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 $U$?

power seriessquare 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