How to determine the degree of the extension $\mathbb{Q}(\sqrt[3] 2,i)$ over $\mathbb{Q}$? and what is the form of these?

Question

How to determine the degree of the extension $\mathbb{Q}(\sqrt[3] 2,i)$ over $\mathbb{Q}$, and what is the form of the elements of $\mathbb{Q}(\sqrt[3] 2,i)$?

I tried to find the polynomial with $\alpha=\sqrt[3] 2+i$ but to no avail. Is there a shorter and simpler way to do it?

Answer 1

Let $\alpha = i\sqrt[3] 2 $.

Prove that $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt[3] 2,i)$.

Find out the minimal polynomial of $\alpha$ over $\mathbb{Q}$.

The basis for $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is $\{1, \alpha ,\cdots, \alpha^5\}$.

So the field has the form: $\mathbb{Q}(\sqrt[3] 2,i) = \{a+b\sqrt[3] 2+ c\sqrt[3] 4+ d i + ei\sqrt[3] 2 + fi\sqrt[3] 4 : a,b,c,d,e,f\in\mathbb{Q}\}$