Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 18J (Galois Theory)

For a monic polynomial with roots $r_1,\dots ,r_n$, its discriminant isHere $\mathrm{Disc}(g)=(u^3-v^3{)}^2$. Both $u^3-v^3$ and the Vandermonde product in the ${\alpha }_i$ are alternating homogeneous cubics. Evaluation at $({\alpha }_1,{\alpha }_2,{\alpha }_3)=(0,1,-1)$ fixes the constant and gives

Using ${\alpha }_1+{\alpha }_2+{\alpha }_3=0$, ${\sum }_{i<j}{\alpha }_i{\alpha }_j=a$, and ${\alpha }_1{\alpha }_2{\alpha }_3=-b$, direct expansion givesConsequentlyandThus

Solve the quadratic resolvent:Choose cube roots with $uv=-3a$. Inverting the discrete Fourier transform givesThese expressions use only square and cube roots. A general cubic is first translated to remove its quadratic term, proving the existence of a formula by radicals.

For an irreducible cubic over $\mathbb{Q}$, the Galois group is $A_3$ when its discriminant is a rational square and $S_3$ otherwise. The rational-root test shows that the first two cubics are irreducible.

For $X^3-21X-22$,is not a square, so the Galois group is $S_3$.

For $X^3-21X-28$,so the Galois group is $A_3\cong C_3$.

Finally,whose remaining roots are $1\pm 3\sqrt{2}$. Its splitting field is $\mathbb{Q}(\sqrt{2})$ and its Galois group is $C_2$.