[Math] Show that $\mathbb{Q}(\sqrt2, \sqrt[3]2)$ is a primitive field extension of $\mathbb{Q}$.

Question

I've tried a method similar to showing that $\mathbb{Q}(\sqrt2, \sqrt3)$ is a primitive field extension, but the cube root of 2 just makes it a nightmare.

Thanks in advance

Answer 1

I think there is a simpler solution.

Let $a = \sqrt{2} \sqrt[3]{2}$.

$a$ is clearly in the field extension and both generators can be generated by it:

$\sqrt{2} = (\frac{a}{2})^{-3}$

$\sqrt[3]{2} = (\frac{a}{2})^{-2}$.