Let $F$ be a finite field. There is an isomorphism of topological groups $\left(\mathrm{Gal}(\overline{F}/F),\circ\right) \cong (\widehat{\mathbb{Z}},+)$. 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 étale $F$-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?

Mostafa Ayaz
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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
  • 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
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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
  • Fair enough! $ $ – Bruno Joyal Nov 26 '13 at 17:05
  • 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
  • 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
  • @lee: This is not convincing. We need ring homomorphisms as transition maps. – Martin Brandenburg Apr 12 '17 at 11:26
  • I believe the answer to your question might lie in galois cohomology. – Chickenmancer Apr 16 '17 at 23:22
  • 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
  • Could someone say what $\widehat{\mathbb{Z}}$ is? – Tanner Strunk Mar 13 '18 at 19:48
  • @TannerStrunk https://en.wikipedia.org/wiki/Profinite_integer – Martin Brandenburg Jul 27 '21 at 09:05

1 Answers1


Yes, there is a natural way of describing the ring structure on $Gal(\overline{F}\backslash F)$, just using the Galois theoretic fact that Galois groups of infinite Galois extensions can be considered as inverse limits of systems of Galois groups of finite extensions.

For simplicity, suppose that $F=\mathbb{F}_p$.

For each $n\in\mathbb{N}$, $Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$ is cyclic of order $n$, generated by the Frobenius $\sigma(x)=x^p$, so there are natural group isomorphisms $\iota_n:\frac{\mathbb{Z}}{n\mathbb{Z}}\cong Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p), r+n\mathbb{Z}\mapsto \sigma^r$.

Therefore $Gal(\mathbb{F}_{p^m}\backslash\mathbb{F}_p)$ carries a natural ring structure given by $\sigma^r*\sigma^s=\sigma^{rs}$, and this makes the maps $\iota_n$ ring isomorphisms.

There are natural surjective transition maps $\nu_{n,m}:Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)\to Gal(\mathbb{F}_{p^m}\backslash\mathbb{F}_p), \alpha\to\alpha|_{\mathbb{F}_{p^m}}$ whenever $m\mid n$, making this collection of Galois groups into an inverse system, and it is a fact in Galois theory that $Gal(\overline{F}\backslash\mathbb{F}_p)$ is the limit of this system.

Therefore, given general $F$-automorphisms $\alpha,\beta\in Gal(\overline{F}\backslash\mathbb{F}_p)$, we can write $\alpha=(\alpha_n)_{n\in\mathbb{N}}$, $\beta=(\beta_n)_{n\in\mathbb{N}}$, where $\alpha_n,\beta_n\in Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$, $\nu_{n,m}(\alpha_n)=\alpha_m$, $\nu_{n,m}(\beta_n)=\beta_m$, and multiplication in $Gal(\overline{F}\backslash\mathbb{F}_p)$ is given by $\alpha*\beta=(\alpha_n*\beta_n)_{n\in\mathbb{N}}$, where multiplication in $Gal(\mathbb{F}_{p^n}\backslash\mathbb{F}_p)$ was defined above.

This is the natural way of describing the ring structure on $Gal(\overline{F}\backslash\mathbb{F}_p)$, and since $\widehat{\mathbb{Z}}$ is identivcally defined as the inverse limit of the system of cyclic groups $\frac{\mathbb{Z}}{n\mathbb{Z}}$, it is equivalent to the ring structure on $\widehat{\mathbb{Z}}$.

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