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

let A be a UFD and K its field of fractions. and where is a monic polynomial. Prove that if f has a root , then in fact


I need some guidance with the proof.
Proof:

which gives

and we observe that *since all the terms on the rhs are multiples of d

but how do we know that ??? from what do we conclude that??

And later in the proof it says '' hence c,d are relatively prime we get that how do we deduce the last part as well?
based on what do we conclude that ?
Any help appreciated!

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.

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 ?

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.

Ah this reformulation helped me a lot! Thank you both