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I frequently talk to people who think of finite fields as arithmetic analogs of punctured discs. This makes some sense: the absolute Galois group of a finite field is the profinite completion of $\mathbb{Z}$. Since the absolute Galois group of a field is its algebraic fundamental group, this gives the feel of punctured disc.

The cohomological dimension of a field is the cohomological dimension of its absolute Galois group. Therefore the cohomological dimension of a finite field is $1$. This agrees with the intuition that finite fields are "like" punctured discs. (Given a projective curve $C$ over $\mathbb{C}$ and a point $P$ on $C$, a small punctured analytic neighborhood of $P$ has dimension $1$ in the sense that it is a neighborhood of a curve.)

The picture gets murky when we get to $\mathbb{Q}^{ab}$. The absolute Galois group of $\mathbb{Q}^{ab}$ is not known, but is conjectured to be profinite free. It is known that the cohomological dimension of $\mathbb{Q}^{ab}$ is $1$. Is there some geometric intuition associated with $\mathbb{Q}^{ab}$? It is surely much more complex than a punctured disc, because its algebraic fundamental group (absolute Galois group) is more complicated.

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It might be better to think of the spectra of finite fields as annuli or circles than as punctured discs. Grothendieck has a quasi-unipotent monodromy theorem for punctured discs, but the Frobenius can act on $\ell$-adic cohomology with eigenvalues that have absolute value not 1. –  S. Carnahan May 16 '12 at 3:33
    
$Q^{ab}$ is $Gal(\bar Q, Q)$ ? I am little confused since I thought that usual natation Q^{ab} - maximal abelian extension of Q , and Gal(Q^ab,Q) is idels by class field theory. –  Alexander Chervov May 16 '12 at 5:06
    
Alex: $\mathbb{Q}^{ab}$ is indeed the maximal abelian extension of $\mathbb{Q}$, but its absolute Galois group is $Gal(\bar{\mathbb{Q}}/\mathbb{Q}^{ab})$ not $Gal(\mathbb{Q}^{ab}/\mathbb{Q})$. –  James D. Taylor May 16 '12 at 16:26
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Carnahan: can you expatiate a little more about your comment? –  James D. Taylor May 16 '12 at 16:26
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If, as conjectured, the Galois group is profinite free on countably many generators, it would be analogous to a wedge of countably many circles, or a plane with infinitely many (non-accumulating) punctures –  Eric Wofsey Dec 1 '12 at 2:25
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