Intersection of a finite Galois extension with simple extensions $f$ is an irreducible polynomial in $\mathbb Q[x]$, $Q \subset E$ is a finite Glaois extension. $\alpha, \beta$ are roots of $f$. Show that if $E$ is the $n$th cyclotomic field then $\mathbb Q(\alpha) \cap E = \mathbb Q (\beta)\cap E$.
First of all, note that $\alpha \in E$ iff $\beta \in E$ because Galois extension are normal. Then we can assume $\alpha$ and $\beta$ are both in $E$. If I can show that given $\sigma \in Aut_{\mathbb Q(\beta)}(E)$, $\sigma$ must fix $\alpha$, then I can show that they are equal by Galois correspondence, but that is not obvious. 
 A: $E/\Bbb{Q}$  is abelian (*) thus $(\Bbb{Q}(a)\cap E)/\Bbb{Q}$ is Galois. Let $\sigma \in Aut(\overline{\Bbb{Q}}/\Bbb{Q})$ such that $\sigma(a)=b $.
$E/\Bbb{Q}$ is Galois thus $\sigma(E) = E$. And $\sigma(\Bbb{Q}(a)) = \sigma(\Bbb{Q})(\sigma(a)) = \Bbb{Q}(b)$. Thus $\sigma(\Bbb{Q}(a)\cap E) = \sigma(\Bbb{Q}(a)) \cap \sigma(E) = \Bbb{Q}(b) \cap E$.
On the other hand $(\Bbb{Q}(a)\cap E)/\Bbb{Q}$ is Galois thus $\sigma(\Bbb{Q}(a)\cap E) =\Bbb{Q}(a)\cap E$.

(*) $E = \Bbb{Q}(\zeta_n)$ where $\zeta_n = e^{2i \pi /n}$. As $\zeta_n$ is a root of $X^n-1$ and $E$ contains all the roots of $X^n-1$ then $E/ \Bbb{Q}$ is Galois.
Any $\rho \in Aut(E/\Bbb{Q})$ must send $\zeta_n$ to another root of $X^n-1$ thus to some $\zeta_n^{m_\rho}$ (with $gcd(n,m_\rho)=1$) so $\rho$ sends $\zeta_n^a$ to $\zeta_n^{am_\rho}$ and $\rho \mapsto  m_\rho$ is an embedding $Aut(E/\Bbb{Q} ) \to \Bbb{Z/n Z}^\times$. Whence $Aut(E/\Bbb{Q} )$ is abelian. 
For any subfield $K \subset E$ with $H = Aut(E/K)$ we'll have that $K = E^H$ (the subfield fixed by $H$) and $\rho(K) = E^{\rho H \rho^{-1}} = E^H = K$ implies $K/\Bbb{Q}$ is Galois (with $Aut(K/\Bbb{Q}) = Aut(E/\Bbb{Q})/H$ abelian).
