my task is simple:
find splitting field $K$ of $x^6-6\in\mathbb{Q}[x]$ and determine the degree $[K:\mathbb{Q}]$.
The roots are obviously $\pm\sqrt[6]{6}$ and $\sqrt[6]{6}/2\left(\pm1\pm i\sqrt{3}\right)$, which are not rational, so I suppose that $K=\mathbb{Q}\left(\sqrt[6]{6},i\sqrt{3}\right)$, but is this indeed the "smallest" field? And the degree of the extesion is $2$?
Thx for help...
The degree of the extension is actually $12$. First note that $K$ must contain $\alpha = \sqrt[6]{6}$. The minimal polynomial of $\alpha$ is $f(x) = x^6 - 6$, thus $[K : \mathbb{Q}] \geq 6$. However, note that the other five roots of $f(x)$ are $$\zeta\alpha, \zeta^2\alpha, \zeta^3\alpha,\zeta^4\alpha,\zeta^5\alpha,$$ where $\zeta$ is a primitive 6th root of unity. It is clear that $K = \mathbb{Q}(\alpha,\zeta)$. Since $\mathbb{Q}(\alpha)$ is a totally real field and four of the other five roots are nonreal, $\mathbb{Q}(\alpha)$ cannot be the splitting field of $f(x)$. Now $[\mathbb{Q}(\zeta) : \mathbb{Q}] = 2$ and $\zeta \not\in \mathbb{Q}(\alpha)$, hence $[K : \mathbb{Q}] = 12$.