Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 114 2 Solution 2026-09-28
Repeated Smith normal form puts a chain complex of finitely generated free abelian groups into its elementary decomposition of a finite free chain complex: one-term summands and two-term summands . Applying reverses a two-term summand but keeps the same multiplication by . Reading its homology gives the universal coefficient theorem for cohomologysplit noncanonically. Thus if with finite, then
The universal coefficient theorem for homology with coefficients givesIf the middle group vanishes for every prime , so does the tensor term. Any nonzero free summand survives for every , and any nonzero finite summand survives for a prime dividing its order. Since integral homology is finitely generated, it must vanish in every degree. This is detection of integral acyclicity modulo primes.
For the displayed complex, writeUsing , one obtainsThus is the mapping cone of in this sign convention. Its long exact sequence in homology shows that is an isomorphism exactly when . If is an isomorphism with coefficients for every prime, then is acyclic for every . Prime-field detection makes integrally acyclic, so the integral long exact sequence gives