most recent 30 from http://mathoverflow.net 2013-06-19T13:36:44Z http://mathoverflow.net/feeds/question/97051 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97051/what-is-the-intuition-for-mathbbqab-having-cohomological-dimension-1 James D. Taylor 2012-05-15T20:57:46Z 2012-05-15T20:57:46Z

<p>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. </p> <p>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.)</p> <p>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. </p>