How many $n$th roots of unity does a finite field $\mathbb{GF}\left(p^k\right)$ have? ($p$ is prime). And then kind of related to that, is there a finite field with exactly $n_1,\ldots,n_N$ $n_1$th,...$n_N$th roots of unity?
closed as too localized by Felipe Voloch, J.C. Ottem, Ben Webster♦ Mar 13 '11 at 23:30This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


I can quickly answer your first question. The multiplicative group of $\mathbb{GF}(p^k)$ is cyclic, let $g$ be a generator. For an element $x$ of the group $x^n=1$ holds iff $x=g^m$ with $nm$ divisible by $p^k1$. The latter is equivalent to $m$ divisible by $(p^k1)/d$, where $d:=\gcd(n,p^k1)$, hence the $n$th roots of unity form the subgroup generated by $g^{(p^k1)/d}$. This subgroup clearly has $d$ elements, so the number of $n$th roots equals $\gcd(n,p^k1)$. EDIT: I did not see KConrad's comment (I was typing slowly). EDIT: The second question is also easy. It is clearly necessary that the $n_i$'s be coprime with $p$. If this condition holds, then there is a $k>0$ such that $p^k1$ is divisible by each of the $n_i$'s, and such a $k$ will do by the above (while no other $k$ will do). 

