# Inverting infinitely many points on an algebraic curve

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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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. – Torsten Ekedahl Nov 28 '10 at 21:25
Torsten, I don't quite see how this is answering the question. Would you mind elaborating a bit? – James D. Taylor Nov 28 '10 at 21:35
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... – Kevin Buzzard Nov 28 '10 at 22:01
...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. – Kevin Buzzard Nov 28 '10 at 22:02