I want to find generators of the field extension $\mathbb{Q}(i,\sqrt[3]{2})$ over $\mathbb{Q}$. Using the primitive element theorem (see here), I get $pi+q\sqrt[3]{2}$, for every nonzero rational $p,q$, is a generator for the field extension (i.e. $\mathbb{Q}(i,\sqrt[3]{2})=\mathbb{Q}(\gamma)$ where $\gamma=pi+q\sqrt[3]{2}$ for any nonzero rational $p,q$).

My question is, are these the only possibilities for the generator, or can there be a generator of different form? How can I show this?

  • $\begingroup$ is it equivalent to $\mathbb{Q}(i,\sqrt[3]{4})$ ? as $\sqrt[3]{4}*\sqrt[3]{4}=2*\sqrt[3]{2}$ $\endgroup$ – Zang MingJie Mar 18 at 7:41
  • $\begingroup$ @ZangMingJie Yes it is the same. So one can see that $pi+q\sqrt[3]{4}$ are also generators of the field, and those are different from the elements listed in the question. $\endgroup$ – lEm Mar 18 at 8:01
  • $\begingroup$ How do you know $\mathbb{Q}(i,\sqrt[3]{2})=\mathbb{Q}(i,\sqrt[3]{4})$? $\supseteq$ holds because $\sqrt[3]{4}=(\sqrt[3]{2})^2$ but for $\subseteq$, how do you show? $\endgroup$ – user500144 Mar 19 at 3:28
  • $\begingroup$ As $\sqrt[3]{2} = (\frac{1}{2}i)(-i)(\sqrt[3]{4})^2$, all factors belong to $\mathbb Q(i, \sqrt[3]{4})$, and field is closed to multiplication. $\endgroup$ – Zang MingJie Mar 19 at 10:22

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