 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-24/3/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

The geometric sum gives

\frac{1 - ζ _{p}^{i}}{1 - ζ _{p}} = 1 + ζ_{p} + \dots + ζ_{p}^{i - 1} \equiv i (mod π_{K}) .

Since

Φ_{p} (1) = p

, its factorization at the nonidentity

p

p = \prod_{i = 1}^{p - 1} (1 - ζ_{p}^{i}) = π_{K}^{p - 1} w, w = \prod_{i = 1}^{p - 1} \frac{1 - ζ _{p}^{i}}{1 - ζ _{p}} .

Each factor is a unit, and Wilson theorem gives

w \equiv (p - 1)! \equiv - 1 (mod π_{K})

u = - w^{- 1}

. Then

u

π_{K}^{p - 1} = - p u, u \in 1 + π_{K} O_{K} .

The minus sign comes from the product of the nonzero elements of the residue field, not from an arbitrary choice of uniformizer.

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