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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 $\gamma$?

Drew Armstrong
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    Write $\gamma=\alpha+x\omega$. Take any $x\in\mathbb{Q}$ that is not of the form $(\alpha-\alpha')/(\omega-\omega')$, where $\alpha'$ is another root of $x^3-2$, the minimal polynomial of $\alpha$, and $\omega'$ is another root of $x^3-1$, the minimal polynomial of $\omega$. – user647486 Mar 22 '19 at 16:40
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    I think $\gamma = \alpha + \omega$ will do the trick here. – Joel Cohen Mar 22 '19 at 17:22
  • @user647486 A minor correction that $x^3 - 1$ is not the minimal polynomial of $\omega$, it is $x^2 + x + 1$. –  Jul 06 '19 at 15:37

2 Answers2

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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.

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The primitive element theorem asserts that for $u \in K$ and $\alpha,\omega$ separables, $\alpha+\omega u$ is a primitive element of $K(\alpha,\omega)/K$ iff $\forall \sigma \in Gal(\overline{K}/K)$, $$\sigma(\alpha)+\sigma(\omega)u = \alpha +\omega u \implies \sigma(\alpha)=\alpha,\sigma(\omega) = \omega$$

only finitely many $u$ don't work because those not working are of the form $u=\frac{ \sigma(\alpha)-\alpha}{\omega - \sigma(\omega)}$.

Here $K=\Bbb{Q}, \alpha= 2^{1/3}, \omega = e^{2i \pi /3}$, from our knowledge of Galois groups $\sigma(\alpha) = \omega^l \alpha, \sigma(\omega) = \omega^m, 3 \nmid m$,

with $u=1$,

for $\sigma(\alpha)+\sigma(\omega)= \alpha +\omega $, if $l\ne 0$ then $\alpha= \frac{\omega^m-\omega}{1-\omega^l} \in \Bbb{Q}(\omega)$ a contradiction, thus $l = 0$ and $\sigma(\alpha) = \alpha,\sigma(\omega)=\omega$.

Whence $\Bbb{Q}(2^{1/3},e^{2i \pi /3})=\Bbb{Q}(2^{1/3}+e^{2i \pi /3})$.

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