= Solution
For $\alpha\ne0$, <Lagrange's theorem for finite groups>[Lagrange's theorem] in the multiplicative <group> of order $q^m-1$ gives $\alpha^{q^m-1}=1$; at zero the desired <identity element>[identity] is immediate. Thus all elements satisfy $\alpha^{q^m}-\alpha=0$.
For an irreducible <vertex degree>[degree]-$m$ <polynomial> $f\in F[X]$, the quotient $L=F[X]/(f)$ is a <field> of <dimension> $m$, hence has $q^m$ elements by part 1. Its element $\alpha=[X]$ therefore satisfies $\alpha^{q^m}-\alpha=0$. The evaluation homomorphism $F[X]\to L$ has kernel $(f)$, so
$$
\boxed{f(X)\mid X^{q^m}-X.}
$$
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