I am stuck on the part in the proof, where they say that $S(h_{i-1}'g_{i-1},h_{i}'g_{i})$ is just $S(g_{i-1},g_{i})$ multiplied by the monomial of multidegree $\alpha - \beta_{i-1,i}$.
One finds that $$\text{LCM}(\text{LT}(h_{i-1}'g_{i-1}),\text{LT}(h_{i}'g_{i})) = \text{LT}(h_{i}'g_{i}) = \text{LT}(h_{i-1}'g_{i-1})$$ (up to a constant) so that $$S(h_{i-1}'g_{i-1},h_{i}'g_{i}) = \frac{h_{i-1}'g_i}{b_{i-1}}-\frac{h_{i}'g_{i}}{b_i}$$ where $b_{i-1},b_{i} \in F$ are just the constants in front of the leading terms of $g_{i-1},g_{i}$. Now this means that $\frac{h_{i-1}'g_{i-1}}{b_{i-1}}$ and $\frac{h_{i}'g_{i}}{b_{i}}$ are monic polynomials of the same multidegree, hence their leading terms cancel out, if I am correct, which is precisely what $S(g_{i},g_{j})$ performs.
What I want to show is that $$\frac{h_{i-1}'g_{i-1}}{b_{i-1}}-\frac{h_{i}'g_{i}}{b_{i}} = k \cdot \Bigg(\frac{\text{LCM}(g_{i-1},g_{i})}{\text{LT}(g_{i-1})}g_i - \frac{\text{LCM}(g_{i-1},g_i)}{\text{LT}(g_i)}g_i\Bigg) \qquad (1)$$ where $k$ is the monomial of multidegree $\alpha-\beta_{i-1,i}$. Here, I suppose they mean that $k$ should be monic, otherwise I don´t understand why they say "the monomial" instead of "a monomial".
We have $\text{LCM}(g_{i-1},g_{i}) = \text{LT}(g_{i-1})r_{i-1} = \text{LT}(g_{i})r_{i}$ for monomials $r_{i-1},r_{i} \in R$ (since the $\text{LCM}(g_{i-1},g_{i})$ is a monomial) so that the expression within the parenthesis on the RHS above reduces to $$r_{i-1}g_{i-1}-r_{i}g_{i}.$$ Since $\text{LCM}(g_{i-1},g_{i})$ is monic, unless $g_{i-1},g_{i}$ are monic, we need the constant in front of $r_{i-1}$ and $r_{i}$ to be such that they are the inverse to the constant for $\text{LT}(g_{i-1})$ and $\text{LT}(g_{i})$, hence we can write $r_{i-1} = \frac{r_{i-1}'}{b_{i-1}}$ and $r_{i} = \frac{r_{i}'}{b_{i}}$ where $r_{i-1}'$ and $r_{i}'$ are monic.
The last step needed (from what I can see) is to show that $\partial(r_{i-1}') = \partial(h_{i-1}')$ and $\partial(r_{i}') = \partial(h_{i}')$, where $\partial(-)$ is the multidegree of the leading term.
We note that if $\partial(\text{LCM}(g_{i-1},g_{i})) = \beta_{i-1,i}$ then $\partial(r_{i-1}'g_{i-1}) = \beta_{i-1,i}$ and $\partial(r_{i}'g_{i}) = \beta_{i-1,i}$ so that if we multiply the expression with $k$ (as in $(1)$) we get $\partial\Big(\frac{kr_{i-1}'g_{i-1}}{b_{i-1}}\Big) = \alpha$ and $\partial\Big(\frac{kr_{i}'g_{i}}{b_{i}}\Big) = \alpha$.
Now this means that $\frac{h_{i-1}'}{b_i} = \frac{k r_{i-1}}{b_i}$ and $\frac{h_{i}'}{b_{i-1}} = \frac{k r_{i}}{b_i}$ and hence $(1)$ holds true, unless I am mistaken.
