the splitting field of $x^3 - x + 1$ over $\mathbb{F}_3$ I am trying to figure out the splitting field of $x^3 - x + 1$ over $\mathbb{F}_3$ .
I know $x^3 - x + 1$ over $\mathbb{F}_3$ is irreducible because there are no root in  $\mathbb{F}_3$.
The spitting field will be the field $\mathbb{F}_3[x]/\langle x^3 -x +1\rangle$（my intuition is that all roots are added to $\mathbb{F}_3$.）
So,the spitting field is isomorphic to $\mathbb{F}_3^3$.

My question is ;
How to formally prove that the spitting field is the field
$\mathbb{F}_3[x]/\langle x^3 -x +1\rangle$

Thank you for your help.
 A: Fix an algebraic closure an take $\omega$ a root of $f = X^3-X+1$ there. Since the Frobenius map is an automorphism, you can check that $\omega + i$ with $i \in \Bbb F_3$ are also roots of $f$.
Hence the splitting field of $f$ is $\Bbb F_3 (\omega)$, of degree $\deg f = 3$ over $\Bbb F_3$, and thus by uniqueness we get $\Bbb F_3(\omega) = \Bbb F_{3^3}$. Now, consider the ring map
$$
\Bbb F_3[X] \to \Bbb F_3(\omega)
$$
sending $X$ to $\omega$.
The kernel of the map is $(f)$. There are a couple of ways to see this but one is to note that $(f) \subset \ker \mathbf{ev}_\omega$ is a maximal ideal, and the kernel a proper submodule since since $X \mapsto \omega \neq 0$. Its
image is $\Bbb F_3[\omega]$, the smallest $\Bbb F_3$-algebra containing $\omega$, hence $\Bbb F_3[X]/(f) \simeq \Bbb F_3[\omega]$.
It is a general fact that if $\alpha$ is algebraic over $k$, then $k(\alpha) = k[\alpha]$. Namely, by the same argument as before one gets $k(\alpha) \supset k[\alpha] = k[X]/(m(\alpha,k))$, which is a field by the irreducibility of the minimal polynomial. Since this field contains $\alpha$, by definition of $k(\alpha)$ we have $k(\alpha) \subset k[\alpha]$, hence the equality.
Also, there's nothing special about $3$ here, it may very well be substituted for any prime $p$ to obtain that the splitting field of $X^p-X+1$ is $\Bbb F_{p^p}$.
