The Stacks project

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

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

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

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. $\square$