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Let $F$ be a finite field. There is an isomorphism of topological groups $(\mathrm{Gal}(\overline{F}/F),\circ) \cong (\widehat{\mathbb{Z}},+)$. It follows that the Galois group carries the structure of a topological ring isomorphic to $\widehat{\mathbb{Z}}$. What does the multiplication $*$ look like? If $\sigma$ is the Frobenius, we have $\sigma^n * \sigma^m = \sigma^{n*m}$, and this describes $*$ completely. Is there any way to give an explicit and natural formula for $\alpha * \beta$ if $\alpha,\beta$ are $F$-automorphisms of $\overline{F}$? Also, is there any more conceptual reason why the Galois group carries the structure of a topological ring (without computing the Galois group)?

Maybe the following is a more precise version of the question: Consider the Galois category $\mathcal{C}$ of finite ├ętale $F$-algebras together with the fiber functor to $\mathsf{FinSet}$. The automorphism group is exactly $\pi_1(\mathrm{Spec}(F))=\widehat{\mathbb{Z}}$. Which additional structure on the Galois category $\mathcal{C}$ is responsible for the ring structure on its automorphism group?

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I mean, the second question seems to follow more naturally since all finite Galois subextensions of $\overline{F}/F$ have Galois groups which are rings--it's merely the fact that the absolute Galois group is the limit of these that gives it the ring structure. So a more poignant question may be "why do finite fields have Galois groups that have a ring structure?" But, us thinking these have a ring structure is more a function of the notation $\mathbb{Z}/n\mathbb{Z}$ then it is a natural ring structure--or so it seems to me. Nice question though, +1. – Alex Youcis Nov 17 '13 at 8:01
I guess, my question is why you'd expect the ring structure to be natural. For example, if someone wrote $\text{Gal}(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})=\mathbb{Z}_p^\times$, you may think that there is no natural ring structure. But, if instead someone had written it as $\mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$, you may ask the same question there. – Alex Youcis Nov 17 '13 at 8:06
You are probably right, why should it be natural, and what should this mean? Actually for every $u \in \widehat{\mathbb{Z}}$ there is a ring structure extending the group structure with unit $\sigma^u$, namely $\sigma^n *' \sigma^m = \sigma^{n+m-u}$. But there is only one ring structure (extending the group structure) with unit $\sigma$. – Martin Brandenburg Nov 17 '13 at 12:32
Dear Martin, this is an interesting question. Have you ever seen this multiplication appear naturally somewhere? Cheers, – Bruno Joyal Nov 20 '13 at 14:21
Actually this question just comes out of curiosity. And I've learned in the last years that it is better not to ignore extra structures. – Martin Brandenburg Nov 21 '13 at 18:50

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