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Math Help - Deg of polynomials

  1. #1
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    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?
    Last edited by tttcomrader; November 10th 2008 at 06:59 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post

    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.
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  3. #3
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    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.
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