Processing math: 100%

Tuesday, May 28, 2013

Alternate Proof of Hilbert's Basis Theorem and Gröbner Bases (9.6.43-44)

Dummit and Foote Abstract Algebra, section 9.6, exercises 43-44:

MathJax TeX Test Page 43. Let S={N | N not a finitely generated monomial ideal} and assume by way of contradiction that S.

(a) Prove S has a maximal element.
(b) Show there exist monomials x,yM such that xyM.
(c) Show there exists M0M 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 IF[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=NCN 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 niNinjNj:⇔NjNi, and observe NiM=(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,yF[x1,...,xn] such that x,yM and xyM. By 9.6.10, there exist monomial terms in x,y not divisible by any mi for M=(m1,...). Showing xy=mM 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 xyM for x composed of terms not divisible by monomials of M. Similarly, xyM. 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 mmmd=m so that, since d1, we have md=mm, i.e. mmmm.

Let x be the minimal monomial by not within M such that there exists another monomial y not within M fulfilling xyM. Revisit M=(m1,...) and let M=(m1,...,x) as before. Since MM 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 ninj for all ij. Assume xT: Then by 9.6.10 since xM we have nix so nid=x for some i, and since niMxM, yet also niMni<xni(dy)M with dyM since d<x as well, we inevitably violate the minimality of x, and we must instead have xT. Rewrite T={n1,...,ns1,x}.

Now let M0=(T{x}). Assume njT{x} yet njM: Since the elements of T are from M generated by {m1,...,x} we must have either minj for some i or xnj; the latter is excluded by the minimalization of T and the former implies njM. Therefore M0M, and evidently M0+(x)=M+(x)=M.

Show M=M0+(x)(M:(x)): () By the definition of (M:(x)), the direction is clear. () We have miMM0+(x), so that either nkmimiM0 or xmimi(x)(M:(x)).

Finally, we see that M0 is finitely generated, and also yM with y(M:(x)) so that (M:(x)) is finitely generated by the maximality of M, and also (x)(M:(x)) and now M0+(x)(M:(x))=M is finitely generated, a contradiction. 

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(gi) for gi elements of I, so that by proposition 24 these gi form a Gröbner basis and as well a finite generating set for the arbitrary ideal I, which is Hilbert's Basis Theorem. 

No comments:

Post a Comment