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Consider the finite field $\mathbf{F}_{p^n}$.

If $3$ divides $p^n - 1$, can it be shown that $\mathbf{F}_{p^n}$ contains a primitive third root of unity?

I am interested in the answer to this question because it arose while trying to prove that every element of $\mathbf{F}_{p^n}$ has a cube root in $\mathbf{F}_{p^{3n}}$. Specifically, I want to use that $a^{p^{3n}} = a$ and $a^{p^n} = a$. If $p^{3n} \equiv 0 \;(\text{mod 3})$ then we're done. Otherwise, $a^{p^{3n}} = a^{p^{3n} - (p^n - 1)}$. If $p^n - 1 \equiv 1 \text{ or } 2 \;(\text{mod}3)$, then we can do this a couple of times and be done. This doesn't work if $3$ divides $p^n - 1$. However, if $F_{p^n}$ contains a primitive third root of unity, then I can be done by using the lemma in the answer here: Is it true that every element of $\mathbb{F}_p$ has an $n$-th root in $\mathbb{F}_{p^n}$?

Since $p^n - 1 = (p-1)(p^{n-1} + p^{n-2} + \dots + 1)$ we can reduce to the case that $3$ divides either factor.

If $3$ divides $p-1$, then the answer is yes because the group of units $(\mathbf{F}_p)^{\times}$ is cyclic of order $p-1$.

What about the case that $3$ divides $p^{n-1} + p^{n-2} + \dots + 1$?

I also know the group of units $(F_{p^n})^{\times}$ is cyclic of order $p^{n-1}(p-1)$.

Thanks for your help.

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Doug
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    I think the group $\mathbb{F}{p^n} ^*$ is cyclic of order $\left|\mathbb{F}{q} ^*\right|=p^n-1$. This should give you the answer. – Joel Cohen Aug 15 '16 at 23:32
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    There's a serious (but common) mistake here. The group of units in $\Bbb F_{p^n}$ is cyclic, yes, but its order is $p^n-1$, not $p^{n-1}(p-1)$. Remember, there's exactly one non-invertible element in a field, by definition. There is another ring, namely $\Bbb Z/p^n\Bbb Z$, whose group of units is cyclic (when $p>2$) of order $p^{n-1}(p-1)$. But not confusing the two rings is very important. – Greg Martin Aug 15 '16 at 23:33
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    Given that $\Bbb F_{p^n}^\times$ is cyclic of order $p^n-1$, your question—whether $\Bbb F_{p^n}$ contains a primitive 3rd root of unity when $3\mid(p^n-1)$—is exactly the same question as whether the cyclic group $C_{p^n-1}$ contains an element of order 3. Do you see why they're the same question? Can you solve the latter question? – Greg Martin Aug 15 '16 at 23:34
  • Oh you're totally right. Thanks. Would anyone like to post this as an official answer? – Doug Aug 15 '16 at 23:35
  • Once I know $3$ divides $p^n - 1$, I understand that $C_{p^n - 1}$ contains an element of order $3$. – Doug Aug 15 '16 at 23:36
  • If you have a third root of unity, how can it be primitive? – abnry Aug 15 '16 at 23:42
  • Primitive means $x^3 = 1$ but $x^n \neq 1$ for $n = 1, 2$. So $1$ is a third root of unity that is not primitive. – Doug Aug 15 '16 at 23:43
  • So, for instance, $e^{2 \pi i / 3}$ and $e^{4 \pi i / 3}$ are the only two primitive third roots of unity in the complex numbers. – Doug Aug 15 '16 at 23:51
  • Related – Viktor Vaughn Aug 16 '16 at 00:19

1 Answers1

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As said in the comments, $|\Bbb F_{p^n}^{\times}|=p^n-1$. Since 3 divides it, Cauchy's Theorem in group theory gives the existence of an element $a$ of order 3 in this group, i.e. such that $a^3=1$ and $a\neq 1$.

Edit: If we know that $\Bbb F_{p^n}^{\times}$ is cyclic, then we know that it has a cyclic subgroup of order 3, which gives the result without using Cauchy's Theorem.

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