Let $p$ and $q$ be odd primes. Let $\Omega$ be the algebraic closure of $\Bbb F_p$. Let $\omega$ be a primitive $q$-th root of unity. Show that $\omega \in \Omega$.

How do I show that? Please help me in this regard.

Thank you very much.

  • 1
    $\begingroup$ This is most likely a misstatement of the question "Prove that there exists a primitive $q$-th root of unity in $\Omega$ if $q \neq p$." Also, the oddness of $p$ and $q$ is utterly irrelevant. $\endgroup$ – darij grinberg Feb 8 at 16:58

By definition $\omega$ is a root of $X^q-1\in\Bbb{F}_p[X]$, and by definition every polynomial in $\Bbb{F}_p[X]$ splits into linear factors in $\Omega[X]$. Hence $X-\omega\in\Omega[X]$ and so $\omega\in\Omega$.

  • $\begingroup$ I have a similar kind of argument. But my instructor said that it doesn't work. Rather he suggested me to apply Fermat's little theorem. Though I don't really find any use of FLT here. Is it of any importance here @Servaes? $\endgroup$ – Dbchatto67 Feb 8 at 16:13
  • $\begingroup$ Why does it not work? Are you using different definitions of primitive root or algebraic closure? I do not see how FLT is relevant here. $\endgroup$ – Servaes Feb 8 at 16:17
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    $\begingroup$ No. I have used the same keywords what you have mentioned to explain this. But I don't really know why did he suggest me to apply Fermat's little theorem here? It follows directly from the definition of the algebraic closure. Why do I use weapons when I can able to fight by hand? It's completely nonsense. $\endgroup$ – Dbchatto67 Feb 8 at 16:19
  • $\begingroup$ Whatever! Thank you so much for your patience @Servaes. $\endgroup$ – Dbchatto67 Feb 8 at 16:26
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    $\begingroup$ By your argument $\Omega$ contains a root of $X^p-1$ but is it primitive? That is, $1$ is a root of any $X^n-1$ but are there other roots, and are they primitive? That is, $X^p-1 = (X-1)^p$ and so there are no primitive $p$-th roots of unity in $\Bbb{F}_p$. Where did you use that $q$ is prime to $p$? $\endgroup$ – Somos Feb 8 at 16:54

As noted in the comment of darij grinberg, all you can do is to prove that a primitive root of $x^{q} - 1$ is in $\Omega$, not a specific one. And then you need $p \ne q$, as elucidated in other comments and answers.

Now the point is that $(x^{q} - 1)' = q x^{q-1} \ne 0$ in $\mathbb{F}_{p}[x]$, as $p \ne q$. Therefore $$\gcd(x^{q} - 1, (x^{q} - 1)') = \gcd(x^{q} - 1, q x^{q-1}) = 1,$$ and thus $x^{q} - 1$ has $q > 1$ distinct roots in any of its splitting field. Any such root $\omega \ne 1$ will be a primitive $q$-th root of unity, as $q$ is prime. And then of course $\Omega$ contains such a splitting field of $x^{q} - 1$.


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