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?
Best Answer
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 $\deg f=2$.