Dimension of $\mathbb{Q(\omega)}$ and minimal polynomial of $\sqrt[3]{2}$

Question

Consider:

$$\omega = \frac{-1}{2} + \frac{\sqrt{3} i}{2}$$

and the simple extension $\mathbb{Q(\omega)}$. Find the dimension of $\mathbb{Q(\omega)}$ and the minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q(\omega)}$;

1. The minimal polynomial of $\omega$ is $f(x) = x^2+x+1$ because $f(\omega) = 0$ and is irreducible over $\mathbb{Q}$. Since:

$$\mathbb{Q(\omega)} \cong \frac{\mathbb{Q[x]}}{x^2+x+1}$$

and $deg(x^2+x+1) = 2$ , the dimension of $\mathbb{Q(\omega)}$ is 2.

2. The minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q(\omega)}$ is $g(x)=x^3-2$ , because $g(\sqrt[3]{2}) = 0$ and it hasn't roots over $\mathbb{Q(\omega)}$

It's right?

Answer 1

The second part could be more rigurous, since sometimes is not easy to prove the irreducibility over extensions like $\mathbb{Q}(\omega)$. Note that $$[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6,$$ since $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, and $2$ and $3$ are coprimes. Now we have by the transitivity of degree that $$[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}(\omega)]\cdot[\mathbb{Q}(\omega):\mathbb{Q}]\implies [\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}(\omega)]=6/2=3.$$