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Let $\omega$ be a primitive $n-$th root of unity.

(i) Show that its powers $\omega^k$, for $k ∈ {1, \ldots, n}$, are all different;

(ii) Deduce that they are precisely all the $n-$th roots of unity.

I know that the powers have to be different as otherwise the order of $ω$ would be less than $n$ so $\omega$ wouldn't be a primitive $n-$th root of unity but I don't know how to prove this rigorously.

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    $\begingroup$ your idea is good: if $\omega^k=\omega^j$ then $\omega^{k \mod n}=\omega^{j \mod n}$ so $k \mod n=j \mod n$ so if $j,k\in\{1,\dots,n\}$ then $j=k$ $\endgroup$ Nov 6, 2019 at 22:25
  • $\begingroup$ why does ω^k(mod n)=ω^j(mod n) imply that k(mod n) = j(mod n) ? $\endgroup$
    – user720013
    Nov 6, 2019 at 22:35
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    $\begingroup$ $\omega^a=\omega^b\implies\omega^{a-b}=1$ but since $\omega^n=1$ but $\omega^k\ne1$ for $0<k\le n$ we have $a\equiv b\mod n$ $\endgroup$ Nov 6, 2019 at 22:38
  • $\begingroup$ oh ok, it took me a bit to understand this as I haven't been taught it yet but it makes sense now $\endgroup$
    – user720013
    Nov 6, 2019 at 22:50

1 Answer 1

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A primitive $n$-th root of unity is an

$\omega \in \Bbb C \tag 1$

such that

$\omega^n = 1, \tag 2$

but

$\omega^k \ne 1, \; 1 \le k < n; \tag 3$

if

$\exists 1 \le j \ne l \le n, \; \omega^j = \omega^l, \tag 4$

we may without loss of generality assume

$j < l; \tag 5$

then from (4),

$\omega^{l - j} = 1; \tag 6$

but

$0 < l - j < n, \tag 7$

in contradiction to (3); thus (4) is false and the $\omega^k$, $1 \le k \le n$, are all distinct.

We note that each $\omega^k$, $1 \le k \le n$, is an $n$-th root of unity, since

$(\omega^k)^n = \omega^{kn} = \omega^{nk} = (\omega^n)^k = 1^k = 1. \tag 8$

In accord with (2) and (8), we see that each $\omega^k$ is a root of

$x^n - 1 \in \Bbb Q[x] \subsetneq \Bbb C[x]; \tag 9$

now since

$\deg (x^n - 1) = n, \tag{10}$

this polynomial has at most $n$ distinct roots; so since

$\vert \{\omega^1 = \omega, \omega^2, \ldots, \omega^{n - 1}, \omega^n = 1\} \vert = n, \tag{11}$

every $n$-th root of unity is in this set; there are no others.

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  • $\begingroup$ You have explained this really well thank you!! $\endgroup$
    – user720013
    Nov 7, 2019 at 15:11
  • $\begingroup$ You are most welcome, my friend. And thanks for the "acceptance"! Cheers! $\endgroup$ Nov 7, 2019 at 15:25

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