## 111.34 Morphisms

An important question is, given a morphism $\pi : X \to S$, whether the morphism has a section or a rational section. Here are some example exercises.

Exercise 111.34.1. Consider the morphism of schemes

$\pi : X = \mathop{\mathrm{Spec}}(\mathbf{C}[x, t, 1/xt]) \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{C}[t]).$

1. Show there does not exist a morphism $\sigma : S \to X$ such that $\pi \circ \sigma = \text{id}_ S$.

2. Show there does exist a nonempty open $U \subset S$ and a morphism $\sigma : U \to X$ such that $\pi \circ \sigma = \text{id}_ U$.

Exercise 111.34.2. Consider the morphism of schemes

$\pi : X = \mathop{\mathrm{Spec}}(\mathbf{C}[x, t]/(x^2 + t)) \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{C}[t]).$

Show there does not exist a nonempty open $U \subset S$ and a morphism $\sigma : U \to X$ such that $\pi \circ \sigma = \text{id}_ U$.

Exercise 111.34.3. Let $A, B, C \in \mathbf{C}[t]$ be nonzero polynomials. Consider the morphism of schemes

$\pi : X = \mathop{\mathrm{Spec}}(\mathbf{C}[x, y, t]/(A + Bx^2 + Cy^2)) \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{C}[t]).$

Show there does exist a nonempty open $U \subset S$ and a morphism $\sigma : U \to X$ such that $\pi \circ \sigma = \text{id}_ U$. (Hint: Symbolically, write $x = X/Z$, $y = Y/Z$ for some $X, Y, Z \in \mathbf{C}[t]$ of degree $\leq d$ for some $d$, and work out the condition that this solves the equation. Then show, using dimension theory, that if $d >> 0$ you can find nonzero $X, Y, Z$ solving the equation.)

Remark 111.34.4. Exercise 111.34.3 is a special case of “Tsen's theorem”. Exercise 111.34.5 shows that the method is limited to low degree equations (conics when the base and fibre have dimension 1).

Exercise 111.34.5. Consider the morphism of schemes

$\pi : X = \mathop{\mathrm{Spec}}(\mathbf{C}[x, y, t] /(1 + t x^3 + t^2 y^3)) \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{C}[t])$

Show there does not exist a nonempty open $U \subset S$ and a morphism $\sigma : U \to X$ such that $\pi \circ \sigma = \text{id}_ U$.

Exercise 111.34.6. Consider the schemes

$X = \mathop{\mathrm{Spec}}(\mathbf{C}[\{ x_ i\} _{i = 1}^{8}, s, t] /(1 + s x_1^3 + s^2 x_2^3 + t x_3^3 + st x_4^3 + s^2t x_5^3 + t^2 x_6^3 + st^2 x_7^3 + s^2t^2 x_8^3))$

and

$S = \mathop{\mathrm{Spec}}(\mathbf{C}[s, t])$

and the morphism of schemes

$\pi : X \longrightarrow S$

Show there does not exist a nonempty open $U \subset S$ and a morphism $\sigma : U \to X$ such that $\pi \circ \sigma = \text{id}_ U$.

Exercise 111.34.7. (For the number theorists.) Give an example of a closed subscheme

$Z \subset \mathop{\mathrm{Spec}}\left({\mathbf Z}[x, \frac{1 }{ x(x-1)(2x-1)}]\right)$

such that the morphism $Z \to \mathop{\mathrm{Spec}}({\mathbf Z})$ is finite and surjective.

Exercise 111.34.8. If you do not like number theory, you can try the variant where you look at

$\mathop{\mathrm{Spec}}\left({\mathbf F}_ p[t, x, \frac{1 }{ x(x-t)(tx-1)}]\right) \longrightarrow \mathop{\mathrm{Spec}}({\mathbf F}_ p[t])$

and you try to find a closed subscheme of the top scheme which maps finite surjectively to the bottom one. (There is a theoretical reason for having a finite ground field here; although it may not be necessary in this particular case.)

Remark 111.34.9. The interpretation of the results of Exercise 111.34.7 and 111.34.8 is that given the morphism $X \to S$ all of whose fibres are nonempty, there exists a finite surjective morphism $S' \to S$ such that the base change $X_{S'} \to S'$ does have a section. This is not a general fact, but it holds if the base is the spectrum of a dedekind ring with finite residue fields at closed points, and the morphism $X \to S$ is flat with geometrically irreducible generic fibre. See Exercise 111.34.10 below for an example where it doesn't work.

Exercise 111.34.10. Prove there exist a $f \in \mathbf{C}[x, t]$ which is not divisible by $t - \alpha$ for any $\alpha \in \mathbf{C}$ such that there does not exist any $Z \subset \mathop{\mathrm{Spec}}(\mathbf{C}[x, t, 1/f])$ which maps finite surjectively to $\mathop{\mathrm{Spec}}(\mathbf{C}[t])$. (I think that $f(x, t) = (xt - 2)(x - t + 3)$ works. To show any candidate has the required property is not so easy I think.)

Exercise 111.34.11. Let $A \to B$ be a finite type ring map. Suppose that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ factors through a closed immersion $\mathop{\mathrm{Spec}}(B) \to \mathbf{P}^ n_ A$ for some $n$. Prove that $A \to B$ is a finite ring map, i.e., that $B$ is finite as an $A$-module. Hint: if $A$ is Noetherian (please just assume this) you can argue using that $H^ i(Z, \mathcal{O}_ Z)$ for $i \in \mathbf{Z}$ is a finite $A$-module for every closed subscheme $Z \subset \mathbf{P}^ n_ A$.

Exercise 111.34.12. Let $k$ be an algebraically closed field. Let $f : X \to Y$ be a morphism of projective varieties such that $f^{-1}(\{ y\} )$ is finite for every closed point $y \in Y$. Prove that $f$ is finite as a morphism of schemes. Hints: (a) being finite is a local property, (b) try to reduce to Exercise 111.34.11, and (c) use a closed immersion $X \to \mathbf{P}^ n_ k$ to get a closed immersion $X \to \mathbf{P}^ n_ Y$ over $Y$.

Comment #2364 by Marco Castronovo on

$\pi\circ\sigma = \operatorname{id}_S$ in part (1) of Exercise 92.33.1 right?

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