I'm currently reading a book on Galois theory, and an example shows how $\mathbb{Q}(\zeta_{24})/\mathbb{Q}$ is a Galois extension with Galois group $C_2\times C_2\times C_2$.
In its proof, the author asserts that $\mathbb{Q}(\zeta_{24}) = \mathbb{Q}(i,\sqrt{2},\sqrt{3})$. I can follow the rest of the proof easily, but I'm unsure about how to write cyclotomic extensions this way (especially in general).
I took a geometric approach: since $24$ is a multiple of $4$, it is obvious that $i\in\mathbb{Q}(\zeta_{24})$. I also figured that since $24$ is a multiple of $8$ that $\mathbb{Q}(\zeta_{24})$ contains the regular octagon (in the complex plane), therefore containing $\frac{1}{2}\sqrt{2}$ and so $\sqrt{2}\in \mathbb{Q}(\zeta_{24})$. Since $6\mid 24$ the same applies for the regular hexagon, so that also $\sqrt{3}\in \mathbb{Q}(\zeta_{24})$.
So, as a matter of fact, this way I found out that at least $\mathbb{Q}(\zeta_{24}) \supseteq \mathbb{Q}(i,\sqrt{2},\sqrt{3})$.
The part at which I'm stuck: how can I be sure of the converse inclusion?
$\frac{\sqrt{2}}2(1+i)\frac12(-1+\sqrt{3}i) \in \mathbb{Q}(i,\sqrt{2},\sqrt{3})$
but this number is a $24$-th root of unity (since it is the product of a cube root and an eighth root)