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When do we use the $n^{th}$ primitive root of unity, $\zeta_{n}$, when trying to find roots of a polynomial?

For example, let $f=x^{3}-2\in \mathbb{Q}$[x]. The roots of $f$, are $\sqrt[3]{2},~\zeta_{3}\sqrt[3]{2},~\zeta_{3}^{2}\sqrt[3]{2}$. I would have only guessed $\sqrt[3]{2}$.

In general, given a field $\mathbb{F}$ and $f\in\mathbb{F}$[x], where, $f=x^{n}-a$, why do we sometimes use the primitive roots of unity?

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If $x^n-a=0$ and $\zeta_n$ is an $n^{th}$ root of unity, then $(\zeta_n^mx)^n-a=0$ as well, always. – Alex Becker Apr 28 '12 at 3:11
@AlexBecker, what is the superscript "m" you are using for zeta? – Edison Apr 28 '12 at 3:15
If you want to solve $x^3 - 2 = 0$, try the substitution $y = \sqrt[3]{2}$ and solve for $y$. You can always do this for equations of the form $x^n - a = 0 $ – Jonathan Apr 28 '12 at 3:18
@Edison An exponent. It's true for all powers of zeta. – Alex Becker Apr 28 '12 at 3:31
up vote 3 down vote accepted

In the real numbers, $x^n - a$ has either $0$, $1$, or $2$ roots (depending on whether $n$ is even and $a$ negative; $n$ is odd or $a=0$; or $n$ is even and $a$ is positive).

In the complex numbers, this polynomial is supposed to have $n$ roots in all cases, except when $a=0$ (in which case, we get the "repeated root" $0$, $n$ times). If $r$ and $s$ are roots, then $r^n = s^n$, so $(r/s)^n = 1$; that is, $r$ and $s$ differ by an $n$th root of unity. Conversely, if $\zeta_n$ is an $n$th root of unity and $rf$ is a root of $x^n-a$, then $(r\zeta_n)^n = r^n\zeta_n^n = r^n(1) = r^n = a$, so $r\zeta_n$ is also a root. So if $r$ is any particular root, and $\zeta$ is a primitive $n$th root of unity, then $r$, $r\zeta$, $r\zeta^2,\ldots,r\zeta^{n-1}$ are all distinct roots of the polynomial $x^n-a$.

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Thank you for the detailed comment. – Edison Apr 28 '12 at 4:44

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