# A generator of a different form than the one given by the primitive element theorem

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?

• is it equivalent to $\mathbb{Q}(i,\sqrt[3]{4})$ ? as $\sqrt[3]{4}*\sqrt[3]{4}=2*\sqrt[3]{2}$ – Zang MingJie Mar 18 at 7:41
• @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. – lEm Mar 18 at 8:01
• 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? – user500144 Mar 19 at 3:28
• 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. – Zang MingJie Mar 19 at 10:22