abstract algebra - Help check my proof that the splitting field of irreducible $f(x)inBbb F_p[x]$ with degree $n$ is $Bbb F_{p^n}$

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?

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:

• That every field extension of degree $n$ of $\Bbb{F}_p$ is isomorphic to $\Bbb{F}_{p^n}$.


• That if two finite subfields of a given field (here $K$) are isomorphic, then they are the same. You quote this as an exercise.

Depending on the context you may or may not want to substantiate these claims.