Extension field of $x^3 - 2$ in $\mathbb Q$

Question

I know this has been asked previous on Stackexchange, but I just want to have something clarified:

I am supposed to find an extension K of Q having all roots of $x^3 - 2$ such that $[K:\mathbb Q] = 6.$

Setting $\alpha = 2^\frac{1}{3}$ and $\omega = e^\frac{2\pi i}{3}$, I can write all the roots in a basis such that the degree is $3$: $$\{\alpha,\alpha \omega, \alpha \omega^2\}$$ I guess this is a field of degree 3? So I am then supposed to add three more random linear independent elements such that $[K:\mathbb Q] = 6$?

Answer 1

Remember that $[K:\mathbb{Q}] = [K:\mathbb{Q(\alpha)}][\mathbb{Q}(\alpha):\mathbb{Q}]$. Now you took a third root of unit $\xi$, and $p(x) = x^2 + x + 1 $ is irreducible in $\mathbb{Q}$ which has $\xi$ as root.

From this what is the degree $[\mathbb Q (\xi): \mathbb Q]$? What does this tell you about $[K:\mathbb{Q(\alpha)}]$?

What you conclude from that?

Answer 2

Hint: The field extension is generated by $\alpha_1=\sqrt[3]{2}$ and $\alpha_2=i\sqrt{3}$

Now use that $[\mathbb Q(\alpha):\mathbb Q]=deg(f_{\alpha})$ where $f_{\alpha}\in \mathbb Q[X]$ is the minimal polynomial of $\alpha$ and the formula $[M:K]=[M:L]\cdot [L:K]$ for field extensions $K\subset L\subset M$