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The main theorem of Galois theory (in the finite case, for convenience) says that

Theorem 1: Let $L/k$ be a finite Galois extension with Galois group $G$. The maps $$M\mapsto \operatorname{Aut}(L|M)\qquad\text{and}\qquad H\mapsto L^H$$ yield an inclusion-reversing bijection between subfields $k\subset M\subset L$ and subgroups $H\subset G$.

This result can be generalized into an equivalence of categories which is known as Grothendieck's Galois theory:

Theorem 2: Let $k$ be a field and fix a separable closure $k_s$. The functor mapping a finite étale $k$-algebra $A$ to the finite set $\hom_k(A,k_s)$ gives an anti-equivalence between the category of finite étale $k$-algebras and the category of finite sets with continuous left $\operatorname{Gal}(k)$-action.

Now, the main theorem of Kummer theory is very similar in style to our Theorem 1. For that, we will say that a finite Galois extension $L/k$ is $n$-Kummer if $k$ contains a primitive $n$-th root of unity and if $\operatorname{Gal}(L/k)$ is abelian of exponent dividing $n$.

Theorem 3: Let $n\geq 1$, $k$ be a field and fix and algebraic closure $\bar{k}$. The maps $$M\mapsto k^\times \cap M^{\times n}/k^{\times n}\qquad\text{and}\qquad A\to k[\sqrt[n]{a}:[a]\in A]$$ yield an inclusion-preserving bijection between the $n$-Kummer extensions $M/k$ contained in $\bar{k}$ and the finite subgroups $A$ os $k^\times/k^{\times n}$.

I wonder if we can write our theorem 3 as an equivalence of categories similar to the one in theorem 2.

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Perhaps this is not what you were looking for, but the bijection of Kummer theory can be realized as a special case of the Galois correspondence and phrased in Grothendieck's language. To see why this is true, we want to reformulate Grothendieck's Galois correspondence just a little bit.

For an extension $L/k$, we'll say that a finite $k$-algebra $A$ is $L$-split if $A\otimes_k L \cong L^{n}$ for some $n$. The finite etale $k$-algebras are then precisely those that are $k^s$-split. Grothendieck's Galois theory can be refined to say that for a separable extension $L/k$, there is an (anti-)equivalence of categories $$\{\text{finite }L\text{-split algebras}\} \longleftrightarrow \{\operatorname{Gal}(L/k)\text{-sets}\}$$ where the equivalence sends $A$ to $\hom_k(A,L)$. The full correspondence occurs when $L=k^s$.

Now Kummer theory is simply this same statement, but we use $L=k^{(n)}$, where $k^{(n)}/k$ is the maximal abelian extension of exponent $n$. Of course this isn't really giving a new proof of Kummer theory; you still need to compute $k^{(n)}$, which is equivalent to classifying extensions of exponent $n$. One way is to apply cohomology to the Kummer exact sequence $$1\longrightarrow \mu_n \longrightarrow (k^s)^\times \overset{x\mapsto x^n}{\longrightarrow} (k^s)^\times \longrightarrow 1$$ which immediately gives an isomorphism $\operatorname{Hom}(\operatorname{Gal}(k^s/k),\mu_n) \cong k^{\times}/k^{\times n}$ (this time this is a homomorphism of groups). But every homomorphism $\operatorname{Gal}(k^s/k)\to \mu_n$ must factor through $\operatorname{Gal}(k^{(n)}/k)$ so from here one can conclude that $\operatorname{Gal}(k^{(n)}/k)\cong k^{\times}/k^{\times n}$. Taking a little more care shows that the element $a\in k^{\times}/k^{\times n}$ maps to the character $\chi_a \in \operatorname{Hom}(\operatorname{Gal}(k^s/k),\mu_n)$ given by $\chi_a(\sigma) = \frac{\sigma \sqrt[n]a}{\sqrt[n]a}$, which has kernel equal to $\operatorname{Gal}(k^s/k(\sqrt[n] a))$. This shows that the bijections given by Grothendieck's equivalence of categories and/or usual Galois theory are the same ones you have in your Theorem 3.

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  • $\begingroup$ Really beautiful answer. +1 $\endgroup$ Commented Apr 21, 2021 at 7:15
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    $\begingroup$ This is really cool! Do you happen to know if we can do something similar with Artin-Schreier extensions? And do you know some reference for all that? $\endgroup$
    – Gabriel
    Commented Apr 21, 2021 at 17:26
  • $\begingroup$ Artin--Schreier theory works the same way, replacing the Kummer exact sequence by $0\to \mathbb F_p\to k^s \to k^s \to 0$ where the map $k^s \to k^s$ is $x\mapsto x^p-x$. $\endgroup$
    – vijay
    Commented Apr 21, 2021 at 19:35

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