# A question about Galois extension

Let $n\geq 2$. Let $K$ be a field which contains $n$ distinct n-th roots of unity. Let $L=K(\sqrt[n]{a})$ with $a\in K$ and $[L:K]=n$. Show: there exists a Galois extension $K\hookrightarrow M$ such that $L\subset M$ and $Gal(M/K)\cong \mathbb{Z}/n^2\mathbb{Z}$ iff $\{x\in K \vert x^n=1\}\subset Norm_{L/K}(L^*).$

I think that the condition says intuitively that the roots of unity are also n-th powers. But so far I can not prove neither of the two implications. Thanks in advance for your help.

If such an extension $M$ exists, then the subgroup of $\Bbb Z/n^2 \Bbb Z$ corresponding to $L$ has to be $n(\Bbb Z/n \Bbb Z)$, so $Gal(M/L)$ is also cyclic of order $n$, and there exists $b \in L$ such that $M = L(\sqrt[n]b)$.
Let $\sigma$ be a generator of $Gal(M/K)$. Then $\sigma(\sqrt[n]b)$ is a $n$th root of $\sigma(b)$, and $\sigma^n = \tau\in Gal(M/L)$, so $\tau(\sqrt[n]b) = \zeta_n\sqrt[n]b$ for some primitive $n$th root of $1$ $\zeta_n$.
Then, $\tau(\sqrt[n]b / \sigma(\sqrt[n]b)) = \tau(\sqrt[n]b)/\sigma(\tau(\sqrt[n]b)) = \sqrt[n]b / \sigma(\sqrt[n]b)$, hence $\sqrt[n]b/\sigma(\sqrt[n]b) = c \in M^\tau = L$. Then $Norm_{L/K}(c) = \prod_{k=0}^{n-1} \sigma^k(\sqrt[n]b)/\sigma^{k+1}(\sqrt[n]b) = \sqrt[n]b/\tau(\sqrt[n]b) = \zeta_n^{-1}$. Therefore a necessary condition is that $\zeta_n$ (and the whole group of $n$th roots of unity) is the norm of an element of $L^*$.
Conversely, suppose $\sigma$ is a generator of $Gal(L/K)$ and that $\zeta_n^{-1} = Norm_{L/K}(c)$ for some $c \in L$. Then $c^n$ is of norm $1$, and by Hilbert's theorem 90, we can find some element $b \in L$ such that $b/\sigma(b) = c^n$. Then picking $M = L(\sqrt[n]b)$, we find that $K \subset M$ is Galois with Galois group generated by $\sigma_M$ induced by $(\sqrt[n]b,c) \mapsto (\sqrt[n]b/c, \sigma(c))$, of order $n^2$.
In particular, if $\zeta_n$ is already an $n$th root in $K$, we can always pick $c = \zeta_{n^2}^{-1} \in K$, $b = \sqrt[n]a$ for some $a \in K$, and $\sigma_M$ becomes the automorphism induced by $\sqrt[n^2]a \mapsto \zeta_{n^2} \sqrt[n^2]a$.