$\mathbb Q$-linear independence of unit vectors on upper half circle

Question

Let $e(\theta)=(\cos(\theta),\sin(\theta))$ and let $C=\big\lbrace e(\theta) \ \big| \ 0\leq \theta < \pi\big\rbrace$. Trivially, $C$ is not linearly independent over $\mathbb R$ (for example, $e(\frac{\pi}{4})+e(\frac{3\pi}{4})=\sqrt{2}e(\frac{\pi}{2})$). But is $C$ linearly independent over $\mathbb Q$ ?

More generally, is $C_{\alpha}=\big\lbrace e(\theta) \ \big| \ 0\leq \theta < \alpha\big\rbrace$ linearly independent over $\mathbb Q$ for some $\alpha >0$ ?

Answer 1

The answer is no. Consider the rational parametrization of the half-circumference $C$: $$P(m)=\left(\frac{1-m^2}{m^2+1},\frac{2m}{m^2+1}\right), \, m\in\mathbb{Q}_+.$$

For any $m\in \mathbb{Q}_+$ one can find $a,b\in \mathbb{Q}$ such that $$P(2m)=aP(m)+bP(0).$$ Thus we also have that $$C_{\alpha}=\big\lbrace e(\theta) \ \big| \ 0\leq \theta < \alpha\big\rbrace$$ is not linearly independent over $\mathbb{Q}$ for any $\alpha>0.$

(I am considering $0\in \mathbb{Q}_+.$)

Answer 2

We have, for example, $e(\frac\pi3) = \frac{1}{2} +\frac{i}{2}\sqrt{3}$ and $e(\frac{2}{3}\pi) = -\frac{1}{2}+\frac{i}{2}\sqrt{3}$, so $e(\frac\pi3) - e(\frac{2}{3}\pi) = 1 = e(0)$. So $C$ is not linearly independent over $\mathbb Q$.