The Stacks project

Lemma 53.24.2. 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. There is a unique morphism (up to unique isomorphism)

\[ c : X \longrightarrow Y \]

of schemes over $k$ having the following properties:

  1. $Y$ is proper over $k$, $\dim (Y) = 1$, the singularities of $Y$ are at-worst-nodal,

  2. $\mathcal{O}_ Y = c_*\mathcal{O}_ X$ and $R^1c_*\mathcal{O}_ X = 0$, and

  3. $\omega _ Y$ is ample on $Y$.

Proof. Existence: A morphism with all the properties listed exists by combining Lemmas 53.22.6 and 53.23.6 as discussed in the introduction to this section. Moreover, we see that it can be written as a composition

\[ X \to X_1 \to X_2 \ldots \to X_ n \to X_{n + 1} \to \ldots \to X_{n + n'} \]

where the first $n$ morphisms are contractions of rational tails and the last $n'$ morphisms are contractions of rational bridges. Note that property (2) holds for each contraction of a rational tail (Example 53.22.1) and contraction of a rational bridge (Example 53.23.1). It is easy to see that this property is inherited by compositions of morphisms.

Uniqueness: Let $c : X \to Y$ be a morphism satisfying conditions (1), (2), and (3). We will show that there is a unique isomorphism $X_{n + n'} \to Y$ compatible with the morphisms $X \to X_{n + n'}$ and $c$.

Before we start the proof we make some observations about $c$. We first observe that the fibres of $c$ are geometrically connected by More on Morphisms, Theorem 37.53.4. In particular $c$ is surjective. For a closed point $y \in Y$ the fibre $X_ y$ satisfies

\[ H^1(X_ y, \mathcal{O}_{X_ y}) = 0 \quad \text{and}\quad H^0(X_ y, \mathcal{O}_{X_ y}) = \kappa (y) \]

The first equality by More on Morphisms, Lemma 37.72.1 and the second by More on Morphisms, Lemma 37.72.4. Thus either $X_ y = x$ where $x$ is the unique point of $X$ mapping to $y$ and has the same residue field as $y$, or $X_ y$ is a $1$-dimensional proper scheme over $\kappa (y)$. Observe that in the second case $X_ y$ is Cohen-Macaulay (Lemma 53.6.1). However, since $X$ is reduced, we see that $X_ y$ must be reduced at all of its generic points (details omitted), and hence $X_ y$ is reduced by Properties, Lemma 28.12.4. It follows that the singularities of $X_ y$ are at-worst-nodal (Lemma 53.19.17). Note that the genus of $X_ y$ is zero (see above). Finally, there are only a finite number of points $y$ where the fibre $X_ y$ has dimension $1$, say $\{ y_1, \ldots , y_ r\} $, and $c^{-1}(Y \setminus \{ y_1, \ldots , y_ r\} )$ maps isomorphically to $Y \setminus \{ y_1, \ldots , y_ r\} $ by $c$. Some details omitted; hint: outside of $\{ y_1, \ldots , y_ r\} $ the morphism $c$ is finite, see Cohomology of Schemes, Lemma 30.21.1.

Let $C \subset X$ be a rational tail. We claim that $c$ maps $C$ to a point. Assume that this is not the case to get a contradiction. Then the image of $C$ is an irreducible component $D \subset Y$. Recall that $H^0(C, \mathcal{O}_ C) = k'$ is a finite separable extension of $k$ and that $C$ has a $k'$-rational point $x$ which is also the unique intersection of $C$ with the “rest” of $X$. We conclude from the general discussion above that $C \setminus \{ x\} \subset c^{-1}(Y \setminus \{ y_1, \ldots , y_ r\} )$ maps isomorphically to an open $V$ of $D$. Let $y = c(x) \in D$. Observe that $y$ is the only point of $D$ meeting the “rest” of $Y$. If $y \not\in \{ y_1, \ldots , y_ r\} $, then $C \cong D$ and it is clear that $D$ is a rational tail of $Y$ which is a contradiction with the ampleness of $\omega _ Y$ (Lemma 53.22.2). Thus $y \in \{ y_1, \ldots , y_ r\} $ and $\dim (X_ y) = 1$. Then $x \in X_ y \cap C$ and $x$ is a smooth point of $X_ y$ and $C$ (Lemma 53.19.17). If $y \in D$ is a singular point of $D$, then $y$ is a node and then $Y = D$ (because there cannot be another component of $Y$ passing through $y$ by Lemma 53.19.17). Then $X = X_ y \cup C$ which means $g = 0$ because it is equal to the genus of $X_ y$ by the discussion in Example 53.22.1; a contradiction. If $y \in D$ is a smooth point of $D$, then $C \to D$ is an isomorphism (because the nonsingular projective model is unique and $C$ and $D$ are birational, see Section 53.2). Then $D$ is a rational tail of $Y$ which is a contradiction with ampleness of $\omega _ Y$.

Assume $n \geq 1$. If $C \subset X$ is the rational tail contracted by $X \to X_1$, then we see that $C$ is mapped to a point of $Y$ by the previous paragraph. Hence $c : X \to Y$ factors through $X \to X_1$ (because $X$ is the pushout of $C$ and $X_1$, see discussion in Example 53.22.1). After replacing $X$ by $X_1$ we have decreased $n$. By induction we may assume $n = 0$, i.e., $X$ does not have a rational tail.

Assume $n = 0$, i.e., $X$ does not have any rational tails. Then $\omega _ X^{\otimes 2}$ and $\omega _ X^{\otimes 3}$ are globally generated by Lemma 53.22.5. It follows that $H^1(X, \omega _ X^{\otimes 3}) = 0$ by Lemma 53.6.4. By Lemma 53.24.1 applied with $m = 3$ we find that $c^*\omega _ Y \cong \omega _ X$. We also have that $\omega _ X = (X \to X_{n'})^*\omega _{X_{n'}}$ by Lemma 53.23.4 and induction. Applying the projection formula for both $c$ and $X \to X_{n'}$ we conclude that

\[ \Gamma (X_{n'}, \omega _{X_{n'}}^{\otimes m}) = \Gamma (X, \omega _ X^{\otimes m}) = \Gamma (Y, \omega _ Y^{\otimes m}) \]

for all $m$. Since $X_{n'}$ and $Y$ are the Proj of the direct sum of these by Morphisms, Lemma 29.43.17 we conclude that there is a canonical isomorphism $X_{n'} = Y$ as desired. We omit the verification that this is the unique isomorphism making the diagram commute. $\square$

Comments (2)

Comment #5463 by Leo on

Let be the union of rational bridges and tails. Is the composite equal to the pushout ? ( is a copy of the base field for each connected component).

It seems that the iterative construction for each in both cases 0E3M and 0E3H is the pushout of along some unstable component mapping to the base field, which one could do in one step as above. This universal property might make it easier to study the stabilization in families if it holds.

A related question is whether stabilization is functorial: If I have an automorphism , does this induce an automorphism of ?

Comment #5679 by on

Dear Leo, some of these questions are answered later in the Stacks project. For example see Section 109.23 for contractions in families of curves. Your final question has a positive answer as is the unique (up to unique isomorphism) morphism satisfying conditions 1), 2), 3). OK?

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