Past exam of the number theory course of the University of Oxford / 2009 / A / 2 / c / iii / Solution

The nonzero cubes modulo thirteen are $\{1,5,8,12\}$, which excludes three. Thus this cubic has no root in its base field and is irreducible. For a root $\alpha$, the extension ${\mathbb{F}}_{13}(\alpha )$ has degree three. The base field already contains the primitive cube root of unity $3$, since $3^3=1$ and $3\ne 1$. Hence the other two roots are $3\alpha$ and $9\alpha$, already in this extension, making it the full splitting field.

The Frobenius endomorphism satisfies ${\alpha }^{13}=\alpha ({\alpha }^3{)}^4=3\alpha$, so it cycles all three roots. Therefore