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):

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$S$ 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.


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    $\begingroup$ If you have to explain why something is trivial, then it isn't trivial. /rant $\endgroup$ – RghtHndSd Nov 16 '14 at 23:38
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    $\begingroup$ 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

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 $\mathbb P^1(\overline{\mathbb Q})$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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