- #1
nonequilibrium
- 1,439
- 2
In a lot of places, I can read that the roots of unity form a cyclic group, however I can find no proofs. Is the reasoning as follows:
Let's work in a field of characteristic zero (I think that's necessary). Let's look at the nth roots of unity, i.e. the solutions of [itex]x^n - 1[/itex]. There are n different roots, since the derivative is [itex]nx^{n-1}[/itex], which is not zero since the characteristic is zero. Now suppose the group of roots is not cyclic, then the exponent of that group is [itex]m < n [/itex]. In that case the group is also the set of solutions of [itex]x^m-1[/itex], however this can only have m solutions. Contradiction.
Let's work in a field of characteristic zero (I think that's necessary). Let's look at the nth roots of unity, i.e. the solutions of [itex]x^n - 1[/itex]. There are n different roots, since the derivative is [itex]nx^{n-1}[/itex], which is not zero since the characteristic is zero. Now suppose the group of roots is not cyclic, then the exponent of that group is [itex]m < n [/itex]. In that case the group is also the set of solutions of [itex]x^m-1[/itex], however this can only have m solutions. Contradiction.