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I know the fundamental group of $\mathbb{Z}$ is trivial, and the prime candidate for finite etale covers involving the $x$'s is the endomorphism sending $x\mapsto x^n$, but that's ramified at all primes dividing $n$, so my guess is that $\pi_1(\text{Spec }\mathbb{Z}((x)))$ is trivial, though I'm not sure.

What about the fundamental group of things like $\text{Spec }\mathbb{Z}((x))\otimes_\mathbb{Z}R$ where $R$ is some ring like $\mathbb{Z}[1/N]$ or $\mathbb{Z}[1/N,e^{2\pi i/N}]$ or $\mathbb{C}$?

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Well, let's do the easiest case: $X=\text{Spec}(\mathbb{C}((x)))$. Since this is a field, we know that $\pi_1(X)=G_{\mathbb{C}((x))}$. But, all extensions of $\mathbb{C}(x))$ are simple:

Claim: All extensions of $\mathbb{C}((x))$ are of the form $\mathbb{C}((t))$ with $t^n=x$ for some $n$.

Proof: Let $K/\mathbb{C}((x))$ be a finite extension, and let $\mathcal{O}_K$ be its ring of integers, with uniformizer $\pi$. Since $\mathbb{C}[[x]]$ has closed residue field, we know that the $f$ of this extension is $1$. This implies that $e=n:=[K:\mathbb{C}((x))]$. Thus, $x=\pi^n u$ for some unit $u\in\mathcal{O}_K^\times$. Now, by Hensel's lemma, the polynomial $T^n-u$ splits into unique linear factors in $\mathcal{O}_K[T]$, and so $u$ has an $n^{\text{th}}$ root $v$ in $\mathcal{O}_K$. Since $\varpi=v\pi$ is another uniformizer, we know that $x=\varpi^n$, where $\pi$ is a uniformizer of $\mathcal{O}_K$. That said, we note that $\mathbb{C}[[x]][\varpi]\subseteq \mathcal{O}_K$ is actually integral, since it's isomorphic to $\mathbb{C}[[T]]$, and thus $\mathcal{O}_K=\mathbb{C}[[x]][\varpi]$. Thus, $K=\mathbb{C}((x))(\varpi)$, which is of the desired form. $\blacksquare$

From this, it's easy to see that

$$\pi_1(X)=G_{\mathbb{C}((x))}=\varprojlim \text{Gal}\left(\mathbb{C}((\sqrt[n]{x}))/\mathbb{C}((x))\right)=\varprojlim \mathbb{Z}/n\mathbb{Z}=\widehat{\mathbb{Z}}$$

I'll come back later to answer the others.

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  • $\begingroup$ Thanks for your answer! ...though perhaps you were assuming that $\mathbb{Z}((x))\otimes_\mathbb{Z}\mathbb{C}$ is the same as $\mathbb{C}((x))$? (they're not the same, mostly because the power series in the former case all have 'bounded denominators') $\endgroup$
    – oxeimon
    Dec 6, 2014 at 9:14
  • $\begingroup$ @oxeimon Of course! How embarrassing :) I'll leave this up for posterity though! This seems like a strange thing to care about then. Looking it up, it seems as though you're choice needn't be Noetherian even if $R$ is, in which case $\pi_1$ gets wonky. Why these rings? $\endgroup$ Dec 6, 2014 at 9:33
  • $\begingroup$ Hey, is there any chance you could take a shot at my original question about the fundamental group of Z((x))? $\endgroup$
    – oxeimon
    Dec 12, 2014 at 11:15

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