I am given this problem in a Algebraic Equations class. I need to prove the following:
$$ \mathbb{F}_{p^n}\mid \mathbb{F}_{p^m} \text{ field extension} \Leftrightarrow m\mid n $$
I have proved the $\Rightarrow$ implication, but I am having trouble proving the other one. This is what I have done:
$ m|n \Rightarrow \exists a \in \mathbb{N}: n=ma $ ($a \in \mathbb{N}$ because $n,m>0$).
Besides, we know that $$\mathbb{F}_{p^n}=\{X^{p^n}-X \text{ polinomials' roots}\},\mathbb{F}_{p^m}=\{X^{p^m}-X \text{ polinomials' roots}\}. $$ Let $u\in \mathbb{F}_{p^m}$. So we have that $u^{p^m}=u$. If we evaluate $X^{p^n}-X=X^{p^{ma}}-X$ on u: $$u^{p^{ma}}-u $$ But I don't know how I should continue... I would be really thankful if someone helped me.
As you do have the explicit description of $\Bbb F_{p^n}$, we show that every root of $X^{p^m} - X$ is indeed a root of $X^{p^n} - X$.
To this end, let $\alpha$ be a root of the former. We have $$\alpha^{p^m} = \alpha.$$ Now raise both sides to the power $p^m$ to obtain $$\alpha^{p^{2m}} = \alpha^{p^m} = \alpha.$$ Continue repeatedly to get $$\alpha^{p^{km}} = \alpha$$ for all $k \in \Bbb N$. In particular, for $k = a = n/m$, the above is true. This gives us that $\alpha^{p^{n}} = \alpha$, as desired.