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

Question

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.

Answer 1

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 $\omega\in\Omega$.

Answer 2

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 $x^{q} - 1$.