My Math Forum unique factorization domain, zero of a polynomial , proof

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 September 10th, 2012, 09:15 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 unique factorization domain, zero of a polynomial , 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? Any help appreciated! My questions are: how do we know that $c^{n}|d$ how do we know that c, d are relatively prime? based on what do we conclude that $\frac{c}{d}\in A$
 September 10th, 2012, 12:33 PM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: unique factorization domain, zero of a polynomial , proo If c and d are not relatively prime then they have a common factor p which can be divided out at any time from $f(x)=x^{n}+a_{n-1}x^{n-1}+....+a_{1}x+a_{0} =( \frac{c}{d})^{n}+a_{n-1}( \frac{c}{d})^{n-1}+....+a_{1}( \frac{c}{d}) +a_{0}$ so it suffices to consider c and d relatively prime. now $c^n$ won't divide $d$ unless $c= \pm 1$ because c and d are relatively prime.
 October 19th, 2012, 07:38 AM #3 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 Re: unique factorization domain, zero of a polynomial , proo thanks, I finally understand it

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