minimal polynomial in Kummer extension Let $n>1$ be an integer. Let $K$ be a field such that $n$ does not divide the characteristic of $K$ and $K$ contains the $n$-th roots of unity. Let $\mu_n\subseteq K$ be the set of $n$-th roots of unity. Consider $\Phi\in\text{Hom}(K^{\times}/K^{\times n},\mu_n)$. Let  $x\in K^{\times}$. Let $\Phi(xK^{\times n})=\zeta_x$ for some $\zeta_x\in\mu_n$. 
Let $x^{1/n}$ be any $n$-th root of $x$. I have a feeling that $x^{1/n}\zeta_x$ and $x^{1/n}$ have the same minimal polynomial over $K$. Is it correct ?
 A: The answer is yea, and I’ll give you an argument which I’m sure is not maximally efficient.
Kummer theory gives us a perfect pairing between the discrete group $K^\times/K^{\times n}$ and the Galois group $G^{K_n}_K$, where by $K_n$ I mean the field generated by all the $n$-th roots of elements of $K$. That is, we have $\varphi:K^\times/K^{\times n}\times G^{K_n}_K\rightarrow\mu_n$, bilinear, such that if $\varphi(a,\sigma)=1$ for all $\sigma$ in the Galois group, then $a$ is an $n$-th power of an element of $K$; and if $\varphi(a,\sigma)=1$ for all $a\in K$, then $\sigma$ is identity.
This means that your $\Phi$, whatever it is, can be identified with an element of the Galois group $G^{K_n}_K$, say $\tau$, which induces $\Phi$ by: for $a\in K^\times$, choose any $b$ with $b^n=a$, then $\Phi(a)=\tau(b)/b$. In particular, your $x$ has $\Phi(x)=\zeta$ and thus for an $n$-th root $y$ of $x$ we see that $\tau(y)=\zeta y$. According to your setup, $\zeta x^{1/n}=\zeta y$ is thus a Galois conjugate of $x^{1/n}=y$, and has the same minimal polynomial.
