## Representation of polynomials

Posted: May 23, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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This is a generalization of the ordinary representation of polynomials:

Problem. Let $R$ be a commutative ring with $1$ and $A \in R[x]$ 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 $A = Q_0+Q_1B+...+Q_nB^n$

SolutionUniqueness 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 that $Q_0 + Q_1B + \cdots + Q_nB^n = 0,$ with $Q_n \neq 0.$ Let $\alpha, \ \beta$ be the leading coefficients of $Q_n$ and $B$ repectively. Then the leading coefficient of $Q_0 + Q_1B + \cdots +Q_nB^n$ is $\alpha \beta^n.$ Thus $\alpha \beta^n = 0.$ Since $\beta$ is a unit, we’ll get $\alpha = 0,$ which contradicts $Q_n \neq 0.$ Therefore $Q_0 = Q_1= \cdots = Q_n=0.$

Existence of the representation : We only need to prove the claim for $A=x^n.$ The proof is by induction over $n.$ It is clear for $n = 0,$ Suppose that the claim is true for any $k < n.$ If $n < \deg B,$ then choose $A=Q_0$ and $Q_1 = \cdots = Q_n=0.$ So we may assume that $n \geq \deg B.$ Let $B=b_mx^m + b_{m-1}x^{m-1}+ \cdots + b_0.$ Therefore, since $b_m$ is a unit, we will have $x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0,$ which will give 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 $x^{n-k}, \ 1 \leq k \leq m,$ to finish the proof.