I am trying to show that $i \notin \mathbb{Q}(5^{\frac{1}{4}}i)$. Here is my attempt:
The minimal polynomial of $5^{\frac{1}{4}}i$ over $\mathbb{Q}$ is $x^4-5$. Considering the evaluation homomorphism $\text{ev}_{\alpha}$ of $\alpha = 5^{\frac{1}{4}}i$ from $\mathbb{Q}[x]$ to $\mathbb{C}$, we can deduce that
$$\frac{\mathbb{Q}[x]}{(x^4-5)} \cong \text{Im } \text{ev}_{\alpha} = \mathbb{Q}(5^{\frac{1}{4}}i)$$.
We thus know that $\mathbb{Q}(5^{\frac{1}{4}}i) = \{k_n\alpha^n+k_{n-1}\alpha^{n-1}+\cdots+k_1\alpha^1+k_0\vert n\in\mathbb{N}, k_i\in\mathbb{Q}\}$. However, from here onwards I struggled to show that $i$ cannot be expressed as a polynomial in $\alpha$.
My second attempt was to show that $\mathbb{Q}(5^{\frac{1}{4}}i)=\{a+b\alpha \vert a,b \in\mathbb{Q}\}$ and from this deduce that $i$ is not in $\mathbb{Q}(5^{\frac{1}{4}}i)$. However, again, I struggled to show the set equality, and even if I assumed the equality it still wasn't clear how to show that we can't express $i$ as $a+b\alpha$ for $a,b \neq 0$.
So I'm wondering if my approaches are correct or whether there are simpler/alternative ways to tackle this. Thanks.