Let $\alpha:=\sqrt[3]{2}\in\mathbb{R}$ and $\omega:=e^{2\pi i/3}\in\mathbb{C}$. Then the splitting field for the polynomial $x^3-2\in\mathbb{Q}[x]$ is $$\mathbb{Q}(\alpha,\omega\alpha,\omega^2\alpha)=\mathbb{Q}(\alpha,\omega).$$
Since $\mathbb{Q}$ has characteristic zero we know from the Primitive Element Theorem that there exists some $\gamma\in\mathbb{Q}(\alpha,\omega)$ with $$\mathbb{Q}(\alpha,\omega)=\mathbb{Q}(\gamma).$$
Question: How can I find a specific example of such an element $\gamma$?
Best Answer
The method outlined by @user647486 in the comments is definitely useful, but as @JoelCohen commented, one can show that $\gamma = \alpha + \omega$ is a primitive element of $\Bbb{Q}(\alpha,\omega)$.
To show this, it suffices to show that $1,\gamma, \gamma^2, \gamma^3$ are $\Bbb{Q}$-linearly independent. By computing the powers, we have (in terms of the basis $\{ 1,\alpha,\alpha^2,\omega,\alpha\omega,\alpha^2\omega\}$ of $\Bbb{Q}(\alpha,\omega)/\Bbb{Q}$): $$ \begin{alignat}{10} 1 &{}={}& 1 &{} {}& &{} {}& &{} {}& &{} {}& &{} {}& \\ \gamma &{}={}& &{} {}& \alpha &{} {}& &{}+{}& \omega &{} {}& &{} {}& \\ \gamma^2 &{}={}& -1 &{} {}& &{}+{}& \alpha^2 &{}-{}& \omega &{}+{}& 2\alpha\omega &{} {}& \\ \gamma^3 &{}={}& 3 &{}-{}& 3\alpha &{} {}& &{} {}& &{}-{}& 3\alpha\omega &{}+{}& 3\alpha^2\omega \end{alignat} $$ These are clearly linearly independent, so we are done.