If $f\in\mathbb{F}_p[x]$ is irreducible and has a root in $\mathbb{F}_{p^n}$, then $f$ splits over $\mathbb{F}_{p^n}$

Question

Let $f(X) \in \mathbb F_p[X]$ irreducible with $p$ prime and assume $\exists \alpha \in \mathbb F_{p^n}: f(\alpha) = 0$ where $n \geq 1$. I then have to prove that $f$ splits over $\mathbb F_{p^n}$.

Some general thoughts are that $\mathbb F_{p^n}$ is the splittingfield of $X^{p^n}-X \in \mathbb F_p[X]$. I can look at $\mathbb F_p(\alpha)$ which must be a subfield of $\mathbb F_{p^n}$ since $\alpha \in \mathbb F_{p^n}$. I further have that $f$ divides $X^{p^n}-X$ since $f$ is irreducible and $\alpha$ is a zero of both polynomials.

I am quite confused. Help please :)

Some more considerations: Assume $f(\beta) = 0$ Then with $f| X^{p^n}-X$ is get $f(X)g(X) = X^{p^n}-X$ so $\beta^{p^n}-\beta = 0$ s.t. $\beta \in \mathbb F_{p^n}$ and thus $f$ splits in $\mathbb F_{p^n}$ ?

Answer 1

You've done all but wrap the whole thing up:

$$f(x)\mid\left(x^{p^n}-x\right)\iff x^{p^n}-x=f(x)g(x)\implies$$

every root of $\,f\,$ is also a root of $\,x^{p^n}-x\,$ and thus...

Answer 2

Another way of reaching the conclusion would be to observe that if $\alpha$ is a root of $f(x)$, then so are $\alpha^p$, $\alpha^{p^2}$,$\ldots$, $\alpha^{p^{n-1}}$ and $\alpha^{p^n}=\alpha$. As $f(x)$ is irreducible and the product $$(x-\alpha)(x-\alpha^p)\cdots(x-\alpha^{p^{n-1}})$$ is in $\mathbb{F}_p[x]$, these must be all the roots of $f(x)$. Hence they are in the field $\mathbb{F}_p[\alpha]$.