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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!

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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 $6\times2$.

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