How can I show that $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$ is a splitting field for $X^3-2=0$?

Question

I want to concretely show that the roots of the polynomial, $\sqrt[3]{2},\zeta_3\sqrt[3]{2},\zeta_3^2\sqrt[3]{2}$ all lie within this field, but the latter two aren't rational multiples of $\sqrt[3]{2}$ and $\zeta_3$.

Answer 1

The definition is this: $$\mathbb{Q}(\sqrt[3]{2}, \zeta_3) = \Bigg\{ \frac{f(\sqrt[3]{2}, \zeta_3)}{g(\sqrt[3]{2}, \zeta_3)} \;\bigg| \; f, g \in \mathbb{Q}[x,y], \; g(\sqrt[3]{2}, \zeta_3) \neq 0\Bigg\}$$

So definitely all the roots of your equation are there.

Answer 2

They don't need being rational multiples of $\sqrt[3]{2}$ and of $\zeta_3$. The field can be described either as $K(\zeta_3)$ or $L(\sqrt[3]{2})$, where $K=\mathbb{Q}(\sqrt[3]{2})$ and $L=\mathbb{Q}(\zeta_3)$. So, for instance, $$ K(\zeta_3)=\alpha+\beta\zeta_3+\gamma\zeta_3^2 $$ where $\alpha,\beta,\gamma\in K$.

The three roots of $X^3-2$ so belong to the field $F=\mathbb{Q}(\sqrt[3]{2},\zeta_3)$, hence $X^3-2$ splits in $F$. You should also prove that this field is generated by the three roots. But $\sqrt[3]{2}$ is already a root and $$ \zeta_3=\frac{\zeta_3^2\sqrt[3]{2}}{\zeta_3\sqrt[3]{2}} $$ so $\zeta_3$ belongs to the field generated by the roots of $X^3-2$.