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

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SUMMARY

Every nonzero polynomial f(x) in the polynomial ring F[x], where F is a field, has a unique monic associate. A polynomial g(x) is considered monic if its leading coefficient is 1. The proof relies on the existence of a unit u in F such that g(x) = u * f(x), where g(x) is monic. The uniqueness is established by demonstrating that if g(x) = u * f(x) and g(x) = u' * f(x) for some other unit u', then u must equal u', confirming that the monic associate is unique.

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  • Understanding of polynomial rings, specifically F[x]
  • Knowledge of fields and units within fields
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  • Basic proof techniques in abstract algebra
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  • Learn about units in fields and their role in polynomial multiplication
  • Explore the concept of monic polynomials and their significance in algebra
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Homework Statement


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.
 
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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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