Consider the polynomial $g(x)=x^4+x^3+x^2+x+1 \in \mathbb{F}_3[x]$. It's possible to show that $g$ is irreducible in $\mathbb{F}_3$. If we let $\alpha$ be a root of $g$, then $\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$, and it can be shown from this that $\alpha^5=1$. There's an interesting result that the relation $g(\alpha)=0$ gives us. If $K$ is the splitting field of $g$, then $5 \big| p^m-1$, where $m$ is the degree of the splitting field over $\mathbb{F}_3$. This comes from Lagrange's theorem, since, if $|K|=p^m$, then $K^*$ is a group under multiplicaition of order $p^m-1$. One can pretty easily check that if $|K|\ge p^4$, since $5$ divides neither $3^2-1$ nor $3^3-1$.
My question is, if we've found the smallest field in which $g$ could split, whether it necessarily does split. That is, given the relation that $\alpha^5=1$, and that $5\big|3^4-1$, whether that is sufficient to assure $\mathbb{F}_{3^4}$ is the splitting field of $g$, or whether the only way to prove $\mathbb{F}_3[x]/(g(x))$ splits $g$ is to actually find the roots of $g$ in terms of $\alpha$.
I think that $\alpha^2$ should also be a root of $g$, but finding the other roots are a little tricky. I believe I've seen it before elsewhere that the splitting field of $g$ is $\mathbb{F}_{3^4}$, and so $g$ should split completely in $\mathbb{F}_3[x]/(g(x))$, but I was hoping for a more eloquent way of proving this than blindly guessing what its roots are.
There is a calculation-free proof that "the smallest field over which $g$ could split" is indeed the splitting field. The key lemma is that a finite field has a unique extension of cardinality $n$ for any $n$.
Indeed, let $F$ be a finite field of cardinality $q$ and let $K$ be a degree $n$ extension of $F$. The cardinality of $K^\times$ is $q^n - 1$, and so every element $K^\times$ satisfies the polynomial $x^{q^n} - x$. This polynomial is separable (the formal derivative is $-1$), so it has $q^n$ distinct roots in $K$ -- but the cardinality of $K$ is $q^n$, so its roots are exactly the elements of $K$. That means that this polynomial splits in $K$ and over no subfield of $K$, and therefore $K$ is the splitting field of this polynomial, which makes it unique up to isomorphism.