Find extension degree

Question

Let $\zeta =e^{\pi i/12}$.

• Find the extension degree of $\mathbb{Q}\leq \mathbb{Q}(\zeta)$
• Show that $\mathbb{Q}(\zeta)=\mathbb{Q}(\sqrt{2} , \sqrt{3} , i)$

$\zeta$ is a root of $x^{24}-1$

$Irr(\zeta, \mathbb{Q})=\Phi_{24}(x)$ which has degree $\phi(24)=8$

So the extension degree is $8$.

Is this ccorrect??

For the second question are we looking a primitive element??

Answer 1

The degree is equal to $8$ in both cases, i.e., $[\mathbb{Q}(\zeta):\mathbb{Q}]=8=[\mathbb{Q}(\sqrt{2},\sqrt{3},i):\mathbb{Q}]$; and one field is contained in the other, see here for details, i.e., the relation between cyclotomic fields and quadratic fields.