Finding the minimal polynomial of $e^{2\pi i/5}$ over $\mathbb Q(\cos(2\pi/5))$ How would one find the minimal polynomial of $e^{2\pi i/5}$ over $\mathbb{Q}(\cos(2{\pi}/5))$?
The $\mathbb{Q}(\cos(2{\pi}/5))$ is what is confusing me the most. I know that $\mathbb{Q}(\cos(2\pi/5))$ is an extension of $\mathbb{Q}$ where we have polynomials with rational numbers and $\cos(2\pi/5)$ as coefficients, but I don't know how to combine with finding the minimal polynomial.
I tried setting $x=e^{2\pi i/5}$ and I know that $x^5=1$. Then I could do $$x^5-1=0 \rightarrow x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$ but after that I don't know how to proceed.
 A: Hint: Using the fact that $$\cos \theta = \frac{e^{i\theta}+e^{-i\theta}}2,$$
we have $\mathbb Q(\cos\frac{2\pi}5)=\mathbb Q(e^{2\pi i/5}+e^{-2\pi i/5})$.
Can you find a polynomial with $x=e^{2i\pi/5}$ as a root, whose coefficients lie in the field $\mathbb Q(e^{2\pi i/5}+e^{-2\pi i/5})=\mathbb Q(x+\frac1x)$?
A: You've got it down to $x^4+x^3+x^2+x+1$ and the remaining question is whether to factor that further within $\mathbb Q(\cos(2\pi/5))$.  In $\mathbb C$, you can get that down to four first-degree factors, in two complex-conjugate pairs, hence two quadratics.  So the question is whether that one of those quadratics whose roots are $\exp(\pm 2\pi i/5)$ has its coefficients in $\mathbb Q(\cos(2\pi/5)$.  The polynomial is
$$
(x - e^{2\pi i/5}) (x - e^{-2\pi i/5}) = x^2 - 2x\cos(2\pi/5) + 1.
$$
And then I think the answer becomes clear.
PS: A quicker way: First find the minimal polynomial of $e^{2\pi i/5}$ over $\mathbb R$ and then ask whether its coefficients are in $\mathbb Q(\cos(2\pi/5))$.  The minimal polynomial over $\mathbb R$ of anything in $\mathbb C\setminus\mathbb R$ is a quadratic polynomial whose other root is the complex conjugate.
