Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 2 1 i Solution Created 2026-10-03 Updated 2026-10-14
Put , and . The multiplication of matrix-coordinate triples is . In particular is central, , and every element has the unique normal form . ConsequentlyThe distinct normal-form monomials are the group basis, so this is an isomorphism, not merely a surjection from a presented algebra. This is the integral Heisenberg group algebra as a skew Laurent ring.
Here is the needed skew Hilbert basis theorem, including its proof. If is a left Noetherian ring and an automorphism, let be a left ideal of , where . DefineThese are left ideals of : adding lifts adds the coefficient, and left multiplication by multiplies its normalized value by . Multiplication of a lift by proves . The chain stabilizes at some . Choose finite generators for each , , and lifts with coefficient .
For of degree , write its normalized leading coefficient as . Thenhas smaller degree: the displayed summands have leading coefficients . For use the generators for without the shift. Induction on degree expresses every element of in terms of the finitely many chosen lifts. Thus is a left Noetherian ring. Passing to the opposite ring replaces by and proves the corresponding right-sided result.
This proof includes the ordinary Hilbert basis theorem when is the identity. Starting with the field , it makes Noetherian; localization at powers of makes Noetherian as well. Finally, for a left ideal in , its intersection with has finitely many generators. For each , some lies in , so is generated by those same elements in the skew Laurent polynomial ring. Right multiplication clears powers for a right ideal. Therefore is a left Noetherian ring and a right Noetherian ring.