Find $[\mathbb{Q}(14^{1/8}) : \mathbb{Q}(14^{1/2})]$ The question asks the following: 

Let $F =  \mathbb{Q}(14^{1/2})$ and $E = \mathbb{Q}(14^{1/8})$, and consider field extensions $\mathbb{Q} \subseteq F \subseteq E$. Find $[E: F]$ and $\text{deg}(\min_F(14^{1/8}))$. 

My difficulty with the problem is that even though I know $(14^{1/8})^4 - \sqrt{14} = 0$, I cannot think of an easy way to prove that is indeed the minimal polynomial over $F$ by proving that it is irreducible. 
So instead I argued the following: $a$ is clearly algebraic over $F$, so $F[14^{1/8}] = F(14^{1/8})$. Since 
$$F(14^{1/8}, 14^{1/2}) = F(14^{1/8}) = E,$$
we have
$$[E:F] = [F[14^{1/8}]: F] = \text{deg}({\min}_F(14^{1/8})).$$ 
It's easy to see that $\{1, 14^{1/8}, 14^{1/4}, 14^{1/2}\}$ forms a basis of $F[14^{1/8}]$ over $F$. So 
$$[F[14^{1/8}]: F] = \text{deg}({\min}_F(14^{1/8})) = 4.$$
Is this a correct alternative to proving $(14^{1/8})^4 - \sqrt{14}$ is minimal polynomial of $14^{1/8}$ over $14^{1/2}$?
 A: Why make things so complicated?
The polynomials $X^8-14$ and $X^2-14$ are both irreducible over $\mathbf{Q}[X]$ by Eisenstein for $p=7$.
Hence $[\mathbf{Q}(\sqrt[8]{14}):\mathbf{Q}]=8$ and $[\mathbf{Q}(\sqrt[2]{14}):\mathbf{Q}]=2$. By the tower law for field extensions,
$$[\mathbf{Q}(\sqrt[8]{14}):\mathbf{Q}(\sqrt[2]{14})]=\tfrac82=4.$$
This degree is by definition equal to the degree of the minimal polynomial of $\sqrt[8]{14}$ over $\mathbf{Q}(\sqrt[2]{14})$.
A: A routine check shows that that $X^4-\sqrt{14}\in F[X]$ is irreducible: It has no roots as
$$(a+b\sqrt{14})^4-\sqrt{14}=0,$$
implies that $a^4+28a^2b^2+196b^4=0$, and hence $a=b=0$, a contradiction. Expanding
$$X^4-\sqrt{14}=(X^2+rX+s)(X^2+uX+v),$$
yields the system of equations
$$r+u=0,\qquad ur+s+v=0,\qquad rv+su=0,\qquad sv=\sqrt{14},$$
with $r,s,u,v\in F$. The first equation shows that $u=-r$ and hence the third becomes
$$r(v-s)=0,$$
so either $r=0$ or $v=s$. If $r=0$ then the second equation becomes $s+v=0$, so either way we have $v=\pm s$. Then $sv=\sqrt{14}$ shows that $s^2=\pm\sqrt{14}$. Write $s=a+b\sqrt{14}$ to get
$$\pm\sqrt{14}=s^2=(a+b\sqrt{14})^2=a^2+14b^2+2ab\sqrt{14},$$
which shows that $a^2+14b^2=0$ and $2ab=\pm1$, so there are clearly no solutions. Hence the polynomial is irreducible.

This can of course be done a lot more easily; showing that $X^4-\sqrt{14}\in F[X]$ is irreducible means showing that $\sqrt{14}$ is not a fourth power in $F$. Then it suffices to show that $\sqrt{14}$ is not a square in $F$; as before
$$(a+b\sqrt{14})^2=\sqrt{14},$$
implies $a^2+14b^2=0$ and $2ab=1$, which has no rational solutions, so the polynomial is irreducible.

Even more quickly, by Gauss' lemma the polynomial is irreducible in $F[X]$ if and only if it is irreducible in $\mathcal{O}[X]$, where $\mathcal{O}$ denotes the ring of integers of $F$. In this ring the ideal $\sqrt{14}\mathcal{O}\subset\mathcal{O}$ is prime, so $X^4-\sqrt{14}$ is Eisenstein at $\sqrt{14}\mathcal{O}$ and hence it is irreducible.

Alternatively you could show that $[\Bbb{Q}(\sqrt[8]{14}):\Bbb{Q}]=8$ and that $[\Bbb{Q}(\sqrt{14}):\Bbb{Q}]=2$ so that
$$[\Bbb{Q}(\sqrt[8]{14}):\Bbb{Q}]=[\Bbb{Q}(\sqrt[8]{14}):\Bbb{Q}(\sqrt{14})]\cdot[\Bbb{Q}(\sqrt{14}):\Bbb{Q}],$$
shows that $[\Bbb{Q}(\sqrt[8]{14}):\Bbb{Q}(\sqrt{14})]=4$.
