 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/20g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The element

ω

satisfies

ω^{2} - ω + 12 = 0.

2

and

3

, this polynomial factors as

X (X - 1)

. Hence

(2) = p_{2} q_{2}, p_{2} = (2, ω), q_{2} = (2, ω - 1),

and

(3) = p_{3} q_{3}, p_{3} = (3, ω), q_{3} = (3, ω - 1),

where the respective ideal norms are

2, 2, 3, 3

Now

N (ω) = 12

, and

ω

lies in

p_{2}

and

p_{3}

but not in their conjugate primes. Similarly

N (2 + ω) = 18

, while

2 + ω

lies in

p_{2}

and

q_{3}

. Comparing norms gives

(ω) = p_{2}^{2} p_{3}, (2 + ω) = p_{2} q_{3}^{2} .

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