In the proof of proposition 13.2.8 of Ireland and rosen they consider some $m$ and $m_0$ such that $m=2m_0$ and $m_0$ is odd. They then state that a primitive $m_0$th root of unity is a primitive $m$th root of unity. I don't see this.

  • $\begingroup$ Use the example of $m=3$. If $\omega$ is a primitive cube root of $1$, then $-\omega$ is a primitive sixth root. Maybe that’s what they meant? $\endgroup$
    – Lubin
    Feb 6, 2014 at 4:00

1 Answer 1


That's not quite what it says.

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Since $m_0$ is odd, $$(-\zeta_{m_0})^{m_0}=(-1)^{m_0}\zeta_{m_0}^{m_0}=(-1)\cdot 1=-1$$ and thus $$(-\zeta_{m_0})^{m}=(-\zeta_{m_0})^{2m_0}=1.$$

To see that in fact $-\zeta_{m_0}$ is a primitive $m$th root of unity, we can proceed as follows:

  • Of course $\zeta_{m_0}^k\neq 1$ for any $1\leq k\leq m_0-1$, and since $m_0$ is odd, it is clear that $\zeta_{m_0}^k\neq -1$ for any $1\leq k\leq m_0-1$, so that $$(-\zeta_{m_0})^k=(-1)^k\zeta_{m_0}^k=\pm\zeta_{m_0}^k\neq 1$$ for any $1\leq k\leq m_0-1$.

  • We already showed above that $(-\zeta_{m_0})^{m_0}=-1\neq 1$.

  • For any $m_0+1\leq k\leq 2m_0-1$, we have $$(-\zeta_{m_0})^k=(-\zeta_{m_0})^{m_0}(-\zeta_{m_0})^{\ell}=(-1)(-\zeta_{m_0})^{\ell}=\pm\zeta_{m_0}^{\ell}\neq 1$$ for $1\leq\ell=k-m_0\leq m_0-1$ (by the first case).

  • 1
    $\begingroup$ Always good to go back to the source! $\endgroup$
    – Lubin
    Feb 6, 2014 at 4:01
  • $\begingroup$ thanks, as you pointed out i misread it-didn't make out the -1 before $\zeta$ $\endgroup$
    – animalcroc
    Feb 6, 2014 at 14:40

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