Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 16I / b

Over ${\mathbb{F}}_3$,Both quadratics have discriminant $2$, a nonsquare in ${\mathbb{F}}_3$, so are irreducible polynomials. Both split in the same quadratic finite field ${\mathbb{F}}_9$. Consequently $\boxed{\mathrm{Gal}(t^4+1/{\mathbb{F}}_3)\cong C_2}$. Its generator is the finite-field Frobenius automorphism $x\mapsto x^3$, acting as a product of two transpositions on the four roots.

The polynomial $t^3-t-2$ has values $3,3,4,2,3$ at $0,1,2,3,4$ respectively, modulo $5$. Thus $t^3-t-2$ has no root and is an irreducible polynomial over ${\mathbb{F}}_5$. One root generates ${\mathbb{F}}_{125}$; all its Frobenius conjugates are there, so this is its splitting field. Hence $\boxed{\mathrm{Gal}(t^3-t-2/{\mathbb{F}}_5)\cong C_3}$, generated by $x\mapsto x^5$, a 3-cycle on the roots.

Over ${\mathbb{F}}_7$, the polynomial $t^4-1$ factors as $(t-1)(t+1)(t^2+1)$. Since $-1$ is not a square modulo $7$, the last factor is an irreducible polynomial. The splitting field is ${\mathbb{F}}_{49}$, so $\boxed{\mathrm{Gal}(t^4-1/{\mathbb{F}}_7)\cong C_2}$. The finite-field Frobenius automorphism fixes $1,-1$ and swaps the other two roots.