Is is true that $\zeta$ has finite order?

Question

Let $\zeta$ be a complex number on the unit circle $\{z\in \mathbb{C}: |z|=1\}$.Suppose that $[\mathbb{Q}(\zeta):\mathbb{Q}] < \infty$.Is it true that $\zeta ^n=1$ for some positive integer $n$?

Answer 1

An example (from which you can deduce many others):

Take the non-real algebraic number $\;z:=1+\sqrt2\,i\;$ and look at

$$\omega:=\frac z{|z|}=\frac1{\sqrt5}\left(1+\sqrt2\,i\right)=e^{i\arctan\sqrt2}$$

Then clearly $\;\omega\in S^1\;$ and algebraic , yet

$$\forall\;n\in\Bbb N\;\;:\;\;\omega^n=e^{ni\arctan\sqrt2}=1\iff n\arctan\sqrt2=2k\pi\;,\;\;k\in\Bbb Z$$

Now read the answers here...and the comment below the accepted one: ArcTan(2) a rational multiple of $\pi$?

Answer 2

Roots of unit are algebraic integers. Take an algebraic number that don't be algebraic integer.

Answer 3

In the spirit of Alex Youcis' comment let $$ z=\frac{2+i}{2-i}=\frac35+\frac45i. $$ Then $[\Bbb{Q}[z]:\Bbb{Q}]=2<\infty.$

The numbers $2+i$ and $2-i$ generate distinct prime ideals in the ring of Gaussian integers $\Bbb{Z}[i]$. But $\Bbb{Z}][i]$ is a unique factorization domain and it only has the four obvious units. If $z^n=1$ then $(2+i)^n=(2-i)^n$ violating unique factorization.