Uniqueness of reduced Groebner basis - proof

asked 2017-11-25 14:54:03 -0600

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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.

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Comments

This question seems more adapted for math.stackexchange.com than for ask.sagemath.org.

slelievre gravatar imageslelievre ( 2017-11-25 20:45:31 -0600 )edit