Let $f_1,\dots, f_s\in k[x_1,\dots, x_n]$ and $LM(f_i)=x^{\alpha(i)}$ the leading monomial of $f_i$. Define
\begin{align*} &Δ_1 = α(1) + \mathbb{Z}^n_{≥0},\\ &Δ_2 = (α(2) + \mathbb{Z}^n_{≥0}) \setminus Δ_1,\\ &\;\vdots\\ &Δ_s = (α(s) + \mathbb{Z}^n_{≥0}) \setminus \left(\bigcup^{s−1}_{i=1} Δ_i \right) ,\\ &\overline{Δ} = \mathbb{Z}^n_{≥0} \setminus \left(\bigcup^{s}_{i=1} Δ_i \right) . \end{align*}
I've been asked several questions about the division algorithm for multivariate polynomials described in Ideals, Varieties, and Algorithms using this construction. One of them, which I've solved:
(c) Show that in the expression $f = q_1 f_1 + \dots + q_s f_s + r$ computed by the division algorithm, for every $i$, every monomial $x^β$ in $q_i$ satisfies $β + α(i) ∈ Δ_i$, and every monomial $x^γ$ in $r$ satisfies $γ ∈ \overline{Δ}$.
But then, there's the following:
Show that there is exactly one expression $f = q_1 f_1 + \dots + q_s f_s + r$ satisfying the properties given in part (c).
How can I show that uniqueness?
So far I've thought that, since $LT(q_1f_1)=LT(f)=LT(q_1'f_1)$, it must be the case that $LT(q_1)=LT(q_1')$, but I don't know what to do with this.
Consider two expressions \begin{align*} f &= q_1f_1+\dots+q_sf_s+r \tag{1}\\ f &= q_1'f_1+\dots+q_s'f_s+r' \tag{2} \end{align*} satisfying (c). Subtracting $(2)$ from $(1)$, we get \begin{equation} 0 = (q_1-q_1')f_1+\dots+(q_s-q_s')f_s+r-r'. \tag{3} \end{equation} Suppose that $q_i \neq q_i'$ for some $1 \leq i \leq s$. The idea now is to expand the products $(q_j-q_j')f_j$ in $(3)$ and to look at the largest monomials appearing before proceeding with the cancellation. More precisely, if we let $x^{\beta(j)}$ denote the leading monomial of $q_j-q_j'$, we want to consider the largest monomial in \begin{equation*} S = \left\{x^{\beta(j)+\alpha(j)}:1 \leq j \leq s\right\}. \end{equation*} Let $1 \leq k \leq s$ be the smallest index for which $x^{\beta(k)+\alpha(k)}$ is the largest monomial in $S$. We claim that there must be an index $k < \ell \leq s$ for which $x^{\beta(\ell)+\alpha(\ell)}$ is also the largest monomial in $S$. Indeed, the overall expression in $(3)$ is $0$, so $x^{\beta(k)+\alpha(k)}$ must appear again in one of the subsequent expansions, say $(q_\ell-q_\ell')f_\ell$, for there to be cancellation (note that it cannot appear in $r-r'$ as all the monomials in $r-r'$ are in $\overline{\Delta}$, so in particular they are not divisible by $x^{\alpha(k)}$). Moreover, it cannot appear as $x^{\beta_\ell+\alpha_\ell}$ for either $\beta_\ell < \beta(\ell)$ or $\alpha_\ell < \alpha(\ell)$, as this would contradict the maximality of $x^{\beta(k)+\alpha(k)}$. We have thus reached a contradiction: $x^{\beta(\ell)}$ is a monomial in either $q_\ell$ or $q_\ell'$ such that $x^{\beta(\ell)+\alpha(\ell)}$ is divisible by $x^{\alpha(k)}$, i.e. $\beta(\ell)+\alpha(\ell) \notin \Delta_\ell$. Therefore, we conclude that $q_i = q_i'$ for all $1 \leq i \leq s$, from which it follows directly that $r = r'$ as well.