How many subfields are there between $\mathbb{Q}$ and $\mathbb{Q}[\sqrt[16]{2}]$ 
Let $\alpha = \sqrt[16]{2}$ be a positive real number and $K = \mathbb{Q}[\alpha]$ be the algebraic extension over $\mathbb{Q}$ by alpha. Find the number of intermediate field $F$ such that $\mathbb{Q} \subseteq F \subseteq K$.

Ok. I have an ugly solution. Suppose that $F$ is such an intermediate field. Consider the irreducible polynomial of $\alpha$ over $F$, say $f(x)$. Clearly $f(x) | x^{12}-2$. Noting that all the fields under consideration are contained in the real field, and $$x^{16}-2 = \left(x-\sqrt[16]{2}\right) \left(x+\sqrt[16]{2}\right) \left(-x^2+2^{9/16}
   x-\sqrt[8]{2}\right) \left(x^2+\sqrt[8]{2}\right) \left(x^2+2^{9/16} x+\sqrt[8]{2}\right)
   \left(-x^4+2^{5/8} x^2-\sqrt[4]{2}\right) \left(x^4+2^{5/8} x^2+\sqrt[4]{2}\right)$$ and observing that all the coefficients are of the form $\alpha^k$, I have concluded that
(*) $F$ should be of the form $\mathbb{Q}[\alpha^t]$, $t=0,1,2,4,8,16$.
My question is the following;

Is there an argument to say (*) without actual fatorization?

or it could be great if we one can prove the following

All the subfields of $\mathbb{Q}[2^{\frac{1}{2^n}}]$ are of the form $\mathbb{Q}[2^{\frac{1}{2^k}}]$, $0 \le k \le n$.

Thanks for your attention.
 A: Let $\alpha = 2^{1/12}$ and let $\zeta_{12}$ be a 12th root of unity.
$\mathbb Q(\zeta_{12}, \alpha)$ is the splitting field of $X^{12}-2$ and has Galois group $V_4 \rtimes C_{12}$. (Calculating this Galois group is rather difficult and involves more work than the factorization you have already done).
By the Galois correspondence, $\mathbb Q(\alpha)$ corresponds to the subgroup $C_{12}$, and its subfields correspond to subgroups of $C_{12}$.
Therefore we can deduce that the subfields are $\mathbb Q(\alpha), \mathbb Q(\alpha^2), \mathbb Q(\alpha^3), \mathbb Q(\alpha^4), \mathbb Q(\alpha^6), \mathbb Q(\alpha^{12}) = \mathbb Q$.

Let $K = \mathbb Q(\alpha)$. We will do one example explicitly. $C_4 \le C_{12}$. By the Galois correspondence the fixed field $|K : K^{C_4}| = |C_4| = 4$.
Let $\sigma \alpha = \zeta_{12} \alpha$ so that $\alpha$ generates $C_{12}$. Then $\sigma^3$ generates our $C_4$ subgroup.
Notice that $\sigma^3 \alpha^4 = (\zeta_{12}^3 \alpha)^4 = \alpha^4$. So $\alpha^4$ is fixed by $\sigma_3$.
Thus $\mathbb Q(\alpha^4)$ is our subfield of index 4 corresponding to $C_4$.
A: We will prove the following.

Let $n$ be a positive integer and $\alpha = \sqrt[n]{2}$. Then the number of subfields of the field $\mathbb{Q}[\alpha]$ is $\tau(n)$ and the subfields are precisely those $\mathbb{Q}\left[ \alpha^d \right]$ where $d|n$.

Let $K = \mathbb{Q}[\alpha]$ let $d$ be a positive divisor of $n$. Let $F = \mathbb{Q}\left[\alpha^d\right]$. We have $[F:\mathbb{Q}]=\frac{n}{d}$ because $x^{\frac{n}{d}}-2$ is irreducible by the Eisenstein criterion. It follows that $[K:F]=d$. Now suppose that $E$ is a subfield of $K$ with $[K:E] =d$. Note that $K=E[\alpha]$. Let $f(x)=irr_{\alpha,E}(x)$. Then $f(x)|x^2-2$ and its degree is $d$. Hence the zeros of $f(x)$ are of the form $\alpha \zeta$ where $\zeta$s are some root of unity. Multiplying all the roots of $f(x)$ we have $\alpha^d Z$ where $Z$ is some root of unity again. Note that all the fields under consideration are some subfields of the field of the real numbers. Therefore we have $Z= \pm 1$. This tells us that $\alpha^d \in E$. Hence we have that $F$ is a subfield of $E$ with $[E:F]=1$ which completes the proof.
