 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-5/3/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Φ (P_{i}) = P_{i + 1}

d (P_{i}) = dim_{F_{p}} P_{i} / P_{i + 1} = lo g_{p} [P_{i} : P_{i + 1}] .

G

is uniform, all these indices equal

[G : P_{2}] = p^{d (G)}

. Hence $d (P_{i}) = d (G)$ for every $i \geq 1$ , proving (i)

\Rightarrow

(ii).

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