$\newcommand{\gal}{\operatorname{Gal}}\newcommand{\char}{\operatorname{char}}$I recently learnt a little about Kummer extension from Professor Borcherd's lecture and from the Wikipedia article. Wikipedia and other sources seemed to be saying that in general, a Galois extension $L/K$ is abelian iff. it is generated by some $n$th root, if one presumes:
- $K$ contains distinct $n$th roots of unity (in particular, $\char K\nmid n$)
- $\gal(L/K)$ is a group of exponent $n$
Wikipedia and Borcherds both prove that Kummer extensions are given by taking $n$th roots for extensions $L/K$ for which $\gal(L/K)\cong\Bbb Z/p\Bbb Z$ for some prime $p$. The most general case of any abelian group is not covered.
I thought I had a solution to this. If $\gal(L/K)$ is taken to be abelian, then all subgroups are normal and all quotients exist. So, in a simple case, let's say $|\gal(L/K)|=p^3$ for some prime $p$. By Cauchy's theorem, there is an element $\sigma_1$ of order $p$, and I can consider $H_1:=\langle\sigma_1\rangle$ a normal subgroup. By a basic theorem of Galois theory, then the fixed field $K^{H_1}$ is a Galois extension of $K$, and $\gal(L/K^{H_1})=H_1$ and $\gal(K^{H_1}/K)\cong\gal(L/K)/H_1$ which has order $p^2$. By the theorem for abelian extensions of order $p$ (automatically cyclic) and presumption on $K$ containing all $p^3$th roots of unity, $K$ also contains all $p$th roots of unity - so does $K^{H_1}$, of course - so $L=K^{H_1}(\sqrt[p]{\gamma_1})$ for some element $\gamma_1\in K^{H_1}$.
We can induct - again by Cauchy's theorem, there is some order $p$ element in $\gal(K^{H_1}/K)$, call it $\sigma_2$, so we can consider the normal subgroup $H_2:=\langle\sigma_2\rangle$. Then $K^{H_2}$ is a Galois extension of $K$, and $K^{H_1}=K^{H_2}(\sqrt[p]{\gamma_2})$ for some element $\gamma_2\in K^{H_2}$, and $\gal(K^{H_2}/K)\cong\gal(K^{H_1}/K)/H_2$ which has order $p$, so is cyclic itself with some generator $\sigma_3$. Finally, $K^{H_2}=K(\sqrt[p]{\gamma_3})$ for some $\gamma_3\in K$, and I conclude: $$L=K^{H_1}(\sqrt[p]{\gamma_1})=\cdots=K(\sqrt[p]{\gamma_3})(\sqrt[p]{\gamma_2})(\sqrt[p]{\gamma_1})$$Somehow we can combine this into a $p^3$-root of some $\gamma\in K$ (edit: actually no we can’t in general) but I'm not completely confident on how to do this. Wikipedia take a large product, but that feels wrong to me. $\gamma_2$ is some combination in $\sqrt[p]{\gamma_3}$, etc., so I suppose one just replaces all instances of $\gamma_{1,2}$ with their expressions in $\gamma_3$, finally leaving one with one single root of size $p^3$.
My main question - is that correct? Does it generalise in the way I hope to, say, when the exponent of $\gal(L/K)$ is some product of primes, say $p_1^3p_2^2$? I think one can use the structure theorem to reduce to the cases where there is only one prime power present, but I'm not sure.
As an aside, I'd appreciate a solution-verification on what I've done below.
I tried to apply this on a specific example, to see if it really worked. Consider $K=\Bbb Q$, $L=\Bbb Q(\alpha=\sqrt{2+\sqrt{2}})$. This has a Galois group which is cyclic of order $4=2^2$, generated by $\sigma$ which maps $\sqrt{2+\sqrt{2}}\mapsto\sqrt{2-\sqrt{2}}$. To begin the induction, I take the subgroup $H_1=\langle\sigma^2\rangle$. We follow the eigenvalue/eigenvector algorithm - the second roots of unity are just $\pm1$, so we take the eigenvector with value $-1$. Take the root $\alpha$ and construct $v_1=\alpha-\sigma^2(\alpha)$. According to the Kummer theory, $v_1^2$ should be an element of $K^{H_1}$. This is easily seen, and in fact $v_1=2\alpha$, so $v_1^2=4\alpha^2=8+4\sqrt{2}$, which agrees with our prior knowledge since we know $K^{H_1}=K(\sqrt{2})$. Then the theory claims $L=K^{H_1}(\sqrt{8+4\sqrt{2}})=K^{H_1}(\sqrt{2+\sqrt{2}})$ by bringing out the factor of $2^2$ as it is irrelevant. So far, so good. The group $\gal(K^{H_1}/K)$ is harder to determine (pretending we don't have the full knowledge already) - according to my notes, it is given by the restriction map $\gal(L/K)\to\gal(K^{H_1}/K)\cong\gal(L/K)/H_1$. Then I think the order two element of this group should just be the restriction of $\sigma|_{K^{H_1}}$. So, following the same process, we pick a value in $K^{H_1}\setminus K$, say $\beta=v_1^2$, and construct $v_2=\beta-\sigma(\beta)$. The claim would be that $v_2^2\in K$, which is true. $v_2=8\sqrt{2}$ and $v_2^2=128$.
Then according to the theory, $L=K(\sqrt{128})(\sqrt{8+4\sqrt{2}})=K(\sqrt{2})(\sqrt{2+\sqrt{2}})$, which is true.
My second question - did I do that correctly? I know the conclusion is correct, but I feels like I've not done something "properly" since I was only able to compute these values with my prior knowledge of all the roots and what $\sigma$ is. Is it even possible to use this theory to find the (nested) radicals without exact knowledge of the elements of the Galois group?