# Does every curve over a number field have infinitely many rational functions of fixed degree

Let $X$ be a curve over a number field $K$ of genus $g\geq 2$. Does there exist an integer $d$ such that $X$ has infinitely many rational functions (i.e., finite morphisms $f:X\to \mathbf{P}^1_K$) of degree $d$?

If yes, can we choose/bound $d$ in terms of $K$ and $g$?

Note. This is not the same question as in the title. The answer to the question in the title is in fact negative as the example below shows for $d=2$.

Note: I want rational functions to be really different. So we mod out by the action of the automorphism group of $\mathbf{P}^1_K$. That is, $f$ and $\sigma\circ f$ are the same if $\sigma$ is an automorphism of $\mathbf{P}^1_K$.

Example 1. We can not have $d=2$. Hyperelliptic maps are unique.

Question 2. How does the answer to this question change if we replace $K$ by an algebraically closed field $k$?

Example 2. Let $X$ be a general curve of odd genus $g\geq 3$ over an algebraically closed field. Then, it has an infinite number of (really different) gonal morphisms. In fact, this family is one-dimensional.

Idea. I think the question can be reduced to a question on $\mathbf{P}^1_K$. In fact, it suffices to show that there are infinitely many (really different) rational functions on $\mathbf{P}^1_K$ of some degree, say $3$. In fact, once you know this, composing some morphism $f:X\to \mathbf{P}^1_K$ with such a rational function gives infinitely many rational functions of degree $d \leq 3 \deg f$. The only problem is then finding (in a controlled way) a morphism $f:X\to \mathbf{P}^1_K$. This is a hard problem, but let's allow finite base change if necessary...

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I think you are looking for rational subextensions of $K(X)$ of index $d$.

First in the field $K(t)$ of rational functions of $\mathbb P^1_K$, there are infinitely many subextensions of index $2$: Consider the $K(t^2+\lambda t+1)$ for $\lambda\in K$. If $t^2+\lambda t + 1\in K(t^2+\mu t+1)$, then $(\lambda - \mu)t\in K(t^2+\mu t+1)$, hence $\lambda - \mu=0$ because otherwise $t\in K(t^2+\mu t +1)$. Therefore we get infinitely many subextension of index $2$ (this has no much to do with number fields, just need $K$ is infinite).

Now apply your idea and you get a positive answer to your question. I don't understand what you mean by a hard problem. Any inclusion $K(x)\subset K(X)$ corresponds to a unique morphism $X\to \mathbb P^1_K$ of degree $[K(X) : K(x)]$.

For a bound on $d$: first there always exists a morphism $X\to \mathbb P^1_K$ of degree $\le 2g-2$ (use the canonical morphism, recall $g\ge 2$). So a rough upper bound is $2(2g-2)$.

Edit When are there infinitely many gonal morphisms (i.e. morphisms to $\mathbb P^1$ of smallest degree), including the case $g=1$ ? Let $f : X\to \mathbb P^1_K$ be given by an $f\in L(D)$ for some effective divisor $D$ of degree $d$ with $\deg f=d$. Suppose $D$ is base point free and if $\dim L(D)\ge 3$, then there are infinitely many gonal morphisms if $K$ is infinite.

Proof. Write $D=\sum_{1\le i\le n} a_i[x_i]$ with $a_i>0$. As $D$ is base point free, $L(D)\ne L(D-[x_i])$. So there exists $f_1\in L(D)\setminus \cup_i L(D-[x_i])$ because $K$ is inifnite. Let $f_2\in L(D) \setminus (K+Kf_1)$ (recall that $\dim L(D)\ge 3$). Let's show that $$K(f_2)\ne K(f_1).$$ Otherwise $f_2=(\alpha f_1+ \beta)/(\gamma f_2+ \delta)$ with $\alpha, \beta, \gamma, \delta\in K$ and $\gamma\ne 0$. So we can write $$af_2+b=1/(\gamma f_1+c), \quad a, b, c\in K, \ a\ne 0.$$ So $1/(\gamma f_1+c)\in L(D)$. As $$(\gamma f_1+c)_{\infty}=(f_1)_{\infty}=D,$$ this is impossible. Let it is easy to see that $K(f_1+\lambda f_2)$, when $\lambda$ runs through $K$, form an infinite family of pairwise distinct rational subextension of $K(X)$ of index $d$.

For genus $1$ curves, the gonality is the maximum of $2$ and the index of the curve. If the gonality $\gamma$ is at least $3$, the result above plus Riemann-Roch show there are infinitely many morphisms of degree $\gamma$ from $X$ to the projective line.

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I was having genus 1 curves in my mind. There, I think, the problem is harder to find a rational function with bounded degree. I mean, if this were not hard, then "index problems" of elliptic curves would be easy. But ok, this answer my question. There are infinitely many rational functions $\mathbf{P}^1_k\to \mathbf{P}^1_k$ of degree $2$ when $k$ is infinite and there always exists a rational function of degree bounded by $(2g-2)$ if $g>1$. Thus, there are infinitely many rational functions of degree bounded by $2(2g-2)$ on $X$, where $X$ is a curve over some infinite field $k$. – Harry May 28 '12 at 16:01
Some questions. Firstly, do the infinite number of index 2 subextensions really give rise to different degree 2 morphisms? I have a feeling that some of them might a priori be the same up to an automorphism. Finally, in the book by Griffiths, Cornalba and Harris one can find that a curve of even genus has only finitely many gonal morphisms. You on the other hand seem to prove the contrary. What am i missing here? (I think you don't mean to say "gonal morphism" but just morphism of degree $d$, right?) – Harry May 29 '12 at 18:07
@Harry, my answer to the first question is yes. You define $f$ and $g$ be the same if $g=\sigma\circ f$ for an automorphism $\sigma$ of the downstaire $\mathbb P^1$. But even if you allow automorphism in the upstaire $P^1$, the result can be shown to be the same. As for gonal morphisms, I mean what I wrote: they are morphisms to $P^1$ of the smallest degree. What is your definition ? – user18119 May 29 '12 at 20:29
Our definition of gonal morphisms is the same. I was just confused (which is completely my fault). Let $\gamma$ be the gonality of $X/K$ and let $f$ be a gonal morphism such that $f$ is given by an element of $L(D)$ with $D$ of degree $\gamma$. Then, you seem to prove that there are infinitely many gonal morphisms if $\dim L(D) \geq 3$ and $D$ is base point free. I guess this condition is not satisfied when $X$ is of even genus (by the Cornalba Harris result...). I can't see why though, directly. – Harry May 29 '12 at 21:06
@Harry, could you indicate a precise reference to Cornalba Harris ? – user18119 May 29 '12 at 21:21