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.


First, I don't think this question deserves the downvotes, as you make a reasonable attempt to replace your problem with a simpler one to understand it, and I understand what you're asking.

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}) $.

  • $\begingroup$ Wow, that's a simple but very important observation. Thank you so much. $\endgroup$ – love_sodam Nov 20 '20 at 18:59

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