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This question is very naive, but that's why I'm asking it.

Say we begin with $\mathbb{A}^1_{\mathbb{C}}$. Let $U$ be the open disc around $0$ of radius $1$. Now invert all the $a$'s not in $U$: $Spec(\mathbb{C}[x][\frac{1}{x-a}]_{a \not \in U})$. Would it be true that $\pi_1$ of it would be trivial? In greater generality, if I pick $U$ to be some (open?) set, would this construction yield a scheme with an algebraic $\pi_1$ which is the profinite completion of that of $U$?

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    $\begingroup$ You can find many non-trivial connected finite étale covers of the affine line minus any finite number of points and those covers give field extensions of the generic point. Hence they don't become trivial after removal of any number of closed points. $\endgroup$
    – Torsten Ekedahl
    Commented Nov 28, 2010 at 21:25
  • $\begingroup$ Torsten, I don't quite see how this is answering the question. Would you mind elaborating a bit? $\endgroup$
    – James D. Taylor
    Commented Nov 28, 2010 at 21:35
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    $\begingroup$ He says that if you consider the cover of the affine line that you get by taking a square root of $x-7$, say, then this is ramified at 7 but unramified on $U$. Topologically above $U$ it's just two discs, but your scheme isn't a very good model for $U$ somehow, because it's in the algebraic category and the only functions on it are rational functions with no poles on $U$. So in particular this square root of $x-7$ cover gives an etale cover of $U$ that isn't trivial: generically it corresponds to the extension of $\mathbf C(x)$ obtained by adjoining a square root of $x-7$ and... $\endgroup$
    – Kevin Buzzard
    Commented Nov 28, 2010 at 22:01
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    $\begingroup$ ...whilst there's a power series square root of this, there ain't a square root of it in $\mathbf C(x)$, so in algebraic geometry the cover is still non-trivial. $\endgroup$
    – Kevin Buzzard
    Commented Nov 28, 2010 at 22:02

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