Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 88 2 iii Solution Created 2026-10-03 Updated 2026-10-14
The dimension assumption supplies a nonzero . The projective index of the Gamma 0 congruence subgroup is twelve: reduction modulo eleven identifies its cosets with the twelve points of . There are two modular cusps, by the preceding classification. There are no effective elliptic fixed points. An order-two fixed point would require a root of modulo eleven, which does not exist; an order-three fixed point would require a root of , equivalent to a nontrivial cube root of unity in a group of order ten, which also does not exist.
The regular-cusp valence formula on a torsion-free modular curve therefore givesAll orders are nonnegative and both cusp orders are at least one. Thus has order exactly one at both modular cusps and has no zeros in the half-plane. One way to justify the degree count here is the genus formula for a modular curve: the same index, cusp and elliptic data give genus one. The invariant differential has cusp order , because is a nonzero constant times . Its total canonical degree therefore makes the sum of the two cusp-adjusted orders and the interior orders zero, giving precisely the count above.
Let . By the product formula it has no zeros in the half-plane; by the cusp-order table it has order twelve at each cusp. Its weight is twenty-four. Therefore is a weight-zero modular function, with neither zeros nor poles anywhere on the compactified modular curve. It is constant by the maximum modulus principle on a compact surface. Normalize so its leading Fourier coefficient at infinity is one. Since , that constant becomes one:This fixes the holomorphic twelfth root by its leading term and proves the weight-two discriminant root at level eleven is a nonzero member, hence a generator, of the given space. Taking an arbitrary pointwise twelfth root without this argument would not establish modularity: it could carry an unwanted character.