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Let $R$ be a ring of characteristic $p$. Let $G$ be the kernel of the natural map $\pi_1^{et} (\operatorname{Spec} R[[x]] [1/x]) \to \pi_1^{et}( \operatorname{Spec} R)$. $G$ has a natural map to $\prod_{l\neq p} \mathbb Z_l(1)$, coming from the etale coverings adjoining the $n$th roots of $x$ for $n$ prime to $p$. Is the kernel of this natural map always a pro-$p$ group?

For $R$ a field, this is standard. But I'm not sure when $R$ is not a field.

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  • $\begingroup$ If $R$ is a henselian regular local ring (in particular, noetherian) then this follows from Abyhankar's Lemma. The next case to wonder about is a normal noetherian domain (maybe henselian), since then "normalization" makes sense and $x$-adic separatedness and completeness is inherited by finitely generated $R[[x]]$-modules (as $R$ is noetherian). Perhaps try to understand that before the case of more general $R$. $\endgroup$
    – user30379
    Apr 4, 2013 at 4:25
  • $\begingroup$ Can you elaborate on the Abyhankar's Lemma argument a bit? As far as I know it's just about local rings of dimension one. I did not mean to imply, in asking the general question, that I was not interested in special cases. I've been thinking mostly about the normal case. For the application I'm considering, regular and Noetherian are fine assumptions, but local is problematic. $\endgroup$
    – Will Sawin
    Apr 4, 2013 at 5:24
  • $\begingroup$ Whoops, the argument I had in mind only shows (via Abyhankar's Lemma) that the kernel has no nontrivial prime-to-$p$ quotients, which is of course much weaker than being pro-$p$, so I have nothing useful to say about your question. More specifically, assuming regularity and excellence but not "henselian local", the maximal prime-to-$p$ quotient of the kernel is the quotient $\prod_{\ell} {\mathbf{Z}}_{\ell}(1)$ that you know. Please talk to experts near you in Princeton to learn about Abhykanar's Lemma beyond the 1-dimensional case (as higher dimensions is where the real strength of it lies). $\endgroup$
    – user30379
    Apr 5, 2013 at 4:34
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    $\begingroup$ Or read SGA 1 XIII section 5 to learn more about Abhyankar's lemma, then talk to experts. $\endgroup$
    – user31960
    Apr 5, 2013 at 13:50

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The answer is no.

Take $R=\mathbb F_2[y]$ and consider the cover defined by the equation $t^3+yt+x=0$. By computing the discriminant, this cover is etale. The inertia group of this cover is a subgroup of $S_3$. By reducing mod $(y)$, it contains a $3$-cycle. By reducing mod $y-1+x^2$, the equation factors into $(t^2-xt+1)(t+x)$, so the inertia group contains a $2$-cycle. Thus the inertia group is full $S_3$.

So $S_3$ is a quotient of $G$. But clearly if $G$ has the property I describe, every finite quotient of it must have a normal $2$-Sylow subgroup, which $S_3$ does not.

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