Prove that $\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\omega)$

Question

Let $\omega=e^{\frac{2\pi i}{3}}$ be the third root of unity. Define $f=X^6+3X^3+3\in\mathbb{Q}$. Let $L$ be the splitting field of $f$ over $\mathbb{Q}$. Let $\alpha\in L$ be a zero of $f$.

I then need to prove that $\alpha^3\in\mathbb{Q}(\omega)$ and that $\omega\in\mathbb{Q}(\alpha).$

I found that $f$ is irreducible and that the zeros of $f$ are $\omega^l\alpha$ and $\omega^l\sqrt[3]{3}/\alpha$ for $l=0,1,2$.

I then observed that $\alpha^3$ is a zero of $P(Y)=Y^2+3Y+3$. The zeros of P(Y) are $-3/2\pm\sqrt{-3}/2$. Therefore $\alpha^3\in\mathbb{Q}(\sqrt{-3})$.

My book then says that that means that $\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\omega)$. Why?

And why does this mean that $\omega\in\mathbb{Q}(\alpha)$?

Answer 1

We have, because of $1+\omega+\omega^2=0$, that $$ \omega=\frac{-1-\sqrt{-3}}{2}, $$ Hence $\Bbb Q(\omega)=\Bbb Q(\sqrt{-3})$, because $\omega\in \Bbb Q(\sqrt{-3})$ and conversely also $\sqrt{-3}\in \Bbb Q(\omega)$.