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Math Help - Degree Problem

  1. #1
    Super Member
    Joined
    Feb 2008
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    535

    Degree Problem

    Let
    f eR[x] where R is a commutative ring. If f has form

    a(sub
    n)x^n + ... + a1x + a0 with an a(subn) != 0 then the degree of f is defined to be n and the leading coefficient is defined to be an.



    If
    f, g eF[x] are non-zero polynomials where F is a field,
    then,

    deg(
    fg) = deg f + deg g

    Justify the above theorem. Explain why the proof does
    not work if the coefficients are in
    Zm where m is composite. Hint: focus on the leading coefficients.

    Any advice??

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  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    You know how to multiply polynomials right?

    Let f(x), g(x) \in R[x]
    f(x)=a_nx^n + a_{n-1}x^{n-1}+... a_1x + a_0
    g(x)=b_mx^m + b_{m-1}x^{m-1}+... b_1x + b_0
    for a_m, a_n \not = 0

    f(x)g(x)= a_nb_mx^{n+m} + ... + a_0b_0

    Now if R is a field, fields have no zero divisors, so since a_m, a_n \not = 0 you know a_nb_m \not = 0 thus the degree of this is the degree of f (n) plus the degree of g (m).

    If the coefficients come from \mathbb{Z}_m where m=pr (it is composite, so it has a prime factor p).

    Consider
    f(x) = px+1
    g(x)=rx+1

    f(x)g(x)=prx^2+(p+r)x+1=mx^2+(p+r)x+1=(p+r)x+1 \in \mathbb{Z}_m
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