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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?

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    $\begingroup$ Every etale cover of $A^1-\{0,1\}$ restricts to $A^1-S$. $\endgroup$
    – Donu Arapura
    Commented Jan 5, 2018 at 18:04
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    $\begingroup$ @DonuArapura that's right. Given $X\to \mathbb{A}^1-\{0,1\}$ we get a finite etale cover $Y\to \mathbb{A}^1-S$ by restricting the former. But $X$ might not be isomorphic to $Y$... $\endgroup$
    – Neeroen
    Commented Jan 5, 2018 at 18:08
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    $\begingroup$ @DonuArapura Yes, that's right. (For example, $X$ could be $\mathbb{A}^1-S$. It certainly happens for some $S$ that $\mathbb{A}^1-S$ is a finite etale cover of $\mathbb{A}^1-\{0,1\}$.) But more complicated examples can be found using modular curves. In any case, I just realized that probably the answer is negative because any finite etale cover of $\mathbb{A}^1-\{0,1\}$ can be defined over a number field. But finite etale covers of $\mathbb{A}^1-S$ might not be definable over a number field if $S$ contains enough transcendental numbers. So, let us assume $S$ contains only algebraic numbers. $\endgroup$
    – Neeroen
    Commented Jan 5, 2018 at 18:23
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    $\begingroup$ What is your ground field (or base)? In characteristic $p$, there is a trick due to Abhyankar; see e.g. Proposition 5.2 (5.2.1 in the arXiv version) of Achinger, Wild ramification and $K(\pi,1)$-spaces. The proposition only concerns étale maps to $\mathbb A^n$; perhaps you can modify it to say something about étale maps to $\mathbb A^1\setminus\{0,1\}$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jan 5, 2018 at 19:22
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    $\begingroup$ @R.vanDobbendeBruyn That's a good point. In characteristic $p>0$, the answer is positive. Indeed, by Prop. 5.2 in Achinger's paper, $\mathbb{A}^1\setminus S$ is a finite etale cover of $\mathbb{A}^1$. Let $f:\mathbb{A}^1-S \to \mathbb{A}^1$ and $g:\mathbb{A}^1-\{0,1\} \to \mathbb{A}^1$ be finite etale maps. Now define $X$ to be (a connected component of) the fibre product $(\mathbb{A}^1-S)\times_{f,\mathbb{A}^1,g} (\mathbb{A}^1-\{0,1\})$. This comes with finite etale maps to $\mathbb{A}^1-S$ and $\mathbb{A}^1-\{0,1\}$. $\endgroup$
    – Ariyan Javanpeykar
    Commented Jan 6, 2018 at 14:41

1 Answer 1

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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 $\mathbb{A}^1_k\setminus \{0,1\}$.

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