$1$, set $c_{ik_{i}}=\frac{\lc_{\sigma}(v)}{\lc_{\sigma}(g_{i})}, w_{ik_{i}}=w, w'_{ik_{i}}=w'$, and replace $v$ by $v-c_{ik_{i}}w_{ik_{i}}g_{i}w'_{ik_{i}}$.

-   Repeat step [(D2)]{} until there is no more $i\in\{1,\dots,s\}$ such that $\lw_{\sigma}(v)$ is a multiple of $\lw_{\sigma}(g_{i})$. If now $v\neq 0$, then replace $p$ by $p+\lm_{\sigma}(v)$ and $v$ by $v-\lm_{\sigma}(v)$, continue with step [(D2)]{}.

-   Return the tuples $(c_{11},w_{11},w'_{11}),\dots,(c_{sk_{s}},w_{sk_{s}},w'_{sk_{s}})$ and $p$.

This is an algorithm which returns tuples $(c_{11},w_{11},w'_{11}),\dots,(c_{sk_{s}},w_{sk_{s}},w'_{sk_{s}})$ and a polynomial $p\in K\langle X\rangle$ such that the following conditions are satisfied.

-   We have $f=\sum^{s}_{i=1}\sum^{k_{i}}_{j=1}c_{ij}w_{ij}g_{i}w'_{ij}+p$.

-   No element of ${\rm Supp}(p)$ is contained in $\langle\lw_{\sigma}(g_{1}),\dots,\lw_{\sigma}(g_{s})\rangle$.

-   For all $i\in\{1,\dots,s\}$ and all $j\in\{1,\dots,k_{i}\}$, we have $\lw_{\sigma}(w_{ij}g_{i}w'_{ij})\leq_{\sigma}\lw_{\sigma}(f)$. If $p\neq 0$, we have $\lw_{\sigma}(p)\leq_{\sigma}\lw_{\sigma}(f)$.

-   For all $i\in\{1,\dots, s\}$ and all $j\in\{1,\dots,k_{i}\}$, we have $\lw_{\sigma}(w_{ij}g_{i}w'_{ij})\notin\langle\lw_{\sigma}(g_{1}),\dots,\lw_{\sigma}(g_{i-1})\rangle$.

Note that the resulting tuples $(c_{11},w_{11},w'_{11}),\dots,(c_{sk_{s}},w_{sk_{s}},w'_{sk_{s}})$ and polynomial $p$ satisfying conditions (a)-(d) are *not* unique. This is due to the fact that in step (D2) of the Division Algorithm there might exist more that one pair $(w,w')$ satisfying $\lw_{\sigma}(v)=w\lw_{\sigma}(g_{i})w'$ (see [@Xiu12], Example 3.2.2). A polynomial $p\in K\langle X\rangle$ obtained in Theorem \[sec2the3\] is called a *normal remainder* of $f$ with respect to $G$ and is denoted by ${\rm NR}_{\sigma,G}(f)$.

For $s\geq 1$, we let $F_s=(K\langle X\rangle\otimes_{K}K\langle X\rangle)^s$ be the free two-sided $K\langle X\rangle$-module of rank $s$ 