Find $\alpha$ s.t. $\mathbb{Q}(i,\sqrt[3]{2})$ is $\mathbb{Q}(\alpha)$

Question

I want to find $\alpha$ s.t. $\mathbb{Q}(i,\sqrt[3]{2})$ is $\mathbb{Q}(\alpha)$, but i'm not sure how to do that. $i^2 \in \mathbb{Q}$ and $\sqrt(3)^3 \in \mathbb{Q}$ and $2$ and $3$ are coprime maybe suggests something.

Answer 1

First, try to convince yourself that your field extension is of degree 6.

Then, consider $\alpha = i+\sqrt[3]{2}$. Then the degree of this element over $\mathbb{Q}$ is 6 or 2 or 3, if we can show it is 6, then we are done.

If the degree is 2, then $\mathbb{Q}(\alpha,i)=\mathbb{Q}(i,\sqrt[3]{2})/\mathbb{Q}$ has degree at most 4, which is absurd.

If the degree is 3, consider all the conjugates of $\alpha$, i.e., $\pm i+\sqrt[3]{2}\omega^j$ where $\omega$ is primitive root of 3 and $j\in\{0,1,2\}$(there are 6 of them). Then if our assumption is true, $(x-x_1)(x-x_2)(x-x_3)\in\mathbb{Q}[x]$ for 3 distinct conjugates $x_i$, which can be easily checked to be impossible.

One alternative way to show this is to follow the construction in the proof of the primitive element theorem, you will probably get the same answer.

Intuitively, one should always try $\alpha+\beta$ or $\alpha\beta$, most of the time both works.

Answer 2

Let $\alpha=i\sqrt[3]{2}$. It is clear that $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(i,\sqrt[3]{2})$. We prove the reverse inclusion.

Since $\alpha^3=-2i$, we have that $i\in \mathbb{Q}(\alpha)$ and therefore $\sqrt[3]{2}\in \mathbb{Q}(\alpha)$.