# Solving problems of the form $[\mathbb Q(\omega):\mathbb Q]$, where $\omega=\cos(\frac {2\theta\pi}\phi)+i\sin(\frac {2\theta\pi}\phi)$.

I am trying to find $[\mathbb Q(\omega):\mathbb Q]$, where $\omega=\cos(\frac {2\theta\pi}\phi)+i\sin(\frac {2\theta\pi}\phi),$ and $\theta,\phi\in\mathbb Z$.

I have tried putting $\omega$ into exponential form, so, $\omega=e^{\frac{2\theta i\pi}\phi}$.

Then considered $\omega^\phi$, so that we have $\omega^\phi=1$.

Thus $\omega$ is a root of the polynomial $x^\phi-1$.

Hence $\omega$ is algebraic over $\mathbb Q$.

I now want to use the fact that if this polynomial is irreducible in $\mathbb Q$, then $[\mathbb Q(\omega):\mathbb Q]=\phi$, since if it is irreducible then it will the the minimal polynomial of $\omega$ over $\mathbb Q$.

But I can't seem to prove that it is, not using Eisenstein's criterion anyway.

I have a feeling that cyclotomic polynomials may be useful, since they are all irreducible, but I can't quite see how to use this fact.

Any help would be appreciated, thanks.

• $x^{\phi} - 1$ is not irreducible, it even has a rational root: $1$. – Paul K Feb 27 '17 at 15:11

## 1 Answer

You're right with your reasoning, as $\omega$ is indeed a root of a cyclotomatic polynomial. To notice this first reduce $\frac{\theta}{\phi}$ to $\frac{\theta_1}{\phi_1}$ such that $(\theta_1, \phi_1) = 1$ and $\phi_1>0$. Then we have that $\omega$ is a root of $\Phi_{\phi_1}$ as

$$\Phi_n = \prod_{1\le k \le n\text{ and }(n,k) = 1} (x - e^{2\pi \frac kn})$$

Obviously from this definition we have that the degree of the polynomial is the number of coprime integers to $\phi_1$ and less than $\phi_1$, which can be found using the Euler's Totient Function.