 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-4/18h/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

K

gives the orbits of the Galois group on the three resolvent roots. If

Q

splits completely, the quotient is trivial and

G = V_{4}

. If

Q

is a linear factor times an irreducible quadratic, the quotient is

C_{2}

and

G = D_{8}

Q

is irreducible, the quotient acts transitively on three roots, so it is

C_{3}

S_{3}

. The product of their pairwise differences is a square root of the nonzero discriminant

Δ

and changes by the sign of the permutation. Thus

Δ

is a square in

K

exactly when the quotient contains only even permutations. The criterion is

factorization and discriminant of Q three linear factors linear times irreducible quadratic irreducible, Δ \in K^{\times 2} irreducible, Δ \in / K^{\times 2} G V_{4} D_{8} A_{4} S_{4}

The assumed containment of

V_{4}

is essential to reconstruct

G

from the quotient by this table.

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