Degree of extension $\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p))/\mathbb{Q}(\cos(2\pi/p))$.

Question

Let $p$ be an odd prime number. I want to compute the degree $$ [\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}(\cos(2\pi/p))]. $$ I already showed that $$ [\mathbb{Q}(\cos(2\pi/p)):\mathbb{Q}] = \frac{p-1}{2}, $$ so it will be helpful to know $$ [\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}]. $$ We know that the polynomial $$ X^2+\cos^2(2\pi/p)-1 \in \mathbb{Q}(\cos(2\pi/p))[X] $$ vanish on $\sin(2\pi/p)$, but I don't how to show that this polynomial is irreducible over $\mathbb{Q}(\cos(2\pi/p))$. I believe that $$ [\mathbb{Q}(\cos(2\pi/p),\sin(2\pi/p)):\mathbb{Q}(\cos(2\pi/p))]=2, $$ but I cannot see why it is possible or not to write $$ \sin(2\pi/p) = \sum_{k=0}^{n} a_k\cos^k(2\pi/p) $$ for some $a_0,a_1,\dots,a_n\in\mathbb{Q}$.

Any help will be appreciated!

Answer 1

With $\zeta_p$ the p-th primitive root of unity we have that $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}]$ is equal to:

$$[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)][\mathbb{Q}(\cos2\pi/p,\sin2\pi/p):\mathbb{Q}(\cos2\pi/p)][\mathbb{Q}(\cos2\pi/p):\mathbb{Q}]$$

Now, since $i$ cannot be in $\mathbb{Q}(\zeta_p)$ (why?), we have that $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\zeta_p)]=2$ because $i$ is the root of the an irreducible quadratic. Since $[\mathbb{Q}(\zeta_p):\mathbb{Q}]=p-1$, we see that the entire expression above is equal to $2(p-1)$. Note that $[\mathbb{Q}(\zeta_p,i):\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)]$ is degree $2$ because $\mathbb{Q}(\zeta_p,i)=\mathbb{Q}(\cos2\pi/p,\sin2\pi/p,i)$ (why?) and $\mathbb{Q}(\cos2\pi/p,\sin2\pi/p)$ is real.

Putting this all together, and by your previous computation we have that $2(p-1)=2[\mathbb{Q}(\cos2\pi/p,\sin2\pi/p):\mathbb{Q}(\cos2\pi/p)]\frac{p-1}{2}$ from which the result follows.

Answer 2

Let $\zeta = \exp(\frac{2\pi i}{4p})$. Then $\zeta$ is a primitive $(4p)$-th root of unity, and since $2p-1$ is coprime to $4p$, it follows that there is a Galois automorphism (let us call it $\sigma$) of ${\mathbb Q}(\zeta)$ such that $\sigma(\zeta)=\zeta^{2p-1}$.

From Euler's formula we deduce

$$ \cos\bigg(\frac{2\pi}{p}\bigg)=\frac{\zeta^4-\zeta^{2p-4}}{2}, \ \sin\bigg(\frac{2\pi}{p}\bigg)=\frac{\zeta^{p-4}-\zeta^{p+4}}{2} \tag{1} $$

Also, straightforward computations show that $$ \sigma(\zeta^4)=-\zeta^{2p-4},\sigma(\zeta^{2p-4})=-\zeta^{4}, \sigma(\zeta^{p-4})=\zeta^{p+4},\sigma(\zeta^{p+4})=\zeta^{p-4}, \tag{2} $$

Combining (1) with (2), we see that $\sigma$ fixes $\cos\big(\frac{2\pi}{p}\big)$ and sends $\sin\big(\frac{2\pi}{p}\big)$ to its opposite. It follows that $\sin\big(\frac{2\pi}{p}\big)\not\in{\mathbb Q}\big(\cos\big(\frac{2\pi}{p}\big)\big)$.