Find the splitting field of a polynomial

Question

The extension $\mathbb{Z}_p \leq \mathbb{F}_{p^n}$ is normal, as the splitting field of the polynomial $f(x)=x^{p^n}-x$ ($\mathbb{Z}_p$ is a perfect field therefore each polynomial is separable).

So, if $a \in \mathbb{F}_{p^n}$, then $q(x)=Irr(a,\mathbb{Z}_p)$ can be splitted over $\mathbb{F}_{p^n}$ (since all the roots are in $ \mathbb{F}_{p^n}$).

How can I find the splitting of $q(x)$ as an expression of powers of $a$??

We have that $q(x)=Irr(a, \mathbb{Z}_p)$, that means that $q(a)=0$.

Since $a \in \mathbb{F}_{p^n}$ it stands that $a^{p^n}=a$.

Therefore, $q(a^{p^n})=0$.

How could I continue??

Answer 1

If $\;a\in\Bbb F_{p^n}\;$ , then $\;\Bbb F_p(a)\subset\Bbb F_{p^n}\;$ , and from here that $\;\Bbb F_p(a)\cong\Bbb F_{p^m}\;$ , with $\;m\mid n\;$ since $\;\Bbb F_{p^m}\;$ subfield of $\;\Bbb F_{p^n}\iff m\mid n\;$ .

As you mentioned, $\;a^{p^m}=a\;$ , but for this I'm not sure what else could "express with powers of $\;a\;$ the splitting field of $\;q(x)\;$" mean .

Answer 2

Are you referring to the fact that if $f(a)=0$, where $f\in\mathbb F_p[x]$, then also $0=(f(a))^p=f(a^p)$? This implies that $q(x)=(x-a)(x-a^p)(x-a^{p^2})\cdots$