I am trying to determine the degree of the field extension $\mathbb{Q}\subset\mathbb{Q}(\sqrt[4]{3},i)$. For this I use the following:
$[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})][\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]$.
First I determine: $[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}] = 4$ (I think I know the reason)
Then I try to determine: $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})]$
$i \in \mathbb{Q}(\sqrt[4]{3},i)$ is the root of $x^2+1 \in\mathbb{Q}(\sqrt[4]{3})[X]$
And $x^2+1$ is the minimal polynomial, because $i\notin\mathbb{Q}(\sqrt[4]{3})$.
=> $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})] \leq 2$
So I claim: $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})] = 1$
It follows $\mathbb{Q}(\sqrt[4]{3},i)=\mathbb{Q}(\sqrt[4]{3})$, coz $i \in\mathbb{C}\setminus\mathbb{R}$ and $\mathbb{Q}(\sqrt[4]{3})\subset\mathbb{R}$
=> $i \notin \mathbb{Q}(\sqrt[4]{3})$
=> $\mathbb{Q}(\sqrt[4]{3},i) \neq \mathbb{Q}(\sqrt[4]{3})$
=> $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})] \neq 1$
Is this correct? How to argue $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})] > 2$?
Thanks in advance