ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sat, 25 Nov 2017 21:54:03 +0100Uniqueness of reduced Groebner basis - proofhttps://ask.sagemath.org/question/39792/uniqueness-of-reduced-groebner-basis-proof/ Hi, can you please help me understand something.
I defined minimal GB like this:
Let G = {g1, . . . , gs} be a GB of an ideal I ⊂ k[x1, . . . , xn].
Then G is a **minimal GB** if and only if for each i = 1, . . . , s, the polynomial LC(gi) = 1 and its leading monomial LM(gi) does not divide LM(gj) for any j different than i.
and I showed that if:
G = {g1, . . . , gs} and H = {h1, . . . , ht} are two minimal GB for I then s = t and, after renumbering as necessary, LT(gi) = LT(hi) for i = 1, . . . , s.
Now I have a definition for reduced GB:
Let G = {g1, . . . , gs} be a GB of an ideal I ⊂ k[x1, . . . , xn].
Then G is a **reduced GB** if and only if for each i = 1, . . . , s LC(gi) = 1 and its leading monomial LM(gi) does not divide
any term of any gj for any j different then i.
Now i want to show **uniqueness** of reduced GB and the proof goes like this:
Suppose that {f1, . . . , fs} and {g1, . . . , gs} are both reduced and ordered so that LT(fi) = LT(gi) for each i.
Consider fi − gi ∈ I. If it is not zero, then its leading term must be a term that appeared in fi or in gi . In either case, this
contradicts the bases being reduced, so in fact fi = gi as claimed.
I don't understand why is there a contradiction with the bases being reduced.PetraSat, 25 Nov 2017 21:54:03 +0100https://ask.sagemath.org/question/39792/