I think it's true that in a field $F$ of characteristic zero there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $\bar{F}$), that is, roots of the polynomial $x^n-1$.
I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?
I also know that in fields of characteristic zero all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.
Best Answer
Suppose we have a primitive $n$th root of unity, $\omega$. Then for all $1 < a \le n$, $(\omega^a)^n = (\omega^n)^a = 1^a = 1$. If $1 \le a < b \le n$ then $\omega^a = \omega^b \implies \omega^{b-a} = 1$, contradicting the assumption that $\omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.
Now, let $\omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = \frac nm$. Then $$\gcd\left(\frac{x^{km}-1}{x^m-1}, x^m - 1\right) \neq 1$$ Now, $\gcd(p(x), q(x)) = \gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = \sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have $$\frac{x^{km} - 1}{x^m-1} - (x^m-1) \sum_{i=0}^{k-2} (k-1-i)x^{im} = \frac{(x^{km} - 1) - (x^m-1)^2 \sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$ But that $r$ was carefully chosen to telescope: $$(x^m-1)^2 \sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$ so that $$\gcd\left(\frac{x^{km}-1}{x^m-1}, x^m - 1\right) = \gcd(k, x^m - 1)$$ This is non-constant iff $k$ is a multiple of the characteristic of $F$.