Monday 25 August 2014

abstract algebra - Degree of Field Extension $mathbb{Q}(sqrt[4]{2}):mathbb{Q}(sqrt{2})$



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).


Answer



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


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