I'm currently studying cyclic Galois extensions (i.e. with a cyclic Galois group), specifically the special case where the characteristic is coprime to the extension degree. So consider some cyclic Galois extension $E/K$ with $n := [E:K]$ and $\mathrm{char}(K) \nmid n$.
Now, I know that if $K$ contains all $n$-th roots of unity (which makes $E/K$ into a special case of a so-called Kummer extension), then the extension is radical, i.e. there is some $\alpha \in E$ with $E = K(\alpha)$ and $\alpha^n \in K$.
Indeed, if given a primitive $n$-th root of unity $\zeta_n \in K$, the above statement can be proven by using Hilbert's Satz 90 on $\zeta_n^{-1}$. Now, I'm guessing that the statement does not hold if we drop the assumption of $K$ containing all $n$-th roots of unity. I have tried to come up with a counterexample, however I have so far not been able to find one, so I would be thankful for any hints on how to get to one.
Specifically, I am looking for a cyclic Galois extension $E/K$ of degree $n = [E:K]$ and with $\mathrm{char}(K) \nmid n$ such that $K$ does not contain all $n$-th roots of unity and $E/K$ is not a radical extension, i.e. there is no $\alpha \in E$ with $E = K(\alpha)$ and $\alpha^n \in K$.
$[E:K]=n$ and $E=K(a)$ with $a^n\in K$ means that $E/K$ is purely radical: $x^n-a^n$ is the minimal polynomial of $a$ and $E/K$ is normal iff $x^n-a^n$ splits completely iff $x^n-1$ splits completely in $E$.
To me $E/K$ is radical means that $E=K(a)$ with $a^m\in K$ for some $m$.
$a$ is a root of $x^m-a^m$ so its minimal polynomial is $\prod_{j=1}^n (x-\zeta_m^{c_j}a)$ for some $c_j$ and $E/K$ is Galois means that the $\zeta_m^{c_j}$ are in $E$.
With $K=\Bbb{Q},E=K(\cos(2\pi / 7))$ this implies that $\zeta_m^{c_j}=\pm 1$ thus in fact $n=2$, a contradiction since $[E:K]=3$. Whence $E/K$ is not radical.
On the other hand $E$ is contained in the tower of purely radical extensions $\Bbb{Q}(\zeta_3,\cos(2\pi / 7))/\Bbb{Q}(\sqrt{-3})/\Bbb{Q}$.