## 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

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

the scheme theoretic intersection $C \cap X'$ is a single reduced point $x$,

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

$C$ 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

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

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

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

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

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

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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