Degree of Field Extension $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ My problem is understanding how we relate field extensions with the same minimum polynomial. I am running into some problems understanding some of the details of the field extension $\mathbb{Q}(2^{\frac{1}{3}})$ over $\mathbb{Q}$ and similarly $\mathbb{Q}(2^{\frac{1}{3}}, \omega)$ over $\mathbb{Q}(2^{\frac{1}{3}})$. From my understanding of the degree of a finite field extension, the degree is equal to the degree of the minimum polynomial for the root $2^{\frac{1}{3}}$. 
Now this has the minimum polynomial $p(x) = x^3 - 2$. So by my understanding of the this definition, we should have $[\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}] = 3$. 
But $p(x)$ has two additional roots which do not exist in the field in $\mathbb{Q}(2^{\frac{1}{3}})$ but do exist in the field extension $\mathbb{Q}(2^{\frac{1}{3}}, \omega)$ where $\omega = e^{\frac{2}{3}\pi i}$. But then the minimum polynomial of $\mathbb{Q}(2^{\frac{1}{3}}, \omega)$ is still our $p(x) = x^3 - 2$. To me, this seems to imply that $[\mathbb{Q}(2^{\frac{1}{3}}, \omega):\mathbb{Q}] = 3$. 
I also know and understand that for finite field extensions that 
$$[\mathbb{Q}(2^{\frac{1}{3}}, \omega):\mathbb{Q}] = [\mathbb{Q}(2^{\frac{1}{3}}, \omega ):\mathbb{Q}(2^{\frac{1}{3}})][\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}]$$
But this seems to imply that when solving that $[\mathbb{Q}(2^{\frac{1}{3}}, \omega ):\mathbb{Q}(2^{\frac{2}{3}})] = 1$ which would also imply that $\mathbb{Q}(2^{\frac{1}{3}}, \omega )=\mathbb{Q}(2^{\frac{2}{3}})$. 
I am almost certain that the last statement is false, but I am unsure where I may be making mistakes with my definitions or applying theorems incorrectly, or if it is just a simple mistake on my part.
Thanks!
 A: The degree of the extension for adding one root is the degree of the minimal polynomial.  That's not the same as the degree of the extension for adding all the roots.  In fact, $\omega$ has minimal polynomial $x^2+x+1$ over $\Bbb{Q}$ and thus also over $\Bbb{Q}(2^{\frac{1}{3}})$ since the degrees are co-prime.  Thus, $\Bbb{Q}(2^{\frac{1}{3}},\omega)$ which is the splitting field of $x^3-2$ over $\Bbb{Q}$ has degree $3\cdot2=6$ over $\Bbb{Q}$.  Splitting fields of irreducible polynomials of degree $n$ have degrees between $n$ and $n!$ over the ground field.
Since the only intermediate fields between $\Bbb{Q}$ and $\Bbb{Q}(2^\frac{1}{3},\omega)$ are $\Bbb{Q}(2^\frac{1}{3})$, $\Bbb{Q}(\omega2^\frac{1}{3})$, $\Bbb{Q}(\omega^22^\frac{1}{3})$, and $\Bbb{Q}(\omega)$, which have dimensions 3, 3, 3, and 2 over $\Bbb{Q}$, "almost" any element of the 6-dimensional space $\Bbb{Q}(2^\frac{1}{3},\omega)$ will have degree $6$ and generate that entire field.  For example, $\omega+ 2^\frac{1}{3}$ will work.
A: You can't directly find the degree of the extension $\mathbb{Q}(\sqrt[3]{2},\omega)$ over $\mathbb{Q}$, by just using the minimal polynomial. You have to consider the minimal polynomial of $\omega$ over $\mathbb{Q}(\sqrt[3]{2})$, which is $x^2+x+1$. Therefore we would have:
$$[\mathbb{Q}(2^{\frac{1}{3}}, \omega):\mathbb{Q}] = [\mathbb{Q}(2^{\frac{1}{3}}, \omega ):\mathbb{Q}(2^{\frac{1}{3}})][\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}] = 6$$
