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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.

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
  • @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
  • Grothendieck's Galois theory has been used in computer science to study the properties of programming languages and to develop new programming languages, such as the programming language Coq. – Furdzik Zbignew Sep 22 '23 at 12:43

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