# Is there a solvable point on any variety over the field of complex rational functions?

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.

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.

• My question has also been answered in a paper of Ambrus Pal, where he explicitly constructs such a variey. See wwwf.imperial.ac.uk/~apal4/publ/solvpoint.pdf Theorem 4.11 and Example 4.12 following it. – Pablo Jan 22 '15 at 13:10
• Let V be a variety (geometrically irreducible and reduced) defined over F(T) for number field F, L/F(T) finite Galois, and P in V(L). What conditions on integers N such that specializing T=N gives geometrically irreducible and reduced V(N)/F and P(N) in V(N) with Gal(F(P(N))/F)=Gal(F(T)(P)/F(T))? How do the conditions relate to F, V, Gal(F(T)(P)/F(T))? – David Lampert Jan 22 '15 at 22:23
• @DavidLampert There seems to be no conditions of this kind. By Hilberts Irreducibility Theorem we know however that there is an abundance of such specializations. – Pablo Jan 23 '15 at 6:08
• @Pablo I expect such N should include a congruence class mod P for P depending on prime divisors and heights of defining coefficients for V,F. Is this evident (not to me) from model theory of arithmetic? – David Lampert Jan 23 '15 at 17:33
• @Pablo Clarifying my last comment: a congruence class mod M for M depending on prime divisors and heights of defining coefficients for V,F,P. – David Lampert Jan 23 '15 at 17:41