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I am trying to generalise the question discussed by me in the following link and answered by lfz

Finite field with $9$th primitive root of unity. 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 we are 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 $p^{\gcd(n,6)} \equiv 1\bmod 9$.

Now QUESTION is can i say that All finite field $GF(p^m)$ has primitive $9$th root of unity where $m=6n+k$ and $n\ne 0?$

Discussion from the following link

Finite field with $9$th primitive root of unity.

neelkanth
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  • Don't copy an answer into your question. Add a link instead. – lhf Mar 13 '18 at 12:26
  • ok i am adding link.... – neelkanth Mar 13 '18 at 12:27
  • @lhf I added link .... Actually you solved the problem already asked but my question was to find for which $GF(p^n)$ contains $9$th primitive root and which does not. But in my question i typed question in different way . Therefore i needed different question... – neelkanth Mar 13 '18 at 12:32

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The finite fields with a primitive $9$th root of unity are exactly the fields such that $\Phi_9(x)=x^6+x^3+1$ factors in linear terms. Assuming $p\neq 3$, the degree of the splitting field of $\Phi_9(x)$ over $\mathbb{F}_{p}$ is the least $h\geq 0$ such that $9\mid\left(p^h-1\right)$. Since $\varphi(9)=6$, any field of the form $\mathbb{F}_{p^{6k}}$ with $p\neq 3$ contains a primitive $9$th root of unity.

Jack D'Aurizio
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  • But these are not only fields that contains $9$th primitive root of unity.... – neelkanth Mar 13 '18 at 13:00
  • In above link it is said all field $\mathbb F_{p^{6n+k}}$ has 9th primitive root of unitumy – neelkanth Mar 13 '18 at 13:12
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    @neelkanth: $$\mathbb{F}{p^{6n}}\subset\mathbb{F}{p^{6n+k}}$$ so if the former contains a primitive $9$th root of unity, so does the latter. – Jack D'Aurizio Mar 13 '18 at 13:15
  • In general, $\mathbb{F}_{p^m}$ contains a primitive $9$th root of unity iff $p^m\equiv 1\pmod{9}$. – Jack D'Aurizio Mar 13 '18 at 13:16
  • @ Jack D'Aurizio Ok....it means that i asked in question is correct ? – neelkanth Mar 13 '18 at 13:25
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    @neelkanth: all the fields $\mathbb{F}{p^{6n+k}}$ with $p\neq 3$ contain a primitive $9$th root of unity, that is correct. But they are not the only ones. $\mathbb{F}{7^3}$ contains a primitive $9$th root of unity too, for instance. – Jack D'Aurizio Mar 13 '18 at 13:32
  • Yes thanks ....can you categories all others such fields? Thank you very much .... – neelkanth Mar 13 '18 at 13:36
  • Or should i asked it in different question form ? – neelkanth Mar 13 '18 at 13:49
  • My answer handles all cases: https://math.stackexchange.com/a/2677759/589 – lhf Mar 13 '18 at 14:05
  • But it does not gives exactly which field .... – neelkanth Mar 13 '18 at 14:22
  • @lhf all i know from the discussion is that all field of the form $\mathbb{F}_{p^{6n+k}}$$(n\ne 0)$ contains primitive $9$th root of unity. But i don't know which types of field does not contains $9$ th primitive root of unity.. – neelkanth Mar 13 '18 at 15:26
  • @neelkanth: read carefully the part about the degree of the splitting field of $\Phi_9$. As already stated, $\mathbb{F}_{p^k}$ contains a primitive $9$th root of unity if and only if $p\neq 3$ and $9\mid(p^k-1)$. – Jack D'Aurizio Mar 13 '18 at 16:03