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

1. ## 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!

2. ## 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.

3. ## Re: unique factorization domain, root of a polynomial, field of fractions proof

Originally Posted by johnsomeone
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 $\frac{c}{d}\in A$?

4. ## 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.

5. ## Re: unique factorization domain, root of a polynomial, field of fractions proof

Ah this reformulation helped me a lot! Thank you both