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Every nonzero f(x) in F[x] has a unique monic associate in F[x]

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    F a field.

    Prove that every nonzero f(x) in F[x] has a unique monic associate in F[x].

    2. The attempt at a solution

    Since F is a field, there exists a unit u in F and a function f(x) in F[x] such that g(x)=uf(x) with g(x) in F[x] a monic associate of f(x).

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    I'm having issues figuring out how to explain that it's unique. I haven't done a proof like that before and I only have the proofs in class to go about doing that.

    I want to say something that then there also exists a unit u' in F such that g(x)=u'f(x) but am i allowed to then say u'f(x)=uf(x) and cancel the f's?

    i just need a little push in the right direction. thanks.
     
  2. jcsd
  3. Mar 12, 2008 #2

    morphism

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    Suppose deg(f)=n and write f(x) = a0 + a1x + ... + anxn. As I'm presuming you already know, the units in F[x] are precisely the nonzero elements in F. (If you don't know this, prove it! This is basically the essence of the problem.) So take any nonzero u in F and look at uf(x); if this is monic, then uan=1. What can we conclude now?
     
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