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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.

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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...

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  • $\begingroup$ Of course, elements are either invertible OR zero divisors. $\endgroup$ – Wesley Strik Oct 19 '18 at 12:36
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    $\begingroup$ @WesleyGroupshaveFeelingsToo ... or neither. $\endgroup$ – Saucy O'Path Oct 19 '18 at 12:43
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    $\begingroup$ @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... $\endgroup$ – DonAntonio Oct 19 '18 at 12:50
  • $\begingroup$ what do you mean by "should be" $\bar{a} \cdot \bar{f} = \bar{0}$? Congruence not equality? $\endgroup$ – Wesley Strik Oct 19 '18 at 13:05
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    $\begingroup$ Yes. So in fact $\;\overline a= a+\langle d\rangle\in R[x]/\langle d\rangle\;$ $\endgroup$ – DonAntonio Oct 19 '18 at 14:23

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