Degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$

Question

Consider $f(x) = x^3 - 2 \in \mathbb{Q}[x]$.

The roots of $f(x)$ are $2^{\frac{1}{3}}, 2^{\frac{1}{3}}\omega, 2^{\frac{1}{3}}{\omega}^2$.

$\mathbb{Q}(2^{\frac{1}{3}},\omega)$ is the minimal splitting field of $f(x)$ over $\mathbb{Q}$.

What is degree of $\mathbb{Q}(2^{\frac{1}{3}},\omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$?

I have been told that the degree is $3$ because it satisfies degree $3$ irreducible polynomial. But how were we able to come to that conclusion?

Answer 1

We know that

$$|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| \leq |\mathbb{Q}(w):\mathbb{Q}| = 2.$$

So $|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| \in \{1,2\}.$ But if $|\mathbb{Q}(2^{1/3},w):\mathbb{Q}(2^{1/3})| = 1$ then $w \in \mathbb{Q}(2^{1/3})$ which of course isn't true.

Explicitly finding a minimal polynomial for $w$ over $\mathbb{Q}(2^{1/3})$ would be doable, but often this approach is relatively painful compared to the above approach. The idea is the use the inequality stated in the first line. (It holds since any polynomial over $\mathbb{Q}$ with $w$ has a root is also a polynomial over $\mathbb{Q}(2^{1/3})$ with $w$ as a root).

Answer 2

I have been told that the degree is 3 because it satisfies degree 3 irreducible polynomial.

The number $\omega$ is a root of the cubic $x^3-1$. But that is not irreducible. One factor is $x^2+x+1$, which $\omega$ is a root of.

The minimal polynomial of $\omega$ (be it over $\mathbb{Q}$ or $\mathbb{Q(2^{1/3})}$) has degree $2$, not $3$.