How do I find the degree of this field extension?

$$ \mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}(\sqrt{2}) $$

I've tried thinking of the larger field as a vector space over the smaller one to find a basis, but I haven't had any luck. There's no tower law to use here, and I don't think I can use a minimal polynomial argument here, because both fields are different (the larger one isn't the smaller one plus some algebraic element).


1 Answer 1


Hint: $\left[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}\right] =\left[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}(\sqrt{2})\right]\cdot \left[\mathbb{Q}(\sqrt{2}):\mathbb{Q}\right]$

  • 1
    $\begingroup$ Ok, so then $\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}$ is degree 4 -- because the minimal polynomial is $x^4 - 2$, and $\mathbb{Q}(\sqrt{2}):\mathbb{Q}$ is degree 2, which lets me conclude mine is degree 2. Thanks! $\endgroup$
    – Moz
    Oct 13, 2016 at 23:35

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