# If $f|d$, then $f$ is invertible in $R[X]/(d)R[X]$

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...
• @WesleyGroupshaveFeelingsToo Well, if I understand correctly what you wrote, then no: there can be elements which are neither. For example, in the ring $\;\Bbb Z\;$ , all the non-zero elements which are not $\;\pm1\;$ are neither zero divisors nor invertible... – DonAntonio Oct 19 '18 at 12:50
• what do you mean by "should be" $\bar{a} \cdot \bar{f} = \bar{0}$? Congruence not equality? – Wesley Strik Oct 19 '18 at 13:05
• Yes. So in fact $\;\overline a= a+\langle d\rangle\in R[x]/\langle d\rangle\;$ – DonAntonio Oct 19 '18 at 14:23