Let $p$ be prime. What is the biggest number $m$ such that $\{1,\zeta_p,\zeta_p^2,...,\zeta_p^m\}$ is linearly independent over $\mathbb{Q}$? Let $p$ be prime. What is the biggest number $m$ such that $\{1,\zeta_p,\zeta_p^2,...,\zeta_p^m\}$ is linearly independent over $\mathbb{Q}$? Here, $\zeta_p$ is a primitive $p$th root of unity, such as $e^{\frac{2\pi i}{p}}$
Here is my attempt:
So I know I can factor $x^p-1=(x-1)(x^{p-1}+x^{p-2}+...+x+1)$. Then I would try to do a quotient
$\frac{\mathbb{Q}[x]}{(x^{p-1}+x^{p-2}+...+x+1)}\cong\mathbb{Q}(\zeta_p)$. 
Thus, everything on the left can be written as a linear combination of from the set $\{1,x,...,x^{p-2}\}$. I conclude that $m=p-2$ is the right number, since in my mind I identified $x$ with $\zeta_p$.
Is this correct?

Let $\alpha$ be an algebraic number with minimum polynomial $f(x)$.
Then $1,\alpha,\ldots,\alpha^m$ are $\Bbb Q$-linearly independent iff
$m<\deg f$.
In the case where $\alpha=\zeta_n$ a primitive $n$-th root of unity, then
the minimum polynomial is $\Phi_n(x)$, the $n$-th cyclotomic polynomial.
Then $\deg\Phi_n=\phi(n)$, the where $\phi$ is the Euler phi-function.
So the maximal $m$ is $\phi(n)-1$.
When $n=p$ is prime, the maximal $m$ is $\phi(p)-1=p-2$, as you say.
In this case $\Phi_p(x)=x^{p-1}+\cdots+x+1$. The proof depends on the cyclotomic
polynomials being irreducible. This is a non-trivial result: for $n=p$
one observes that $\Phi_p(x+1)$ satisfies Eisenstein's criterion.
