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.


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