Unique factorization domain, roots of a polynomial, abstract algebra

[rayman123]

Homework Statement

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!

Homework Statement

how do we know that $$c^{n}|d$$

2. Homework Statement
how do we know that c, d are relatively prime?

3. Homework Statement
based on what do we conclude that $$\frac{c}{d}\in A$$

 

[lippyka]

?1. Since the polynomial f is monic, the leading coefficient is 1 and so a_n-1 = 0. Therefore, c^n = -a_{n-1}c^{n-1}d - ... - a_1cd^{n-1} - a_0d^n = -a_0d^n. This means that d divides c^n. 2. To show that c and d are relatively prime, we can use the fact that if two integers a and b are not relatively prime, then there exists an integer m greater than 1 such that m divides both a and b. Since c^n = -a_0d^n, if c and d are not relatively prime, then there exists an integer m greater than 1 such that m divides both c^n and d. However, since c^n = -a_0d^n, this implies that m would have to divide both -a_0 and d, which is impossible since -a_0 and d are integers. Therefore, c and d must be relatively prime. 3. Since c and d are relatively prime, we can conclude that \frac{c}{d} is an integer and thus \frac{c}{d} \in A.