Clarification about a proof regarding a sum of polynomials being expressed as a linear combination of S-polynomials 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. 
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.
 A: $1)$ The lemma asserts that we have
\begin{equation*}
\sum_{i=1}^sp_i = \sum_{1 \leq j,l \leq s}a_{jl}S(p_j,p_l)
\end{equation*}
for some coefficients $a_{jl} \in k$. Then the proof tells us that something even stronger is true, namely we can take all of the coefficients $a_{jl}$ with $1 \leq l \leq s-1$ to be $0$.
$2)$ This is a little bit more subtle. Let $p_1,\dots,p_s \in k[x_1,\dots,x_n]$ be polynomials such that
\begin{equation*}
\max(\text{multideg}(p_i)) = \delta \quad\text{and}\quad \text{multideg}\left(\sum_{i=1}^sp_i\right) < \delta.
\end{equation*}
Without loss of generality, suppose that the $p_i$ are ordered so that $p_1,\dots,p_r$ are of multidegree $\delta$ while $p_{r+1},\dots,p_s$ are all of multidegree less than $\delta$. It follows that $\text{multideg}(\sum_{i=1}^rp_i) < \delta$, but now all the polynomials involved are of multidegree $\delta$, so we can apply the lemma to conclude that $\sum_{i=1}^rp_i$ is a $k$-linear combination of the $S$-polynomials $S(p_j,p_l)$ for $1 \leq j,l \leq r$. Unfortunately, this is pretty much all we can say in general. Indeed, it may happen that
\begin{equation*}
\text{multideg}\left(\sum_{i=r+1}^sp_i\right) = \max_{r+1 \leq i \leq s}(\text{multideg}(p_i)),
\end{equation*}
in which case we can't apply the lemma to write $\sum_{i=r+1}^sp_i$ as a $k$-linear combination of the $S$-polynomials $S(p_j,p_l)$ for $r+1 \leq j,l \leq s$. Of course, this observation alone is not enough to conclude that the lemma fails in general, but it certainly suggests so.
Building on the observation above, here is an example illustrating how the lemma can fail when the $p_i$ have different multidegrees. Consider the three polynomials
\begin{equation*}
p_1 = x^2+xy,\quad p_2 = -x^2+y^2 \quad\text{and}\quad p_3 = x.
\end{equation*}
Computing the $S$-polynomials (either using the lex or the grlex order), we get
\begin{equation*}
S(p_1,p_2) = xy+y^2,\quad S(p_1,p_3) = xy \quad\text{and}\quad S(p_2,p_3) = -y^2.
\end{equation*}
As expected, $p_1+p_2 = S(p_1,p_2)$. However, $p_1+p_2+p_3$ cannot be written as a linear combination of $S(p_1,p_2)$, $S(p_1,p_3)$ and $S(p_2,p_3)$. This is because the subspace
\begin{equation*}
\text{span}\{S(p_1,p_2),S(p_1,p_3),S(p_2,p_3)\}
\end{equation*}
is generated by $xy$ and $y^2$, so it does not contain $p_3$. In fact, even the ideal generated by $S(p_1,p_2)$, $S(p_1,p_3)$ and $S(p_2,p_3)$ does not contain $p_3$, since all its nonzero elements have total degree at least $2$.
