# Prove that a primitive $q$-th root of unity is in the algebraic closure of $\Bbb F_p$

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 $$\omega \in \Omega$$.

Thank you very much.

• 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. – 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 $$\omega\in\Omega$$.
• 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 $p$? – 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 $$x^{q} - 1$$.