# Kummer theory as an equivalence of categories

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.

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.
• 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 $x\mapsto x^p-x$. Commented Apr 21, 2021 at 19:35