Let $K = \mathbb{C}(T)$ be the field of complex rational functions in one variable, and let $V$ be a variety defined over $K$. 


**Must $V$ have a solvable point?**


The variety $V$ is assumed geometrically irreducible.


A solvable point is a point in $V(L)$ where $L/K$ is a finite Galois extension with $\mathrm{Gal}(L/K)$ solvable.



Is there a solvable point on any variety over the field of complex rational functions?
No, $V$ need not have a solvable point. The proofs I know construct such $V$ via deformation theory. The basic idea is in my paper.


MR2579389 (2011g:14095) Reviewed   

Starr, Jason Michael(1-SUNYS)   

A pencil of Enriques surfaces of index one with no section. (English summary)   

Algebra Number Theory 3 (2009), no. 6, 637–652.   

14J28 (14D06 14G05) 


Let $d$ and $n$ be positive integers with $d>n$. Let $H= \mathbb{P}\text{Sym}^d(V^\vee)$ be the projective space parameterizing degree $d$ hypersurfaces in $\mathbb{P}V = \mathbb{P}^n$. Let $$\mathcal{X}\subset H\times \mathbb{P}V$$ be the universal hypersurface. Let $S\subset H$ be the subvariety parameterizing hypersurfaces that are unions of $d$ hyperplanes; this is the same as the image of the natural morphism from the Segre variety $$(\mathbb{P}(V^\vee))^d \to \mathbb{P}\text{Sym}^d(V^\vee)$$ induced by the multiplication map. Let $$\phi: \mathbb{P}^1 \to S,$$ be a sufficiently general morphism such that $\phi^\*\mathcal{O}\_H(1)$ has degree at least $d$. 


It is straightforward to compute that the pullback $$\mathcal{X}\_\phi := \mathbb{P}^1\times\_{\phi,H} \mathcal{X}$$ has no solvable multisection over $\mathbb{P}^1$ as soon as $d \geq 5$. Simply consider the monodromy among the components of the strata in the stratification of the geometric generic fiber as a normal crossings variety. Of course the geometric generic fiber is *not* irreducible, as required. 


However, since the base field is the *uncountable* field $\mathbb{C}$, for a sufficiently general deformation of the image curve $\phi(\mathbb{P}^1)$ inside $H$, the same result holds true. Using the valuative criterion of properness, and analysis of the finite flat covers of curves, solvable multisections specialize to solvable multisections. Thus, if a sufficiently general deformation always has solvable multisection, since there are countably many parameter spaces for solvable multisections and since the parameter space $M$ for deformations of $\phi(\mathbb{P}^1)$ has uncountably many closed points, it follows that the original family $\mathcal{X}\_\phi$ also has a solvable multisection. 


In fact, a similar argument works over fields that have large transcendence degree over their prime subfield. However, that excludes fields such as finite fields. That is why this argument does not apply to global function fields. However, as you know, Ambrus Pal does prove the result over function fields of surfaces over finite fields, the next best result possible.


