0
$\begingroup$

This question already has an answer here:

Show that for $k \in \mathbb{Z}$:

Is $\zeta_n$ a primitive $n$-th root of unity, then $\zeta_n^k$ is primitive if and only if $(k,n) = 1$.

I only need the backwards direction:

$\zeta_n^k$ is primitive $\Rightarrow$ $(k,n) = 1$

$\endgroup$

marked as duplicate by Dietrich Burde abstract-algebra Jan 30 at 11:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

-1
$\begingroup$

First of all notice that $\zeta_n^k$ is of the form $$ e^{2\pi i\frac kn} $$ and the hypotesis of being primitive means that $n$ is the smallest (positive) integer such that $(\zeta_n^k)^n=1$. If by contradiction $(k,n)=d>1$ then you can write $k=ad$ and $n=bd$, with $a,b\in\Bbb Z$ and $a,b\ge1$. Thus $$ \zeta_n^k=e^{2\pi i\frac {ad}{bd}}=e^{2\pi i\frac d{b}}. $$ From here we deduce that $(\zeta_n^k)^b=1$ and since $b<n$ (in fact $0\le k\le n-1$) our root cannot be primitive.

$\endgroup$
  • $\begingroup$ $(3,9)=3$ so whats $a,b$ here? It should be $1 \leq a,b$ but is your argument then still working? $\endgroup$ – user625682 Jan 30 at 12:09
  • $\begingroup$ The key fact is that, since $k\le n-1$ then the common divisor between $k$ and $n$ strictly divides $n$, that is $n=bd$ with both $b$ and $d$ (wlog positive) and $<n$. In your case $k=3$ and $n=9$, thus we can take $b=d=3$ (I have edited $a,b\ge1$ and in this case $a=1$). $\endgroup$ – Joe Jan 30 at 13:29
  • $\begingroup$ The question is tagged abstract-algebra and nothing about it suggests that the field in question is a subfield or an extension of $\mathbb{C}$, so an approach which argues on the basis of primitive roots in $\mathbb{C}$ doesn't really answer the question. $\endgroup$ – Peter Taylor Jan 30 at 14:40