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).

$\endgroup$

1 Answer 1

6
$\begingroup$

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
  • 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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .