Why algebraic closure of $\Bbb F_p$ is gained by adjoining primitive $n$-th ($n$ is coprime to $p$) root of unity to $\Bbb F_p$?

Question

Why algebraic closure of $\Bbb F_p$ is gained by adjoining primitive $n$-th ($n$ is coprime to $p$) root of unity to $\Bbb F_p$?

I know algebraic closure of $\Bbb F_p$ is union of $\Bbb F_{p^n}$ where $n$ runs $n≧1$.

But I don't know why the titled statement holds.

$\Bbb F_{p^n}＝\Bbb F_{p} (μ_{p^n-1})$, where $μ_{p^n-1}$ is $p^n$-th roof of unity, but I cannot proceed from here.

Thank you in advance.

Answer 1

We have that $\Bbb F_{p^n}\leq \Bbb F_{p^m}$ iff $n\mid m$. Then for any two such fields there is another field of the same type containing both. Hence, if we take the union of all $\Bbb F_{p^n}$ with $n\geq 1$, such an union has a natural field structure, let us call such an union $\Bbb F$. This is equivalent to the construction you're refering to.

Any element that is algebraic over $\Bbb F$ is algebraic over some $\Bbb F_{p^n}$ by construction. Hence, since the only finite splitting extensions of $\Bbb F_{p^n}$ are finite fields of the same type, we get that such element was already in $\Bbb F$.

Answer 2

Try this approach:

1. Every element of $\overline{\Bbb F_p}$ is in a finite extension of $\Bbb F_p$. Therefore:
2. Every element of $\overline{\Bbb F_p}$ is torsion.
3. No element of $\overline{\Bbb F_p}$ has precise order divisible by $p$.
4. If $\zeta\in \overline{\Bbb F_p}$ and $\zeta^n=1$, then $\zeta$ is in the group generated by a given primitive $n$-th root of unity $\zeta_n$, and therefore is in $\Bbb F_p(\zeta_n)$.

I think that should do it.