Consider the Belyi's theorem:

If a smooth projective curve $X$ is defined over $\overline{\mathbb Q}$, then there exists a finite morphism $X\longrightarrow\mathbb P^1(\mathbb C)$ with at most $3$ branch points.

For the proof, one takes a morphism $f:X\longrightarrow \mathbb P^1(\mathbb C)$ defined over $\overline{\mathbb Q}$ and then uses some "techniques" to reduce the branch locus to a set of three points. These "techniques" involve only morphisms from $\mathbb P^1(\mathbb C)$ to $\mathbb P^1(\mathbb C)$ so they don't require the fundamental hypothesis "$X$ is defined over $\overline{\mathbb Q}$". Now I don't understand where this hypothesis is really used, maybe it only ensures that we can choose such a morphism $f$ defined over $\overline{\mathbb Q}$? It seems that this is the only point where it is needed.

Look for example at the following excerpts taken respectively from Serre's "Lectures on the Mordell-Weil theorem" and Szamuely's "Galois Groups and Fundamental Groups":

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Thanks in advance

  • 1
    $\begingroup$ This argument is written out at math.umn.edu/~garrett/m/number_theory/belyi.pdf $\endgroup$ Commented Apr 11, 2014 at 19:51
  • $\begingroup$ note that this hypothesis is essential: a morphism ramified at 0, 1, and $\infty$ is determined by combinatorial data, and can be shown to be defined over $\overline{\mathbb{Q}}$ $\endgroup$ Commented Mar 10, 2016 at 16:37

1 Answer 1


In the process of reducing the branch points, you need to know that they are algebraic. So you start with a cover $f:X\longrightarrow \mathbb P^1(\mathbb C)$ which has algebraic branch points.That such an $f$ exists is an easy consequence of that $X$ is defined over the algebraic closure of the rationals.

  • $\begingroup$ Ok, so my intuition was right. Many thanks. $\endgroup$
    – Dubious
    Commented Apr 11, 2014 at 19:36

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