 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/20g/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

β

g (T) = T^{3} - 5 T^{2} + 11 T - 12,

and part (b) showed that

O_{K} = Z [β]

3

\overline{g} (T) = T (T^{2} + T - 1) .

The quadratic has no root in

F_{3}

, so it is irreducible. The Dedekind factorization theorem therefore gives

(3) = (3, β) (3, β^{2} + β - 1) .

The two factors are distinct prime ideal, with norms

3

and

3^{2} = 9

, respectively.

2

\overline{g} (T) = T (T^{2} + T + 1),

and the quadratic is irreducible over

F_{2}

. Hence

(2) = (2, β) (2, β^{2} + β + 1) .

This is a product of two distinct proper prime ideals, so

(2)

 New to topics? Read the docs here!