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Grobner Bases - Second question on D&F Proposition 24

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote Section 9.6 Polynomials in Several Variables Over a Field and Grobner Bases.

I have a second problem (see previous post for first problem) understanding a step in the proof of Proposition 24, Page 322 of D&F

Proposition 24 reads as follows:

Proposition 24. Fix a monomial ordering on [TEX] R= F[x_1, ... , x_n] [/TEX] and let I be a non-zero ideal in R

(1) If [TEX] g_1, ... , g_m [/TEX] are any elements of I such that [TEX] LT(I) = (LT(g_1), ... ... LT(g_m) )[/TEX]

then [TEX] \{ g_1, ... , g_m \} [/TEX] is a Grobner Basis for I

(2) The ideal I has a Grobner Basis


The proof of Proposition 24 begins as follows:


Proof: Suppose [TEX] g_1, ... , g_m \in I [/TEX] with [TEX] LT(I) = (LT(g_1), ... ... LT(g_m) ) [/TEX] .

We need to see that [TEX] g_1, ... , g_m [/TEX] generate the ideal I.

If [TEX] f \in I [/TEX] use general polynomial division to write [TEX] f = \sum q_i g_i + r [/TEX] where no non-zero term in the remainder r is divisible by any [TEX] LT(g_i) [/TEX]

Since [TEX] f \in I [/TEX], also [TEX] r \in I [/TEX], which means LT9r) is in LT(I).

But then LT(r) would be divisible by one of [TEX] LT(g_1), ... ... LT(g_m) [/TEX], which is a contradiction unless r= 0 ... ... etc etc

... ... ...

My problem is with the last (bold) statement - as follows:

We have that [TEX] LT(I) = (LT(g_1), ... ... LT(g_m) )[/TEX]

Now since [TEX] r \in LT(I) [/TEX], then we have that r is a finite sum of the form

[TEX] r = r_1 LT(g_1) + r_2 LT(g_2) + ... ... + r_m LT(g_m) [/TEX] ... ... ... (*)

where [TEX] LT(g_i) \in I [/TEX] and [TEX] r_i \in R [/TEX]

But surely (*) is NOT guaranteed to be divisible by [TEX] LT(g_i) [/TEX] for some i

Can someone please clarify this issue?

Peter

[This has also been posted on MHF]