Let $f(x)\in\Bbb F_p[x]$ be irreducible with degree $n$. I come up with my proof that the splitting field of $f(x)$ over $\Bbb F_p$ is exactly $\Bbb F_{p^n}$.
Let $K$ be the splitting field of $f(x)$ over $\Bbb F_p$. Let all roots of $f(x)$ be $\alpha_1,~\alpha_2,~\cdots,~\alpha_n$. Since the minimal polynomial of $\alpha_1$ is $f(x)$, $[\Bbb F_p(\alpha_1):\Bbb F_p]=n$. And then $\Bbb F_p(\alpha_1)\cong \Bbb F_{p^n}$. Using the same arguements on $\alpha_2,~\cdots,~\alpha_n$, we get that $\Bbb F_p(\alpha_1)\cong \Bbb F_{p^n}\cong\Bbb F_p(\alpha_2)\cong\Bbb F_p(\alpha_3)\cong\cdots\Bbb F_p(\alpha_n)$. And notice that all $\Bbb F_p(\alpha_i)$'s are contained in $K$. Thus by a theorem in finite field (the introduction book of Hungerford had listed it as an exercise), $\Bbb F_p(\alpha_1)=\Bbb F_p(\alpha_2)=\cdots=\Bbb F_p(\alpha_n)$.
Therefore, $\Bbb F_p(\alpha_1)$ contains all of the roots of $f(x)$, and so $\Bbb F_p(\alpha_1)$ is the splitting field of $f(x)$ over $\Bbb F_p$ (i.e. $\Bbb F_p(\alpha_1)=K$). And $\Bbb F_p(\alpha_1)$ is isomorphic to $\Bbb F_{p^n}$. QED.
Is the proof correct?
Best Answer
Your proof is fine. You may want to qualify that $f$ is the minimal polynomial of $\alpha_i$ over $\Bbb{F}_p$.
Also, your proof implicitly uses the following two claims:
Depending on the context you may or may not want to substantiate these claims.