To find a primitive $n$-th root of unity in a field $F_q$ of size $q$, one takes the smallest positive integer $m$ such that $q^m \equiv 1 \bmod n$ and finds a primitive $n$-th root of unity in $\mathbb{F}_{q^m}$ by the formula $$\beta = \alpha^{\frac{q^m-1}{n}},$$ where $\alpha$ is a primitive element of $\mathbb{F}_{q^m}$.

But what is the reasoning behind this formula?

  • $\begingroup$ It's just group theory: if $G$ is a cyclic group of order $N$, with generator $g$, and $d|N$, the power $g^{N/d}$ has order $d$ in $G$. For you, $G = F_{q^m}^\times$. $\endgroup$ – KCd Jan 2 '15 at 17:29
  • $\begingroup$ Read this question, its answers and the comments on the question and answers. $\endgroup$ – Dilip Sarwate Jan 2 '15 at 17:33
  • $\begingroup$ ok, thank you, I think i get it $\endgroup$ – Isostorm Jan 2 '15 at 18:26

Hint: in any group, $$\text{ord}(\alpha^k) = \frac{\text{ord}(\alpha)}{\text{gcd}(\text{ord}(\alpha), k)}.$$

  • $\begingroup$ Why not just: $\text{ord}(\alpha^k) = \frac{\text{ord}(\alpha)}{ k}$. $\endgroup$ – Isostorm Jan 2 '15 at 18:28
  • $\begingroup$ The expression on the right doesn't always make sense. Let's see, mod 5 the number 2 has order 4, so is the order of $2^3 \bmod 5$ equal to $4/3$? $\endgroup$ – KCd Jan 2 '15 at 18:45

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