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Mar 14, 2011 at 12:55 answer added Sebastian Petersen timeline score: 1
Jan 31, 2011 at 23:41 comment added Minhyong Kim Donu's argument works by the invariance of the etale $pi_1$ under base-change of algebraically closed fields in characteristic zero. I believe the required specialization argument can be found in Serre's lectures on Galois theory. The key point is that $\pi_1(X\times Y)=\pi_1(X)\times \pi_1(Y)$, which fails in positive characteristic. An amusing (or vexing) fact is that an 'algebraic proof', in any reasonable sense, of finite-generation is unknown even for curves.
Jan 31, 2011 at 21:37 comment added Donu Arapura Martin, in general, you can choose alg. closed fields $K\supset K_0\subset \mathbb{C}$ so that $X$ is defined over $K_0$. You would need to verify that the (tame) fundamental group is unchanged in the process. This may require digging through SGA and/or supplying a proof.
Jan 31, 2011 at 20:25 comment added Martin Brandenburg @Donu: This only covers the case $K = \mathbb{C}$, right? Or can we generalize this using model theory?
Jan 31, 2011 at 18:09 comment added Donu Arapura Mariano, I don't know of a purely algebraic argument, but it would certainly be interesting. As an aside, I might point out that the "wild" part of $\pi_1(\matbb{A}^1)$ in positive characteristic need not be finitely generated.
Jan 31, 2011 at 16:35 comment added Mariano Suárez-Álvarez @Donu: can it be proved 'purely algebraicly'?
Jan 31, 2011 at 16:21 comment added Donu Arapura Yes. Use the finite generation for the topological fund. grp (by using finite triangulability of complex varieties) + the comparison theorem SGA1 exp XII cor 5.2.
Jan 31, 2011 at 16:11 history asked Sebastian Petersen CC BY-SA 2.5