Why is $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p$), $p$ a prime number, topologically cyclic? I understand that the topological generator is supposed to be $\text{Frob}_p$, the automorphism of $\overline{\mathbb{F}_p}$ which sends $x$ to $x^p$, and that it would be enough to say some power of this map lies in every basic open set of $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p$), i.e. in every coset of $\text{ Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_{p^l}$) where $l$ is a positive integer. It's clear that $(\text{Frob}_p)^l$ is in $\text{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_{p^l}$), but I'm having trouble thinking about the cosets. Is there a better way to think about this?
 A: Don’t think about cosets.
Rather, just as you get a better and better view of a microbe by using higher and higher-power microscopes, you get better and better understanding of an element of the total Galois group by seeing what it does to larger and larger extensions of your base field. But no matter how big your finite extension $K$ of $\Bbb F_p$ is, $\text{Frob}_p$ is the generator of $\text{Gal}^K_{\Bbb F_p}$.
Continuing on these lines, what does it mean for $\sigma=\text{Frob}_p$ to generate the absolute Galois group? It means that for any given $\tau$ in the group, no matter what open subgroup $S$ about the identity you choose, there’s a power of $\sigma$ that’s $S$-close to $\tau$. Well, “$\sigma^m$ and $\tau$ are $S$-close” means $\tau^{-1}\sigma^m\in S$, means $\tau^{-1}\sigma^m$ is identity on the (finite) fixed field of $S$. Thus you’re saying that any $\tau$ in the absolute Galois group is power-of-Frobenius on each finite extension of $\Bbb F_p$. And that’s just the fact that every $\text{Gal}^K_{\Bbb F_p}$ is cyclic, generated by Frobenius, which we all know.
