All the numbers I mentioned below are integers.
Question:
In $\mathbb{Z}_m[x]$, if $f(c_1) = 0$ and $f(c_2) = 0$, it does not always follow that $(x - c_1)(x - c_2) \mid f(x).$ What hypothesis on $c_1, c_2$ is needed to make that true?
I think when $c_1$ and $c_2$ are in different congruent class, the statement is true. But I don't know how to rigorously prove it.
My attempt: Suppose $f(x)$ is expresses in its simpliest form (i.e. all the coefficients are between [0,m-1]). Then $f(c_1)-ms_1,f(c_2)-ms_2=0$ for some integers $s_1,s_2$. I want to find a polynomial $g(x)$ such that $g(c_1)=s_1,g(c_2)=s_2$, where all the coefficients of $g(x)$ are integers. Thus, We can apply the Factor Theorem in $\mathbb{Z}[x]$ to $f(x)+g(x)m$.
However, I don't know whether my idea is correct, and I don't know how to find such $g(x)$.
The usual division algorithm works in any polynomial ring for monic divisors. This gives $$f(x) = (x-c_2)g(x)$$ To repeat this step again, you need $g(c_1)=0$ but all we know is $$0=f(c_1)=(c_1-c_2)g(c_1)$$ This will imply $g(c_1)=0$ if $c_1-c_2\in\mathbb{Z}_m$ is not a zero divisor, so $$\gcd(c_1-c_2,m)=1$$ is probably the most simple condition.