unique factorization domain, root of a polynomial, field of fractions proof

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

I need some guidance with the proof.

Proof:

$\displaystyle f(\alpha)=0\Rightarrow c^{n}+a_{n-1}c^{n-1}d+...+a_{1}cd^{n-1}+a_{0}d^{n}=0$

which gives

$\displaystyle c^{n}=-a_{n-1}c^{n-1}d-...-a_{1}cd^{n-1}-a_{0}d^{n}$

and we observe that $\displaystyle d|c^{n}$*since all the terms on the rhs are multiples of d

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

And later in the proof it says '' hence c,d are relatively prime we get that $\displaystyle \frac{c}{d}\in A$ how do we deduce the last part as well?

based on what do we conclude that $\displaystyle \frac{c}{d}\in A$?

Any help appreciated!

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.

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

Quote:

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 $\displaystyle \frac{c}{d}\in A$?

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.

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

Ah this reformulation helped me a lot! Thank you both :)