Existence of a unique subfield of degree dividing the degree of a given irreducible polynomial Let $n$ be a positive integer and $d$ a positive integer that divides $n$. Let $\alpha \in \mathbb{R}$ be the root of the polynomial $x^{n} - 2 \in \mathbb{Q}[x]$. Prove that there is precisely one subfield $F$ of $\mathbb{Q}(\alpha)$ with $[F: \mathbb{Q}] = d$. 
Let $G = \mathrm{Gal}(K/\mathbb{Q})$, where $K = \mathbb{Q} \left (\alpha, e^{\frac{2\pi i}{n}} \right)$. There exists a subgroup $H$ by the duality of the Fundamental Theorem of Galois theory such that $K^{H} = \mathbb{Q}(\alpha)$. Choose an automorphism $\sigma$ that fixes $\mathbb{Q}(e^{\frac{2\pi i}{n}})$, but moves $\alpha$ to $e^{\frac{2\pi i}{n}}\alpha$. Then $\langle \sigma \rangle$ is a normal subgroup of $G$. $\langle \sigma \rangle H = G$ since $K^{\langle \sigma \rangle H} = \mathbb{Q}$. It follows that if $x \in G$, then $x = \sigma^{i}h$. 
Let $F \subset \mathbb{Q}(\alpha)$ such that $[F: \mathbb{Q}] = d|n$. Let $T \leq G$ such that $K^{T} = F$. Since $K^{H} = \mathbb{Q}(\alpha) \supset F = K^{T}$, it follows that $H \leq T$. Since every element $x$ of $G$ is of the form $x = \sigma^{i}h$, we have $xT = \sigma^{i}hT = \sigma^{i}T$. Let $r$ be the smallest positive integer such that $\sigma^{r} \in T$. Then $\langle \sigma^{r}, H \rangle = T \implies F = \mathbb{Q}(\alpha^{m})$, where $m = \frac{n}{\mathrm{gcd}(r, n) }$. We also have $[\mathbb{Q}(\alpha^{m}): \mathbb{Q}] = \frac{n}{\mathrm{gcd}(m,n)}$. So that $m = \frac{n}{d}$ is determined by $d$, so $F$ is uniquely determined. 
Is there a field theoretic solution that doesn't use Galois' machinery? Is this a specific case of a general phenomena?
 A: First, take $F' = \mathbb{Q}(\alpha^{n/d})$, where $d$ divides $n$. As $X^n-2 = X^n-\alpha^n$ is irreducible over $\mathbb{Q}$, the minimal polynomial of $\alpha^{n/d}$ over $\mathbb{Q}$ is $X^d-2$. This shows that $[\mathbb{Q}(\alpha^{n/d}):\mathbb{Q}] = d$, and obviously $\mathbb{Q}(\alpha^{n/d}) \subset \mathbb{Q}(\alpha)$.
Conversely, let $F$ be a subfield of $\mathbb{Q}(\alpha)$ such that $[F:\mathbb{Q}]=d$.
Note that $F(\alpha) = \mathbb{Q}(\alpha)$ : indeed $\mathbb{Q}(\alpha)$ does contain $F$ and $\alpha$, and is the smallest such field (as $\mathbb{Q} \subset F$). Thus we have $[F(\alpha):F] = [\mathbb{Q}(\alpha):F] = [\mathbb{Q}(\alpha):\mathbb{Q}] / [F:\mathbb{Q}] = n/d$. Now consider $\Pi_{F,\alpha}$ the minimal polynomial of $\alpha$ over $F$ ; we just showed its degree is $n/d$, and it divides $\Pi_{\mathbb{Q},\alpha} = X^n-\alpha^n$. Its constant coefficient $\Pi_{F,\alpha}(0)$ is then a product of $n/d$ roots of $X^n-\alpha^n$, and is real (as $F \subset \mathbb{Q}(\alpha) \subset \mathbb{R}$), so $\Pi_{F,\alpha}(0) = \pm \alpha^{n/d}$.
Therefore, $\alpha^{n/d} \in F$, so $F'=\mathbb{Q}(\alpha^{n/d}) \subset F$ ; comparing the degrees, it yields $F = F'$.
