Integral domains, polynomials and division We know that :

If $K$ is a field and $P\in K[X]$ is a polynomial with coefficients in $K,$ then $\alpha\in K$ is a root of $P$ if and only if $(X-\alpha)$ divides $P$. 

My question is the following : does this hold for polynomials with coefficients in an integral domain $R,$ and if it doesn't can you give me a counterexample ? I tried to think about the field of fractions, but no managed to find an answer in the general case. Thank you for all your answers !
 A: As already indicated in the comments the answer is yes. The easiest way to see this is to note that
$$x^k-\alpha^k=(x-\alpha)(x^{k-1}+\alpha x^{k-2}+\alpha^2x^{k-3}+\cdots+\alpha^{k-1})$$
so for $f(x)=c_nx^n+\cdots+c_1x+c_0$ it is clear that 
$$f(x)-f(\alpha)=c_n(x^n-\alpha^n)+\cdots+c_1(x-\alpha)$$
is divisible by $x-\alpha$. Therefore, $f(x)$ is divisible by $x-\alpha$ precisely when $f(\alpha)=0$.
A: Yes it does, let $D$ be an integral domain with a unit.
Then it can be shown via induction (over the degree of the polynomial) that generalized division holds.
In other words, given $P$ a polynomial and $D$ a monic polynomial,  polynomials $Q$ and $R$ exist so that:
$P=QD+R$, where the degree of $R$ is less than the degree of $D$.

Going back to our problem, let $P$ be the polynomial, and let $\alpha$ be the proposed root.
We let $D=(x-\alpha)$ and apply generalized division to get:
$P=(x-\alpha)Q+c$, where $c$ is in the integral domain (because the degree is $1$).
Clearly when we evaluate the polynomial at $x$ we get:
$P(x)=0+c$. So $\alpha$ is a root of $P$ if and only if $c=0$, if and only if $(x-\alpha)$ divides $P$.
