Facts About General Roots of Unity (13.6.1-4)

Dummit and Foote Abstract Algebra, section 13.6, exercises 1-4:

1. Suppose $m$ and $n$ are relatively prime integers. Prove $\zeta_m \zeta_n$ is a primitive $mn^{th}$ root of unity.

2. Let $d~|~n$. Prove $\zeta_n^d$ is a primitive $(n/d)^{th}$ root of unity.

3. Prove that if a field contains the $n^{th}$ roots of unity for odd $n$ then it also contains the $2n^{th}$ roots of unity.

4. Prove that if $n=p^km$ for $p~\not \mid~m$ then there are precisely $m$ distinct $n^{th}$ roots of unity over a field of characteristic $p$.

Proof: (1) Let the field in question be of characteristic $p$ or $0$, and assume without loss that $p~\not \mid~n$. Further assume that $p~\not \mid m$ or that the characteristic is $0$. Then by (4) we may assume $\zeta_m,\zeta_n$ are of orders $m,n$ respectively. We see $(\zeta_m\zeta_n)^{mn}=1$ so the order of $\zeta_m\zeta_n$ is $≤mn$. As well, $(\zeta_m\zeta_n)^m=\zeta_n^m$ is of order $n$, so $n$ divides the order of $\zeta_m\zeta_n$. Similarly $m$ divides this order and $\zeta_m\zeta_n$ is of order $mn$ and thus a primitive $mn^{th}$ root of unity (still keeping mind of the case of characteristic $p$, as the $mn$ zeros of $x^{mn}-1$ are all distinct).

Now assume exclusively characteristic $p$ and $m=p^km'$ for $p~\not \mid~m'$. Then $x^m-1=(x^{m'}-1)^{p^k}$ and so $\zeta_m$ is of order $m'$ and $\zeta_m\zeta_n$ is of order $m'n$. Finally, we see $x^{mn}-1=(x^{m'n}-1)^{p^k}$ so there are exactly $m'n$ distinct solutions and thus $\zeta_m\zeta_n$ is a primitive $mn^{th}$ root.

(2) Suppose characteristic $0$. Then clearly $\zeta_n^d$ is of order $≤n/d$ and at least of this order, so is a primitive root. Suppose characteristic $p$ and write $d=p^jd'$ and $n=p^kn'$ for maximal $j,k$. Then $x^{n/d}-1=(x^{n'/d'}-1)^{p^{k-j}}$ has $n'/d'$ solutions, so since $\zeta_n$ is of order $n'$ by (4) and thus $\zeta_n^d$ of order $n'/d'$, we thus have $\zeta_n^d$ is primitive.

(3) Since $-1$ is the $2^{nd}$ root of unity, by (1) $-\zeta_n$ are the $2n^{th}$ roots of unity.

(4) We have $x^n-1=(x^m-1)^{p^k}$ where $D_x x^m-1 = mx \neq 0$ has no zeros in common with $x^m-1$, so there are $m$ distinct roots of $x^m-1$ and thus of $x^n-1$.$\square$