# Find the splitting field $K$ of $x^{12}-9$ over $\mathbb{Q}$ and determine $[K:\mathbb{Q}]$.

Find the splitting field $$K$$ of $$x^{12}-9$$ over $$\mathbb{Q}$$ and determine $$[K:\mathbb{Q}]$$.

My approach: First we can factor it $$x^{12}-9 = (x^6-3)(x^6+3)$$ so that the first factor gives us that $$\sqrt[6]{3}$$ and $$\zeta_6$$ (primitive 6th root of unity) should be in $$K$$. I'm not sure about the second factor. The plan is to take the 6th root we would have $$\sqrt[6]{3}$$ and $$\sqrt[6]{i}$$. Write $$i = e^{i\pi/2}$$, then $$\sqrt[6]{i} = e^{i\pi/12}$$ and I got stuck.

Thanks for any help!

The zeros of $$x^{12}-9$$ are the $$\zeta^k\sqrt[6]3$$ where $$\zeta=\exp(\pi i/6)=\frac12(\sqrt3+i)$$. By Eisenstein's criterion, $$|\Bbb Q(\sqrt[6]3):\Bbb Q|=6$$. Also adjoining $$\zeta$$ to $$\Bbb Q(\sqrt[6]3)$$ is the same as adjoining $$i$$, so gives a quadratic extension. So $$K$$ has degree $$6\times2$$.