We work over the field of complex numbers. (But remarks in characteristic $p$ are very welcome.)


Let $S$ be a finite set of points in $\mathbb{A}^1$ containing $0$ and $1$. [Edit: Assume $S$ contains only algebraic numbers.]



> 
> 
> > 
> > Do $\mathbb{A}^1-S$ and $\mathbb{A}^1-\{0,1\}$ have a finite etale cover in common? 
> > 
> > 
> > 
> 
> 
> 


That is, does there exist a curve $X$, a finite etale morphism $X\to \mathbb{A}^1-S$ and a finite etale morphism $X\to \mathbb{A}^1-\{0,1\}$?


For which $S$ is the answer positive?



Do $\mathbb{A}^1-S$ and $\mathbb{A}^1-\{0,1\}$ have a finite etale cover in common?
The answer is positive if and only if $\mathbb{A}^1\setminus S$ is an arithmetic curve, i.e., $\pi\_1(\mathbb{A}^1\setminus S)\subset \mathrm{Aut}(\mathbb{H}) = PSL\_2(\mathbb{R})$ is an arithmetic subgroup. 


This however does not happen "very often". Let me be more precise.


Note that the Euler characteristic of $\mathbb{A}^1\setminus S$ equals $1-\# S$.
For any fixed integer $e$, there are only finitely many isomorphism classes of arithmetic curves $X$ with Euler characteristic $e$ by Takeuchi's theorem; see Theorem 2.1 in <https://projecteuclid.org/download/pdf_1/euclid.jmsj/1230396454>


Thus, if you fix an integer $n$, there are only finitely many $\mathbb{A}^1\setminus S$ with $\# S = n$ which share a common finite etale cover with $\mathbb{A}^1\setminus \{0,1\}$.


In characteristic $p>0$, the answer is positive (over any field $k$). Indeed, let $k$ be a field of characteristic $p>0$. Then, by Prop. 5.2 in Achinger's paper *Wild ramification and $K(\pi,1)$-spaces*, $\mathbb{A}^1\_k\setminus S$ is a finite etale cover of $\mathbb{A}^1\_k$. Let $f:\mathbb{A}^1\_k\setminus S \to \mathbb{A}^1\_k$ and $g:\mathbb{A}^1\_k\setminus\{0,1\} \to \mathbb{A}^1\_k$ be finite etale maps. Now define $X$ to be (a connected component of) the fibre product $\mathbb{A}^1\_k\setminus S\times\_{f,\mathbb{A}^1\_k,g} \mathbb{A}^1\_k\setminus\{0,1\}$. This comes with finite etale maps to $\mathbb{A}^1\_k\setminus S$ and $\mathbb{A}^1\_k\setminus \{0,1\}$.


