Ring structure on the Galois group of a finite field

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, intrinsically?

Well, if $$\sigma$$ is the Frobenius, we have $$\sigma^n * \sigma^m = \sigma^{n \cdot m}$$ for all $$n,m \in \mathbb{Z}$$, and this describes $$*$$ completely. But is there any way to give an explicit and intrinsic 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 latter question using Grothendieck's Galois theory: Consider the Galois category $$\mathcal{C}$$ of finite étale $$F$$-algebras together with the fiber functor $$\mathcal{C} \to \mathsf{FinSet}$$. The automorphism group is exactly $$\pi_1(\mathrm{Spec}(F))=\widehat{\mathbb{Z}}$$. So we may ask:

Which additional structure on a Galois category is responsible for the ring structure on its automorphism group?

Here is an idea: Grothendieck's main theorem of Galois theory states that $$G \mapsto G{-}\mathsf{FinSet}$$ is an anti-equivalence of categories from profinite groups to Galois categories (with their fiber functors) -- right? The category of profinite groups has finite products (easy), so there are finite coproducts of Galois categories. But how do we describe these, intrinsically? We have $$G{-}\mathsf{FinSet} \sqcup H{-}\mathsf{FinSet} = (G \times H){-}\mathsf{FinSet}$$ for example. The connection to the question is as follows: The anti-equivalence above induces an anti-equivalence of monoids with respect to the product. So there is an anti-equivalence of categories between topological rings and comonoids of Galois categories, the latter being equipped with some kind of functor $$\mathcal{C} \to \mathcal{C} \sqcup \mathcal{C}$$ etc. So this seems to be the additional structure I am looking for. And the original question asks to give an explicit functor for the special case $$\mathcal{C} =$$ finite étale $$F$$-algebras.

• 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. Nov 17, 2013 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. Nov 17, 2013 at 8:06
• 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. Nov 21, 2013 at 18:50
• I believe the answer to your question might lie in galois cohomology. Apr 16, 2017 at 23:22
• The "ring structure" is not compatible between $Gal(\overline{\Bbb{F}}_p/\Bbb{F}_p)$ and $Gal(\overline{\Bbb{F}}_{p^2}/\Bbb{F}_{p^2})$ (in the former it is $\phi_{p^2}*\phi_{p^2}=4\phi_{p^2}$ in the latter it becomes $\phi_{p^2}*\phi_{p^2}=\phi_{p^2}$). This should make it clear that it is not given by something canonical/natural. Jun 6 at 22:04