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Let $D = {\rm Sp}\, \mathbb{C}_p\langle x\rangle$ be the affinoid unit disc over $\mathbb{C}_p$. Is there an example of a connected finite etale cover of $D$ whose restriction to the "unit circle" ${\rm Sp}\, \mathbb{C}_p\langle x, x^{-1}\rangle \subseteq D$ is disconnected?

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  • $\begingroup$ Just to point out something obvious, but maybe could be useful to someone trying to construct a counter example: I'm pretty sure the cover needs to not come from the special fiber of a normal integral model. In particular, any of the strange covers of $\mathbb{A}^1_{\overline{F}_p}$ won't help. $\endgroup$ May 26, 2020 at 15:00
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    $\begingroup$ Does the proof of 7.5 here pdfs.semanticscholar.org/4f28/… show that the answer is no? $\endgroup$ May 26, 2020 at 17:14
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    $\begingroup$ @AriyanJavanpeykar Does the statement itself not imply the claim since the Gauss point is contained in the unit circle and the usual fact about fundamental groups :"surjective iff remains connected upon pullback"? $\endgroup$ May 26, 2020 at 17:58
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    $\begingroup$ @AriyanJavanpeykar thank you, this answers the question, at least if "becomes disconnected" is replaced with "splits completely". To record the argument here: let $D' = {\rm Sp}\, \mathbb{C}_p\langle x^{-1}\rangle$ so that $\{D, D'\}$ is an admissible covering of $\mathbb{P}^1$ and $D\cap D'$ is the unit circle. Given a finite etale cover which splits completely over $D\cap D'$, we can extend it to a finite etale cover of $\mathbb{P}^1$ by pasting in disjoint copies of $D'$. As $\mathbb{P}^1$ is (algebraically) simply connected, the extended cover is trivial, and hence so is the original one. $\endgroup$ May 26, 2020 at 19:48

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$\newcommand{\Sp}{\mathrm{Sp}\,}\newcommand{\cO}{\mathcal{O}}\newcommand{\fm}{\mathrm{m}}$There seem to exist such examples and this does not contradict de Jong's argument because his proof only shows that the homomorphism $G:=\pi_1(\Sp C\langle x^{\pm 1}\rangle)\to H:=\pi_1(\Sp C\langle x\rangle)$ satisfies the following property: if an action of $H$ on a finite set becomes trivial when restricted to $G$ then it is trivial. Equivalently, the normal subgroup generated by the image of $G$ is equal to the whole of $H$.

Take $$A= C\langle x,y\rangle/(y^{p+1}-xy-p^{1/2})$$ The discriminant of this polynomial in $y$ is $-(p+1)^{p+1}p^{p/2}-p^px^{p+1}$ up to a unit and is invertible in $C\langle x\rangle$. Therefore, $C\langle x\rangle \to A$ is a finite etale extension.

To check that $\Sp A$ is connected it is enough to show that the polynomial $y^{p+1}-xy+p^{1/2}$ is irreducible in $C\langle x\rangle[y]$. If $y^{p+1}-xy-p^{1/2}=f_1(y)\cdot f_2(y)$ is a factorization into a product of monic polynomials then both $f_1,f_2$ have to lie in $\cO_C\langle x\rangle[y]$ so their reductions modulo the maximal ideal $\fm_C\subset\cO_C$ provide a factorization of $(y^{p}-x)y$. Hence, we may and will assume that $\deg f_1=1$ and $y^{p+1}-xy-p^{1/2}$ has a root $f(x)\in \cO_C\langle x\rangle$. A root has to have $f(0)^{p+1}=p^{1/2}$ and, taking the derivative of the equation $f(x)^{p+1}-xf(x)+p^{1/2}=0$ we also get $(p+1)f'(x)f(x)^p-f(x)-xf'(x)=0$ so $f'(0)=\frac{1}{p+1}f(0)^{1-p}\not\in \cO_C$ which is a contradiction. Therefore, $\Sp A\to D$ is a connected finite etale cover.

On the other hand, this cover splits over $C\langle x,x^{-1}\rangle $: we can find a root of $y^{p+1}-xy-p^{1/2}=0$ by the Hensel's lemma arguing by induction on $n$: suppose that $y_n\in \cO_C/p^{n/2}[x^{\pm 1}]$ is a root of this equation modulo $p^{n/2}$ that reduces to $0$ mod $p^{1/2}$. To lift it over $p^{(n+1)/2}$ it is enough to be able to supply, for any given $a\in\cO_C/p^{1/2}[x^{\pm 1}]$, an element $z$ such that $(p+1)y_n^p\cdot z-x\cdot z=a$ in $\cO_C/{p^{1/2}}[x^{\pm 1}]$. This is possible because $y_n^p-x$ is a sum of a nilpotent and invertible element, hence is invertible.


This example is motivated by the behavior of the $p$-torsion in a family of elliptic curves: given such family $\mathcal{E}\to S$ over a $p$-adic formal scheme $S$ over $\cO_K$ let $S_0(p)\to S_K$ be the etale covering of the generic fiber parametrizing $1$-dimensional $\mathbb{F}_p$ subspaces in $\mathcal{E}_K[p]$. When restricted to the generic fiber of the ordinary locus $S_{ord}$ this covering admits a section by the theory of canonical subgroup but it need not have one over the whole of $S_K$. In other words the restriction of the representation $\pi_1(S_K)\to GL_2(\mathbb{F}_p)$ to $\pi_1(S_{ord})$ lands inside a Borel subgroup.

Using Katz-Mazur-Drinfeld integral model the etale cover $S_0(p)\to S_K$ extends to a flat cover of $S$ itself and the special fiber splits into two irreducible components exactly as in the example above.

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