I had to find $a$ such that $\mathbb Q(2^{1/4}):\mathbb Q$ is equal to $\mathbb Q(\sqrt 2)(a)$ But i found that $a=2^{1/4}$, is that right?
For it i used that $\mathbb Q(\sqrt 2) =$ { $a+b\sqrt 2$} and then tried with $2^{1/4}$. I found { $a+b\sqrt 2+ c2^{1/4}+d\sqrt 2 + e\sqrt 2 + f \sqrt 2*2^{1/4} + g \sqrt 2* 2^{1/4} + 2h$ }
On
In the proof for multiplicativity or degrees in towers, we find that when $F/E$ has basis $\{\alpha_\lambda\}$, $E/K$ has basis $\{\beta_\mu\}$, then $F/K$ has basis $\{\alpha_\lambda\beta_\mu\}$.
But in this case n this case, $\Bbb Q(\sqrt{2})(\sqrt[4]{2}) = \Bbb Q(\sqrt{2})$. Which has basis over $\Bbb Q$ all linear combinations of the powers of $\sqrt{2}$ with coefficients in $\Bbb Q$, from the $0$ power up to one less than the degree of the minimal polynomial for $\sqrt{2}$ over $\Bbb Q$, which is 2. So it as basis $\{1, \sqrt{2}\}$ over $\Bbb Q$.
The extension $\Bbb Q(\sqrt 2)(2^{1/4})$ is $$ \{a + b2^{1/4}\mid a, b\in \Bbb Q(\sqrt 2)\} $$ Rewriting that in terms of $\Bbb Q$ we get $$ \{(a_1+a_2\sqrt 2) + (b_1 + b_2\sqrt 2)2^{1/4}\mid a_i, b_i\in \Bbb Q\}\\ \{a_1+a_2\sqrt 2 + b_12^{1/4} + b_22^{1/4}\sqrt 2\mid a_i, b_i\in \Bbb Q\} $$ In general, if $\Bbb Q\subseteq \Bbb Q(x)\subseteq \Bbb Q(y)$ are field extensions, then we always have $\Bbb Q(x)(y) = \Bbb Q(y)$ (and there is nothing special about $\Bbb Q$ here, or the fact that the extensions have single generators).