Computing $[L:K]$ (minimal polynomial)

Question

Let $L=\mathbb{Q}(\sqrt[4]{3},i)$ and $K=\mathbb{Q}(i)$

Determine $[L:K]$.

My way was:

$[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(i)]\cdot [\mathbb{Q}(i):\mathbb{Q}]$

The minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}$ is $x^4-3$.

$\sqrt[4]{3} \notin \mathbb{Q}(i)$, so $x^4-3$ is also the minimal polynomial of $\sqrt[4]{3}$ over $\mathbb{Q}(i)$.

$\Rightarrow \mathrm{deg}(x^4-3)=[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(i)]=4$

Is this way right or is there something to improve?

Answer 1

Try instead $$ [\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}]= [\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})] \cdot [\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}] = 2 \cdot 4 $$ Here, $[\mathbb{Q}(\sqrt[4]{3},i):\mathbb{Q}(\sqrt[4]{3})]=[\mathbb{Q}(\sqrt[4]{3})(i):\mathbb{Q}(\sqrt[4]{3})]=2$ because $\mathbb{Q}(\sqrt[4]{3}) \subseteq \mathbb{R}$.

Answer 2

Your argument is not correct, as a degree 4 polynomial can have no zeroes and still fail to be irreducible. Thus $\sqrt[4] {3} \notin \Bbb Q(i) $ is not sufficient to conclude that $x^4-3$ is the minimal polynomial over $\Bbb Q(i) $