(a) Prove S has a maximal element.
(b) Show there exist monomials x,y∉M such that xy∈M.
(c) Show there exists M0⊆M with M0 a finitely generated monomial ideal such that M0+(x)=M+(x) and M=M0+(x)(M:(x)). Deduce M is finitely generated, forcing S=∅.
44. If ∅≠I⊆F[x1,..,xn] is an ideal, prove LT(I) is finitely generated. Conclude I has a Gröbner basis and deduce Hilbert's Basis Theorem.
Proof: (a) Let C be a typical chain totally ordered by inclusion. Since M=⋃N∈CN is a monomial ideal by 9.6.39, by Zorn's Lemma it suffices to show it is infinitely generated; assume the contrary, so that M=(n1,...,nr). Let ni be maximal with regard to the ordering ni∈Ni≥nj∈Nj:⇔Nj⊆Ni, and observe Ni⊆M=(n1,...,nr)⊆Ni so that Ni=M is finitely generated, a contradiction.
(b) By 9.6.42(a), the only prime monomial ideals are finitely generated. Therefore there exist polynomials x′,y′∈F[x1,...,xn] such that x′,y′∉M and x′y′∈M. By 9.6.10, there exist monomial terms in x′,y′ not divisible by any mi for M=(m1,...). Showing x′y′=m∈M and expanding x′ to subtract the products of y′ and monomial terms of x′ divisible by monomials of M (which are elements of M), we obtain xy′∈M for x composed of terms not divisible by monomials of M. Similarly, xy∈M. Now LT(xy)=LT(x)LT(y)∈LT(M)⊆M (the latter ⊆ being due again to 9.6.10) so that LT(x) and LT(y) are monomials not within M whose product is within M.
(c) Fix a monomial ordering ≤. First note that m∣m′⇒md=m′ so that, since d≥1, we have md=m′≥m, i.e. m∣m′⇒m≤m′.
Let x be the minimal monomial by ≤ not within M such that there exists another monomial y not within M fulfilling xy∈M. Revisit M=(m1,...) and let M∗=(m1,...,x) as before. Since M⊂M∗ by the maximality of M we must have M∗=(T) for some finite subset of monomials T={q1,...,qt}. Reformat this generating subset so that it is minimal, by excluding a generator n if M∗=(T∖{n}). This process is seen to terminate, and in part we are left with T={n1,...,ns} where ni∤ for all i≠j. Assume x∉T: Then by 9.6.10 since x∈M^* we have n_i \mid x so n_id=x for some i, and since n_i∈M⇒x∈M, yet also n_i∉M⇒n_i < x \land n_i(dy)∈M with dy∉M since d < x as well, we inevitably violate the minimality of x, and we must instead have x∈T. Rewrite T=\{n_1,...,n_{s-1},x\}.
Now let M_0=(T \setminus \{x\}). Assume n_j∈T \setminus \{x\} yet n_j∉M: Since the elements of T are from M^* generated by \{m_1,...,x\} we must have either m_i \mid n_j for some i or x \mid n_j; the latter is excluded by the minimalization of T and the former implies n_j∈M. Therefore M_0 \subseteq M, and evidently M_0+(x)=M+(x)=M^*.
Show M=M_0+(x)(M : (x)): (\supseteq) By the definition of (M : (x)), the direction is clear. (\subseteq) We have m_i∈M \subseteq M_0+(x), so that either n_k \mid m_i ⇒ m_i∈M_0 or x \mid m_i ⇒ m_i ∈ (x)(M : (x)).
Finally, we see that M_0 is finitely generated, and also y∉M with y∈(M : (x)) so that (M : (x)) is finitely generated by the maximality of M, and also (x)(M : (x)) and now M_0+(x)(M : (x))=M is finitely generated, a contradiction.~\square
44. Since LT(I) is a monomial ideal, by the previous it is finitely generated. In the same way as in 9.6.1, we may generate each of the elements of the finite generating set of LT(I) by a finite number of finite combinations of LT(g_i) for g_i elements of I, so that by proposition 24 these g_i form a Gröbner basis and as well a finite generating set for the arbitrary ideal I, which is Hilbert's Basis Theorem.~\square
No comments:
Post a Comment