The following interesting question came up in a discussion I was having with Alex Wright.

Suppose given a branched cover C -> P^1 with four branch points.  It's not hard to see that the field of definition of C has transcendence degree at most 1 over $\mathbf{\bar{Q}}$.   

Which leads one to ask:

Give an example of a field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, and a geometrically connected curve C/K, such that C *does not* admit a branched covering to P^1 with four branch points.

Some remarks:

$\bullet$ If you replace "four branch points" with "three branch points" and "transcendence degree 1" with "transcendence degree 0," the nonexistence of such an example is Belyi's theorem.

$\bullet$ There is no obstruction coming from the choice of K; a theorem of Diaz, Donagi, and Harbater guarantees that for any field K of transcendence degree 1 over $\mathbf{\bar{Q}}$, there exists a geometrically connected curve C/K which admits a 4-branched cover to P^1.

$\bullet$ There is a somewhat subtle issue which doesn't arise in the Belyi case:  a 4-branched cover C/K->P^1/K yields a map from Spec K to $\mathcal{M}_{{0,4}}$ 

("forget the cover, remember the branch points") which, after passage to the generic point gives you a choice of inclusion of the function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$ into K.

One can either consider K as an abstract field, or as a fixed extension of the rational function field $\mathbf{\bar{Q}}(\mathcal{M}_{0,4})$, which places a stronger condition on C.  E.G. if you ask that this inclusion be an isomorphism, you are requiring that C be a 4-branched cover which is "determined uniquely by its branch points," like y^2 = x(x-1)(x-t).