Given the splitting field $L$ of $f(X)=X^6+3X^3+3$ over $\mathbb{Q}$. Let $\omega$ be a third primitive root of unity, so $\omega^3=1, \omega\neq 1$. Suppose $\omega^k\alpha,\omega^l\sqrt[3]{3}/\alpha\in L$, with $l,k\in\{0,1,2\}$ are the six zeros of $f$. Then I proved that $L=\mathbb{Q}(\alpha, \sqrt[3]{3})$.
I know $\alpha^3\in\mathbb{Q}(\omega)$.
Let $\sqrt[3]{3}\notin\mathbb{Q}(\alpha)$. Show $\text{Gal}(L/\mathbb{Q}(\omega))\cong C_3 \times C_3$.
My reader starts by picking $\sigma\in\text{Gal}(L/\mathbb{Q}(\omega))$. It then concludes that there are $k,l\in\{0,1,2\}$ such that $\sigma(\sqrt[3]{3})=\omega^k\sqrt[3]{3}$ and $\sigma(\alpha)=\omega^l\alpha$. Which are nine possibilities.
I know that elements of $\text{Gal}(L/\mathbb{Q}(\omega))$ permute zeros of $f$, but why in this way? Why can't we have $\sigma(\alpha)=\omega^k\sqrt[3]{3}$ and $\sigma(\sqrt[3]{3})=\omega^l\alpha$.
It then concludes that those nine possibilities are all elements of $\text{Gal}(L/\mathbb{Q}(\omega))$ since $[L:\mathbb{Q}(\omega)]=9$. But I thought $[L : \mathbb{Q}]=[L:\mathbb{Q}(\omega)][\mathbb{Q}(\omega):\mathbb{Q}]$. I have previously proven that $[L:\mathbb{Q}]=18$. But then, to get to the 9, we must have $[\mathbb{Q}(\omega):\mathbb{Q}]=2$. But I thought that the minimal polynomial of $\omega$ would be $X^3-1$, because then $\omega^3-1=1-1=0$, which would mean that $[\mathbb{Q}(\omega):\mathbb{Q}]=3\neq 2$.
If $\alpha$ is a root of $X^6+3X^3+3$, then so are $\omega\alpha, \omega^2\alpha , \sqrt[3]3/\alpha, \omega\sqrt[3]3/\alpha, $ and $\omega^2\sqrt[3]3/\alpha,$ where $\omega^3=1$ but $\omega\ne1$, i.e., $\dfrac{\omega^3-1}{\omega-1}=\omega^2+\omega+1=0.\;$ ($\omega$ is a primitive cube root of unity.)
The map you suggested, which takes $\alpha$ to $\sqrt[3]3$, is not an automorphism,
because $\alpha$ is a root of $X^6+3X^3+3$, but $\sqrt[3]3$ is not.
Finally, as indicated, the minimal polynomial for $\omega$ is a quadratic, so indeed $[\mathbb Q(\omega):\mathbb Q]=2$.