Is $F = \{a+b\sqrt[4]{2} \mid a,b\in\mathbb{Q}\}$ a subfield of $\mathbb{R}$, the field of real numbers?

Question

I'm having a hard time understanding if this will be a subfield of the real numbers.

Answer 1

If it is, then $\;\sqrt[4]2\cdot\sqrt[4]2=\sqrt2\in F\;$ ,right? But then there are rational $\;a,b\;$ such that

$$\sqrt2=a+b\sqrt[4]2\implies \sqrt2-a=b\sqrt[4]2\implies 2+a^2-2\sqrt2\,a=b^2\sqrt2...$$

Can you get a contradiction from the above?