I have an extension $\mathbb{Q}(5^{1/4}, i)$, and I want to show that $4^{1/4}$ is not contained in it.
(I hope what I am trying to prove is true!)
Anyways, my natural starting point is to assume for a contradiction that there exist polynomials $p(x,y)$ and $q(x,y)$ both over $\mathbb{Q}$ such that $4^{1/4} = \frac{p(5^{1/4},i)}{q(5^{1/4},i)}$.
But I have no clue what to do from here, yet I suspect there is a standard technique for showing things like this? (assuming it's true)
Edit:
I have an extension, which I want to show is not normal, so I want to show that it contains $\mathbb{Q}(5^{1/4}, i)$, which is not normal over $\mathbb{Q}$, by the following argument:
First observe that $\mathbb{Q}(5^{1/4}, i)$ contains a root of the irreducible polynomial $p(x) = x^{4} + 20$ (namely $5^{1/4} + 5^{1/4}i$) $over $\mathbb{Q}.
Another root of this polynomial is \sqrt{2}(5)^{1/4}i$, which is not in $\mathbb{Q}(5^{1/4}, i)$, since $\sqrt{2}\notin\mathbb{Q}(5^{1/4}, i).
Therefore the extension is not normal over \mathbb{Q}$.
Is my reasoning correct?