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Math Help - Field question Divisibility

  1. #1
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    Field question Divisibility

    Let f(x), g(x) be in F(x). Show that if g(x)|f(x) and f(x)|g(x), then f(x)=kg(x) for some k in F.

    Since g(x)|f(x), then f(x)=g(x)r(x) for some r(x) in F(x).
    Similarily, since f(x)|g(x), then g(x)=f(x)s(x)
    So f(x)=f(x)s(x)r(x)
    I don't know where to go from here.
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Let f(x), g(x) be in F(x). Show that if g(x)|f(x) and f(x)|g(x), then f(x)=kg(x) for some k in F.

    Since g(x)|f(x), then f(x)=g(x)r(x) for some r(x) in F(x).
    Similarily, since f(x)|g(x), then g(x)=f(x)s(x)
    So f(x)=f(x)s(x)r(x)
    I don't know where to go from here.
    You are almost there Notice that

    \displaystyle f(x)=f(x)s(x)r(x) \iff f(x)[1-s(x)r(x)]=0

    Since \mathbb{F}[x] is an integral domain the zero product principle applies so you get

    \displaystyle 1-s(x)r(x)=0 \iff s(x)r(x)=1 this tells us that s(x), r(x) must be units, and that they are inverses of each other. What are the only invertible elements of \mathbb{F}[x]?
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  3. #3
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    s(x) and r(x) are the only invertible elements?
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    Quote Originally Posted by kathrynmath View Post
    s(x) and r(x) are the only invertible elements?
    No. Given any field \mathbb{F} polynomials over the field \mathbb{F}[x] what are the invertible elements?

    This may help. Does 2x have an inverse is \mathbb{R}[x]. Does there exist an f(x) suchthat
    2xf(x)=1? Think about this and good luck.
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    Well for that, we have 2x(1/2x^-1)=1
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  6. #6
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    Quote Originally Posted by kathrynmath View Post
    Well for that, we have 2x(1/2x^-1)=1
    Is x^{-1}\in\mathbb{R}[x]?
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