5
$\begingroup$

Let $K$ be an algebraically closed field of characteristic zero. Let $X/K$ be a smooth variety. Is it true that the \'etale fundamental group $\pi_1(X)$ is topologically finitely generated.

I know that the answer is ``yes'' in the following two cases:

1) $X/K$ is proper.

2) $\dim(X)=1$. (In this case we can write down a presentation of $\pi_1(X)$ very explicitly.)

Can there anything go wrong in the general case? (It would be wonderful to have a reference which one can simply quote.)

Improve this question
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Donu Arapura
    Commented Jan 31, 2011 at 16:21
  • $\begingroup$ @Donu: can it be proved 'purely algebraicly'? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 31, 2011 at 16:35
  • $\begingroup$ 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. $\endgroup$
    – Donu Arapura
    Commented Jan 31, 2011 at 18:09
  • $\begingroup$ @Donu: This only covers the case $K = \mathbb{C}$, right? Or can we generalize this using model theory? $\endgroup$
    – Martin Brandenburg
    Commented Jan 31, 2011 at 20:25
  • 4
    $\begingroup$ 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. $\endgroup$
    – Minhyong Kim
    Commented Jan 31, 2011 at 23:41

1 Answer 1

Reset to default
1
$\begingroup$

Thanks, Dono Arapura, for explaining how one proves this statement, using the Riemann existance theorem in SGA1.

I mention for the sake of completeness: Today in the morning, I found that there is a SGA-reference for my question as it stands above: SGA 7.1, II.2.3.1. (I was surprised, that it is SGA 7.1, not SGA 1.)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .