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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?

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closed as too localized by Felipe Voloch, J.C. Ottem, Ben Webster Mar 13 '11 at 23:30

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The group of nonzero elements is cyclic of order p^k-1, so your question is the same as: in a cyclic group of order N (=p^k-1), how many solutions are there to the equation x^n = 1? This is enough of a hint. Now review your knowledge of cyclic groups. –  KConrad Mar 13 '11 at 20:02
    
Anadim- MO is intended for questions of a research level. Thus, questions which are covered in standard undergraduate courses (in this case, Galois theory) are a better fit for math.stackexchange.com –  Ben Webster Mar 13 '11 at 23:32
    
thank you all. It didn't seem to me that it was undegrad level, so I thought I'd go for MO. I 'll be more careful next time :). –  Anadim Mar 14 '11 at 0:10
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1 Answer 1

up vote 3 down vote accepted

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 $p^k-1$. The latter is equivalent to $m$ divisible by $(p^k-1)/d$, where $d:=\gcd(n,p^k-1)$, hence the $n$-th roots of unity form the subgroup generated by $g^{(p^k-1)/d}$. This subgroup clearly has $d$ elements, so the number of $n$-th roots equals $\gcd(n,p^k-1)$.

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 $p^k-1$ is divisible by each of the $n_i$'s, and such a $k$ will do by the above (while no other $k$ will do).

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