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.