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Question

Let X be a smooth, projective curve over the algebraic closure of ℚ. Let f:X->ℙ1 be a meromorphic function. Assume that the zeros and the poles are defined over some number field, K. Then does this imply that cf is defined over K, for some c?

If so, do we have to assume that the zeros and poles are individually defined over K, or would this work if they are collectively defined over K as well?

Clarification

By being collectively defined I mean that there's some K-model of X, XK, and a closed subscheme YK of XK, such that after base change YK becomes the ramification locus of f.

Thoughts

Let D:=(f). Obviously, H0(O(D),X) is 1 dimensional. DK:=(fK) will also be degree 0, so all that's left to show is that H0(O(DK),XK) is also nonzero. This smacks of some invariance of cohomology theorem. I couldn't quite find the right one to use.

If this line of argument works, this seems to imply that the ramification locus may be defined merely collectively.

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It is not sufficient that the subscheme of poles and zeroes is defined together over $K$, as the example of the function $(z+i)/(z-i):\mathbb P^1 \to \mathbb P^1$, defined only over $\mathbb Q(i)$, illustrates.

If poles of $f$, which form together a subscheme $D_\infty$, are defined over $K$, then the line bundle $\mathcal O(D_\infty)$ is defined over $K$. $f$ can be defined as the section of this line bundle that has the most poles, so it should be defined over $K$ as well.

More formally, suppose $f$ is not defined over $K$ (where $K$ is a separable extension of base field, which is $\mathbb Q$ here anyway). Then for some element $\sigma$ of Galois group of base field $f$ and $\sigma f$ would be sections of $\mathcal O(D_\infty)$ that have the same poles and zeroes, so their ratio would be a holomorphic function on $X$, thus constant.

Therefore every $\sigma$ must act as a multiplication by constant. It remains to select any source point $x$ and divide the function by $f(x)$. After this, action of $\sigma$ will have to be multiplication by 1, thus a multiple of $f$ is defined over $K$ indeed.

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  • $\begingroup$ Oh, darn. Of course that's what I meant. Up to constant. $\endgroup$
    – H. Hasson
    Commented Jan 5, 2010 at 21:40
  • $\begingroup$ I changed it in the body of the question. Aha, I see. Oh, excellent. Wonderful. $\endgroup$
    – H. Hasson
    Commented Jan 5, 2010 at 22:01
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    $\begingroup$ The operation of dividing $f$ by the value $f(x)$ at a distinguished point $x$ does not commute with the operation of $\sigma$. So this proof is not quite right. Instead, writing $\sigma(f) = c_ {\sigma} f$ for all $\sigma \in G_K$, where $c_ {\sigma} \in \overline{K}^{\times}$, we see that $\sigma \mapsto c_ {\sigma}$ lies in ${\rm{H}}^1(K, \mathbf{G}_ m) = 1$, so $c_ {\sigma} = \sigma(a)/a$ for some $a \in \overline{K}^{\times}$. Then $(1/a)f$ is Galois-invariant and hence defined over $K$. $\endgroup$
    – Boyarsky
    Commented Jun 15, 2010 at 13:13

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