Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-29/3/i/solution
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 29 3 i Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Here is the usual one-dimensional commutative formal group law. Normalize the valuation by , and put . We show explicitly that for any integer ,This subgroup will have finite index in .
Set . Since terms of total degree at least two, . Its inverse has integral coefficients, say with . Define the formal logarithm over byDifferentiate associativity with respect to at zero. This gives , soEvaluating at proves the formal identity .
The coefficients satisfy . Thus the series converges on , since for the valuations of its terms are at least . More quantitatively,Indeed, if then . If , every nonlinear term has valuation strictly greater than . Write . For , use to obtainThe extra valuation is integral and positive because for all .
For any , solve by iterating . This map preserves the complete ball and improves differences by at least one valuation step. The contraction mapping theorem gives a unique solution. Therefore is a bijection of this ball onto itself, and the formal identity, now evaluated in convergent series, makes it a group isomorphism. Multiplication by then identifies the additive target with .
Integral coefficients in and its formal inverse ensure that every is a subgroup. Reduction modulo respects the group law; two parameters represent the same coset of exactly when they are congruent modulo . If the residue field has elements, the quotient consequently has elements. This proves both the finite-index assertion and the claimed additive structure, with an explicit depth that also works in ramified extensions and when .
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