Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 75 5 Solution Created 2026-10-03 Updated 2026-10-14
Let , where the bar denotes the effective projective image. This is the degree of the modular curve projection, rather than the full index when is not in .
Reduction modulo is onto . One proof uses the Chinese remainder theorem and elementary matrices: over , a unimodular first column has a unit entry, so elementary row operations reduce a determinant-one matrix to the identity. Each elementary matrix lifts integrally. Over , the first column has choices and the second has choices with prescribed determinant, giving . Each subsequent prime-power lift has choices, because the determinant-one linearized condition is one trace equation on four entries. HenceFor , the subgroup contains ; for it does not. ConsequentlyFor there are no effective elliptic stabilizer subgroups in the subgroup. Indeed, write . The determinant condition gives . A noncentral elliptic element in has trace or , none congruent to modulo for . Thus the projective action is torsion-free.
All cusp widths are . At infinity, lies in the effective subgroup exactly when is divisible by : the alternative can only occur for , and for it imposes the same condition. Every rational modular cusp is a modular translate of infinity, and the subgroup is normal, so the same width holds there. The sum of modular cusp analytic ramification indices equals , giving modular cusps.
The base is a sphere. For instance is a meromorphic function on it with a single simple pole at the modular cusp and no other poles, using the nonvanishing of proved above. It therefore defines a degree-one map to the Riemann sphere, making the base genus zero.
The only branch values of are the two elliptic orbits and the modular cusp. For , the numbers of points above them and their analytic ramification indices are respectively with index two, with index three, and with index . Applying the Riemann-Hurwitz formula directly,This is the elliptic and cusp ramification for principal level calculation. Level one is the identity projection. The complete result isFor example the genera for are . Keeping the exceptional projective index at level two is essential; applying the factor one-half there would give a nonintegral answer. This agrees with the principal congruence modular curve genus formula obtained from these branch counts.