Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 25 5 Solution Created 2026-10-03 Updated 2026-10-14
Fix a positive integer and an integer coprime to . We use Dirichlet characters modulo , extended by zero on integers not coprime to , and write for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums are therefore . Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue .
The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus . If a nonreal Dirichlet character had , its distinct conjugate would also have a zero there, since . These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that as , contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves for every nonprincipal Dirichlet character.
The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since . For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value ; the Euler-product branch differs from it by a fixed integral multiple of on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as , so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo contains infinitely many primes. The argument also covers and , when there need not be a nonreal character.
For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any , choose a fixed so large that the relative error in is at most for . The initial integral is a constant, while the remaining integral lies betweenThe boundary term is . Divide by , let , and then let . We obtainOnly a relative asymptotic is deduced from the given information; an additive error, or uniformity for a growing modulus , would require additional estimates.