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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.

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    I'm curious why you would use $(-1)^{2/5}$. It doesn't seem appropriate or efficient. I mean that is just saying $(1)^{1/5}$, and here we run into the issue of what do you mean by that? A priori, that could just mean $1$. You really ought to use a letter like $\omega$ and say that it is a primitive fifth root of unity. Or use $e^{\frac{2\pi i}{5}}$. – 2017-02-21
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    @Ken Duna edited – 2017-02-21
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    is in it just by the definition of $\mathbb{Q}(3^{1/5}, \omega)$? – 2017-02-21

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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

$$ [\mathbf Q(3^{1/5}, 7^{1/5}, \zeta_5) : \mathbf Q] = 100 $$

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    how do you know the extension $\mathbb{Q}(\zeta_5)$ over $\mathbb{Q}$ is of degree 4? – 2017-02-21
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    @usere5225321 It's a standard result that the degree of $ \mathbf Q(\zeta_n) $ over $ \mathbf Q $ is $ \varphi(n) $. In this case, you may note that the minimal polynomial of $ \zeta_5 $ is $ X^4 + X^3 + X^2 + X + 1 $, which becomes Eisenstein at $ 5 $ when one substitutes $ X \to X+1 $. – 2017-02-21
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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 $\mathbb{Q}$.

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    $ \beta^5 \notin \mathbf Q(\alpha) $ is false - $ 7 $ is most certainly in $ \mathbf Q(\alpha) $... – 2017-02-21
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    @Starfall, You are absolutely correct. I meant $\beta\not\in\mathbb{Q}(\alpha)$. Don't know why I typed $\beta^5$. – 2017-02-22