Help check my proof that the splitting field of irreducible $f(x)\in\Bbb F_p[x]$ 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?
 A: 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.
A: See here for a related answer.
I will present here a proof using Galois extension, which is inspired from the answer I linked (maybe the user Uncountable thought of it too).
We start with a definition of Galois extension.

Claim: If $E/k$ is a finite extension of fields, the following
  statements are equivalent:
i) $E$ is a splitting field of some separable polynomial $f(x)\in k[x]$.
ii) Every irreducible $p(x)\in k[x]$ having one root in $E$ is separable and splits in $E[x]$.

A field extension is called a Galois extension if it satisfies any of the equivalent conditions above. For a proof, see Rotman's "Advanced Modern Algebra", page 223-224, Theorem 4.34. Now back to our proof.
Since $\mathbb{F}_{p^n}$ is a splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$, and $x^{p^n}-x$ is a separable polynomial, hence $\mathbb{F}_{p^n}/\mathbb{F}_p$ is a Galois extension. Identitying $\mathbb{F}_p(\alpha_1)$ with $\mathbb{F}_{p^n}$ (you have proved they are isomorphic), we conclude that $f(x)$ have one root in the Galois extension $\mathbb{F}_{p^n}$. Then $f(x)$ splits in $\mathbb{F}_{p^n}[x]$, the conclusion easily followed.
