Let $K$ be a field of characteristic $0$ containing $n$-th root of unity. Then, why $K(a^{1/n})＝K(b^{1/n})$ is equivalent to $b∈a(K^×)^n$?

Question

Let $K$ be a field of characteristic $0$ containing $n$-th root of unity. Then, why $K(a^{1/n})＝K(b^{1/n})$ is equivalent to $b∈a(K^×)^n$? I can prove this tediously, but I heard this result directly follows from Kummer theory. How can I interpret this fact by Kummer theory ?

Answer 1

Kummer theory (at least one of its versions) states that, in this setting, for any finite abelian Galois extension $L/K$ with $n \cdot Gal(L/K)=0$, $(\sigma,u) \in Gal(L/K) \times (K^{\times} \cap (L^{\times})^n)/((K^{\times})^n)\longmapsto \sigma(u)/u \in \mu_n$ is a perfect pairing

Now, consider a finite subgroup $S$ of $K^{\times}/(K^{\times})^n$, the extension $L=K(S^{1/n})$ is finite Galois of exponent dividing $n$ and $N_L=(K^{\times} \cap L^n)/((K^{\times})^n) \supset S$. We restrict the Kummer pairing to $Gal(L/K) \times S \rightarrow \mu_n$, and we can (elementarily) see that $Gal(L/K) \rightarrow Hom(S,\mu_n)$ is injective. It follows that $|N_L|=|Gal(L/K)|=|S| \leq |N_L|$ so that $|S|=|N_L|$ and finally $S=N_L$.

That statement can also be considered as a part of Kummer theory, by the way. I guess I just wanted to show an example of using such a Galois-theoretic statement as above.

So, in this case, let $a,b \in K^{\times}$, and assume $L:=K(a^{1/n})=K(b^{1/n})$. Then by the previous paragraph, $a$ and $b$ generate the same subgroup of $K^{\times}/(K^{\times})^n$, and the conclusion (corrected by GreginGre) follows.