If $\sqrt[3]{2}+\omega\in \Bbb{K}$, prove that $\sqrt[3]{2}\in \Bbb{K}$

Question

Let $K=\Bbb{Q}(\sqrt[3]{2}+\omega)$ be a field extension of $\Bbb{Q}$ where $\omega$ is a root of $x^2+x+1$. Prove that $\sqrt[3]{2}\in K$

How do I prove this? I am at a loss. I did it by a very long method, involving solving for coefficients. I'm sure there is a better way.

My strategy: I calculated $(\sqrt[3]{2}+\omega), (\sqrt[3]{2}+\omega)^2, (\sqrt[3]{2}+\omega)^3, \frac{1}{(\sqrt[3]{2}+\omega)}, \frac{1}{(\sqrt[3]{2}+\omega)^2}$. Then I determined the coefficients I would need to multiply each by to finally get $2^{1/3}$

Answer 1

Let $L$ be the splitting field of $X^3 - 2$ over $\mathbb{Q}$. The roots of this polynomial are $\sqrt[3]{2}$, $\omega \sqrt[3]{2}$, $\omega^2 \sqrt[3]{2}$. Therefore $L = \mathbb{Q}(\sqrt[3]{2}, \omega)$. Clearly, $[L:\mathbb{Q}(\sqrt[3]{2})] = 2$, and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}] = 3$. Hence $[L:\mathbb{Q}] = 6$.

It will be enough to show that $\mathbb{Q}(\sqrt[3]{2} + \omega) = L$. This will follow if we can show that, of the six automorphisms of $L$, the element $\sqrt[3]{2} + \omega$ is fixed only by the identity. These automorphisms are given by $\sqrt[3]{2} \mapsto \sqrt[3]{2}, \omega \sqrt[3]{2}, \omega^2 \sqrt[3]{2}; \omega \mapsto \omega, \omega^2$.

Thus it is enough to check that $$\sqrt[3]{2} + \omega \not\in \{ \omega \sqrt[3]{2} + \omega, \omega^2 \sqrt[3]{2} + \omega, \sqrt[3]{2} + \omega^2, \omega \sqrt[3]{2} + \omega^2, \omega^2 \sqrt[3]{2} + \omega^2 \}.$$

For the first three elements, this is obvious because their difference with $\sqrt[3]{2} + \omega$ is obviously nonzero. The fourth element is $\omega(\sqrt[3]{2} + \omega)$. The fifth element has modulus $\sqrt[3]{2} + 1$.