# $[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$

My original question is

$$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}):\Bbb Q]=p_1p_2p_3$$ ? where $$p_1,p_2,p_3,q_1,q_2,q_3$$ are all distinct prime numbers.

But I would rather change the question to $$[\Bbb Q(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}):\Bbb Q(\sqrt[p_1]{q_1})] = p_2$$ i.e. $$x^{p_2}-q_2$$ is irreducible over $$\Bbb Q(\sqrt[p_1]{q_1})$$. Any hints or ideas?

Edit: I don't know whether the above statement is true or not. But I guess it's true but I have no idea how to prove it.

Here is a hint to help you think about your simpler problem, and I believe it will also get you to an answer to your original question. Consider the field lattice diagram consisting of $$F = \mathbb{Q}$$, the two extensions $$E_1 = \mathbb{Q}(\sqrt[p_1]{q_1})$$ and $$E_2 = \mathbb{Q}(\sqrt[p_2]{q_2}$$), and the full field $$K = \mathbb{Q}(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$$. Then (assuming they are non-trivial) these extensions $$E_1/F$$ and $$E_2/F$$ have index $$p_1$$ and $$p_2$$. Now here is the punchline: $$[K : F] = [K : E_1][E_1 : F] = [K:E_2][E_2:F]$$ i.e. $$[K:F]$$ is divisible by both $$p_1$$ and $$p_2$$. This means $$[K : F]$$ is divisible by $$p_1 p_2$$ (!). I believe you can take it from here to conclude that $$K = \mathbb{Q}(\sqrt[p_1]{q_1})(\sqrt[p_2]{q_2})$$ is a degree-$$p_2$$ extension of $$E_1 = \mathbb{Q}(\sqrt[p_1]{q_1})$$.