1. ## Deg of polynomials

Let $A \in R[x]$ have degree $n \geq 0$ and let $B \in R[x]$ have degree at least 1. Prove that if the leading coefficient of B is a unit of R, then there exist unique polynomials $Q_0,Q_1,...,Q_n \in R[x]$ such that $deg Q_i < deg B$ for all i and $A = Q_0+Q_1B+...+Q_nB^n$.

I'm a bit threw off by this problem, if B is a unit of R, what would that do?

Let $A \in R[x]$ have degree $n \geq 0$ and let $B \in R[x]$ have degree at least 1. Prove that if the leading coefficient of B is a unit of R, then there exist unique polynomials

$Q_0,Q_1,...,Q_n \in R[x]$ such that $deg Q_i < deg B$ for all i and $A = Q_0+Q_1B+...+Q_nB^n$.

I'm a bit threw off by this problem, if B is a unit of R, what would that do?
this result is an interesting generalization of the representation of polynomials. it looks so natural that i don't know why i've never thought about it myself?! so i'm going to have to thank you

for sharing it with us!

uniqueness of the representation: since the leading coefficient of $B$ is a unit, for any $C \in R[x]$ we have $\deg (BC)=\deg B + \deg C.$ now suppose $Q_0 + Q_1B + \cdots + Q_nB^n = 0,$ with $Q_n \neq 0.$

let $\alpha$ and $\beta$ be the leading coefficients of $Q_n$ and $B$ repectively. then the leading coefficient of $Q_0 + Q_1B + \cdots +Q_nB^n$ is $\alpha \beta^n.$ why? thus $\alpha \beta^n = 0.$ since $\beta$ is a unit, we'll get $\alpha = 0,$ which

contradicts $Q_n \neq 0.$ thus: $Q_0 = Q_1= \cdots = Q_n=0.$

existence of the representation: we only need to prove the claim for $A=x^n.$ why? proof by induction over $n$: if n = 0, then A = 1 and we put $Q_0=1.$ then $A=Q_0.$ so suppose the claim holds

for any $x^k$ with k < n. if $n < \deg B,$ then just choose $A=Q_0$ and $Q_1 = \cdots = Q_n=0.$ so we may assume that $n \geq \deg B.$ let $B=b_mx^m + b_{m-1}x^{m-1}+ \cdots + b_0.$ since $b_m$ is a unit, we'll have:

$x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0,$ which gives us: $x^n = b_m^{-1}x^{n-m}B - b_m^{-1} b_{m-1}x^{n-1} - \cdots - b_m^{-1}b_0 x^{n-m}.$ now apply the induction hypothesis to each term $x^{n-k}, \ 1 \leq k \leq m,$ in the RHS

to finish the proof.

3. Here is another way to prove existence, though it is not general. It assumes that $R$ is a field (and so the division algorithm holds). But it gives us an algorithmic way to find the expansion for our favorite rings such as $\mathbb{Q},\mathbb{R},\mathbb{C}$. It also seems to parallel the general way of expressing numbers in bases for positive integers.

Let $A\in R[x]$ with $B\in R[x]$ some fixed polynomial with $\deg B \geq 1$.
Find the smallest $n\in \mathbb{N}$ such that $\deg B^n \leq \deg A$ and $\deg B^{n+1} > A$.
By division algorithm we get, $A = Q_n B^n + R_n$.
Notice that $\deg Q_n < \deg B$ and $\deg R_n < \deg B^n$.
By division algorithm we get, $R_n = Q_{n-1}B^{n-1} + R_{n-1}$.
Notice that $\deg Q_{n-1} < \deg B$ and $\deg R_{n-1} < \deg B^{n-1}$.
Continuing in this manner we get, $R_1 = Q_1B + R_0$ where $\deg Q_1 < \deg B$ and $\deg R_0 = \deg B^1$.
Set $Q_0 = R_0$.

It follows that,
$A = Q_nB^n + R_n = Q_nB^n + Q_{n-1}B^{n-1}+R_{n-1} = ... = Q_nB^n + Q_{n-1}B^{n-1}+...+Q_1B+Q_0$.