# Looking at the intermediate fields of $\Bbb{Q}_7 = \Bbb{Q}(\omega)/\Bbb{Q}$ where $\omega = e^{i2\pi/7}$ .

From Basic Abstract Algebra (Robert Ash):

The question that I'm concerned with is number 3, but I will write problems 1 and 2 as well, since they are all related...

We now do a detailed analysis of subgroups and intermediate fields associated with the cyclotomic extension $\Bbb{Q}_7 = \Bbb{Q}(\omega)/\Bbb{Q}$ where $\omega = e^{i2\pi/7}$ is a primitive $7^{th}$ root of unity. The Galois group G consists of automorphisms $\sigma_i, i= 1,2,3,4,5,6,$ where $\sigma_i(\omega) = \omega^i$.

1. Show that $\sigma_3$ generates the cyclic group G.

2. Show that the subgroups of G are $\langle 1 \rangle$ (order 1), $\langle \sigma_6 \rangle$ (order 2), $\langle \sigma_2 \rangle$ (order 3), and $G = \langle \sigma_3 \rangle$ (order 6).

3. The fixed field of $\langle 1 \rangle$ is $\Bbb{Q}_7$ and the fixed field of G is $\Bbb{Q}$. Let $K$ be the fixed field of $\langle \sigma_6 \rangle$. Show that $\omega + \omega^{-1} \in K$, and deduce that $K = \Bbb{Q}(\omega + \omega^{-1}) = \Bbb{Q}(\cos 2\pi/7)$

$\sigma_6(\omega + \omega^6) = \omega^6 + \omega^{36} = \omega + \omega^6$, so $\omega + \omega^6 \in K$. Now $\omega + \omega^6 = \omega + \omega^{-1} = 2 \cos(2 \pi)/7$, so $\omega$ satisfies a quadratic equation over $\Bbb{Q}(\cos 2\pi/7)$. By (3.1.9),

$$[\Bbb{Q}_7 : \Bbb{Q}] = [\Bbb{Q}_7 : K][K : \Bbb{Q}(\cos 2\pi/7)][\Bbb{Q}(\cos 2\pi/7) : \Bbb{Q}]$$

where the term on the left is 6, the first term on the right is $| \langle \sigma_6 \rangle|=2$, and the second term on the right is (by the above remarks) 1 or 2. But $[K : \Bbb{Q}(\cos 2\pi/7)]$ cannot be 2 (since 6 is not a multiple of 4), so we must have $K= \Bbb{Q}(\cos 2\pi/7).$

My Question

1) We know that $\omega^k = \cos( 2\pi k/n) + i\sin(2 \pi k/n)$, so

$$\omega + \omega^6 = \cos( 2\pi/7) + i\sin(2\pi/7) + \cos( 12\pi/7) + i\sin(12\pi/7)$$

I was trying to figure out how this was equal to $2 \cos 2\pi/7$. I tried looking up the basic properties of cosine and sine in case I had forgotten some of them...but I still can't see how these are equal.

2) How do we know that $\omega$ satisfies a quadratic equation over $\Bbb{Q}(\cos 2\pi/7)$? We can only know that if we have $$\cos( 4\pi/7) + i\sin(4\pi/7) \in \Bbb{Q}(\cos 2\pi/7),$$ right?

For your first question, use the fact that $2\pi=\frac{14\pi}7,$ together with the results (obtained through periodic and even/odd properties of the sine and cosine functions):$$\cos(2\pi-z)=\cos(-z)=\cos(z)\\\sin(2\pi-z)=\sin(-z)=-\sin(z)$$
For the second, note that $\omega+\omega^{-1}=2\cos\frac{2\pi}7$ is equivalent to $$\omega^2-\left(2\cos\frac{2\pi}7\right)\omega+1=0.$$ Thus, $\omega$ satisfies $$x^2-\left(2\cos\frac{2\pi}7\right)x+1=0,$$ a quadratic with coefficients in $\Bbb Q(\cos\frac{2\pi}7).$