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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, 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.

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    $\begingroup$ 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. $\endgroup$
    – Alex Youcis
    Nov 17, 2013 at 8:01
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    $\begingroup$ 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. $\endgroup$
    – Alex Youcis
    Nov 17, 2013 at 8:06
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    $\begingroup$ 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. $\endgroup$
    – Martin Brandenburg
    Nov 21, 2013 at 18:50
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    $\begingroup$ 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. $\endgroup$
    – reuns
    Jun 6, 2022 at 22:04
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    $\begingroup$ As @Chickenmancer, comments, perhaps the (cup product) ring structure in cohomology is a potential answer. Its existence is mildly surprising, after all. But I don't have anything more substantial to say about possibilities. :) $\endgroup$
    – paul garrett
    Jun 6, 2022 at 22:49

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