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I know how to find $[\mathbb{Q}(\cos( 2\pi / n )) : \mathbb{Q}]$ but for this I am lost! I have been working a looong time on this problem, any help would be greatly appreciated.

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closed as unclear what you're asking by Derek Holt, Najib Idrissi, user1729, Claude Leibovici, Davide Giraudo Jun 6 at 9:29

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

4  
What is $\xi$ in this case? Also is $i$ the imaginary number? –  DanZimm Jun 6 at 2:06
    
A related question with some discussion. –  Jyrki Lahtonen Jun 6 at 6:26
    
It would be helpful if you stated the question in entirety in the body of your post...(Also, what do you mean by $[P:R]$? Index?) –  user1729 Jun 6 at 8:27

1 Answer 1

I'm unsure what you mean by

when $i \in \mathbb{Q}(\xi)$ and when $i \not\in \mathbb{Q}(\xi)$

but I believe I can provide some help to finding $[\mathbb{Q}(\sin(2 \pi / 5)) : \mathbb{Q}]$.

Hint:

Take different powers of $\sin(2 \pi / 5)$ and create a linear combination of them with arbitrary constants in $\mathbb{Q}$. See if you can find constants so that the linear combination equals $0$. If so this is your minimal polynomial for $\sin(2 \pi /5)$

For example one way to find $[\mathbb{Q}(\sqrt{2}+\sqrt{3}) : \mathbb{Q}]$ is by doing the following:

$$ (\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions for } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions for } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^3 + a_2(\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \text{ has no solutions with } a_i \in \mathbb{Q} \\ (\sqrt{2} + \sqrt{3})^4 + a_3(\sqrt{2} + \sqrt{3})^3 + a_2(\sqrt{2} + \sqrt{3})^2 + a_1(\sqrt{2} + \sqrt{3})^1 + a_0 = 0 \\\text{ has no solutions with } a_i \in \mathbb{Q} $$ since if we notice $$ (\sqrt{2} + \sqrt{3})^2 = 5 + 2\sqrt{2}\sqrt{3} \\ (\sqrt{2} + \sqrt{3})^3 = (\sqrt{2} + \sqrt{3})(5 + 2\sqrt{2}\sqrt{3}) = 5\sqrt{2} + 5\sqrt{3} + 4 \sqrt{3} + 6\sqrt{2} = 11\sqrt{2} + 9\sqrt{3} \\ (\sqrt{2} + \sqrt{3})^4 = (\sqrt{2} + \sqrt{3})(11\sqrt{2} + 9\sqrt{3}) = 49 + 20 \sqrt{2}\sqrt{3} $$ so then we need to solve $$ 49 + 20\sqrt{2}\sqrt{3} + 11a_3\sqrt{2} + 9a_3\sqrt{3} + 5a_2 + 2a_2\sqrt{2}\sqrt{3} + a_1\sqrt{2} + a_1\sqrt{3} + a_0 = 0 $$ with $a_i \in \mathbb{Q}$. This gives us the following equations: $$ 49 + 5a_2 + a_0 = 0 \\ 20 + 2a_2 = 0 \\ 11 a_3 + a_1 = 0 \\ 9a_3 + a_1 = 0 $$ so we get that $a_1 = a_3 = 0, a_2 = -10$ and $$ 49 + 5(-10) + a_0 = 0 \implies a_0 = 1 $$ so that the minimal polynomial of $\sqrt{2} + \sqrt{3}$ is $p(x) = x^4 - 10x^2 + 1 = 0$ thus $$ [\mathbb{Q}(\sqrt{2} + \sqrt{3}) : \mathbb{Q}] = 4 $$ Do you see how to apply this process to your polynomial?

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