I'm reading this proof from Ideals, Varieties, and Algorithms by David A. Cox, Donal O'Shea, and John Little. You can find an online version here.

This is Lemma 5 of Chapter 2, page 85. enter image description here

From my understanding, what it is saying is that if you have some polynomials with the same multidegree and you sum them together and get a polynomial with a smaller multidegree, then this sum can be expressed as some linear combination of S-polynomials. The 'cancellation' of terms of the highest multidegree in a sense happens in the S-polynomial. Correct me if I'm wrong.

I don't have a problem with the proof per se, but I have some questions that need clarification:

1) The proof seems to imply that the second S-polynomial has to be fixed in the summation. In $\sum_{i=0}^{s-1}d_iS(p_i,p_s)$ on the last line of page 85, the summation is summed over $i$, so the $p_s$ term is unvaried. So when the proof says expressed as a linear combination of S-polynomials, do they mean any S-polynomial generated by $2$ completely different S-polynomials, or does one of the polynomials generating the S-polynomial has to stay constant? If the former case is true, how can we prove it?

2) As an extension: would the following lemma still apply if the polynomials, $p_i$ have different multidegrees? I feel like it should, but I have no idea how to start the proof.

  • $\begingroup$ What is an $S$-polynomial? $\endgroup$ – Servaes Apr 12 at 15:29
  • $\begingroup$ An S-polynomial is defined here: maths.qmul.ac.uk/~whitty/LSBU/MathsStudyGroup/Buchberger.pdf or you can look inside the book given in my post. The intuition is that the leading monomials inside 2 given polynomials will cancel out. Hope it helps. Unfortunately, the actual definition is quite hard to write out in this comment without reference to prior terminology. $\endgroup$ – Yip Jung Hon Apr 12 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.