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Thread: Deg of polynomials

  1. #1
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    Deg of polynomials

    Let $\displaystyle A \in R[x] $ have degree $\displaystyle n \geq 0 $ and let $\displaystyle 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 $\displaystyle Q_0,Q_1,...,Q_n \in R[x] $ such that $\displaystyle deg Q_i < deg B $ for all i and $\displaystyle A = Q_0+Q_1B+...+Q_nB^n$.

    I'm a bit threw off by this problem, if B is a unit of R, what would that do?
    Last edited by tttcomrader; Nov 10th 2008 at 06:59 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post

    Let $\displaystyle A \in R[x] $ have degree $\displaystyle n \geq 0 $ and let $\displaystyle 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

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

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

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

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

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

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

    $\displaystyle x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0,$ which gives us: $\displaystyle x^n = b_m^{-1}x^{n-m}B - b_m^{-1} b_{m-1}x^{n-1} - \cdots - b_m^{-1}b_0 x^{n-m}.$ now apply the induction hypothesis to each term $\displaystyle x^{n-k}, \ 1 \leq k \leq m,$ 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 $\displaystyle R$ 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 $\displaystyle \mathbb{Q},\mathbb{R},\mathbb{C}$. It also seems to parallel the general way of expressing numbers in bases for positive integers.

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

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