splitting field of $(x^5-3)(x^5-7)$ and its degree over $\mathbb{Q}$ Im am trying to find the splitting field of $(x^5-3)(x^5-7)$ and its degree over $\mathbb{Q}$.
What I have so far:
Let $\omega$ be a primitive fifth root of unity
The splitting field is $\mathbb{Q}(3^{1/5},\omega,7^{1/5})$.
By the tower lawm we have :
$[\mathbb{Q}(3^{1/5},(-1)^{2/5},7^{1/5}):\mathbb{Q}]=[\mathbb{Q}(3^{1/5},\omega,7^{1/5}):\mathbb{Q}(3^{1/5},\omega)][\mathbb{Q}(3^{1/5},\omega): \mathbb{Q}(3^{1/5})][\mathbb{Q}(3^{1/5}):\mathbb{Q}]$ 
But, clearly, 
$[\mathbb{Q}(3^{1/5}):\mathbb{Q}]$=5 .
I want to find $A=[\mathbb{Q}(3^{1/5},\omega,7^{1/5}):\mathbb{Q}(3^{1/5},\omega)]$ and $B=[\mathbb{Q}(3^{1/5},\omega): \mathbb{Q}(3^{1/5})]$
We have that $7^{1/5} \not \in \mathbb{Q}(3^{1/5},\omega)$. Since $x^5-7$ is irreducible on $\mathbb{Q}$ by eisenstein, and $\mathbb{Q}$ is in $\mathbb{Q}(3^{1/5},\omega)$,then I thought maybe this meant A|5, i.e. A=5 since  $7^{1/5} \not \in \mathbb{Q}(3^{1/5},\omega)$.
Also, how do I calculate B?
I cannot use Galois groups since I have not seen them in class.
 A: We first recall the result that given any field $ F $ and any prime $ p $, $ X^p - a $ is irreducible in $ F[X] $ if and only if $ a $ is not a $ p $th power in $ F $.
It's clear that $ [\mathbf Q(3^{1/5})/\mathbf Q] = 5 $. If $ X^5 - 7 $ were reducible over $ \mathbf Q(3^{1/5}) $, we would have $ \mathbf Q(3^{1/5}) = \mathbf Q(7^{1/5}) $ by degree considerations, but this is impossible; as there is an embedding $ \mathbf Q(7^{1/5}) \to \mathbf Q_3 $ but no embedding $ \mathbf Q(3^{1/5}) \to \mathbf Q_3 $. It follows that
$$ [\mathbf Q(3^{1/5}, 7^{1/5}):\mathbf Q] = 25 $$
Finally, since the extensions $ \mathbf Q(\zeta_5)/\mathbf Q $ and $ \mathbf Q(3^{1/5}, 7^{1/5})/\mathbf Q $ have coprime degrees, they are linearly disjoint. It follows that
$$ [\mathbf Q(3^{1/5}, 7^{1/5}, \zeta_5) : \mathbf Q] = 100 $$
A: This is what I would do.  Let $\alpha:=\sqrt[5]{3}$, $\beta:=\sqrt[5]{7}$, and $\omega:=\exp\left(\frac{2\pi\text{i}}{5}\right)$.  I would look at $\mathbb{Q}(\alpha,\beta)$.  Since $\beta^5\notin\mathbb{Q}(\alpha)$, it follows that $\big[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)\big]=5$.  Now, note that $\mathbb{Q}(\omega)$ is an extension of degree $4$ over $\mathbb{Q}$.  So, $4$ must divide $\big[\mathbb{Q}(\alpha,\beta,\omega):\mathbb{Q}\big]$.  This means $\mathbb{Q}(\alpha,\beta,\omega)$ is an extension of degree $4\times 5\times 5=100$ over $\mathbb{Q}$.
