Grobner Bases - Second question on D&F Proposition 24

[Math Amateur]

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 $$R= F[x_1, ... , x_n]$$ and let I be a non-zero ideal in R

(1) If $$g_1, ... , g_m$$ are any elements of I such that $$LT(I) = (LT(g_1), ... ... LT(g_m) )$$

then $$\{ g_1, ... , g_m \}$$ is a Grobner Basis for I

(2) The ideal I has a Grobner BasisThe proof of Proposition 24 begins as follows:Proof: Suppose $$g_1, ... , g_m \in I$$ with $$LT(I) = (LT(g_1), ... ... LT(g_m) )$$ .

We need to see that $$g_1, ... , g_m$$ generate the ideal I.

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

Since $$f \in I$$, also $$r \in I$$, which means LT9r) is in LT(I).

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

... ... ...

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

We have that $$LT(I) = (LT(g_1), ... ... LT(g_m) )$$

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

$$r = r_1 LT(g_1) + r_2 LT(g_2) + ... ... + r_m LT(g_m)$$ ... ... ... (*)

where $$LT(g_i) \in I$$ and $$r_i \in R$$

But surely (*) is NOT guaranteed to be divisible by $$LT(g_i)$$ for some i

Can someone please clarify this issue?

Peter

[This has also been posted on MHF]

 

[jvicens]

Dear Peter,

Thank you for bringing this issue to our attention. After reviewing the proof of Proposition 24 in Dummit and Foote, we have identified where the confusion lies.

The statement in question, "But then LT(r) would be divisible by one of LT(g_1), ... ... LT(g_m), which is a contradiction unless r=0," is not entirely accurate. The correct statement should be, "But then LT(r) would be divisible by one of LT(g_1), ... ... LT(g_m), which contradicts our assumption that no non-zero term in the remainder r is divisible by any LT(g_i)."

In other words, the contradiction arises from the fact that LT(r) is both in LT(I) and not divisible by any LT(g_i), which goes against our initial assumption.

We apologize for any confusion this may have caused and hope this clarifies the issue for you. If you have any further questions or concerns, please do not hesitate to reach out.