Show that $p=2^k+1$

Question

When $p$ is an odd prime and $a=Re \left ( e^{\frac{2 \pi i}{p}} \right)$ then $[\mathbb{Q}(a) : \mathbb{Q}]=\frac{p-1}{2}$.

Let $\theta = \frac{2 \pi}{p}$.

If $\sin{\theta}$ is a constructable number show that $p=2^k+1$, $k \in \mathbb{N}$.

So that $\sin{\theta}$ is a constructable number, the degree of $Irr(\sin{\theta}, \mathbb{Q})$ should be a power of $2$.

Isn't $\sin{\theta}$ the imaginary part of $e^{\frac{2 \pi }{p}}$ ?? How can we use $[\mathbb{Q}(a):\mathbb{Q}(a)]$ ??

Answer 1

If a complex number on the unit circle or any of its components is constructible (by compass and straightedge), all three are constructible. Each construction creates a point in either the earlier extension or some quadratic extension, and so if $a$ is constructible the degree of the final extension $\mathbb{Q}(a)$ over $\mathbb{Q}$ is a power of 2. Since $[\mathbb{Q}(a):\mathbb{Q}] = \frac{p-1}{2}$, you get the result you are looking for.