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I know the etale fundamental group of $\mathbb{Z}$ is trivial. For algebraically closed fields $K$, the etale fundamental group of $K((x))$ is $\hat{\mathbb{Z}}$, since all covers in this case are obtained by adjoining $n$th roots of $x$. However, adjoining $n$th roots isnt etale over $\mathbb{Z}$...so I'm beginning to suspect that the etale fundamental group of $\mathbb{Z}((x)) := \mathbb{Z}[[x]][x^{-1}]$ is trivial, though I'm far from sure...

What about the fundamental group of Spec $\mathbb{Z}[[x]]$ or Spec $\mathbb{Z}((x))\otimes_\mathbb{Z} R$ where $R$ is a relatively 'nice' $\mathbb{Z}$-algebra? (I'm thinking of $R = \mathbb{Z}[1/n]$ or some ring of integers of a number field). Are there any tools for calculating fundamental groups of fiber products over a simply connected space (ie, $\mathbb{Z}$)?

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  • $\begingroup$ There is a non-trivial automorphism of $\mathbb{Z}((x^{1/2}))$. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 12, 2014 at 12:00
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    $\begingroup$ @S.Carnahan: $\mathbb{Z}((x^{1/2}))$ is not étale over $\mathbb{Z}((x))$. To see this, tensor with $\mathbb{F}_2$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 12, 2014 at 15:58
  • $\begingroup$ @LaurentMoret-Bailly Yes I meant Z[[x]][1/x] (edited now). Thanks $\endgroup$
    – Will Chen
    Commented Dec 12, 2014 at 16:30
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    $\begingroup$ Let $X\rightarrow{\rm{Spec}}(\mathbf{Z}(\!(x)\!))$ be finite etale and connected, so $X={\rm{Spec}}(A)$ for a domain $A$ finite etale over $\mathbf{Z}(\!(x)\!)$. Let $A'$ be the module-finite normalization of $\mathbf{Z}[\![x]\!]$ in $A$. Then $A$ is tame at the characteristic-0 codimension-1 generic point of $x=0$, so by Abyhankar's Lemma $A$ is dominated by $\mathbf{Z}[\![x^{1/e}]\!]$ for some $e>0$. Equality is forced for some such $e$ by considering $x$-adic valuation on fraction fields. If $e>1$ then reduction mod $p$ for prime $p|e$ gives a contradiction. QED $\endgroup$
    – user74230
    Commented Dec 12, 2014 at 17:46
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    $\begingroup$ @LaurentMoret-Bailly: Any finite extension of $\mathbf{Q}[\![x]\!]$ is of form $K[\![x^{1/e}]\!]$, so I'm not sure the version of A's Lemma you have in mind. The one in Freitag-Kiehl A.1.11 gives that if $B$ is the integral closure of $\mathbf{Z}[\![x]\!]$ in the Galois closure of the generic fiber of $X$ then its strict henselization at any $t$ over $x=0$ is $A^{\rm{sh}}[x^{1/e(t)}]$ for $e(t)>0$ that is $e$ from the generic point of $x=0$, so the normalization $B'$ of $\mathbf{Z}[\![x^{1/e}]\!]$ in $L[Y]/(Y^e-x)$ is finite etale, so split. Any connected component of Spec($B'$) does the job. $\endgroup$
    – user74230
    Commented Dec 13, 2014 at 15:27

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