Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 18 1 iii Solution Created 2026-10-03 Updated 2026-10-14
We prove the reverse-inclusion well-quasi-ordering of monomial ideals, equivalently of all upward-closed subsets of , by induction on . For the sets are tails and the empty set; reverse inclusion orders their endpoints as , a well-order. The case , if needed for the induction, is just a finite family.
For an upset , let . These are upsets of and satisfy . They stabilize. Indeed the union has a finite minimal basis by Dickson's lemma: an infinite set of minimal elements would be an antichain in the componentwise well-quasi-ordering of . Each basis element appears in some slice, and after the largest of these finitely many indices the slice contains the entire upset .
Choose a stabilization index , and encode by the finite word and the final slice . The induction hypothesis makes slices a well-quasi-ordering by reverse inclusion. Higman's lemma, proved in Question 3 below, makes their finite words a well-quasi-ordering; finite product closure of well-quasi-orderings also compares the final slices. Thus in every infinite sequence of upsets there are earlier and later with a strictly increasing matching of the earlier pre-stabilization slices into the later ones, such thatSince and the -slices increase, for . For , . Therefore , completing the induction.
Now choose one finite support for each term of an arbitrary infinite sequence in the finitely generated incline. The result supplies with , and the preceding absorption calculation gives . Hence every finitely generated incline is well-quasi-ordered by the specified reverse absorption order. The proof allows all additional algebraic identifications among the generators.
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 18 1 ii Solution Created 2026-10-03 Updated 2026-10-14
Suppose the incline is generated by . Repeated distributivity, followed by commutativity and additive idempotence, expresses every element as a finite nonempty sumNo multiplicative identity is assumed: a zero exponent means that the corresponding generator is omitted, and every displayed monomial has at least one factor. Different supports may represent the same element; uniqueness is unnecessary.
Associate to its upward-closed exponent setIf , each monomial of is either equal to, or a multiplicative extension of, a monomial of . It is absorbed by that earlier monomial, and hence by the full sum . Adding the terms of one at a time givesFor equal exponent vectors the first absorption is additive idempotence; for strictly larger vectors it is the incline's multiplicative absorption. Thus the issue is to prove that exponent upsets are well-quasi-ordered by reverse inclusion, not merely to apply Dickson's lemma to individual exponents.