Theorem (Single factor theorem) Let $R$ be a commutative ring, and let $P\in R[X]$, where $R[X]$ is the polynomial ring over the indeterminate $X$. Suppose $P(\alpha)=0$. Then $(X-\alpha)$ divides $P(X)$.
If $R$ is a field, then I know the following proof:
Proof (for a field): Using polynomial long division, we can write:
$$ P(X)=Q(X)(X-\alpha)+r $$
where $r\in R$ is a constant, since it has to have strictly lower degree than $(X-\alpha)$. But then, putting $X=\alpha$, we get:
$$ 0=0+r $$
So $r=0$, and we win. $\Box$
The proof above doesn't work over an arbitrary ring, because we cannot always do polynomial long division over an arbitrary ring. However, I have read in a number of places (including on this site) that the theorem is true when $R$ is an arbitrary commutative ring. I haven't been able to find anything other than the proof for fields outlined above, either online or in Chrystal's Algebra. One idea I had was to try and use Gauss's lemma, but I'd like to know if there is a 'standard' proof of this fact.