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Thread: Degree Problem

  1. #1
    Super Member
    Feb 2008

    Degree Problem

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

    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.

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

    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
    Dec 2008
    Iowa City, IA
    You know how to multiply polynomials right?

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

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

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

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

    f(x) = px+1

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