9
$\begingroup$

I'm reading the article "Belyi's theorem for complex surfaces - Gabino Gonzalez Diez" and there are few lines of a certain proof that I don't understand (the author claims that all is trivial):

enter image description here

enter image description here

Notations:

$S$ is a projective surface $S$ embedded in some $\mathbb P^n$ and we say that it is defined over a number field if $S\cong \{g_1=0,\ldots,g_m=0\}$ for polynomials $g_i$ with coefficients in a finite extension of $\mathbb Q$.

Here a Lefschetz pencil is referred as the rational function $f$ from $S$ to $\mathbb P^1$ induced by the hyperplane sections $\{S_\lambda=H_\lambda\cap S\}$ ($\{H_\lambda\}$ is a pencil of hyperplanes in $\mathbb P^n$)

The critical points of a Lefschetz pencil are the finite singular points (nodes) of the hyperplane sections and the critical values are the images of these points through $f$.


I know the Bertini's theorem and moreover I know how to construct a Lefschetz pencil. The rational map $f$ sends a point $x\in S$ in $f(x)=\lambda$ iff $x\in S_\lambda$, but I don't understand why if $S$ is defined over a number field then $f(x)\in\mathbb P^1\left(\overline{\mathbb Q}\right)$ for any critical point $x$.

I hope the question is clear; I've tried to explain in minor space as possible the framework and the notations, but if you want more explainations I'll edit the question.

Thanks.

$\endgroup$
  • 1
    $\begingroup$ If you have to explain why something is trivial, then it isn't trivial. /rant $\endgroup$ – RghtHndSd Nov 16 '14 at 23:38
  • 2
    $\begingroup$ I think its just that if you take a pencil defined over $\bar{\mathbb{Q}}$ which you can do by Bertini since $\bar{\mathbb{Q}}$ is algebraically closed, then the condition of being a critical value is given by a polynomial equation on $\mathbb{P}^1$ with $\bar{\mathbb{Q}}$ coefficients since the equations of the pencil have $\bar{\mathbb{Q}}$ coefficients. Then $\bar{\mathbb{Q}}$ is algebraically closed so the critical values are defined over $\bar{\mathbb{Q}}$ because they satisfy a polynomial equation over $\bar{\mathbb{Q}}$. $\endgroup$ – Dori Bejleri Nov 17 '14 at 0:36
  • $\begingroup$ @ Dori Bejleri But in this way the pencil of hyperplanes is $\{H_\lambda\}_{\lambda\in\mathbb P^1(\overline{\mathbb Q})}$. I need a pencil where $\lambda$ ranges in $\mathbb P^1(\mathbb C)$. $\endgroup$ – manifold Nov 17 '14 at 6:59
  • $\begingroup$ ... to be more precise I don't understand why if $S$ is defined over $\overline {\mathbb Q}$, then $S=\bigcup_{\lambda\in\mathbb P^1(\overline {\mathbb Q})} S\cap H_\lambda$. What about the hyperplanes of the type $H_r$ where $r\in\mathbb P^1(\mathbb C)\setminus\mathbb P^1(\overline {\mathbb Q})$? $\endgroup$ – manifold Nov 17 '14 at 7:24
2
$\begingroup$

If $S$ can be defined over $\overline{\mathbb Q}$, then there exists a model $S_0$ for $S$ over $\overline{\mathbb Q}$ (with respect to some embedding of $\overline{\mathbb Q} \to \mathbb C$).

Choose your Lefschetz pencil on $S_0$ and note that the critical points of this Lefschetz pencil lie in $\mathbb P^1(\overline{\mathbb Q})$; see the comment of Dori Bejleri above.

The Lefschetz pencil on $S$ is the Lefschetz pencil on $S_0$ base-changed to $\mathbb C$. The critical points of this Lefschetz pencil are those of the Lefschetz pencil on $S_0$, so they still lie in $\mathbb P^1(\overline{\mathbb Q})$.

$\endgroup$
  • $\begingroup$ Ok clear answer but there is still a question. why when I change the base of the Lefschetz pencil I don't add any critical point? $\endgroup$ – manifold Nov 17 '14 at 9:40
  • 1
    $\begingroup$ That's by the stability of smooth morphisms under base-change. See Liu's book on algebraic geometry, chapters 2,3 and 4. $\endgroup$ – Ariyan Javanpeykar Nov 17 '14 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.