= Splitting degree of roots of unity over a finite field
{title2=$\min\{s\geq1:m\mid q^s-1\}=\operatorname{ord}_m(q)$}
With $m$ the characteristic-prime-free part of $n$, all <roots of unity over a finite field> lie in $\mathbb F_{q^s}$ exactly when $m\mid q^s-1$. This follows from the cyclicity of the <multiplicative group of a finite field>. For $m=1$ the least positive degree is one. A primitive root of order $m$ has the same <field> degree, because its Frobenius orbit closes exactly when $m\mid q^s-1$.
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