# What are non-canonical generators of $\hat{\mathbb{Z}}$ (resp. the absolute Galois group of a finite field)?

We define $\hat{\mathbb{Z}}$ as the inverse limit $\varprojlim \mathbb{Z}/n \mathbb{Z}$. One can also show that $\hat{\mathbb{Z}}$ is isomorphic to the absolute Galois group $G_k = \operatorname{Gal}(\bar{k}/k)$ of a finite field $k \simeq \mathbb{F}_q$. Literature mentions that there two generators of the absolute Galois group of any finite extension of $k$, namely:

• the arithmetic Frobenius automorphism $x \mapsto x^q$,
• the inverse of the arithmetic Frobenius automorphism, the so-called geometric Frobenius automorphism.

These two elements have infinite order, so the subgroup generated by them are isomorphic to $\mathbb{Z}$ which can be considered to be a subgroup of $\hat{\mathbb{Z}}$. Moreover, these two elements are so-called topological generators of $G_k$, but I do not know yet what this is supposed to mean. Anyway, I guess one of these mentioned facts is the reason we call the Frobenius elements canonical generators.

Questions: What exactly are topological/canonical generators? Are there non-canonical generators of $G_k$ resp. $\hat{\mathbb{Z}}$? If yes, how would they look like?

It would be also great if you could give me an example of such a generator. Thanks for helping!

• Why do you call these things generators? They do not generate $\hat{\mathbb Z}$, only the cyclic dense subgroup $\mathbb Z$ as you said. – D_S Aug 23 '18 at 13:08
• As for generators of the group $\hat{\mathbb Z}$ itself, it is uncountable and isomorphic to $\prod\limits_p \mathbb Z_p$, so there is no generating set you can write down. – D_S Aug 23 '18 at 13:09
• @D_S: You are right, thanks! I looked it up again and noticed that the Frobenius elements are the generators of the absolute Galois group of any finite extension of $k$. Moreover, the Frobenius automorphisms are so-called topological generators of $G_k$ and I do not really know that this means. I will fix my question now... – Diglett Aug 23 '18 at 13:15
• Very brief answer: $\hat{\mathbb{Z}}$ has a topology, and every element is a limit of the subgroup generated by Frobenius. – Hurkyl Aug 23 '18 at 13:27
• Also, the inverse of Frobenius is automatically in the subgroup generated by Frobenius, so that's redundant. Unless, that is, you mean to consider $\hat{\mathbb{Z}}$ as a monoid or maybe a semigroup rather than as a group. – Hurkyl Aug 23 '18 at 13:28

You asked what a non-canonical generator might be. Here’s one answer. Remember that a topological generator $\lambda$ of a group $G$ is just an element that generates a dense subgroup of $G$. Thus, if $G$ is commutative, written additively, you’re asking for the closure of $\Bbb Z\lambda$ to be $G$.
If you think of $\hat{\Bbb Z}$ as the direct product $\prod_p\Bbb Z_p$, then the Frobenius can be thought of as the $\infty$-tuple $(1,1,1,\cdots)$, all componentes being $1$. Its inverse has all components equal to $-1$. But each factor $\Bbb Z_p$ has a veritable profusion of (topological) generators, in fact any (multiplicative) unit of $\Bbb Z_p$ is a topological generator of the additive group $\Bbb Z_p$.
So, I guess that the topological generators of $\hat{\Bbb Z}$ are the $\infty$-tuples where the $p$-th component is a unit of $\Bbb Z_p$.