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

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?

• 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 7 at 10:50
• See this question. – Dietrich Burde May 7 at 10:57
• 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 7 at 11:03

## 2 Answers

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 $$\deg f=2$$.

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

• A bit nitpicky, but for $p=2$ it is actually true, if only because $2!=2$. – Sebastian Schoennenbeck May 7 at 11:25
• @SebastianSchoennenbeck True, thank you. I have edited. – Saucy O'Path May 7 at 12:09