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Math Help - Polynomials with integer coefficients - the GCD identity

  1. #1
    Super Member Bernhard's Avatar
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    Polynomials with integer coefficients - the GCD identity

    I am reading Anderson and Feil on the Factorization of Polynomials - Section 5.3 Polynomials with Integer Coefficients

    A&F point out that some of the therems they have proved for  \mathbb{Q} [x] are false if we restrict ourselves to polynomials with integer coefficients.

    For example they point out that the Divsion Theorem is false for  \mathbb{Z} [x] .

    Also the GCD identity fails in  \mathbb{Z} [x] .

    They then ask the reader the polynomials 2 and x in  \mathbb{Z} [x] pointing out the 1 is the GCD for these polynomials. A&F then ask the reader to prove that we cannot write 1 as a linear combination of 2 and x

    Can anyone help me with a rigorous and formal proof of this.

    Peter
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  2. #2
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    Re: Polynomials with integer coefficients - the GCD identity

    suppose we COULD.

    we then have 1 = 2f(x) + xg(x), for some polynomials f(x),g(x) in Z[x].

    suppose that deg(f) = m, and deg(g) = n. let k = max(m,n). then we can write:

    f(x) = a_0 + a_1x + \cdots + a_kx^k, g(x) = b_0 + b_1x + \cdots + b_kx^k (some of the coefficients may be 0).

    then:

    1 = 2a_0 + (2a_1 + b_0)x + \cdots + (2a_k + b_{k-1})x^k + b_kx^{k+1}.

    since these are equal polynomials, we must have equal constant terms, so 1 = 2a_0 for some integer a_0, which cannot happen.
    Thanks from Bernhard
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Polynomials with integer coefficients - the GCD identity

    Excellent ... yes, so obvious when you see how ... :-)

    Thanks so much ....

    Peter
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