Generalization of the legendre symbol Ive been working through David Cox's book primes of the form $x^2+ny^2$, im stuck on problem 5.13, pg 118 and was wondering if i could get some hints, the problem is as follows:
K is a number field that contains an nth root of unity $\zeta$, let $O_{K}$ be the ring of integers of K. let $a \in O_{K}$ and let $p$ be a prime ideal in $O_{K}$ such that $na \not \in p$
a) prove that $1,\zeta,\dots,\zeta^{n-1}$ are distinct roots modulo $p$. The hint here is it show that $x^n-1$ is separable modulo $p$
b) use a) to prove $n | N(p)-1$ (here N(p) is Norm of $p$)
c) show that $a^{(N(p)-1)/n}$ is congruent to a unique root of unity modulo $p$.
There are more parts but this is where i need help. For part a) i wanted to use the fact that is gcd(f(x),f'(x))=1 then its separable, but im not sure how this works modulo $p$, or does it not really make a difference?
For part b) im not quite sure where the $N(p)-1$ comes from.
Thank you very much
 A: I am going to use $\mathfrak{p}$ where you used $p$, just because I might have need to talk about the prime number $p$ which generates the ideal $\mathfrak{p} \cap \mathbb{Q}$.
For part a, note that in any characteristic, it's true that $f$ is separable iff $\text{gcd}(f(x), f'(x)) = 1$. Now $na \not\in \mathfrak{p}$ and in particular we have that $n \not \in \mathfrak{p}$. It follows that $f'(x) = nx^{n-1} \neq 0$ mod $\mathfrak{p}$ and this polynomial clearly has no common factors with $x^n - 1$ (i.e. $x$ is not a factor of $x^n - 1$). 
Since the polynomial is separable mod $\mathfrak{p}$, its roots are distinct mod $\mathfrak{p}$. 
For part b, consider the abelian group $(\mathcal{O}/\mathfrak{p})^\times$. Its order is $\mathbf{N}\mathfrak{p} - 1$ (this is the definition of the norm of an ideal). By part a, the subgroup generated by $\zeta$ has order $n$.
For part c, note that $(a^{\frac{\mathbf{N}\mathfrak{p} - 1}{n}})^n$ is 1 in $(\mathcal{O}/\mathfrak{p})^\times$ (again because $\mathbf{N}\mathfrak{p} - 1$ is the order of that group); the uniqueness is just a restatement of part a (if this guy were equivalent to two distinct roots of unity, then they'd be equivalent to each other).
