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Math Help - unique factorization domain, root of a polynomial, field of fractions proof

  1. #1
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    unique factorization domain, root of a polynomial, field of fractions proof

    let A be a UFD and K its field of fractions. and f\in A[x] where f(x)=x^{n}+a_{n-1}x^{n-1}+....+a_{1}x+a_{0} is a monic polynomial. Prove that if f has a root \alpha=\frac{c}{d}\in K, K=Frac(A) then in fact \alpha\in A


    I need some guidance with the proof.
    Proof:
    f(\alpha)=0\Rightarrow c^{n}+a_{n-1}c^{n-1}d+...+a_{1}cd^{n-1}+a_{0}d^{n}=0
    which gives
    c^{n}=-a_{n-1}c^{n-1}d-...-a_{1}cd^{n-1}-a_{0}d^{n}
    and we observe that d|c^{n}*since all the terms on the rhs are multiples of d

    but how do we know that c^{n}|d??? from what do we conclude that??

    And later in the proof it says '' hence c,d are relatively prime we get that \frac{c}{d}\in A how do we deduce the last part as well?
    based on what do we conclude that \frac{c}{d}\in A?
    Any help appreciated!
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    Re: unique factorization domain, root of a polynomial, field of fractions proof

    Aren't you choosing c and d such that gcd(c, d) = 1 to begin with? (I don't have your proof in front of me.) If they had a common divisor, you could cancel it so that they then didn't, since the fraction field is, up to isomorphism, from an equivalence relation basically defined by such cancellation. And since A is a UFD, you can always find a GCD. So d divides c^n forces d to be a unit right there, and you're done.
    Since I can't see the proof you're referring to, I can't say for sure.
    Thanks from rayman
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    Re: unique factorization domain, root of a polynomial, field of fractions proof

    Quote Originally Posted by johnsomeone View Post
    Aren't you choosing c and d such that gcd(c, d) = 1 to begin with? (I don't have your proof in front of me.) If they had a common divisor, you could cancel it so that they then didn't, since the fraction field is, up to isomorphism, from an equivalence relation basically defined by such cancellation. And since A is a UFD, you can always find a GCD. So d divides c^n forces d to be a unit right there, and you're done.
    Since I can't see the proof you're referring to, I can't say for sure.
    yes, sorry the hypothesis is that c and d are relatively prime.
    Could you explained based on what then \frac{c}{d}\in A?
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    Re: unique factorization domain, root of a polynomial, field of fractions proof

    What john is saying is that if c and d are relatively prime, and if d divides c^n, then d must be a unit, meaning that the inverse of d is in A, so that c/d is basically c multiplied to the inverse of A, which will be in A.
    Thanks from rayman
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    Re: unique factorization domain, root of a polynomial, field of fractions proof

    Ah this reformulation helped me a lot! Thank you both
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