Describe the elements of the extension $\mathbb{Q}(2^{1/4})$ over the field $\mathbb{Q}(\sqrt{2})$ 
Describe the elements of the extension $\mathbb{Q}(2^{1/4})$ over the field $\mathbb{Q}(\sqrt{2})$.

I am not sure where to start.
 A: This proof will work very similarly to describing the elements of $\mathbb{Q}\sqrt{2}$:
When looking at $\mathbb{Q}({\sqrt{2}})$ over $\mathbb{Q}$, we see that the minimal polynomial is $x^2-2$, and we can describe the elements of the extension field with basis $\{\sqrt{2},1\}$ (more generally, for some field F(a) $\approx F[x]/\langle p(x)\rangle$, the elements are described via basis $\{1, a, a^2,...,a^{n-1}\}$, where $p(x)$ has degree $n$.
Now, looking at $\mathbb{Q}(2^{1/4})$ over $\mathbb{Q}(\sqrt{2})$, we see that $x^2 - \sqrt{2}$ will be the minimal polynomial- we have $\sqrt{2}$ in our base field, and $2^{1/4}$ is a zero of said polynomial. So we need a basis with respect to $2^{1/4}$, and it should have dimension 2.We know that $\mathbb{Q}(2^{1/4})$ will have basis of the form above, where $a = 2^{1/4}$, so the basis will be $\{2^{1/4},1\}$, and so every element can be described by $c_1(2^{1/4})+c_2,$ where $c_1, c_2$ are in $\mathbb{Q}(\sqrt{2})$.
We see this will also give us $2^{3/4}$, if $c_1 = \sqrt{2}$.
A: If $k \subseteq \Omega$ are fields, and $x^2 \in k$ for some $x \in \Omega$, not in $k$, then $k[x]$ is a degree two field extension of $k$.  Hence the linearly independent set $1,x$ is a basis for $k[x]$ over $k$.
You can apply this principle with $k = \mathbf Q(\sqrt{2}), \Omega = \mathbf C$ and $x = \sqrt[4]{2}$.
