A solution for $|\mathbb{Q}(\sqrt[m]{p}, \sqrt[n]{q}):\mathbb{Q}|$; field extension degree generalizaation I first read about the following problem. 

Prove: $|\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}) : \mathbb{Q}| =8$

So I could prove this statement by showing $\sqrt[4]{2} \notin \mathbb{Q}(\sqrt{3})$ using this trick. I am wondering if there is a more generalized proof - if the problem were to solve for

$|\mathbb{Q}(\sqrt[m]{p}, \sqrt[n]{q}):\mathbb{Q}|$ (Say with $p,q$ prime). 

How does one go about?
 A: When $p\neq q$, the extension has degree $nm$ as you might expect. When $p=q$, the extension has degree $\mathsf{lcm}(n,m)$. To see this, use Karpilovsky's theorem (see this answer for more details)
A: For the case $p\ne q$ the order is $mn$ and we prove it here. For primes $p,q$ and positive integers $m,n$ we take the following basis for the extended field: 
$$\Large v_{ij}={p^{i\over m}}{q^{j\over n}}\qquad i\in\lbrace1,...,m\rbrace,j\in\lbrace1,...,n\rbrace$$
so we can represent $x\in\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}) : \mathbb{Q}$ as a linear combination of $v_{ij}$'s as below:
$$\Large x=\Sigma_{i,j}{a_{ij}}{v_{ij}}$$
which can also be written as following:
$$\Large x=\Sigma_{i}\Sigma_{j}{a_{ij}}{v_{ij}}=\Sigma_{i}{p^{i\over m}}\Sigma_{j}{a_{ij}}{q^{j\over n}}$$
Now to prove $v_{ij}$'s are really basis must show that:
$$\Large x=0\to a_{ij}=0\qquad\forall i,j$$
Before that we try to prove a lemma. First define $b_i=\Sigma_{j}{a_{ij}}{q^{j\over n}}$.
Lemma: $\large\Sigma_{i}{b_i}{p^{i\over m}}=0$ yields to $\large b_i=0\qquad\forall j$.
Proof of Lemma: Let $v={p^{1\over m}}$. Then we have $\large\Sigma_{i=1}^{m}{b_i}{v^i}=0$ where $b_i$'s are quotient and $\large{{v^j}\over{v^k}}\notin\Bbb Q\qquad \forall j,k\in\lbrace1,...,n\rbrace$.
Multiplying $v$ to both side of the equation leads us to:
$$\large\Sigma_{i}{b_i}{v^{i+1}}=0$$
from $\large v^n=q$ by substitution and two equations we arrived at we can finally obtain:
$$(b_2b_np-b_1^2)v^2+(b_3b_np-b_1b_2)v^3+...+(b_n^2p-b_1b_{n-1})v^n=0$$ or $$(b_2b_np-b_1^2)v+(b_3b_np-b_1b_2)v^2+...+(b_n^2p-b_1b_{n-1})v^{n-1}=0$$
where new coefficients are quotient either and the order of equation has reduced to $n-1$. By iterating this process we can show that all of those coefficients are zero $\bullet$
Since we have shown $\large b_i=\Sigma_{j}{a_{ij}}{q^{j\over n}}=0\qquad \forall i$ then we can apply this lemma again on ${a_{ij}}$'s for any arbitrary $i\in\lbrace1,...,m\rbrace$ and deduce $a_{ij}=0\qquad \forall i,j$
This is what we wanted show so we prove that the basis is linearly independent and the order of extended field is then $\large mn$.
If $p=q$ let's define $m=m^{'}d$ and $n=n^{'}d$ where $d=g.c.d.(m,n)$ and $g.c.d.(m^{'},n^{'})=1$. Therefore we have:
$$\Large v_{ij}={p^{i\over {m^{'}d}}}{p^{j\over {n^{'}d}}}={p^{{in^{'}+jm^{'}}\over{m^{'}n^{'}d}}}\qquad i\in\lbrace1,...,m\rbrace,j\in\lbrace1,...,n\rbrace$$
In this case we easily see that the order is $\large {{mn}\over{g.c.d.(m,n)}}$.
