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?

  • $\begingroup$ I cannot say whether it is correct, because I cannot guess exactly which theorem for finite fields you are using, and whether you really understand this theorem. If you want to solidify the last step of the proof, which is that the $\alpha_i$ all lie in the same field (not just in isomorphic ones), I would say something like: $\mathbb{F}_{p^n}$ is the splitting field of the (reducible) polynomial $x^{p^n}-x$. Each $\alpha_i$ is a root of this polynomial, hence lies in $\mathbb{F}_{p^n}$. Therefore $K\subseteq\mathbb{F}_{p^n}$. The other inclusion follows from $[K:\mathbb{F}_p]\geq n$. $\endgroup$ – Uncountable Apr 28 '19 at 15:59

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.

  • $\begingroup$ For the reasoning of "$f$ is the mini. poly. of $\alpha_i$ over $\Bbb F_p$", is it valid to just say that since $f(\alpha_i)=0$ and $f(x)$ is irreducible over $\Bbb F_p$? And for the second bullet you mentioned, actually I haven't think about the proof of it. Hmm.. is it a super nontrivial result (i.e. perhaps requires many technical detail in it)? $\endgroup$ – Eric Apr 28 '19 at 16:08
  • $\begingroup$ @Eric For the minimal polynomial; yes, that's fine. For the second point; you quote an exercise, but I don't know what the exercise is. I just took a guess at what it could be. $\endgroup$ – Servaes Apr 28 '19 at 16:25
  • $\begingroup$ Oh yes, this is exactly the exercise in Hungerford's text. I mean I haven't done this exercise yet. Does it require a nontrivial proof? $\endgroup$ – Eric Apr 28 '19 at 16:29
  • $\begingroup$ @Eric The canonical proof is easy to verify when you see it, but not easy to come up with. $\endgroup$ – Servaes Apr 28 '19 at 19:03

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.


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