TITLE: Show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$
QUESTION [11 upvotes]: I want to show that $\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})$.
I think it would be easier to prove it using the following:
$\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{3})$. Then $\sqrt{2}\notin\mathbb{Q}(\sqrt{3})$ so $\sqrt{2}\in \mathbb{Q}(\sqrt[4]{3})\Longleftrightarrow \sqrt{2}=a+b\sqrt[4]{3}$, with $a,b\in\mathbb{Q}(\sqrt{3})$. 
I tried simplifying the above equation, but I did not get anything.

REPLY [2 votes]: It is enough to show $\sqrt[4]{3} \not \in \mathbb{Q}(\sqrt{2})$. Why?

$\sqrt[4]{3} = a+b\sqrt{2} \Rightarrow \sqrt{3} = a^2+2\sqrt{2}ab+2b^2 \Rightarrow \frac{\sqrt{3}}{2ab}-\frac{a^2+2b^2}{2ab}=\sqrt{2} \Rightarrow \sqrt{2} \in \mathbb{Q}(\sqrt{3})$. 
If you haven't proven this already, observe that if $\sqrt{2}=c+d\sqrt{3}$ then we have the following,
$2=(c^2+3d^2)+(2cd\sqrt{3}) \Rightarrow c=d=0$ since if not we would have $\sqrt{3} \in \mathbb{Q}$.