## 53.22 Contracting rational tails

In this section we discuss the simplest possible case of contracting a scheme to improve positivity properties of its canonical sheaf.

Example 53.22.1 (Contracting a rational tail). 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 tail 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 intersection $C \cap X'$ is a single reduced point $x$,

3. $H^0(C, \mathcal{O}_ C)$ maps isomorphically to the residue field of $x$, and

4. $C$ has genus zero.

Since there are at least two irreducible components of $X$ passing through $x$, we conclude that $x$ is a node. Set $k' = H^0(C, \mathcal{O}_ C) = \kappa (x)$. Then $k'/k$ is a finite separable extension of fields (Lemma 53.19.7). There is a canonical morphism

$c : X \longrightarrow X'$

inducing the identity on $X'$ and mapping $C$ to $x \in X'$ via the canonical morphism $C \to \mathop{\mathrm{Spec}}(k') = x$. This follows from Morphisms, Lemma 29.4.6 since $X$ is the scheme theoretic union of $C$ and $X'$ (as $X$ is reduced). Moreover, we claim that

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

To see this, denote $i_ C : C \to X$, $i_{X'} : X' \to X$ and $i_ x : x \to X$ the embeddings and use the exact sequence

$0 \to \mathcal{O}_ X \to i_{C, *}\mathcal{O}_ C \oplus i_{X', *}\mathcal{O}_{X'} \to i_{x, *}\kappa (x) \to 0$

of Morphisms, Lemma 29.4.6. Looking at the long exact sequence of higher direct images, it follows that it suffices to show $H^0(C, \mathcal{O}_ C) = k'$ and $H^1(C, \mathcal{O}_ C) = 0$ which follows from the assumptions. Observe that $X'$ is also a proper scheme over $k$, of dimension $1$ whose singularities are at-worst-nodal (Lemma 53.19.17) has $H^0(X', \mathcal{O}_{X'}) = k$, and $X'$ has the same genus as $X$. We will say $c : X \to X'$ is the contraction of a rational tail.

Lemma 53.22.2. 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 tail (Example 53.22.1). 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 $k'$ as in the example we see that $\deg (\omega _ C) = -2[k' : k]$ as $C \cong \mathbf{P}^1_{k'}$ by Proposition 53.10.4 and $\deg (C \cap X') = [k' : k]$. Hence $\deg (\omega _ X|_ C) = -[k' : k]$ which is negative. $\square$

Lemma 53.22.3. 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 tail (Example 53.22.1). For any field extension $K/k$ the base change $C_ K \subset X_ K$ is a finite disjoint union of rational tails.

Proof. Let $x \in C$ and $k' = \kappa (x)$ be as in the example. Observe that $C \cong \mathbf{P}^1_{k'}$ by Proposition 53.10.4. 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 $K'_ i/K$. Set $C_ i = \mathbf{P}^1_{K'_ i}$ and denote $x_ i \in C_ i$ the inverse image of $x$. Then $C_ K = \coprod C_ i$ and $X'_ K \cap C_ i = x_ i$ as desired. $\square$

Lemma 53.22.4. 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. If $X$ does not have a rational tail (Example 53.22.1), then for every reduced connected closed subscheme $Y \subset X$, $Y \not= X$ of dimension $1$ we have $\deg (\omega _ X|_ Y) \geq \dim _ k H^1(Y, \mathcal{O}_ Y)$.

Proof. Let $Y \subset X$ be as in the statement. Then $k' = H^0(Y, \mathcal{O}_ Y)$ is a field and a finite extension of $k$ and $[k' : k]$ divides all numerical invariants below associated to $Y$ and coherent sheaves on $Y$, see Varieties, Lemma 33.44.10. Let $Z \subset X$ be 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

$0 \to \omega _ Y \to \omega _ X|_ Y \to \mathcal{O}_{Y \cap Z} \to 0$

See Lemma 53.4.6. We conclude that

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

Hence, if the lemma is false, then

$2[k' : k] > \deg (Y \cap Z) + \dim _ k H^1(Y, \mathcal{O}_ Y)$

Since $Y \cap Z$ is nonempty and by the divisiblity mentioned above, this can happen only if $Y \cap Z$ is a single $k'$-rational point of the smooth locus of $Y$ and $H^1(Y, \mathcal{O}_ Y) = 0$. If $Y$ is irreducible, then this implies $Y$ is a rational tail. If $Y$ is reducible, then since $\deg (\omega _ X|_ Y) = -[k' : k]$ we find there is some irreducible component $C$ of $Y$ such that $\deg (\omega _ X|_ C) < 0$, see Varieties, Lemma 33.44.6. Then the analysis above applied to $C$ gives that $C$ is a rational tail. $\square$

Lemma 53.22.5. 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 the singularities of $X$ are at-worst-nodal. Assume $X$ does not have a rational tail (Example 53.22.1). If

1. the genus of $X$ is $0$, then $X$ is isomorphic to an irreducible plane conic and $\omega _ X^{\otimes -1}$ is very ample,

2. the genus of $X$ is $1$, then $\omega _ X \cong \mathcal{O}_ X$,

3. the genus of $X$ is $\geq 2$, then $\omega _ X^{\otimes m}$ is globally generated for $m \geq 2$.

Proof. By Lemma 53.19.16 we find that $X$ is Gorenstein, i.e., $\omega _ X$ is an invertible $\mathcal{O}_ X$-module.

If the genus of $X$ is zero, then $\deg (\omega _ X) < 0$, hence if $X$ has more than one irreducible component, we get a contradiction with Lemma 53.22.4. In the irreducible case we see that $X$ is isomorphic to an irreducible plane conic and $\omega _ X^{\otimes -1}$ is very ample by Lemma 53.10.3.

If the genus of $X$ is $1$, then $\omega _ X$ has a global section and $\deg (\omega _ X|_ C) = 0$ for all irreducible components. Namely, $\deg (\omega _ X|_ C) \geq 0$ for all irreducible components $C$ by Lemma 53.22.4, the sum of these numbers is $0$ by Lemma 53.8.3, and we can apply Varieties, Lemma 33.44.6. Then $\omega _ X \cong \mathcal{O}_ X$ by Varieties, Lemma 33.44.13.

Assume the genus $g$ of $X$ is greater than or equal to $2$. If $X$ is irreducible, then we are done by Lemma 53.21.3. Assume $X$ reducible. By Lemma 53.22.4 the inequalities of Lemma 53.21.6 hold for every $Y \subset X$ as in the statement, except for $Y = X$. Analyzing the proof of Lemma 53.21.6 we see that (in the reducible case) the only inequality used for $Y = X$ are

$\deg (\omega _ X^{\otimes m}) > -2 \chi (\mathcal{O}_ X) \quad \text{and}\quad \deg (\omega _ X^{\otimes m}) + \chi (\mathcal{O}_ X) > \dim _ k H^1(X, \mathcal{O}_ X)$

Since these both hold under the assumption $g \geq 2$ and $m \geq 2$ we win. $\square$

Lemma 53.22.6. Let $k$ be a field. Let $X$ be a proper scheme over $k$ of dimension $1$ with $H^0(X, \mathcal{O}_ X) = k$. Assume the singularities of $X$ are at-worst-nodal. Consider a sequence

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

of contractions of rational tails (Example 53.22.1) until none are left. Then

1. if the genus of $X$ is $0$, then $X'$ is an irreducible plane conic,

2. if the genus of $X$ is $1$, then $\omega _{X'} \cong \mathcal{O}_ X$,

3. if the genus of $X$ is $> 1$, then $\omega _{X'}^{\otimes m}$ is globally generated for $m \geq 2$.

If the genus of $X$ is $\geq 1$, then 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 tails until there are none left. Then we see that (1), (2), (3) hold by Lemma 53.22.5.

Uniqueness. To see that $f : X \to X'$ is independent of the choices made, it suffices to show: any rational tail $C \subset X$ is mapped to a point by $X \to X'$; some details omitted. If not, then we can find a section $s \in \Gamma (X', \omega _{X'}^{\otimes 2})$ which does not vanish in the generic point of the irreducible component $f(C)$. Since in each of the contractions $X_ i \to X_{i + 1}$ we have a section $X_{i + 1} \to X_ i$, there is a section $X' \to X$ of $f$. Then we have an exact sequence

$0 \to \omega _{X'} \to \omega _ X \to \omega _ X|_{X''} \to 0$

where $X'' \subset X$ is the union of the irreducible components contracted by $f$. See Lemma 53.4.6. Thus we get a map $\omega _{X'}^{\otimes 2} \to \omega _ X^{\otimes 2}$ and we can take the image of $s$ to get a section of $\omega _ X^{\otimes 2}$ not vanishing in the generic point of $C$. This is a contradiction with the fact that the restriction of $\omega _ X$ to a rational tail has negative degree (Lemma 53.22.2).

The statement on base field extensions follows from Lemma 53.22.3. Some details omitted. $\square$

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