Let \(F = \mathbb{F}_q\) be a finite field with \(q\) elements. Let \(n\) be a positive integer relatively prime to \(q\) (\(\gcd(n, q) = 1\)), and let \(\omega\) be a primitive \(n\)-th root of unity in an algebraic closure \(\overline{F}\). Let \(E = F(\omega) = F[\omega]\) and \(k = [E : F]\).
Prove that \(n\) divides \(q^k - 1\).
Let \(m = \operatorname{ord}_n(q)\) be the multiplicative order of \(q\) in the group \((\mathbb{Z}/n\mathbb{Z})^\times\). Prove that \(m\) divides \(k\).
Prove that \(m = k\).
Solution
Goal: Prove that the extension degree \([F(\omega) : F]\) for a primitive \(n\)-th root of unity over \(\mathbb{F}_q\) equals the multiplicative order of \(q\) modulo \(n\).
1. Part (a): \(n\) divides \(q^k - 1\).
1.1. The field \(E = F(\omega)\) is a finite field extension of \(F = \mathbb{F}_q\) of degree \([E : F] = k\).
1.2. The cardinality of \(E\) is \(|E| = |F|^k = q^k\).
1.3. The multiplicative group of units \(E^\times = E \setminus \{0\}\) has order \(|E^\times| = q^k - 1\).
1.4. Since \(\omega\) is a primitive \(n\)-th root of unity, \(\omega \in E^\times\) and the order of \(\omega\) in the group \(E^\times\) is \(\operatorname{ord}(\omega) = n\).
1.5. By Lagrange’s Theorem for finite groups, the order of any element divides the order of the group.
1.6. Thus \(n\) divides \(q^k - 1\).
2. Part (b): \(m\) divides \(k\).
2.1. By Part (a), \(n \mid (q^k - 1)\), which means \(q^k \equiv 1 \pmod n\).
2.2. Since \(\gcd(n, q) = 1\), the residue class \([q]_n \in (\mathbb{Z}/n\mathbb{Z})^\times\) has finite order \(m = \operatorname{ord}_n(q)\).
2.3. In any group, \(g^k = e\) if and only if the order \(\operatorname{ord}(g)\) divides \(k\).
2.4. Applying this to \([q]_n \in (\mathbb{Z}/n\mathbb{Z})^\times\) yields \(m \mid k\).
3. Part (c): \(m = k\).
3.1. Since \(m = \operatorname{ord}_n(q)\), \(q^m \equiv 1 \pmod n\), so \(n\) divides \(q^m - 1\).
3.2. Let \(K = \mathbb{F}_{q^m}\) be the unique subfield of \(\overline{F}\) with \(q^m\) elements, which is an extension of \(F = \mathbb{F}_q\) of degree \([K : F] = m\).
3.3. The multiplicative group \(K^\times\) is cyclic of order \(q^m - 1\).
3.4. Because \(n\) divides \(q^m - 1\), the cyclic group \(K^\times\) contains a cyclic subgroup of order \(n\).
3.5. The elements of this subgroup are the \(n\) distinct roots of \(x^n - 1\) in \(K\) (which are distinct because \(\gcd(n, q) = 1\) ensures \(\frac{d}{dx}(x^n - 1) = n x^{n-1} \ne 0\)).
3.6. Since \(\omega\) is a primitive \(n\)-th root of unity, \(\omega\) is a root of \(x^n - 1\), and therefore \(\omega \in K\).
3.7. Since \(F \subseteq K\) and \(\omega \in K\), the subfield generated by \(F\) and \(\omega\) satisfies\[E = F(\omega) \subseteq K.\]
3.8. By the Tower Law for field extensions:\[m = [K : F] = [K : E] \cdot [E : F] = [K : E] \cdot k.\]
3.9. This implies \(k \le m\) (and \(k \mid m\)).
3.10. Combining \(m \mid k\) (from Part (b)) and \(k \le m\) with \(k, m \ge 1\) forces \(m = k\).
4. Conclusion:
$n \mid (q^k - 1)$, $m \mid k$, and $[F(\omega) : F] = m = \operatorname{ord}_n(q)$.