0
$\begingroup$

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).

$\endgroup$
5
$\begingroup$

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

$\endgroup$
  • 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 '16 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.