I like to find those finite field $GF(p^n)$ which contains primitive $9$th root of unity. One thing is clear that $GF(p^n)^*$ is cyclic group of size $p^n-1. $ So for $9$th primitive root we must have $9/(p^n-1).$ i.e. $$p^n-1\equiv 0\mod9$$ so for primes of the form $p=9n+1$ finite field $GF(p^n)$ must have $9$th primitive root of unity. Please help me to find all prime $p$ so that $GF(p^n)$ has $9$th primitive root of unity. Thanks.
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Clearly you need $p\neq3$. Whether $GF(p^n)$ has a ninth primitive root will depend on the residue class of $p$ modulo $9$ and on the residue class of $n$ modulo $\phi(9)=6$. See the answer by lhf for the details. – Jyrki Lahtonen Mar 05 '18 at 21:00
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A $9$th primitive root of unity is an element of order $9$.
In a cyclic group of order $m$, there is an element of order $d$ iff $d$ divides $m$.
Therefore, as you have found, $GF(p^n)^\times$ has an element of order $9$ iff $p^n\equiv 1\bmod 9$.
Since $p$ cannot be $3$, we have $\gcd(p,9)=1$ and so $p^6\equiv 1 \bmod 9$ and $1 \equiv p^n\equiv p^{n \bmod 6}\bmod 9$.
If you're only interested in $p$ but not in $n$, then the answer is: all $p\ne 3$.
If you're interested in which $n$ work for which $p$, then:
$n \equiv 1 \bmod 6$ implies $p \equiv 1 \bmod 9$
$n \equiv 2 \bmod 6$ implies $p^2 \equiv 1 \bmod 9$, that is, $p \equiv \pm1 \bmod 18$
$n \equiv 3 \bmod 6$ implies $p^3 \equiv 1 \bmod 9$, that is, $p \equiv 1 \bmod 6$
$n \equiv 6 \bmod 6$ implies $p^6 \equiv 1 \bmod 9$, that is, $p \equiv \pm1 \bmod 3$
The cases $n \equiv 4,5 \bmod 6$ reduce to $n \equiv 2,1 \bmod 6$ because $p^6\equiv 1\equiv p^n\bmod 9$ implies $p^{\gcd(n,6)} \equiv 1\bmod 9$.
lhf
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@lhf Can i say that field $GF(p^m)$ always has $9$th root of unity where $m=6n+k$ with $n\ne 0$? – neelkanth Mar 13 '18 at 12:19
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https://math.stackexchange.com/questions/2689253/all-finite-field-gfpm-has-primitive-9th-root-of-unity – neelkanth Mar 13 '18 at 12:24