_{X'}$ because this is true for contraction of a rational bridge. Thus the projection formula says that $f_*f^*\mathcal{L} = \mathcal{L}$ for all invertible $\mathcal{O}_{X'}$-modules $\mathcal{L}$. Hence

$\Gamma (X', \omega _{X'}^{\otimes m}) = \Gamma (X, \omega _ X^{\otimes m})$

for all $m$. Since $X'$ is the Proj of the direct sum of these by Morphisms, Lemma 29.43.17 we conclude that the morphism $X \to X'$ is completely canonical.

Let $K/k$ be an extension of fields, then $\omega _{X_ K}$ is the pullback of $\omega _ X$ (Lemma 53.4.4) and we have $\Gamma (X, \omega _ X^{\otimes m}) \otimes _ k K$ is equal to $\Gamma (X_ K, \omega _{X_ K}^{\otimes m})$ by Cohomology of Schemes, Lemma 30.5.2. Thus formation of $f : X \to X'$ commutes with base change by $K/k$ by the arguments given above. Some details omitted. $\square$

Comment #4549 by Jonas Ehrhard on

The intro text here is exactly the same as in the previous section. Both claim to be "the simplest possible case" of cantractions.

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