If $d$ divides $n$ and $\alpha$ is a root of $X^n-2$, then there is only one subfield of $\mathbb{Q}(\alpha)$ of index $d$ over $\mathbb{Q}$. I'm studying for quals which are in a week, so I'm trying to quickly go through a ton of problems which in happier circumstances I would gladly dedicate the time to figure out myself.  Let $\alpha$ be a root of $X^n - 2 \in \mathbb{Q}[X]$, and $d$ a divisor of $n$.  Then there is exactly one subfield $F \subseteq \mathbb{Q}(\alpha)$ for which $[F : \mathbb{Q}] = d$.
It's obvious the subfield in question should be $\mathbb{Q}(\alpha^{\frac{n}{d}})$, but why is it the only such subfield?  Can someone please give me a (not too big) hint?  
 A: Let $L = \mathbb{Q}(\alpha)$, and $F = \mathbb{Q}(\alpha^{\frac{n}{d}})$.  It is not difficult to see that $F$ has degree $d$ over $\mathbb{Q}$, and we want to show that it is the only subfield of $L/\mathbb{Q}$ with this property.  
We will assume that $\alpha$ is real.  If $E$ is another subfield of $L$ which has degree $d$ over $\mathbb{Q}$, let's look at $N_{L/E}(\alpha)$.  If $\zeta$ is a primitive $n$th root of unity, then $$N_{L/\mathbb{Q}}(\alpha) = \prod\limits_{i=0}^{n-1} \zeta^{i}\alpha$$ To compute $N_{L/E}(\alpha)$, we will only be multiplying $n/d$ components of the above product together, so $$N_{L/E}(\alpha) = \zeta^k \alpha^{\frac{n}{d}}$$  for some integer $k$.  So $\zeta^k \alpha^{\frac{n}{d}} \in E$, with $E \subseteq \mathbb{R}$, so $\alpha^{\frac{n}{d}} \in E$.  Therefore $E \subseteq F$, hence $E = F$ since these fields have the same degree over $\mathbb{Q}$.
The case where $\alpha$ is not real is the same as the case where $L = \mathbb{Q}(\alpha \zeta^k)$ for some $k$.  But this field is isomorphic to $\mathbb{Q}(\alpha)$, so the layout of its subfields is the same as the case we just proved.  
