The Stacks project

Proof. The fibres of $c$ are geometrically connected by More on Morphisms, Theorem 37.53.4. In particular $c$ is surjective. There are finitely many closed points $y = y_1, \ldots , y_ r$ of $Y$ where $X_ y$ has dimension $1$ and over $Y \setminus \{ y_1, \ldots , y_ r\} $ the morphism $c$ is an isomorphism. 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 us carefully construct a map $b : c^*\omega _ Y \to \omega _ X$. Denote $f : X \to \mathop{\mathrm{Spec}}(k)$ and $g : Y \to \mathop{\mathrm{Spec}}(k)$ the structure morphisms. We have $f^!k = \omega _ X[1]$ and $g^!k = \omega _ Y[1]$, see Lemma 53.4.1 and its proof. Then $f^! = c^! \circ g^!$ and hence $c^!\omega _ Y = \omega _ X$. Thus there is a functorial isomorphism

for coherent $\mathcal{O}_ X$-modules $\mathcal{F}$ by definition of $c^!$¹. This isomorphism is induced by a trace map $t : Rc_*\omega _ X \to \omega _ Y$ (the counit of the adjunction). By the projection formula (Cohomology, Lemma 20.54.2) the canonical map $a : \omega _ Y \to Rc_*c^*\omega _ Y$ is an isomorphism. Combining the above we see there is a canonical map $b : c^*\omega _ Y \to \omega _ X$ such that

In particular, if we restrict $b$ to $c^{-1}(Y \setminus \{ y_1, \ldots , y_ r\} )$ then it is an isomorphism (because it is a map between invertible modules whose composition with another gives the isomorphism $a^{-1}$).

Choose $m \in \mathbf{Z}$ as in (3) consider the map

This map is injective because $Y$ is reduced and by the last property of $b$ mentioned in its construction. By Riemann-Roch (Lemma 53.5.2) we have $\chi (X, \omega _ X^{\otimes m}) =\chi (Y, \omega _ Y^{\otimes m})$. Thus

and we conclude $b^{\otimes m}$ induces an isomorphism on global sections. So $b^{\otimes m} : c^*\omega _ Y^{\otimes m} \to \omega _ X^{\otimes m}$ is surjective as generators of $\omega _ X^{\otimes m}$ are in the image. Hence $b^{\otimes m}$ is an isomorphism. Thus $b$ is an isomorphism. $\square$

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. $\square$

Proof. Combining Varieties, Lemma 33.44.15 and Lemmas 53.22.2 and 53.23.2 we see that $X$ contains no rational tails or bridges. Then we see that $\omega _ X^{\otimes 3}$ is globally generated by Lemma 53.22.6. Choose a $k$-basis $s_0, \ldots , s_ n$ of $H^0(X, \omega _ X^{\otimes 3})$. We get a morphism

See Constructions, Section 27.13. The lemma asserts that this morphism is a closed immersion. To check this we may replace $k$ by its algebraic closure, see Descent, Lemma 35.23.19. Thus we may assume $k$ is algebraically closed.

Assume $k$ is algebraically closed. We will use Varieties, Lemma 33.23.2 to prove the lemma. Let $Z \subset X$ be a closed subscheme of degree $2$ over $Z$ with ideal sheaf $\mathcal{I} \subset \mathcal{O}_ X$. We have to show that

is surjective. Thus it suffices to show that $H^1(X, \mathcal{I}\mathcal{L}) = 0$. To do this we will use Lemma 53.21.6. Thus it suffices to show that

for every reduced connected closed subscheme $Y \subset X$. Since $k$ is algebraically closed and $Y$ connected and reduced we have $H^0(Y, \mathcal{O}_ Y) = k$ (Varieties, Lemma 33.9.3). Hence $\chi (Y, \mathcal{O}_ Y) = 1 - \dim H^1(Y, \mathcal{O}_ Y)$. Thus we have to show

which is true by Lemma 53.22.4 except possibly if $Y = X$ or if $\deg (\omega _ X|_ Y) = 0$. Since $\omega _ X$ is ample the second possibility does not occur (see first lemma cited in this proof). Finally, if $Y = X$ we can use Riemann-Roch (Lemma 53.5.2) and the fact that $g \geq 2$ to see that the inquality holds. The same argument with $Z = \emptyset $ shows that $H^1(X, \omega _ X^{\otimes 3}) = 0$. $\square$