Splitting field of x^3-2 as a simple extension Is there any elegant way to show that $\mathbb Q(\sqrt[3]{2}, w)=\mathbb Q(\sqrt[3]{2}+w)$, where $w=e^{i\frac{2\pi}{3}}$.
I was thinking to show that $ 9+9 x+3 x^3+6 x^4+3 x^5+x^6$ is the minimal polynomial of $\sqrt[3]{2}+w$, but checking irreducibility seems a bit tedious.
 A: Let $\beta=2^{1/3}$; we know that $\beta^3=2$ and $w^2=-w-1$. Take $\{1,w,\beta,\beta w,\beta^2,\beta^2w \}$ as the basis of a vector space. Taking $\alpha=\beta + w$, we can write each power $\alpha^0,\ldots,\alpha^5$ as a column vector on the left of the bar in the matrix:
$\left[\begin{array}{cccccc|cc} 1 & 0 & -1 & 3 & 0 & -24 & 0 & 0 \\ 0 & 1 & -1 & 0 & 12 & -24 & 1 & 0 \\ 0 & 1 & 0 & -3 & 6 & 0 & 0 & 1 \\ 0 & 0 & 2 & -3 & 0 & 18 & 0 & 0 \\ 0 & 0 & 1 & 0 & -6 & 12 & 0 & 0 \\ 0 & 0 & 0 & 3 & -6 & 0 & 0 & 0 \end{array}\right]$,
and the two vectors on the right represent $w$ and $\beta$. If the augmented matrix is consistent, then $w$ and $\beta$ can be expressed as polynomials in $\alpha$, and are therefore in the desired field.
It turns out the matrix is consistent (the coefficient matrix is invertible), and we get:
$w=\frac{\alpha^5+\alpha^4+2\alpha^3-6\alpha^2+12}{-6} \\
\beta=\alpha-w$
This might be not at all what you meant by elegant. For a more theory-based approach, try checking out proofs of the Primitive Element Theorem. I've seen one that should fit this situation, and essentially show that $\alpha$ is a primitive element for the field $\mathbb{Q}(\beta,w)$.
