Degree of a splitting field of an irreducible polynomial over $\mathbb{F}_p$

Question

Is it true that whenever $p$ is an odd prime, and $f$ an irreducible polynomial of degree $p$ in $\mathbb{F}_p$, then the splitting field of $f$, denoted $L$, satisfies $[L:\mathbb{F}_p] = p!$ ?

I know that since $L$ is a splitting field for $f$ over $\mathbb{F}_p$, $[L : \mathbb{F}_p] \leq (\text{deg}f)!$, but I'm not sure in what circumstances equality holds? Furthermore, what's the significance (if any) of the degree of $f$ being equal to $p$, the same as the characteristic of the field $\mathbb{F}_p$ over which it is irreducible?

Answer 1

If $L$ is a finite extension of $\mathbb{F}_p$, then $|L|=p^m$ and $L$ is a splitting field of $x^{p^m}-x$.

Therefore, if a polynomial $f(x)\in\mathbb{F}_p[x]$ has a root in $L$, it splits completely over $L$. Hence $\mathbb{F}_p[\alpha]$, where $\alpha$ is a root of the irreducible $f(x)\in\mathbb{F}_p[x]$, is already a splitting field for $f$ and $[\mathbb{F}_p[\alpha]:\mathbb{F}_p]=\deg f$.

The only cases in which we have $[\mathbb{F}_p[\alpha]:\mathbb{F}_p]=(\deg f)!$ are $\deg f=1$ or $\deg f=2$.

Answer 2

It never does (since $p> 2$). Given an algebraic closure $\overline{\Bbb F_p}$ and some $m\in\Bbb N$, there is only one subfield of $K\subseteq \overline{\Bbb F_p}$ such that $[K:\Bbb F_p]=m$, which we may unambiguously call $\Bbb F_{p^m}$. In your case ($p=m$, but it could be anything), any root of $\xi$ of $f$ must be contained in $\Bbb F_{p^p}$ and, conversely, $\Bbb F_p[\xi]=\Bbb F_{p^p}$.