I am trying to prove the following theorem
Theorem: Let $(R, +, \cdot)$ be a unital commutative ring and let $x$ be a symbol. Let $a(x), b(x) \in R[x]$ and suppose that $b(x) \neq 0_{R[x]}$ and that $L.c(b(x))$ is a unit in $(R, +, \cdot)$. Then there exists a unique $q(x), r(x), \in R[x]$ such that $a(x) = b(x) \odot q(x) \oplus r(x)$ and that $r(x) = 0_{R[x]}$ or $\partial(r(x)) < \partial(b(x))$
Note that $\odot$ refers to the multiplication operation, and $\oplus$ the addition operation in the polynomial ring $R[x]$. and $L.c(b(x))$ refers to the leading coefficient of $b(x)$
In class we handled the existence in the case where $a(x) = 0_{R[x]}$ and in the case where $a(x) \neq 0_{R[x]}$ and $\partial (a(x)) < \partial (b(x))$.
We proved the existence in the case where $a(x) \neq 0_{R[x]}$ and $\partial (a(x)) \geq \partial (b(x))$ by induction on the degree of $a(x)$. However I have an issue following a few steps in the proof.
The proof of the base case $\partial(a(x)) = 0$ is fine to me. The problem I have is after the induction step. An excerpt of the proof from the induction step onwards goes as follows:
Now for the induction step let us select an arbitrary $k \in \{0, 1, 2, \ldots\}$. let us assume the existence of such polynomials $q(x), r(x)$ of the desired kind is known whenever $\partial(a(x)) \leq k$
Let us now assume that $\partial(a(x)) = k+1$. We may abbrevitae $a(x) = a_0 + a_1x^1 + \cdots + a_kx^k + a_{k+1}x^{k+1}$ and we may also abbreviate $b(x) = b_0 + b_1x + \cdots + b_{m-1}x^{m-1} + b_mx^m$ where $m = \partial(b(x))$. Since $\partial (a(x)) \geq \partial (b(x))$ we have that $k+1 - m \geq 0$
Let $a'(x) = a(x) - b_m^{-1}\cdot x^{k+1-m} \odot b(x)$. By keeping track of the arithmetic in $R[x]$ we note that the co-efficients of $x^{k+1}$ in $a'(x)$ is $0_R$. Thus either $a'(x) = 0_{R[x]}$ or $\partial(a'(x)) \leq k$
From the part "keeping track of the arithmetic in $R[x]$..." I can't follow the proof
When I try to do the arithmetic in $R[x]$ this is what I get:
Firstly we have $b_m{-1}\cdot x^{k+1-m} = 0_Rx^0 +0_Rx^1 +0_Rx^2 + .. + b_m^{-1}x^{k+1-m} + 0_Rx^{k+2-m} + 0_Rx^{k+3-m} + \cdots$
Then computing $b_m^{-1}\cdot x^{k+1-m} \odot b(x)$ I got (leaving out the lengthy calculations) $$b_m^{-1}\cdot x^{k+1-m} \odot b(x) = (b_m^{-1}b_0)x^{k+1-m}+ (b_m^{-1}b_1)x^{k+2-m} + \cdots + (b_m^{-1}b_{2m-k-1})x^{2m-k}$$
and I certainly don't see how the co-efficients of $x^{k+1}$ in $a'(x)$ is $0_R$
What am I doing wrong here? I also don't see how we can conclude that $a'(x) = 0_{R[x]}$ or $\partial(a'(x)) \leq k$