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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 absolute Galois group carries the structure of a topological ring isomorphic to $\widehat{\mathbb{Z}}$.

The multiplication looks as follows. 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. This makes me wonder:

Can we describe the multiplication $*$ intrinsically?

I mean is there any formula for $\alpha * \beta$ if $\alpha,\beta$ are $F$-automorphisms of $\overline{F}$ which doesn't just use their classification as above?

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). 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.

Martin Brandenburg
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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 08:01
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    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 08:06
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    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
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    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
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    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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    Fair enough! $ $ – Bruno Joyal Nov 26 '13 at 17:05
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    The Galois splitting field is all the prime elements of multiplicative inverses to unity. This would be seen nicely as a arithmetic modulo function combine with the prime factorisation function. It is also a nice pre-empt to unsolvability of quintic radicals. (How Abelian groups necessitate the limits of cyclic rings to the power of 5) – McTaffy Dec 08 '16 at 10:45
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    Whenever you have an abelian extension $L/K$, the Galois group will have a ring structure coming from finite level. Because on each finite level you have isomorphism looks like $Gal(K'/K)=\prod (Z/n_i)$. – lee Feb 04 '17 at 03:40
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    @lee: This is not convincing. We need ring homomorphisms as transition maps. – Martin Brandenburg Apr 12 '17 at 11:26
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    I believe the answer to your question might lie in galois cohomology. – Chickenmancer Apr 16 '17 at 23:22
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    If the continuity is linear then the difference equation representing the topology should have a stable point or a singular curve. This singular curve will be the kernel of the galois splitting field. – McTaffy Jul 24 '17 at 15:19
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    Could someone say what $\widehat{\mathbb{Z}}$ is? – Tanner Strunk Mar 13 '18 at 19:48
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    @TannerStrunk https://en.wikipedia.org/wiki/Profinite_integer – Martin Brandenburg Jul 27 '21 at 09:05
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    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. – reuns Jun 06 '22 at 22:04
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    @reuns, very good point, in my opinion. On another hand, sometimes "off by something" becomes ok at some cohomological or derived-what's-it level. Still, I tend to agree that the fact that $\mathbb A/\mathbb Q\approx \mathbb R/\mathbb Z\times \prod_p \mathbb Z_p$, as topological groups, and that the $\mathbb Z_p$'s are rings, does not seem to say much. Maybe something about endomorphisms? Dunno. – paul garrett Jun 06 '22 at 22:19
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    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. :) – paul garrett Jun 06 '22 at 22:49

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