Let $k$ be a field and let $X$ be a hyperbolic curve over $k$.

Then, there are only finitely many hyperbolic curves $Y$ over $k$ dominated by $X$.

I know this statement holds over $k=\mathbf{C}$. In particular, it holds over $k=\overline{\mathbf{Q}}$.

Does it hold over any field $k$?

What about a number field?

  • $\begingroup$ What is a hyperbolic curve over $k$ ? $\endgroup$ – Georges Elencwajg Oct 22 '12 at 18:42
  • $\begingroup$ A smooth projective geometrically connected curve of genus $\geq 2$. $\endgroup$ – Harry Oct 22 '12 at 19:05
  • $\begingroup$ Thanks. Where did you see that definition? $\endgroup$ – Georges Elencwajg Oct 22 '12 at 19:22
  • $\begingroup$ I can't remember...I think sometimes people also use it for an integral curve whose normalization is a smooth projective geometrically connected curve of genus $\geq 2$. $\endgroup$ – Harry Oct 22 '12 at 19:32

The answer is yes if your curves are projective, see E. Kani: Bounds on the number of nonrational subfields of a function field. Invent. Math. 85 (1986), no. 1, 185–198.

Over an algebraically closed field, a hyperbolic curve is either a non-empty open subset of a projective curve of genus $g\ge 2$, or an elliptic curve minus at least one point, or the projective line minus at least three points. Equivalently, this is a smooth geometrically connected curve (not necessarily projective) with finite automomorphism group.

  • $\begingroup$ A very important word here is separable. In characteristic $p > 0$, you can pre- or postcompose with arbitrary powers of Frobenius to get infinitely many morphisms. The cited article only covers separable maps. $\endgroup$ – Remy Nov 6 '17 at 7:01

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