Let $\omega = \cos{\frac{2\pi}{10}}+i\sin{\frac{2\pi}{10}}$. Let $K = \mathbb{Q}(\omega^2)$ and $L = \mathbb{Q}(\omega)$. Then
- $[L : \mathbb{Q}] = 10.$
- $[L : K] = 2$.
- $[K : \mathbb{Q}] = 4$.
- $L = K$.
$\omega^5=-1$. Hence $\omega$ satisfies $x^5+1=0=(x+1)(x^4-x^3+x^2-x+1)$. So the minimal polynomial of $\omega$ over $\mathbb{Q}$ is $(x^4-x^3+x^2-x+1)$. $[L:\mathbb{Q}]=4$
Now $(\omega^2)^5=1$.Hence $\omega$ satisfies $x^{5}-1=0\implies (x-1)(x^4+x^3+x^2+x+1)$. So the minimal polynomial of $\omega$ over $\mathbb{Q}$ is $(x^4+x^3+x^2+x+1)$. $[K:\mathbb{Q}] = 4$
Now $ [L : \mathbb{Q}] = [L : K] [K : \mathbb{Q}] \implies 4 = 4[L : K] \implies [L : K] = 1 \implies L = K. $
So only $3,4$ are correct. Can someone please verify my answer.