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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.

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    $\begingroup$ Well, to begin with, you should probably tell us what you understand $\Bbb Q\left(\cos\frac{2\pi}5\right)$ to mean. More generally, it's good to include your thoughts and efforts. It helps us gauge your experience level and pinpoint possible areas of misunderstanding, so we can better tailor our answers to fit your needs. $\endgroup$ Aug 7, 2015 at 18:31
  • $\begingroup$ Thanks, I'll consider that for future questions. Right now I already edited a bit. $\endgroup$
    – user255368
    Aug 7, 2015 at 18:38
  • $\begingroup$ You should not ask the same question again: math.stackexchange.com/questions/1390074/…. $\endgroup$ Aug 8, 2015 at 18:49

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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 $\mathbb Q(e^{2\pi i/5}+e^{-2\pi i/5})=\mathbb Q(x+\frac1x)$?

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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.

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  • $\begingroup$ Thanks Michael! In general, what would be a good way to approach these types of problems? $\endgroup$
    – user255368
    Aug 7, 2015 at 19:35
  • $\begingroup$ @Cesar : Merely recalling and applying some basic definitions is what I did. ${}\qquad{}$ $\endgroup$ Aug 8, 2015 at 23:37

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