Let $R$ be a field and $f$ and $d$ be polynomials in $R[X]$. If $f|d$, then $f$ is invertible in $R[X]/(d)R[X]$.
I tried to either prove of disprove this statement, but so far I haven't been able to find a counterexample. I was thinking if $f|d$ then for some polynomial $a$ we can write:
$$ a \cdot f= d$$
Now since $d$ is congruent $0$ we can write:
$$ a \cdot f \equiv 0 \pmod{d}$$ But I don't know how to now deduce that $f$ may or may not be invertible, for invertibility I actually want something like $$a \cdot f =1.$$ I haven't used yet that we are dealing with a field.
When you reached $\;a\cdot f=0\;$ , which in fact should be $\;\overline a\cdot\overline f=\overline 0\;$ (in the quotient ring), you already got a contradiction: since $\;\overline f\;$ is a divisor of zero , it cannot be invertible...