Let $f ∈ \mathbb{F}_p[x]$ be an irreducible polynomial. Show that $f$ splits into linear factors in $\mathbb{F}_{p^{\deg(f)}}$.

Question

As the title explains, I'm trying to answer the following question

Let $f ∈ \mathbb{F}_p[x]$ be an irreducible polynomial. Show that $f$ splits into linear factors in $\mathbb{F}_{p^{\deg(f)}}$.

This is the last part of a long question I'm working on which I've attached a screenshot of so you can see the other results I'm working with/see if any of them are helpful.

I can't see what to do so I'd appreciate any help you could offer.

Answer 1

Let $n=\deg(f)$. If $f(x)$ is irreducible of degree $n$, then $f(x)\mid x^{p^{n}}-x$. $\mathbb{F}_{p^{n}}$ is the splitting field over $\mathbb{F}_{p}$ of $x^{p^{n}}-x$, a separable polynomial (use the fact $\mathbb{F}_{p^{n}}^{\star}$ is a group and use Lagrange's theorem to see that), $x^{p^{n}}-x=\prod\limits_{a \in \mathbb{F}_p^{n}}(x-a)$. Hence $f(x)$ splits into distinct linear factors in $\mathbb{F}_{p_{n}}$.

One other way, maybe more appropriate with your sequence of questions : Let $\alpha \in \overline{\mathbb{F}_{p}}$ be a root of $f(x)$. Then $[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}]=n$. Note that $\mathbb{F}_{p^{n}}$ is $\mathbb{F}_p$ isomorphic to $\mathbb{F}_{p}(\alpha)$, so $\mathbb{F}_{p^{n}}$ contains a root of $f(x)$. But $\mathbb{F}_{p^{n}}$ is a Galois extension of $\mathbb{F}_{p}$, any irreducible polynomial in $\mathbb{F}_{p}$ having one root in...

Answer 2

Another approach:

The polynomial $f$ has no root in $\mathbf F_p$. Consider the field $\;\mathbf F_p[X]/(f(X))$: it is (isomorphic to) $\mathbf F_{p^{\mkern2mu\deg f}}$ and $\xi=X+(f(X))$ is a root of $f(X)$.

Furthermore; all images of $\xi$ by the Galois group $\operatorname{Gal}\bigl(\mathbf F_{p^{\mkern2mu\deg f}}/\mathbf F_p\bigr)$ are also roots. This Galois groups is cyclic, of order $\deg f$ and generated by the Frobenius morphism $x\mapsto x^p$. It is easy to see that all images of $\xi\,$: $\;\xi,\: \xi^p ,\:\xi^{p^2},\dots,\:\xi^{p^{\mkern 2mu\deg f -1}}$, are distinct, hence $f(X)$ splits as $$f(X)=C(X-\xi)(X-\xi^p)\bigl(X-\xi^{p^2}\bigr)\dots(X-\xi^{p^{\mkern 2mu\deg f-1}}\bigr),$$ where $C$ is the leading coefficient of $f(X)$.