Find the minimal polynomial of $e^{i\frac{\pi}{6}} - \sqrt{3}$ over $\mathbb{Q}$

Question

I have problems with finding the minimal polynomial of $e^{i\frac{\pi}{6}} -\sqrt{3}$ over $\mathbb{Q}$. Let $\omega = e^{i\frac{\pi}{6}}$. I have already shown that $[\mathbb{Q}(\omega):\mathbb{Q}] = 4$ and that $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 2$, so by degree formula it follows that $[\mathbb{Q}(\omega - \sqrt{3}):\mathbb{Q}] \geq 4$ and it can not be $6$.

Hints are greatly appreciated!

Answer 1

But what is $e^{i\pi/6}-\sqrt3$? $$e^{i\pi/6}=\frac{\sqrt3}2+\frac12i$$ Thefefore $$e^{i\pi/6}-\sqrt3=-\frac{\sqrt3}2+\frac12i=e^{5i\pi/6}$$ and the minimal polynomial is $\Phi_{12}(x)$ or $x^4-x^2+1$.