Let $F$ be a field of characteristic not dividing $n$ containing the $n^{th}$ rooth of unity and let $K$ be a cyclic extension of degree $d$ dividing $n$. Then $K = F(\sqrt[n]{a})$ for some nonzero $a \in F$. Let $\sigma$ be a generator of the cyclic group $Gal(K/F)$.
So, I'm asked to prove the following
- Show that $\sigma(\sqrt[n]{a}) = \zeta \sqrt[n]{a}$ for some primitive $d^{th}$ rooth of unity $\zeta$.
I tried some stuff but it got me nowhere and now I'm stuck...
Note that $\sqrt[n]a$ satisfies $X^n-a$ over $F$, so its minimal polynomial over $F$ divides $X^n-a$. The roots of this polynomial are $\{\zeta\sqrt[n]a\mid \zeta^n=1\}$ (note that they are all distinct as $char(F)\nmid n$), so the roots of the minimal polynomial are a subset of this set. Since $\sigma(\sqrt[n]a)$ is a root of the minimal polynomial, we now know that $\sigma(\sqrt[n]a)= \zeta\sqrt[n]a$ for some $\zeta$ such that $\zeta^n=1$. Since $\zeta\in F$ and $\sigma^d=id_K$, we have $\sqrt[n]a= \sigma^d(\sqrt[n]a)=\zeta^d\sqrt[n]a$, so $\zeta^d=1$, i.e. $\zeta$ is a $d$-th root of unity. On the other hand, since Galois group acts transitively on the roots of the minimal polynomial and there are $d$ distinct roots, we cannot have $\sigma^k(\sqrt[n]a)=\sqrt[n]a$, i.e. $\zeta^k\sqrt[n]a=\sqrt[n]a$, for $1\leqslant k<d$, so $\zeta^k\neq 1$ for $a\leqslant k<d$. Therefore $\zeta$ is a primitive $d$-th root of unity.