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Thread: Polynomials

  1. #1
    Apr 2010


    Assistance and help on this would be great! thank you

    Let F be a field. Let a(x), b(x) exist in F[x] be polynomials such that a(x)≠ 0 or b(x) ≠ 0.
    Let d(x)=gcd(a(x), b(x), that is, d(x) is the monic polynomial in F[x] of highest degree such
    that d(x)|a(x) and d(x)| b(x). Suppose that d1(x) exist in F[x] is a monic polynomial such that
    d1(x)=a(x)u(x) + b(x)v(x) for some u(x), v(x) that exist in F[x], d1(x)| a(x) and d1(x)| b(x)
    Prove that d(x) = d1(x).
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  2. #2
    Member HappyJoe's Avatar
    Sep 2010
    What results do you know in connection with this? See if you can follow the following argument with your current knowledge.

    Since $\displaystyle d_1(x) | a(x)$ and $\displaystyle d_1(x) | b(x)$, and since $\displaystyle d(x)$ is the greatest divisor of both $\displaystyle a(x)$ and $\displaystyle b(x)$, we have $\displaystyle d_1(x) | d(x)$.

    Conversely, since $\displaystyle d(x)$ divides both $\displaystyle a(x)$ and $\displaystyle b(x)$, we have that $\displaystyle d(x)$ divides $\displaystyle a(x)u(x)+b(x)v(x)$, i.e. $\displaystyle d(x) | d_1(x)$. In conclusion, $\displaystyle d(x) = \lambda d_1(x)$ for some scalar $\displaystyle \lambda$. Since both $\displaystyle d(x)$ and $\displaystyle d_1(x)$ are monic, the scalar is equal to 1, hence $\displaystyle d(x)=d_1(x)$.
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