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...


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.

  • $\begingroup$ 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 $k$. $\endgroup$ – Harry May 28 '12 at 16:01
  • $\begingroup$ 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?) $\endgroup$ – Harry May 29 '12 at 18:07
  • $\begingroup$ @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 ? $\endgroup$ – user18119 May 29 '12 at 20:29
  • $\begingroup$ 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. $\endgroup$ – Harry May 29 '12 at 21:06
  • $\begingroup$ @Harry, could you indicate a precise reference to Cornalba Harris ? $\endgroup$ – user18119 May 29 '12 at 21:21

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.