About primes and cyclotomic extensions I have the following problem

Let $p\geq3$ a prime. Show that $\mathbb{Q}(\sqrt[p]{p})$ is not contained in any cyclotomic extension. 

I don't know how to start the problem. Any hint or help will be appreciated !
Thanks in advance.
 A: We use a lemma:

Let $p$ be a prime number, $k\in \mathbb{Q}$, if $x^p-k$ has no rational root, then $x^p-k$ is irreducible over $\mathbb{Q}[x]$

Assume $\sqrt[p]{k} \in \mathbb{Q}(\zeta_n)$. Since the extension $\mathbb{Q}(\zeta_n)$ is normal over $\mathbb{Q}$ and $\sqrt[p]{k}$ is a root of the polynomial $x^p-k$, the lemma says the polynomial $x^p-k$ splits completely in $\mathbb{Q}(\zeta_n)$, hence $\zeta_p \in \mathbb{Q}(\zeta_n)$. Consider the chain of extensions:
$$
F:=\mathbb{Q}\quad \subset \quad L:=\mathbb{Q}(\zeta_p, \sqrt[p]{k}) \quad \subset \quad K:=\mathbb{Q}(\zeta_n)$$
Both extensions $K/F$ and $L/F$ are Galois, the Galois group for $K/F$ is abelian of order $\varphi(n)$, while the Galois group for $L/F$ has order $p(p-1)$, it is a group which is not abelian when $p\geq 3$, (more specifically, it is the general affine group over $\mathbb{F}_p$), a contradiction, hence $\sqrt[p]{k} \notin \mathbb{Q}(\zeta_n)$.
A: Note for $n\geq 3,\, \Bbb{Q}[\zeta_n]$ is complex and for any $n$ is normal with abelian galois group.  Suppose $\sqrt[p]{p} \in \Bbb{Q}[\zeta_n]$.  Since $\sqrt[p]{p}$ is real, it is contained in the fixed field of complex conjugation, call it $K$.  As $Gal(\Bbb{Q}[\zeta_n])$ is abelian, $K$ is galois hence must be normal.  But if $p\geq 3$, $K$ doesn't contain the roots of $x^p-p$ conjugate to $\sqrt[p]{p}$, namely $\zeta_p\sqrt[p]{p},\, \zeta_p^2\sqrt[p]{p},\dots$ since the roots are complex, so it can't be normal.  Hence $\sqrt[p]{p} \not \in \Bbb{Q}[\zeta_n]$ for any $p\geq 3$
A: *

*$Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \sim \mathbb{Z}_n^\times$ is an abelian (and Galois) extension. Thus for any field $F \subseteq \mathbb{Q}(\zeta_n)$, $Gal(F/\mathbb{Q})$ is a subgroup of $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ and $  F/\mathbb{Q}$ is an abelian extension.

*Let $K = \mathbb{Q}(\sqrt[p]{p},\zeta_p)$.
$[\mathbb{Q}(\zeta_p):\mathbb{Q}]= p-1$ and $[\mathbb{Q}(\sqrt[p]{p}):\mathbb{Q}]= p$ so
$[K:\mathbb{Q}] = p(p-1)$ and its Galois group $Gal(K/\mathbb{Q})$ has elements of the form $$\sigma_{a,b}(\zeta_p^l \sqrt[p]{p}) = \sigma_{a,b}(\zeta_p^l)\sigma_{a,b}( \sqrt[p]{p})=\zeta_p^{al}\zeta_p^b \sqrt[p]{p}, \qquad a \in (\mathbb{Z}/p\mathbb{Z})^\times,b \in \mathbb{Z}/p\mathbb{Z}$$
and hence for $p \ge 3$ :
$$\sigma_{2,1}( \sigma_{1,2}(\zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+4+1} \sqrt[p]{p} \ne \sigma_{1,2}(\sigma_{2,1}( \zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+3} \sqrt[p]{p}$$
Therefore $Gal(K/\mathbb{Q})$ is not an abelian group so neither $K$ nor $\mathbb{Q}(\sqrt[p]{p})$ is contained in $\mathbb{Q}(\zeta_n)$.
