Past exam of the rings and modules course of the University of Oxford 2023 2 b i Solution 2026-10-25
For , let have and for . If and , then can be nonzero only where or is nonzero, and can be nonzero only where is nonzero. These are finite sets. Since , this proves that is a submodule.
Every has the finite expressionEvaluating at coordinate recovers the coefficient of , so the expression is unique. Thus is a basis of a module for , andis a free module. By contrast, the ambient module of integer sequences allows arbitrary integer sequences.