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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?

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    I thought due to the Frobenius automorphism, already $\mathbb{F}_p[x]/\langle f \rangle$ would be the splitting field? Am I wrong? Do you have an example where the degree is equal to $p!$ and bigger than $p$ (i.e. $p \neq 2$)? – Dirk May 07 '19 at 10:50
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    See this question. – Dietrich Burde May 07 '19 at 10:57
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    Given one root $a$ of $f \in \Bbb{F}q[x]$ monic irreducible then $f(x) = \prod{m=1}^{\deg(f)} (x-a^{q^m})$ (proof : $ \Bbb{F}_q$ is the subfield of $\overline{ \Bbb{F}_q}$ fixed by the automorphism $c \mapsto c^q$) – reuns May 07 '19 at 11:03

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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, $\Bbb F_p[\xi]=\Bbb F_{p^p}$.

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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 $\deg f=2$.

egreg
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