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?

  • 3
    $\begingroup$ 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. $\endgroup$ – Alex Becker Apr 28 '12 at 3:11
  • $\begingroup$ @AlexBecker, what is the superscript "m" you are using for zeta? $\endgroup$ – Edison Apr 28 '12 at 3:15
  • $\begingroup$ 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 $ $\endgroup$ – Jonathan Apr 28 '12 at 3:18
  • $\begingroup$ @Edison An exponent. It's true for all powers of zeta. $\endgroup$ – Alex Becker Apr 28 '12 at 3:31

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 $x^n-a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.